Nitrobenzene is reduced with Sn/HCl to give A. A is treated with ethanoic anhydride to give B. B is brominated with Br2 in ethanoic acid to give the major monobromo product C. C is hydrolysed with dilute HCl (reflux) and then neutralised with NaOH to give D. D is

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Nitrobenzene is reduced with Sn/HCl to give A. A is treated with ethanoic anhydride to give B. B is brominated with $Br X_{2}$ in ethanoic acid to give the major monobromo product C. C is hydrolysed with dilute HCl (reflux) and then neutralised with NaOH to give D. D is

A2,4,6-tribromoaniline
B3-bromoaniline
C2-bromoaniline
D4-bromoaniline

Answer: D

A is aniline and B is acetanilide. The acetamido group is a moderately activating ortho/para director and, being bulky, gives mainly the para product, 4-bromoacetanilide (C). Hydrolysis removes the acetyl group to give 4-bromoaniline.

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Nitrobenzene is reduced with Sn/HCl to give A. A is treated with ethanoic anhydride to give B. B is brominated with BrX2​ in ethanoic acid to give the major monobromo product C. C is hydrolysed with dilute HCl (reflux) and then neutralised with NaOH to give D. D is

Answer: D

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Nitrobenzene is reduced with Sn/HCl to give A. A is treated with ethanoic anhydride to give B. B is brominated with $Br X_{2}$ in ethanoic acid to give the major monobromo product C. C is hydrolysed with dilute HCl (reflux) and then neutralised with NaOH to give D. D is