What is obtained when nitrobenzene is treated sequentially with (i) ${N H}_{4} C l / Z n$ dust and (ii) $H_{2} {S O}_{4} / {N a}_{2} {C r}_{2} O_{7}$ ?

Benzene

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Nitrosobenzene

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m-Chlorobenzene

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p-Chlorobenzene

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Solution

The correct option is C Nitrosobenzene
When Nitrobenzene first reacts with $N H_{4} C l / Z n$ it forms phenyl hydroxylamine.
When this reacts with $H_{2} S O_{4} / N a_{2} C r_{2} O_{7}$ it gives Nitroso benzene as a product.

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Similar questions

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

N H_{4} C l

H C H O + K_{2} {C r}_{2} O_{7} + H_{2} {S O}_{4} ⟶ ?

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