H-C `equiv` C - H + 2HCl `rarr`Final product, final product will be:

Options

A.
B. `CH_(2) = `CHCl
C. `underset(Cl)underset(|)(CH_(2)) - underset(Cl)(underset(|)(CH_(2))`
D. none of these

Answer

To solve the question regarding the reaction of an alkyne with 2 moles of HCl, we will follow a step-by-step approach to understand the mechanism and identify the final product.

### Step 1: Identify the Reactant
We start with an alkyne, which has the general formula C≡C. Alkynes are characterized by having a triple bond between two carbon atoms.

### Step 2: Reaction with HCl
When the alkyne reacts with HCl, it undergoes an electrophilic addition reaction. The HCl dissociates into H⁺ (proton) and Cl⁻ (chloride ion). The alkyne acts as a nucleophile and donates its electrons to the H⁺ ion, leading to the formation of a carbocation.

### Step 3: Formation of Carbocation
The addition of H⁺ to one of the carbon atoms of the alkyne results in the formation of a carbocation. The structure of the carbocation can be represented as follows:
- One carbon atom has an H attached, and the other carbon atom has a positive charge (C⁺).

### Step 4: Attack by Chloride Ion
The chloride ion (Cl⁻) from HCl then attacks the electron-deficient carbocation, leading to the formation of a vinyl halide. The structure of the vinyl halide will have a double bond between the two carbon atoms, with one of the carbons bonded to a chlorine atom and the other to a hydrogen atom.

### Step 5: Second Equivalent of HCl
Now, since we have another equivalent of HCl, the vinyl halide formed can react again with HCl. The process is similar: the vinyl halide will again donate electrons to H⁺, forming a new carbocation.

### Step 6: Formation of Germinal Dihalide
The chloride ion will attack this new carbocation, leading to the formation of a germinal dihalide. The final product will have two chlorine atoms attached to the same carbon atom, along with the hydrogen atom on the other carbon.

### Final Product
The final product of this reaction is a germinal dichloride, which can be represented as:
- C-Cl
- |
- C-Cl
- |
- H

Thus, the final product is a germinal dihalide (dichloride).

### Summary of Steps:
1. Start with an alkyne (C≡C).
2. React with HCl to form a carbocation.
3. Chloride ion attacks the carbocation to form a vinyl halide.
4. React vinyl halide with another HCl to form a second carbocation.
5. Chloride ion attacks the second carbocation to form a germinal dichloride.