JEE Challenger
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ChemHardAdv 2017 • Paper 2

Identify Final Product S in Sequential Reaction of Ethyl Ester

Comprehension Passage

Reaction of compound P with excess CH3MgBr\text{CH}_3\text{MgBr} in (C2H5)2O(\text{C}_2\text{H}_5)_2\text{O} followed by addition of H2O\text{H}_2\text{O} yields Q. Treating compound Q with H2SO4\text{H}_2\text{SO}_4 at 0∘C0^\circ\text{C} forms R. Reacting R with CH3COCl\text{CH}_3\text{COCl} in the presence of anhydrous AlCl3\text{AlCl}_3 in CH2Cl2\text{CH}_2\text{Cl}_2, followed by treatment with H2O\text{H}_2\text{O}, produces compound S. [Here, Et\text{Et} in compound P represents an ethyl group]

Determine the structure corresponding to product S:

Question Diagram 1

Options

A
Option A
Correct
B
Option B
C
Option C
D
Option D

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsGrignard ReagentElectrophilic Aromatic Substitution

Step-by-Step Solution

To determine the structure of the final product S, let us trace the sequential chemical reactions step-by-step:

Step 1: Conversion of Compound P to Q (Grignard Reaction)

  • Compound P is ethyl 3-(4-(tert-butyl)phenyl)propanoate, which contains an ester functional group (−CO2Et-\text{CO}_2\text{Et}).
  • Reaction with excess methylmagnesium bromide (CH3MgBr\text{CH}_3\text{MgBr}) followed by hydrolysis (H2O\text{H}_2\text{O}) results in nucleophilic addition of two methyl groups to the carbonyl carbon of the ester, yielding a tertiary alcohol.
  • Thus, compound Q is 4-(4-(tert-butyl)phenyl)-2-methylbutan-2-ol: Ar−CH2−CH2−CO2Et→2. H2O1. excess CH3MgBrAr−CH2−CH2−C(OH)(CH3)2\text{Ar}-\text{CH}_2-\text{CH}_2-\text{CO}_2\text{Et} \xrightarrow[2.\ \text{H}_2\text{O}]{1.\ \text{excess }\text{CH}_3\text{MgBr}} \text{Ar}-\text{CH}_2-\text{CH}_2-\text{C}(\text{OH})(\text{CH}_3)_2

Step 2: Conversion of Compound Q to R (Intramolecular Friedel-Crafts Alkylation)

  • Treatment of tertiary alcohol Q with concentrated sulfuric acid (H2SO4\text{H}_2\text{SO}_4) at 0∘C0^\circ\text{C} results in the protonation of the hydroxyl group and elimination of a water molecule to generate a stable tertiary carbocation: Ar−CH2−CH2−C(OH)(CH3)2→H+Ar−CH2−CH2−C+(CH3)2+H2O\text{Ar}-\text{CH}_2-\text{CH}_2-\text{C}(\text{OH})(\text{CH}_3)_2 \xrightarrow{\text{H}^+} \text{Ar}-\text{CH}_2-\text{CH}_2-\text{C}^+(\text{CH}_3)_2 + \text{H}_2\text{O}
  • This tertiary carbocation undergoes an intramolecular Friedel-Crafts alkylation attacking the benzene ring at the position ortho to the alkyl chain, giving a 5-membered fused ring system (an indane derivative).
  • Thus, compound R is 6-(tert-butyl)-1,1-dimethylindane.

Step 3: Conversion of Compound R to S (Friedel-Crafts Acylation)

  • Reaction of R with acetyl chloride (CH3COCl\text{CH}_3\text{COCl}) in the presence of anhydrous AlCl3\text{AlCl}_3 causes electrophilic aromatic substitution (Friedel-Crafts acylation).
  • The aromatic ring has three open positions for electrophilic attack: C4, C5, and C7.
    • Positions C5 and C7 are sterically hindered due to the adjacent bulky tert-butyl group at C6 and the quaternary carbon at C1.
    • Position C4 is ortho to the less hindered −CH2−-\text{CH}_2- carbon (C3) of the fused ring and meta to the tert-butyl group.
  • Therefore, electrophilic substitution by the acetylium ion (CH3CO+\text{CH}_3\text{CO}^+) occurs selectively at position C4 to minimize steric repulsion.

Hence, the correct structure corresponding to product S is given in Option A.

Correct Option: [A]

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