Reaction of ethyl formate with excess of `CH_(3)Mgl ` followed by hydrolysis gives .

Options

A. n-propyl alcohol
B. ethanal
C. propanal
D. isopropyl alcohol

Explanation

To solve the question regarding the reaction of ethyl formate with excess of CH₃MgI followed by hydrolysis, we will go through the steps systematically.

### Step 1: Identify the Reactants
The reactants are ethyl formate and excess CH₃MgI (methyl magnesium iodide). Ethyl formate has the structure:
\[ \text{HCOOCH}_2\text{CH}_3 \]
This can be represented as:
\[ \text{H}_2\text{C}=O\text{O}\text{C}_2\text{H}_5 \]

### Step 2: First Nucleophilic Attack
When ethyl formate reacts with CH₃MgI, the nucleophile (CH₃⁻) attacks the carbonyl carbon of ethyl formate. The carbonyl group (C=O) is polarized, making the carbon atom electrophilic. The reaction can be illustrated as follows:

1. The nucleophile (CH₃⁻) attacks the carbonyl carbon:
\[ \text{HCOOCH}_2\text{CH}_3 + \text{CH}_3\text{MgI} \rightarrow \text{H}_2\text{C}(O^-) \text{C}(CH_3)(OCH_2CH_3) \]

2. This results in the formation of a tetrahedral intermediate where the oxygen becomes negatively charged (alkoxide ion).

### Step 3: Second Nucleophilic Attack
Since we have excess CH₃MgI, another molecule of CH₃MgI will attack the carbonyl carbon of the tetrahedral intermediate formed in the previous step:

1. The second CH₃⁻ attacks the carbon atom:
\[ \text{H}_2\text{C}(O^-) \text{C}(CH_3)(OCH_2CH_3) + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3\text{C}(O^-) \text{C}(CH_3)(OCH_2CH_3) \]

2. This results in the formation of a new alkoxide intermediate.

### Step 4: Hydrolysis
The final step is hydrolysis of the alkoxide intermediate. When we add water (or dilute acid), the alkoxide will be converted into the corresponding alcohol:

1. Hydrolysis of the alkoxide:
\[ \text{CH}_3\text{C}(O^-) \text{C}(CH_3)(OCH_2CH_3) + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{C}(OH) \text{C}(CH_3)(OCH_2CH_3) \]

2. The product formed is isopropyl alcohol (2-propanol):
\[ \text{CH}_3\text{CHOH}\text{CH}_3 \]

### Final Answer
The product of the reaction of ethyl formate with excess CH₃MgI followed by hydrolysis is **isopropyl alcohol**.