During reduction of carbonyl compounds by `H_(2)N NH_(2)` and `KOH`, the first intermediane is :

Options

A. `PC -= N`
B. `RCONH_(2)`
C. `RCH = NH`
D. `RCH = "NNH"_(2)`

Explanation

To solve the question regarding the first intermediate formed during the reduction of carbonyl compounds by hydrazine (H₂N-NH₂) and KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction in question is a type of reduction known as the Wolff-Kishner reduction. This process involves the conversion of carbonyl compounds (aldehydes or ketones) into hydrocarbons.

2. **Identifying the Reactants**:
The key reactants in this reaction are:
- A carbonyl compound (which can be an aldehyde or a ketone).
- Hydrazine (H₂N-NH₂).
- Potassium hydroxide (KOH).

3. **Formation of the Hydrazone**:
When a carbonyl compound reacts with hydrazine, a condensation reaction occurs. This leads to the formation of a hydrazone intermediate. The reaction can be represented as follows:
\[ RCHO + H₂N-NH₂ \rightarrow RCH=NNH₂ + H₂O \]
Here, RCHO represents the carbonyl compound (aldehyde), and RCH=NNH₂ is the hydrazone.

4. **Role of KOH**:
The next step involves the treatment of the hydrazone with KOH in a high boiling solvent (such as ethylene glycol). This step leads to the reduction of the hydrazone to a hydrocarbon, releasing nitrogen gas (N₂) in the process:
\[ RCH=NNH₂ + KOH \rightarrow RCH₃ + N₂ + KOH \]

5. **Identifying the First Intermediate**:
The first intermediate formed in this reaction is the hydrazone (RCH=NNH₂). Therefore, the answer to the question is that the first intermediate during the reduction of carbonyl compounds by hydrazine and KOH is a hydrazone.

### Final Answer:
The first intermediate is **hydrazone (RCH=NNH₂)**.

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