`underset((A))(MeCH)=O+underset((B))(H_(2)NOH) overset(HCOONa)underset(Delta)rarr (C )+H_(2)O` Product (C ) is:

Options

A.
B.
C. `Me-C-=N`
D.

Explanation

To solve the problem, we need to analyze the reaction between the given aldehyde (MeCH=O, which is acetaldehyde) and hydroxylamine (H2NOH) in the presence of sodium formate (HCOONa) and heat. The product (C) we are looking for is formed through a series of steps involving the formation of an oxime and then a subsequent reaction leading to a nitrile.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are acetaldehyde (MeCH=O) and hydroxylamine (H2NOH).

2. **Formation of Oxime**:
- When acetaldehyde reacts with hydroxylamine, it undergoes nucleophilic addition. The nitrogen atom of hydroxylamine attacks the carbonyl carbon of acetaldehyde.
- This results in the formation of an intermediate where the nitrogen is bonded to the carbon and the hydroxyl group (OH) is attached to the nitrogen, leading to the formation of an oxime.
- The structure of the oxime formed is CH3C(=N-OH)H.

3. **Dehydration of Oxime**:
- Upon heating, the oxime undergoes dehydration (loss of water). The hydroxyl group (OH) from the oxime and a hydrogen atom from the adjacent carbon are eliminated, forming a double bond between carbon and nitrogen.
- This results in the formation of a nitrile (specifically, ethane nitrile or acetonitrile).

4. **Formation of Product (C)**:
- The final product (C) is ethane nitrile (or acetonitrile), represented as CH3C≡N.

5. **Overall Reaction**:
- The overall reaction can be summarized as:
\[ \text{MeCH=O} + \text{H}_2\text{NOH} \xrightarrow{\text{HCOONa}, \Delta} \text{C (CH}_3\text{C}\equiv\text{N}) + \text{H}_2\text{O} \]

### Final Answer:
The product (C) is **Ethane nitrile (CH3C≡N)**.

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