Which one of the following is the major product of the given reaction?

Solution

Chemistry ma’am · 4 steps

One Grignard for the nitrile, one for the ketone; then the alkene forms in conjugation with the ring.

1. Two reactive sites. The molecule has two electrophilic carbons: the nitrile carbon and the ring ketone carbon. One CH₃MgBr adds to each. The nitrile gives a negatively charged imine salt, which does not take a second Grignard.
2. Acid workup. H₃O⁺ hydrolyses the imine to a ketone, so -CN becomes -COCH₃. The ring alkoxide becomes a tertiary alcohol, and that carbon now carries both OH and a new CH₃.
3. Dehydration. Hot H₂SO₄ removes water and forms a tertiary carbocation. A proton can leave from either neighbouring ring carbon, and both choices give an alkene with four carbon groups. So conjugation decides: the alkene next to the benzene ring is conjugated with it and is more stable.

   Option	Group on benzene	Ring C=C	Verdict
   (a)	-COCH₃	next to benzene, conjugated	major product
   (b)	-CN	next to benzene, conjugated	nitrile left unreacted
   (c)	-CN	away from benzene, isolated	both features wrong
   (d)	-COCH₃	away from benzene, isolated	less stable alkene

4. So. The product has an acetyl group in place of CN and a C=C conjugated with the aromatic ring. That is option (a).

The idea to keep. Count the Grignard equivalents and give each one a target. A nitrile turns into a ketone after hydrolysis. In acid dehydration, the alkene that ends up conjugated with a benzene ring wins.

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