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 question, we will break down the reaction sequence step by step.

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

### Step 2: Reaction with \( \text{Cu}_2\text{Cl}_2 \)
When two moles of acetylene react in the presence of \( \text{Cu}_2\text{Cl}_2 \), a coupling reaction occurs. The acidic hydrogen from one acetylene molecule is lost, and the remaining carbon atoms form a double bond. The product formed (A) can be represented as:
\[ A: \text{CH}_2 = \text{CH} - \text{C} \equiv \text{CH} \]
This represents a compound with a double bond between the first two carbon atoms and a triple bond with the last carbon.

### Step 3: Reaction with HCl
Next, product A is treated with hydrochloric acid (HCl). In this reaction, the \( \text{H}^+ \) from HCl will attack the triple bond because it is more reactive than the double bond. The reaction can be illustrated as follows:
1. The \( \text{H}^+ \) attacks the carbon in the triple bond, creating a carbocation.
2. The carbocation is then attacked by \( \text{Cl}^- \).

The resulting product (B) after this step can be represented as:
\[ B: \text{CH}_2 = \text{CH} - \text{CCl} = \text{CH}_2 \]
This shows that the chlorine atom has added to the carbon that was part of the triple bond.

### Final Product
Thus, the final product B is:
\[ \text{CH}_2 = \text{CH} - \text{CCl} = \text{CH}_2 \]

### Summary of Products
- Product A: \( \text{CH}_2 = \text{CH} - \text{C} \equiv \text{CH} \)
- Product B: \( \text{CH}_2 = \text{CH} - \text{CCl} = \text{CH}_2 \)