I. Choose the best answer:<\/span><\/p>\nQuestion 1.
\nThe correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is
\na) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
\nb) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
\nc) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
\nd) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
\nAnswer:
\nb) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.<\/p>\n
Question 2.
\nC2<\/sub>H5<\/sub> Br + 2Na
C4<\/sub>H10<\/sub> + 2NaBr.
\nThe above reaction is an example of which of the following
\na) Reimer Tiemann reaction
\nb) Wurtz reaction
\nc) Aldol condensation
\nd) Hoffmann reaction
\nAnswer:
\nb) Wurtz reaction<\/p>\nQuestion 3.
\nAn alkyl bromide (A) reacts with sodium in ether to form 4, 5 – diethyloctane, the compound (A) is
\na) CH3<\/sub>(CH2<\/sub>)3<\/sub>Br<\/p>\nb) CH3<\/sub>(CH2<\/sub>)5<\/sub>Br<\/p>\nc) CH3<\/sub>(CH2<\/sub>)3<\/sub> CH(Br)CH3<\/sub><\/p>\nd) 
\nAnswer:
\nd)
<\/p>\n
Question 4.
\nThe C – H bond and C – C bond in ethane are formed by which of the following types of overlap
\na) sp3<\/sup> – s and sp3<\/sup> – sp3<\/sup>
\nb) sp2<\/sup> – s and sp2<\/sup> – sp2<\/sup>
\nc) sp – sp and sp – sp
\nd) p – s and p – p
\nAnswer:
\na) sp3<\/sup> – s and sp3<\/sup> – sp3<\/sup><\/p>\nQuestion 5.
\nIn the following reaction,
\n
<\/p>\n

\nAnswer:
\nc)
<\/p>\n
<\/p>\n
Question 6.
\nWhich of the following is optically active
\na) 2 – methyl pentane
\nb) citric acid
\nc) Glycerol
\nd) none of these
\nAnswer:
\na) 2 – methyl pentane<\/p>\n
Question 7.
\nThe compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are
\na) CH4<\/sub> and H2<\/sub>
\nb) CH4<\/sub> and CO2<\/sub>
\nc) C2<\/sub>H6<\/sub> and CO2<\/sub>
\nd) C2<\/sub>H4<\/sub> and Cl2<\/sub>
\nAnswer:
\nc) C2<\/sub>H6<\/sub> and CO2<\/sub><\/p>\nQuestion 8.
\nThe general formula for cyclo alkanes
\na) Cn<\/sub>Hn<\/sub>
\nb) Cn<\/sub>H2n<\/sub>
\nC) Cn<\/sub>H2n-2<\/sub>
\nd) Cn<\/sub>H2n + 2<\/sub>
\nAnswer:
\nb) Cn<\/sub>H2n<\/sub><\/p>\nQuestion 9.
\nThe compound that will react most readily with gaseous bromine has the formula
\na) C3<\/sub>H6<\/sub>
\nb) C2<\/sub>H2<\/sub>
\nc) C4<\/sub>H10<\/sub>
\nd) C2<\/sub>H4<\/sub>
\nAnswer:
\na) C3<\/sub>H6<\/sub><\/p>\nQuestion 10.
\nWhich of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction
\na) 
\nb) CH3<\/sub> – CH2<\/sub> – CH2<\/sub> – OH
\nc) H2<\/sub>C = C = O
\nd) CH3<\/sub> – CH2<\/sub> – CH2<\/sub>Br
\nAnswer:
\nc) H2<\/sub>C = C = O<\/p>\n
<\/p>\n
Question 11.
\nWhich among the following alkenes on reductive ozonolysis produces only propanone?
\na) 2 – Methyl propene
\nb) 2 – Methyl but – 2 – ene
\nc) 2, 3 – Dimethyl but – 1- ene
\nd) 2, 3 – Dimethyl but – 2 – ene
\nAnswer:
\nd) 2, 3 – Dimethyl but – 2 – ene<\/p>\n
Question 12.
\nThe major product formed when 2 – bromo – 2 – methyl butane is refluxed with ethanolic KOH is
\na) 2 – methylbut – 2- ene
\nb) 2 – methyl butan – 1 – ol
\nc) 2 – methyl but – 1 – ene
\nd) 2 – methyl butan – 2- ol
\nAnswer:
\na) 2 – methylbut – 2- ene<\/p>\n
Question 13.
\nMajor product of the below mentioned reaction is, (CH3<\/sub>)2<\/sub> C = CH2<\/sub> 
\na) 2 – chloro – 1 – iode – 2 – methyl propane
\nb) 1 – chloro – 2 – iodo – 2 – methyl propane
\nc) 1, 2 – dichloro – 2 – methyl propane
\nd) 1, 2 – diiodo – 2 – methyl propane
\nAnswer:
\na) 2 – chloro – 1 – iode – 2 – methyl propane<\/p>\nQuestion 14.
\nThe IUPAC name of the following compound is
\n
\na) trans – 2- chloro – 3- iodo – 2- pentane
\nb) cis – 3- iodo – 4 chloro – 3 – pentane
\nc) trans – 3 – iodo – 4 – chloro – 3 – pentene
\nd) cis – 2 – chloro – 3 iodo – 2 – pentene
\nAnswer:
\na) trans – 2- chloro – 3- iodo – 2- pentane<\/p>\n
Question 15.
\nCis – 2- butene and trans – 2 – butene are
\na) conformational isomers
\nb) structural isomers
\nc) configurational isomers
\nd) optical isomers
\nAnswer:
\nc) configurational isomers<\/p>\n
<\/p>\n
Question 16.
\nIdentify the compound (A) in the following reaction<\/p>\n
<\/p>\n

\nAnswer:
\na)
<\/p>\n
Question 17.
\n
CH \u2261 CH , where A is,
\na) Zn
\nb) conc. H2<\/sub>SO4<\/sub>
\nc) alc. KOH
\nd) dil. H2<\/sub>SO4<\/sub>
\nAnswer:
\nc) alc. KOH<\/p>\nQuestion 18.
\nConsider the nitration of benzene using mixed con H2<\/sub>SO4<\/sub> and HNO3<\/sub> if a large quantity of KHSO4<\/sub> is added to the mixture, the rate of nitration will be
\na) unchanged
\nb) doubled
\nc) faster
\nd) slower
\nAnswer:
\nd) slower<\/p>\nQuestion 19.
\nIn which of the following molecules, all atoms are co-planar
\na)
<\/p>\n
b)
<\/p>\n
c) 
\nd) both (a) and (b)
\nAnswer:
\nd) both (a) and (b)<\/p>\n
Question 20.
\nPropyne on passing through red hot iron tube gives
\na)
<\/p>\n
b)
<\/p>\n
c) 
\nd) none of these
\nAnswer:
\na)
<\/p>\n
<\/p>\n
Question 21.
\n
is<\/p>\n
a)
<\/p>\n
b)
<\/p>\n
c) both (a) and (b)<\/p>\n
d) 
\nAnswer:
\nd)
<\/p>\n
Question 22.
\nWhich one of the following is non aromatic?
\na)
<\/p>\n
b)
<\/p>\n
c)
<\/p>\n
d) 
\nAnswer:
\nd)
<\/p>\n
Question 23.
\nWhich of the following compounds will not undergo Friedal – crafts reaction easily?
\na) Nitro benzene
\nb) Toluene
\nc) Cumene
\nd) Xylene
\nAnswer:
\na) Nitro benzene<\/p>\n
Question 24.
\nSome meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
\na) – COOH
\nb) – NO2<\/sub>
\nc) -C \u2261 N
\nd) -SO3<\/sub>H
\nAnswer:
\nb) – NO2<\/sub><\/p>\nQuestion 25.
\nWhich of the following can be used as the halide component for friedal – crafts reaction?
\na) Chloro benzene
\nb) Bromo benzene
\nc) Chloro ethene
\nd) Isopropyl chloride
\nAnswer:
\nd) Isopropyl chloride<\/p>\n
<\/p>\n
Question 26.
\nAn alkane isobtainedbydecarboxylation of sodium propionate. Same alkane can be prepared by
\na) Catalytic hydrogenation of propene
\nb) action of sodium metal on iodomethane
\nc) reduction of 1 – chloro propane
\nd) reduction of bromomethane
\nAnswer:
\nb) action of sodium metal on iodomethane<\/p>\n
Question 27.
\nWhich of the following is aliphatic saturated hydrocarbon
\na) C8<\/sub>H18<\/sub>
\nb) C9<\/sub>H18<\/sub>
\nc) C8<\/sub>H14<\/sub>
\nd) All of these
\nAnswer:
\na) C8<\/sub>H18<\/sub><\/p>\nQuestion 28.
\nIdentify the compound \u2018Z\u2019 in the following reaction
\n
\na) Formaldehyde
\nb) Acetaldehyde
\nc) Formic acid
\nd) none of these
\nAnswer:
\na) Formaldehyde<\/p>\n
Question 29.
\nPeroxide effect (Kharasch effect) can be studied in case of
\na) Oct – 4 – ene
\nb) hex – 3 – ene
\nc) pent – 1 – ene
\nd) but – 2 – ene
\nAnswer:
\nc) pent – 1 – ene<\/p>\n
Question 30.
\n2 – butyne on chlorination gives
\na) 1 – chloro butane
\nb) 1, 2 – dichloro butane
\nc) 1, 1, 2, 2 – tetrachlorobutane
\nd) 2, 2, 3, 3 – tetra chloro butane
\nAnswer:
\nd) 2, 2, 3, 3 – tetra chloro butane<\/p>\n
<\/p>\n
II. Write brief answer to the following questions:<\/span><\/p>\nQuestion 31.
\nGive IUPAC names for the following compounds
\n1) CH3<\/sub> – CH = CH – CH = CH – C \u2261 C – CH3<\/sub><\/p>\n2)
<\/p>\n
3) (CH3<\/sub>)3<\/sub> C – C \u2261 C – CH(CH3<\/sub>)2<\/sub><\/p>\n4) ethyl isopropyl acetylene<\/p>\n
5) CH \u2261 C – C \u2261 C – C \u2261 CH
\nAnswer:
\n1)
<\/p>\n
2)
<\/p>\n
3)
<\/p>\n
4) ethyl isopropyl acetylene
\n
<\/p>\n
5)
<\/p>\n
Question 32.
\nIdentify the compound A, B, C and D in the following series of reactions
\n
\nAnswer:
\n
<\/p>\n
Question 33.
\nWrite short notes on ortho, para directors in aromatic electrophilic substitution reactions.
\nAnswer:
\nAll the activating groups are \u2018ortho-para\u2019 directors.
\nExample: – OH, – NH2<\/sub>, -NHR, -NHCOCH3<\/sub>, -OCH3<\/sub> -CH3<\/sub> – C2<\/sub>H5<\/sub> etc.
\nLet us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.
\n
<\/p>\nIn 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 \u2018ortho\u2019 and \u2018para positions and hence -OH group is an ortho-para director and activator.<\/p>\n
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.<\/p>\n
<\/p>\n
Question 34.
\nHow is propyne prepared from an alkene dihalide?
\nAnswer:
\n
<\/p>\n
Question 35.
\nAn alkylhalide with molecular formula C6<\/sub>H13<\/sub>Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C6<\/sub>H12<\/sub>. On reductive ozonolysis, X and Y gave four compounds CH3<\/sub>COCH3<\/sub>, CH3<\/sub>CHO, CH3<\/sub>CH,CHO and (CH3<\/sub>)2<\/sub> CHCHO. Find the alkyihalide.
\nAnswer:
\n
<\/p>\nQuestion 36.
\nDescribe the mechanism of Nitration of benzene.
\nAnswer:
\nIt is prepared by the action of a mixture of con. HNO3<\/sub> and con. H2<\/sub>SO4<\/sub> (nitrating mixture) on benzene maintaining the temperature below 333 K.
\nNitration:
\n
<\/p>\nSulphuric acid generates the electrophile – NO2<\/sub>+<\/sup>, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction.
\n
<\/p>\nThe generation of nitronium ion
\nH2<\/sub>SO4<\/sub> + HONO2<\/sub> \u2192
+ HSO4<\/sub>–<\/sup>
\n
<\/p>\nTo the nitronium ion (being an electron deficient species) the \u03c0 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.
\n
<\/p>\n
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<\/sub>–<\/sup>, so that stable aromatic system is formed.
\n
<\/p>\nQuestion 37.
\nHow does Huckel rule help to decide the aromatic character of a compound.
\nAnswer:
\nIn 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
\nThere were two objections:
\nBenzene forms only one ortho disubstituted products whereas the Kekule\u2019s structure predicts two o-di substituted products as shown below.
\n
<\/p>\n
Kekule\u2019s 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.
\n
<\/p>\n
Resonance description of benzene:
\nThe 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\u2019s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
\nThe structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.
\n
<\/p>\n
Spectroscopic measurements:
\nSpectroscopic measurements show that benzene is planar and all of its carbon- carbon bonds are of equal length 1.40 \u00c5. This value lies between carbon-carbon single bond length 1.54 \u00c5 and carbon- carbon double bond length 1.34 \u00c5.<\/p>\n
Molecular orbital structure:
\nThe structure of benzene is best de\u00acscribed in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp2<\/sup> hybridized. Six sp2<\/sup> 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<\/sup> hybrid orbitals of carbon forms six C – C sigma bonds.
\n
\nFormation of Sigma bond in benzene<\/p>\nAll the \u03c3 bonds in benzene lie in one plane with bond angle 120\u00b0. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 \u03c0- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization, strong \u03c0-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.<\/p>\n
<\/p>\n
All carbon atoms have the delocalized \u03c0 MO is formed by p-orbitais the overlap of six p-orbitals.<\/p>\n
Representation of benzene:
\nHence, there are three ways in which benzene can be represented.
\n
<\/p>\n
<\/p>\n
Question 38.
\nSuggest the route for the preparation of the following from benzene.
\n1) 3 – chloro nitrobenzene
\n2) 4 – chlorotoluene
\n3) Bromo benzene
\n4) m – dinitro benzene
\nAnswer:
\n1) 3 – chloro nitrobenzene
\n
<\/p>\n
2) 4 – chlorotoluene
\n
<\/p>\n
3) Bromo benzene
\n
<\/p>\n
4) m – dinitro benzene
\n
<\/p>\n
Question 39.
\nSuggest a simple chemical test to distinguish propane and propene.
\nAnswer:
\nPropene decolourises Br2<\/sub>\/H2<\/sub>O it forms dibromo compound but propane does not react with Br2<\/sub>\/ H2<\/sub>O.<\/p>\nQuestion 40.
\nWhat happens when Isobutylene is treated with acidified potassium permanganate?
\nAnswer:
\n
<\/p>\n
Question 41.
\nHow will you convert ethyl chloride into
\n(i) ethane
\n(ii) n – butane
\nAnswer:
\n(i) ethane
\n
<\/p>\n
(ii) n – butane
\n
<\/p>\n
<\/p>\n
Question 42.
\nDescribe the conformers of n – butane.
\nAnswer:
\nConformations of n-Butane:
\nn-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group<\/p>\n
Eclipsed conformation:
\nIn 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.<\/p>\n
Anti or staggered form:
\nIn 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.<\/p>\n
The following potentially energy diagram shows the relative stabilities of various conformers of n-butane.
\n
<\/p>\n
Question 43.
\nWrite the chemical equations for combustion of propane.
\nAnswer:
\nC3<\/sub>H8<\/sub>(g) + 5O2<\/sub>(g) ) \u2192 3CO2<\/sub>(g) + 4H2<\/sub>O(l)
\npropane<\/p>\nQuestion 44.
\nExplain Markovnikoff\u2019s rule with suitable example.
\nAnswer:
\nMarkovnikoff \u2018s rule:
\n\u201cWhen 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\u201d. 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.<\/p>\n
Addition of water: (Hydration of alkenes)
\nNormally, 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.
\n
<\/p>\n
Question 45.
\nWhat happens when ethylene is passed ‘ through cold dilute alkaline potassium permanganate.
\nAnswer:
\nCH2<\/sub> = CH + H2<\/sub>O + (O)
CH2<\/sub>OH – CH2<\/sub>OH
\nEthylene\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Ethylene glycol<\/p>\n
<\/p>\n
Question 46.
\nWrite the structures of following alkanes.
\n1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
\n2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
\n3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
\nAnswer:
\n1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
\n
<\/p>\n
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
\n
<\/p>\n
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
\n
<\/p>\n
Question 47.
\nHow will you prepare propane from a sodium salt of fatty acid?
\nAnswer:
\n
<\/p>\n
Question 48.
\n
\nIdentify A and B
\nAnswer:
\nA)
<\/p>\n
b)
<\/p>\n
Question 49.
\nComplete the following :<\/p>\n
i) 2 – butyne\u00a0
<\/p>\n
ii) CH2<\/sub> = CH2<\/sub>
<\/p>\niii)
<\/p>\n
iv) CaC2<\/sub> 
\nAnswer:
\ni) 2 – butyne 
\n
<\/p>\nii) CH2<\/sub> = CH2<\/sub> 
\n
<\/p>\niii) 
\n
<\/p>\n
iv) CaC2<\/sub> 
\nCaC2<\/sub> + 2H2<\/sub>O \u2192 Ca(OH)2<\/sub> + C2<\/sub>H2<\/sub>\u2191
\nCalcium Carbide\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Acetylene<\/p>\nQuestion 50.
\nHow will distinguish 1 – butyne and 2 – butyne?
\nAnswer:
\n
\n1 – butyne reacts with ammoniacal AgNO3<\/sub> solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO3<\/sub> solution.<\/p>\n
<\/p>\n
11th Chemistry Guide Hydrocarbons Additional Questions and Answers<\/h3>\n
I. Choose the best answer:<\/span><\/p>\nQuestion 1.
\nStatement-I: Methane. ethane, prop\u00e8ne, and butane are alkane group compounds.
\nStatement-II: They are obeying C2<\/sub>H2n+2<\/sub> + formula and each member differs from it proceeding member by a CH2<\/sub> group.
\n(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.
\n(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I
\n(e) Stat\u00ebment-I is correct but statement-II is wrong.
\n(d) Statement-I is wrong but statement-II is correct.
\nAnswer:
\n(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.<\/p>\nQuestion 2.
\nAlkanes are also known as
\na) olefins
\nb) unsaturated aliphatic hydrocarbons
\nc) saturated aromatic hydrocarbon
\nd) paraffin
\nAnswer:
\nd) paraffin<\/p>\n
Question 3.
\nWhich one of the following shows three possible isomeric structures\u2019?
\n(a) C4<\/sub>H10<\/sub>
\n(b) C5<\/sub>H12<\/sub>
\n(c) C6<\/sub>H12<\/sub>
\n(d) C3<\/sub>H4<\/sub>
\nAnswer:
\n(b) C5<\/sub>H12<\/sub><\/p>\nQuestion 4.
\nThe gas supplied in cylinders for cooking is
\na) marsh gas
\nb) LPG
\nc) mixture CH4<\/sub> and C2<\/sub>H6<\/sub>
\nd) mixture of ethane and propane
\nAnswer:
\nd) mixture of ethane and propane<\/p>\nQuestion 5.
\nWhich of the following compound cannot be prepared by Kolbe\u2019s electrolytic method\u2019?
\n(a) CH3<\/sub>-CH3<\/sub>
\n(b) CH4<\/sub>
\n(c) CH2<\/sub> = CH2<\/sub>
\n(d) CH = CH
\nAnswer:
\n(b) CH4<\/sub><\/p>\n
<\/p>\n
Question 6.
\nSoda-lime is
\na) NaOH
\nb) NaOH + CaO
\nc) CaO
\nd) Na2<\/sub>CO3<\/sub>
\nAnswer:
\nb) NaOH + CaO<\/p>\nQuestion 7.
\nStatement – I : Alkenes shows both structural and geometrical isomerism.
\nStatement – II : Because of the presence of double bond.
\n(a) Statement – I and II are correct and statement-II is correct explanation of statement – I.
\n(b) Statement – I and II are correct but statement-II is not correct explanation of statement – I
\n(c) Statement – I is correct but statement – II is wrong.
\n(d) Statement – II is wrong but statement – II is correct.
\nAnswer:
\n(a) Statement – I and II are correct and statement – II is correct explanation of statement – I.<\/p>\n
Question 8.
\nHydrocarbon which is liquid at room temperature is
\na) Pentane
\nb) Butane
\nc) Propane
\nd) Ethane
\nAnswer:
\na) Pentane<\/p>\n
Question 9.
\nPyrolysis of Methane and respectively are
\na) Exothermic & Endothermic
\nb) Endothermic & Exothermic
\nc) Both are endothermic
\nd) Both are exothermic
\nAnswer:
\nc) Both are endothermic<\/p>\n
Question 10.
\nPeroxide effect is not observed in ………..
\n(a) HCl
\n(b) HI
\n(c) HBr
\n(d) Both (a) and (b)
\nAnswer:
\n(d) Both (a) and (b)<\/p>\n
<\/p>\n
Question 11.
\nFinal products of complete oxidation of hydrocarbon is
\na) Acid
\nb) Dihydric alcohol
\nc) Aldehyde
\nd) H2<\/sub>O + CO2<\/sub>
\nAnswer:
\nd) H2<\/sub>O + CO2<\/sub><\/p>\nQuestion 12.
\nIsomerisation in alkane can be brought by using
\na) Al2<\/sub>O3<\/sub>
\nb) Fe2<\/sub>O3<\/sub>
\nc) Anh.AlCl3<\/sub>\/ HCl at 200\u00b0C
\nd) Cone. H2<\/sub>SO4<\/sub>
\nAnswer:
\nc) Anh.AlCl3<\/sub>\/ HCl at 200\u00b0C<\/p>\nQuestion 13.
\nIn aromatization of n – hexane, the catalyst used is
\na) Cr2<\/sub>O3<\/sub>
\nb) V2<\/sub>O5<\/sub>
\nc) Mo2<\/sub>O3<\/sub>
\nd) All
\nAnswer:
\nd) All<\/p>\nQuestion 14.
\nThe most oxidized form of ethane is
\na) CO2<\/sub>
\nb) HCHO
\nc) HCOOH
\nd) CH3<\/sub>COOH
\nAnswer:
\na) CO2<\/sub><\/p>\nQuestion 15.
\nThe following substance is used as anti-knocking compound
\na) TEL
\nb) Lead tetrachloride
\nc) Lead acetate
\nd) C2<\/sub>H2<\/sub>PbCl
\nAnswer:
\na) TEL<\/p>\n
<\/p>\n
Question 16.
\nThe molecular formula of benzene is ………….
\n(a) C6<\/sub>H6<\/sub>
\n(b) C6<\/sub>H5<\/sub>
\n(c) C7<\/sub>H8<\/sub>
\n(d) CH4<\/sub>