A compound $^{'} A^{'}$ formula of $C_{3} H_{6} C l_{2}$ on reaction with alkali can give $^{'} B^{'}$ of formula $C_{3} H_{6} O$ or $^{'} C^{'}$ of formula $C_{3} H_{4} .^{'} B^{'}$ on oxidation gave a compound of the formula $C_{3} H_{6} O_{2} .^{'} C^{'}$ with dilute $H_{2} S O_{4}$ containing $H g^{2 +}$ ion gave $^{'} D^{'}$ of formula $C_{3} H_{6} O$ , which with bromine and alkali gave the sodium salt of $C_{2} H_{4} O_{2}$ . Then $^{'} A^{'}$ is :

C H_{3} C H_{2} C H C l_{2}

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C H_{3} C C l_{2} C H_{3}

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C H_{2} C l C H_{2} C H_{2} C l

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C H_{3} C H (C l) C H_{2} C l

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Solution

The correct option is A $C H_{3} C H_{2} C H C l_{2}$

The compound A is 1,1-dichloropropane $C H_{3} C H_{2} C H C l_{2}$ .
It reacts with alkali to form propionaldehyde (via gem diol) or propyne (via dehydrochlorination).

Propionaldehyde on oxidation gives propionic acid. Propyne on treatment with sulphuric acid in presence of mecurous ions give acetone.

Acetone is a methyl ketone and on treatment with bromine in presence of alkali undergoes haloform reaction to form sodium salt of acetic acid..

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A compound $A$ of formula $C_{3} H_{6} C l_{2}$ on reaction with alkali can give $B$ of formula $C_{3} H_{6} O$ or $C$ of formula $C_{3} H_{4}$ . $B$ on oxidation gave a compound of the formula $C_{3} H_{6} O_{2}$ . $C$ with dilute $H_{2} S O_{4}$ containing $H g^{2 +}$ ion gave $D$ of formula $C_{3} H_{6} O$ , which with bromine and $N a O H$ gave the sodium salt of $C_{2} H_{4} O_{2}$ . Then $A$ is: