`1-`phenylethanol can be prepared by reaction of benzaldehyde with

Options

A. `CH_(3)Br`
B. Zn and HCI
C. `CH_(3) ` Br and Na
D. `CH_(3) ` I and Mg

Answer

To prepare 1-phenylethanol from benzaldehyde, we can follow these steps:

### Step 1: Understand the Structures
1. **Identify the structure of 1-phenylethanol**: It consists of a benzene ring (C6H5) attached to a carbon (C) that has a hydroxyl group (OH) and a methyl group (CH3). The structure can be represented as:
\[ \text{C6H5-CH(OH)-CH3} \]

2. **Identify the structure of benzaldehyde**: Benzaldehyde has a benzene ring with a carbonyl group (C=O) attached, represented as:
\[ \text{C6H5-CHO} \]

### Step 2: Use of Grignard Reagent
3. **Select the appropriate Grignard reagent**: To convert benzaldehyde to 1-phenylethanol, we need a Grignard reagent. A suitable Grignard reagent is methylmagnesium iodide (CH3MgI).

### Step 3: Reaction Mechanism
4. **Nucleophilic attack**: The nucleophilic methyl group (CH3-) from the Grignard reagent will attack the electrophilic carbon of the carbonyl group in benzaldehyde. This results in the formation of an alkoxide intermediate:
\[ \text{C6H5-CHO} + \text{CH3MgI} \rightarrow \text{C6H5-CH(OMgI)-CH3} \]

### Step 4: Hydrolysis
5. **Hydrolysis of the alkoxide**: The alkoxide intermediate is then hydrolyzed by adding water (H2O). The oxygen in the alkoxide will abstract a proton from water, leading to the formation of 1-phenylethanol:
\[ \text{C6H5-CH(OMgI)-CH3} + \text{H2O} \rightarrow \text{C6H5-CH(OH)-CH3} + \text{Mg(OH)I} \]

### Step 5: Conclusion
6. **Final product**: The final product is 1-phenylethanol, confirming that the reaction of benzaldehyde with methylmagnesium iodide (Grignard reagent) yields the desired compound.

### Summary
Thus, 1-phenylethanol can be prepared by the reaction of benzaldehyde with methylmagnesium iodide (CH3MgI).

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