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JEE Main 2025 Apr 3 Shift 2, Chemistry Q17: Preparation of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

JEE Main2025Apr 3, Shift 2Chemistry
Q.

The sequence from the following that would result in giving predominantly 3, 4, 5 Tribromoaniline is :

  1. A

    1

  2. B

    2

  3. C

    3

  4. D

    4

Solution

The target 3,4,5-tribromoaniline has NH at C-1 and three Br atoms on the opposite side of the ring. To get this pattern, the amino group must come in after the bromines are placed, and a blocking/directing strategy is needed because NH otherwise directs ortho/para to itself.

Starting from p-nitroaniline (option 3):

Step 1: Br/acetic acid (excess) brominates the ring at the two positions ortho to the strongly activating NH that are also meta to the NO. This installs Br at C-3 and C-5 (relative to NH at C-1).

Step 2: Diazotisation of the NH with NaNO/HCl, followed by Sandmeyer with CuBr, converts the amino group at C-1 into Br at C-1. The ring now has Br at the position originally occupied by NH, plus the two earlier Brs at the 3- and 5-positions (i.e. flanking the NO), giving 3,4,5-tribromonitrobenzene relative to the NO numbering.

Step 3: Sn/HCl reduces the NO to NH, giving 3,4,5-tribromoaniline.

The other sequences either give the wrong substitution pattern (2,4,6-tribromoaniline from option 4) or do not deliver three bromines flanking and para to the amine.

Option (3) is the correct sequence.

Concept behind this question

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