Carboxylic Acids and their Derivatives - Result Question 16

####2. In the reaction,

$$ CH _3 COOH \xrightarrow{LiAlH _4} A \xrightarrow{PCl _5} B \xrightarrow{\text { Alc. } KOH} C $$

The product $C$ is

(2014 Main)

(a) acetaldehyde

(b) acetylene

(c) ethylene

(d) acetyl chloride

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Answer:

Correct Answer: 2. (b)

Solution:

  1. This problem is based on successive reduction, chlorination and elimination reaction. To solve such problem, use the function of the given reagents.

(i) $LiAlH _4$ causes reduction

(ii) $PCl _5$ causes chlorination

(iii) Alc. $KOH$ causes elimination reaction

$CH _3 COOH \xrightarrow{LiAlH _4} CH _3 CH _2 OH$

(A)

$$ \xrightarrow{PCl _5} \underset{(B)}{CH _3 CH _2 Cl} \xrightarrow[-HCl]{\mathrm{Alc. \textrm {KOH }}} \underset{\substack{(C) \ \text { Ethylene }}}{CH _2}=CH _2 $$



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