Major product of the following reaction is

Solution

Chemistry ma’am · 3 steps

Excess Grignard hits both groups: the ester takes two methyls, the nitrile takes only one.

1. Ester takes two. CH₃MgBr adds to the ester C=O and CH₃O⁻ leaves, giving a ketone. That ketone is more reactive than the ester, so a second CH₃ adds at once. After H₃O⁺ you get the tertiary alcohol -C(CH₃)₂OH.
2. Nitrile takes one. CH₃MgBr adds once to C≡N and forms an imine salt, C=N-MgBr. This anion is negatively charged, so a second Grignard cannot attack it, even in excess. H₃O⁺ then hydrolyses the imine to a ketone, -COCH₃.

   Group	With excess CH₃MgBr	After H₃O⁺
   -COOCH₃	two CH₃ add, gives an alkoxide	-C(CH₃)₂OH, a tertiary alcohol
   -C≡N	one CH₃ adds, gives an imine salt	-COCH₃, a methyl ketone

3. So. The ketone from the nitrile appears only during workup, after the Grignard has been destroyed, so it survives. The major product is 4-(2-hydroxypropan-2-yl)acetophenone, option (b).

The idea to keep. Ester plus excess RMgX gives a tertiary alcohol, because the ketone formed on the way reacts again. Nitrile plus RMgX gives a ketone, because the imine salt stops the reaction until water arrives.

Answer from the NTA final answer key. Solution by PrepWiser.