JEE Advanced 2017 · Paper 2 · Chemistry Q17

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ChemistrySingle correct+3Question 35 of 54

Passage

PARAGRAPH 2

The reaction of compound P with $\ce{CH3MgBr}$ (excess) in $\ce{(C2H5)2O}$ followed by addition of $\ce{H2O}$ gives Q. The compound Q on treatment with $\ce{H2SO4}$ at 0°C gives R. The reaction of R with $\ce{CH3COCl}$ in the presence of anhydrous $\ce{AlCl3}$ in $\ce{CH2Cl2}$ followed by treatment with $\ce{H2O}$ produces compound S. [Et in compound P is ethyl group]

P → Q → R → S

P: ethyl 3-(4-tert-butylphenyl)propanoate, (H₃C)₃C-C₆H₄-CH₂CH₂CO₂Et (para), converted in turn to Q, R and S.

The product S is

6-tert-Butyl-1,1-dimethylindan with an acetyl group at C-4

1-(5-tert-Butyl-3-methyl-1H-inden-4-yl)ethanone: an indene with CH₃ on the double bond, acetyl and tert-butyl on the benzene ring

2-tert-Butyl-5-(3-methylbut-2-enyl)acetophenone (acetyl ortho to tert-butyl, prenyl chain para to tert-butyl)

A benzene ring bearing tert-butyl, SO₃H (ortho to tert-butyl), a CH₂CH₂COCH₃ chain (para to tert-butyl) and COCH₃ (ortho to the chain)

Answer and worked solution

Official answer: A

P → Q. Excess $\ce{CH3MgBr}$ adds twice to the ester, giving the tertiary alcohol $\ce{Ar-CH2CH2C(OH)(CH3)2}$ , with Ar = 4-tert-butylphenyl.

Q → R. Cold $\ce{H2SO4}$ gives the tertiary cation, which attacks the ring ortho to the side chain (intramolecular Friedel-Crafts alkylation). This closes a five-membered ring: R is 6-tert-butyl-1,1-dimethylindane.

R → S. Friedel-Crafts acylation with $\ce{CH3COCl/AlCl3}$ goes to the free ring position that is ortho to the ring $\ce{CH2}$ and meta to tert-butyl (C-4). The other positions are hindered by tert-butyl or by the gem-dimethyl group.

S is 4-acetyl-6-tert-butyl-1,1-dimethylindane: structure (A).

(A)

Topics:Aromatic hydrocarbons and electrophilic substitution Alcohols Acid derivatives

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JEE Advanced 2017 · Paper 2, Chemistry question 17