Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 13 Hydrocarbons Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

11th Chemistry Guide Hydrocarbons Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is
a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 2.
C₂H₅ Br + 2Na C₄H₁₀ + 2NaBr.
The above reaction is an example of which of the following
a) Reimer Tiemann reaction
b) Wurtz reaction
c) Aldol condensation
d) Hoffmann reaction
Answer:
b) Wurtz reaction

Question 3.
An alkyl bromide (A) reacts with sodium in ether to form 4, 5 – diethyloctane, the compound (A) is
a) CH₃(CH₂)₃Br

b) CH₃(CH₂)₅Br

c) CH₃(CH₂)₃ CH(Br)CH₃

d)
Answer:
d)

Question 4.
The C – H bond and C – C bond in ethane are formed by which of the following types of overlap
a) sp³ – s and sp³ – sp³
b) sp² – s and sp² – sp²
c) sp – sp and sp – sp
d) p – s and p – p
Answer:
a) sp³ – s and sp³ – sp³

Question 5.
In the following reaction,

Answer:
c)

Question 6.
Which of the following is optically active
a) 2 – methyl pentane
b) citric acid
c) Glycerol
d) none of these
Answer:
a) 2 – methyl pentane

Question 7.
The compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are
a) CH₄ and H₂
b) CH₄ and CO₂
c) C₂H₆ and CO₂
d) C₂H₄ and Cl₂
Answer:
c) C₂H₆ and CO₂

Question 8.
The general formula for cyclo alkanes
a) C_nH_n
b) C_nH_2n
C) C_nH_2n-2
d) C_nH_{2n + 2}
Answer:
b) C_nH_2n

Question 9.
The compound that will react most readily with gaseous bromine has the formula
a) C₃H₆
b) C₂H₂
c) C₄H₁₀
d) C₂H₄
Answer:
a) C₃H₆

Question 10.
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction
a)
b) CH₃ – CH₂ – CH₂ – OH
c) H₂C = C = O
d) CH₃ – CH₂ – CH₂Br
Answer:
c) H₂C = C = O

Question 11.
Which among the following alkenes on reductive ozonolysis produces only propanone?
a) 2 – Methyl propene
b) 2 – Methyl but – 2 – ene
c) 2, 3 – Dimethyl but – 1- ene
d) 2, 3 – Dimethyl but – 2 – ene
Answer:
d) 2, 3 – Dimethyl but – 2 – ene

Question 12.
The major product formed when 2 – bromo – 2 – methyl butane is refluxed with ethanolic KOH is
a) 2 – methylbut – 2- ene
b) 2 – methyl butan – 1 – ol
c) 2 – methyl but – 1 – ene
d) 2 – methyl butan – 2- ol
Answer:
a) 2 – methylbut – 2- ene

Question 13.
Major product of the below mentioned reaction is, (CH₃)₂ C = CH₂
a) 2 – chloro – 1 – iode – 2 – methyl propane
b) 1 – chloro – 2 – iodo – 2 – methyl propane
c) 1, 2 – dichloro – 2 – methyl propane
d) 1, 2 – diiodo – 2 – methyl propane
Answer:
a) 2 – chloro – 1 – iode – 2 – methyl propane

Question 14.
The IUPAC name of the following compound is

a) trans – 2- chloro – 3- iodo – 2- pentane
b) cis – 3- iodo – 4 chloro – 3 – pentane
c) trans – 3 – iodo – 4 – chloro – 3 – pentene
d) cis – 2 – chloro – 3 iodo – 2 – pentene
Answer:
a) trans – 2- chloro – 3- iodo – 2- pentane

Question 15.
Cis – 2- butene and trans – 2 – butene are
a) conformational isomers
b) structural isomers
c) configurational isomers
d) optical isomers
Answer:
c) configurational isomers

Question 16.
Identify the compound (A) in the following reaction

Answer:
a)

Question 17.
CH ≡ CH , where A is,
a) Zn
b) conc. H₂SO₄
c) alc. KOH
d) dil. H₂SO₄
Answer:
c) alc. KOH

Question 18.
Consider the nitration of benzene using mixed con H₂SO₄ and HNO₃ if a large quantity of KHSO₄ is added to the mixture, the rate of nitration will be
a) unchanged
b) doubled
c) faster
d) slower
Answer:
d) slower

Question 19.
In which of the following molecules, all atoms are co-planar
a)

b)

c)
d) both (a) and (b)
Answer:
d) both (a) and (b)

Question 20.
Propyne on passing through red hot iron tube gives
a)

b)

c)
d) none of these
Answer:
a)

Question 21.
is

a)

b)

c) both (a) and (b)

d)
Answer:
d)

Question 22.
Which one of the following is non aromatic?
a)

b)

c)

d)
Answer:
d)

Question 23.
Which of the following compounds will not undergo Friedal – crafts reaction easily?
a) Nitro benzene
b) Toluene
c) Cumene
d) Xylene
Answer:
a) Nitro benzene

Question 24.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
a) – COOH
b) – NO₂
c) -C ≡ N
d) -SO₃H
Answer:
b) – NO₂

Question 25.
Which of the following can be used as the halide component for friedal – crafts reaction?
a) Chloro benzene
b) Bromo benzene
c) Chloro ethene
d) Isopropyl chloride
Answer:
d) Isopropyl chloride

Question 26.
An alkane isobtainedbydecarboxylation of sodium propionate. Same alkane can be prepared by
a) Catalytic hydrogenation of propene
b) action of sodium metal on iodomethane
c) reduction of 1 – chloro propane
d) reduction of bromomethane
Answer:
b) action of sodium metal on iodomethane

Question 27.
Which of the following is aliphatic saturated hydrocarbon
a) C₈H₁₈
b) C₉H₁₈
c) C₈H₁₄
d) All of these
Answer:
a) C₈H₁₈

Question 28.
Identify the compound ‘Z’ in the following reaction

a) Formaldehyde
b) Acetaldehyde
c) Formic acid
d) none of these
Answer:
a) Formaldehyde

Question 29.
Peroxide effect (Kharasch effect) can be studied in case of
a) Oct – 4 – ene
b) hex – 3 – ene
c) pent – 1 – ene
d) but – 2 – ene
Answer:
c) pent – 1 – ene

Question 30.
2 – butyne on chlorination gives
a) 1 – chloro butane
b) 1, 2 – dichloro butane
c) 1, 1, 2, 2 – tetrachlorobutane
d) 2, 2, 3, 3 – tetra chloro butane
Answer:
d) 2, 2, 3, 3 – tetra chloro butane

II. Write brief answer to the following questions:

Question 31.
Give IUPAC names for the following compounds
1) CH₃ – CH = CH – CH = CH – C ≡ C – CH₃

2)

3) (CH₃)₃ C – C ≡ C – CH(CH₃)₂

4) ethyl isopropyl acetylene

5) CH ≡ C – C ≡ C – C ≡ CH
Answer:
1)

2)

3)

4) ethyl isopropyl acetylene

5)

Question 32.
Identify the compound A, B, C and D in the following series of reactions

Answer:

Question 33.
Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
Answer:
All the activating groups are ‘ortho-para’ directors.
Example: – OH, – NH₂, -NHR, -NHCOCH₃, -OCH₃ -CH₃ – C₂H₅ etc.
Let us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.

In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. The electron density at ortho and para positions increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at ‘ortho’ and ‘para positions and hence -OH group is an ortho-para director and activator.

In aryl halides, the strong -I effect of the halogens (electron withdrawing tendency) decreases the electron density of benzene ring, thereby deactivating for electrophilic attack. However the presence of lone pair on halogens involved in the resonance with pi electrons of benzene ring, increases electron density at ortho and para position. Hence the halogen group is an ortho-para director and deactivator.

Question 34.
How is propyne prepared from an alkene dihalide?
Answer:

Question 35.
An alkylhalide with molecular formula C₆H₁₃Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH,CHO and (CH₃)₂ CHCHO. Find the alkyihalide.
Answer:

Question 36.
Describe the mechanism of Nitration of benzene.
Answer:
It is prepared by the action of a mixture of con. HNO₃ and con. H₂SO₄ (nitrating mixture) on benzene maintaining the temperature below 333 K.
Nitration:

Sulphuric acid generates the electrophile – NO₂⁺, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction.

The generation of nitronium ion
H₂SO₄ + HONO₂ → + HSO₄^–

To the nitronium ion (being an electron deficient species) the π bond of benzene, donates a pair of electrons forming a 6-bond. A species with a + ve charge is formed as an intermediate. This is called ‘arenium ion’ and is stabilised by Resonance.

In the last step, the hydrogen atom attached to the carbon carrying the nitro group is pulled out as a proton, by the Lewis base HSO₄^–, so that stable aromatic system is formed.

Question 37.
How does Huckel rule help to decide the aromatic character of a compound.
Answer:
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
Benzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.

Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) which are in rapid equilibrium.

Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

Spectroscopic measurements:
Spectroscopic measurements show that benzene is planar and all of its carbon- carbon bonds are of equal length 1.40 Å. This value lies between carbon-carbon single bond length 1.54 Å and carbon- carbon double bond length 1.34 Å.

Molecular orbital structure:
The structure of benzene is best de¬scribed in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp² hybridized. Six sp² hybrid orbitals of carbon linearly overlap with six 1 s orbitals of hydrogen atoms to form six C – H sigma bonds. Overlap between the remaining sp² hybrid orbitals of carbon forms six C – C sigma bonds.

Formation of Sigma bond in benzene

All the σ bonds in benzene lie in one plane with bond angle 120°. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 π- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization, strong π-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.

All carbon atoms have the delocalized π MO is formed by p-orbitais the overlap of six p-orbitals.

Representation of benzene:
Hence, there are three ways in which benzene can be represented.

Question 38.
Suggest the route for the preparation of the following from benzene.
1) 3 – chloro nitrobenzene
2) 4 – chlorotoluene
3) Bromo benzene
4) m – dinitro benzene
Answer:
1) 3 – chloro nitrobenzene

2) 4 – chlorotoluene

3) Bromo benzene

4) m – dinitro benzene

Question 39.
Suggest a simple chemical test to distinguish propane and propene.
Answer:
Propene decolourises Br₂/H₂O it forms dibromo compound but propane does not react with Br₂/ H₂O.

Question 40.
What happens when Isobutylene is treated with acidified potassium permanganate?
Answer:

Question 41.
How will you convert ethyl chloride into
(i) ethane
(ii) n – butane
Answer:
(i) ethane

(ii) n – butane

Question 42.
Describe the conformers of n – butane.
Answer:
Conformations of n-Butane:
n-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group

Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is maximum repulsion between them and it is the least stable conformer.

Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. And it is the most stable conformer.

The following potentially energy diagram shows the relative stabilities of various conformers of n-butane.

Question 43.
Write the chemical equations for combustion of propane.
Answer:
C₃H₈(g) + 5O₂(g) ) → 3CO₂(g) + 4H₂O(l)
propane

Question 44.
Explain Markovnikoff’s rule with suitable example.
Answer:
Markovnikoff ‘s rule:
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon that has more number of hydrogen and halogen add to the carbon having fewer hydrogen”. This rule can also be stated as in the addition reaction of alkene/alkyne, the most electro negative part of the reagent adds on to the least hydrogen attached doubly bonded carbon.

Addition of water: (Hydration of alkenes)
Normally, water does not react with alkenes. In the presence of concentrated sulphuric acid, alkenes react with water to form alcohols. This reaction follows carbocation mechanism and Markovnikoff’s rule.

Question 45.
What happens when ethylene is passed ‘ through cold dilute alkaline potassium permanganate.
Answer:
CH₂ = CH + H₂O + (O) CH₂OH – CH₂OH
Ethylene                                                                Ethylene glycol

Question 46.
Write the structures of following alkanes.
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
Answer:
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane

2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane

3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:

Question 48.

Identify A and B
Answer:
A)

b)

Question 49.
Complete the following :

i) 2 – butyne 

ii) CH₂ = CH₂

iii)

iv) CaC₂
Answer:
i) 2 – butyne

ii) CH₂ = CH₂

iii)

iv) CaC₂
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂↑
Calcium Carbide                  Acetylene

Question 50.
How will distinguish 1 – butyne and 2 – butyne?
Answer:

1 – butyne reacts with ammoniacal AgNO₃ solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO₃ solution.

11th Chemistry Guide Hydrocarbons Additional Questions and Answers

I. Choose the best answer:

Question 1.
Statement-I: Methane. ethane, propène, and butane are alkane group compounds.
Statement-II: They are obeying C₂H_2n+2 + formula and each member differs from it proceeding member by a CH₂ group.
(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I
(e) Statëment-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 2.
Alkanes are also known as
a) olefins
b) unsaturated aliphatic hydrocarbons
c) saturated aromatic hydrocarbon
d) paraffin
Answer:
d) paraffin

Question 3.
Which one of the following shows three possible isomeric structures’?
(a) C₄H₁₀
(b) C₅H₁₂
(c) C₆H₁₂
(d) C₃H₄
Answer:
(b) C₅H₁₂

Question 4.
The gas supplied in cylinders for cooking is
a) marsh gas
b) LPG
c) mixture CH₄ and C₂H₆
d) mixture of ethane and propane
Answer:
d) mixture of ethane and propane

Question 5.
Which of the following compound cannot be prepared by Kolbe’s electrolytic method’?
(a) CH₃-CH₃
(b) CH₄
(c) CH₂ = CH₂
(d) CH = CH
Answer:
(b) CH₄

Question 6.
Soda-lime is
a) NaOH
b) NaOH + CaO
c) CaO
d) Na₂CO₃
Answer:
b) NaOH + CaO

Question 7.
Statement – I : Alkenes shows both structural and geometrical isomerism.
Statement – II : Because of the presence of double bond.
(a) Statement – I and II are correct and statement-II is correct explanation of statement – I.
(b) Statement – I and II are correct but statement-II is not correct explanation of statement – I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – II is wrong but statement – II is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I.

Question 8.
Hydrocarbon which is liquid at room temperature is
a) Pentane
b) Butane
c) Propane
d) Ethane
Answer:
a) Pentane

Question 9.
Pyrolysis of Methane and respectively are
a) Exothermic & Endothermic
b) Endothermic & Exothermic
c) Both are endothermic
d) Both are exothermic
Answer:
c) Both are endothermic

Question 10.
Peroxide effect is not observed in ………..
(a) HCl
(b) HI
(c) HBr
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 11.
Final products of complete oxidation of hydrocarbon is
a) Acid
b) Dihydric alcohol
c) Aldehyde
d) H₂O + CO₂
Answer:
d) H₂O + CO₂

Question 12.
Isomerisation in alkane can be brought by using
a) Al₂O₃
b) Fe₂O₃
c) Anh.AlCl₃/ HCl at 200°C
d) Cone. H₂SO₄
Answer:
c) Anh.AlCl₃/ HCl at 200°C

Question 13.
In aromatization of n – hexane, the catalyst used is
a) Cr₂O₃
b) V₂O₅
c) Mo₂O₃
d) All
Answer:
d) All

Question 14.
The most oxidized form of ethane is
a) CO₂
b) HCHO
c) HCOOH
d) CH₃COOH
Answer:
a) CO₂

Question 15.
The following substance is used as anti-knocking compound
a) TEL
b) Lead tetrachloride
c) Lead acetate
d) C₂H₂PbCl
Answer:
a) TEL

Question 16.
The molecular formula of benzene is ………….
(a) C₆H₆
(b) C₆H₅
(c) C₇H₈
(d) CH₄
Answer:
(a) C₆H₆

Question 17.
The most stable conformation of ethane is
a) Eclipsed
b) Skew
c) Staggered
d) All are equally stable
Answer:
c) Staggered

Question 18.
IUPAC name of the following compound

a) 3 – methyl hexane
b) 3 – Ethyl pentane
c) 2, 3 – dimethyl pentane
d) 2, 2 – dimethyl pentane
Answer:
a) 3 – methyl hexane

Question 19.
The number of sigma bonds formed in ethane by the overlapping of sp3 – sp3 orbitals
a) 7
b) 5
c) 1
d) 4
Answer:
c) 1

Question 20.
Which one of the following is not an ortho-para director?
(a) – NO₂
(b) – CH₃
(c) – OH
(d) – C₂H₅
Answer:
(a) – NO₂

Question 21.
The correct IUPAC name of the following alkane is

a) 3, 6 – diethyl – 2- methyloctane
b) 5 – isopropyl – 3- ethyloctane
c) 3 – ethyl – 5- isopropyloctane
d) 3 – isopropyl – 6 – ethyloctane
Answer:
a) 3, 6 – diethyl – 2- methyloctane

Question 22.
In the Wurtz reaction, n-hexane is obtained from
a) n – propyl chloride
b) n – butyl chloride
c) Ethyl chloride
d) isopropyl chloride
Answer:
a) n – propyl chloride

Question 23.
When sodium acetate is heated with soda lime the reaction is called
a) Dehydration
b) Decarboxylation
c) Dehydrogenation
d) Dehydrohalogenation
Answer:
b) Decarboxylation

Question 24.
The following substance reacts with water to give ethane
a) CH₄
b) C₂H₅MgBr
c) C₂H₄OH
d) C₂H₅OC₂H₅
Answer:
b) C₂H₅MgBr

Question 25.
Reaction of ROH with R’ MgX produces
a) RH
b) R’H
c) R – R
d) R’ – R”
Answer:
b) R’H

Question 26.
The solvent used in Wurtz reacton is
a) C₂H₅OH(aq)
b) CH₃COOH
c) H₂O
d) C₂H₅OC₂H₅(dry)
Answer:
d) C₂H₅OC₂H₅(dry)

Question 27.
Which of the following is less reactive than benzene towards electrophilic substitution reactions?
(a) Nitrobenzene
(b) Aniline
(c) Bromobenzene
(d) Chlorobenzene
Answer:
(a) Nitrobenzene

Question 28.
The compound with the highest boiling point is
a) n-Hexane
b) n – Pentane
c) 2, 2 – dimethyl propane
d) 2 – methyl butane
Answer:
a) n-Hexane

Question 29.
The increasing order of reduction of alkyl halides with zinc and dilute HCl is
a) R – Cl < R – I < R – Br
b) R – Cl < R – Br < R – I
c) R – I < R – Br < R – Cl
d) R – Br < R – I < R – Cl
Answer:
b) R – Cl < R – Br < R – I

Question 30.
The volume of oxygen required for the complete combustion of 4 lit of ethane is
a) 4 lit
b) 8 lit
c) 12 lit
d) 14 lit
Answer:
d) 14 lit

Question 31.
The dihedral angle between the hydrogen atoms of 2 methyl groups in the staggered conformation of ethane is
a) 0°
b) 60°
c) 120°
d) 240°
Answer:
b) 60°

Question 32.
The distances between the hydrogen nuclei in staggered and eclipsed form in ethane respectively are
a) 2.55 Å & 2.29 Å
b) 1.54 Å & 1.34 Å
c) 3.5 Å & 2.5 Å
d) 2.29 Å & 2.55 Å
Answer:
a) 2.55 Å & 2.29 Å

Question 33.
Energy barrier between staggered and eclipsed form in ethane is
a) 0.6 kcal / mole
b) 2.9 kcal / mole
c) 12 kcal / mole
d) 14 cal / mole
Answer:
b) 2.9 kcal / mole

Question 34.
IUPAC name of the following compound

a) 3 – Ethyl, 5 – methyl heptane
b) 5 – Ethyl, 3 – methyl heptane
c) 2 – Ethyl, 5 – methyl heptane
d) 4 – Ethyl, 5 – methyl heptane
Answer:
a) 3 – Ethyl, 5 – methyl heptane

Question 35.
Ethylene is converted to ethane in the presence of Ni at 300°C. In this reaction the hybridization of carbon changes from
a) sp to sp²
b) sp² to sp³
c) sp³ to sp
d) sp to sp³
Answer:
b) sp² to sp³

Question 36.
In which of the following reactions in the preparation of ethane a new C – C bond is formed
a) Sabatier – Senderen’s reaction
b) Reduction of ethyl iodide
c) Decarboxylation
d) Kolbe’s electrolysis
Answer:
d) Kolbe’s electrolysis

Question 37.
Select the correct statements
a) eclipsed and staggered ethanes give different products on reaction with chlorine in presence of light
b) the conformational isomers can be isolated at room temperature
c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°
d) steric strain is minimum in gauche form of n – butane
Answer:
c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°

Question 38.
The fully eclipsed conformation of n-butane is least stable due to the presence of
a) bond opposition strain only
b) steric strain only
c) bond opposition strain as well as steric strain
d) no strain is present in the molecule
Answer:
c) bond opposition strain as well as steric strain

Question 39.
Vinyl group among the following is
a) (CH₃)₂ CH –
b) HC ≡ C –
c) H₂C = CH – CH₂ –
d) CH₂ = CH –
Answer:
d) CH₂ = CH –

Question 40.
The alkene that exhibits geometrical isomerism is
a) propene
b) 2 – methyl propene
c) 2 – butene
d) 2 – methyl – 2 – butene
Answer:
c) 2 – butene

Question 41.

a) 5 – methylocta – 1, 3, 5, 7 – tetraene
b) 4 – methylocta – 1, 3, 5, 7 – tetraene
c) 4 – butenylocta -1, 3- diene
d) octa – 1, 5 – diene
Answer:
b) 4 – methylocta – 1, 3, 5, 7 – tetraene

Question 42.
CH₂ = C (CH₂CH₂CH₃)₂
a) 2 – Propyl pent – 1 – ene
b) 2 – Propyl pent – 2- ene
c) 2 – Propyl pent – 3 – ene
d) 3 – Propyl pent – 1- ene
Answer:
a) 2 – Propyl pent – 1 – ene

Question 43.
The number of sigma (σ) and pi (π) bonds in the following structure are
a) σ bonds – 33 π bonds – 2
b) σ bonds – 22 π bonds – 2
c) σ bonds – 42 π bonds – 2
d) σ bonds – 40 π bonds – 3
Answer:
a) σ bonds – 33 π bonds – 2

Question 44.
In dehydrohalogenation, hydrogen and halogen are removed from
a) the same carbon atom
b) from adjacent carbon atoms
c) from isolate carbon atoms
d) from any two carbon atoms
Answer:
b) from adjacent carbon atoms

Question 45.
Ethylene readily undergoes the following type of reaction.
a) Elimination
b) Addition
c) Rearrangement
d) Substitution
Answer:
b) Addition

Question 46.
Baeyer’s reagent is
a) Aqueous bromine solution
b) Neutral permanganate solution
c) Acidified permanganate solution
d) Alkaline potassium permanganate solution
Answer:
d) Alkaline potassium permanganate solution

Question 47.
Baeyer’s reagent oxidizes ethylene to
a) Ethylene chlorohydrin
b) Ethyl alcohol
c) CO₂ and H₂O
d) Ethane – 1, 2 – diol
Answer:
d) Ethane – 1, 2 – diol

Question 48.
On reductive ozonolysis ethylene gives
a) Aldehyde
b) Ketone
c) Carboxylic acid
d) Ether
Answer:
a) Aldehyde

Question 49.
Polythene is obtained by the polymerization of
a) Styrene
b) A mixture of ethylene and styrene
c) Acetylene
d) Ethene
Answer:
c) Acetylene

Question 50.
Polytetrafluoroethylene is commercially known as
a) Teflon
b) Freon
c) Lewisite
d) Westron
Answer:
a) Teflon

Question 51.
Polythens is
a) (- H₂C = CH₂ -)_n
b) (- HC = CH – )_n
c) (- H₃C – CH₃ – )_n
d) (- H₂C – CH₂ -)_n
Answer:
d) (- H₂C – CH₂ -)_n

Question 52.
When ethanol vapours are passed over alumina heated at 350°C, the main product obtained is
a) C₂H₆
b) C₂H₄
c) C₂H₂
d) C₂H₅OC₂H₅
Answer:
b) C₂H₄

Question 53.
Dehydrohalogenation of ethyl chloride in presence of ale. KOH produces the following
a) HC ≡ CH + KCl + H₂O
b) CH₄ + KCl + H₂O
c) CH₂ = CH₂ + KCl + H₂O
d) C₂H₄ + HCl
Answer:
c) CH₂ = CH₂ + KCl + H₂O

Question 54.
Ethylene is prepared by
a) Dehalogenation of chloroform
b) Pyrolysis of ethane at 450°C
c) Dehydration of methanol with Al₂O₃ / 350°C
d) Methyl chloride on reduction
Answer:
b) Pyrolysis of ethane at 450°C

Question 55.
The peroxide effect involves
a) Ionic mechanism
b) Free – radical mechanism
c) Heterolytic fission of double bond
d) Homolytic fission of double bond
Answer:
b) Free – radical mechanism

Question 56.
In which of the following will Kharasch effect operate?
a) CH₃ – CH₂ – CH = CH₂ + HCl
b) CH₃ – CH₂ – CH = CH₂ + HBr
c) CH₃ – CH = CH – CH₃ + HBr
d) CH₃ – CH₂ – CH = CH₂ + HI
Answer:
b) CH₃ – CH₂ – CH = CH₂ + HBr

Question 57.
Anti Markownikoff addition of HBr is not observed in
a) Propene
b) Butene – 1
c) Butene – 2
d) pentene – 2
Answer:
c) Butene – 2

Question 58.
Conditions used for the formation of ethylene glycol from ethylene
a) bromine water
b) cold alkaline KMnO₄
c) dil H₂SO₄, 60°C
d) Ag / 200°C
Answer:
b) cold alkaline KMnO₄

Question 59.
The olefin which on ozonolysis gives CH₃CH₂CHO and CH₃CHO is
a) 1 – butene
b) 2 – butene
c) 1 – pentene
d) 2 – pentene
Answer:
d) 2 – pentene

Question 60.
In the following reaction, A and B respectively are,

a) C₂H₄ and alcoholic KOH / ∆
b) C₂H₅Cl and aqueous KOH / ∆
c) C₂H₅OH and aq KOH / ∆
d) C₂H₂ and Br₂
Answer:
a) C₂H₄ and alcoholic KOH / ∆

Question 61.
The number of possible alkynes with molecular formula C₅H₈ is
a) 3
b) 4
c) 5
d) 6
Answer:
a) 3

Question 62.
The number of open chain structural isomers possible with molecular formula C5H8 is
a) 7
b) 6
c) 5
d) 4
Answer:
a) 7

Question 63.
Alkynes exhibit.
a) Chain isomerism
b) Position isomerism
c) Functional isomerism
d) All the above
Answer:
d) All the above

Question 64.
The IUPAC name of the compound having the formula CH ≡ C – CH = CH₂
a) Butene – 2 – ye
b) But – 2- yne – 3 – ene
c) 3 – butane 1 – ene
d) But – 1 – ene – 3 – yne
Answer:
d) But – 1 – ene – 3 – yne

Question 65.
Acetylene can be obtained by the electrolysis of the following compound
a) Potassium fumerate
b) Potassium succinate
c) Potassium acetate
d) Potassium formate
Answer:
a) Potassium fumerate

Question 66.
The gas obtained when ethylene chloride reacts with alcoholic potash and sodamide is
a) C₂H₄
b) C₂H₆
c) C₂H₂
d) C₂H₅Cl
Answer:
c) C₂H₂

Question 67.
PVC is the polymer of the following
a) Ethyl chloride
b) Vinyl Chloride
c) Allyl Chloride
d) Ethynyl chloride
Answer:
b) Vinyl Chloride

Question 68.
Which of the following possess acidic hydrogen
a) C₂H₆
b) C₂H₄
c) C₂H₂
d) CH₄
Answer:
c) C₂H₂

Question 69.
Hydrocarbon which gives an oxyacetylene flame
a) ethane
b) ethene
c) ethyne
d) ethanol
Answer:
c) ethyne

Question 70.
The isomer of propyne
a) Allene
b) Propene
c) Cyclopropane
d) Propane
Answer:
a) Allene

Question 71.
Bond angle between C – C in alkyne
a) 109°.28′
b) 120°
c) 180°
d) 60°
Answer:
c) 180°

Question 72.
The molecule having linear structure is
a) Methane
b) Ethylene
c) Acetylene
d) Water
Answer:
c) Acetylene

Question 73.
The C – C bond length is shortest in
a) C₂H₆
b) C₃H₈
c) C₆H₆
d) C₂H₄
Answer:
b) C₃H₈

Question 74.
The hydrolysis of Mg₂C₃ produces
a) acetylene
b) propyne
c) butyne
d) ethylene
Answer:
b) propyne

Question 75.
1 – pentyne and 2 – pentyne can be distinguished by
a) Silver mirror test
b) Iodoform test
c) Addition of H₂
d) Baeyers test
Answer:
a) Silver mirror test

Question 76.
Acetylene on reaction with silver nitrate shows
a) Oxidizing property
b) Reducing property
c) Basic nature
d) Acidic nature
Answer:
d) Acidic nature

Question 77.
The acidic nature of hydrogens in acetylene cannot be explained by the reaction with
a) Sodium metal
b) Ammonical cuprous chloride solution
c) Ammonical silver nitrate solution
d) HCN
Answer:
d) HCN

Question 78.
Westron is the solvent obtained by the reaction of chlorine with
a) Ethylene
b) Ethyne
c) Ethane
d) Methane
Answer:
b) Ethyne

Question 79.
CH ≡ CH is
a) CH₂ = CH – CH = CH₂
b) HC ≡ C – CH₃
c) CH₂ = CH – CH₃
d) CH₃ – CH₂ – CH₃
Answer:
b) HC ≡ C – CH₃

Question 80.
Hydration of ethyne to ethanal takes place through the formation of
a) CH₃CH(OH)₂
b) CH₂ = CHOH
c) CH₂ = CHO^–
d) CH ≡ C^–
Answer:
b) CH₂ = CHOH

Question 81.
B is
a) Acetylene
b) Acetaldehyde
c) Acetone
d) Acetic acid
Answer:
b) Acetaldehyde

Question 82.
is
a) Ethyl chloride
b) 1, 2 dichloro ethene
c) Vinyl chloride
d) Ethylidine chloride
Answer:
c) Vinyl chloride

Question 83.
The Polymer ‘B’ is
a) orlon
b) PVC
c) nylon
d) Teflon
Answer:
b) PVC

Question 84.
Benzene is ______ molecule.
a) Tetrahedral
b) Planar
c) Trigonal
d) Square planar
Answer:
b) Planar

Question 85.
Bond length of C – C in benzene.
a) 1.34 Å
b) 1.39 Å
c) 1.54 Å
d) 1.20 Å
Answer:
c) 1.54 Å

Question 86.
The resonance energy Benzene is
a) 36 kcal / mol
b) 85.8 kJ / mole
c) 150.48 kJ / mole
d) Both (a) and (c)
Answer:
d) Both (a) and (c)

Question 87.
In Huckel’s (4n + 2) π rule for aromaticity, ‘n’ represents
a) Number of carbon atoms
b) Number of rings
c) Whole number
d) Fractional number (or) integer (or) zero
Answer:
c) Whole number

Question 88.
Coal tar is obtained as a by product during
a) Destructive distillation of wood
b) Destructive distillation of coal
c) Destructive distillation of bones
d) steam distillation of light oil
Answer:
b) Destructive distillation of coal

Question 89.
Gammaxene is ________ isomer of benzene hexa chloride.
a) α
b) β
c) γ
d) δ
Answer:
c) γ

Question 90.
The empirical formula of benzene and acetylene is/are
a) CH₂, CH
b) CH₂, CH₂
c) CH, CH
d) CH₃, CH₃
Answer:
c) CH, CH

Question 91.
Chemical name of the insecticide gammaxene is
a) DDT
b) Benzene hexa chloride
c) Chloral
d) Hexa chloro ethane
Answer:
b) Benzene hexa chloride

Question 92.
Which is non benzenoidal aromatic compound?
a) Benzene
b) Pyridine
c) Toluene
d) Phenol
Answer:
b) Pyridine

Question 93.
Benzene is purified by
a) distillation
b) fractional distillation
c) Evaporation
d) sublimation
Answer:
b) fractional distillation

Question 94.
Preparation of benzene from phenol is
a) Reduction
b) Oxidation
c) Addition
d) Dehydrogenation
Answer:
a) Reduction

Question 95.
The true statement about benzene is
a) Because of unsaturation benzene easily undergoes addition reactions
b) There are two types C – C bonds in benzene molecule
c) There is a cyclic delocalization of pi – electrons in benzene
d) Mono substitution of benzene gives three isomeric products
Answer:
c) There is a cyclic delocalization of pi – electrons in benzene

Question 96.
– COOH group in electrophilic substitution directs the incoming group to
a) o – position
b) p – position
c) m – position
d) o – and p – position
Answer:
c) m – position

Question 97.
Which of the following is not meta directing group?
a) -SO₃H
b) – NO₂
c) – CN
d) – NH₂
Answer:
d) – NH₂

Question 98.
Which among the following is very strong o – , p – directing group?
a) – Cl
b) – OR
c) – NH₂
d) – NHR
Answer:
d) – NHR

Question 99.
Cyclo butadiene is said to be
a) Aromatic
b) Aliphatic
c) anti aromatic
d) heterocyclic
Answer:
c) anti aromatic

Question 100.
In the reaction
the attacking species is
a) Cl₂
b) Cl⁺
c) Cl^–
d) FeCl₄^–
Answer:
b) Cl⁺

Question 101.
The electrophile in Acetylation of Benzene

Answer:
b)

Question 102.
The ratio of sigma and pi bonds in benzene is
a) 4 : 1
b) 2 : 3
c) 6 : 1
d) 1 : 1
Answer:
a) 4 : 1

Question 103.
Benzene reacts with _______ to yield acetophenone.
a) CH₃COCl + AlCl₃
b) C₆H₅COCl + AlCl₃
c) RCOCl + AlCl₃
d) C₂H₅COCl + AlCl₃
Answer:
a) CH₃COCl + AlCl₃

Question 104.
The order of activites of the various O and P director is
a) – O^– > – OH > – OCOCH₃ > – COCH₃
b) – OH > – O^– > – OCOCH₃ > – COCH₃
c) – OH > – O^– > – COCH₃ > – OCOCH₃
d) -O^– > – COCH₃ > -OCOCH₃ > -OH
Answer:
a) – O^– > – OH > – OCOCH₃ > – COCH₃

II. Very short question and answers (2 Marks):

Question 1.
What are alkanes? Give example.
Answer:
Alkanes are saturated hydrocarbons represented by the general formula C_nH_{2n + 2} where ‘n’ is the number of carbon atoms in the molecule. Methane CH₄, is- the first member of alkane family. The successive members are ethane C₂H₆, propane C₃H₈,butane C₄H₁₀, pentane C₅H₁₂ and so on. It is evident that each member differs from its proceeding or succeeding member by a -CH₂ group.

Question 2.
Give the IUPAC name of the following alkane.
a)

b)
Answer:
a)

b)

Question 3.
Draw the structural formula for 4,5 – diethyl – 3, 4, 5 – trimethyl octane.
Answer:

Question 4.
How is methane prepared from sodium acetate?
Answer:
When a mixture of sodium salt of carboxylic acid and soda lime (sodium hydroxide + calcium oxide) is heated, alkane is formed. The alkane formed has one carbon atom less than carboxylic acid. This process of eliminating carboxylic
group is known as decarboxylation.
CH₃COONa + NaOH CH₄ + Na₂CO₃
Sodium acetate                            Methane

Question 5.
Write a short note on Kolbe’s Electrolytic method.
Answer:
When sodium or potassium salt of carboxylic acid is electrolyzed, a higher alkane is formed. The decarboxylative dimerization of two carboxylic acid occurs. This method is suitable for preparing symmetrical alkanes(R-R).

Question 6.
How will you convert chloro propane into propane?
Answer:
When chloro propane is reduced with Zn / HCl it gives propane.
Example:

Question 7.
Explain the combustion reaction of alkane with suitable example.
Answer:
A combustion reaction is a chemical reaction between a substances and oxygen with evolution of heat and light (usually as a flame). In the presence of sufficient oxygen, alkanes undergoes combustion when ignited and produces carbondioxide and water.

The combustion reaction is expressed as follows.
Example:
CH₄ + 2O₂ → CO₂ + 2H₂O
∆H° = -890.4 KJ

When alkanes burn in insufficient supply of oxygen, they form carbonmonoxide and carbon black.
2CH₄ + 3O₂ 2CO + 4H₂O
CH₄ + O₂ → C + 2H₂O

Question 8.
How are the following compound prepared by the Aromatisation method?
(i) Benzene
(ii) Toluene
Answer:
(i) Benzene
n-Hexane passed over Cr₂O₃ supported on alumina at 873 k gives benzene.

(ii) Toluene

Question 9.
What happens when methane reacts with steam?
Answer:
Methane reacts with steam at 1273K in the presence of Nickel and decomposes to form carbon monoxide and hydrogen gas.
CH₄(g) + H₂O(g) CO(g) + 3H₂
Production of H₂ gas from methane is known as steam reforming process and it is a well-established industrial process for the production of H₂ gas from hydrocarbons.

Question 10.
Explain the Isomerisation reaction with suitable example.
Answer:
Isomerisation is a chemical process by which a compound is transformed into any its isomeric forms. Normal alkanes can be converted into branched alkanes in the presence of AlCl₃ and HCl at 298 k.

This process is of great industrial importance. The quality of gasoline is improved by isomerizing its components.

Question 11.
What is isomerism? Mention the types of isomerism?
Answer:
The phenomenon in which the same molecular formula may exhibit different structural arrangements is called isomerism.
There are two types of isomerism. namely.

• Structural isomerism
• Stereoisomerism

Question 12.
Explain the Geometrical isomerism of alkene with suitable example.
Answer:
It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.
If the similar groups lie on the same side, then the geometrical isomers are called Cis-isomers. When the similar groups lie on the opposite side, it is called a Trans isomer.
Example:
The geometrical isomers of 2-Butene is expressed as follows

Question 13.
Write the test for Alkene.
Answer:
Bromine in water is reddish brown colour. When small about of bromine water is added to an alkene, the solution is decolourised as it forms dibromo compound. So, this is the characteristic test for unsaturated compounds.
Example:

Question 14.
Mention uses of Alkenes.
Answer:

• Alkenes find many diverse applications in industry. They are used as starting materials in the synthesis of alcohols, plastics, liquors, detergents and fuels
• Ethene is the most important organic feed stock in the polymer industry. E.g,’ PVC, Sarans and polyethylene. These polymer are used in the manufacture of floor tiles, shoe soles, synthetic fibres,raincoats, pipes etc.

Question 15.
How is propyne prepared from propylidine chloride?
Answer:

Question 16.
How is ethyne prepared from CaC₂?
Answer:
Ethyne can be manufactured in large scale
by action of calcium carbide with water.

Calcium carbide needed for this reaction is prepared by heating quick lime and coke in an electric furance at 3273 K
CaO + 3C CaC₂ + CO

Question 17.
Write the reaction of propyne using Br₂.
Answer:

When Br₂ in CCl₄ (Reddishbrown) is added to an aikyne, the colour is decolounsed. This is the test for unsaturation.

Question 18.
Mention the uses of alkynes.
Answer:

1. Acetylene is used in oxy acetylene torch used for welding and cutting metals.
2. It is used for manufacture of PVC, polyvinyl acetate, polyvinyl ether, orlon and neoprene rubbers.

III. Short question and answers (3 Marks):

Question 1.
How are the following conversions carried out?
(i) propene → propane
(ii) ethene → ethane
(iii) prop – 1 – yne → propane
Answer:
(i) propene → propane
CH₃ – CH = CH₂ + H₂ CH₃ – CH₂ – CH₃
propene                                     propane

(ii) ethene → ethane
CH₂ = CH₂ + H₂ CH₃ – CH₃
ethene                               ethane

(iii) prop – 1 – yne → propane
CH₃ – C ≡ CH + 2H₂ CH₃ – CH₂ – CH₃
prop – 1 – yne                            propane

Question 2.
write the IUPAC name of the following compounds.

Answer:

Question 3.
Write the structural formula for the following compounds.
(i) 3 – Ethyl – 4, 5 – dipropyl octane
(ii) 2, 3 – Dimethyl pentane
(iii) 4 – Ethyl – 2,7 – Dimethyl octane
Answer:
(i) 3 – Ethyl – 4, 5 – dipropyl octane

(ii) 2, 3 – Dimethyl pentane

(iii) 4 – Ethyl – 2,7 – Dimethyl octane

Question 4.
Write a short note on
(i) Wurtz reaction
(ii) Corey – House Mechanism
Answer:
(i) Wurtz reaction:
When a solution of halo alkanes in dry ether is treated with sodium metal, higher alkanes are produced. This reaction is used to prepare higher alkanes with even number of carbon atoms.
Example:
CH₃ – Br + 2Na + Br – CH₃ CH₃ – CH₃ + NaBr
methyl bromide                                        ethane

(ii)Corey-House Mechanism:
An alkyl halide and lithium di alkyl cuprate are reacted to give higher alkane.
Example:
CH₃CH₂Br + (CH₃)₂LiCu → CH₃CH₂CH₃ + CH₃Cu + LiBr
Ethyl bromide

Question 5.
How is methane prepared from Grignard reagent?
Answer:
Methyl chloride reacts with magnesium in presence of dry ether gives methyl magnesium chloride. Then methyl magnesium chloride reacts with water to give methane.
CH₃ – Cl + Mg CH₃MgCl
chloromethane                 methyl magnesium chloride

CH₃MgCl + H₂O → CH₄ + Mg(OH)Cl
methane

Question 6.
What is meant by pyrolysis? Explain the pyrolysis reaction of Ethane and propane.
Answer:
Pyrolysis is defined as the thermal decomposition of organic compound into smaller fragments in the absence of air through the application of heat. ‘Pyro’ means ‘fire’ and ‘lysis’ means ‘separating’. Pyrolysis of alkanes also named as cracking.

In the absence of air, when alkane vapours are passed through red-hot metal it breaks down into simpler hydrocarbons.

1)

2) 2CH₃ – CH₃   CH₂ = CH₂ + 2CH₄
Ethane                                 Ethylene

Question 7.
Mention the uses of alkanes.
Answer:
The exothermic nature of alkane combustion reaction explains the extensive use of alkanes as fuels. Methane present in natural gas is used in home heating. Mixture of propane and butane are known as LPG gas which is used for domestic cooking purpose. GASOLINE is a complex mixture of many hydrocarbons used as a fuel for internal combustion engines.

Carbon black is used in the manufacture of ink, printer ink and black pigments. It is also used s fillers.

Question 8.
Write all possible structural isomers with the molecular formula C₄H₁₀ and name them.
Answer:
(i) CH₃ – CH = CH – CH₃ 2 – butene
(ii) CH₂ = CH – CH₂ – CH₃ 1 – butene
(iii) 2 – mehyl – 1 – propene
structures (i) 8s (ii) are position isomers, structures (i) & (iii), (ii) 8s (iii) are chain isomers.

Question 9.
How is ethene prepared by Kolbe’s electrolytic method?
Answer:
When an aqueous solution of potassium succinate is electrolyzed between two platinum electrodes, ethene is produced at the anode.

Question 10.
How are the following compounds prepared by ozonolysis method?
(i) Formaldehyde
(ii) Acetaldehyde
Answer:
(i) Formaldehyde:

(ii) Acetaldehyde:

Question 11.
How ozone reacts with 2 – methyl propene?
Answer:

When 2 – methyl propene reacts with ozone to give ozonoid. Ozonide is treated with Zn/H₂O gives acetone.

Question 12.
How is acetylene prepared from ethylene?
Answer:
This process involves two steps:
(i) Halogenation of alkenes to form vicinal dihalides
(ii) Dehalogenation of vicinal dihalides to form alkynes.

Question 13.
How is acetylene prepared by Kolbe’s electrolytic method?
Answer:
Electrolysis of sodium or potassium salt of maleic or fumaric acid yields alkynes.

Question 14.
Write Ozonolysis reaction of Propyne?
Answer:

Question 15.
How is BHC prepared? Give its uses.
Answer:
Chlorination of Benzene:
Benzene reacts with three molecules of Cl₂ in the presence of sun light or UV light to yield Benzene Hexa Chloride (BHC) C₆H₆Cl₆. This is known as gammaxane or Lindane which is a powerful insecticide.

Question 16.
How propane is prepared form 1, 2 – dichloro propane?
Answer:

Question 17.
Explain the polymerization reaction of alkenes.
Answer:
A polymer is a large molecule formed by the combination of larger number of small molecules. The process is known as polymensation. Alkenes undergo polymerisation at high temperature and pressure, in the presence of a catalyst.
Example:
red hot

Question 18.
Write a note on acidic nature of Alkynes.
Answer:
An alkyne shows acidic nature only if it contains terminal hydrogen. This can be explained by considering sp hybrid orbitals of carbon atom in alkynes. The percentage of S-character of sp hybrid orbital (50%) is more than sp² hybrid orbital of alkene (33%) and sp³ hybrid orbital of alkane (25%). Because of this, Carbon becomes more electronegative facilitating donation of H+ ions to bases. So hydrogen attached to triply bonded carbon atoms is acidic but not the others.

Question 19.
Write a short note on
(i) Wurtz – Fittig reactions
(ii) Friedel Craft’s reaction
Answer:
(i) Wurtz – Fittig reactions:
When a solution of bromo benzene and iodo methane in dry ether is treated with metallic sodium, toluene is formed.

(ii) Friedel Craft’s reaction:
When benzene is treated with methyl chloride in the presence of anhydrous aluminium chloride, toluene is formed.

IV. Long Question and answers (5 Marks):

Question 1.
How to draw structural formula for given IUPAC name with suitable example.
Answer:
After you learn the rules for naming alkanes, it is relatively easy to reverse the procedure and translate the name of an alkane into a structural formula. The example below show how this is done.
Let us draw the structural formula for
3 – ethyl – 2, 3 – dimethyl pentane
Step 1:
The parent hydrocarbon is pentane. Draw the chain of five carbon atoms and number it. .

Step 2:
Complete the carbon skeleton by attaching the alkyl group as they are specified in the name. An ethyl group is attached to carbon 3 and two methyl groups are attached to carbon 2 and 3.

Step 3:
Add hydrogen atoms to the carbon skeleton so that each carbon atoms has four bonds.

Question 2.
Explain the conformations of ethane.
Answer:
The two tetrahedral methyl groups can rotate about the carbon – carbon bond axis yielding several arrangements called conformers. The extreme conformations are staggered and eclipsed conformation. There can be number of other arrangements between staggered and eclipsed forms and their arrangements are known as skew forms.
Eclipsed conformation:

In this conformation, the hydrogen’s of one carbon are directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.

Staggered conformation:

In this conformation, the hydrogen’s of both the carbon atoms are far apart from each other. The repulsion between the atoms is minimum and it is the most stable conformation.

Skew conformation:

The infinite numbers of possible intermediate conformations between the two extreme conformations are referred as skew conformations. The stabilities of various conformations of ethane are:
Staggered > Skew > Eclipsed
The potential energy difference between the staggered and eclipsed conformation of ethane is around 12.5 kJmol^-1. The various conformations can be represented by new man projection formula.

Eclipsed Skew Staggered newmann projection formula for Ethane.

Question 3.
Explain the steps involved in the mechanism of halogenations of alkane.
Halogenation:
A Halogenation reaction is the chemical reaction between an alkane and halogen in which one or more hydrogen atoms are substituted by the halogens.

Chlorination and Bromination are two widely used halogenation reactions. Fluorination is too quick and iodination too slow. Methane reacts with chlorine in the presence of light or when heated as follows.

Mechanism:
The reaction proceeds through the free radical chain mechanism. This mechanism is characterized by three steps initiation, propagation and termination.

(i) Chain Initiation:
The chain is initiated by UV light leading to homolytic fission of chlorine molecules into free radicals (chlorine atoms).

Here we choose Cl – Cl bond for fission because C – C & C – H bonds are stronger than Cl – Cl.

ii) Propagation:
It proceeds as follows
a) Chlorine free radial attacks the methane molecule and breaks the C – H bond resulting in the generation of methyl free radical

b) The methyl free radical thus obtained attacks the second molecule of chlorine to give chloromethane (CH₃Cl) and a chlorine free radical as follows.

c) This chlorine free radical then cycles back to step a) and both step a) and b) are repeated many times and thus chain of reaction is set up.

iii) Chain termination:
After sometimes, the reactions stops due to consumption of reactant and the chain is terminated by the combination of free radicals.

Question 4.
Write the IUPAC name of the following compounds.
1) CH₃ – CH = CH₂
2) CH₃ – CH = CH – CH₃
3)
4)
Answer:

Question 5.
Write the IUPAC name of the following compounds.

Answer:
1)

2)

3)

4)

Question 6.
Draw the structures for the following alkenes.
(i) 6 – Bromo – 2, 3 – dimethyl – 2 – hexene
(ii) 5 – Bromo – 4 – chloro – 1 -heptene
(iii) 2, 5 – dimethyl – 4 – octene
(iv) 4 – Methyl – 2 – pentene
Answer:
(i) 6- Bromo – 2, 3-dimethyl – 2 -hexene

(ii) 5 – Bromo – 4 – chloro – 1 -heptene

(iii) 2, 5 – dimethyl – 4 – octene

(iv) 4 – Methyl – 2 – pentene

Question 7.
Draw the structure and write down the IUPAC name for the isomerism exhibited by the molecular formulae:
(i) C₅H₁₂ – Pentene (3 isomers)
(ii) C₆H₁₄ – Hexene (5 isomers)
(iii) C₅H₁₂ – Pentene (3 isomers)
Answer:
(i) C₅H₁₂ – Pentene
a) CH₃ – CH₂ – CH₂ – CH = CH₂
1 – Pentene

b)
2 – methyl – 2 – butene

c) CH₃CH = CH – CH₂CH₃
2 – pentene

(ii) C₆H₁₄ – Hexene
a) CH₃ – CH₂ – CH₂ – CH₂ – CH = CH₂
1 – hexane

b) CH₃ – CH₂ – CH₂ – CH = CH – CH₃
2 – hexane

c) CH₃ – CH₂ – CH = CH – CH₃ – CH₂
3 – hexane

d)

e)

Question 8.
How are the following conversions carried out?
(i) ethanol → ethene
(ii) 2- butyne cis → 2 – butene and trans – 2 – butane
(iii) 1 – bromopropane → prop – 1 – ene
(iv) 1 – 2 – dibromoethane – ethene
Answer:
1) ethanol → ethene
C₂H₅OH H₂C = H₂C
ethanol ethane

(ii) 2 – butyne → cis – 2 – butene
H₃C – C ≡ C – CH₃ + H₂

2 – butane → trans – 2 – butane

(iii) 1 – bromopropane → prop – 1 – ene
H₃C – CH₂ – CH₂ – Br H₃C – CH = CH₂ + KBr + H₂O
1 – bromopropane → prop – 1 – ene

(iv) 1 – 2 – dibromoethane – ethene

1 – 2 – dibromoethane – ethene

Question 9.
Write the mechanism involved in the addition of HBr to alkene in the presence of organic peroxide.
Answer:
Anti-Markovnikoff’s Rule (Or) Peroxide Effect (Or) KharaschAddition:
The addition of HBr to an alkene in the presence of organic peroxide takes place in the direction opposite to that predicted by Markovnikoff’s rule.
CH₃ – CH = CH₂ + HBr CH₃ – CH₂ – CH₃ – CH₂ – CH₂ – Br
prop – 1 – ene                                                          1 – bromopropane

Mechanism:
The reaction proceeds via free radical mechanism.

Step 1:
The week O – O single bond linkages of peroxides undergoes homolytic cleavage to generate free radical.

Step 2:
The radicals abstract H from HBr thus generating bromine radical.

Step 3:
The Bromine radical adds to the double bond in the way to form more stable alkyl free radical.

Step 4:
Addition of HBr to secondary free radical

Step 5:
Addition of HBr to primary free radical

Question 10.
How are the following conversions carried out?
Answer:
(i) Propene → 2 – Propanol
(ii) ethene → ethylene glycol
(iii) 2 – butene → Acetic acid
(iv) 2 – methyl – 1 – propene → Acetone
Answer:
(i) Propene → 2 – Propanol

(ii) ethene → ethylene glycol

(iii) 2 – butene → Acetic acid

(iv) 2 – methyl – 1 – propene → Acetone

Question 11.
An organic compound (A) C₂H₄ decolourises bromine water. (A) on reaction with chlorine gives (B) A reacts with HBr to give (C). Identity (A), (B), (C). Explain th. reactions.
Answer:

Result:
(A) – ethylene
(B) – 1, 2 – dichloroethane
(C) – ethyl bromide

Question 12.
How are the following conversions carried out?
(i) 2 – butyne → 2, 2 – dichlorobutane
(ii) 1 – butyne → 2, 2 – dibromobutane
(iii) ethyne → acetaldehyde
(iv) Propyne → propan – 2 – one
Answer:
(i) 2 – butyne → 2, 2 – dichlorobutane

(ii) 1 – butyne → 2, 2 – dibromobutane

(iii) ethyne → acetaldehyde

(iv) Propyne → propan – 2 – one

Question 13.
How are the following compounds prepared from ethyne?
(i) Formic acid
(ii) Vinyl ethylene
(iii) Benzene
Answer:
(i) Formic acid

(ii) Vinyl ethylene
2CH ≡ CH CH₂ = CH – C ≡ CH (Vinyl ethylene)

(iii) Benzene

Question 14.
Explain the Huckel’s rule of Aromaticity.
Answer:
Huckel proposed that aromaticity is a function of electronic structure. A compound may be aromatic, if it obeys the following rules (Huckel’s rule)
i) The molecule must be co-planar
ii) Complete delocalization of n electron in the ring
iii) Presence of (4n+2) n electrons in the ring where n is an integer (n = 0,1,2….)
This is known as Huckle’s rule.
Some of the examples which obey Huckel rule.

1. The benzene ring is planar with delocalized π electrons (6)
It obeys Huckel’s rule because 4n + 2 = (4 × 1) + 2 = 6π electrons.
Hence, it is an aromatic compound

2. It has planar ring structure with delocalized π electrons (10).
Applying Huckle’s rule (4 × 2) + 2 = 10π electrons.
Hence it is an aromatic compound

Naphthalene

3. It has a planar ring structure with delocalized π electrons(14).
Applying Huckle’s rule (4 × 3) + 2 = 14π electrons.

4. It has planar ring structure but the π electrons are not delocalized.
Applying Huckle’s rule (4 × 1) + 2 = 6π electrons.
But it has only 4electrons. So it is not an aromatic compound

5. It has a tub shaped ring structure. It does not obey Huckel’s rule.
Applying Huckle’s rule 4n + 2 = 8π electrons if n = 2
So it is not an aromatic compound.

Question 15.
Explain the Kekule’s structure of benzene.
Answer:
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
(i) Benzene forms only one orthodisubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.

Presence of double bond between the substituents

Presence of single bond between the substituents

(ii) Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) in rapid equilibrium.

Question 16.
Explain the resonance in benzene.
Answer:
The phenomenon in which two or more structures can be written for a substance which has identical portions of atoms in called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alter nature structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.

The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

Question 17.
Write the industrial preparation of benzene from coal tar.
Answer:
Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds namely benzene, toluene, xylene in the temperature range of 350 to 443 K. These vapours are collected at the upper part of the fractionating column

Question 18.
How is benzene prepared from
(i) Acetylene
(ii) Decarboxylation
(iii) Phenol
Answer:
(i) From acetylene
Acetylene on passing through a red -hot tube trimerises to give benzene. We havealready studied this concept in polymerization of alkynes.

(ii) Decarboxylation of aromatic acid.
When sodium benzonate is heated with sodalime, benzene vapours distil over.
C₆H₅COONa + NaOH C₆H₆ + Na₂CO₃

(iii) Preparation of benzene from Phenol
When phenol vapours are passed over zinc dust, then it is reduced to benzene.
C₆H₆OH + Zn  C₆H₆ + ZnO

Question 19.
Explain Electrophilic substitution reaction of benzene.
Answer:
(i) nitration:
When benzene is heated at 330K with a nitrating mixture (Con. HNO₃ + Con. H₂SO₄), nitro benzene is formed by replacing of hydrogen atom by nitronium ion NO₂⁺ (electrophile).

Concentrated H₂SO₄ is added to produce nitronium ion NO₂⁺.

(ii) Halogenation:
Benzene reacts with halogens (X₂ = Cl₂, Br₂,) in the presence of Lewis acid such as FeCl₃, FeBr₃ orAlCl₃ and give corresponding halo benzene.

(iii) Sulphonation:
Benzene reacts with fuming sulphuric acid (Con. H₂SO₄ + SO₃) and gives benzene sulphonic acid. The electrophile SO₃ is a molecule.

(iv) Friedel craft’s alkylation: (methylation)
When benzene is treated with a alkyl halide in the presence of only AlCl₃, alkyl benzene is formed.

(v) Friedel craft’s acylatlon: (Acetylation)
When benzene is treated with acetyichioride in the presence of AlCl₃, acyl benzene is formed.

Question 20.
How is benzene converted into
a) Cyclo hezane
b) maleic anhydride
c) 1, 4 cyclo hexadiene
Answer:
a) Cyclo hexane:
Benzene reacts with hydrogen in the presence of Platinum or Palladium to yield Cyclohexane. This is known as hydrogenation.

b) maleic anhydride:
Although benzene is very stable to strong oxidizing agents, it quickly undergoes vapour phase oxidation by passing its vapour mixed with oxygen over V₂O₅ at 773k. The ring breaks to give maleic anhydride.

c) 1, 4 cyclo hexadiene:
Benzene can be reduced to non- conjugated dienes 1, 4 – cyclohexadiene by treatment with Na or Li in a mixture of liquid ammonia and alcohol. It is the convenient method to prepare cyclicdienes.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 8 Excretion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

11th Bio Zoology Guide Excretion Text Book Back Questions and Answers

Part – I

Question 1.
Arrange the following structures in the order that a drop of water entering the nephron would encounter them.
a) Afferent arteriole
b) Bowman’s Capsule
c) Collecting duct
d) Distal tubule
e) Glomerulus
f) Loop of Henle
g) Proximal tubule
h) Renal Pelvis
Answer:
(a), (e), (b), (g), (f), (d), (c), (h)

Question 2.
Name the three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule. What components of the blood are usually excluded by these layers.
Answer:
The three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule are

1. Glomerular capillary endothelium
2. Basal lamina or basement membrane
3. Epithelium of Bowman’s capsule. Blood corpuscles and plasma protein are excluded by these layers.

Question 3.
What forces promote glomerular filtration? What forces oppose them? What is meant by net filtration pressure?
Answer:

• Glomerulus hydrostatic pressure
• Glomerulus pressure
• Opposing pressure: Colloidal osmotic pressure, Capsular hydrostatic pressure
• Net filtration pressure = Glomerular hydrostatic pressure – (Colloidal osmotic pressure + capsular hydrostatic pressure.

Question 4.
Identify the following structures and explain their significance in renal physiology?
Answer:
1. Juxtaglomerular apparatus:
Juxtaglomerular apparatus is a specialized tissue in the afferent arteriole of the nephron that consists of macula densa and granular cells. The macula densa cells sense distal tubular flow and affect afferent arteriole diameter. The granular cells secrete an enzyme called renin. It plays an important role in reabsorption of water, Na⁺ and excretion of K⁺.

2. Podocytes:
The visceral layer of the Bowman’s capsule is made up of epithelial cells called podocytes. The podocytes end in foot processes which cling to the basement membrane of the glomerulus. The openings between the foot processes are called filtration slits. It is important for glomerular filtration.

3. Sphincters in the bladders:
Sphincter muscles in the bladder controls the flow of urine from the bladder. When urinary bladder is filled with urine, it stretches and stimulates the central nervous system through the sensory neurons of the parasympathetic nervous system and brings about contraction of the bladder.

Simultaneously, somatic motor neurons induce the sphincters to close. Smooth muscles contracts resulting in the opening of the internal sphincters passively and relaxing the external sphincter. When the stimulatory and inhibitory controls exceed the threshold, the sphincter opens and the urine is expelled out.

4. Renal cortex:
The outer portion of the kidney is the renal cortex. It contains renal corpuscles and the proximal and distal tubules. It is thin and fibrous.

Question 5.
In which segment of the nephron most of the re-absorption of substances takes place?
Answer:
In the proximal convoluted tubule.

Question 6.
When a molecule or ion is reabsorbed from the lumen of the nephron, Where does it go?
Answer:
When a molecule or ion is reabsorbed from the lumen of the nephron, it goes to the bloodstream through an efferent arteriole which carries blood away from the glomerulus. If a solute is filtered and not reabsorbed from the tubule, it goes along with urine.

Question 7.
Which segment is the site of secretion and regulated reabsorption of ions and pH homeostasis?
Answer:

• 1. Distal tubule; 2. Collecting duct
• In the renal tubules H⁺ and NH⁴ are secreted and liberated into the tubules and excreted through
  urine thus maintaining acid base balance.
• For each H⁺ ions from the liberated filtrate one Na⁺ is reabsorbed in the tubules.
• The secreted HCO₃, PO₃ and NH₃ combines to form carbonic acid and phosphoric acids.
• Thus the H⁺ are maintained and their reabsorption is prevented.

Question 8.
What solute is normally present in the body to estimate GFR in humans?
Answer:

• The glomerulus filtrate consists of water glucose amino acids creatinine protein salts and urea.
• These solutes decide the glomerular filtration rate.

Question 9.
Which part of the autonomic nervous system is involved in the micturition process?
Answer:
Parasympathetic nervous system.

Question 10.
If the afferent arteriole of the nephron constricts, what happens to the GFR in that nephron? If the efferent arteriole constricts what happens to the GFR in that nephron? Assume that no autoregulation takes place.
Answer:

• Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
• Because afferent arteriole is wider than efferent arteriole.
• The constriction of this does not affect the rate of filtration.

Question 11.
The concentration of urine depends upon which part of the nephron.
Answer:
a) Bowman’s capsule
b) Length of Henle’s loop
c) PCT
d) network of capillaries arising from the glomerulus.
Answer:
b) Length of Henle’s loop

Question 12.
If Henle’s loop were absent from mammalian nephron, which one of the following is to be expected?
a) There will be no urine formation
b) There will be hardly any change in the quality and quantity of urine formed.
c) The urine will be more concentrated
d) The urine will be more dilute
Answer:
d) The urine will be more dilute

Question 13.
What will happen if the stretch receptors of the urinary bladder wall are totally removed?
a) Micturition will continue
b) Urine will continue to collect normally in the bladder
c) There will be micturition
d) Urine will not collect in the bladder
Answer:
a) Micturition will continue

Question 14.
The end product of the Ornithine cycle is
a) Carbon dioxide
b) Uric acid
c) Urea
d) Ammonia
Answer:
c) Urea

Question 15.
Identify the wrong match.

a) Bowman’s capsule	Glomerular filtration
b) DCT	Absorption of glucose
c) Henle’s loop	Concentration of urine
d) PCT	Absorption of Na+ and K+ ions

Answer:
b) DCT – Absorption of glucose

Question 16.
Podocytes are the cells present on the
a) Outer wall of Bowman’s capsule
b) Inner wall of Bowman’s capsule
c) Neck of the nephron
d) Wall glomerular capillaries
Answer:
b) Inner wall of Bowman’s capsule

Question 17.
Glomerular filtrate contains
a) Blood without blood cells and proteins
b) Plasma without sugar
c) Blood with proteins but without cells
d) Blood without urea
Answer:
c) Blood with proteins but without cells

Question 18.
Kidney stones are produced due to the deposition of uric acid and Silicates
b) Minerals
c) Calcium Carbonate
d) Calcium oxalate
Answer:
d) Calcium oxalate

Question 19.
Animal requiring the minimum amount of water to produce urine is
a) Ureotelic
b) Ammonotelic
c) Uricotelic
d) Chemotelic
Answer:
c) Uricotelic

Question 20.
Aldosterone acts at the distal convoluted tubule and collecting duct resulting in the absorption of water through
a) Aquaporins
b) Spectrins
c) GLUT
d) Chloride channels
Answer:
a) Aquaporins

Question 21.
The hormone which helps in the reabsorption of water in kidney tubules is
a) Cholecystokinin
b) Angiotensin Il
c) Antidiuretic hormone
d) Pancreozymin
Answer:
c) Antidiuretic hormone

Question 22.
Malpighian tubules remove excretory products from
a) Mouth
b) Oesophagus
c) Haemolymph
d) Alimentary canal
Answer:
c) Haemolymph

Question 23.
Identify the biological term excretion, glomerulus, urinary bladder, glomerular filtrate, ureters, urine, Bowman’s capsule, urinary system, reabsorption, micturition, osmosis, proteins.
Answer:

• A liquid which gathers in the bladder – Urine
• Produced when blood is filtered in a Bowman’s capsule – Glomerular filtrate
• The temporary storage of urine – Urinary bladder
• A ball of intertwined capillaries – Glomerulus
• Removal of unwanted substances from the body – Excretion
• Each contains a glomerulus – Bowman’s Capsule
• Carry urine from the kidneys to the bladder – Ureter
• The scientific term for urination – Micturition
• Regulation of water and dissolved substances in the blood and tissue fluid – Osmoregulation
• Consists of the kidneys ureters and bladder Excretory system
• Removal of useful substances from glomerular filtrate reabsorption
• What solute does blood contain that is not present in the glomerular filtrate? – Plasma Protein

Question 24.
With regards to toxicity and the need for dilution in water how different are ureotelic and ureocotelic excretions? Give examples of animals that use these types of excretions?
Answer:

• The type of nitrogenous end product (Urea or Uric acid or ammonia) of an animal excretion depends upon the habitat of the animal.
• For the Excretion of Ammonia, more water is needed.
• Animals that excrete most of their nitrogen in the form of ammonia are called ammonites.
• Fishes Amphibians. Uric acid can be eliminated with a minimum loss of water and is called Uricoteles.
• Reptiles, birds insects, landrails. Mammals and terrestrial amphibians excrete urea and are called ureoteles.
  Nitrogenous

Question 25.
Differentiate protonephridia from metanephridia.
Answer:

Protonephridia	Metanephridia
1. Proto – first it is a network of dead-end tubules lacking internal opening. (E.g) Platyhelminthes	Meta – after They have a tubular internal opening called nephrostome. (E.g) Earthworm
2. Flame cells are excretory structures.	The excretory products are filters and selectively reabsorbed.
3. Excretory product excretes through nephridiopore	The excretory product excretes through the nephridiopore.
4. Excretory structure are usually osmo regulators	Excretory structures are osmo regulators and helps in excretion.

Question 26.
What is the nitrogenous waste produced by amphibian larvae and by adult animals?
Answer:
The adult amphibian’s excretory product is urea. The larval form of amphibian’s excretory product is ammonia.

Question 27.
How is urea formed in the human body?
Answer:
Urine formation involves three main processes namely Glomerular Alteration, tubular reabsorption and tubular secretion.

I. Glomerular filtration:

• Blood enters the kidney from the renal artery into the glomerulus.
• Blood is composed of water colloidal proteins, sugars and nitrogenous end product.
• The filteration is started in the glomerulus. The fluid that leaves the glomerular capillaries and enters the Bowman’s capsule is called the glomerular filtrate. It formed 170-180 litre within 24 hours.
• Glomerular membrane has large surface area. Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
• This is because afferent arteriole is wider than efferent arteriole and the glomerular hydrostatic pressure is around 55 mm Hg.
• Molecules larger than 5mm are barred from entering the tubule.

The two opposing forces are contributed by the plasma proteins in the capillaries.

1. Colloidal osmotic pressure – 30 mm Hg
2. Hydrostatic pressure -15 mm Hg

Both pressures combine. 30 mm Hg + 15mm Hg = 45mm Hg

• The net filtration pressure of 10mm Hg is responsible for the renal filtration.
• Net filtration pressure – Glomerular
• Hydrostatic Pressure – Colloidal Osmotic pressure + Capsular hydrostatic pressure.

Net filteration pressure = 55 mm Hg – (30mm Hg +15mm Hg) = 10mm Hg

Substance	Concentration in blood Plasm / g dm-3	Concentration in glomerular filtrate / g dm-3
Water	900	900
Proteins	80.	0.05
Aminoacids	0.5	0.5
Glucose	1.0	1.0

Substance	Concentration in blood PIasm /gdm-3	Concentration glomerular filtrate/gdm-3
Urea	0.3	0.3
Uric Acid	0.04	0.04
Creatinine	0.01	0.01
Inorganic ions (mainly (Na+, K+and Cl–)	7.2	7.2

II. Tubular reabsorption:

• The volume of filtrate formed per day is around 170 – 180l.
• Urine released is around 1.51 per day. 99% of the glomerular filtrate is reabsorbed by the renal tubules.

Reabsorption in proximal convoluted tubule:

• Glucose lactate amino acids Na in the filtrate are reabsorbed in the PCT.
• Sodium is reabsorbed by active transport through a sodium-potassium pump in the PCT.
• Small amounts of urea and uric acid are reabsorbed.

Reabsorption in Henle Loop:

• Descending limb of Henle’s loop is permeable to water due to the presence of aqua porlns but not permeable to salts.
• Water is lost in the descending limb. Hence Na and Cl get concentrated in the filtrate.
• Ascending limb of Henle’s loop is impermeable to water but permeable to Na⁺, Cl^– andK⁺.

Distal Convoluted Tubule:

• It recovers water and secretes potassium into tubule.
• Na⁺, Cl^-, and water remain in the DCT.
• Reabsorption of HCO₃ takes place to regulate the blood pH.
• Collecting the tubule is permeable to water potassium ions are actively transported into the tubule and Na⁺ to produce concentrated urine.

Tubular Secretion:

• Once it enters the collecting duct water is absorbed and concentrated hypertonic urine is formed.
• For every H secreted into the tubular filtrate, a Na⁺ is absorbed by the tubular cell.
• The H⁺ secreted combines with HCO⁺₃, HPO^–_3, and NH⁺ and gets fixed as H2CO₃, H2PO_3, and NH4⁺.
• Since H⁺ gets fixed in the fluid reabsorption of H⁺is prevented.

Question 28.
Differentiate cortical from medullary nephrons.
Answer:
In the renal tubules, proximal convoluted tubule and distal convoluted tubule are situated in the cortical region of the kidney and the loop of Henle is in the medullary region.

Cortical Nephron	Medullary Nephron
1. The loop of Henle is too short and extends only very little into the medulla.	Nephrons have a long loop of Henle that run deep in to the medulla.
2. There is no vasa recta.	Vasa recta is present.

Question 29.
What vessels carry blood to the kidneys? Is this blood arterial or venous?
Answer:
Renal artery right and left artery. The blood is arterial.

Question 30.
Which vessels drain filtered blood from the kidneys?
Answer:
Renal vein. The filtered blood is taken from the kidney to the inferior vena cava through the renal vein.

Question 31.
What is tubular secretion? Name the substances secreted through the renal tubules?
Answer:
The movement of substances such as H⁺, K⁺, NH₄⁺, creatinine and organic acids from the peritubular capillaries into the tubular fluid, the filtrate is called Tubular secretion.

Question 32.
How are the kidneys involved in controlling blood volume? How is the volume of blood in the body related to arterial pressure?
Answer:

• When the volume of blood decreases the flow pressure, decreases.
• This can be sensed by the hypothalamus and osmoreceptors are stimulated and the antidiuretic hormone is secreted from the neurohypophysis.
• The aquaporins in the proximal convoluted tubules and collecting tubule reabsorb water. Hence the blood volume increases and the blood pressure increases.

Question 33.
Name the three main hormones that are involved in the regulation of the renal function?
Answer:

• Renin
• Angio Tensin I
• Angiotensin II

Question 34.
What is the function of the antidiuretic hormone? Where is it produced and What stimuli increases or decreases its secretion?
Answer:

• ADH which is also called vasopressin or Antidiuretic hormone is secreted from the neurohypophysis.
• Fluid loss or if blood pressure increases the osmoreceptors of the hypothalamus is stimulated and hence neurohypophysis is stimulated and secretes ADH.
• When the fluid level and pressure are maintained due to the negative feedback mechanism ADH secretion stops.

Question 35.
What is the effect of aldosterone on kidneys and where is it produced?
Answer:

• Due to the stimulation of Angiotensin II the adrenal cortex secretes aldosterone. That causes reabsorption of Na⁺, K⁺ excretion, and absorption of water from distal convoluted tubule and collecting tubule.
• This mechanism is known as Renin – Angiotensin Aldosterone System.

Question 36.
Explain the heart’s role in secreting a hormone that regulates renal function? What hormone is this?
Answer:

• Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or factor (ANF) travels to the kidney where H increases Na+ excretion and increases the blood flow to the glomerulus.
• Acting on the afferent glomerular arterioles as a vasodilator or an efferent arterioles as a vasoconstrictor.
• The first identical natri uretic hormone is atrial natriuretic horome of heart.

Part – II.

11th Bio Zoology Guide Excretion Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
The elimination of ……………………….. requires a large amount of water.
(a) Urea
(b) Uric acid
(c) Ammonia
(d) creatinine
Answer:
(c) ammonia

Question 2.
Whether the following statements are true or false find the correct sequence. Find the correct sequence.
I) According to the environmental changes organism change their osmotic concentration and are called osmo confirmers.
II) Osmo regulators maintain their internal osmotic concentration irrespective of their external osmotic environment.
III) The Euryhaline animals are able to tolerate only narrow fluctuations in the salt concentration?
IV) The steno haline animals can tolerate wide fluctuations in the salt concentration.
Series:
a) I – False, II – True, III – False, IV – True
b) I – True, II – True, III – False, IV- False
c) I – False, II – True, III – False, IV – True
d) I – False, II – False, III – True, IV – True
Answer:
b) I – True, II – True, III – False, IV- False

Question 3.
Solenocytes are the specialized cells for excretion in
(a) Flatworms
(b) Molluscs
(c) Insects
(d) Amphioxus
Answer:
(d) amphioxus

Question 4.
Find out the odd one out.
a) Ammonoteles – Aquatic Amphibians
b) Urico teles – Birds
c) Ureoteles – Land amphibians
d) Ureoteles – Earthworm (When it is in water)
Answer:
d) Ureoteles – Earthworm (When it is in water)

Question 5.
…………………… have antennal glands or green glands which perform an excretory function.
(a) Insects
(b) Annelids
(c) Crustaceans
(d) Flatworms
Answer:
(c) Crustaceans

Question 6.
What type of Urine is formed in organism with long Henle’s loop?
a) Isotonic
b) less concentrated
c) Concentrated
d) None of the above
Answer:
c) Concentrated

Question 7.
In the marine organism the kidney with glomerulus produce ………………. type of concentrated urine.
a) Concentrated than body fluid
b) Equal to the concentration of body
c) Less than the concentration of body fluids
d) None of the above
Answer:
b) Equal to the concentration of body

Question 8.
Aglomerlar kidneys of marine fishes produce little urine that is ………………………. to the body fluid.
(a) Hypotonic
(b) Hyperosmotic
(c) Isoosmotic
(d) None of the above
Answer:
(c) Isoosmotic

Question 9.
What is the outer covering of the kidney?
a) Renal fascia perirenal fat capsule pleura
b) Renal fascia perirenal fat capsule peri cardial membrane
c) Renal fascia, Peri renal fat capsule meninges
d) Renal fascia perirenal fat capsule fibrous capsule
Answer:
d) Renal fascia perirenal fat capsule fibrous capsule

Question 10.
What is meant by renal corpuscle?
a) Glomerulus and Bowman’s Capsule
b) Glomerulus and Malpighian capsule
c) Glomerulus and nephron
d) Glomerulus and Henle’s loop
Answer:
d) Glomerulus and Henle’s loop

Question 11.
Regarding renal tubules find the correct sequences.
a) Glomerulus Bowman’s capsule Malpighian capsule uriniferous tubules.
b) Proximal convoluted tubules thick descending loop thin ascending limb distal convoluted tubule.
c) Proximal convoluted tubule, Henle’s loop. Distal convoluted tubule.
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.
Answer:
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.

Question 12.
Glucose, amino acids, Na+, and water in the filtrate are reabsorbed in the
(a) descending limb of Henle’s loop
(b) ascending limb of Henle’s loop
(c) proximal convoluted tubule
(d) distal convoluted tubule
Answer:
(c) Proximal convoluted tubule

Question 13.
What is the pressure in the afferent arthery of Glomerulus?
a) 55 mm Hg
b) 50 mm Hg
c) 57 mm Hg
d) 58 mm Hg
Answer:
a) 55 mm Hg

Question 14.
What is the amount of filteration of the glomerulus in 24 hours?
a) 200l
b) 180l
c)190l
d) 170l
Answer:
b) 180l

Question 15.
The pH value of human urine is ……………
(a) 7.5
(b) 6.0
(c) 4.3
(d) 9.5
Answer:
(a) 6.0

Question 16.
What is the pressure in glomerulus?
a) 50 mm Hg
b) 60 mm Hg
c) 55 mm Hg
d) 62 mm Hg
Answer:
c) 55 mm Hg

Question 17.
The glomerular pressure encounter ………….. of colloidal osmotic pressure and …………… of hydrostatic pressure.
a) 30mmHg;15mmHg
b) 15mmHg;30mmHg
c) 40 mm Hg; 30 mm Hg
d) 30mmHg;40mmHg
Answer:
a) 30mmHg;15mmHg

Question 18.
What is the net filteration pressure?
a) 55mmHg – (15mmHg + 30mmHg) = 10mm Hg
b) 55 mm Hg – (30 mm Hg +15 mm Hg) =10 mm Hg
c) 55 mm Hg – (35 mm Hg + 5 mm Hg) = 15 mm Hg
d) 55 mm Hg – (5 mm Hg + 35 mm Hg) = 15 mm Hg
Answer:
b) 55 mm Hg – (30 mm Hg + 15 mm Hg) =10 mm Hg

Question 19.
What is the amount of filtrate formed in one minute?
a) 100 ml -125 ml
b) 100 ml -150 ml
c) 120 ml -125 ml
d) 130 ml -140 ml
Answer:
c) 120 ml -125 ml

Question 20.
Through hemodialysis, ……………………….. can be removed from the blood.
(a) Ketone bodies
(b) Glucose
(c) Amino acids
(d) Urea
Answer:
(d) Urea

Question 21.
What is the amount of urine excreted in a day?
a) 1.5l
b) 1l
c) 2l
d) 2.5l
Answer:
a) 1.5l

Question 22.
Which of the following hormones are secreted by the kidney?
a) Renin
b) Gastrin
c) Calcitriol
d) secretin
a) (ii) and (iii)
b) (i) and (iii)
c) (ii) and (iv)
d) (i) and (ii)
Answer:
b) (i) and (iii)

Question 23.
Where is aquaporins present?
a) Proximal convoluted tubule
b) Distal convoluted tubule and ascending limb of Henle
c) The descending limb of Henle and distal convoluted tubule
d) All the above
Answer:
b) Distal convoluted tubule and ascending limb of Henle

Question 24.
Name the hormone that helpsin the reabsorption of water in distal convoluted tubule.
a) Vasopressin
b) Serotonin
c) Oxytocin
d) All the above
Answer:
a) Vasopressin

Question 25.
Where is Renin synthesized in Kidney?
a) Afferent arteriole
b) Efferent arteriole
c) Vasarecta
d) Collecting duct
Answer:
a) Afferent arteriole

Question 26.
Name the hormone that helpsin the conversion of plasma protein angioten- sinogen into angiotensin I
a) Renin
b) Vasopressin
c) ADH
d) STH
Answer:
a) Renin

Question 27.
What is micturition?
a) Excretion of urine from the urinary bladder
b) The formation of urine in the glomerulus
c) Urine formation in distal convoluted tubule
d) the absorption of urine in Bowman’s capsule
Answer:
a) Excretion of urine from the urinary bladder

Question 28.
According to the food that maneats the pH of urine ……………… to ……………. can be altered.
a) pH 4.8-7.5
b) PH 4.9-7.9
c) pH 4.5-8.0
d) pH 4.4 -7.4
Answer:
c) pH 4.5-8.0

Question 29.
What is the reason for the yellow colour of the urine?
a) Urochrome
b) Haemoerythrin
c) Haemocyanin
d) None
Answer:
a) Urochrome

Question 30.
What is the normal pH of urine?
a) pH 6.0
b) pH 5.3
c) pH 6.4
d) pH 5.9
Answer:
a) pH 6.0

Question 31.
What is the amount of CO₂ released from lungs in a day?
a) 20l
b) 19l
c) 18l
d) 119l
Answer:
c) 18l

Question 32.
Confirm
Statement A: The important function of sweat gland is to cool the body.
Statement B: The sweat glands excrete sodi¬um chloride urea and lactic acid.
a) Statement A-True B-True
b) Statement A-True B-False
c) Statement A-FalseB-True
d) Statement A – False B – False
Answer:
a) Statement A-True B-True

Question 33.
Confirmation:
Statement S: When kidney fails suddenly there is more chance of recovery.
Statement T: In the chronic kidney failure there may not be any chance of recovery.
a) Statement S-True T-True
b) Statement S-True T-False
c) Statement S-True T-True
d) Statement S – True T – False
Answer:
c) Statement S-True T-True

Question 34.
Confirmation:
Statement A: The kidney infection leads to inflammation of bladder and kidney
Statement B: Urination with pain, urinary urgency bloodytinged urine.
a) Statement A – True Statement B explain the symptom of A.
b) A – True – The statement B does not explains the statement A
c) A and B are false
d) A False B – True
Answer:
a) Statement A – True Statement B explain the symptom of A.

Question 35.
What is the normal urea level in the blood
a) 17-30mg/100ml
b) 30-35mg/ 100ml
c) 10-15 mg/100 ml
d) 5-10 mg/100 ml
Answer:
a) 17-30mg/100ml

Question 36.
Find the wrong pair.
a) Renal stone – nephrolithiasis
b) Urine – Urochrome
c) Deficiency of ADH – Urine output decreases
d) Skin- Lactic acid excretion
Answer:
c) Deficiency of ADH – Urine output decreases

Question 37.
Confirmation:
Statement A: Bright’s disease is due to the infection of streptococcus in children
statement B: There is inflammation of glomerulus.
a) Statement A and Bare false
b) Statement A True B explain the A
c) A True B False
d) A False B – True
Answer:
b) Statement A True B explain the A

Question 38.
Find the wrong pair.
a) Heporin – Anticoagulating factor
b) Glomerulonephritis – Accumulation of water in the body
c) Primary kidney – Meso nephridia
d) Uremia – Increase in the blood urea level
Answer:
c) Primary kidney – Meso nephridia

Question 39.
How much urine can be stored up in the bladder?
a) 300-600 ml
b) 200-300 ml
c) 400-700 ml
d) 500-800 ml
Answer:
a) 300-600 ml

Question 40.
For how much time urine can be held in the bladder?
a) 6 hours
b) 2 hours
c) 5 hours
d) 3 hours
Answer:
c) 5 hours

Question 41.
The urinary bladder is made up this muscle ……………………….
a) Detrusor muscle
b) Striated muscle
c) Sphincter muscle
d) None of the above
Answer:
a) Detrusor muscle

Question 42.
Match the following:

1. Steno haline	I Shark
2. Eurihaline	II Otter
3. Osmo regulators	III Goldfish
4. Osrno confirmers	IV Salmon

a) 1-IV 2-III 3-II 4-1
b) 1 -II 2-II 3-IV 4-1
c) 1 -III 2-IV 3-II 4-1
d) 1-1 2-II 3-IV 4-III
Answer:
c) 1 – III, 2- IV, 3 – II, 4 – I

Question 43.
Find out the less toxic waste among the following:
a) Urea
b) Uric acid
c) Ammonia
d) Creatinine
Answer:
a) Urea

Question 44.
Find out the ammonotelic organisms?
a) Reptiles
b) Birds
c) Aquatic amphibians
d) Frog
Answer:
c) Aquatic amphibians

Question 45.
Find out the wrong pair.
a) Rennette cells – Annelida
b) Molluscs – Metanephridia
c) Amphioxus – Mesonephridia
d) Tapeworm – Flame cells
Answer:
c) Amphioxus – Mesonephridia

Question 46.
Find the excretory structure of prawn?
a) Malpighian tubules
b) Green glands
c) Rennette cells
d) Peyer gland
Answer:
b) Green glands

Question 47.
Kidney with noglomerulus form this types of urine?
a) Hypertonic
b) Isotonic
c) Hypotonic
d) Neutrotonic
Answer:
c) Hypotonic

Question 48.
Match the following:

1. Conical tissue masses	a) Columns of Bertini
2. Extension in renal	b) Renal pelvis
3. Broad part of Hilum	c) Calyces
4. Projection of pelvis	d) Medullary pyramids

a) 1-d 2-a 3-b 4-c
b) 1-a 2-b 3-c 4-d
c) 1-b 2-a 3-c 4-d
d) 1-a 2-c 3-b 4-d
Answer:
a) 1-d 2-a 3-b 4-c

Question 49.
How much urine can be held in the urinary bladder?
a) 300-800ml
b)300-600ml
c) 100-400ml
d)200-300ml
Answer:
b)300-600ml

Question 50.
The muscles of urinary bladder is called by this name?
a) sphincter muscle
b) Detrusor muscle
c) constricting muscle
d) Peristaltic muscle
Answer:
b) Detrusor muscle

Question 51.
Name the nephron where the Henle’s loop is short?
a) Medullary nephron
b) Cortical nephron
c) Juxta medullary nephron
d) Medulla nephron
Answer:
b) Cortical nephron

Question 52.
The vessel which runs parallel to the loop of Henle is called by this name.
a) Efferent artery
b) Afferent artery
c) Vasarecta
d) Vasanervosa
Answer:
c) Vasarecta

Question 53.
What is the process of urea formation?
a) Ross cycle
b) Ornithine cycle
c) Urea cycle
d) b and c
Answer:
b) Ornithine cycle

Question 54.
What is the pressure in the afferent arteriole?
a) 35 mm Hg
b) 55 mm Hg
c) 20 mm Hg
d) 40 mm Hg
Answer:
b) 55 mm Hg

Question 55.
What is the filtrate entered into the Bowman’s capsule from glomerulus?
a) Secondary urine
b) Primary urine
c) Tertiary urine
d) Quarternary urine
Answer:
b) Primary urine

Question 56.
Find the correct option
Assertion: The glomerular filtrate resembles the blood
Reason: The glomerulus filtered the blood received from efferent artery
a) Assertion True Reason True
b) Assertion False Reason False
c) Assertion True The Reason does not explains the statement
d) Assertion True Reason explains A
Answer:
c) Assertion True The Reason does not explains the statement

Question 57.
What is the fate of water in the descending limb of Henle?
a) The concentrations of Na reduces
b) The concentration of Cl^– ions reduces
c) Na and Cl^– gets concentrated in the filtrate.
d) Formation of Hypotonic solution.
Answer:
c) Na and Cl^– gets concentrated in the filtrate.

Question 58.
How can the pH of blood be regulated?
a) Due to glucose reabsorption
b) Due to the reabsorption of HCO^–₃
c) Due to the reabsorption of Na⁺
d) Due to the reabsorption of Cl^–
a) 1-2 II-4 III-3 IV-1
b) 1-1 II-2 III-3 IV-4
c) 1-4 II-3 III-2 IV-1
d) 1-2 II-l III-3 IV-4
Answer:
b) 1-1 II-2 III-3 IV-4

Question 59.
Match the following
Answer:

I. Potassium	1. Descending limb of Renie
II. Water	2. Active transport
III. Glucose	3. Proximal convoluted tubule
IV. Na, cl, lc	4. Aqua porin

Question 60.
How is osma regulation in medulla maintained?
a) Juxtamedullary
b) Counter current
c) Positive current exchanger
d) Negative control
Answer:
b) Counter current

Question 61.
Find the wrong pair.
a) Vasa recta – Proximal convoluted tubule
b) ADH – Neuro hypophysis
c) Specialized nephron tissue – Juxtaglomerular apparatus
d) Renin – Glanularcell
Answer:
a) Vasa recta – Proximal convoluted tubule

Question 62.
This synthesizes angio tenginogen.
a) Kidney
b) Liver
c) Malpighian tubule
d) Glomerular
Answer:
b) Liver

Question 63.
This increases the absorption of sodium ions.
a) Renin
b) Angiotensin I
c) Angio tensin II
d) Angio tensinogen
Answer:
c) Angiotensin II

Question 64.
Name the hormone that decreases the blood pressure
a) Angiotensin I
b) Atrial natriuretic peptide
c) Vasopressin
d) ADH
Answer:
b) Atrial natriuretic peptide

Question 65.
This decreases the excretion of Renin?
a) Aldosterone
b) vasopressin
c) Atrial natriuretic peptide
d) Ventrical natriuretic peptide
Answer:
c) Atrial natriuretic peptide

Question 66.
What is the pH of Urine?
a) 6.5
b) 7.0
c) 6.0
d) 8
Answer:
c) 6.0

Question 67.
What is the reason for the yellow colour of urine?
a) Uro chrome
b) Cytochrome
c) Chlorochrome
d) Phytochrome
Answer:
a) Uro chrome

Question 68.
The increase in the level of urea
a) Uremia
b) ureomia
c) Uricmia
d) Ketosis
Answer:
a) Uremia

Question 69.
The accumulation of salt in the blood.
a) Poly urea
b) Oligo urea
c) Polydipsia
d) Polyphagia
Answer:
b) Oligo urea

Two marks

II. Very short answer 

Question 1.
What is osmotic regulation?
Answer:
Osmotic regulation is the control of tissue osmotic pressure which acts as a driving force for the movement of water across biological membranes.

Question 2.
What is meant by Eury haline animals?
Answer:
Eury haline animals are able to tolerate wide fluctuations in the salt concentrations. (eg) Salmons Tilapia.

Question 3.
Define excretion?
Answer:
The process by which the body gets rid of the nitrogenous waste products of protein metabolism is called excretion.

Question 4.
What is meant by Ionic regulation?
Answer:
It is the control of ionic composition of body fluids.

Question 5.
What are the nitrogenous waste formed due to the degeneration of amino acid?
Answer:

• Ammonia
• Urea
• Uric acid

Question 6.
Name some nitrogenous waste product produced by various animals?
Answer:
Some of the nitrogenous wastes produced by various animals other than ammonia, urea and uric acid are:
Jrimethyl amine oxide (TMO) in marine teleosts, guanine in spiders, hippuric acid in mammals, reptiles and other nitrogenous wastes include allantonin, allantoic acid, omithuric acid, creatinine, creatine, purines, pyramidines and pterines.

Question 7.
What is meant by renal hilum?
Answer:
The centre of the inner concave surface of the kidney has a notch called the renal hilum through which ureter blood vessels and nerves innervate.

Question 8.
What is meant by malpighian capsule or renal corpuscle?
Answer:
The Bowman’s capsule and the glomerulus together constitute the renal corpuscle.

Question 9.
What is the difference between nephron present in reptiles and mammals?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop. Mammals have a long Henle’s loop. Reptiles produce hypotonic urine whereas mammals produce hypertonic urine.

Question 10.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little into the medulla and are called cortical nephrons.

Question 11.
What is meant by Juxta medullary nephrons?
Answer:
Some nephrons have very long loop of Henle that run deep in to the medulla and are called Juxta medullary nephrons.

Question 12.
What is vasa recta?
Answer:
The efferent arteriole serving the juxtamedullary nephron forms bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Question 13.
List the three important process of urine formation?
Answer:

1. Glomerulus filtration
2. Tubular reabsorption
3. Tubular secretion

Question 14.
What is meant by Glomerulus filtrate? What is its composition?
Answer:
The blood comes to the glomerulus are filtered and enters the Bowman’s capsule is called glomerular filtrate
Composition Water glucose amino acids and nitrogenous wastes.

Question 15.
What is meant by Net filtration pressure?
Answer:
The two opposing forces against the glomerular blood pressure

1. Collodial osmotic pressure is 30 mm Hg
2. Capsular hydrostatic pressure is 15mm Hg

Net filtration pressure- Glomerular
Hydrostatic pressure – Colloidal
Osmotic pressure + Capsular hydrostatic pressure = 55 mm Hg – 30 mm Hg + 15 mm Hg = 10 mm Hg

Question 16.
What is meant by Glomerular filtration rate?
Answer:
It is the volume of filtrate formed in a minute in all nephrons of both the kidneys.

Question 17.
What is the amount of glomerular filtrate?
Answer:
In adults the GFR is 120 -125 ml per minute.

Question 18.
What is meant by primary filtrate?
Answer:
The filtrate enters from glomerulus to the Bowman’s capsule is called as primary filtrate.

Question 19.
Why glomercular filtrate resembles blood plasma?
Answer:
In the glomerular filtrate all the contents that present in the blood except the plasma protein is present.

Question 20.
How much filtrate is formed in one day?
Answer:
The amount of filtrate formed in a day is 170 to 180l.

Question 21.
What is meant by selective permeability?
Answer:

• Some substances present in the glomerular filtrate is essential for our body.
• Hence these molecules are reabsorbed in a tubules. This process is called as selective reabsorption.

Question 22.
Name the process in which selective reabsorption is taking place?
Answer:

• Passive transport
• Active transport
• Diffusion
• Osmosis

Question 23.
What are aqua porins?
Answer:
Aquaporins are membrane transport proteins that allow water to move across the epithelial cells.

Question 24.
What is Micturition?
Answer:
The process of release of urine from the bladder is called micturition or urination.

Question 25.
What is meant by Isotonic solution?
Answer:
Isotonic condition of a solution indicates no passage of water across the membrane separating two such solution.

Question 26.
What is meant by Hypotonic solution?
Answer:
The solution in which there is a loss of water then that solution is a hypotonic solution.

Question 27.
What is meant by hypertonic solution?
Answer:
When two solutions A and B are separated by a semi permeable membrane when water move from solution A to B across the membrane then the B solution is hypertonic and the solution A where the water loses is known as hypotonic solution.

Question 28.
Name the organ that the angiotensin II acton?
Answer:

• Heart
• Kidney
• Brain
• Adrenal cortex

Question 29.
What are the symptoms of diabetes mellitus?
Answer:

• Excess glucose and ketone bodies in the urine
• Poly dipsia – Excessive drinking of water
• Polyurea – Excretion of large quantities of urea
• Polyphagia – Excessive appetite

Question 30.
Name the organ that excrete nitrogen other than kidney?
Answer:

• Lungs
• Liver
• Skin

Question 31.
What are the significance of sweat glands?
Answer:

• Sweat produced by the sweat glands helps to cool the body.
• It excretes Na⁺ and Cl^– small quantities of urea and lactate.

Question 32.
What is meant by nephrolithiasis?
Answer:
It is the formation of hard stone like masses in the renal tubules of renal pelvis.

Question 33.
What is meant by phleothotomy or lithotripsy?
Answer:
Renal stones can be removed by the technique pyleothotomy or lithotripsy.

Question 34.
What is meant by Bright’s disease?
Answer:
The inflammation of the glomeruli of both kidney of children due to the streptococcal infection is called as Bright’s disease.

Question 35.
What is meant by haemodialysis?
Answer:
The process of removing toxic urea from the blood of renal failure patients is known as haemodialysis.

Question 36.
Why females are prone to urinary tract infection than men?
Answer:
Females are prone to recurring urinary tract infection as they have shorter urethra.

Question 37.
Why men are finding difficult to urinate in their old age?
Answer:
With age prostate in males may enlarge which forces urethra to tighten restricting a normal urinary flow.

Question 38.
What is the change in Urine formation when there is a deficiency of ADH?
Answer:
When there is a deficiency of ADG the reabsorption of water from the proximal convoluted tubule decreases leads to dilute urine formation.

Question 39.
Why there is a increase in the body fluid when we drink large volume of water with out eating anything salty?
Answer:

• When we drink or eat salty products the Na+ enters into the body fluids.
• The sodium ions helps in the reabsorption of water
• But when we drink only water as there is no sodium ions the tubules cannot reabsorb water.
• Hence there is an increase in the urine output.

Question 40.
Write a short note on Haemodialysis?
Answer:
Malfunction of the kidneys can lead to accumulation of urea and other toxic substances, leading to kidney failure. In such patients, toxic urea can be removed from the blood by a process called hemodialysis. A dialyzing machine or an artificial kidney is connected to the patient’s body. A dialyzing machine consists of a long cellulose tube surrounded by the dialyzing fluid in a water bath.

The patient’s blood is drawn from a convenient artery and pumped into the dialyzing unit after adding an anticoagulant like heparin. The tiny pores in the dialysis tube allow small molecules such as glucose, salts, and urea to enter the water bath, whereas blood cells and protein molecules do not enter these pores.

This stage is similar to the filtration process in the glomerulus. The dialyzing liquid in the water bath consists of a solution of salt and sugar in the correct proportion in order to prevent loss of glucose and essential salts from the blood. The cleared blood is then pumped back to the body through a vein.

Question 41.
Name the different process that maintains water level? When there is a severe loss of water in the body?
Answer:

• The blood vessels supplies to skin constricts and thus there is a decrease in the secretion of sweat prevents loss of water.
• There is a reduction in the glomerular blood pressure and the rate of filtration decreases.
• The reaborption of water in the proximal distal convoluted tubules increase.
• There is absorption of water from the small intestine and large intestine and thus increases the water content in the blood.

Question 42.
What is meant by Osmolarity?
Answer:

• The solute concentration of a solution of water is known as osmolarity,
• The unit is millosmoles / litre (mOsm /l)

Question 43.
What is meant by aquaporins? What are its functions?
Answer:
Aquaporins are water permeable channels.
Functions
It helps in allowing water to move across the epithelial cells in relation to the osmotic difference from the lumen to the interstitial fluid.

Question 44.
How can we measure that there is a efficient glomerular filtration?
Answer:
If the renal clearance is equal to the glomerular filtration rate with little reabsoption and secretion. Then we know the kidney is functioning efficiently.

Question 45.
What is the Significance of having long and short Henle’s loop of Nephrons?
Answer:

• The main function of Henle’s loop is to reabsorb water from filtrate.
• If the length of the loop is longer then there is more reabsorption of water and if the lengh of Henle’s loop is shorter then the reabsorption of water is less.

Question 46.
Give notes on Capillary to capsule.
Answer:
Bloodcells and most blood proteins are too big to cross the capsular membrane into the capsule space. But the membrane’s slits and pores allow through water mineral salts polypeptides and other small molecules including waste such as urea ammonia and creatinine.

Question 47.
Give short notes on Blood enters the glomerulus.
Answer:
Blood flows from renal arteriole into the knot of capillaries. It enters at pressure which will force water and other out of the capillaries into the capsular space.

Question 48.
Give notes on filteration in proximal convoluted tubule?
Answer:
Proximal tubule is nearer to the Bowman’s capsule. This region allows much water to be reabsorbed into the capillaries and surronding fluids as well as glucose mineral salts and other useful substances.

Question 49.
Give notes on filteration in peritubular capillaries.
Answer:
It is also called the vasarecta this network reabsorbs upto 99 percent of the water in the tubule as well as various other substance using active pumps it also moves sodium from the blood to the tubule.

Question 50.
Give notes on filteration in Henle’s loop (Ascending)
Answer:

• As the loop of the Henle dipsin to the renal medcula more water moves from the tubule into the blood as well as small amounts of salts and some urea and creatinine.
• Some acids and amines may move into the tubule in which ammonia cango in both the direction.

Question 51.
Give notes on filteration in distal tubule?
Answer:

• Distal tubule is far from capsule. This region may see water go in or out of the tubule depending on the concentration of water already in the tubule/ while hydrogen and potassium ions move to regulate both blood and urine pH.
• Acids amines and ammonia compounds may also transported into the tubule.

Question 52.
Give notes on filteration in collecing duct.
Answer:

• Fine adjustment of urine composition continues into the collecting duct system.
• About 5 percent of all the water and sodium being reabsorbed into the blood is recovered here.

Question 53.
Give notes on venousflow?
Answer:

• Blood flowing away from the nephrons carries 99 % of its orginal water.
• 98% of its sodium calcium and cholrides and about 40% of its urea.

Question 54.
We are not consuming urea. But in our body area is produced. Why?
Answer:
Through Arnithine cycle in the lives the nitrogenous waste formed due to the breakdown of amino acid creates urea.

Question 55.
What is meant by Ionic regulation?
Answer:
It is the control of the ionic composition of body fluids.

Question 56.
What are stenohaline animals?
Answer:
They can tolerate only narrow fluctuations in the salt concentration. Ex: Goldfish

Question 57.
What are Euryhaline animals?
Answer:

• They are able to tolerate wide fluctuations in the salt concentrations. Ex: Artemia Salmons.
• Acids amines and ammonia compounds may also transported into the tubule.

Question 58.
List the nitrogenous wastes
Answer:
Ammonia urea uricacid.

Question 59.
What are the other nitrogenous wastes of protein metabolism?
Answer:
Alantonin Alantoic acid, Ornithuriacid creatinine creatine purines pyramidines and pterines.

Question 60.
Which are called ammoneteles?
Answer:
Animals that excreate most of its nitrogen inthe form of ammonia are called ammonoteles.

Question 61.
What are called as uricoteles?
Answer:
Animals that excrete uric acid crystals with minimum loss of water are called uricoteles.

Question 62.
What are ureoteles?
Answer:
Mammals and terrestrial amphibians mainly excrete urea are called ureoteles.

Question 63.
How is Reptiles produced lesshypotonic urine?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henles loop and hence produce very little hypotonic urine.

Question 64.
How is mammals produced concentrated urine?
Answer:
Mammalian kidneys produce concentrated urine due to the presence of Henle’s loop.

Question 65.
What are the three coverings of kidney?
Answer:

• Renal facia
• Perirenal fat capsule
• Fibrous capsule

Question 66.
What are medullary pyramids?
Answer:
The medulla is divided into a few conical tissue masses called medullary pyramids.

Question 67.
What is meant by renal columns of Bertini?
Answer:
The part of cortex that extends in between the medullary pyramids is the renal columns of Bertini.

Question 68.
What is meant by renal pelvis?
Answer:
A broad funnel shaped space inner to the hilum is called renal pelvis.

Question 69.
What is calyces?
Answer:
The projection in the pelvis is called calyces.

Question 70.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little in to the medulla and are called cortical nephrons.

Question 71.
What is meant by juxta medullary nephron?
Answer:
Some nephrons have very long loop of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 72.
What is the functions of aquaporins?
Answer:
This helps in allow water to move across the epithelial cells in relation to the osmotic difference from the human to the interstitial fluid.

Question 73.
What is meant by juxtaglomerular apparatus?
Answer:
It is a specialized tissue in the afferent arteriole of the nephron that consists of maculadensa and granular cells.

Question 74.
What is meant by micturition?
Answer:
The process of release of urine from the bladder is called micturition.

Question 75.
What are the symptoms of diabetes mellitus?
Answer:
Presence of glucose and ketone bodies in the urine.

Question 76.
How is lung acting as a excretory organ?
Answer:
Lungs remove large quantities of carbondixide 181 / day and significant quantities of water every day

Question 77.
What is meant by renal clearnace?
Answer:
The amount of solute passing from the urine in a given period of time is renal clearance.

Question 78.
How can we estimate the efficiency of kidney?
Answer:

• The renal clearance is equal to the glomerular filtrate then there is efficient filtration with little reabsorption and secretion.
• Thus we can estimate the clearance is equal.

Three Marks

Short Answer –

Question 1.
a) List the ma j or nitrogenous wastes.
b) What are other wastes formed during protein metabolism?
Answer:
a) 1. Major nitrogenous wastes.

• ammonia
• Urea
• Uricacid

b) Other Wastes:

• Trimethy lamine oxide TMO
• Quanine
• Allantonin
• Creatinine
• Creatine
• Purines

Question 2.
Give notes on ammonoteles uricoteles and ureoteles.
Answer:
Ammonoteles:

• Animals that excrete most of its nitrogen in the form of ammonia are called ammonoteles.
  (e.g) fishes Amphibians aquatic insects.
• In bony fishes ammonia diffuses out acrossthe body surface.

Uricoteles:

• Animals which excrete uricacid crystals with a minimum loss of water is called.
• Uricoteles (e.g) Reptiles Birds land snails and insects.

Ureoteles:

• Animals which excrete urea as a nitrogenous wastes are called ureoteles
• (e.g) Mammals, terrestrial amphibians.

Question 3.
Nephrons are the functional and structural unit of kidney’s. What is the relationship between glomerulus, Henle’s loop and urine formation?
Answer:

• Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop and produce hypotonic urine (dilute)
• Mammalian kidneys produce concentrated urine due to the presence of long Henle’s loop.
• Aglomerular kidneys of marine fishes produce little urine that is iso osmotic to the body fluid.
• Amphibians and fresh water fishlack Henle’s loop hence produce dilute urine.

Question 4.
Differentiate the cortical nephron fron juxta medullary nephron
Answer:

Question 5.
Give Short notes on capillary bed of the nephron:
Answer:
The first capillary bed of the nephron is the glomerulus.
The other is peritubular capillaries.
1. Glomerulus:

Blood enters into the glomerulus through afferent arteriole and drained by the efferent arteriole.

2. Peritubular capillaries:

• The efferent arteriole forms a fine cappillary network around the renal tubule called the peritubular capillaries.
• The efferent arteriole serving the juxta medullary nephrons forms bundles of long straight vessel called vasa recta.
• Vasa recta is absent in cortical nephrons.

Question 6.
What happens to the filtrate that comes to the proximal convoluted tubule? (or) Explain about reabsorption?
Answer:

• In the proximal convoluted tubule glucose lacticacid aminoacid sodium ions are reabsorbed.
• Sodium is reabsorbed – potassium pump in the proximal convoluted tubule.
• Small amounts of urea and uric acid are also reabsorbed.

Question 7.
What happen to the filtrate that comes to the Henle’s loop? (or) Explain the reabsorptionin the Henle’s loop?
Answer:

Descending Loop:
The aquaporin present in the descending limb of Henle permeable to water but not permeable to salts.
Hence Na⁺ and cl^– gets concentrated in the filtrate.

Ascending Limb:
It is impermeable to water but permeable to solutes like Na⁺cl^– and K⁺.

Question 8.
Give an account of tubular reabsorption?
Answer:
The volume of filterate formed perday is 170-180 is and the urine released in a day is 1.5l

• Nearly 99 % of the glomerular filtrate is reabsorbed by the renal tubules. It is called selective reabsorption.
• Reabsorption is taken place by the tubular epithelial cells in different segments of the nephron by active transport or passive transport diffusion and osmosis.

Question 9.
What is happenning to the filtrate in distal convoluted tubule (or) Reabsorption taking place here?
Answer:

• Depending on the body’s need the reabsorption taking place here and is regulated by hormones.
• Reabsorption of bicarbonate HCO₃^–  takes place to regulate the blood pH.
• Homestasis of K⁺ and Na⁺ in the blood is also regulated in this region.

Question 10.
Name the structures that regulate the functioning of kidney?
Answer:

• Hypothalamus
• Juxta glomerular apparatus
• Heart

Question 11.
What is meant by diabetes incipidus?
Answer:
If there is deficiency or absence of ADH that leads to dilute urine called diabetes incipidus.
Symptoms.

• Excessive thirst
• Excretion of large quantities of dilute urine.
• FaIl in blood pressure.

Question 12.
Give notes on juxta glomerular apparatus?
Answer:

• Specialized tissue in the afferent arteriole of ncphron is the juxta glomerular apparatus.
• It consists of macula densa and granular cells.
• The macula densa cells sense distal tubular flow and affect afferent alteriole diameter.
• The granular cells secrete renin.

Question 13.
Why female are prone to urinary tract infections? (Urethritis)
Answer:

• Female’s urethra is very short and its external opening is close to the analopening.
• Hence improper toilet habits can easily carry faecal bacteria into the urethra.
• The urethral mucusa is continuous with the urianary tract and the inflammation of the urethra is called urethriti’s.

Question 14.
What is cystitis?
Answer:
The urinary tract infection leads to inflammation of bladder called cystiti’s.
Symptoms:

• Painful urination
• Urinary Urgency
• Cloudy or bloodtingedurine
• Back pain head ache offen occurs

Question 15.
What is meant by renal failure? What are its types?
Answer:
When the kidney fails to excrete wastes may lead to accumulation of urea with marked reduction in the out put called renal failure.
Types
Acute renal failure
Chronic renal failure

Question 16.
Why the chronic renal failure is dangerous than acute renal failure?
Answer:

• Though the kidney stops its function abruptly there are chances for recovery of kidney function in acute renal failure.
• But in chronic failure there is a progressive loss of function of the nephron which gradually decreases the function of kidney.

Question 17.
What is meant by glomerulo nephritis or Bright’s disease? What are its symptoms?
Answer:
Inflammation of the glomerulus of both the kidneys due to the strepto coccal infection in children is called as Bright’s disease Symptoms

• Haematuria
• Proteinuria
• Salt and water retention – Oligouria (Low urine out put)
• Hypertension and Pulmonaryoedema

Question 18.
a) What is meant by kidney transplantations.
b) Where is graft kidney received from?
c) What are the steps to be taken to avoid graft rejection?
Answer:

• Transfer of healthy kidney from one person (donor) to another person with kidney failure is called kidney transplantation.
• The donated kidney may be taken from a healthy person who is declared brain death or from sibling or close relatives.
• Immuno supressive drugs are administered to the patient to avoid tissue rejection.

Question 19.
a) Name the hormone that the dipsticks contain which tests urine?
b) Which colour indicates the presence of glucose in the urine?
Answer:

• Glucose oxidase and peroxidase.
• Brown coloured compound is produced.

Question 20.
What are Osmo confirmers?
Answer:
Osmo confirmers are able to change their internal osmotic concentration with change in external environments asin marine and sharks. molluscs

Question 21.
What are Osmo regulators?
Answer:
They maintain their internal osmotic concentration irrespective of their external osmotic environment, (eg) Otters.

Question 22.
List the excretory structures of different organisms.
Answer:

• Protonephridia
• Meta nephridia
• Flame cells – Platy helminthes
• Rennette cells – Nematodes
• Malpighian tubules – Insects
• Greenglands – Prawns

Question 23.
What is meant by filtration slits?
Answer:

• The visceral layer of glomerulus is made of epithelial cells called podocytes and ends in foot processes which cling to the basement membrane of the glomerulus.
• The openings between the foot processes are called filtration slits.

Question 24.
Draw the diagram of ornithine cycle?
Answer:

Question 25.
Why there is a pressure reduction when the blood goes through efferent arteriole?
Answer:

• Blood enters the glomerulus faster with greater force through afferent arteriole.
• Because the afferent arteriole is broader than efferent arteriole that is why the pressure reduces when it goes through the efferent arteriole.

Question 26.
What are the changes taking place in our body when there is a fluid loss?
Answer:

• The osmo receptors in the hypothalamus is stimulated.
• The neurohypophysis is stimulated and anti diuretic hormone is liberated.
• The aquaporins in the tubuler are increased and water is reabsorbed and enters into the interstitial cell and the water loss is rectified.

Question 27.
How is skin acted as a excretory organ?
Answer:

• Skin excretes Na⁺ and Cl^– small quantities of urea and lactate.
• Sebaceous glands eliminated certain substances like steroids, hydrocarbons and waxes.

Question 28.
What is meant by urethritis?
Answer:
The urethral mucosa is continuous with the urinary tract. The infection in the urethra is called urethritis.

Question 29.
What is meant by Cystitis?
Answer:
The infection in the urethra can ascend the tract to cause bladder inflammation called cystitis.

Question 30.
What is meant by Phelitis or Pyelone phritis?
Answer:
The bladder infection ascend to the renal inflammation called pyelitis or pyelonephritis.

Question 31.
What are the two types of renal failure?
Answer:

1. Acute renal failure
2. Chronic renal failure

1. Acute renal failure

• In acute renal failure the kidney stops its function abruptly.
• There are chances for recovery of kidney function.

2. Chronic renal failure
In chronic renal failure there is a progressive loss of function of the nephron which gradually decreases the function of kidneys.

Question 32.
What is meant by Uremia?
Answer:
Uremia is characterized by increase in Urea uric acid and creatinine in blood.

Question 33.
What is meant by Nephrolithiasis?
Answer:
It is a formation of hard stone like masses in the renal tubules of renal pelvis.

Question 34.
How is water excess taken through drinking too much fruit juice regulated?
Answer:

• When we drink much fruit juice the osmo receptors in hypothalamus is not stimulated and hence the secretion of vaso pressin from neuro hypophysis is reduced.
• The aquaporin escapes from collecting duct to cytoplasm and hence water reabsorption is prevented and formed dilute urine.

Question 35.
What is uremia?
Answer:
Uremia is a condition in which there is a increase level of urea uric acid and creatinine in blood.

Question 36.
What is the amount of Urea present in the blood?
Answer:

• The level of urea in the blood is 17 -30 mg /100ml.
• In chronic kidney failure there is 10 times increase in urea level.

Five marks

IV. Detailed Answers –

Question 1.
Name the different excretory structure and different organisms.
Answer:
1. Invertebrate – Protonephridia / Meta nephridia
2. Platvhelminthes – Flamecells
3.  Amphioxues – Solenocytes
4 Nematodes – Rennette cells
5. Annelida – Metanephridia
6. Insects – Malpighian tubules
7. Prawn / Crustaceans – Green glands / Antenna! glands

Question 2.
The concentration of urine depends on the structure of nephron? Explain?
Answer:

1. In the reptiles the glomerulus is reduced or there may be no glomerulus and 1-lenle’s loop and hence produces dilute urine (chypotonic).
2. In the mammals the long Henle’s loop produces concentrated urine (hypertonic)
3. A glomerular kidneys of marine fishes produce little urine that is isoosmotic to the body fluid.
4. Amphibians and freshwater fish lack Henle’s loop hence produce dilute urine.

Question 3.
Draw the following diagram and mark the following parts.
Answer:
A) Aorta
B) Renal vein
C) Ureter
D) Urinary bladder

Question 4.
a. What is the weight of kidneys? What are its outer coverings?
b. Draw the L.S of kidney and name the parts.
c. Explain the internal structure of kidney
Answer:
a. Each kidney weighs an average of 120 – 170 gms.
The outer layer of the kidney is covered by three layers of supportive tissue namely renal fascia perirenal fat capsule fibrous capsule.

b. Draw the LS of kidney and name the parts.

c. Internal Structure of kidney

• The longitudinal section of kidney shows an outer cortex inner medulla and pelvis.
• The inner concave surface of the kidney is renal hilum through which ureter blood vessels and nerves enter.
• Inner to the hium is a funnel shaped renal pelvis with projection called calyces.
• The calyces collect the urine and empties in to the ureter.
• The medulla consists of conical tissues called medullary pyramids or renal pyramids.

Question 5.
a) What is the structural and functional unit of kidney.
b)Draw the diagram of nephron and name the parts.
c) Give short notes on Malpighian body or Renal Corpuscle.
Answer:
a) Hie structural and functional unit of kidney is nephron. It is composed of Malpighian body or renal corpuscle and Urine ferous tubule.
b) Structure of nephron

c. Malpighian body/Renal Corpuscle.

• The Bowman’s capsule and the glomerulus together constitutes Malpighian corpuscle.
• Bowman’s capsule is made up of two layers. It contains blood vessels called glomerules.
• The endothelial of glomerulus has many pores. The viscral layers of glomerulus is made of epithelial cells called podocytes.
• Tire podocytes end in foot processes which cling to the basement membrance of the glomerulus.
• The openings between the foot processes are called filtration slits.

Question 6.
a) Give the different regions of renal tubule.
b) Where is renal tubule present in the kidney?
c) How is renal tubules differentiated depending on the Henle’s loop?
Answer:
a)
1. proximal convoluted tubule.
2. Henle’s loop
a. Thindescending limb of Henle’s loop.
b. Thick ascending limb.

2. Distal convoluted tubules
The distal convoluced tubules opens in to acollecting duct.
Several collecting ducts fuse to form papillary duct that delivers urine in to the calvces which opens into the renal pelvis.
b) The PCT and DCT are situated in the cortical region of the kidney.
The loop of Henle is in the medulla region.
c) The loop of Henle is too short and extends only very little into the medulla and are called cortical nephron.
Some nephrons have very long loops of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 7.
a) Give notes on capillaries of nephron
b) Give an account of blood vessels of glomerulus.
c) What is vasa recta? Where is it seen?
Answer:
Capillaries of nephron
1. Glomerulus capillary bed
2. Peritubular capillaries

1. Glomerular capillarybed

• It consists of afferent and efferent arteriole.
• The afferent arteriole is broader than efferent arteriole.
• The efferent arteriole that comes out of the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries.

Vasa recta
The efferent arteriole serving the juxta medullary nephrons form bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Question 8.
a) Describe the three important process in the urine formation?
b) Give an account on glomerular filtration
Answer:
a. Main Processes

• Clornerular Filtration
• Tubular reabsorption
• Tubular secretion

b. Glomerular filtration

• Blood enters the kidney from the renal artery.
• Blood enters the glomerulus faster with greater force through the afferent arterioles because the afferent arteriole is wider than efferent arteriole.
• The glomerular hydrostatic pressure is around 55 mm Hg.
• The colloidal osmotic pressure is 30 mm Hg. Capsular hydrostatic pressure is 15 mm Hg. The net pressure of 10 mm Hg is responsible for the renal filtration.
• Next filtration pressure 55 mm Hg – (30 mm Hg +15 mm Hg) 10 mm Hg.
• The kidneys produce 1801 of glomerular filtrate 24 hours and 120 -125 ml/min.
• This filtrate will enter into the Bowman’s capsule and called primary urine. It contains more water colloidal protein, glucose salts and nitrogenous waste.

Question 9.
a) Why glomerular filtrate resembles blood plasma?
b) Tabulate the concentration of substances in the blood plasma and in the glomerular filtrate.
Answer:
As the glomerular filtrate forms it contain all the substances except plasma protein of blood. Hence it resembles blood.

Substance	Concentration in blood Plasma / g dm-3	Concentration in glomerular filtrate / g dm-3
Water	900	900
Proteins	80.0	0.05
Aminoacids	0.5	0.5
Glucose	1.0	1.0
Urea	0.3	0.3
Uric Acid	0.04	0.04
Creatinine	0.01	0.01
Inorganic ions (mainly (Na+, K+ and Cl– )	7.2	7.2

Question 10.
a) What is meant by Tubular Secretion?
b) Give an account of tubular secretion of nephron.
Answer:
Tubular Secretion:

• The collecting tubule of nephron secrete H⁺ NH⁴ , Creatinine and Organic acid and liberated into the tubules and excreted through urine.
• Most of the water is absorbed in the proximal convoluted tubule.
• Na^– is exchanged for water in the loop of Henle.
• The hypotonic fluid enters the distal convoluted tubule.
• Substances such as urea and salts pass from peritubular blood into the cells as distal convoluted tubule and then to collecting duct.
• Water is absorbed and concentrated hypertonic urine is form ed.
• For every H⁺ secreted into the tubular filtrate a Na⁺ is absorbed by the tubular cell.
• The H secreted combines with HCO₃, HPO₃ and NH₃ and gets fixed as carbonic acid CH2CO₃ and Phosphoric acid CH2PO4
• Since H⁺ gets fixed in the fluid reabsorption of H⁺ is prevented.

Question 11.
a) In which process concentrated urine is formed?
b) How is concentrated urine formed in the Hen le’s loop.
Answer:
a) The major function of Henle’s loop is to concentrate Na⁺ and Cl^–.

• There is low osmolarity near the cortex and high osmolarity towards the medulla.
• This osmolarity in the medulla is due to the presence of the solutes transporters and is maintained by
  the arrangement of the loop of Henle collecting duct and vasa recta.
• The osmolarity of interstitial fluid is 300 m Osm.
• The Henle’s loop create a countercurrent multiplier.

• As the fluid enters the descending limb water moves from the lumen into the inter stitial fluid the osmolarity reduces.
• To counteract this dilution the region of the ascending limb actively pumps solutes from the lumen into the interstitial fluid and the osmolarity increases to about 1200 m OSM in medula.

b) The vasa recta maintains the medullary osmotic gradient via counter current exchanger.

• The counter current exchanger of vasa recta preserves the medullary gradient while removing reabsorbed water and solutes.
• The vasa recta leaves the kidney at the junction between the cortex and medulla.
• When the blood leaves the efferent arteriole and enters vasa recta the osmolarity in the medulla increases (1200 rnOsm) and result in passive up take of solutes and loss of water.
• As the blood enters the cortex the osmolarity in the blood decreases and the blood loses solutes and gain water to form concentrated urine.

Question 12.
How is vasa recta helps in producing concentrated urine?
Answer:
Vasa recta maintains the medullary osmotic gradient via counter current exchanger.

• Vasa recta preserves the medullary gradient while removing reabsorbed water and solutes through counter current exchanges.
• The vasa recta leave the kidney at the junction between the cortex and medulla.
• The interstitial fluid at this point is iso – osmotic to blood.
• When the blood leaves the efferent arteriole and enters the vasa recta the osmalarity in the medulla increases (1200 mOsm) and results in passive up take of solutes and loss of water.
• As the blood enters the cortex the osmolarity in the blood decreases (300mOsm) and the blood loses solutes and gains water to form concentrated urine.
• Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Question 13.
a) What are the structures that regulate kidney functioning?
b) What is the role of ADH in Urine formation.
c) What are the symptoms of diabetes insipidus.
Answer:
a) The structures that regulate kidney functioning

• Hypothalamus
• Juxtaglomerular apparatus
• Heart

b) The functions of ADH

• When there is excessive loss of fluid from the body or when there is an increase in the blood pressure the osmoreceptors of the hypothalamus is stimulated.
• The osmoreceptors stimulate the neurohypophysis to secrete an antidiuretic hormone (AOH) or vasopressin.
• ADH facilitates reabsorption of water by increasing the number of aquaporins on the cell surface of the distal convoluted tubule and collecting duct and prevents excessloss of water.

Question 14.
a) Name the cell that secretes the enzyme renin.
b) Where is granular cell present?
c) What is the role of renin in Osmoregulation.
Answer:
a) Renin is secreted by granular cells.
b) Granular cells are present in the afferent arteriole.
c) The role of renin

• A fall in glomerular blood flow blood pressure and filtration rate can activate granular cells of juxtaglomerular cells to release renin.
• Renin converts the plasma protein angiotensinogen into angiotensis I and angiotensin II.
• Angiotensis II stimulates Na⁺ reabsorption in the proximal convoluted tubule by vasoconstriction of the blood vessels and increases the glomerular blood pressure.
• Angiotensis II stimulates adrenal cortex to secrete aldosterone that causes reabsorption of Na⁺ ,K⁺ excretion and absorption of water.
• This increases the glomerular blood pressure and glomerular filtration rate.
• Hence renin regulates osmoregulation.

Question 15.
a) Where is atrial natriuretic peptide liberated from?
b) Write its significance in short.
Answer:
a) This is liberated from atrium of heart.
b) Use of Atrial natriuretic peptide

• It increases Na excretion and increases the blood flow to the glomerulus.
• It acts on the afferent glomerular arteriole as a vaso dilator or an efferent arteriole as a vaso constrictor.
• It reduces aldosterone from adrenal cortex and renin secretion.
• Thus decreases the angiotensin II.
• The atrial natri uretic factor acts antagonistically to renin angiotensin system aldosterone and vasopressin.

Question 16.
a) What is micturition?
b) How is central nervous system regulates urination?
Answer:
a) The process of release of Urine from the bladder is called micturition or urination.
b) Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till it receives a signal from the central nervous system.

• The stretch receptors present inthe urinary bladder are stimulated when it gets filled with urine.
• At the same time the internal sphincters opens and relaxing the external sphincter.
• The sphincter opens and the urine is expelled out.

Question 17.
Answer for the following questions.
a) What is the average excretion of an adult human?
b) What is the pH of Urine
c) How much pH is differed due to the food we eat?
d) What is the reason for the yellow colour of urine?
e) How much urea is excreted in a day?
f) If there is more glucose, and ketone what does it indicates?
Answer:

• 1.5l
• pH = 6
• pH = 4.5-8
• Uro chrome
• 25 – 30 g
• Diabetes melitus

Question 18.
Answer for the following question in excretion by other organs.
a) What are the other structures that excrete nitrogenous wastes rather than kidney?
b) How much CO₂ is excreted through lungs?
c) What are the wastes excreted by digestive systems? ’
d) What are the glands that excrete waste through skin?
e) What is the main function of sweat glands?
f) What is the second important function of sweat gland?
g) Name the substance excreted by sebaceous glands.
h) Name the substances excreted by sebaceous glands.
i) Name the waste excreted through saliva.
Answer:

• Lungs, Liver and skin
• 18 Litre, much Water
• Bilirubin, Biliverdin, Cholesterol, Vitamins and drugs.
• Sweat glands and sebaceous glands
• To cool the body
• Na⁺ and Cl^–, small quantities of Urea and Lactate excretion.
• Sebum
• Steroid Hydrocarbon and wax.
• Nitrogenous wastes

Question 19.
Draw the schematic representations of renin hormone in the regulation of body fluid concentration.
Answer:

Question 20.
a) What is meant by Haemodialysis?
b) Why is it called an artifical kidney?
c) Give an account of Haemodialysis.
Answer:
a) in the patients of kidney failure toxic urea can be removed from the blood by a process called haemodialysis.
b) The dialyzing machine is a artificial kidney.
c) Hemodialysis

• A dialyzing machine consists of a long cellulose tube surrounded by the dialysing fluid in a water bath.
• A patient’s blood is drawn from a convenient artery and pumped into the dialysing unit after adding an anticoagulant like heparin.
• The tiny pores in the dialysis tube allow small molecules such as glucose salts and urea to enter into the water bath.
• Whereas blood cells and protein molecules do not enter these pores the cleared blood is then pumped back to the body through a vein.

Question 21.
Give notes on kidney transplantation?
Answer:

• This involves transfer of healthy kidney from one person (donor) to another person who is with kidney failure.
• The donated kidney may be taken from a healthy person who is declared brain dead or from sibling or close relatives to minimize the chances of rejection by the immune system of the host.
• Immuno suppressive drugs are usually administered to the patient to avoid tissue rejection.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 11 Fundamentals of Organic Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
Select the molecule which has only one π bond.
a) CH₃ – CH = CH – CH₃
b) CH₃ – CH = CH – CHO
c) CH₃ – CH = CH – COOH
d) All of these
Answer:
a) CH₃ – CH = CH – CH₃

Question 2.
In the hydrocarbon the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence.
a) sp, sp, sp³, sp², sp³
b) sp², sp, sp³, sp², sp³
c) sp, sp, sp², sp, sp³
d) none of these
Answer:
a) sp, sp, sp³, sp², sp³

Question 3.
The general formula for alkadiene is
a) C_nH_2n
b) C_nH_{2n – 1}
c) C_nH_{2n – 2}
d) C_nH_{n – 2}
Answer:
c) C_nH_{2n – 2}

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept – 2 – ene is

Answer:
a)

Question 5.
The IUPAC name of the compound is

a) 2, 3 – Dimethylheptane
b) 3 – methyl – 4 – ethyloctane
c) 5 – ethyl – 6- methyloctane
d) 4 – Ethyl – 3 methyloctane
Answer:
d) 4 – Ethyl – 3 methyloctane

Question 6.
Which one of the following names does not fit a real name?
a) 3 – Methyl – 3- hexanone
b) 4 – Methyl – 3 – hexanone
c) 3 – Methyl – 3 hexanol
d) 2 – Methyl cyclo hexanone
Answer:
a) 3 – Methyl – 3- hexanone

Question 7.
The IUPAC name of the compound CH₃ – CH = CH – C ≡ CH is
a) Pent – 4- yn – 2 – ene
b) Pent – 3- en – 1- yne
c) Pent – 2 – en – 4 – yne
d) Pent – 1 yn – 3 – ene
Answer:
b) Pent – 3- en – 1- yne

Question 8.
IUPAC name of  is
a) 3, 4, 4 – Trimethylheptane
b) 2 – Ethyl – 3, 3, – dimethyl heptane
c) 3, 4, 4 – Trimethyloctane
d) 2 – Butyl – 2 – methyl – 3 ethyl – butane
Answer:
c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of
is
a) 2,4,4 – Trimethylpent – 2 – ene
b) 2,4,4 – Trimethylpent – 3 – ene
c) 2,2,4 – Trimethylpent – 3 – ene
d) 2,2,4 – Trirnethylpent – 2 – ene
Answer:
c) 2,2,4 – Trimethylpent – 3 – ene

Question 10.
The IUPAC name of the compound is
a) 3 – Ethyl – 2 – hexene
b) 3 – Propyl – 3 – hexene
c) 4 – Ethyl – 4 – hexene
d) 3 – Propyl – 2 – hexene
Answer:
a) 3 – Ethyl – 2 – hexene

Question 11.
The IUPAC name of the compound
is
a) 2 – Hydroxypropionic acid
b) 2 – Hydroxy Propanoic acid
c) Propan – 2 – ol – 1 – oic acid
d) 1 – Carboxyethanol
Answer:
b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of is
a) 2 – Bromo – 3- methyl butanoic acid
b) 2 – methyl – 3 bromo butanoic acid
c) 3 – Bromo – 2 – methylbutanoic acid
d) 3 – Bromo – 2, 3 – dimethyl propanoic acid
Answer:
c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is
a) CH₃ – CH₂ – CH₂ – CH₂

b)

c)

d)
Answer:
c)

Question 14.
The number of stereoisomers of 1, 2 – dihydroxy cyclopentane
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

Question 15.
Which of the following is optically active?
a) 3 – Chloropentane
b) 2- Chloro propane
c) Meso – tartaric acid
d) Glucose
Answer:
d) Glucose

Question 16.
The isomer of ethanol is
a) acetaldehyde
b) dimethyl ether
c) acetone
d) methyl carbinol
Answer:
b) dimethyl ether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C₃H₆O?
a) 4
b) 5
c) 9
d) 10
Answer:
c) 9

Question 18.
Which one of the following shows functional isomerism?
a) ethylene
b) Propane
c) ethanol
d) CH₂Cl₂
Answer:
c) ethanol

Question 19.
are
a) resonating structure
b) tautomers
c) Optical isomers
d) Conformers
Answer:
b) tautomers

Question 20.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
a) Fe₃[Fe(CN)₆]₂
b) Fe₄ [ Fe(CN)₆]₃
c) Fe₄[Fe(CN)₆]₂
d) Fe₃[Fe(CN)₆]₃
Answer:
b) Fe₄ [ Fe(CN)₆]₃

Question 21.
Lassaigne’s test for the detection of nitrogen fails in
a) H₂N – CO – NH.NH₂.HCl
b) NH₂ – NH₂.HCl
c) C₆H₅ – NH – NH₂.HCl
d) C₆H₅CONH₂
Answer:
c) C₆H₅ – NH – NH₂.HCl

Question 22.
Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
a) NH₂ NH₂ HCl and ClCH₂ – CHO
b) NH₂ CS NH₂ and CH₃ – CH₂Cl
c) NH₂ CH₂ COOH and NH₂ CONH₂
d) C₆H₅NH₂ and ClCH₂ – CHO
Answer:
d) C₆H₅NH₂ and ClCH₂ – CHO

Question 23.
Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of
a) [Fe (CN)₅ NO]^3-
b) [Fe (NO)₅ CN]⁺
c) [Fe (CN)₅ NOS]^4-
d) [Fe (CN)₅ NOS]^3-
Answer:
c) [Fe (CN)₅ NOS]^4-

Question 24.
An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to
a) 46 %
b) 34 %
c) 3.4 %
d) 4.6 %
Answer:
b) 34 %

Question 25.
A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50mL of 0.5 M H₂SO₄ The remaining acid after neutralization by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is.
a) 14 %
b) 28 %
c) 42 %
d) 56 %
Answer:
b) 28 %

Question 26.
In an organic compound, phosphorus is estimated as
a) Mg₂P₂O₇
b) Mg₃(PO₄)₂
c) H₃PO₄
d) P₂O ₅
Answer:
a) Mg₂P₂O₇

Question 27.
Ortho and para – nitro phenol can be separated by
a) azeotropic distillation
b) destructive distillation
c) steam distillation
d) cannot be separated
Answer:
c) steam distillation

Question 28.
The purity of an organic – compound is determined by
a) Chromatography
b) Crystallization
c) melting or boiling point
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by
a) distillation at atmospheric pressure
b) distillation under reduced pressure
c) fractional distillation
d) steam distillation
Answer:
b) distillation under reduced pressure

Question 30.
Assertion:

Reason:
The principal functional group gets lowest number followed by double bond (or) triple bond.
a) both the assertion and reason are true and the reason is the correct explanation of assertion.
b) both assertion and reason are true and the reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both the assertion and reason are false
Answer:
a) both the assertion and reason are true and the reason is the correct explanation of assertion.

II. Write brief answers to the following questions:

Question 31.
Give the general characteristics of organic compounds.
Answer:
All organic compounds have the following characteristic properties.

1. They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc…
2. Many of the organic compounds are inflammable (except CCl₄). They possess low boiling and melting points due to their covalent nature.
3. Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.

Question 32.
Describe the classification of organic compounds based on their structure.
Answer:

Question 33.
Write a note on homologous series.
Answer:
Homologous series:
A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH₂ group is called homologous series.
Example:
Alkanes:
Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈) etc..
Alcohols:
Methanol (CH₃OH), Ethanol (C₂H₅OH) Propanol (C₃H₇OH) etc…

Question 34.
What is meant by a functional group?
Identify the functional group in the following compounds.
a) acetaldehyde
b) oxalic acid
e) dimethyl ether
d) methylamine
Answer:
Functional group:
An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties.
a) acetaldehyde → – CHO
b) oxalic acid → – COOH
c) di methyl ether → – O –
d) methyiamine → – NH₂

Question 35.
Give the general formula for the following classes of organic compounds
a) Aliphatic monohydric alcohol
b) Aliphatic ketones
c) Aliphatic amines
Answer:
a) Aliphatic monohydric alcohol → C_nH_{2n + 2}O
b) Aliphatic ketones → C_nH_2nO
c) Aliphatic amines → C_nH_{3n + 2}N

Question 36.
Write the molecular formula of the first six members of homologous series of nitre alkanes.
Answer:

• CH₃NO₂
• C₂H₅NO₂
• C₃H₇NO₂
• C₄H₉NO₂
• C₅H₁₁NO₂
• C₆H₁₃NO₂

Question 37.
Write the molecular formula and possible structural formula of the first four members of homologous series of carboxylic acids.
Answer:

Question 38.
Give the IUPAC names of the following compounds.
i) (CH₃)₂CH – CH₂ – CH(CH₃) – CH(CH₃)₂

ii)

iii) CH₃ – O – CH₃

iv)

v) CH₂ = CH – CH – CH₂

vi)

vii)

viii)

ix)

x)

xi)

xii)

xiii)

xiv)

xv)

xvi)
Answer:
(i) 2, 3, 5 – trimethyihexane
(ii) 2 – bromo – 3 – methylbutane
(iii) methoxymethane
(iv) 2 – hydroxybutanal
(y) buta – 1 ,3 – diene
(vi) 4 – chloropent – 2 – yne
(vii) 1 – bromobut – 2 – ene
(viii) 5 – oxohexanoic acid
(ix) 3 – ethyl – 4 – ethenylheptane
(x) 2, 4, 4 – trimethylpent – 2 – ene
(xi) 2 – methyl -1 – phenyipropan – 1 -amine
(xii) 2, 2 – dimethyl – 4oxopentanenitrile
(xiii) 2 – ethoxypropane
(xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – bromo – 2 – methylpentanal
(xvi) Acetophenone

Question 39.
Give the structure for the following compound
(i) 3 – ethyl – 2 methyl – 1 – pentene
(ii) 1, 3, 5 – Trimethyl cyclohex – 1 – ene
(iii) tetry butyl iodide
(iv) 3 – Chlorobutanal
(V) 3 – Chlorobutanol
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2 – dimethyl – 1 – chloropropane
(viii) 3 – methylbut -1- ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1, 3 – diene
(xi) 1, 3 – Dimethylcyclohexane
(xii) 3 – Chlorobut – 1 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde
Answer:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
This method involves the conversion of covalently bonded N, S or halogen present in the organic compounds to corresponding water soluble ions in the form of sodium salts For this purpose a freshly cut piece of Na of the size of a paper, dried it by pressing between the folds of a filter paper is taken in a fusion tube in an iron tand clamping it just near the upper end and it is gently heated.

When it melts to a shiñing globule, put a pinch of the organic compound on it. Heat the tube with the tip of the flame till all reaction ceases and it becomes red hot. Now plunges it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striding against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for qualitative analysis of nitrogen, sulfur and halogens present in organic compounds.

Test for Nitrogen:
If nitrogen is present it gets converted to sodium cyanide which on reaction with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl gives a Prussian blue color or green color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the excess of NaOH on FeSO₄ which would otherwise mark the Prussian blue precipitate. The following reaction takes part in the formation of Prussian blue.

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Answer:
Estimation of halogens (Carius method):
A known mass of the organic compound is heated with fuming HNO₃ along with AgNO₃. C, H & S gets oxidized to CO₂, H₂O, SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated.

A known mass of the substance is taken along with fuming HNO₃ and AgNO₃ is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made.

Calculation:
Weight of the organic compound: w g
Weight of AgCl precipitate = a g
143. 5 g of AgCl contains 35.5 g of Cl
∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a
w g Organic compound gives a g AgCl
Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\)
Let Weight of silver Bromide be ‘b’ g
188 g of AgBr contains 80 g of Br
∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br
w g Organic compound gives b g AgBr
Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
Let Weight of silver Iodide be ‘c’ g
235 g of AgI contains 127 g of I
∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I
w g Organic compound gives c g AgI
percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)

Question 42.
Give a brief description of the principles of
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.

2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.

(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.

(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.

(d) In column chromatography, the above principle is carried out in a long glass column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of ‘ paper acts as an adsorbent. This method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.

A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase. The solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots . on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Structural isomerism:
This type of isomers have same molecular formula but differ in their bonding sequence.

(a) Chain or nuclear or skeletal isomerism:
These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched).

(b) Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton, but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.

(c) Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
Molecular formula C₃H₆O
CH₃ – CH₂ – CHO
propanal (aldehyde group)

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

Question 46.
Briefly explain geametrical isomerism in alkene by considering 2-butene as an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.

In alkenes, the carbon-carbon double bond is sp² hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The π bond is formed by the head on overlap of sp² hybrid orbitals. The n bond is formed by the side-wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible. This restriction of rotation about C – C double bond is responsible for geometrical isomerism in alkenes.

These two compounds are termed as geometrical isomers and are distinguished from each other by the terms c is and trans. The c is isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis- trans isomerism.

The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the n bond so that rotation about a bond becomes possible. Upon cooling, the reformation of the n bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butene.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Solution:
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g

Percentage of carbon:
44 g of carbondioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \(\frac{12 \times 0.88}{44}\) g
0.30 g substance contains,
carbon = \(\frac{12 \times 0.88}{44}\) g

100 g substance Contains \(\frac{12 \times 0.88}{44}\) × \(\frac{100}{0.30}\) = 80 g of carbon
Percentage of carbon = 80 %

Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \(\frac{2 \times 0.54}{18}\) g
0.30 g of substance contains hydrogen = \(\frac{2 \times 0.54}{18 \times 0.30}\) g
100 g of substance contains = \(\frac{2 \times 0.54}{18 \times 0.30}\) × 100 g = 20 g of hydrogen
Percentage of hydrogen = 20 %

Question 48.
The ammonia evolved form 0.20 g of an organic compound by Kjeldahl method neutralized 15 ml of N / 20 sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%

Question 49.
0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Solution:
Mass of the substance taken = 0.32 g
Mass of BaSO₄ formed = 0.466 g
Molecular mass of BaSO₄ = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of BaSO₄ = \(\frac{0.466 \times 32}{233}\)

Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %

Question 50.
0.24 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Solution:
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143. 5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) × 0.287
Percentage of chlorine = \(\frac{35.5}{143.5} \times \frac{0.287}{0.24}\) × 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Solution:
Volume of N₂ at NTP = \(\frac{V \times P}{t+273} \times \frac{273}{760}\)
= V₀ ml
Substituting the various values in the above equation,
V₀ = \(\frac{20.7 \times 760}{288} \times \frac{273}{760}\) = 19.62 ml

weight of 19.62 ml of Nitrogen = \(\frac{28}{22400}\) × 19.62 g

∴ Percentage of Nitrogen = \(\frac{28}{22400}\) × 19.62 × \(\frac{100}{0.35}\)
= 4.9 %

11th Chemistry Guide Fundamentals of Organic Chemistry Additional Questions and Answers

I. Choose the correct answer:

Question 1.
Organic compounds can be formed by
a) Plants only
b) Animals only
c) Plants and Animals
d) Plants, animals and can be synthesized in the laboratory
Answer:
d) Plants, animals and can be synthesized in the laboratory

Question 2.
Which of the following is not an organic compound?
(a) DNA
(b) Lipid
(c) Glycogen
(d) Bronze
Answer:
(d) Bronze
Solution:
It is an alloy and a mixture of metals and all others are organic compounds.

Question 3.
The vital force theory was proposed by
a) Wohler
b) Berthlot
c) Berzelius
d) Kolbe
Answer:
c) Berzelius

Question 4.
Which of the following is an example of heterocyclic aromatic compound?
(a) THF
(b) Cyclopropane
(c) Pyridine
(d) Azulene
Answer:
(c) Pyridine

Question 5.
The first organic compound was synthesized in the laboratory by
a) Wohler
b) Kolbe
c) Berzelius
d) Neil Barthlot
Answer:
a) Wohler

Question 6.
Which of the following pair is an example of aromatic compounds?
(a) Benzene, Toluene
(b) Cyclopropane, Cyclobuane
(c) Pyridine, Pyrrole
(d) Propane, Butane
Answer:
(a) Benzene, Toluene

Question 7.
Marsh gas mainly contains
a) C₂H₂
b) C₂H₄
c) CH₄
d) C₂H₆
Answer:
c) CH₄

Question 8.
Hybridization at 2nd carbon in CH₂ = CH – CH₃ is
a) sp
b) sp²
c) sp³
d) sp³d
Answer:
b) sp²

Question 9.
Number of possible position isomers for Dichlorobenzene is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 10.
Which of the following represents thiol?
(a) -SH
(b) -OH
(c) -SR
(d) -SCN
Answer:
(a) -SH

Question 11.
Alkanols and Alkoxyalkanes are
a) Functional isomers
b) Keto – enol tautomers
c) Geometrical isomers
d) Not isomers at all
Answer:
a) Functional isomers

Question 12.
n – propyl alcohol and isopropyl alcohol are examples of
a) Position isomerism
b) Chain isomerism
c) Tautomerism
d) Geometrical isomerism
Answer:
a) Position isomerism

Question 13.
The number of structural alcoholic isomers for C₄H₁₀O is
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 14.
Cycloalkanes are isomeric with
a) Alkadienes
b) Alkynes
c) Aromatic compounds
d) Olefins
Answer:
d) Olefins

Question 15.
Which one of the following is called benzylchloride?
(a) C₆H₅CH₂Cl
(b) C₆H₅CHCl₂
(c) C₆H₅CCl₃
(d) C₆H₅Cl
Answer:
(a) C₆H₅CH₂Cl

Question 16.
Diethyl ether and n – propyl methyl ether are
a) Metamers
b) Chain isomers
c) Geometrical isomers
d) Position isomers
Answer:
a) Metamers

Question 17.
The total number of structural isomers for the compound of the formula C₄H₁₀O is
a) 7
b) 6
c) 4
d) 3
Answer:
a) 7

Question 18.
The number of primary alcoholic isomers with the formula C₄H₁₀O is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 19.
Which colour is formed in Lassaigne’s test for nitrogen?
(a) Purple
(b) Black
(c) Prussian blue
(d) Violet
Answer:
(c) Prussian blue

Question 20.
The number of isomeric amines possible for the formula C₃H₉N
a) 4
b) 3
c) 5
d) 6
Answer:
a) 4

Question 21.
Which hybrid orbitals are involved in the CH₃ – CH = CH – CH₃ compound
a) sp and sp³
b) sp² and sp³
c) sp and sp²
d) only sp³
Answer:
b) sp² and sp³

Question 22.
Which of the following bonds is strongest?
a)

b) > C = C <

c)

d) – C – C –
Answer:
a)

Question 23.
Identify the colour formed when Lassigne’s extract of sulphur containing organic compound is mixed with sodium nitroprusside solution?
(a) Prussian blue
(b) Black
(c) Green
(d) Purple
Answer:
(d) Purple

Question 24.
Which of the following is an aromatic compound
a) Phenol
b) Naphthalene
c) Pyridine
d) All
Answer:
d) All

Question 25.
Which one of the following is not identified by Lassaigne’s test?
(a) nitrogen
(b) sulphur
(c) halogens
(d) phosphorous
Answer:
(d) phosphorous

Question 26.
Which is an alicyclic compound?
a) benzene
b) cyclohexane
c) pyridine
d) pyrrole
Answer:
b) cyclohexane

Question 27.
Which of the following is not a cyclic compound?
a) Anthracene
b) Pyrrole
c) Phenol
d) Isobutylene
Answer:
d) Isobutylene

Question 28.
Functional group present in amides is
a) – COOH
b) – NH₂
c) – CONH₂
d) – COO –
Answer:
c) – CONH₂

Question 29.
Which of the following is used as moisture absorbent?
(a) Potash
(b) Soda
(c) Conc. H₂SO₄
(d) Na₃CO₄
Answer:
(c) Conc. H₂SO₄

Question 30.
IUPAC name of methyl cyanide is
a) Cyano methane
b) Ethanenitrile
c) Methane nitrile
d) Methyl – n – butylamine
Answer:
b) Ethanenitrile

Question 31.
The correct IUPAC name of is
a) 1, 2 – diethyl butene
b) 2 – ethyl – 3- methyl pentene
c) 3, 4 – dimethyl hex – 3 – ene
d) 2, 3 – dimethyl pent – 2 – ene
Answer:
d) 2, 3 – dimethyl pent – 2 – ene

Question 32.
Which of the following is not purified by the sublimation method?
(a) Camphor
(b) Benzoic acid
(c) Naphthalene
(d) Nitrobenzene
Answer:
(d) nitrobenzene

Question 33.
IUPAC name of CH ≡ C – CH = CH₂ is
a) but – 3 – ene – 1 – yne
b) but – 1 – ene – 3 – yne
c) but – 1 – yne – 3 – ene
d) but – 3 – yne – 1 – ene
Answer:
b) but – 1 – ene – 3 – yne

Question 34.
The IUPAC name of is
a) 4 – hydroxy – 2 – methyl pentanal
b) 2 – hydroxy – 4 – methyl pentanal
c) 4 – hydroxy – 2 – methyl pentanol
d) 2 – hydroxy – 4 – methyl pentanol
Answer:
a) 4 – hydroxy – 2 – methyl pentanal

Question 35.
3 – methyl penta -1, 3- diene is
a) CH₂ = CH (CH₂)₂ CH₃
b) CH₂ = CHCH (CH₃) CH₂CH₃
c) CH₃CH = C(CH₃)CH = CH₂
d) CH₃ = C = CH (CH₃)₂
Answer:
c) CH₃CH = C(CH₃)CH = CH₂

Question 36.
The structural formula of methyl amino methane is
a) (CH₃)₂ CH NH₂
b) (CH₃)₃ N
c) (CH₃)₂ NH
d) CH₃NH₂
Answer:
c) (CH₃)₂ NH

Question 37.
Which of the following compounds gives curdy white precipitate in Lassaigne’s test?
(a) CH₃Br
(b) C₂H₅I
(c) CH₃Cl
(d) C₆H₅NO₂
Answer:
(c) CH₃Cl

Question 38.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) 1 – Butanol
c) 2 – Methyl – 2 – propanol
d) Butanone
Answer:
d) Butanone

Question 39.
Which of the following pairs of compounds are tautomers?
a) Propanol & propanone
b) Ethanol & vinyl alcohol
c) Ethanol & allyl alcohol
d) Vinyl alcohol & ethanal
Answer:
d) Vinyl alcohol & ethanal

Question 40.
Which of the following compounds does not have any tertiary hydrogen atoms?
a) (CH₃)₃ CCH₂ CH₃
b) (CH₃)₂ CHCH₂ CH₃
c) (CH₃)₂ CHCH (CH₃)₂
d) (CH₃)₃ CH
Answer:
a) (CH₃)₃ CCH₂ CH₃

Question 41.
The IUPAC name of is
a) 2 – Methyl – 2 – butenoic acid
b) 3 – Methyl – 3 – butenoic acid
c) 3 – Methyl – 2 – butenoic acid
d) 2 – Methyl – 3 – butenoic acid
Answer:
c) 3 – Methyl – 2 – butenoic acid

Question 42.
Which of the following reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound?
(a) NaOH
(b) FeCl₃
(c) H₂SO₄
(d) NH₄OH
Answer:
(d) NH₄OH

Question 43.
The IUPAC name of the Compound CH₃ – CH(OH) – COOH is
a) Lactic acid
b) 2 – Hydroxy propanoic acid
c) 3 – Hydroxy propanoic acid
d) Carboxy propanol
Answer:
b) 2 – Hydroxy propanoic acid

Question 44.
The IUPAC name of the compound is
a) 2 – Ethyl – ethyl acetate
b) Ethyl – 3 – methy1butanoate
c) Ethyl – 2 – methyl butanoate
d) 2- methyl butanoic acid
Answer:
c) Ethyl – 2 – methyl butanoate

Question 45.
The IUPAC name of the given compound is
a) 2,2 — Dimethyl butane
b) lsohexane
c) 2, 3 – Dimethyl butane
d) Di isohexane
Answer:
c) 2, 3 – Dimethyl butane

Question 46.
The IUPAC name of the given compound is
a) Octyl cyclopentane
b) 3 – cyclopentyl octane
c) Cyclopentane octane
d) 6 – cyclopentyl octane
Answer:
b) 3 – cyclopentyl octane

Question 47.
The IUPAC name of is
a) but – 2- ene – 2,3 – diol
b) pent – 2- ene – 2,3 – diol
c) 2 – methyl but – 2 – ene – 2,3 – diol
d) hex – 2- ene – 2,3 – diol
Answer:
b) pent – 2- ene – 2,3 – diol

Question 48.
The structure of 3-bromoprop-1-ene is
a)

b) CH₃ – CH = CH – Br

c)

d) Br – CH₂ – CH ≡ CH₂
Answer:
d) Br – CH₂ – CH ≡ CH₂

Question 49.
Neo-heptyl alcohol is correctly represented as
a)

b)

c)

d)
Answer:
c)

Question 50.
The number of dibromo derivatives possible for propane is
a) 2
b) 3
c) 1
d) 4
Answer:
d) 4

Question 51.
The number of aromatic isomers possible for C₇H₈O is
a) 2
b) 3
c) 4
d) 5
Answer:
d) 5

Question 52.
Isomers of propanoic acid are
a) HCOOC₂H₅ and CH₃COOCH₃
b) H – COOC₂H₅ and C₃H₇COOH
c) CH₃COOCH₃ and C₃H₇OH
d) C₃H₇OH and CH₃COCH₃
Answer:
a) HCOOC₂H₅ and CH₃COOCH₃

Question 53.
IUPAC name of CH₃ – CH (OCH₃) – CH₂ – NH₂
a) 2-methoxy propanamine
b) 1-amino – 2-methoxy propane
c) 1-amino – 2-methyl – 2-methoxy ethane
d) 1 – methoxy- 2-amino propane
Answer:
a) 2-methoxy propanamine

Question 54.
IUPAC name of is
a) 3 – cyanopentane – 1, 5 – dinitrile
b) Propane – 1, 2, 3-tri nitrile
c) 1, 2, 3-tri cyano propane
d) Propane 1, 2, 3-tricarbonitrile
Answer:
d) Propane 1, 2, 3-tricarbonitrile

Question 55.
The IUPAC name of the compound is
a) 3 – Carboxylic pentane – 1,5 -dioic acid
b) Propane – 1, 2, 3 – trioic acid
c) 1, 2, 3- tricarboxylic propane
d) Propane – 1,2, 3 – tricarboxylic acid
Answer:
b) Propane – 1, 2, 3 – trioic acid

Question 56.
The IUPAC name of the following compound
CH₃ – C(CH₃)₂ – CH₂ – CH = CH₂ is
a) 2, 2 – Dimethyl – 4 – pentene
b) 4, 4 – Dimethyl – 1 – pentene
c) 1, 1, 1 – trimethyl – 3 – butene
d) 4, 4, 4 – trimethyl – 1 – butene
Answer:
b) 4, 4 – Dimethyl – 1 – pentene

Question 57.
The IUPAC name of is
a) 2, 4 – Dimethyl pentan – 2 – ol
b) 2, 4 – Dimethyl pentan – 4 – ol
c) 2,2 – Dimethyl butan – 2- ol
d) Butan – 2 – ol
Answer:
a) 2, 4 – Dimethyl pentan – 2 – ol

Question 58.
The IUPAC name the compound is
a) Butane – 2, 3, 4 – triol
b) Butane – 1,2, 3 – triol
c) Pentane – 1, 2, 3 – triol
d) 2, 3 dihydroxy butanol
Answer:
b) Butane – 1,2, 3 – triol

Question 59.
The IUPAC name the compound is
a) 3, 3 – Dimethyl – 1 – cyclohexanol
b) 1, 1- Dimethyl – 3 – hydroxy cyclohexane
c) 3, 3 – Dimethyl – 1 – hydroxy cyclohexane
d) 1, 1 – Dimethyl – 3 – cyclohexanol
Answer:
a) 3, 3 – Dimethyl – 1 – cyclohexanol

Question 60.
The IUPAC name of the compound is
a) 2 – Ethylprop – 2 – en – 1 – ol
b) 2 – Hydroxymethyl butan – 1- ol
c) 2 – Methylene butan – 1 – ol
d) 2 – Ethyl -3 hydroxyprop – 1 – ene
Answer:
a) 2 – Ethylprop – 2 – en – 1 – ol

Question 61.
The number of possible alkynes with molecular formula C₅H₈ is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 62.
What is the IUPAC name of the following is
a) 3 – chloro cyclo hexa – 1, 5 – diene
b) 5 – chloro cyclo hexa – 1, 3 – diene
c) 1 – chloro cyclo hexa – 2, 5 – diene
d) 2 – chloro cyclo hexa – 1, 4 – diene
Answer:
b) 5 – chloro cyclo hexa – 1, 3 – diene

Question 63.
What is the IUPAC name of the following is
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al
b) 4 – hydroxy cyclohex – 1 – ene – 3 – al
C) 2 – hydroxy cyclohex – 5 – ene – 1 – al
d) 1 – formyl cyclohex – 5 – ene – 2 – ol
Answer:
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al

Question 64.
What is the IUPAC name of the following?

a) 1 – chloro – 2 – bromo – 4 – nitrobenzene
b) 1 – bromo – 2- chloro – 4 – mtrobenzene
c) 3 – bromo – 4 – chloro – nitrobenzene
d) 2 – bromo – 1 – chloro – 4- nitrobenzene
Answer:
d) 2 – bromo – 1 – chloro – 4- nitrobenzene

Question 65.
What is the IUPAC name of the following?

a) Ethenyl cyclo pentane
b) cyclopentyl ethene
c) cyclopentyl ethylene
d) vinyl cyclopentane
Answer:
b) cyclopentyl ethene

Question 66.
The hybridization of carbon atoms in C – C single bond is HC ≡ C – CH = CH₂ is
a) sp³ – sp³
b) sp² – sp³
c) sp – sp²
d) sp³ – sp
Answer:
c) sp – sp²

Question 67.
The correct IUPAC name of the compound is
a) 1, 4 – Butane dioicacid
b) Ethane – 1, 2 – dicarboxylic acid
c) Succinic acid
d) 1, 2 – Ethane dioic acid
Answer:
a) 1, 4 – Butane dioic acid

Question 68.
Correct statements about CH₃ – CH₂ – CN is
a) common name of the compound is ethylcyanide.
b) IUPAC name of the compound propane – 1 – nitrile.
c) secondary suffix of the compound is nitrile.
d) IUPAC name of the compound is ethane nitrile.
Answer:
a) common name of the compound is ethylcyanide.

Question 69.
Tautomerism is shown by
a) R – C ≡ N
b) R – NO₂
c) R – OH
d) R – COOH
Answer:
b) R – NO₂

Question 70.
Stereo isomers have different
a) Molecular mass
b) Molecular formula
c) Structural formula
d) Configuration
Answer:
d) Configuration

Question 71.
Geometrical isomerism may be exhibited by compounds having atleast.
a) One double bond
b) One triple bond
c) One asymmetric carbon
d) One polar bond
Answer:
a) One double bond

Question 72.
The prefixes syn – and anti – are used to denote
a) structural isomers
b) conformational isomers
c) geometrical isomers
d) optical isomers
Answer:
c) geometrical isomers

Question 73.
d – tartaric acid and l – tartaric acid are
a) geometrical isomers
b) conformers
c) enantiomers
d) diastereomers
Answer:
c) enantiomers

Question 74.
The method of separation of enantiomers from racemic mixture is known as
a) inversion
b) recemisation
c) resolution
d) asymmetric synthesis
Answer:
c) resolution

Question 75.
Racemic mixture is optically inactive due to
a) internal compensation
b) external compensation
c) inversion
d) plane of symmetry
Answer:
c) inversion

Question 76.
Meso isomers are possible when the organic compound contains
a) one asymmetric carbon
b) two or more dissimilar asymmetric carbons
c) similar asymmetric carbons
d) unsaturation
Answer:
c) similar asymmetric carbons

Question 77.
Optical inactivity of meso isomer is due to
a) element of symmetry and element of asymmetry
b) internal compensation
c) due to lack of asymmetric carbon
d) External compensation
Answer:
b) internal compensation

Question 78.
A racemic mixture is a mixture of
a) meso and its isomers
b) d and l isomers of same compound in equimolar proportions
c) d and l isomers of same compound in different proportions
d) mixture of d and meso isomers
Answer:
b) d and 1 isomers of same compound in equimolar proportions

Question 79.
Which of the following is optically active?
a) HOOC – CH₂ – COOH
b) CH₃ – CO – COOH
c) CH₃ – CH(OH) – COOH
d) CH₃ – CH₂ – COOH
Answer:
c) CH₃ – CH(OH) – COOH

Question 80.
Which of the following is optically active?
a) n – propanal
b) 2 – chlorobutane
c) n – butanal
d) 3 – pentanol
Answer:
b) 2 – chlorobutane

Question 81.
The number of optical enantiomers of tartaric acid
a) 3
b) 2
c) 4
d) 1
Answer:
b) 2

Question 82.
Geometrical isomers differ in
a) position of substituents
b) Position of double bond
c) C – C bond length
d) Spatial arrangement of groups
Answer:
d) Spatial arrangement of groups

Question 83.
Which of the following exhibit cis – trans isomerism
a) propene
b) 1 – butene
c) 2 – butene
d) benzene
Answer:
c) 2 – butene

Question 84.
Which of the following show geometrical isomerism?
a) CH₃CH = CHCH₃
b) (CH₃)₂C = CH₂
c) C₂H₅CH = CH₂
d) CH₃CH = CH₂
Answer:
a) CH₃CH = CHCH₃

Question 85.
Which of the following does not show geometrical isomerism?
a) 1, 2 – dichloro – 1 – pentene
b) 1, 3 – dichloro – 2 – pentene
c) 1, 1 – dichloro – 1 – pentene
d) 1, 4 – dichloro – 2 – pentene
Answer:
c) 1, 1 – dichloro – 1 – pentene

Question 86.
Geometrical isomerism is not shown by
a)

b)

c) CH₂ = C(CI) CH₃

d) CH₃ – CH = CH – CH = CH₂
Answer:
c) CH₂ = C(CI) CH₃

Question 87.
Which of the following is optically active?
a) Glycerine
b) Acetaldehyde
c) Glyceraldehyde
d) Acetone
Answer:
c) Glyceraldehyde

Question 88.
The minimum number of C atoms for a hydrocarbon to exhibit optical isomerism
a) 4
b) 5
c) 6
d) 7
Answer:
d) 7

Question 89.
Which of the following can form cis – trans isomers?
a) C₂H₅Br
b) (CH)₂(COOH)₂
c) CH₃CHO
d) (CH₂)₂COOH
Answer:
b) (CH)₂(COOH)₂

Question 90.
No.of geometrical isomers possible for the compound CH₃ – CH = CH – CH = CH – C₂H₅
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 91.
The number of geometrical isomers of CH₃ – CH = CH – CH = CH – CH = CHCl is
a) 2
b) 4
c) 6
d) 8
Answer:
d) 8

Question 92.
Minimum number of C atoms for an alkene hydrocarbon, that shows geometrical & optical isomerism both
a) 5
b) 6
c) 7
d) 8
Answer:
c) 7

Question 93.
Among the following compounds which exhibits optical isomerism?
a) propanol
b) 2 – propanol
c) 1 – butanol
d) 2 – butanol
Answer:
d) 2 – butanol

Question 94.
Which of the mesoisomer?
a) CH₂OHCHOHCHO
b) CH₂OHCHOHCHOHCHO
c) HOOCCHOHCHOHCOOH
d) HOH₂CCHOHCHOHCOOH
Answer:
c) HOOCCHOHCHOHCOOH

Question 95.
d – tartaric acid and l – tartaric acid can be separated by
a) Salt formation
b) Fractional distillation
c) Fractional crystallization
d) Chromatography
Answer:
a) Salt formation

Question 96.
Paper chromatography is
a) Adsorption chromatography
b) partition chromatography
c) Ion exchange chromatography
d) all of these
Answer:
b) partition chromatography

Question 97.
Simple distillation can be used to separate liquids which differ in their boiling points at least by
a) 5°C
b) 10°C
c) 40 – 50°C
d) 100°C
Answer:
c) 40 – 50°C

Question 98.
In adsorption chromatography mobile phase will be
a) Only solid
b) Only liquid
c) Only gas
d) Liquid as well as gas
Answer:
d) Liquid as well as gas

Question 99.
Which of the following can be used as adsorbent in adsorption chromatography?
a) Silica gel
b) Alumina
c) Cellulose powder
d) All of these
Answer:
d) All of these

Question 100.
Two substances when separated out on the basis of their extent of adsorption by one material, the phenomenon is called
a) Chromatography
b) Crystallization
c) Sublimation
d) Steam distillation
Answer:
a) Chromatography

Question 101.
Chromatographic technique is used for the separation of
a) Camphor
b) Alcohol & Water
c) Acetone and Methanol
d) Plant pigments
Answer:
d) Plant pigments

Question 102.
In column chromatography stationary phase is
a) only solid
b) only liquid
c) only gas
d) All of these
Answer:
a) only solid

Question 103.
Which of the following method is used for the purification of solids?
a) Distillation under reduced pressure
b) Distillation
c) Strain distillation
d) Sublimation
Answer:
d) Sublimation

Question 104.
Vacuum distillation is used to purify liquids which
a) are highly volatile
b) are explosive in nature
c) soluble in water
d) decomposes below their B.P’s
Answer:
c) soluble in water

Question 105.
Impure Napthalene is purified by
a) Fractional crystallization
b) Fractional distillation
C) solvent extraction
d) sublimation
Answer:
d) sublimation

Question 106.
A very common adsorbent used in coloun^n chromatography is
a) Powdered charcoal
b) Alumina
c) Chalk
d) Sodium carbonate
Answer:
b) Alumina

Question 107.
Simple distillation of liquids involves simultaneously
a) Vapourisation and condensation
b) Condensation and vapourisation
c) Vapourisation and sublimation
d) Sublimation and condensation
Answer:
a) Vapourisation and condensation

Question 108.
The latest technique for the purification of organic compounds is
a) Fractional distillation
b) Chromatography
c) Vacuum distillation
d) Crystallization
Answer:
b) Chromatography

Question 109.
Fixed melting point of an organic compound informs
a) Purity of an organic compound
b) Conductivity of compound
c) Chemical nature of compound
d) Whether the compound is liquid or gas
Answer:
a) Purity of an organic compound

Question 110.
Lassaigne’s test is used in qualitative analysis to detect
a) Nitrogen
b) Sulphur
c) Chlorine
d) All of these
Answer:
d) All of these

Question 111.
In Lassaigne’s method organic compound is fused with
a) Sodium metal
b) Zinc dust
c) Sodium carbonate and Zinc dust
d) Calcium metal
Answer:
a) Sodium metal

Question 112.
Presence of nitrogen in organic compound in Lassaigne’s extract as
a) Nitrogen gas
b) NH₃
c) NO
d) CN^–
Answer:
d) CN^–

Question 113.
Medium of Sodium extract is
a) Neutral
b) Basic
c) Acidic
d) Depends on organic compound
Answer:
b) Basic

Question 114.
H₂O vapours on passing through anhydrous CuSO₄ turns it to
a) Green
b) Blue
c) Violet
d) White
Answer:
b) Blue

Question 115.
When a nitrogenous organic compound is fused with sodium, the nitrogen present in the compound is converted into
a) Sodium Nitrate
b) Sodium nitrite
c) Sodamide
d) Sodium cyanide
Answer:
d) Sodium cyanide

Question 116.
In the Lassainge’s test the Sulphur present in the organic compound first changes into
a) Na₂SO₃
b) CS₂
c) Na₂SO₄
d) Na₂S
Answer:
d) Na₂S

Question 117.
Which of the following elements in an organic compound cannot be detected by Lassaigne’s test?
a) N
b) S
c) Cl
d) H
Answer:
d) H

Question 118.
A compound which does not give a positive result in the Lassaigne’s test for nitrogen is
a) Urea
b) Hydroxyl amine
c) Glycine
d) Phenylhydrazine
Answer:
b) Hydroxyl amine

Question 119.
Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of
a) N
b) S
c) O
d) Cl
Answer:
b) S

Question 120.
The presence of halogen in an organic compound is detected by
a) Iodoform test
b) Silver nitrate test
c) Beilstein’s test
d) Million’s test
Answer:
c) Beilstein’s test

Question 121.
The Beilstein’s test in a rapid test used for . organic compound^ to^ detect
a) Phosphorous
b) Sulphur
c) Halogens
d) Nitrogen
Answer:
c) Halogens

Question 122.
Liebig’s method is used for the estimation of
a) Nitrogen
b) Sulphur
c) Carbon and hydrogen
d) Halogens
Answer:
c) Carbon and hydrogen

Question 123.
In Kjeldahl’s method of estimation of nitrogen, copper sulphate act as
a) Oxidizing agent
b) reducing agent
c) Catalytic agent
d) Hydrolysing agent
Answer:
c) Catalytic agent

Question 124.
Percentage of carbon in an organic compound is determined by
a) Duma’s method
b) Kjeldahl’s method
c) Carius method
d) Liebig’s method
Answer:
d) Liebig’s method

Question 125.
Halogen can be estimated by
a) Duma’s method
b) Carius method
c) Leibig’s method
d) All of these
Answer:
b) Carius method

Question 126.
In Garius method halogens are estimated
a) X₂
b) BaX₂
c) PbX₂
d) AgX
Answer:
d) AgX

Question 127.
In Duma’s method nitrogen in organic compound is estimated in the form of
a) N₂
b) NO
c) NH₃
d) N₂O₅
Answer:
a) N₂

Question 128.
In Kjeldahl’s method to estimate nitrogen, compound is heated with conc.H₂SO₄ in presence of
a) CaSO₄
b) (NH₄)₂SO₄
c) CuSO₄
d) P₂O₅
Answer:
c) CuSO₄

Question 129.
In organic compounds, Sulphur is estimated as
a) BaSO₄
b) BaCl₂
c) Ba₃(PO₄)₂
d) H₂SO₄
Answer:
a) BaSO₄

Question 130.
In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is the amount of C0„ formed then
a) %C = \(\frac{12 \times x}{16 \times w}\)

b) %C = \(\frac{12}{44} \times \frac{\mathrm{w}}{\mathrm{x}} \times 100\)

c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

d) %C = \(\frac{12}{44} \times \frac{x}{w}\)
Answer:
c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

Question 131.
In Dumas method for estimating nitrogen in organic compound, the gas finally collected is
a) N₂
b) NO
c) NH₃
d) H₂
Answer:
a) N₂

Question 132.
In Dumas method, the gas which is collected in Nitrometer is
a) N₂
b) NO
c) NH₃
d) H₂
Answer:
a) N₂

Question 133.
In Kjeldahl’s method, the nitrogen presence is estimated as
a) N₂
b) NH₃
c) NO₂
d) N₂O₃
Answer:
b) NH₃

Question 134.
In Kjeldahl’s method, nitrogen present in the organic compound is first converted into
a) NH₃
b) (NH₄)₂SO₄
c) N₂
d) NO
Answer:
b) (NH₄)₂SO₄

Question 135.
In Liebig’s method for the estimation of C and H, the combustion tube is passed over
a) CuO pellets
b) Copper turnings
c) Iron fillings
d) Zinc – copper couple
Answer:
a) CuO pellets

Question 136.
Which gas is introduced into the combustion tube in Liebig’s method?
a) Pure and dry CO₂
b) Pure and dry N₂
c) Pure and dry O₂
d) Pure and dry He
Answer:
c) Pure and dry O₂

Question 137.
Chromatographic techniques of purification can be used for
a) Coloured compounds
b) Liquids
c) Solids
d) All of these
Answer:
d) All of these

Question 138.
Two substances when separated on the basis of partition co – efficient between two liquid phase, then the technique is known as
a) column chromatography
b) Paper chromatography
c) GLC
d) TLC
Answer:
b) Paper chromatography

Question 139.
Ortho and para nitro phenols can be separated by
a) crystallization
b) distillation
c) sublimation
d) solvent extraction
Answer:
b) distillation

Question 140.
In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is
a) Equal to atmospheric pressure
b) Less than atmospheric pressure
c) More than atmospheric pressure
d) Exactly half of the atmospheric pressure
Answer:
a) Equal to atmospheric pressure

Question 141.
Organic compound is fused with metallic sodium for testing nitrogen, sulphur, and halogens because
a) To make the solution alkaline
b) To convert into elemental state of nitrogen, sulphur, and halogens
c) To convert covalent compound into ionic compound
d) To decrease fusion temperature
Answer:
c) To convert covalent compound into ionic compound

Question 142.
Sodium extract gives blood red colour when treated with FeCl₃, Formation of blood-red colour confirms the presence of
a) Only nitrogen
b) Only sulphur
c) Only halogens
d) Both Nitrogen and Sulphur
Answer:
d) Both Nitrogen and Sulphur

Question 143.
The compound not formed in the positive test for nitrogen with the Lussaigne’s solution of an organic compound is
a) Fe₄[Fe(CN)₆]₃
b) Na₃[Fe(CN)₆]
c) Fe(CN)₃
d) Na₃[Fe(CN)₅NOS]
a) b, c, d
b) a, b
c) a, b, c
d) a only
Answer:
a) b, c, d

Question 144.
The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of
a) Na₄[Fe(CN)₆]
b) Fe₃[Fe(CN)₆]₂
c) Fe₂[Fe(CN)₆]
d) Fe₄[Fe(CN)₆]₃
Answer:
d) Fe₄[Fe(CN)₆]₃

Question 145.
Which of the following compounds will answer Lassaigne’s test for nitrogen?
a) NH₂NH₂
b) NH₂OH
c) NaCN
d) NaNO₃
Answer:
c) NaCN

Question 146.
In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is
a) 28
b) 38
c) 18
d) 48
Answer:
a) 28

Question 147.
0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be
a) 49.32
b) 59.32
c) 29.32
d) 98.64
Answer:
a) 49.32

Question 148.
In an estimation of S by Carius method 0.217 g of the compound gave 0.5825 g of BaSO₄. Percentage of S is
a) 36.78 %
b) 35.50 %
c) 36.48 %
d) 35.69 %
Answer:
a) 36.78 %

II. Very short question and answers (2 Marks):

Question 1.
What are Acyclic compounds? Give suitable example.
Answer:
These are the compounds in which carbon atoms are linked to form open chain (straight or branched). These compounds may be saturated (all single bonds) or unsaturated (multiple bonds).
Example:

These acyclic compounds are known as open chain or aliphatic compounds.

Question 2.
What arc Alicyclic Compounds? Give suitable example.
Answer:
These are saturated or unsaturated carbocyclic compounds which resemble the corresponding acylic compounds in their properties.

Question 3.
What are Aromatic heterocyclic Compounds? Give example.
Answer:
These are the heterocyclic compounds which possess aromaticity and resemble, the corresponding aromatic compounds in most of their properties. These are also called non-benzenoid aromatic compounds.

Question 4.
What is functional group? Give example.
Answer:
An atom or group of atoms within a molecule that shows a characteristics set of physical and chemical properties.
Example:
(i) – NH₂ – amines
(ii) = NH – Imines
(iii)

Question 5.
Almost all compounds of carbon form covalent bonds. Give reason.
Answer:
Carbon (Z = 6) have electronic configuration of is² 2s² 2p². It is not possible for the carbon to form either C⁴⁺ or C^4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, form four covalent bonds.

Question 6.
What is Metamerism? With suitable examples.
Answer:
Metamerism:
This type of isomerism is a special kind of structural isomerism arises due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to the either side of the same functional group and having same molecular formula. This isomerism is shown by compounds having functional group such as ethers, ketones, esters and secondary amines between two alkyl groups.
C₄H₁₀O
CH₃ – O – C₃H₇ – Methyl propyl ether 1 – methoxypropane

C₂H₅ – O – C₂H₅ – diethyl ether ethoxyethane

– methyl isopropyl ether 2 – methoxypropane.

Question 7.
What is meant by stereochemistry?
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This branch of chemistry dealing with the study of the three-dimensional nature (spactial arrangement) of molecules is known as stereochemistry. The metabolic activities in living organisms, natural synthesis and drug synthesis involve various stereoisomers.

Question 8.
What is enantiomerism?
Answer:
An optically active substance may exist in two or more isomeric forms which have same physical & chemical properties but differ in terms of direction of rotation of plane-polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenomenon is known as enantiomerism.
Example: d and l lactic acid.

Question 9.
What are the conditions for an organic compound is said to be optically active?
Answer:
(1) The molecule must contains at least one chiral or Asymmetric carbon atom.
(2) The object molecule should not be super impossable with its own mirror image.

Question 10.
How will you detect phosphorus present in the given organic compound?
Answer:
Test for phosphorous:
A solid compound is strongly heated with a mixture of Na₂CO₃ & KNO₃. Phosphorous present in the compound is oxidized to sodium phosphate. The residue is extracted with water and boiled with Conc. HNO₃. A solution of ammonium molybdate is added to the above solution. A canary yellow coloration or precipitate shows the presence of phosphorous.

Question 11.
Explain the sublimation process for the purification of organic compounds.
Answer:
Few substances like benzoic acid, naphthalene and camphor when heated pass directly from solid to vapor without melting (ie liquid). On cooling the vapours will give back solids. Such phenomenon is called sublimation. It is a useful technique to separate volatile and non-volatile solid. It has limited application because only a few substance will sublime.
Example:
naphthalene, benzoic acid.

Question 12.
What is the need for purifying an organic compound?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds they must be in a pure state.

III. Short question and answers (3 Marks):

Question 1.
How are naphthalene and camphor purified?
Answer:
1. Naphthalcne, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non-volatile impurities.

2. Substances to he purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resuJting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 2.
What are Alicycic heterocyclic Compounds? Give example.
Answer:
These heterocyclic compounds resemble the corresponding aliphatic compounds in most of their properties.
Example:

Question 3.
Write the IUPAC name of the following compounds.
a)

b)

c)

d)
Answer:
a)

b)

c)

d)

Question 4.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 5.
Write the IUPAC name of the following compounds.
a)
b)

c)

d)
Answer:
a)

b)

c)

d)

Question 6.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 7.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 8.
Classify the following compounds based on the structure.
i) CH ≡ C – CH₂ – C ≡ CH

ii) CH₃ – CH₂ – CH₂ – CH₂ – CH₃

iii)

iv)
Answer:
i) CH ≡ C – CH₂ – C ≡ CH is unsaturated open chain compound

ii) CH₃ – CH₂ – CH₂ – CH₂ – CH₃ is saturated open chain compound

iii) is aromatic benzenoid compound

iv) is alicyclic compound

Question 9.
Give two examples for each of the following type of organic compounds.
(i) non-benzenoid aromatic
(ii) aromatic heterocyclic
(iii) alicyclic
(iv) aliphatic open chain
Answer:
(i) non-benzenoid aromatic

(ii) aromatic heterocyclic

(iii) alicyclic

(iv) aliphatic open chain

Question 10.
Explain how nitrobenzene and benzene arc separated and purified. (or) How will you separate the mixture of diethyl ether and ethanol?
Answer:
Distillation:

• The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver.
• This method ¡s to purify liquids from non-volatile impurities and used for separating the constituents of a liquid mixture which differ in their boiling points.
• In this simple distillation process, liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified.
• e.g., the mixture of C₆H₅NO₂ nitrobenzene (h.p. 484 K) and C₆H₆ benzene (b.p. 354 K) can be urified and separated. Similarly the mixture of diethyL ether (b.p. 308K) and ethyl alcohol (h.p. 351 K) can he purified and separated.

Question 11.
In an estimation of sulphur by carius method. 0.2175 g of the substance gave 0.5825 g of BaSO₄ calculate the percentage composition of S in the compound.
Solution:
Weight of organic compound = 0.2175 g
Weight of BaSO₄ = 0.5825 g
233 g of BaSO₄ contains = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175}\right)\)
0.5825 g of BaSO₄ contains
Percentage of S = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100\right)\)
= 36.78 %

Question 12.
0.16 g of an organic compound was heated in a carius tube and H₂SO₄ acid formed
was precipitated with BaCl₂. The mass of BaSO₄ was 0.35 g. Find the percentage of
sulphur [30.04]
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of sulphur = \(\left(\frac{32}{233} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{32}{233} \times \frac{0.35}{0.16} \times 100\right)\)
= 30.04 %

Question 13.
0.284 g of an organic substance gave 0.287 g AgCl in a Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Solution:
Weight of the organic substance = 0.284 g
Weight of AgCl is = 0.287 g
143.5 g of AgCl contains = 35.5 g of chlorine
0.287 g of AgCl contains = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284}\right)\)

% of chlorine is = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100\right)\)
= 24.56 %

Question 14.
0.185 g of an organic compound when treated with Conc. HNO₃ and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
(Ag = 108, Br = 80)
Answer:
Weight of organic substance (w) = 0.185 g
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\left(\frac{80}{188} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{80}{188} \times \frac{0.32}{0.185} \times 100\right)\)
= 73.6%

Question 15.
0.40 g of a iodo – substituted organic compound gave 0.235 g of AgI by carlus method. Calculate the percentage of iodine in the compound. (Ag = 108 I = 127).
Solution:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
Percentage of iodine = \(\left(\frac{127}{235} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{127}{235} \times \frac{0.235}{0.40} \times 100\right)\)
= 31.75%

Question 16.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of phosphorous in the compound.
Solution:
Weight of an organic compound = 0.24 g
Weight of Mg₂P₂O₇ = 0.66 g
222 g of Mg₂P₂O₇ contains = 62 g of P
0.66 g contains = \(\frac{62}{222} \times \frac{0.66}{0.24}\)

Percentage of p = \(\frac{80}{180} \times \frac{0.66}{0.24}\) = 76.80%

Question 17.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound.
Solution:
Weight of organic substance (w) = 0.33 g
Weight of Mg₂P₂O₇ (x) = 0.397 g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{0.397}{0.33} \times 100\) = 33. 59 %

Question 18.
Explain simple distillation process with suitable example.
Answer:
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. This method is called simple distillation. Liquids with large difference in boiling point(about 40k) and do not decompose under ordinary pressure can be purified by simple distillation.
Example:
The mixture of C₆H₅NO₂ (b.p 484K) &
C₆H₆ (354 K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 35 K).

Question 19.
How will you purify an organic compound by differential extraction process?
Answer:
The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction. When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel. The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl₃).

The organic solvent immiscible with water will form a separate layer and the contents are shaken gently. The solute being more soluble in the organic solvent is transfered to it. The solvent layer is then separated by opening the tap of the separating funnel, and the substance is recovered.

Question 20.
Explain the steam distillation process for purifying organic compound.
Answer:
This method is applicable for solids and liquids. If the compound to be steam distilled the compound should not decompose at the steam temperature, should have a fairly high vapour pressure at 373 K, it should be insoluble in water; the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round-bottom flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam. The mixture in the flask is heated and then a current of steam passed in to it. The vapours of the compound mix up with steam and escape into the condenser.

The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers, also in the manufacture of aniline and turpentine oil.

Question 21.
Write a note on azeotropic distillation with suitable example.
Answer:
These are the mixture of liquids that cannot be separated by fractional distillation. The mixtures that can be purified only by azeotropic distillation are called as azeotropes. These azeotropes are constant boiling mixtures, which distil as a single compound at a fixed temperature. Ethanol and water are the most common examples of azeotropic mixture in the ratio of 95.87 : 4.13.

In this method apart from azeotropic mixture a third component like C₆H₆, CCl₄, ether, glycerol, glycol which act as a dehydrating agent depress the partial pressure of one component so that the boiling point of that component is raised sufficiently and thus other component will distill over.

Dehydrating agents like C₆H₆, CCl₄ have low boiling points and reduce the partial vapour pressure of alcohol more than that of water whereas glycerol & glycol, etc. have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 22.
What is Ring chain isomerism? Give an example.
Answer:
In this type of isomerism, compounds having same molecular formula but differ terms of bonding of carbon atom to form open-chain and cyclic structures.

Question 23.
Briefly explain geometrical isomerism in oximes with suitable example.
Ans:
Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here ‘syn’ and ‘anti’ are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and -OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.
For eg:

IV. Long Question and answers (5 Marks):

Question 1.
Write the structure of the following compounds.
a) Cinnamic acid
b) Lactic acid
c) Phthalic acid
d) Tartaric acid
d) Benzoic acid
Answer:
a) Cinnamic acid

b) Lactic acid

c) Phthalic acid

d) Tartaric acid

d) Benzoic acid
C₆H₅COOH

Question 2.
Explain the methods for the representation of structure of organic compounds with suitable example.
Answer:
The structure of an organic compound can he represented using any one of the below mentioned methods.

1. Lewis structure or dot structure,
2. Dash structure or line bond structure,
3. Condensed structure
4. Bond line structure

We know how to draw the Lewis structure for a molecule. The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a Lewis structure. A single line or dash represents single a covalent bond, double line represents double bond (1 σ bond, 2 π bond) and a triple line represents triple bond (1 σ bond, 2 π bond). Lone pair of electrons on hetero atoms may or may not be shown. This represents the complete structural formula.

This structural formula can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.

For further simplification, organic chemists use another way of representing the structures in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are shown in a zigzag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. These representations can be easily understood by the following illustration.

Question 3.
Explain the molecular model method for the representation of structure of organic compounds.
Answer:
Molecular models:

Molecular models are physical devices that are used for a better visualisation and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available.
(i) Frame work model
(ii) Ball and stick model &
(iii) space filling model.

In the frame work model only the bonds connecting the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of the atom.

In the ball and stick model, both the atoms and the bonds are shown. Ball represent atoms and the stick a bond. Compounds containing C = C can be best represented by using springs in place of sticks and this model is termed as ball and spring model.

The space filling model emphasizes the relative size of each atom based on its Vander Waals radius.

Question 4.
Explain briefly
(i) Fisher projection
(ii) sawhorse projection
(iii) Newman projection formula with neat example.
Answer:
(i) Fisher projection formula:
This is a method of representing three dimensional structures in two dimension. In this method, the chiral atom(s) lies in the plane of paper. The horizontal substituents are pointing towards the observer and the vertical substituents are away from the observer. Fisher projection formula for tartaric acid is given below.

Sawhorse projection formula:
Here the bond between two carbon atoms is drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back. The Fischer projection inadequately portrays the spatial relationship between ligands attached to adjacent atoms. The sawhorse projection attempts to clarify the relative location of the groups.

Newman projection formula:
In this method the molecules are viewed from the front along the carbon-carbon bond axis. The two carbon atom forming a bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point whereas the carbon lying further from the eye is represented by the origin of the circle. Therefore, the C-H bonds of the front carbon are depicted from the circle while C-H bonds of the back carbon are drawn from the circumference of the circle with an angle of 120° to each other.

Question 5.
What is Tautomerism? Explain different types of Tautomerism with suitable examples.
Answer:
Tautomerism:
It is a special type of functional isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus, generally hydrogen. The two different structures are known as tautomers. There are several types of tautomerism and the two important types are dyad and triad systems.

Dyad system:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.
Example:

In this example, the hydrogen atom oscillates between carbon & nitrogen atom.

Triad system:
In this system, the hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 migration of hydrogen atom from one polyvalent atom to other within the molecule. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol-form. The polyvalent atoms involved are one oxygen and two carbon atoms. Enolisation is a process in which keto-form is converted to enol form. Both tautomeric forms are not equally stable. The less stable form is known as the labile form.

Question 6.
How is sulphur detected by Lassaigne test?
Answer:
a) To a portion of the lassaigne’s extract, add freshly prepared sodium nitroprusside solution. A deep violet colouration is obtained. This test is also used to detect S^2- in inorganic salt analysis.
Na₂S + Na₂ [Fe(CN₅) NO] → Na₄ [Fe (CN₅) NOS]
sodium nitroprusside

b) Acidify another portion of lassaigne’s extract with acetic acid and add lead acetate solution. A black precipitate is obtained.
(CH₃COO)₂Pb + Na₂S → PbS↓ + 2CH₃COONa
(black ppt)

c) Oxidation test:
The organic substances are fused with a mixture of KNO₃ and Na₂CO₃. The sulphur, if present is oxidized to sulphate.
Na₂CO₃ + S + 3O + Na₂SO₄ + CO₂
The fused mass is extracted with water, acidified with HCl and then BaCl₂ solution is added to it. A white precipitate indicates the pressure of sulphur.
BaCl₂ + Na₂SO₄ -> BaSO₄ + 2NaCl

Question 7.
How is halogen detected by using the Lassaigne test?
Answer:
To another portion of the Lassaigne’s filtrate add dil HNO₃ warm gently and add AgNO₃
Solution:
a) Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
b) Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
c) Appearance of a yellow precipitate insoluble in ammonia solution indicates the presence of iodine.

Na + NaX ( where X = Cl, Br, I)
from organic compound
NaX + AgNO₃ → AgX + NaNO₃

If N or S is present in the compound along with the halogen, we might obtain NaCN and Na₂S in the solution, which interfere with the detection of the halogen in the AgNO₃ test Therefore we boil the lassaignes extract with HNO₃ which decomposes NaCN and Na₂S as

NaCN + HNO₃ NaNO₃ + HCN T
Na₂S + 2HNO₃ 2NaNO₃ + H2S
NaCN + AgNO₃ AgCN + NaNO₃
white ppt confusing with AgCl
Na₂S + AgNO₃ → Ag₂S ↓ + NaNO₃
black ppt

Question 8.
Explain the various steps involved in the crystallization method.
Answer:
Most solid organic compounds are purified by the crystallization method. This process is carried out by the following steps.
1. Selection of solvent:
rganic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:
The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any colored substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:
The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Question 9.
Explain the Carius method for the estimation of sulphur in an organic compound.
Answer:
Estimation of sulphur Carius method:
A known mass of the organic substance is heated strongly with fuming HNO₃, C & H get oxidized to CO₂ & H₂O while sulphur is oxidized to sulphuric acid as per the following reaction.

C CO₂
2H H₂O
S → SO₂  H₂SO₄

The resulting solution is treated with excess of BaCl₂ solution H₂SO₄ present in the solution in quantitatively converted into BaSO₄, from the mass of BaSO₄, the mass of sulphur and hence the percentage of sulphur in the compound can be calculated.

Procedure:
A known mass of the organic compound is taken in clean carius tube and added a few mL of fuming HNO₃. The tube is the sealed. It is then placed in an iron tube and heated for about 5 hours. The tube is allowed to cool to temperature and a small hole is made to allow gases produced inside to escape.

The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker H₂SO₄ acid formed as a result of the reaction is converted to BaSO₄. The precipitate of BaSO₄ is filtered, washed, dried and weighed. From the mass of BaSO₄, percentage of S is found.

Mass of the organic compound = w g
Mass of the BaSO₄ formed = x g
233 g of BaSO₄ contains = 32 g of sulphur

∴ x g of BaSO₄ contain = \(\left(\frac{32}{233} \times \frac{x}{w}\right)\) g of S

Percentage of sulphur = (\(\left(\frac{32}{233} \times \frac{x}{w}\right)\) × 100) %

Question 10.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g
Percentage of hydrogen
18 g of water contain 2 g of hydrogen
0.039 of water contain = \(\left(\frac{2}{18} \times \frac{0.039}{0.26}\right)\) of hydrogen

% of hydrogen = \(\left(\frac{0.039}{0.26} \times \frac{2}{18} \times 100\right)\) = 1.66%
Percentage of Carbon
44 g of CO₂ contain 12 g of C
0.245 g of CO₂ contains \(\left(\frac{12}{44} \times \frac{0.245}{0.26}\right)\) g of C

% of Carbon = \(\left(\frac{12}{44} \times \frac{0.245}{0.26} \times 100\right)\) = 25.69 %

Question 11.
0.2346 of an organic compound yielded C, H & 0 0.2754 g of H₂O and 0.4488 g CO₂. Calculate the % composition.
Solution:
Weight of organic substance w = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO₂ (y), = 0.4488 g
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\)

= \(\left(\frac{12}{44} \times \frac{0.4488}{0.2346} \times 100\right)\) = 52.17 %

Percentage of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{2}{18} \times \frac{0.2754}{0.2346} \times 100\right)\) = 13. 04%
Percentage of oxygen= [100 – (52.17 + 13.04)]
= 100 – 65.21 = 34.79 %

Question 12.
Explain the estimation of phosphours by Carius method.
Answer:
Carius method:
A known mass of the organic compound (w) containing phosphorous is heated with fuming HNO in a sealed tube where C is converted into CO₂ and H to H₂O. phosphorous present in organic compound is oxidized to phosphoric acid which is precipitated, as ammonium phosphomolybdate by heating with Cone. HNO₃ and then adding ammonium molybdate.

H₃PO₄ + 12(NH₄)₂MoO₄ + 21 HNO₃  (NH₄)₃PO₄ + 21 NH₄NO₃ + 12HNO₃

The precipitate of ammonium phosphomolybdate thus formed is filtered washed, dried, and weighed. In an alternative method, the phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture (a mixture containing MgCl₂, NH₄Cl, and ammonia) This ppt is washed, dried, and ignited to get magnesium pyrophosphate which is washed, dried a weighed. The following are the reaction that takes place.

By knowing the mass of the organic compound and the, mass of ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of P is calculated.

Mass of organic compound = w g
Weight of ammonium phosphomolybdate = x g
Weight of magnesium pyrophosphate = y g
Mole mass of (NH₄)₃PO₄. 12MoO₃ is = 1877 g
[3 × (14 + 4) + 31 + 4(16)]+ 12(96 + 3 × 16)
Molar mass of Mg₃P₂O₇ is 222 g
(2 × 24) + (31 × 2) + (7 × 16)
1877 g of (NH₄)₃PO₄. 12 MoO₃ contains 31 g of P
X g of (NH₄)₃PO₄. 12 MoO₃ in w g of organic
compound contains \(\frac{31}{1877} \times \frac{x}{w}\) of phosphorous
Percentage of phosphorous = \(\frac{31}{1877} \times \frac{x}{w}\) (or)
227 of Mg₂P₂O₇ contains 62 g of P
y g of Mg₂P₂O₂ in w g of Organic compound contains = \(\frac{62}{227} \times \frac{y}{w}\) of P
Percentage of phosphorous = \(\frac{62}{227} \times \frac{y}{w}\)%

Question 13.
Explain the estimation of nitrogen by the Dumas method.
Answer:
1. Dumas method:
This method is based upon the fact that nitrogenous compound when heated with cupric oxide in an atmosphere of CO₂ yields free nitrogen. Thus
C_x H_y N_z + (2x + y/2) CuO → x CO₂ + y/2 H₂O Z/2 N₂ +(2X + y/2)Cu.

Traces of oxide of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing overheated copper spiral. The apparatus used in the Dumas method consists of a CO₂ generator, combustion tube, Schiff’s nitrometer.

CO₂ generator:
CO₂ needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube or by the action of dil. HCl on marble in a Kipps apparatus. The gas is passed through the combustion tube after being dried by bubbling through Cone. H₂SO₄ contained in a Drechel bottle.

Combustion Tube:
The combustion tube is heated in a furnace is charged with
a) A roll of oxidized copper gauze to prevent the back diffusion of the products of combustion and to heat the organic substance mixed with CuO by radiation
b) a weighed amount of the organic substance mixed with an excess of CuO,
c) a layer, of course, CuO packed in about 2/3 of the entire length of the tube and kept in position by a loose asbestos plug on either side; this oxidizes the organic vapors passing through it, and
d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion to nitrogen.

Schiff’s nitro meter:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with a considerable excess of CO₂ It is estimated by passing a nitrometer when CO₂ is absorbed by KOH and the nitrogen gets collected in the upper part of the graduated tube.

Procedure:
To start with the tap of the nitrometer is left open. CO₂ is passed through the combustion tube to expels the air in it. When the gas bubbles rising through, the potash solution fails to reach the top of it and is completely absorbed it shows that only CO₂ is coming and that all air has been expelled from the combustion tube. The nitrometer is then filled with KOH solution by lowering the reservoir and the tap is closed. The combustion tube is now heated in the furnace and the temperature rises gradually.

The nitrogen set free from the compound collects in the nitrometer. When the combustion is complete a strong current of CO₂ is sent through, the apparatus in order to sweep the last trace of nitrogen from it. The volume of the gas gets collected is noted after adjusting the reservoir so that the solution in it and the graduated tube is the same. The atmospheric pressure and the temperature are also recorded.

Calculation:
Weight of the substance taken = w g
Volume of nitrogen = V₁ L
Room Temperature = T₁ K
Atmospheric Pressure = P mm of Hg
Aqueous tension at room temperature = P’ mm of Hg
Pressure of diy nitrogen = (P – P’) = P’ mm of Hg.
Let P₀V₀ and T₀ be the pressure, volume, and temperature respectively of dry nitrogen at STP,
Then,

Calculation of percentage of nitrogen, 22.4 L of N₂ at STP weight 28 g of N2
∴ V₀ L of N₂ at S.T.P weigh = \(\frac{28}{22.4}\) × V₀
Wg Organic compound contain = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) of nitrogen

∴ percentage of nitrogen = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) × 100

Question 14.
0.1688 g when analyzed by the Dumas method yield 31.7 ml of moist nitrogen measured at 14°C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C = 12 mm)
Solution:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V₁) = 31.7 mL
= 31.7 × 10^-3 L
Temperature (T¹) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C =12 mm of Hg
∴ Pressure of dry nitrogen = (P – P¹)
= 758 – 12 = 746 mm of Hg

Question 15.
Explain the estimation of nitrogen by Kjeldahl’s method.
Answer:
This method is carried much more easily than the Dumas method. It is used largely in the analysis of foods and fertilizers. Kjeldahl’s method is based on the fact that when an organic compound containing nitrogen is heated with Conc. H₂SO₄, the nitrogen in it is quantitatively converted to ammonium sulphate. The resultant liquid is then treated with excess alkali and then liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia (and hence nitrogen) is determined by finding the amount of acid neutralized by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance (0.3 to 0.5 g) is placed in a special long-necked Kjeldahl flask made of pyrex glass. About 25 mL of Cone. H₂SO₄ together with a little K₂SO₄ and CuSO ₄ (catalyst) is added to it the flask is loosely stoppered by a glass bulb and heated gently in an inclined position. The heating is continued till the brown color of the liquid first produced, disappears leaving the contents clear as before. At this point all the nitrogen in the substance is converted to (NH₄)₂SO₄.

The Kjeldahl flask is then cooled and its contents are diluted with some distilled water and then carefully transferred into a 1 lit round bottom flask. An excess NaOH solution is poured down the side of the flask and it is fitted with a Kjeldahl trap and a water condenser. The lower end of the condenser dips in a measured volume of excess the H₂SO₄ solution. The liquid in the round bottom flask is then heated and the liberated ammonia is distilled into sulphuric acid. The Kjeldahl trap serves to retain any alkali splashed up on vigorous boiling.

When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = w g
Volume of H₂SO₄ required for the complete neutralization of evolved NH₃ = V ml.
Strength of H2S04 used to neutralize NH₃ = N
Let the Volume and the strength of NH₃ formed are V₁ and N₁ respectively.
We know that V₁N₁ = VN
The amount of nitrogen present in the = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times \mathrm{w}}\)
Percentage of Nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times \mathrm{w}}\right)\) × 100 %

Question 16.
0.6 g of an organic compound was kjeldalised and NH3 evolved was absorbed into 50 mL of semi-normal solution of H₂SO₄. The residual acid solution was diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac{\mathbf{N}}{20}\) NaOH solution for complete neutralization. Calculate the % of N in the compound.
Solution:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20ml of diluted solution of unreacted sulphuric acid was neutralized by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric aicd = \(\frac{35 \times 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with Organic compound = V₁ mL
Strength of the diluted H₂SO₄ = 0.5 N
Volume of the diluted H₂SO₄ = 150 mL
Strength of the diluted H₂SO₄ = 0.0875 N
V₁ = \(\frac{150 \times 0.0875}{0.5}\) = 26.25 ml
Volume of H₂SO₄ consumed by ammonia = 50 – 26.25 = 23.75 ml
23.75 ml of 0.5 N H₂SO₄ = 23.75 ml of 0.5 N NH₃
The amount of Nitrogen present in the 0.6 g of organic compound = \(\frac{14 g}{1000 m L \times 1 N}\) × 23.75 × 0.5 N = 0.166 g
Percentage of Nitrogen = \(\frac{0.166}{0.6}\) × 100 = 27.66%

Question 17.
Explain the various steps involved in the purification of organic compounds by crystallization process.
Answer:
(i) Selection of solvent:
Most of the organic substances being covalent do not dissolve in polar solvents like water, hence selection of solvent (suitable) becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is just sufficient to dissolve the solute (ie) organic compound. If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with other solvents like benzene, ether, acetone, and alcohol till the most suitably one is sorted out.

(ii) Preparation of solution:
The quantity to be purified is taken in a conical flash fitted with a reflux condenser. The solvent selected in first stage is also taken along with solute and the quantity of the solvent just enough to dissolve the whole solid on boiling. Small amount of are animal char-coal can be added before boiling to decolorize any colored substance. The heating may be done over a wire gauze or water bath depending upon the nature of liquid (ie) whether the solvent is low boiling or high boiling.

(iii) Filtration of hot solution:
The solution so obtained is filtered through a fluted filter paper placed in a hot water funnel.

(iv) Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper, the pure solid substance separate as crystal. When copious amount of crystal has been obtained, then the crystallization is complete. If the rate of crystallization is slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of the pure compounds to the solution.

(v) Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration. Filtration is done under reduced pressure using a Bucher funnel. When the whole of the mother liquor has been drained into the filtration flask, the crystals are washed with small quantities of the pure cold solvent and then dried.

Question 18.
How will you purify an organic compound by Thin layer chromatography?
Answer:
This method is another type of adsorption chromatography with this method it is possible to separate even minute quantities of mixtures. A sheet of a glass is coated with a thin layer of adsorbent (cellulose, silica gel or alumina). This sheet of glass is called chromoplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluent (solvent). The eluent is drawn up to the adsorbent layer by capillary action.

The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retardation factors (ie) Rf value
Rf = \(\frac{\text { Distance moved by the substance from baseline }(\mathrm{x})}{\text { Distance moved by the solvent from baseline }(\mathrm{y})}\)

The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under UV light or in another method using iodine crystals or by using an appropriate reagent.