In the given reaction `C_6H_5-overset(O)overset(||)C-CH_3overset((i)C_2H_8MgBr)underset((ii)H_2"O"//H^+)rarr(X)` X will be

Options

A. `1^@` - alcohol
B. `2^@` - alcohol
C. Optically inactive `3^@` - alcohol
D. Optically active `3^@` - alcohol

Answer

To solve the given reaction step by step, we will analyze the components involved and the transformations that occur.

### Step 1: Identify the Reactants
The reactant given is acetophenone, which can be represented as:
\[ C_6H_5C(=O)CH_3 \]
This compound consists of a phenyl group (C6H5) attached to a carbonyl group (C=O) and a methyl group (CH3).

### Step 2: Understand the Grignard Reagent
The Grignard reagent used in the reaction is ethyl magnesium bromide, represented as:
\[ C_2H_5MgBr \]
Grignard reagents are strong nucleophiles and react with electrophilic centers, such as the carbonyl carbon in acetophenone.

### Step 3: Nucleophilic Addition Reaction
When the Grignard reagent reacts with acetophenone, the nucleophilic ethyl group (C2H5) attacks the electrophilic carbon of the carbonyl group. This leads to the formation of a tetrahedral intermediate:
\[ C_6H_5C(OH)(C_2H_5)CH_3 \]
At this stage, the carbon that was part of the carbonyl group now has an -OH group, an ethyl group, and a methyl group attached to it.

### Step 4: Hydrolysis of the Intermediate
The next step involves hydrolysis, which typically involves the addition of water (H2O) and an acid (H+). The hydrolysis converts the intermediate into a stable alcohol:
\[ C_6H_5C(OH)(C_2H_5)CH_3 \]
This compound is a tertiary alcohol because the carbon bearing the -OH group is attached to three other carbon atoms (one from the ethyl group, one from the methyl group, and one from the phenyl group).

### Step 5: Determine Chirality
The carbon that is bonded to the -OH group is also bonded to three different groups (C6H5, C2H5, and CH3), making it a chiral center. Therefore, this compound is optically active.

### Conclusion
The final product X is an optically active tertiary alcohol.

### Answer
The correct answer is: **Optically active tertiary alcohol (Option 4)**.

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