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NCERT Exemplar · Q55

Q.How will you carry out the following conversions, both starting from aniline (C6H5NH2)?

(i) aniline is to be converted into 3,5-dibromonitrobenzene (a benzene ring with –NO2 at position 1 and –Br at positions 3 and 5);
(ii) aniline is to be converted into 3,5-dibromo-4-iodonitrobenzene (a benzene ring with –NO2 at position 1, –Br at positions 3 and 5, and –I at position 4).
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Both targets share the intermediate 2,6-dibromo-4-nitroaniline, built by using the amino group to place –NO2 (para) and –Br (both ortho). The amino group is then diazotised and either removed (H3PO2 → H) to give the 3,5-dibromo product, or replaced by iodine (KI) to give the 3,5-dibromo-4-iodo product.

Common intermediate – 2,6-dibromo-4-nitroaniline

  1. Aniline + (CH3CO)2O → acetanilide (protect the –NH2).
  2. Acetanilide + HNO3/H2SO4 → p-nitroacetanilide (–NO2 enters para to the acetamido group).
  3. p-Nitroacetanilide + H3O+ → p-nitroaniline (–NH2 at C1, –NO2 at C4).
  4. p-Nitroaniline + 2 Br2 → 2,6-dibromo-4-nitroaniline. The strongly activating –NH2 directs Br to both ortho positions (C2, C6); the para position is blocked by –NO2.

Conversion (i) – to 3,5-dibromonitrobenzene

  1. Diazotise: 2,6-dibromo-4-nitroaniline + NaNO2/HCl, 273–278 K → diazonium salt.
  2. Deaminate: + H3PO2/H2O → –N2+ (C1) is replaced by –H. Renumbering from the nitro group, the two bromines are at positions 3 and 5 → 3,5-dibromonitrobenzene.

Conversion (ii) – to 3,5-dibromo-4-iodonitrobenzene

  1. Diazotise as above. …

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