Q15 · 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: Medium
The products A and B in the following reaction sequence are: 
LDA 
A $\xrightarrow[\mathrm{CuCl,\ DMF\text{-}H_2O}]{\mathrm{PdCl_2,\ O_2}}$ B $\xrightarrow[2.\ \mathrm{H_3O^+}]{1.\ \mathrm{NaOEt}}$ 
LDA A $\xrightarrow[\mathrm{CuCl,\ DMF\text{-}H_2O}]{\mathrm{PdCl_2,\ O_2}}$ B $\xrightarrow[2.\ \mathrm{H_3O^+}]{1.\ \mathrm{NaOEt}}$ (a)A = 
B = 
B = (b)A = 
B = 
B = (c)A = 
B = 
B = (d)A = 
B = 
B = Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
LDA/allyl bromide gives 2-allylcyclohexanone (A); Wacker oxidation converts the terminal alkene into the methyl ketone (B); NaOEt intramolecular aldol condensation gives the bicyclic enone (a).
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