Complete the reaction with suitable reagents and compounds.

(A) - C H_{4}, (B) - C_{6} H_{5} C H O, (C) - C_{6} H_{5} C O O N a, (D) - C H_{4}

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(A) - C H_{3} C l, (B) - C_{6} H_{5} C O O H, (C) - C_{6} H_{5} C O O N a, (D) - a l c . K O H

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(A) - C H_{3} C l, (B) - C_{6} H_{5} C O O H, (C) - C_{6} H_{5} C O O N a, (D) - N a O H + C a O

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(A) - C H_{3} C O C l, (B) - C_{6} H_{5} C H O, (C) - C_{6} H_{5} C O O N a, (D) - A l C l_{3}

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Solution

The correct option is C $(A) - C H_{3} C l, (B) - C_{6} H_{5} C O O H, (C) - C_{6} H_{5} C O O N a, (D) - N a O H + C a O$
The first reaction is Friedel-crafts reaction and the product is toluene, therefore $A$ is $C H_{3} C l$ .

$C_{6} H_{6} C H_{3} C l - --- \to A l {C l}_{3} C_{6} H_{5} - C H_{3}$

Toluene, on oxidation forms benzoic acid $(B)$ .
$C_{6} H_{5} - C H_{3} o x i d a t i o n - ---- \to C_{6} H_{5} - C O O H$

Benzoic acid, on further reaction with $N a O H$ forms sodium benzoate $(C)$ .
$C_{6} H_{5} - C O O H N a O H - --- \to C_{6} H_{5} - C O O N a Sodium benzoate$

To convert sodium benzoate in benzene, soda lime $(D)$ can be used as reagent.
$C_{6} H_{5} - C O O N a N a O H + C a O - ------- \to - {N a}_{2} C O_{3} C_{6} H_{6}$

Hence, $C$ is correct answer.

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