Carboxylic Acids and their Derivatives 1 Question 1
2. In the reaction,
$$ \mathrm{CH}{3} \mathrm{COOH} \xrightarrow{\mathrm{LiAlH}{4}} A \xrightarrow{\mathrm{PCl}_{5}} B \xrightarrow{\text { Alc. } \mathrm{KOH}} C $$
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) $\mathrm{LiAlH}_{4}$ causes reduction
(ii) $\mathrm{PCl}_{5}$ causes chlorination
(iii) Alc. $\mathrm{KOH}$ causes elimination reaction
$\mathrm{CH}{3} \mathrm{COOH} \xrightarrow{\mathrm{LiAlH}{4}} \mathrm{CH}{3} \mathrm{CH}{2} \mathrm{OH}$
(A)
$$ \xrightarrow{\mathrm{PCl}{5}} \underset{(B)}{\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{Cl}} \xrightarrow[-\mathrm{HCl}]{\mathrm{Alc. \textrm {KOH }}} \underset{\substack{(C) \ \text { Ethylene }}}{\mathrm{CH}{2}}=\mathrm{CH}_{2} $$
