Q23 · CSIR NET Chemistry, December 2014
Paper: CSIR NET December 2014 · Subject: Organic Chemistry · Chapter: Reagents, Reaction Mechanisms & Named Reactions · Topic: Reagents, Reaction Mechanisms & Named Reactions – General · Marks: 2 · Difficulty: Easy
The products A and B in the following reaction sequence are 
2. $\mathrm{ICH_2CH_2CH_2Cl}$ $\mathrm{A}\ \xrightarrow[\text{2. }\mathrm{H_3O^{+}}]{\text{1. MeMgBr, }\mathrm{Et_2O}}\ \mathrm{B}$
2. $\mathrm{ICH_2CH_2CH_2Cl}$ $\mathrm{A}\ \xrightarrow[\text{2. }\mathrm{H_3O^{+}}]{\text{1. MeMgBr, }\mathrm{Et_2O}}\ \mathrm{B}$(a)$\mathrm{A}=$ 
$\mathrm{B}=$ 
$\mathrm{B}=$ (b)$\mathrm{A}=$ 
$\mathrm{B}=$ 
$\mathrm{B}=$ (c)$\mathrm{A}=$ 
$\mathrm{B}=$ 
$\mathrm{B}=$ (d)$\mathrm{A}=$ 
$\mathrm{B}=$ 
$\mathrm{B}=$ Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
LDA deprotonates the α′ (C-6) position of the vinylogous ester and the iodide alkylates it (A, chloropropyl retained); MeMgBr adds 1,2 to the ketone and aqueous acid hydrolyses/eliminates the enol ether to the 3-methylcyclohexenone (c).
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