Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.
The compound R is
(A)(see structure)✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)
Answer: (A)
Solution
Compound O (o-diethylbenzene) is oxidized by KMnO4/H+ to phthalic acid (P). Heating with ammonia converts it to phthalimide (Q). The Hofmann bromamide degradation (Br2/NaOH) converts the imide to o-phenylenediamine (R), which is structure (A).
JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option