Question

How could each of the following compounds be prepared from cyclohexanone?

a.

b.

c.

Short answer

a.

b.

c.

Step-by-step solution

Step-by-Step SolutionStep 1: Alkylation of the α-carbon in the presence of a strong base (LDA)

The process of alkylation of the $\alpha$-carbon is done by removing protons present on the$\alpha$-carbon using a strong base (LDA) and THF solvent. A strong base forms an enolate ion, and then the addition of alkyl halide to the enolate ion takes place.

Compound (a) preparation from cyclohexanone

a) The materials for preparing the given compound are cyclohexanone (symmetrical ketone) and 1-bromopropane (alkyl halide). An enolate ion is formed when a cyclohexanone is treated with a strong base LDA and THF solvent. After this, an alkyl halide is added to the enolate ion, where the alkyl group of 1-bromopropane is added to the $\mathrm{\alpha }$-carbon of the cyclohexanone compound.

Formation of alkylated $\mathrm{\alpha }$-carbon compound

Compound (b) preparation from cyclohexanone

(b) The given compound is prepared when 3-bromopropene is treated with cyclohexanone (symmetrical ketone). A strong base LDA and THF solvent are added to cyclohexanone, making an enolate ion. In the next step, 3-bromopropene is added to the enolate ion, and the alkyl group of 3-bromopropene is added to the $\alpha$-carbon of the enolate ion.

Formation of alkylated $\mathrm{\alpha }$-carbon compound

Compound (c) preparation from cyclohexanone

(c) The given compound is prepared when benzyl bromide is treated with cyclohexanone (symmetrical ketone). A strong base LDA and THF solvent are added to cyclohexanone, making an enolate ion. After this, benzyl bromide is added to enolate ion, where the alkyl group of benzyl bromide is added to the $\alpha$-carbon of the cyclohexanone compound.

Formation of alkylated $\alpha$-carbon compound