The major product of the following reaction is : \[ \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCO}_{2} \mathrm{CH}_{3} \xrightarrow{\mathrm{LiAlH}_{4}} \]

(B) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{OH}$

(A) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$

(C) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}$

(D) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

The major product of the following reaction is :

CH_{3} CH = CHCO_{2} CH_{3} LiAlH_{4}

(A) $CH_{3} CH_{2} CH_{2} CHO$
(B) $CH_{3} CH = CHCH_{2} OH$ ✓ Correct
(C) $CH_{3} CH_{2} CH_{2} CO_{2} CH_{3}$
(D) $CH_{3} CH_{2} CH_{2} CH_{2} OH$

Answer: (B)

Solution

LiAlH X_{4}

is a hydride donor that reduces the ester carbonyl but leaves an isolated carbon-carbon double bond untouched. So

CH X_{3} CH = CHCO X_{2} CH X_{3}

loses only its

- CO X_{2} CH X_{3}

group, which becomes

- CH X_{2} OH

, and the product is the unsaturated alcohol

CH X_{3} CH = CHCH X_{2} OH

with its double bond intact.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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