JEE Advanced 2016 · Paper 2 · Chemistry Q35

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ChemistrySingle correct+3Question 35 of 54

Passage

PARAGRAPH 2

Treatment of compound O with $\ce{KMnO4/H+}$ gave P, which on heating with ammonia gave Q. The compound Q on treatment with $\ce{Br2/NaOH}$ 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.

Compound O: a benzene ring carrying two adjacent n-propyl side chains (1,2-dipropylbenzene).

The compound R is

Benzene-1,2-diamine (o-phenylenediamine): a benzene ring with two adjacent NH₂ groups.

Phthaloyl dibromide: a benzene ring with two adjacent C(=O)Br groups.

N,N′-Dibromophthalamide: a benzene ring with two adjacent C(=O)NHBr groups.

N-Bromophthalimide: phthalimide with Br on the ring nitrogen.

Answer and worked solution

Official answer: A

Sequence.

O (1,2-dipropylbenzene) is oxidised by KMnO₄/H⁺ to P, phthalic acid.
Heating with ammonia gives Q, phthalamide.
Hofmann bromamide degradation ( $\ce{Br2/NaOH}$ ) of both amide groups gives R, benzene-1,2-diamine.

(Strong heating of phthalamide gives S, phthalimide; see Q36.)

(A)

Topics:Amines Acid derivatives

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JEE Advanced 2016 · Paper 2, Chemistry question 35