Acetyl bromide reacts with excess of `CH_(3)MgI` followed by
treatment with a saturated solution of `NH_(4)Cl` gives:

Options

A. 2-Methylpropan-2-ol
B. Acetamide
C. Acetone
D. Acetyl iodide

Explanation

To solve the problem of what product is formed when acetyl bromide reacts with excess CH3MgI followed by treatment with a saturated solution of NH4Cl, we can break it down into steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Acetyl bromide (CH3C(O)Br)
- Methylmagnesium iodide (CH3MgI)

### Step 2: Nucleophilic Attack
Methylmagnesium iodide (CH3MgI) is a Grignard reagent and acts as a nucleophile. It will attack the carbonyl carbon of acetyl bromide. The reaction can be represented as follows:

\[ \text{CH}_3C(O)Br + \text{CH}_3MgI \rightarrow \text{CH}_3C(O)MgBr + \text{CH}_3I \]

In this step, the Grignard reagent adds to the carbonyl group, forming a magnesium alkoxide intermediate.

### Step 3: Second Nucleophilic Attack
Since we have excess CH3MgI, another molecule of CH3MgI will attack the carbonyl carbon of the formed magnesium alkoxide:

\[ \text{CH}_3C(O)MgBr + \text{CH}_3MgI \rightarrow \text{CH}_3C(O)(CH_3)MgBr \]

This results in the formation of a tertiary alkoxide.

### Step 4: Hydrolysis with NH4Cl
Next, we treat the reaction mixture with a saturated solution of NH4Cl. This step is a hydrolysis step where the alkoxide is converted into an alcohol:

\[ \text{CH}_3C(O)(CH_3)MgBr + NH_4Cl \rightarrow \text{CH}_3C(OH)(CH_3) + MgBrCl \]

### Step 5: Identify the Product
The final product after hydrolysis is 2-methylpropan-2-ol (also known as tert-butyl alcohol):

\[ \text{Product: } \text{2-methylpropan-2-ol} \]

### Conclusion
Thus, the final product of the reaction is 2-methylpropan-2-ol.

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