Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)
S→ major product(s): 1) Sn / HCl 2) Br2/H2O (excess) 3) NaNO2, HCl / 273-278 K 4) H3PO2
(A)(see structure)
(B)(see structure)
(C)(see structure)
(D)(see structure)✓ Correct
Answer: (D)
Solution
Aniline reacts with mixed acid to give m-nitroaniline (P, 51%), p-nitroaniline (Q, 47%), and o-nitroaniline (R, 2%). The reaction sequence involves: (1) acetylation to protect the amino group, (2) bromination at ortho positions, (3) deprotection, (4) diazotization, (5) reduction to remove the diazo group. The major product P (m-nitroaniline) undergoes this sequence. After Sn/HCl reduction of the nitro group to amino, bromination in excess Br2/H2O gives tribromination at positions 2, 4, and 6. Diazotization followed by H3PO2 removes the amino group, giving 1,3,5-tribromobenzene (option D).
JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options