Question

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

 

• A. A
• B. B
• C. C
• D. D

Hint:

When a molecule has multiple functional groups that can react with a reagent like Grignard, check the stoichiometry. Using multiple equivalents suggests all reactive sites will undergo reaction. Remember that Grignard reagents convert nitriles to ketones (after hydrolysis) and ketones to tertiary alcohols.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a multi-step synthesis involving a starting material with two functional groups (a ketone and a nitrile) that react with a Grignard reagent, followed by dehydration. We need to predict the final major product.
Step 2: Detailed Explanation:
Step (i) & (ii): Reaction with 2CH\(_3\)MgBr followed by H\(_3\)O\(^+\) workup.
The starting material has two electrophilic sites: the carbonyl carbon of the ketone and the carbon of the nitrile group (–C\(\equiv\)N). Grignard reagents react with both. Since two equivalents are used, both groups will react.
- Reaction at the Ketone: The first equivalent of CH\(_3\)MgBr will attack the ketone carbonyl carbon. The workup step (H\(_3\)O\(^+\)) protonates the resulting alkoxide to form a tertiary alcohol.
- Reaction at the Nitrile: The second equivalent of CH\(_3\)MgBr attacks the nitrile carbon. The initial adduct, upon hydrolysis with H\(_3\)O\(^+\), forms an imine which is then further hydrolyzed to a ketone. The nitrile group (–CN) is converted into a methyl ketone group (–C(=O)CH\(_3\)).
The intermediate product after these two steps has a tertiary alcohol where the original ketone was, and an acetyl group where the original nitrile was.
Step (iii): H\(_2\)SO\(_4\), heat.
This step causes the acid-catalyzed dehydration of the tertiary alcohol formed in the previous step.
1. The –OH group is protonated by the acid to form a good leaving group, –OH\(_2\)\(^+\).
2. The leaving group departs, forming a stable tertiary, benzylic carbocation.
3. A proton is eliminated from an adjacent carbon to form an alkene. Elimination of a proton from the more substituted adjacent carbon (Zaitsev's rule) leads to the more stable, more substituted alkene. In this case, removing a proton from the adjacent carbon in the ring forms an endocyclic double bond. This double bond is also conjugated with the benzene ring, making it the highly favored major product.
4. Identifying the Final Product:
The final product will have:
- The original nitrile group converted to an acetyl group (–COCH\(_3\)).
- The original ketone group converted to a C=C double bond within the six-membered ring, with a methyl group attached to one of the carbons of the double bond.
This structure corresponds exactly to option (B).
Step 3: Final Answer:
The major product of the given reaction sequence is the structure shown in option (B).