JEE Chemistry Alcohols Phenols and Ethers MCQs Set 06

Answer: (d) \( \ce{C4H9MgI} \) and propanone, followed by hydrolysis

Explanation:
1. 3-Methylhexan-3-ol is \( \ce{CH3CH2C(OH)(CH3)CH2CH2CH3} \), a tertiary alcohol with groups CH3, C2H5 and C3H7 on the carbinol carbon.
2. The first three combinations give these three groups. \( \ce{C4H9MgI} \) with propanone gives 2-methylhexan-2-ol, so the last cannot be used.

Teacher's Note:
1. Count the carbons.
2. Grignard + ketone.

Answer: (b) Reduction of ester with \( \ce{Na/C2H5OH} \)

Explanation:
1. The Bouveault-Blanc reduction is the reduction of esters to alcohols using sodium in ethanol.
2. The others are Rosenmund, LiAlH4 and Clemmensen reductions.

Teacher's Note:
1. Named reaction.
2. Dissolving metal reduction.

Answer: (b) benzene-1,2,4-triol

Explanation:
1. Excess HI cleaves each aryl methyl ether into a phenol and methyl iodide; the aryl carbon-oxygen bond does not break.
2. The product is benzene-1,2,4-triol.

Teacher's Note:
1. Ar-O bond has partial double bond character.
2. Methyl iodide forms.

Answer: (d) All of these

Explanation:
1. \( \ce{LiAlH4} \) followed by water, \( \ce{NaBH4} \) in ethanol, and \( \ce{BH3.THF} \) followed by acid workup all reduce a ketone to the secondary alcohol.
2. So all three reagents work.

Teacher's Note:
1. C=O to CH-OH.
2. Hydride reagents.

Answer: (b) 2,3-dimethylbutan-2-ol

Explanation:
1. Protonation of 3,3-dimethylbut-1-ene gives a secondary carbocation, which undergoes a methyl shift to the more stable tertiary carbocation.
2. Water adds to it to give 2,3-dimethylbutan-2-ol.

Teacher's Note:
1. Rearrangement.
2. Tertiary cation is more stable.

Answer: (a) \( \ce{CH3OH} \)

Explanation:
1. Water gas (\( \ce{CO + H2} \)) is converted to methanol at high pressure with ZnO-Cr2O3 catalyst.
2. \( \ce{CO + 2H2 -> CH3OH} \).

Teacher's Note:
1. Industrial synthesis.
2. Methanol.

Answer: (a) 2-methylcyclohexanone

Explanation:
1. The group \( \ce{CH(OH)CH3} \) gives a positive iodoform reaction and is converted into a carboxylate (the carbon chain loses CHI3).
2. On acidification a beta-keto acid forms, which loses \( \ce{CO2} \) on heating to give 2-methylcyclohexanone.

Teacher's Note:
1. Iodoform followed by decarboxylation.
2. Beta-keto acids decarboxylate.

Answer: (d) acetic ester and ethanoic acid both

Explanation:
1. Ethanol reacts with acetic anhydride to give ethyl acetate and ethanoic acid.
2. \( \ce{(CH3CO)2O + C2H5OH -> CH3COOC2H5 + CH3COOH} \).

Teacher's Note:
1. Acetylation.
2. Both products form.

Answer: (d) tetrahydrofuran

Explanation:
1. The alkoxide formed in alcoholic KOH attacks the carbon bearing chlorine within the same molecule.
2. This intramolecular Williamson reaction gives a five-membered ring, tetrahydrofuran.

Teacher's Note:
1. Five-membered ring formation is fast.
2. Intramolecular SN2.

Answer: (c) sulphonic ester

Explanation:
1. Hinsberg's reagent is benzenesulphonyl chloride; with phenol it gives the sulphonic ester (phenyl benzenesulphonate).
2. It gives a sulphonamide with amines.

Teacher's Note:
1. Ar-SO2Cl.
2. O-sulphonylation.

Answer: (c) \( \ce{CH3CH2CH2OH} \) and \( \ce{CH3Br} \)

Explanation:
1. Methyl propyl ether is cleaved by HBr through protonation followed by SN2 attack of bromide at the less hindered alkyl group, the methyl group.
2. Products are \( \ce{CH3Br} \) and propan-1-ol.

Teacher's Note:
1. Smaller alkyl group is attacked.
2. SN2 mechanism.

Answer: (d) propan-2-ol

Explanation:
1. Red colour with ceric ammonium nitrate shows an alcohol; no CO2 with NaHCO3 means it is not an acid.
2. A positive iodoform test requires CH3CH(OH)-, so P is propan-2-ol.

Teacher's Note:
1. Iodoform test for CH3CH(OH).
2. C3H8O alcohols.

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

Explanation:
1. Phenol does not dehydrate since its C-O bond has partial double bond character.
2. Cyclohexanol dehydrates to an alkene, allylic alcohols form resonance stabilised cations more easily, and cyclohexa-2,5-dien-1-ol gives aromatic benzene. Order: I < II < III < IV.

Teacher's Note:
1. Ease depends on the cation or product stability.
2. Aromatisation drives the last.

Answer: (c) 5-bromo-2-hydroxy-4-methylbenzaldehyde

Explanation:
1. Reimer-Tiemann formylation puts the CHO group ortho to OH at the less hindered position, giving 2-hydroxy-4-methylbenzaldehyde.
2. In bromination, bromine goes para to the strongly activating OH group, giving 5-bromo-2-hydroxy-4-methylbenzaldehyde.

Teacher's Note:
1. OH directs.
2. Para to OH is free.