Identify the major organic product formed in the following reaction.

The reactant is a molecule containing both a ketone group ($-C(=O)-$ ) and a cyano group ($-C\equiv N$).

Grignard reagents ($CH_3MgBr$) are strong nucleophiles and will react with both electrophilic centers: the carbonyl carbon and the cyano carbon.

Reaction at the ketone group: The methyl nucleophile ($CH_3^-$) attacks the carbonyl carbon, opening the $C=O$ double bond to form a tetrahedral alkoxide intermediate. Upon acidic workup ($H_3O^+$), this alkoxide is protonated to form a tertiary alcohol. Since the original ketone has an ethyl group and a propyl chain attached, the addition of a methyl group results in a tertiary alcohol with ethyl, methyl, and the rest of the chain attached to the carbon.

Reaction at the cyano group: The methyl nucleophile ($CH_3^-$) attacks the electrophilic carbon of the nitrile group, forming an imine anion intermediate ($-C(CH_3)=N^-MgBr^+$). Upon acidic hydrolysis ($H_3O^+$), this imine intermediate is converted into a ketone. The resulting ketone will have a methyl group attached to the former cyano carbon, forming a methyl ketone.

Since the Grignard reagent is in excess, both functional groups will react completely.

The ketone on the right side of the molecule becomes a tertiary alcohol (with a new methyl group added).

The cyano group on the left side of the molecule is converted to a methyl ketone.

Therefore, the final product will have a tertiary alcohol at one end and a methyl ketone at the other end. Looking at the options, structure (3) correctly shows the tertiary alcohol where the original ketone was (with a new methyl group) and the methyl ketone where the original cyano group was.