IIT-JEE 2009 · Paper 1 · Chemistry Q18

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ChemistrySingle correct+4 / -1Question 18 of 60

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Paragraph for Question Nos. 16 to 18

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

$\mathbf P\ \xrightarrow[\text{2. }\ce{H+, H2O}\ \ \text{3. }\ce{H2SO4},\ \Delta]{\text{1. MeMgBr}}\ \mathbf Q\ \xrightarrow[\text{2. Zn, }\ce{H2O}]{\text{1. }\ce{O3}}\ \mathbf R\ \xrightarrow[\text{2. }\Delta]{\text{1. }\ce{OH-}}\ \mathbf S$

The structure of the product S is

4-Methylnaphthalen-1(4H)-one: a benzo-fused cyclohexadienone with C=O and a CH(Me) para to it.

1,1-Dimethylnaphthalen-2(1H)-one: the ring C=O is flanked by a C(Me)₂ attached to the benzene ring and a CH=CH.

4,4-Dimethylnaphthalen-1(4H)-one: C=O on the benzo carbon and C(Me)₂ para to it, joined by CH=CH.

2-Methylnaphthalen-1(4H)-one: C=O on the benzo carbon, Me on the adjacent ring C=C, CH₂ at C4.

Answer and worked solution

Official answer: B

Intramolecular aldol. The only enolisable position is the methyl of the ketone ( $\ce{CMe2}$ has no H). Its enolate attacks the aromatic CHO, closing a six-membered ring; dehydration gives a C=C conjugated with both the ketone and the ring:
$\text{S}=1,1\text{-dimethylnaphthalen-2(1}H\text{)-one}.$

(B)

Topics:Aldol and related condensations

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IIT-JEE 2009 · Paper 1, Chemistry question 18