The reaction of phenylacetylene with one equivalent of methyl magnesium bromide followed by reaction with benzaldehyde provides

Options

A. (a)
B. (b)
C. (c)
D. (d)

Answer

To solve the question regarding the reaction of phenylacetylene with one equivalent of methyl magnesium bromide followed by reaction with benzaldehyde, we can break down the process step by step.

### Step 1: Understand the Reactants
- **Phenylacetylene (C6H5C≡CH)**: This compound has a phenyl group (C6H5) attached to a terminal alkyne (C≡CH).
- **Methyl magnesium bromide (CH3MgBr)**: This is a Grignard reagent, which acts as a strong nucleophile.

### Step 2: Reaction with Grignard Reagent
- When phenylacetylene reacts with one equivalent of methyl magnesium bromide, the nucleophilic carbon (C) in the Grignard reagent attacks the acidic hydrogen (H) of the terminal alkyne.
- This results in the formation of a phenylacetylide anion (C6H5C≡C^−) and the release of methane (CH4) upon proton abstraction.

**Reaction:**
\[ \text{C6H5C≡CH} + \text{CH3MgBr} \rightarrow \text{C6H5C≡C}^- + \text{CH4} + \text{MgBr}^+ \]

### Step 3: Reaction with Benzaldehyde
- The phenylacetylide anion (C6H5C≡C^−) is now a strong nucleophile and will react with benzaldehyde (C6H5CHO).
- The nucleophile attacks the electrophilic carbon of the carbonyl group in benzaldehyde.

**Reaction:**
\[ \text{C6H5C≡C}^- + \text{C6H5CHO} \rightarrow \text{C6H5C≡C(OH)C6H5}^- \]

### Step 4: Protonation
- The resulting alkoxide intermediate (C6H5C≡C(OH)C6H5^−) can be protonated by adding water (H2O) or an acid (H+).
- This leads to the formation of the final product, which is a propargylic alcohol.

**Final Product:**
\[ \text{C6H5C≡C(OH)C6H5} \]

### Conclusion
The final product of the reaction between phenylacetylene, one equivalent of methyl magnesium bromide, and benzaldehyde is **1,3-diphenyl-2-propen-1-ol**.

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