In the given reaction `CH_3-CN+CH_3CHOoverset(NaNH_2//NH_3(l))rarr(X),` (X) will be

Options

A. `CH_3-overset(OH)overset(|)(CH)-CH_2-CN`
B. `{:(CH_3-C=NH),(" "|),(" "CH_2-CHO):}`
C. `CH_3-CH=CH-CN`
D. `CH_3-CH=CH-CHO`

Answer

To solve the given reaction \( CH_3-CN + CH_3CHO \overset{NaNH_2/NH_3(l)}{\rightarrow} (X) \), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are methyl cyanide (also known as acetonitrile, \( CH_3-CN \)) and acetaldehyde (\( CH_3CHO \)).

### Step 2: Understand the Role of Sodium Amide
Sodium amide (\( NaNH_2 \)) is a strong base. In this reaction, it will deprotonate the most acidic hydrogen in the reactants. The most acidic hydrogen here is the one on the \( CH_2 \) group of the acetaldehyde.

### Step 3: Deprotonation
The sodium amide will remove the acidic proton from the \( CH_2 \) of acetaldehyde, resulting in the formation of an enolate ion. The reaction can be represented as follows:
\[ CH_3CHO + NaNH_2 \rightarrow CH_3C(O^-) + NH_3 \]

### Step 4: Nucleophilic Attack
The enolate ion formed will then act as a nucleophile and attack the carbonyl carbon of the acetaldehyde. The reaction can be illustrated as:
\[ CH_3C(O^-) + CH_3CN \rightarrow CH_3CH(CN)(OH) \]

### Step 5: Protonation
After the nucleophilic attack, the resulting alkoxide ion will be protonated by ammonia (\( NH_3 \)), leading to the formation of the final product:
\[ CH_3CH(OH)C(NH_2) \]

### Final Product
Thus, the product \( (X) \) of the reaction is:
\[ X = CH_3CH(OH)C(NH_2) \]

### Conclusion
The final answer is \( CH_3CH(OH)C(NH_2) \).

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