In the following reaction `CH_(3)CHO+NH_(2)*NH_(2)*toAoverset(B)toCH_(3)CH_(3)+N_(2)` Identify A and B .

Options

A. `CH_(3)CH=N NH_(2)andC_(2)H_(5)ONa`
B. `CH_(3)CH_(2)-NH_(2)andC_(2)H_(5)ONa`
C. `CH_(3)-HN-NH-CH_(3)andC_(2)H_(5)OH`
D. `CH_(3)CH_(2)NH_(2)andC_(2)H_(5)OH`

Answer

To solve the question, we need to identify the products A and B in the reaction:

\[ \text{CH}_3\text{CHO} + \text{NH}_2\text{NH}_2 \xrightarrow{B} \text{CH}_3\text{CH}_3 + \text{N}_2 \]

### Step-by-step Solution:

1. **Identify the Reactants**:
- The reactants are acetaldehyde (CH₃CHO) and hydrazine (NH₂NH₂).

2. **Reaction of Acetaldehyde with Hydrazine**:
- When acetaldehyde reacts with hydrazine, it forms a hydrazone. The reaction can be represented as:
\[ \text{CH}_3\text{CHO} + \text{NH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{C}(=N-NH_2)\text{H} \]
- Here, A is the hydrazone product, which is CH₃C(=N-NH₂)H.

3. **Identify the Reaction Conditions**:
- The product A (hydrazone) undergoes a reaction known as the Wolff-Kishner reduction. This requires a strong base (B) to convert the hydrazone into an alkane.

4. **Wolff-Kishner Reduction**:
- The hydrazone reacts with a strong base (commonly sodium ethoxide, C₂H₅O⁻Na⁺) to eliminate nitrogen gas (N₂) and form the corresponding alkane (CH₃CH₃).
- The reaction can be summarized as:
\[ \text{CH}_3\text{C}(=N-NH_2)\text{H} \xrightarrow{B} \text{CH}_3\text{CH}_3 + \text{N}_2 \]

5. **Conclusion**:
- Therefore, A is the hydrazone (CH₃C(=N-NH₂)H) and B is the strong base (C₂H₅O⁻Na⁺).

### Final Answer:
- A = CH₃C(=N-NH₂)H (hydrazone)
- B = C₂H₅O⁻Na⁺ (sodium ethoxide)