Above conversion can be achieved by :

L i A l H_{4}

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N a B H_{4}

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H_{2} / N i

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C r O_{3}

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Solution

The correct option is A $N a B H_{4}$

(Refer to Image)
$N a B H_{4}$ is used as reducing agents to convert aldehydes, ketones into alcohols. It does not affect $- N O_{2}$ groups. While, $L i A l H_{4}$ converts $- N O_{2}$ into $- N H_{2}$ which is not favourable here. $H_{2} / N i$ also converts $- N O_{2}$ to $N H_{2}$ and $C r O_{3}$ cannot be used because, it is an oxidizing agent.

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C H_{3} C l \to C H_{4}

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