What are X and Y in the following sequence?
`CH_(2)=CHBroverset(NaNH_(2))rarrXunderset(H_(2)O,333K)overset(Hg^(2+)//H^(+))rarrY`

Options

A. `{:("X","Y"),("ethane","ethanol"):}`
B. `{:("X","Y"),("ethyne","acetic acid"):}`
C. `{:("X","Y"),("ethyne","ethanal"):}`
D. `{:("X","Y"),("ethylamine","ethanal"):}`

Answer

To solve the question regarding the sequence of reactions involving `CH₂=CHBr`, we will break down the steps systematically to identify the compounds X and Y.

### Step-by-Step Solution:

1. **Identify the Starting Compound**:
The starting compound is `CH₂=CHBr`, which is vinyl bromide (or bromoethene).

2. **First Reaction with NaNH₂**:
When `CH₂=CHBr` reacts with sodium amide (NaNH₂), the bromine atom is replaced by a hydrogen atom. This is a dehydrohalogenation reaction that leads to the formation of a triple bond.
- The reaction can be represented as:
\[ CH₂=CHBr + NaNH₂ \rightarrow CH≡CH + NaBr + NH₃ \]
- The product formed here is `CH≡CH`, which is acetylene (or ethyne).
- Therefore, **X = CH≡CH**.

3. **Second Reaction with Hg²⁺ and H⁺**:
Next, acetylene (`CH≡CH`) undergoes a reaction with mercuric ions (Hg²⁺) in the presence of acid (H⁺) at 333 K. This reaction typically leads to the formation of an enol intermediate.
- The reaction can be summarized as:
\[ CH≡CH + Hg^{2+} + H^+ \rightarrow CH_2=CH-Hg^{2+} \rightarrow CH_2=CHOH \]
- The enol formed is unstable and quickly rearranges to form a more stable ketone or aldehyde. In this case, it forms ethanol (CH₃CHOH).
- Therefore, **Y = CH₃CHOH**, which is ethanol.

### Final Answer:
- **X = CH≡CH (Acetylene)**
- **Y = CH₃CHOH (Ethanol)**