Question

How could you prepare the following ketones by reaction of an acid chloride with a lithium diorganocopper reagent?

Short answer

Ketones can be prepared by the reaction of acid chlorides with Gilman reagent that is organometal (Copper and lithium) compounds. The general formula of this reagent is ${\mathbf{R}}_{\mathbf{2}}\mathbf{CuLi}$.

Step-by-step solution

Formation of ketones

Ketones can be prepared by the reaction of acid chlorides with Gilman reagent that is organometal (Copper and lithium) compounds. The general formula of this reagent is ${\mathbf{R}}_{\mathbf{2}}\mathbf{CuLi}$.

Subpart (a) 

The structure given is 2-methyl-1-phenylpropan-1-one. It can be prepared by using

Benzoyl chloride and ${\left({\left({\mathrm{CH}}_3\right)}_2\mathrm{CH}\right)}_2\mathrm{CuLi}$. The chemical equation is as shown below:

Subpart (b)

The ketone given is hex-1-ene-3-one. By using Gilman reagent it can be prepared by the following way: