In the reaction given below `H-C -= C - H + H - C -= C - H underset(NH_4Cl)overset(Cu_2Cl)(rarr)(x)`, 'X' will be

Options

A. `CH_2 = CH - C -= CH`
B. `CU - C -= C - Cu`
C. `CH -= C - Cu`
D. `CH -= C - C -= CH`

Answer

To solve the given reaction, we start with the reactants and analyze the process step by step.

### Step 1: Identify the Reactants
The reactants in the reaction are two molecules of ethyne (acetylene), represented as:
\[ \text{H-C} \equiv \text{C-H} \]

### Step 2: Understand the Role of the Reagents
The reaction takes place in the presence of ammonium chloride (\( \text{NH}_4\text{Cl} \)) and copper(I) chloride (\( \text{Cu}_2\text{Cl} \)). The \( \text{Cu}_2\text{Cl} \) acts as a catalyst and facilitates the reaction.

### Step 3: Mechanism of the Reaction
1. **Formation of a Carbanion:** The chloride ion (\( \text{Cl}^- \)) from \( \text{NH}_4\text{Cl} \) attacks one of the carbon atoms in the ethyne molecule, resulting in the formation of a carbanion:
\[ \text{H-C} \equiv \text{C}^- \text{ + Cl}^+ \rightarrow \text{H-C} \equiv \text{C}^- \text{ + Cu}^+ \]

2. **Nucleophilic Attack:** The carbanion then attacks another ethyne molecule. The carbon atom of the carbanion forms a bond with the carbon atom of the second ethyne molecule. This results in the formation of a new carbon-carbon bond:
\[ \text{H-C} \equiv \text{C}^- + \text{H-C} \equiv \text{C} \rightarrow \text{H-C} \equiv \text{C-C} \equiv \text{C-H} \]

3. **Elimination of HCl:** During this process, a molecule of HCl is eliminated, leading to the formation of the final product.

### Step 4: Write the Final Product
The final product \( X \) formed from the reaction is:
\[ \text{H-C} \equiv \text{C-C} \equiv \text{C-H} \]
This is 1,3-butadiene.

### Conclusion
Thus, the product \( X \) in the reaction is:
\[ \text{X} = \text{1,3-butadiene} \]

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