Step 1: Understanding the Concept:
Nitriles (\( \text{R-CN} \)) react with Grignard reagents (\( \text{R'MgX} \)) to form an imine salt intermediate, which upon acidic hydrolysis yields a ketone. Step 2: Detailed Explanation:
1. Addition: Methyl magnesium bromide (\( \text{CH}_{3}\text{MgBr} \)) adds to the triple bond of the cyanide (\( -\text{C} \equiv \text{N} \)) group in m-methoxybenzonitrile.
- The nucleophilic \( \text{CH}_{3}^{-} \) attacks the carbon, and \( \text{MgBr}^{+} \) associates with the nitrogen.
- Intermediate Q: \( \text{m-CH}_{3}\text{O-C}_{6}\text{H}_{4}\text{-C(CH}_{3}\text{)=NMgBr} \).
2. Hydrolysis: Acidic hydrolysis (\( \text{H}_{3}\text{O}^{+} \)) converts the imine intermediate into a carbonyl group.
\[ \text{R-C(R')=NMgBr} + \text{H}_{3}\text{O}^{+} \longrightarrow \text{R-CO-R'} + \text{NH}_{4}^{+} + \text{Mg}^{2+} + \text{Br}^{-} \]
- Here, \( \text{R} = \text{m-methoxyphenyl} \) and \( \text{R'} = \text{methyl} \).
- Product P: 3-methoxyacetophenone. Step 3: Final Answer:
The product P is 3-methoxyacetophenone.
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