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 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 are acetonitrile (\( CH_3-CN \)) and acetaldehyde (\( CH_3CHO \)).

### Step 2: Role of Sodium Amide (NaNH2)
Sodium amide (\( NaNH_2 \)) is a strong base. It will abstract a proton from the acetonitrile (\( CH_3-CN \)), converting it into an anion. The reaction can be represented as:
\[ CH_3-CN + NaNH_2 \rightarrow CH_3-CN^- + Na^+ + NH_3 \]

### Step 3: Nucleophilic Attack
The anion (\( CH_3-CN^- \)) formed will act as a nucleophile and attack the carbonyl carbon of acetaldehyde (\( CH_3CHO \)). The carbonyl carbon is electrophilic due to the partial positive charge on it. The reaction can be represented as:
\[ CH_3-CN^- + CH_3CHO \rightarrow CH_3-C(OH)(CN)CH_3 \]

### Step 4: Proton Transfer
After the nucleophilic attack, a proton transfer occurs. The hydroxyl group (OH) formed will take a proton from the surrounding medium, leading to the formation of a compound with a carbon-carbon bond and a hydroxyl group:
\[ CH_3-C(OH)(CN)CH_3 \]

### Step 5: Elimination of Water
Upon heating, the compound can undergo dehydration (loss of water) to form an unsaturated compound. The elimination of water (H and OH) from the intermediate will yield:
\[ X = CH_3-CH=CH-CN \]

### Conclusion
Thus, the final product \( X \) is:
\[ X = CH_3-CH=CH-CN \]