The equivalent weight of nitrobenzene in the reaction: $C_{6} H_{5} N O_{2} \to C_{6} H_{5} N H_{2}$ is:

M

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\frac{M}{2}

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\frac{M}{4}

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\frac{M}{6}

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Solution

The correct option is D $\frac{M}{6}$
The reduction of one molecule of nitrobenzene to aniline requires $6$ hydrogen atoms.

The reaction is as follows:

$C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$

Thus, the equivalent weight of nitrobenzene is $\frac{M}{6}$ .

Here, $M$ represents the molecular weight of nitrobenzene.

Hence, the correct option is $D$

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