The sequence from the following that would result in giving predominantly 3,4,5-Tribromoaniline is :
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Answer
(3) [figure]
Solution
(3): 4-nitroaniline; -NH2 strongly activates and directs Br to both its ortho positions; then -NH2 is replaced by Br through the diazonium salt (Sandmeyer reaction) and -NO2 is reduced to -NH2:
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The product is 3,4,5-tribromoaniline.
Other sequences:
(1) Bromination of nitrobenzene gives m-bromonitrobenzene 3-bromoaniline after reduction.
(2) Bromobenzene does not react with NH3 under ordinary conditions (C–Br bond of haloarene has partial double bond character).
(4) Aniline with bromine water gives 2,4,6-tribromoaniline.
Answer: (3)