JEE Chemistry Alcohols Phenols and Ethers MCQs Set 30

Answer: (c) The carbon–oxygen bond length in methanol, CH₃OH, is shorter than that of the C–O bond in phenol

Explanation:
1. In phenol the lone pair on oxygen is delocalised into the ring, giving the C–O bond partial double-bond character, so it is shorter (about 136 pm) than the C–O bond of methanol (about 142 pm).
2. The other three statements are correct: alcohols are weaker acids than water, +I effect of alkyl groups makes tertiary alcohols the weakest acids, and the C–O–H angle is 108.9°.
3. Answer: (c)

Teacher's Note:
1. Phenol C–O bond is shorter because of resonance.
2. Alkyl groups decrease acidity of alcohols.

Answer: (d) butan-2-ol and 2-methylbutan-2-ol

Explanation:
1. Reaction with HBr proceeds through carbocations (S_N1), so reactivity is 3° > 2° > 1°.
2. In (d) the second alcohol is tertiary and the first is secondary, so the second is more reactive.
3. In the other pairs the first alcohol is the more reactive (tertiary against secondary in (a) and (b), secondary against primary in (c)).
4. Answer: (d)

Teacher's Note:
1. Rate follows carbocation stability.
2. Tertiary alcohols react fastest.

Answer: (a) NaBD₄ in CH₃OH

Explanation:
1. The deuteride ion from NaBD₄ adds to the carbonyl carbon, which puts D on carbon.
2. The alkoxide formed is protonated by the solvent CH₃OH, which puts ordinary H on oxygen.
3. The other reagents introduce H on carbon or D on oxygen.
4. Answer: (a)

Teacher's Note:
1. Hydride (deuteride) goes to carbon.
2. Solvent proton goes to oxygen.

Answer: (b) an alkene will be the main product

Explanation:
1. The alkoxide is a strong base and the tertiary halide is hindered to S_N2 attack.
2. E2 elimination dominates, giving an alkene.
3. The correct synthesis uses the tertiary alkoxide with a primary alkyl halide.
4. Answer: (b)

Teacher's Note:
1. Tertiary halide + strong base: elimination.
2. Choose the primary halide for Williamson.

Answer: (b) CH₂=CHCH₂Br + C₆H₅ONa

Explanation:
1. Williamson synthesis combines a sodium alkoxide or phenoxide with an alkyl halide.
2. Allyl bromide is a reactive primary halide, so with sodium phenoxide it gives C₆H₅OCH₂CH=CH₂.
3. Aryl and vinyl halides do not undergo S_N2 substitution, which rules out the other options.
4. Answer: (b)

Teacher's Note:
1. Use the reactive halide (allyl) and the aryl-oxide.
2. Aryl halides are unreactive in S_N2.

Answer: (a) MnO₂

Explanation:
1. Activated manganese dioxide selectively oxidises allylic and benzylic alcohols to aldehydes without attacking the double bond.
2. KMnO₄ and OsO₄ attack the C=C, and H₂O₂ does not oxidise it to acrolein.
3. Answer: (a)

Teacher's Note:
1. MnO₂: selective for allylic OH.
2. Acrolein is CH₂=CH–CHO.

Answer: (d) III, IV

Explanation:
1. Periodic acid cleaves the C–C bond between two carbons carrying OH groups (vicinal diols), or OH and C=O.
2. Glycerol and ethane-1,2-diol are vicinal diols and are cleaved.
3. Propane-1,3-diol has OH groups on non-adjacent carbons, and 1-methoxypropan-2-ol has only one OH, so neither is cleaved.
4. Answer: (d)

Teacher's Note:
1. HIO₄ needs adjacent –OH groups.
2. 1,3-Diols are not cleaved.

Answer: (c) o-nitrophenol

Explanation:
1. Only acids stronger than carbonic acid (pK_a about 6.4) dissolve in NaHCO₃ with evolution of CO₂.
2. Picric acid (pK_a about 0.4), benzoic acid and benzenesulphonic acid are stronger, but o-nitrophenol (pK_a about 7.2) is weaker and does not dissolve.
3. Answer: (c)

Teacher's Note:
1. Strong acids liberate CO₂ from bicarbonate.
2. Simple nitrophenols are weaker than carbonic acid.

Answer: (c) –COCH₃

Explanation:
1. An acetyl group is electron withdrawing by both the −M and −I effects, so it delocalises the negative charge from the phenoxide oxygen.
2. Methyl and methoxy are electron-donating and destabilise the ion, and –CH₂OH has little effect.
3. Answer: (c)

Teacher's Note:
1. Electron-withdrawing group stabilises the phenoxide.
2. Stronger stabilisation means a stronger acid.

Answer: (a) III > I > IV > II

Explanation:
1. The reaction goes through the benzyl carbocation, so electron-donating groups speed it up and electron-withdrawing groups slow it down.
2. Order of stabilisation: –OCH₃ (+M) > –H > –Br (weak −I) > –NO₂ (strong −M, −I).
3. Answer: (a)

Teacher's Note:
1. Electron donors stabilise the cation.
2. Nitro group destabilises it most.

Answer: (a) Reimer–Tiemann reaction

Explanation:
1. Dichlorocarbene, generated from chloroform and alkali, attacks the ortho position of the phenoxide ion.
2. Hydrolysis then gives salicylaldehyde (o-hydroxybenzaldehyde).
3. Answer: (a)

Teacher's Note:
1. Reagent: CHCl₃ + aq. NaOH.
2. Electrophile: :CCl₂.

Answer: (d) (CH₃)₂CHOH + CH₃MgBr

Explanation:
1. Active metals (Na, K, Al with heating) react with alcohols to liberate hydrogen.
2. With the Grignard reagent the alcohol protonates the carbanion and methane is evolved: (CH₃)₂CHOH + CH₃MgBr → CH₄ + (CH₃)₂CHOMgBr.
3. Answer: (d)

Teacher's Note:
1. Grignard reagent + active hydrogen gives an alkane.
2. Hydrogen comes only from reaction with metals.

Answer: (d) HOCH₂CH₂CH₂CH₂OH

Explanation:
1. 1 mol of H₂ needs two active hydrogen atoms per molecule, so X must contain two –OH groups.
2. Butane-1,4-diol gives one mole of H₂ per mole, while a monohydric alcohol gives only 0.5 mol.
3. Answer: (d)

Teacher's Note:
1. 2 –OH groups: 1 mol H₂.
2. One –OH: 0.5 mol H₂.

Answer: (c) PBr₃; then (i) Mg, ether, (ii) D₂O

Explanation:
1. PBr₃ converts the alcohol to 2-bromobutane.
2. Mg in ether gives the Grignard reagent, and D₂O protonates (deuterates) the carbanion carbon to give CH₃CH₂CHDCH₃.
3. Dehydration/D₂ would add two D atoms, and NaOD/D₂O would give an alcohol or alkene.
4. Answer: (c)

Teacher's Note:
1. Grignard reagent + D₂O replaces MgBr by D.
2. Only one D is introduced.

Answer: (c) CH₃CH₂MgBr + CH₃CH₂COCH₂CH₃

Explanation:
1. 3-Ethylpentan-3-ol is (C₂H₅)₃COH, which has three ethyl groups on the carbinol carbon.
2. Ethylmagnesium bromide adds to pentan-3-one (CH₃CH₂COCH₂CH₃) to give it after hydrolysis.
3. Answer: (c)

Teacher's Note:
1. Count the groups on the carbinol carbon.
2. Ketone + Grignard gives a tertiary alcohol.

Answer: (a) CH₃OH, H₂SO₄ (catalytic)

Explanation:
1. Under acidic conditions the protonated epoxide opens at the more substituted carbon, which can better carry the positive charge.
2. Methanol attacks the tertiary carbon, giving HOCH₂C(OCH₃)(CH₃)₂.
3. Under basic conditions (CH₃O⁻) the attack is on the less hindered CH₂ carbon, giving the other isomer.
4. Answer: (a)

Teacher's Note:
1. Acid: attack at the more substituted carbon.
2. Base: attack at the less hindered carbon.

Answer: (d) sec-butyl alcohol

Explanation:
1. Reaction with sodium shows an –OH group, so it is not diethyl ether.
2. A carbonyl compound that does not reduce Tollens' reagent is a ketone, which comes from a secondary alcohol: butan-2-ol (sec-butyl alcohol) gives butanone.
3. Answer: (d)

Teacher's Note:
1. Secondary alcohol gives ketone.
2. Primary alcohol gives aldehyde, which reduces Tollens' reagent.

Answer: (c) 10.22 g

Explanation:
1. Phenol, C₆H₅OH, has molar mass 94 g mol⁻¹, so 2 g is 2/94 = 0.02128 mol.
2. Each mole needs 3 mol of Br₂ (three positions), that is 0.0638 mol.
3. Mass of Br₂ = 0.0638 × 160 = 10.2 g.
4. Answer: (c)

Teacher's Note:
1. Three Br₂ per phenol.
2. Molar mass of Br₂ is 160 g.

Answer: (b) tert-butyl bromide + sodium ethoxide

Explanation:
1. Sodium ethoxide is a strong base and tert-butyl bromide is tertiary, so E2 elimination gives isobutene instead of the ether.
2. The ether can be made from sodium tert-butoxide with ethyl bromide (primary halide), or by acid-catalysed routes through the tert-butyl cation.
3. Answer: (b)

Teacher's Note:
1. Tertiary halide + alkoxide gives alkene.
2. Primary halide + tertiary alkoxide gives the ether.

Answer: (c) boron trifluoride

Explanation:
1. Diethyl ether has lone pairs on oxygen and acts as a Lewis base, so it forms an addition complex with the Lewis acid BF₃ (Et₂O·BF₃).
2. It does not react with dilute acids or alkalis, and sodium is used to dry ether.
3. Answer: (c)

Teacher's Note:
1. Lewis base + Lewis acid gives an adduct.
2. Sodium wire is used to dry ether.

Answer: (a) CH₃OH, H⁺ (catalyst), heat

Explanation:
1. An aldehyde reacts with two equivalents of an alcohol in the presence of an acid catalyst to give an acetal.
2. So butanal with methanol and a trace of acid gives 1,1-dimethoxybutane (dimethyl acetal).
3. Answer: (a)

Teacher's Note:
1. Aldehyde + 2 ROH + H⁺ gives acetal.
2. Water would give the reverse reaction.