Consider the following reactions,
The compound [P] is :
The compound [P] is :
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The Correct Option is B
Solution and Explanation
[P] + \(\mathrm{Br_2/H_2O}\) gives \(\mathrm{C_7H_6NBr_3}\). The parent \(\mathrm{C_7H_9N}\) (toluidine) has lost 3 H and gained 3 Br, i.e. 3 ring positions are substituted. \(\mathrm{NH_2}\) activates only its two ortho and one para positions.
o-Toluidine (option 1): free activated positions are only C4 and C6, so only 2 Br. p-Toluidine (option 4): free activated positions are C2 and C6, so only 2 Br.
m-Toluidine (option 2): positions 2, 4 and 6 are all free and activated, so 2,4,6-tribromo-3-methylaniline \(\mathrm{C_7H_6NBr_3}\) forms.
Hence [P] = m-toluidine.
Answer: (2) m-toluidine (3-methylaniline)
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