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