JEE Chemistry Alcohols Phenols and Ethers MCQs Set 26

Answer: (b) phenol

Explanation:
1. Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide.
2. Acid hydrolysis of the hydroperoxide gives phenol and acetone.

Teacher's Note:
1. One mole of acetone is formed for every mole of phenol.

Answer: (d) 4-methylphenol (p-cresol)

Explanation:
1. The benzylic C–H of the isopropyl group is oxidised to a hydroperoxide.
2. Hydrolysis removes the side chain as acetone and leaves OH where the isopropyl group was, para to the methyl group.

Teacher's Note:
1. The OH ends up on the carbon that carried the isopropyl group.

Answer: (a) phenol

Explanation:
1. This is Dow's process: Cl is replaced by ONa, and acidification frees the phenol.
2. C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O; then C₆H₅ONa + HCl → C₆H₅OH + NaCl.

Teacher's Note:
1. Aryl halides resist ordinary substitution, so drastic conditions are needed.

Answer: (c) 4-methylphenol

Explanation:
1. Alkali fusion replaces the –SO₃H group by –OH at the same carbon.
2. The OH therefore takes the para position relative to the methyl group.

Teacher's Note:
1. Fusion of sulphonates with NaOH is a commercial route to phenols.

Answer: (d) 4-ethylphenol

Explanation:
1. Soda lime removes the carboxylate group as Na₂CO₃ and puts H in its place.
2. The OH and the ethyl group stay para to each other, so 4-ethylphenol results.

Teacher's Note:
1. Decarboxylation keeps the OH group; only COONa is lost.

Answer: (b) 2,4,6-trichlorophenol

Explanation:
1. In a polar solvent the OH group strongly activates both ortho positions and the para position.
2. All three positions are substituted, giving 2,4,6-trichlorophenol.

Teacher's Note:
1. With Br₂/H₂O the same pattern gives a white precipitate of 2,4,6-tribromophenol.

Answer: (c) 2-bromo-4-methylphenol

Explanation:
1. A non-polar solvent at low temperature limits the reaction to mono-substitution.
2. The para position is blocked, so Br enters a position ortho to OH.

Teacher's Note:
1. Br₂/CS₂ gives the mono-bromo product; Br₂/H₂O gives the tribromo product.

Answer: (a) 2-hydroxy-5-methylbenzaldehyde

Explanation:
1. Dichlorocarbene from CHCl₃/NaOH attacks the position ortho to the phenoxide oxygen.
2. Hydrolysis converts the –CHCl₂ group into –CHO, giving the ortho-hydroxy aldehyde.

Teacher's Note:
1. Reimer-Tiemann formylation gives mainly the ortho product.

Answer: (b) CCl₄ / NaOH / H⁺

Explanation:
1. Phenol with CCl₄ and NaOH forms a trichloromethyl derivative that hydrolyses to –COOH.
2. So the sequence is CCl₄ and NaOH, followed by acid.

Teacher's Note:
1. CHCl₃ gives the aldehyde; CCl₄ gives the acid.

Answer: (d) acetic anhydride, (CH₃CO)₂O

Explanation:
1. Aspirin is the acetyl ester of the phenolic OH group of salicylic acid.
2. Acetylation is done with acetic anhydride or acetyl chloride.

Teacher's Note:
1. Propanoic anhydride would give the propanoyl ester instead.

Answer: (a) 2-hydroxy-1-naphthaldehyde

Explanation:
1. Reimer-Tiemann formylation introduces CHO next to the OH group.
2. In 2-naphthol the most reactive adjacent position is C1, giving 2-hydroxy-1-naphthaldehyde.

Teacher's Note:
1. CHCl₃ gives –CHO; CCl₄ would give –COOH.

Answer: (c) 4-cyclohexylphenol

Explanation:
1. Kolbe-Schmidt carboxylation introduces COOH ortho to the OH group of the phenoxide.
2. So the starting phenol already carries the cyclohexyl group para to OH.

Teacher's Note:
1. The phenoxide is formed first with NaOH, then CO₂ is added.

Answer: (d) propanone (acetone)

Explanation:
1. The product is a tertiary alcohol, so a ketone is needed.
2. The carbinol carbon carries CH₃, CH₃ and C₂H₅, so the ketone is CH₃COCH₃ and the Grignard supplies C₂H₅.

Teacher's Note:
1. Methanal gives 1°, other aldehydes 2°, ketones 3° alcohols.

Answer: (b) 1-phenylpropan-1-ol

Explanation:
1. The phenyl group adds to the carbonyl carbon of the aldehyde.
2. C₂H₅CHO gives C₂H₅–CH(OH)–C₆H₅ after hydrolysis.

Teacher's Note:
1. Aldehydes other than methanal give secondary alcohols.

Answer: (c) methanal (HCHO)

Explanation:
1. Only HCHO has two H atoms on the carbonyl carbon.
2. R–MgX adds to HCHO to give R–CH₂OH, a primary alcohol.

Teacher's Note:
1. Ethanal gives 2° alcohols; ketones give 3° alcohols.

Answer: (a) cis-cyclohexane-1,2-diol

Explanation:
1. OsO₄ adds two OH groups to the same face of the double bond (syn addition).
2. Reduction of the osmate ester with NaHSO₃ releases cis-cyclohexane-1,2-diol.

Teacher's Note:
1. Syn dihydroxylation of a cyclic alkene gives a cis-diol.

Answer: (b) Baeyer's reagent (cold dilute alkaline KMnO₄)

Explanation:
1. Cold dilute alkaline KMnO₄ hydroxylates a C=C bond to a vicinal diol.
2. HCl gives 2-chlorobutane, dilute acid gives butan-2-ol and H₂/Ni gives butane.

Teacher's Note:
1. The purple colour of Baeyer's reagent is discharged by alkenes.

Answer: (d) both OsO₄/NaHSO₃ and cold dilute alkaline KMnO₄

Explanation:
1. Both reagents deliver the two OH groups to the same face of the double bond.
2. A peracid followed by H₃O⁺ gives the trans-diol (anti addition) instead.

Teacher's Note:
1. Syn: OsO₄ or cold KMnO₄. Anti: peracid then H₃O⁺.

Answer: (a) butane-2,3-diol

Explanation:
1. The peracid forms an epoxide across C2–C3.
2. Acid-catalysed ring opening places one OH on each of the two carbons, giving butane-2,3-diol.

Teacher's Note:
1. The ring opening is anti.