What is the principal product(s) formed when 1 mol of methylmagnesium iodide reacts with 1 mol p-hydroxyacetophenone?
Methylmagnesium iodide (MeMgI) is basically the same thing as a Grignard reagent, just with an iodide instead of a bromide. It can act as either a base or a nucleophile. p-Hydroxyacetophenone is a benzene ring with an acetyl (CH3CO) group on carbon 1 and an OH group on carbon 4. It can act as either an electrophile due to the carbonyl on the acetyl group or an acid due to the proton of the hydroxyl group. When the two are reacted in equimolar fashion, you will get a mixture of two products. One will be the deprotonation of the hydroxyl by the MeMgI to make the MgI salt of the hydroxy anion (with methane CH4 as the byproduct). The other product will be the addition of the methyl group to the carbon of the carbonyl group to make the tetrahedral tertiary alcohol product. The deprotonated product will probably predominate, but both of these products will form.