JEE Chemistry Alcohols Phenols and Ethers MCQs Set 29

Answer: (b) n-Propyl alcohol

Explanation:
1. A = ethyl iodide, B = ethylmagnesium iodide.
2. Formaldehyde with a Grignard reagent gives a primary alcohol with one more carbon: D = \( \ce{CH3CH2CH2OH} \).

Teacher's Note:
1. HCHO plus RMgX gives RCH2OH.

Answer: (a) (a-q), (b-r), (c-p), (d-s)

Explanation:
1. Phenol gives a violet colour with neutral ferric chloride and a white precipitate of 2,4,6-tribromophenol with bromine water.
2. Phenol is weaker than carbonic acid, so no reaction with \( \ce{NaHCO3} \); picric acid is stronger and liberates \( \ce{CO2} \).

Teacher's Note:
1. Compare acidity with H2CO3.

Answer: (b) \( \ce{Na} \) with dry ether

Explanation:
1. X is bromomethane (Hunsdiecker product).
2. \( \ce{LiAlH4} \) and \( \ce{Zn/HCl} \) reduce \( \ce{CH3Br} \) to methane.
3. Sodium with ether (Wurtz reaction) would give ethane, not methane.

Teacher's Note:
1. Wurtz coupling doubles the carbon chain.

Answer: (b) Cumene (isopropylbenzene) and acetone

Explanation:
1. The n-propyl carbocation rearranges to the more stable isopropyl carbocation, so P = cumene.
2. Cumene hydroperoxide on acid treatment gives phenol and acetone (Q).

Teacher's Note:
1. Friedel-Crafts alkylation with n-propyl chloride gives isopropylbenzene.

Answer: (b) Mg in dry ether, then ethanal followed by \( \ce{H3O+} \)

Explanation:
1. \( \ce{C6H5Br} \) forms the Grignard reagent \( \ce{C6H5MgBr} \).
2. Reaction with \( \ce{CH3CHO} \) and hydrolysis gives the secondary alcohol \( \ce{C6H5CH(OH)CH3} \).

Teacher's Note:
1. Aldehyde plus Grignard gives a secondary alcohol.

Answer: (a) 2-Nitrophenol

Explanation:
1. NaHCO3 dissolves acids stronger than carbonic acid (pKa about 6.4) with evolution of \( \ce{CO2} \).
2. Sulfonic acid, benzoic acid and picric acid are stronger acids. 2-Nitrophenol (pKa about 7.2) is weaker than carbonic acid.

Teacher's Note:
1. Only acids stronger than H2CO3 react with NaHCO3.

Answer: (d) (i) \( \ce{BH3}\), THF; (ii) \( \ce{H2O2/OH-} \)

Explanation:
1. Hydroboration-oxidation gives the primary alcohol 3-methylbutan-1-ol, which gives a primary chloride; \( \ce{CH3ONa} \) then gives the ether by \( S_N2 \) in good yield.
2. The other methods give secondary or tertiary alcohols (after rearrangement), whose chlorides would mainly undergo elimination with methoxide.

Teacher's Note:
1. Primary halides favour Williamson ether synthesis.

Answer: (c) a nucleophile

Explanation:
1. The oxygen of methanol uses its lone pair to attack the protonated carbonyl carbon of the acid.
2. Therefore methanol acts as a nucleophile; the \( ^{18}\)O stays in the ester.

Teacher's Note:
1. Attacks the carbon using a lone pair means nucleophile.

Answer: (c) 2-Fluorobutane with alcoholic KOH gives but-2-ene as the major product

Explanation:
1. Alkyl fluorides give the Hofmann product (the less substituted alkene), so but-1-ene is major.
2. The other three reactions are correct: internal Cannizzaro, cleavage of the aryl alkyl ether, and haloform-type cleavage of chloral.

Teacher's Note:
1. Fluoride is a poor leaving group and follows Hofmann elimination.

Answer: (c) Benzyl alcohol and phenylacetic acid

Explanation:
1. \( \ce{PCl5} \) converts \( \ce{C6H5CH2OH} \) into \( \ce{C6H5CH2Cl} \).
2. Hydrolysis of \( \ce{C6H5CH2CN} \) gives phenylacetic acid, \( \ce{C6H5CH2COOH} \).

Teacher's Note:
1. Benzyl cyanide hydrolyses to phenylacetic acid.

Answer: (c) Butan-2-ol

Explanation:
1. Na forms the alkoxide, which displaces Cl intramolecularly giving propylene oxide (A).
2. The Grignard reagent opens the epoxide at the less hindered carbon: \( \ce{CH3CH(OH)CH2CH3} \) (butan-2-ol).

Teacher's Note:
1. Grignard opens an epoxide at the less hindered carbon.