Q.Which of the following compounds on reaction with methylmagnesium bromide will give a tertiary alcohol?

Step 1. Methylmagnesium bromide adding to an ALDEHYDE (benzaldehyde, acetaldehyde) gives a SECONDARY alcohol (one new C-C bond, but the carbonyl carbon already carried only one alkyl/aryl group and one H).

Step 2. Adding to a carboxylic ACID (propanoic acid) merely deprotonates the acid (Grignard reagents are destroyed by any acidic proton) -- no addition to the carbonyl occurs, so no alcohol at all forms from one equivalent.

Step 3. Adding to an ESTER (methyl propanoate) proceeds in two stages: the first equivalent of CH3MgBr displaces methoxide to give a ketone (butan-2-one) in situ; a SECOND equivalent of the Grignard then adds to that ketone's carbonyl carbon.

Step 4. The tetrahedral alkoxide from that second addition, on acid hydrolysis, gives a TERTIARY alcohol (2-methylbutan-2-ol), since the carbonyl carbon now carries three carbon substituents.

Count how many carbon substituents will sit on the resulting alcohol's carbinol carbon: an ester reacting with excess Grignard is the only listed option that builds up to three.

• Assuming any Grignard addition to a carbonyl compound gives a tertiary alcohol -- only an ester (or an acid chloride/anhydride) reacting with EXCESS Grignard does; aldehydes give secondary alcohols, formaldehyde gives primary.
• Forgetting that a carboxylic acid quenches the Grignard reagent as a base before any addition can occur.