SolutionsOrganolithiums, Grignard reagents and cuprates
Exercise 1
What product is obtained by treating methyl benzoate with (a) two equivalents of methylmagnesium bromide and (b) one equivalent? How would you prepare acetophenone from benzoic acid with a Grignard reagent?
(a) 2-Phenylpropan-2-ol: the first addition gives acetophenone, more reactive than the ester, which consumes the second equivalent. (b) A mixture of the tertiary alcohol and unreacted ester, with very little ketone, for the same reason. For acetophenone, benzoic acid is converted into the Weinreb amide (for example, via the acid chloride and MeNH(OMe)·HCl) and this is treated with MeMgBr: the chelated intermediate does not release the ketone until hydrolysis.
Exercise 2
Predict the major product of the reaction of cyclohex-2-enone with (a) Ph2CuLi and (b) PhLi, followed by hydrolysis. What would happen if the enolate formed in (a) were treated with methyl iodide before hydrolysis?
(a) 3-Phenylcyclohexanone, by conjugate addition of the soft cuprate nucleophile. (b) 1-Phenylcyclohex-2-en-1-ol, by 1,2-addition of the hard, highly reactive organolithium to the carbonyl. If the lithium enolate from (a) is alkylated with MeI, 2-methyl-3-phenylcyclohexanone is obtained, mainly with the two substituents trans: the electrophile enters from the face opposite the phenyl group.