The major product of the following reaction is

The correct option is Option 1.

Step-by-Step Explanation:

1. Analysis of the Reactant and Reagents:
The starting material shown in the image (image_0.webp) is 4-oxo-4-phenylbutanenitrile:
$C_6{H}_5-CO-C{H}_2-C{H}_2-C\equiv N$
This compound contains two distinct electrophilic functional groups:
(a) A ketone group ($-CO-$) adjacent to the phenyl ring.
(b) A nitrile group ($-C\equiv N$) at the terminal position.

The reaction utilizes excess methylmagnesium bromide ($C{H}_{3}MgBr$), which is a strong nucleophile (source of $C{H}_3^{-}$).

2. Reaction at the Ketone Carbonyl Group:
Ketones are highly electrophilic and react readily with Grignard reagents. One equivalent of $C{H}_{3}MgBr$ attacks the ketone carbonyl carbon:
$C_6{H}_5-CO-C{H}_2-C{H}_2-CN+C{H}_{3}MgBr\to C_6{H}_5-C\left(OMgBr\right)\left(C{H}_3\right)-C{H}_2-C{H}_2-CN$

3. Reaction at the Nitrile Group:
Because the Grignard reagent is in excess, a second equivalent reacts with the nitrile group ($-C\equiv N$). The nucleophilic methyl group attacks the cyano carbon, forming an imine magnesium salt:
$C_6{H}_5-C\left(OMgBr\right)\left(C{H}_3\right)-C{H}_2-C{H}_2-C(=NMgBr)C{H}_3$

4. Acidic Hydrolysis:
Upon adding aqueous acid ($H_3{O}^{+}$):
• The magnesium alkoxide intermediate is protonated to form a tertiary alcohol ($-OH$).
• The imine magnesium salt is hydrolyzed to yield a methyl ketone ($-COC{H}_3$).
$C_6{H}_5-C\left(OH\right)\left(C{H}_3\right)-C{H}_2-C{H}_2-CO-C{H}_3$

This matches the structure shown in Option 1 (image_1.webp).