JEE Challenger
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Identification of Product S in Organic Reaction Sequence

Comprehension Passage

Paragraph for Question Nos. 16 to 18

A carbonyl compound P\mathbf{P}, which gives positive iodoform test, undergoes reaction with MeMgBr\mathrm{MeMgBr} followed by dehydration to give an olefin Q\mathbf{Q}. Ozonolysis of Q\mathbf{Q} leads to a dicarbonyl compound R\mathbf{R}, which undergoes intramolecular aldol reaction to give predominantly S\mathbf{S}.

Passage Diagram 0

The structure of the product S\mathbf{S} is

Options

A
Option A Diagram 1
B
Option B Diagram 1
Correct
C
Option C Diagram 1
D
Option D Diagram 1

Step-by-Step Solution

To determine the structure of the product S\mathbf{S}, we analyze the reaction sequence and the mechanism involved:

  1. Identification of the Key Intermediate:
    The reaction pathway proceeds via an ortho-disubstituted benzene intermediate bearing a formyl group and a methyl ketone group separated by a quaternary carbon with two methyl groups:
    o-OHC−C6H4−C(CH3)2COCH3o\text{-}\mathrm{OHC-C_6H_4-C(CH_3)_2COCH_3}
    (This keto-aldehyde is typically generated via the reductive ozonolysis of a substituted indene derivative, such as 1,1,2-trimethyl-1H1H-indene).

  2. Intramolecular Aldol Condensation:
    When this intermediate is treated with a base (OH−\mathrm{OH^-}):

    • Enolization occurs exclusively at the methyl group of the acetyl moiety, as the adjacent benzylic carbon is quaternary (−C(CH3)2−-\mathrm{C(CH_3)_2}-) and possesses no α\alpha-hydrogens:
      −C(CH3)2COCH3→OH−−C(CH3)2COCH2−\mathrm{-C(CH_3)_2COCH_3 \xrightarrow{OH^-} -C(CH_3)_2COCH_2^-}
    • The enolate nucleophile undergoes an intramolecular nucleophilic addition to the more electrophilic benzylic aldehyde carbonyl group (−CHO-CHO), forming a stable six-membered ring β\beta-hydroxy ketone intermediate.
  3. Dehydration to Product S\mathbf{S}:

    • Under the reaction conditions, the β\beta-hydroxy ketone undergoes base-catalyzed dehydration (−H2O-\mathrm{H_2O}) to form an α,β\alpha,\beta-unsaturated ketone.
    • The resulting double bond is formed between the benzylic carbon and the α\alpha-carbon, which establishes an extended conjugated system involving the aromatic benzene ring, the carbon-carbon double bond, and the carbonyl group:
      Ar−CH=CH−C(=O)−C(CH3)2−\mathrm{Ar-CH=CH-C(=O)-C(CH_3)_2-}
  4. Conclusion:
    The product obtained is 4,4-dimethylnaphthalen-2(4H)-one, where:

    • The carbonyl group (C=OC=\mathrm{O}) is adjacent to the gem-dimethyl group (−C(CH3)2−-\mathrm{C(CH_3)_2}-).
    • The double bond is in conjugation with both the carbonyl group and the benzene ring.

This uniquely corresponds to the structure shown in option (B).

Correct Answer: B