Grignard reagent and acyl chloride react to give ketone as the product. But, when excess Grignard reagent is available tertiary alcohol is given as the product when water is added to reactant mixture at final stage.
First, add excess grignard reagent (CH3MgBr) to the CH3COCl to react with each other. Then add water to the solution to produce tertiary alcohol.
As first product, propanone (ketone) is given and then propanone again reacts with CH3MgBr and water to give 2-methylprop-2-ol as the final product.
One acyl chloride molecule reacts with two grignard reagents to produce one tertiary alcohol molecule.
You will see, two methyl groups are connected to the acetyl chloride molecule when tertiary alcohol is given. So more CH3MgBr is required.
As methyl magnesium bromide, ethyl magnesium bromide also reacts with acetyl chloride to give a tertiary alcohol.
3-methyl-pent-3-ol is given as the product. You have to add excess CH3CH2MgBr to get the tertiary alcohol.
Questions
Methanol can be used to prepare methyl magnesium bromide (CH3MgBr) as below.
CH3OH is treated with concentrated HBr to produce CH3MgBr. Then CH3Br is treated with Mg an dry ether to prepare methyl magnesium bromide (CH3MgBr).
Ethanol is oxidized to ethanoic acid by strong oxidizing agents. Then ethanoic acid is treated with PCl33, to prepare CH3COCl.
Acetyl chloride (CH3COCl) reacts with grignard reagent and produce tertiary alcohol when H2O is added in the final stage. If we denote grignard reagent as, RMgBr. Product will be CH3C(R)2OH.