In the reaction sequence
`CH -= CH + CH -=CH overset(Cu_2Cl_2)(rarr)(A), 'B' overset(HCl)(rarr)(B)`
Will be

Options

A. `CH_3-overset(Cl)overset(|)(CH)-C-=CH`
B. `CH_2=overset(Cl)overset(|)(C )-CH=CH_2`
C. `CH_2=CH-C-=C-Cl`
D. `CH_3-CH_2-C-=C-Cl`

Answer

To solve the given reaction sequence, we will follow these steps:

### Step 1: Identify the reactants
The reactants in the reaction are two moles of acetylene (ethyne), represented as:
\[ \text{CH} \equiv \text{CH} + \text{CH} \equiv \text{CH} \]

### Step 2: Understand the role of the catalyst
The catalyst used in the reaction is \( \text{Cu}_2\text{Cl}_2 \). This catalyst facilitates the coupling of the two acetylene molecules.

### Step 3: Reaction of acetylene in the presence of \( \text{Cu}_2\text{Cl}_2 \)
When two moles of acetylene react in the presence of \( \text{Cu}_2\text{Cl}_2 \), they undergo a coupling reaction. The terminal hydrogen of one acetylene molecule attacks the carbon of the triple bond of the other acetylene molecule. This results in the formation of a new compound, which is:
\[ \text{A} = \text{CH} = \text{C} = \text{CH} \]
This compound is buta-1,3-diene.

### Step 4: Reaction of compound A with HCl
Next, we take the product \( \text{A} \) and react it with hydrochloric acid (HCl). The reaction can be represented as:
\[ \text{A} + \text{HCl} \rightarrow \text{B} \]

### Step 5: Determine the product B
In the presence of HCl, the double bond in buta-1,3-diene will react with the hydrogen ion (H⁺) from HCl. The more stable carbocation will form, and the chloride ion (Cl⁻) will attach to the more substituted carbon. The product formed will be:
\[ \text{B} = \text{CH}_2 = \text{CHCl} - \text{CH}_2 \]
This compound is 3-chlorobut-1-ene.

### Final Answer
The final product \( \text{B} \) is 3-chlorobut-1-ene.

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