Discussion
Nitrobenzene is treated with Sn / HCl to give a compound (X) which on treatment with NaNO2 and HCl at 278 K gives compound (Y). When the compound (Y) is treated with Cu / HBr , compound 'Z' is obtained. The compound 'Z' is
Answer: D
Compound Z is Bromobenzene.
The sequence is three standard steps:
- Sn and HCl reduce nitrobenzene to aniline: X is C6H5NH2.
- NaNO2 and HCl at 273-278 K diazotise aniline: Y is benzenediazonium chloride, C6H5N2+Cl-.
- Copper powder with HBr replaces the diazonium group by bromine with loss of nitrogen (the Gattermann reaction): Z is C6H5Br.
Benzene diazonium bromide is not formed; the diazonium salt is the intermediate Y, and it is decomposed in the last step. Phenol would come from warming Y with water, benzene from reduction with H3PO2, and chlorobenzene from Cu/HCl rather than Cu/HBr.
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