Which of the following are intermediates in the reaction of excess of `CH_(3) MgBr " with" C_(6)H_(5)COOC_(2)H_(5)` to make 2 - phenyl -2- propanol ?
A. `C_(6)H_(5) - overset("OMgBr") overset(|) underset(CH_(3))`
B. `C_(6)H_(5) - overset(O) overset(||) - CH_(3)`
C. `C_(6)H_(5) - overset("OMgBr") overset(|) underset(CH_(3)) underset(|) C- CH_(3)`

Options

A. A and B
B. A,B and C
C. A and C
D. B and C

Answer

To solve the question regarding the intermediates formed in the reaction of excess methyl magnesium bromide (CH₃MgBr) with ethyl benzoate (C₆H₅COOC₂H₅) to produce 2-phenyl-2-propanol, we will follow the reaction mechanism step by step.

### Step 1: Nucleophilic Attack
1. The reaction begins with the nucleophilic attack of the methyl magnesium bromide (CH₃MgBr) on the carbonyl carbon of ethyl benzoate (C₆H₅COOC₂H₅).
2. The carbonyl carbon is electrophilic, and the nucleophile (CH₃⁻) attacks it, leading to the formation of a tetrahedral intermediate.

**Intermediate Formed:**
- The tetrahedral intermediate can be represented as:
\[ C_6H_5 - O^-(MgBr) - CH_3 - C_2H_5 \]

### Step 2: Leaving Group Departure
3. In the next step, the ethoxy group (C₂H₅O⁻) leaves, and the electron density returns to form a carbonyl bond, resulting in the formation of a ketone.

**Intermediate Formed:**
- The resulting ketone can be represented as:
\[ C_6H_5 - C(=O) - CH_3 \]

### Step 3: Second Nucleophilic Attack
4. The ketone formed in the previous step now undergoes another nucleophilic attack by another equivalent of methyl magnesium bromide (CH₃MgBr).
5. This leads to the formation of another tetrahedral intermediate.

**Intermediate Formed:**
- The new tetrahedral intermediate can be represented as:
\[ C_6H_5 - O^-(MgBr) - CH_3 - CH_3 \]

### Step 4: Protonation
6. Finally, the intermediate is protonated (usually by adding water or another proton source) to yield the final product, 2-phenyl-2-propanol.

### Summary of Intermediates
- The intermediates formed during the reaction are:
1. **Intermediate A**: \( C_6H_5 - O^-(MgBr) - CH_3 - C_2H_5 \)
2. **Intermediate B**: \( C_6H_5 - C(=O) - CH_3 \)
3. **Intermediate C**: \( C_6H_5 - O^-(MgBr) - CH_3 - CH_3 \)

### Conclusion
All three intermediates A, B, and C are formed during the reaction. Therefore, the correct answer is that all three options (A, B, and C) are intermediates in the reaction.