In the given reaction `CH_3-CH_2-C-=Noverset((i)CH_3MgCl)rarr[X]` [X] will be

Options

A. `CH_3-CH_2-overset(O)overset(||)C-CH_2-CH_3`
B. `CH_3-CH_2-overset(O)overset(||)C-CH_3`
C. `C_2H_5-underset(CH_3)underset(|)overset(OH)overset(|)C-C_2H_5OH`
D. `C_2H_5-underset(CH_3)underset(|)overset(OH)overset(|)C-CH_5`

Answer

To solve the given reaction `CH3-CH2-C≡N + CH3MgCl → [X]`, we will follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- A nitrile compound: `CH3-CH2-C≡N` (propionitrile)
- A Grignard reagent: `CH3MgCl`

### Step 2: Understand the Reaction Mechanism
Grignard reagents react with nitriles to form intermediate imines or amines. In this case, the Grignard reagent will attack the carbon atom of the nitrile group (C≡N).

### Step 3: Nucleophilic Attack
The nucleophilic carbon from the Grignard reagent (`CH3-`) will attack the carbon of the nitrile group:
- The reaction can be represented as:
\[ CH3-CH2-C≡N + CH3MgCl \rightarrow CH3-CH2-C(CH3)=N^- \]
This results in the formation of an intermediate with a double bond between carbon and nitrogen.

### Step 4: Protonation of the Intermediate
Next, we will add water (H3O+) to hydrolyze the intermediate:
- The nitrogen atom will be protonated, and the double bond will rearrange:
\[ CH3-CH2-C(CH3)=N^- + H2O \rightarrow CH3-CH2-C(OH)(CH3)-NH2 \]

### Step 5: Final Product Formation
After the protonation, the nitrogen will lose a proton (NH3), leading to the formation of a ketone:
- The final product can be represented as:
\[ CH3-CH2-C(=O)-CH3 \]
This is 2-pentanone.

### Final Answer
Thus, the product [X] is:
\[ [X] = CH3-CH2-C(=O)-CH3 \]

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