2<\/sub>
\nSolution:
\n
<\/p>\n
<\/p>\n
Question 10.
\nThe correct IUPAC name of the compound,
\n
\n(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol
\n(b) 2, 3 – dimethyl – 4 – chloropentan – 1 – ol
\n(c) 2, 3, 4 – trimethyl – 4 – chiorobutan – 1 – ol
\n(d) 4 – chioro – 2, 3, 4 – trimethyl pentan – 1 – ol
\nAnswer:
\n(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol<\/p>\n
Question 11.
\nAssertion: Phenol is more acidic than ethanol
\nReason: Phenoxide ion is resonance stabilized
\n(a) if both assertion and reason are true and reason is the correct explanation of assertion.
\n(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
\n(c) assertion is true but reason is false
\n(d) both assertion and reason are false.
\nAnswer:
\nif both assertion and reason are true and reason is the correct explanation of assertion.<\/p>\n
Question 12.
\nIn the reaction Ethanol
\n
\nis ………………
\n(a) ethane
\n(b) ethoxyethane
\n(c) ethylbisuiphite
\n(d) ethanol
\nAnswer:
\n(d) ethanol
\nSolution:
\n
<\/p>\n
<\/p>\n
Question 13.
\nThe reaction
\n
\ncan be classified as
\n(a) dehydration
\n(b) Williams on alcohol synthesis
\n(c) Williamson ether synthesis
\n(d) dehydrogenation of alcohol
\nAnswer:
\n(c) Williamson ether synthesis
\nSolution:
\nCyclic alcohol \u2192 sodium cyclic alkoxide \u2192 Williamson ether synthesis<\/p>\n
Question 14.
\nIsoprophylbcnzene on air oxidation in the presence of dilute acid gives …………
\n(a) C6<\/sub>H5<\/sub>COOH
\n(b) C6<\/sub>H5<\/sub>COCH3<\/sub>
\n(c) C6<\/sub>H5<\/sub>COC6<\/sub>H5<\/sub>
\n(d) C6<\/sub>H5<\/sub> – OH
\nAnswer:
\n(a) C6<\/sub>H5<\/sub> – OH (phenol)<\/p>\nQuestion 15.
\nAssertion: Phenol is more reactive than benzene towards electrophilic substitution reaction
\nReason: In the case of phenol. the intermediate arenium ion is more stabilized by resonance.
\n(a) if both assertion and reason are true and reason is the correct explanation of assertion.
\n(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
\n(c) assertion is true but reason is false
\n(d) both assertion and reason are false,.
\nAnswer:
\n(a) if both assertion and reason are true and reason is the correct explanation of assertion.<\/p>\n
<\/p>\n
Question 16.
\nHO CH2<\/sub> CH2<\/sub> – OH on heating with periodic acid gives ………..
\n(a) methanoic acid
\n(b) Glyoxal
\n(c) methanol
\n(d) CO2<\/sub>
\nAnswer:
\n(c) methanol<\/p>\nQuestion 17.
\nWhich of the following compound can be used as artireeze in automobile radiators?
\n(a) methanol
\n(b) ethanol
\n(c) Neopentyl alcohol
\n(d) ethan -1, 2-diol
\nAnswer:
\n(d) ethan -1, 2-diol<\/p>\n
Question 18.
\nThe reaction
\n
\nis an example of …………..
\n(a) Wurtz reaction
\n(b) cyclic reaction
\n(c) Williamson reaction
\n(d) Kolbe reactions
\nAnswer:
\n(c) Kolbe reactions<\/p>\n
<\/p>\n
Question 19.
\nOne mole of an organic compound (A) with the formula C3<\/sub>H8<\/sub>O reacts completely with two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound (A) is ……………
\n(a) propan – 2 – ol
\n(b) propan- 1- ol
\n(c) ethoxy ethane
\n(d) methoxy ethane
\nAnswer:
\n(d) methoxy ethane
\nSolution:
\n
<\/p>\n
<\/p>\n
Question 20.
\nAmong the following ethers which one will produce methyl alcohol on treatment with hot HI?
\n
\nAnswer:
\na
\nSolution:
\n
<\/p>\n
<\/p>\n
Question 21.
\nWilliamson synthesis of preparing dimethyl ether is a \/ an
\n(a) SN1<\/sup> reactions
\n(b) SN2<\/sup> reaction
\n(c) electrophilic addition
\n(d) electrophilic substitution
\nAnswer:
\n(b) SN2<\/sup> reaction<\/p>\nQuestion 22.
\nOn reacting with neutral ferric chloride, phenol gives
\n(a) red colour
\n(b) violet colour
\n(c) dark green colour
\n(d) no colouration
\nAnswer:
\n(b) violet colour<\/p>\n
II. Short Answer<\/span><\/p>\nQuestion 1.
\nIdentIfy the product (s) is\/are formed when 1 – methoxy propane is heated with excess HI. Name the mechanism involved in the reaction.
\nAnswer:
\n
\nEthers having primary alkyl group undergo \\(\\mathrm{S}_{\\mathrm{N}}^{2}\\) reaction<\/p>\n
Question 2.
\nDraw the major product formed when 1 – ethoxyprop – 1 – ene is heated with one equivalent of HI
\nAnswer:
\n
<\/p>\n
Question 3.
\nSuggest a suitable reagent to prepare secondary alcohol with an identical groups using a Grignard reagent.
\nAnswer:
\n
<\/p>\n
<\/p>\n
Question 4.
\nWhat is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis
\nAnswer:
\n
<\/p>\n
Question 5.
\nPredict the major product, when 2-methyl but – 2 – ene is converted into alcohol in each of the following methods.<\/p>\n
\n- Acid-catalyzed hydration<\/li>\n
- Hydroboration<\/li>\n
- Hydroxylation using bayers reagent<\/li>\n<\/ol>\n
Answer:
\n
<\/p>\n
<\/p>\n
Question 6.
\nArrange the following in the increasing order of their boiling point and give a reason for your ordering<\/p>\n
\n- Butan – 2 – ol, Butan – 1 – SI, 2 – methylpropan – 2 – ol<\/li>\n
- Propan – 1 – ol, propan – 1, 2, 3 – triol, propan – 1, 3 – diol, propan – 2 – ol<\/li>\n<\/ol>\n
Answer:
\n1. Boiling points increases regularly as the molecular mass increases due to a corresponding increase in their Van der Waal’s force of attraction. Among isomeric alcohols, 2\u00b0 – alcohols have lower boiling points than 1\u00b0 – alcohols due to a corresponding decrease in the extent<\/p>\n
of H-bonding because of steric hindrance. Thus the boiling point of Butan – 2 – ol is lower than that of Butan – 1 – ol. Overall increasing order of boiling points is, 2 – methyl propane – 2 – ol < Butan – 2 – ol < Butan – 1 – ol<\/p>\n
2. 2\u00b0-alcohols have lower boiling points than 1\u00b0 – alcohols due to a corresponding decrease in the extent of H – bonding because of steric hindrance. Therefore Propan – 1 – ol has higher boiling point than Propan – 2 – ol. The hydrogen group increases, boiling point also increases. Overall increasing order of boiling points is, propan – 2 – ol < Propan – 1 – ol < propan – 1, 3 – diol < propan -1, 2, 3 – triol<\/p>\n
Question 7.
\nCan we use nucleophiles such as NH3<\/sub>, CH3<\/sub>O for the Nucleophilic substitution of alcohols
\nAnswer:
\n1. Increasing order of nucleophilicity,
\nNH3<\/sub> < – OH\u2295<\/sup> < CH3<\/sub>O\u2296-<\/sup><\/p>\n2. Higher electron density will increase the nucleophilicity.<\/p>\n
3. Negatively charged species are almost always more nucleophiles than neutral species.<\/p>\n
4. RO\u2296<\/sup> has an alkyl group attached, allowing a greater amount of polarizability. This means oxygen\u2019s lone pairs will be more readily available to reach in RO\u2296<\/sup> than in OH\u2296<\/sup>. Hence CH3<\/sub>O – is the better nucleophile for the nucleophilic substitution of alcohols. NH3<\/sub> cannot act as nucleophiles for the nucleophilic substitution of alcohols.<\/p>\n
<\/p>\n
Question 8.
\nIs it possible to oxidise t – butyl alcohol using acidified dichromate to form a carbonyic compound.
\nAnswer:
\n3\u00b0 – alcohols do not undergo oxidation reaction under normal conditions, but at elevated temperature, under strong oxidising agent cleavage of C – C bond takes place to give a mixture of carboxylic acid.<\/p>\n
Yes, it is possible. t – butyl alcohol is readily oxidsing in acidic solution (K2<\/sub>Cr2<\/sub>O7<\/sub> \/ H2<\/sub>SO4<\/sub>) to a mixture of a ketone and an acid each containing lesser number of carbon atoms than the original alcohol. The oxidation presumably occur via alkenes formed through dehydration of alcohols under acidic conditions.
\n
<\/p>\nQuestion 9.
\nWhat happens when 1 – phenyl ethanol is treated with acidified KMnO4<\/sub>.
\nAnswer:
\n1 – phenyl ethanol reacts with acidified KMnO4<\/sub> to give Acetophenone.
\n
<\/p>\nQuestion 10.
\nWrite the mechanism of acid catalysed dehydration of ethanol to give ethene.
\nAnswer:
\n
<\/p>\n
Question 11.
\nHow is phenol prepared form<\/p>\n
\n- chloro benzene<\/li>\n
- isopropyl benzene<\/li>\n<\/ol>\n
Answer:
\n
<\/p>\n
<\/p>\n
Question 12.
\nExplain Kolbe\u2019s reaction
\nAnswer:
\nKolbe\u2019s (or) Kolbe\u2019s Schmitt reaction:
\nIn this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO2<\/sub>. Treatment of sodium phenoxide with CO2<\/sub> at 400K, 4 -7 bar pressure followed by acid hydrolysis gives salicylic acid.
\n
<\/p>\nQuestion 13.
\nWrites the chemical equation for Williamson synthesis of 2 – ethoxy – 2 – methyl pentane starting from ethanol and 2 – methyl pentan – 2 – ol
\nAnswer:
\nA tertiary alkoxide and primary alkyl halide easily undergo williamson ether synthesis
\n
<\/p>\n
Question 14.
\nWrite the structure of the aldehyde, carboxylic acid and ester that yield 4 – methylpent – 2 – en – 1 – ol.
\nAnswer:
\n
<\/p>\n
<\/p>\n
Question 15.
\nWhat is metamerism? Give the structure and IUPAC name of metamers of 2 – methoxy propane
\nAnswer:
\nMetamerism:
\nIt is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.<\/p>\n
<\/p>\n
Question 16.
\nHow are the following conversions effected<\/p>\n
\n- benzyl chlorjde to benzyl alcohol<\/li>\n
- benzyl alcohol to benzoic acid<\/li>\n<\/ol>\n
Answer:
\n
<\/p>\n
Question 17.
\nComplete the following reactions
\n
\nAnswer:
\n
<\/p>\n
<\/p>\n
Question 18.
\nO.44g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm3<\/sup> of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.
\n
\nCn<\/sub>H2n+1<\/sub>+OH
\n\u21d2 n x 12+ (2n + 1) x 1 + 1 x 16 + 1 x 1 = 88
\n12n + 2n + 1 + 16 + 1 = 88
\n14n + 18 = 88
\n14n = 88 – 18
\n14n = 70
\nn = 70\/14 = 5
\nAnswer:
\n
<\/p>\nQuestion 19.
\nComplete the following reactions
\n
\nAnswer:
\n

\n(According to Saytzeff s rule, during intramolecular dehydration, if there is a possibility to form C = C bond at different locations, the preferred location is the one that gives the more substituted alkene je, the stable alkene).<\/p>\n
Question 20.
\nPhenol is distilled with Zn dust gives (A) followed by Friedel – crafts alkylation with propyl chloride to give a compound B, B on oxidation gives (C). Identify A,B and C.
\nAnswer:
\n
\nNote:
\nCarbon directly attached to the aromatic ri\u00f1g is called benzylic carbon.
\nIf there is hydrogen attached to benzylic carbon it will undergo oxidation.<\/p>\n
Question 21.
\n
\nIdentify A, B, C, D and write the complete equation.
\nAnswer:
\n
<\/p>\n
<\/p>\n
Question 22.
\nWhat will be the product for the following reaction
\n
\nAnswer:
\n
<\/p>\n
Question 23.
\nHow will you convert acetylene into n – butyl alcohol.
\nAnswer:
\n
<\/p>\n
Question 24.
\nPredict the product A, B, X and Y in the following sequence of reaction
\n
\nAnswer:
\n
<\/p>\n
<\/p>\n
Question 25.
\n3,3 – dimethyl butane – 2 – ol on treatment with conc. H2<\/sub>SO4<\/sub> to give tetramethyl ethylene as a major product. Suggest suitable mechanisms.
\nAnswer:
\n
<\/p>\nIII. Evaluate yourself<\/span><\/p>\nQuestion 1.
\nClassify the following alcohols as 10<\/sup>, 20<\/sup>, and 30<\/sup> and give their IUPAC Names.
\n
\nAnswer:<\/p>\n\n\n\na 2\u00b0 alcohol<\/td>\n | 5- bromo -5- methyl -3 hexanol<\/td>\n<\/tr>\n |
\nb 3\u00b0 alcohol<\/td>\n | 3 – ethyl -3- pentanol<\/td>\n<\/tr>\n |
\nc 2\u00b0 alcohol<\/td>\n | 3- chiorobut -3- en – 1 – o!<\/td>\n<\/tr>\n |
\nd 2\u00b0alcohol<\/td>\n | 6-bromohept-3-en-2-ol<\/td>\n<\/tr>\n |
\ne 3\u00b0 alcohol<\/td>\n | 3- phenyl -3 – pentanol<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n <\/p>\n
Question 2. \nWrite all the possible isomers of alcohol having the molecular formula C15<\/sub>H12<\/sub>O and their IUPAC names. \nAnswer: \nEight isomers are possible for C15<\/sub>H12<\/sub>O. They are, \n <\/p>\n <\/p>\n
Question 3. \nSuggest a suitable carbonyl compound for the preparation of pent – 2 – en – 1 ol using LiAlH4<\/sub>. \nAnswer: \n <\/p>\nQuestion 4. \n2 – methylpropan – 2 – ene \n \nAnswer: \n <\/p>\n <\/p>\n
Question 5. \nHow will you prepare the following using a Grignard reagent?<\/p>\n \n- t – butyl alcohol<\/li>\n
- allyl alcohol<\/li>\n<\/ol>\n
Answer: \n <\/p>\n Question 6. \nIdentify the products in the following reactions. Write their IUPAC names and mention the mechanism involved in the reactions. \n \nAnswer: \n <\/p>\n <\/p>\n
Question 7. \nWhat is the major product obtained when 2, 3 – dimethyl pentan – 3 – ol is heated in the presence of H2<\/sub>SO4<\/sub> \nAnswer: \n <\/p>\nQuestion 8. \nWhich of the following set of reactants will give 1 – methoxy – 4 – nitrobenzene. \n \nAnswer: \n <\/p>\n Question 9. \nWhat happens when m – cresol is treated with an acidic solution of sodium dichromate? \nAnswer: \n <\/p>\n Question 10. \nWhen phenol is treated with propan – 2 – ol in the presence of HF, Friedel – Craft reaction takes place. Identify the products. \nAnswer: \n <\/p>\n <\/p>\n
Question 11. \nGiven the IUPAC name for the following ethers and classify them as simple or mixed. \nAnswer: \n <\/p>\n Question 12. \n1. Which of the following reaction will give 1 – methoxy – 4 – nitrobenzene.<\/p>\n \n- 4 – nitro – 1 – bromobenzene + sodium methoxide.<\/li>\n
- 4 – nitrosodium phenoxide + bromomethane<\/li>\n<\/ol>\n
Answer: \n4-nitrosodium phenoxide + bromo methane \u2192 1 – methoxy – 4 – nitrobenzene \n \nReason: 4 – nitro -1- brornobnzene + sodium methoxide can not be used, because bromine is strongly bonded to carbon atom of hcnzcne ring and it can not be replaced easily.<\/p>\n Question 13. \nArrange the following compounds in the increasing order of their acid strength. propan – 1 – ol, 2, 4, 6 – trinitroptienol, 3 – nitrophenol, 3,5 – dinitrophenol, phenol, 4 – methyl phenol. \nAnswer: \nPhenols are stronger acids than alcohols because the phenoxide ion left after the removal of proton is stabilized by resonance while the alkoxide ion left after the removal of a proton from alcohol is not stabilized. Thus propan – 1 – ol is much weaker acid than any phenol.<\/p>\n Thus propan- 1 – ol is a much weaker acid than any phenol. We know that electron-donating groups decrease the acidic character and stronger is the electron-donating group, weaker is the phenol.<\/p>\n Compare to propan – 1 – ol, 4 – methyl phenol is a stronger acidic character. But comparing phenol and 4-methyl phenol, phenol is stronger acidic. Since electron-withdrawing groups increase the acidic character of phenols and the effect is more pronounced at the para position than at the meta position.<\/p>\n Therefore 4 – nitrophenol is a stronger acid than 3 – nitro phenol. Further as the number of electron-withdrawing groups increases the acidic strength further increases. Therefore 2, 4, 6 – trinitrophenol is a stronger acid than 3, 5 – dinitrophenol. It may be noted here that although the two nitro groups in 3, 5 – dinitrophenol are at m – position with respect to OH group,<\/p>\n their combined effect is however greater than one nitro group at p – position. Therefore 3, 5 – dinitrophenol is a stronger acid than 4-nitro phenol. Thus, the overall increasing order of acid strength is. Propan – 1 – 01 < 4 – methyl phenol < phenol < 3 – nitrophenol < 3, 5 – dinitrophenol < 2, 4, 6 – trinitrophenol.<\/p>\n <\/p>\n
Question 14. \n1 mole of HI is allowed to react with t – butyl methylether. Identify the product and write down the mechanism of the reaction. \nAnswer: \n \nThe reaction involves the protonation of oxygen which is followed by the SN1 mechanism. \nThe products are t-butvl iodide and methyl alcohol.<\/p>\n 12th Chemistry Guide Hydroxy Compounds and Ethers Additional Questions and Answers<\/h3>\nPart \u2013 II \u2013 Additional Questions<\/span><\/p>\nI. Choose the correct answer<\/span><\/p>\nQuestion 1. \nIn ethanol the – OH group is attached to hybridised carbon \na) sp \nb) sp2 <\/sup> \nc) sp3<\/sup> \nd) dsp2<\/sup> \nAns : \nc) sp3<\/sup><\/p>\nQuestion 2. \nIn Vinyl alcohol the -OH group is attached to hvbridised carbon \na) sp \nb) sp2 <\/sup> \nc) sp3<\/sup> \nd) dsp2<\/sup> \nAns : \nb) sp2 <\/sup><\/p>\n <\/p>\n
Question 3. \nSorbitol is \na) monohydric alcohol \nb) dihydric alcohol \nc) trihydric alcohol \nd) hexa hydric alcohol \nAnswer: \nd) hexa hydric alcohol<\/p>\n Question 4. \nThe IUPAC name of sorbitol is \na) Ethan -1, 2 – diol \nb) Propan – 1, 2, 3 – triol \nc) Hexan – 1, 2, 3, 4, 5, 6 – hexol \nd) Ethenol \nAnswer: \nc) Hexan – 1, 2, 3,4, 5, 6 – hexol<\/p>\n Question 5. \n \na) prop – 2 – en – 1 – ol \nb) 2 – methyl but – 3 – en – 2 ol \nc) 3 – methyl but – 1 – en – 3 – ol \nd) 1,1 – di methyl prop – 2 – en – 1 – ol \nAnswer: \nb) 2 – methyl but – 3 – en – 2 ol<\/p>\n Question 6. \nThe C – O – H bond angle in methanol is \na) 107\u00b0 \nb) 109\u00b0.5\u00b0 \nc) 108.9\u00b0 \nd) 104\u00b0 \nAnswer: \nc) 108.9\u00b0<\/p>\n <\/p>\n
Question 7. \nCH3<\/sub>-CH2<\/sub>-a + NaOH \u2192 CH3<\/sub>-CH2<\/sub>-OH + NaCl This reaction follows mechanism \na) \\(S_{N}^{1}[latex] \nb) [latex]S_{N}^{2}\\) \nc) E1<\/sup> \nd) E2<\/sup> \nAnswer: \nb) \\(S_{N}^{2}\\)<\/p>\nQuestion 8. \nConversion of isobutyl chloride and isopropyl chloride into isobutyl alcohol and isopropyl alcohol respectively using dilute aquous NaOH follow the mechnisms \n \nAnswer: \nc) \nReason : Isobutylchloride – Primary alkyl halide – \\(S_{N}^{2}\\) mechanism. \nIso propyl chloride – Secondary alkyl halide – Sjsj mechanism. \n\\(S_{N}^{1}\\)<\/p>\n <\/p>\n
Question 9. \nAddition of water across the double bond \nof an unsymmetrical alkene follows \na) Saytzeffs rule \nb) Markownikoff’s rule \nc) anti Markownikoff’s rule \nd) Popoff’s rule \nAnswer: \nb) Markownikoff\u2019s rule<\/p>\n Question 10. \nHydroboration of an alkene follows \na) Saytzeff’s rule \nb) Markownikoff’s rule \nc) anti Markownikoff’s rule \nd) Popoff’s rule \nAnswer: \nc) anti – Markownikoff’s rule<\/p>\n Question 11. \nAddition of water to 2-methyl propene in \npresence of cone, sulphuric acid gives \na) 2-propanol \nb) 2 – methyl – 2 – propanol \nc) 2 – methyl – 1 – propanol \nd) 2 – butanol \nAnswer: \nb) 2 – methyl – 2 – propanol \n <\/p>\n Question 12. \n2 – methyl propene reacts with diborane followed by H2O2 in presence of NaOH gives \na) 2 – proponol \nb) 2 – methyl – 2 – propanol \nc) 2 – methyl – 1 – propanol \nd) 2 – butanol \nAnswer: \nc) 2 – methyl -1 – propanol \n <\/p>\n <\/p>\n
Question 13. \nTo prepare a primary alcohol a Grignard reagent must be reacted with \na) HCHO \nb) RCHO \nc) RCOR \nd) RNH2<\/sub> \nAnswer: \na) HCHO<\/p>\nQuestion 14. \nTo prepare a secondary alcohol a \nGrignard reagent must be reacted with \na) HCHO \nb) RCHO \nc) RCOR \nd) RNH2<\/sub> \nAnswer: \nb) RCHO<\/p>\nQuestion 15. \nTo prepare a tertiary alcohol a Grignard reagent must be reacted with \na) HCHO \nb) RCHO \nc) RCOR \nd) RNH2<\/sub> \nAns: \nb) RCOR<\/p>\nQuestion 16. \nPhenyl magnesium bromide reacts with formaldehyde followed by hydrolysis gives \na) phenol \nb) Benzyl alcohol \nc) Benzaldehyde \nd) Benzoic acid \nAnswer: \nb) Benzyl alcohol \n <\/p>\n <\/p>\n
Question 17. \nName the product obtained when 1 mole of methyl magnesium bromide reacts with ethyl methanoate \na) propan -2- ol \nb) propan – 1 – ol \nc) ethanal \nd) ethanol \nAnswer: \nc) ethanal \n <\/p>\n Question 18. \n \nThe reducing agent X is \na) Ramey Ni \nb) NaBH4<\/sub> \nc) LiA\/H4<\/sub> \nd) Na – Hg \nAnswer: \nb) NaBH4<\/sub><\/p>\nQuestion 19. \nThe best reducing agent to prepare unsaturated alcohols from unsaturated carbonyl compounds is \na) Raney Ni \nb) NaBH4<\/sub> \nc) LiA\/H4<\/sub> \nd) both (b) & (c) \nAnswer: \nd) both (b) & (c)<\/p>\nQuestion 20. \nEthene is converted into ethylene glycol by treating with \na) acidified potassium dichromate \nb) alkaline potassium permanganate \nc) Chromyl chloride \nd) nitric acid \nAnswer: \nb) alkaline potassium permanganate<\/p>\n <\/p>\n
Question 21. \nCold dilute alkaline potassium permanganate is called as \na) Fenton’s reagent \nb) Tollen’s reagent \nc) Baeyer’s reagent \nd) Schiff’s reagent \nAnswer: \nc) Baeyer’s reagent<\/p>\n Question 22. \nGlyceryl palmitate on alkaline hydrolysis gives \na) glycerol and soap \nb) glycerol and sodium palmitate \nc) glycerol and sodium hexadentate \nd) all the above \nAnswer: \nd) all the above<\/p>\n Question 23. \nThe preparation of glycerol and soap from oils and fats is known as \na) esterification \nb) saponification \nc) hydroboration \nd) transesterification \nAnswer: \nb) saponification<\/p>\n <\/p>\n
Question 24. \nLucas reagent is \na) dil HCl & ZnCl2<\/sub> \nb) Cone. HCl & anhydrous ZnCl2<\/sub> \nc) dil H2<\/sub>SO4<\/sub> & AlCl3<\/sub> \nd) Conc. H2SO4 & anhydrous AlCl3 \nAnswer: \nb) Cone. HCl & anhydrous ZnCl2<\/sub><\/p>\nQuestion 25. \nWhich among the following will give immediate turbidity with Lucas reagent? \na) propan – 2 – ol \nb) propan – 1 – ol \nc) 2 – methyl propan – 2 – ol \nd) 2 – methyl propan – 1 – ol \nAnswer: \nc) 2 – methyl propan – 2 – ol \nHint: A teritary alcohol gives immediate turbidity<\/p>\n Question 26. \nIn Victor Meyer\u2019s test, the alcohol which \ngives blue colour is \na) propan – 2 – ol \nb) propan – 1 – ol \nc) 2 – methyl propan – 2 – ol \nd) 2 – methyl propan – 1 – ol \nAnswer: \na) propan – 2 – ol \nHint: A secondary alcohol gives blue colour.<\/p>\n <\/p>\n
Question 27. \nIn the given equation \n \n \nAnswer: \nc<\/p>\n Question 28. \nAmong butyl alcohols which one have the lowest boiling point? \na) n-butyl alcohol \nb) isobutyl alcohol \nc) sec-butyl alcohol \nd) ter – butyl alcohol \nAnswer: \nd) ter – butyl alcohol Hint: Increasing order of boiling point is 30 < 2\u00b0 < 1\u00b0<\/p>\n Question 29. \nWhich among the alcohol has a higher boiling point? \na) n-butyl alcohol \nb) n-propyl alcohol \nc) ethyl alcohol \nd) methyl alcohol \nAnswer: \na) n-butyl alcohol \nHint: As molecular weight of the alcohol increases, boiling point increases.<\/p>\n Question 30. \nWhich among the following is less soluble in water? \na) n-butyl alcohol \nb) n-propyl alcohol \nc) ethyl alcohol \nd) methyl alcohol \nAnswer: \na) n-butyl alcohol Hint: As the molecular weight increases, the solubility of alcohol decreases.<\/p>\n <\/p>\n
Question 31. \nConversion of alcohols into alkyl halide is an example of \na) Nucleophilic addition \nb) Nucleophilic substitution \nc) Electrophilic addition \nd) Electrophilic substitution \nAnswer: \nb) Nucleophilic substitution<\/p>\n Question 32. \nConversion of 2 – methyl – 1 – propanol into 2 – methyl – 1 – bromopropane is …………. reaction \na) SN2<\/sup> \nb) SN1<\/sup> \nc) E2<\/sup> \nd) E1<\/sup> \nAnswer: \na) SN2<\/sup><\/p>\nQuestion 33. \nPrimary alcohols undergo dehydration by mechanism \na) SN2<\/sup> \nb) SN1<\/sup> \nc) E2<\/sup> \nd) E1<\/sup> \nAns: \nd) E1<\/sup><\/p>\nQuestion 34. \n2 – methyl – 2 – propanol when reacted with cone. H2<\/sub>SO4<\/sub> gives 2 – methyl propene. This reaction follows mechanism \na) SN2<\/sup> \nb) SN1<\/sup> \nc) E2<\/sup> \nd) E1<\/sup> \nAnswer: \nd) E1<\/sup><\/p>\n <\/p>\n
Question 35. \nDuring intramolecular dehydration of 3, 3 – dimethyl – 2 – butanol the major product obtained is \na) 2, 3 – dimethyl but – 1 – ene \nb) 2, 3 – dimethyl but – 2 – ene \nc) 3, 3 – dimethyl but – 1 – ene \nd) 3, 3 – dimethyl but – 2 – ene \nAnswer: \nb) 2, 3 – dimethyl but – 2 – ene<\/p>\n Question 36. \nThe major product obtained when phenol reacts with con.H2<\/sub>SO4<\/sub> at 280 K is \na) Salicyclic acid \nb) picric acid \nc) o-phenol sulphonic acid \nd) p-phenol sulphonic acid \nAnswer: \nc) o-phenol sulphonic acid<\/p>\nQuestion 37. \nDuring intramolecular dehydration of alcohols say tzeff’s rule favours the formation of \na) unstable alkenes \nb) less substituted alkenes \nc) more substituted alkenes \nd) none of the above \nAnswer: \nc) more substituted alkenes<\/p>\n Question 38. \nTo stop the oxidation reaction of alcohols at the aldehyde \/ ketone stage…………..is used as an oxidising agent \na) Pottassium permanganate \nb) Py ridinium chloro chromate (PCC) \nc) Potassium di chromate \nd) Sodium di chromate \nAnswer: \nb) pyridinium chloro chromate (PCC)<\/p>\n <\/p>\n
Question 39. \nIn Swern oxidation of alcohols into aldehydes\/ketones the oxidising agent used is \na) Pyridinium chlorochromate (PCC) \nb) Dimethyl sulfoxide (DMSO) \nc) Alkaline potassium permanganate (Baeyer\u2019s reagent) \nd) Ferrous sulphate \/ H2O2 (Fenton’s reagent) \nAnswer: \nb) Dimethyl sulfoxide (DMSO)<\/p>\n Question 40. \nTo detoxify the alcohol produced in animals by the fermentaion of food, the oxidising agent used is \na) ADH \nb) ADP \nc) NAD \nd) ATP \nAnswer: \nc) NAD<\/p>\n Question 41. \nThe catalyst which catalyses the oxidation of toxic alcohols into non-toxic aldehyde in animals is \na) ADH \nb) ADP \nc) NAD \nd) ATP \nAnswer: \na) ADH<\/p>\n <\/p>\n
Question 42. \nWhen ethan-1, 2-diol is heated with anhydrous ZnCl2<\/sub> under pressure it gives \na) Ethanol \nb) Ethanal \nc) Ethanoic acid \nd) Ethene \nAns: \nb) Ethanal<\/p>\nQuestion 43. \nGlycerol can be oxidised to meso oxalic acid by \na) dii HNO3<\/sub> \nb) HIO4<\/sub> \nc) Bismuth nitrate \nd) Fenton\u2019s reagent \nAnswer: \nc) Bismuth nitrate<\/p>\nQuestion 44. \nThe alcohol used in the manufacture of dynamite is \na) methanol \nb) ethanol \nc) ethv lene glycol \nd) glycerol \nAnswer: \nd) glycerol<\/p>\n Question 45. \nThe correct order of acidic nature of alcohols is \na) Ethanol < propan – 2 – ol < 2 – methyl propan – 2- ol \nb) Propan – 2 – ol < Ethanol < 2 – methyl propan \u2014 2 \u2014 ol \nc) 2 – methyl propan – 2 – ol < Ethanol < propan – 2 – ol \nd) 2 – methyl propan – 2 \u2014 ol Answer: \nd) 2 – methyl propan – 2 – ol<\/p>\n <\/p>\n
Question 46. \n1, 3 – dihydroxy benzene is commonly \nknown as \na) Cresol \nb) Catechol \nc) Resorcinol \nd) Quinol \nAnswer: \nc) Resorcinol<\/p>\n Question 47. \nThe JUPAC name of pyrogallol is \na) 1, 2, 3 \u2014 trthydroxy benzene \nh) 1, 2, 4 \u2014 trihydroxy beneze \nc) 1, 3, 5 \u2014 trihydroxv benzene \nd) 1, 4, 5 \u2014 tri hydroxy benzene . \nAnswer: \na) 1, 2, 3 \u2014 trihydroxy benzene<\/p>\n Question 48. \nOricinol is \na) 1, 2 – dihydroxy benzene \nh) 3 \u2014 methyl phenol : \nc) 3, 5 \u2014 dihydroxy toluene \nd) 3, 5 \u2014 dimethyl toluene \nAnswer: \nc) 3, 5 – dihydroxy toluene<\/p>\n <\/p>\n
Question 49. \nWhich of the following reaction will give ether? \n \nAns: \nd) All of the above<\/p>\n Question 50. \nPhenol on oxidation with acidified K2<\/sub>Cr2<\/sub>O7<\/sub>gives \na) 1, 4 – dihydroxy benzene \nb) 1,4- benzoquinone \nc) cyclohexanol \nd) cumene \nAnswer: \nb) 1,4- benzoquinone<\/p>\nQuestion 51. \nWhich among the following is a simple ether? \na) 1 – methoxy propane \nb) 2 – methoxy propane \nc) methoxy methane \nd) methoxv benzene \nAnswer: \nc) methoxy methane<\/p>\n Question 52. \nThe oxygen atom of ether is \nhybridised \na) sp \nb) sp2<\/sup> \nc) sp3<\/sup> \nd) dsp2<\/sup> \nAnswer: \nc) sp3<\/sup><\/p>\n <\/p>\n
Question 53. \nWilliamson\u2019s synthesis is an example of \na) nucleophilic addition \nb) nucleophilic substitution \nc) electrophilic addition \nd) electrophilic substitution \nAnswer: \nb) nucleophilic substitution<\/p>\n Question 54. \nWhich among the following is more reactive \ntowards ethers? \na) HF \nb) HCl \nc) HBr \nd) HI \nAnswer: \nd) HI<\/p>\n Question 55. \nThe products obtained when methoxy \nethane reacted with one mole of HI are \na) methanol & iodoethane \nh) iodomethane & ethanol \nc) iodomethane & jodo ethane \nd) Methanol & ethanol \nAnswer: \nb) iodomethane & ethanol<\/p>\n <\/p>\n
Question 56. \nWhich is used as a precursor to the synthesis of perfumes and insecticide pheromones? \na) Phenol \nb) Phenoxy methane \nc) methoxy benzene \nd) Ethoxy benzene \nAnswer: \nc) methoxy benzene<\/p>\n Question 57. \nAccording to Lewis concept of acids and bases an ether is \na) acidic \nb) basic \nc) neutral \nd) amphoteric \nAnswer: \nb) basic<\/p>\n Question 58. \n \na) CH3<\/sub>COCH3<\/sub> \nb) C2<\/sub>H5<\/sub>OH \nc) CH3<\/sub>-CH(OH)CH3<\/sub> \nd) CH3<\/sub>CHO \nAnswer: \nc) CH3<\/sub>-CH(OH)CH3<\/sub> \n <\/p>\nQuestion 59. \n \na)Propan-1,3-diol \nb) Ethan -1,2- diol \nc) Propan -1,2,3- Triol \nd) Ethanal \nAnswer: \nb) Ethan -1,2- diol \n <\/p>\n Question 60. \nWhich of the following has lowest boiling point? \na) phenol \nb) O-nitro phenol \nc) m-nitro phenol \nd) p-nitro phenol \nAnswer: \nb) O-nitro phenol \nHint: O – nitrophenol has intramolecular hydrogen bonding while others intermolecular hydrogen bonding.<\/p>\n <\/p>\n
II. Assertion and Reason<\/span><\/p>\nQuestion 1. \nAssertion : Tertiary alcohols undergo dehydration more readily than primary alcohol. \nReason : Tertiary alcohols are less acidic than primary alcohol. \na) Both assertion and reason are true and reason is the correct explanation of assertion \nb) Both assertion and reason are true but reason is not the correct explanation of assertion \nc) Assertion is true but reason is false. \nd) Both assertion and reason are false. \nAnswer: \nb) Both assertion and reason are true but reason is not the correct explanation of assertion<\/p>\n <\/p>\n
Question 2. \nAssertion (A) : Alcohols can act as Bronsted acids \nReason (R) : It is due to the presence of unshared electron pairs on oxygen which make them proton acceptors. \na) Both (A) and (R) are correct, R explain A. \nb) Both (A) and (R) are correct, R does not explain A. \nc) (A) is correct but R is correct. \nd) (A) is wrong but R is correct. \nAnswer: \nd) (A) is wrong but R is correct Correct Reason (A) : Alcohols can act as Bronsted bases.<\/p>\n Question 3. \nAssertion (A) : Phenol is more acidic than aliphatic alcohols \nReason (R) : Alkyl substituted phenols show a decreased acidity due to the electron releasing +1 effect of alkyl group. \na) Both (A) and (R) are correct, R explain A. \nb) Both (A) and (R) are correct, R does not explain A. \nc) (A) is correct but R is correct. \nd) (A) is wrong but R is correct. \nAnswer: \nb) Both (A) and (R) are correct, R does not explain A. \nCorrect Reason (R) : The phenoxide ion is more stabilised by resonance than phenol.<\/p>\n Question 4. \nAssertion (A): Orthonitro phenol is slightly soluble in water and more volatile whereas p – nitro phenol is more soluble in water and less volatile. \nReason (R) : Orthonitro phenol forms intramolecular hydrogen bonding and p-nitro phenol forms intermolecular hydrogen bonding. \na) Both (A) and (R) are correct, R explains A. \nb) Both (A) and (R) are correct, R does not explain A. \nc) (A) is correct but R is correct. \nd) (A) is wrong but R is correct. \nAnswer: \na) Both (A) and (R) are correct, R explains A.<\/p>\n <\/p>\n
III. Pick out the correct statements<\/span><\/p>\nQuestion 1. \n(i) The structure of – OH group attached to a sp3<\/sup>– hybridised carbon in alcohol is similar to the structure of – OH group attached to a hydrogen in water. \n(ii) Due to lone pair – lone pair repulsion, the – C – O – H bond angle in methanol is reduced to 104\u00b0 from the regular tetrahedral bond angle of 109.5\u00b0. \n(iii) With Grignard reagent formaldehyde gives primary alcohol and other aldehydes give secondary alcohols. \n(iv) Reaction of Grignard reagent with aldehydes and ketones to form alcohols is an example for nucleophilic substitution reaction. \na) (i) & (ii) \nb) (ii) & (iii) \nc) (i) & (iii) \nd) (i) & (iv) \nAnswer: \nc) (i) & (iii) \nCorrect statements : (ii) Due to lone pair – lone pair repulsion, the – C – O – H bond anlge in methanol is reduced to 108.9\u00b0 from the regular tetrahedral bond angle of 109.5\u00b0. \n(iv) Reaction of Grignard reagent with adlehydes and ketones to form alcohols is an example for nucleophilic addition reaction.<\/p>\nQuestion 2. \n(i) Lower alcohols are waxy solids and the higher members are colourless liquids. \n(ii) Due to the presence of intermolecular hydrogen bonding alcohols have higher boiling point than the corresponding alkanes, aldehydes, ethers etc. \n(iii) Due to the formation of intermolecular hydrogen bonding with water lower alcohols are highly soluble in water. \n(iv) Among isomeric alcohols primary alcohols have lower boiling point and tertiary alcohols have higher boiling points. \na) (i) & (ii) \nb) (ii) & (iii) \nc) (iii) & (iv) \nd) (i) & (iv) \nAnswer: \nb) (ii) & (iii) \nCorrect statements : (i) Lower alcohols are colourless liquids and the higher members are waxy solids. \n(iv) Among isomeric alcohols primary alcohols have higher boiling point and the tertiary alcohols have lower boiling points.<\/p>\n <\/p>\n
Question 3. \n(i) In Swern Oxidation DMSO, oxalyl chloride and triethylamine are used to convert alcohols into aldehydes \/ ketones. \n(ii) Vapours of primary alcohol passed over heated copper at 573 K undergoes dehydration. . \n(iii) Methanol reacts with ethanoic acid in presence of an acid to form ethylethanoate. \n(iv) Vapours of tertiary alcohols react with heated copper at 573 K to give alkenes. \na) (i) & (ii) \nb) (ii) & (iii) \nc) (iii) & (iv) \nd) (i) & (iv) \nAnswer: \nd) (i) & (iv) \nCorrect statements : (ii) Vapours of primary alcohol passed over heated copper at 573 K undergoes dehydrogenation. \n(iii) Methanol reacts with ethanoic acid in presence of an acid to form methyl ethanoate. (or) \nEthanol reacts with ethanoic acid in presence of an acid to form ethyl ethanoate.<\/p>\n <\/p>\n
Question 4. \n(i) Glycerol contains two primary alcoholic group and one secondry alcoholic group. \n(ii) Oxidation of glycerol with cone. HNO3<\/sub> gives mainly formic acid. |