In the reaction sequence `CH_3-C-=C-Hoverset(NaNH_2)rarr[X]overset(HCHO)rarr[Y]`
[Y] will be

Options

A. `CH_3-CH=CH_2`
B. `CH_3-C-=C-CH_2OH`
C. `CH_3-C-=C-CHO`
D. `CH_3-=C-COONa`

Answer

To solve the problem step by step, we will analyze the reaction sequence provided:

1. **Identify the starting compound**: The starting compound is CH₃-C≡C-H (1-butyne).

2. **Reaction with NaNH₂**:
- Sodium amide (NaNH₂) acts as a strong base. It will abstract the acidic hydrogen from the terminal alkyne (the hydrogen attached to the carbon with the triple bond).
- The acidic hydrogen in this case is the one attached to the terminal carbon (the CH at the end of the triple bond).
- The reaction can be represented as:
\[ \text{CH}_3-\text{C} \equiv \text{C}-\text{H} + \text{NaNH}_2 \rightarrow \text{CH}_3-\text{C} \equiv \text{C}^- + \text{Na}^+ + \text{NH}_3 \]
- This gives us the intermediate compound **X**: CH₃-C≡C⁻ (the anion).

3. **Reaction with HCHO (formaldehyde)**:
- The anion (X) will now react with formaldehyde (HCHO). The nucleophilic carbon of the alkyne will attack the electrophilic carbon of formaldehyde.
- The reaction can be represented as:
\[ \text{CH}_3-\text{C} \equiv \text{C}^- + \text{HCHO} \rightarrow \text{CH}_3-\text{C} \equiv \text{C}-\text{CH}_2\text{OH}^- \]
- After the attack, the double bond shifts, and we will have a compound with a hydroxymethyl group (-CH₂OH) attached to the terminal carbon of the triple bond.
- The final product **Y** is: CH₃-C≡C-CH₂OH.

4. **Final Product**:
- The compound Y formed after the reaction sequence is **CH₃-C≡C-CH₂OH** (which is 3-buten-1-ol).

### Final Answer:
The compound Y will be **CH₃-C≡C-CH₂OH**.