Q7 · CSIR-NET Chemistry, December 2015

Paper: CSIR-NET December 2015 · Subject: Organic Chemistry · Chapter: Reagents, Reaction Mechanisms & Named Reactions · Topic: Reagents, Reaction Mechanisms & Named Reactions – General · Marks: 2 · Difficulty: Medium

The structures of A and B in the following reaction sequence are Question structure 1 of 5, ABC26OR0265

(a)

(b)

B =

(c)

(d)

B =

Answer

Answer: C ✓ checked by 4AB · confidence medium

The source book printed B; on checking, C is correct — see the explanation.

Restores the printed key C. Cope elimination is a syn (Ei) elimination. In the menthyl N-oxide (left), cis β-H atoms are available at both C-2 and C-6, and the reported product mix favours 2-menthene (about 64:36 over 3-menthene). In the neomenthyl N-oxide (right), the C-2 H is trans to N–O, so only 2-menthene can form. A = B = 2-menthene. The earlier override to B assumed the trisubstituted 3-menthene is major for A, which conflicts with the known Cope selectivity.

Explanation

Cope elimination is a syn (Ei) elimination through a five-membered cyclic transition state, so it needs a β-H cis to the N-oxide. Left (menthyl N-oxide: N and i-Pr trans): cis β-H atoms exist at both C-2 and C-6. As in other Cope eliminations, the less hindered CH₂ hydrogen is removed preferentially, so 2-menthene (alkene away from the isopropyl carbon) is the major product A (about 64:36 over 3-menthene). Right (neomenthyl N-oxide: N and i-Pr cis): the C-2 H is trans to N–O, so elimination can occur only towards C-6, giving 2-menthene exclusively as B. Hence A = B = 2-menthene. Answer: (c).

Study loop for Reagents, Reaction Mechanisms & Named Reactions

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