JEE Challenger
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ChemMedAdv 2009 • Paper 1

Identification of Carbonyl Compound P in Multistep Synthesis

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 carbonyl compound P\mathbf{P} is

Options

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

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesGrignard Reagent

Step-by-Step Solution

To determine the structure of the carbonyl compound P\mathbf{P}, we analyze the reaction steps:

  1. Iodoform Test:
    A positive iodoform test indicates the presence of a methyl carbonyl group, −COCH3-\mathrm{COCH_3} (or a −CH(OH)CH3-\mathrm{CH(OH)CH_3} group).

    • In option (A): PhCH=CH−COCH3\mathrm{PhCH=CH-COCH_3} (contains −COCH3-\mathrm{COCH_3})
    • In option (B): PhCH=C(CH3)−COCH3\mathrm{PhCH=C(CH_3)-COCH_3} (contains −COCH3-\mathrm{COCH_3})
    • In option (C): PhCH=CH−COCH2CH3\mathrm{PhCH=CH-COCH_2CH_3} (does not contain −COCH3-\mathrm{COCH_3}, negative test)
    • In option (D): Ph−CO−CH=CH−CH3\mathrm{Ph-CO-CH=CH-CH_3} (does not contain −COCH3-\mathrm{COCH_3}, negative test)

    Thus, options (C) and (D) are eliminated.

  2. Reaction with MeMgBr\mathrm{MeMgBr} and Dehydration:
    Starting with compound P\mathbf{P} from option (B), PhCH=C(CH3)−COCH3\mathrm{PhCH=C(CH_3)-COCH_3}:

    • Nucleophilic addition of MeMgBr\mathrm{MeMgBr} to the carbonyl carbon gives a tertiary alcohol:
      PhCH=C(CH3)−COCH3→MeMgBrPhCH=C(CH3)−C(OH)(CH3)2\mathrm{PhCH=C(CH_3)-COCH_3} \xrightarrow{\mathrm{MeMgBr}} \mathrm{PhCH=C(CH_3)-C(OH)(CH_3)_2}
    • Upon heating with acid (H2SO4,Δ\mathrm{H_2SO_4, \Delta}), protonation of the hydroxyl group followed by loss of water yields a resonance-stabilized allylic carbocation:
      PhCH=C(CH3)−C+(CH3)2⟷Ph−C+H−C(CH3)=C(CH3)2\mathrm{PhCH=C(CH_3)-\overset{+}{C}(CH_3)_2} \longleftrightarrow \mathrm{Ph-\overset{+}{C}H-C(CH_3)=C(CH_3)_2}
    • The electrophilic carbocation undergoes an intramolecular Friedel-Crafts alkylation with the aromatic ring to form a five-membered indane core:
      The ortho-position of the benzene ring attacks the tertiary carbocation center, followed by deprotonation to yield an indene derivative Q\mathbf{Q}:
      Q=1,1,2-trimethyl-1H-indene\mathbf{Q} = \text{1,1,2-trimethyl-1}H\text{-indene}
      which contains a gem-dimethyl group (−C(CH3)2−\mathrm{-C(CH_3)_2-}) and a methyl-substituted double bond.

  3. Ozonolysis to R\mathbf{R}:

    • Reductive ozonolysis (O3/Zn,H2O\mathrm{O_3 / Zn, H_2O}) of the double bond in Q\mathbf{Q} cleaves the C=C\mathrm{C=C} bond, giving the dicarbonyl compound R\mathbf{R}:
      R=2−(2-methyl-1-oxopropan-2-yl)benzaldehyde (containing both −CHO and −C(CH3)2COCH3 groups)\mathbf{R} = 2-(2\text{-methyl-1-oxopropan-2-yl})benzaldehyde \text{ (containing both } -\mathrm{CHO} \text{ and } -\mathrm{C(CH_3)_2COCH_3} \text{ groups)}
      This is consistent with the subsequent intramolecular aldol condensation leading to S\mathbf{S}.

Therefore, the structure of carbonyl compound P\mathbf{P} corresponds to (B).

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