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 step by step, we will analyze the reactants and the conditions provided.

### Step 1: Identify the Reactants
The reactant is 1-bromopropene, which can be represented as:
\[ \text{CH}_2\text{Br-CH}=\text{CH}_2 \]

### Step 2: Understand the Reagents and Conditions
The reaction involves sodium amide (NaNH₂) and heat. Sodium amide is a strong base and can deprotonate acidic hydrogens, leading to the formation of a nucleophile.

### Step 3: Deprotonation
In the presence of NaNH₂, the hydrogen atom adjacent to the bromine (the allylic hydrogen) can be removed. This results in the formation of a carbanion:
\[ \text{CH}_2\text{Br-CH}=\text{CH}_2 \xrightarrow{\text{NaNH}_2} \text{CH}_2\text{C}^- \text{H-CH}=\text{CH}_2 + \text{NaBr} \]

### Step 4: Elimination Reaction
The carbanion formed can then undergo an elimination reaction. The bromine atom leaves, and a double bond is formed, resulting in the final product:
\[ \text{CH}_2\text{C}^- \text{H-CH}=\text{CH}_2 \rightarrow \text{CH}_3\text{C}=\text{CH} \]

### Step 5: Final Product
The final product (X) is propene (or propylene):
\[ \text{X} = \text{CH}_3\text{C}=\text{CH} \]

### Summary of the Reaction
The overall reaction can be summarized as:
\[ \text{CH}_2\text{Br-CH}=\text{CH}_2 + \text{NaNH}_2 \xrightarrow{\Delta} \text{CH}_3\text{C}=\text{CH} + \text{NaBr} \]

### Final Answer
Thus, the product X is:
\[ \text{X} = \text{CH}_3\text{C}=\text{CH} \]

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