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 need to analyze the given reaction sequence:

1. **Identify the starting material**: The starting compound is an alkyne, specifically 1-butyne (CH₃-C≡C-H).

2. **Reaction with NaNH₂**: Sodium amide (NaNH₂) is a strong base and will deprotonate the terminal alkyne. The acidic hydrogen on the terminal carbon (the one attached to the triple bond) will be removed, resulting in the formation of a carbanion (negatively charged carbon).

- **Reaction**:
\[ CH₃-C≡C-H + NaNH₂ \rightarrow CH₃-C≡C^{-}Na^{+} + NH₃ \]
- **Product [X]**: The product after this step is the sodium salt of the carbanion, which can be represented as CH₃-C≡C⁻.

3. **Reaction with HCHO (formaldehyde)**: The carbanion formed in the previous step is a nucleophile and will attack the carbonyl carbon of formaldehyde (HCHO). This results in the formation of an alkoxide intermediate.

- **Reaction**:
\[ CH₃-C≡C^{-} + HCHO \rightarrow CH₃-C≡C-CH₂-O^{-} \]
- **Intermediate**: The alkoxide (CH₃-C≡C-CH₂-O⁻) is formed.

4. **Protonation of the alkoxide**: The alkoxide can then be protonated by adding a proton (H⁺) to form the final product.

- **Reaction**:
\[ CH₃-C≡C-CH₂-O^{-} + H^{+} \rightarrow CH₃-C≡C-CH₂-OH \]
- **Final Product [Y]**: The final product after protonation is an alcohol, specifically 1-butyne-2-ol (CH₃-C≡C-CH₂-OH).

Thus, the final answer for [Y] is **1-butyne-2-ol (CH₃-C≡C-CH₂-OH)**.