In the given reaction `overset(Br)overset(|)(CH_2)-CH=CH_2overset(NaNH_2//Delta)rarr(X)` X will be

Options

A. `CH_3-C-=CH`
B. `CH_2=C=CH_2`
C. `CH_3-C-=overset(-o+)("CNa")`
D. `CH_2=C=overset(!o+)(CHNa)`

Answer

To solve the given reaction, we need to analyze the transformation of 3-bromoprop-1-ene (the compound represented as `Br-CH2-CH=CH2`) when treated with sodium amide (NaNH2) under heat (Δ).

### Step-by-Step Solution:

1. **Identify the starting compound**: The starting compound is 3-bromoprop-1-ene, which can be represented as:
\[ \text{Br-CH}_2-\text{C}=\text{CH}_2 \]

2. **Reaction with NaNH2**: Sodium amide (NaNH2) is a strong base. It will first deprotonate the terminal hydrogen (the hydrogen attached to the CH2 group) adjacent to the bromine:
\[ \text{Br-CH}_2-\text{C}=\text{CH}_2 \xrightarrow{\text{NaNH}_2} \text{C}=\text{C} \text{(with Br leaving)} \]
This results in the formation of a double bond between the two carbon atoms where the bromine was attached.

3. **Formation of a triple bond**: The next step involves the deprotonation of the remaining hydrogen on the carbon adjacent to the double bond. NaNH2 will abstract this hydrogen, leading to the formation of a triple bond:
\[ \text{C}=\text{C} \xrightarrow{\text{NaNH}_2} \text{C} \equiv \text{C}^- \]
This results in the formation of a carbanion.

4. **Final product**: The final product after the complete reaction will be:
\[ \text{C} \equiv \text{C}^- \text{(with Na}^+ \text{ counterion)} \]
This can be represented as:
\[ \text{Na}^+ \text{C} \equiv \text{C}^- \]

5. **Conclusion**: The final product X is sodium acetylide (or sodium propyne), represented as:
\[ \text{Na}^+ \text{C} \equiv \text{C}^- \]

### Final Answer:
The product X is \(\text{Na}^+ \text{C} \equiv \text{C}^-\).