The major product of the following reaction is `CH_(3)C-=CH underset((ii) DI)overset((i) DCl(1 "equiv")) to`

Options

A. `CH_(3)CH(Cl)CHD(I)`
B. `CH_(3)C(I)(Cl)CHD_(2)`
C. `CH_(3)CD_(2)CH(Cl)(I)`
D. `CH_(3)CD(I)CHD(Cl)`

Explanation

To solve the problem, we need to analyze the reaction of the alkyne \( \text{CH}_3C \equiv CH \) with one equivalent of DCl (deuterium chloride). The reaction proceeds through a series of steps involving the formation of a carbocation intermediate and subsequent addition of D and Cl.

### Step-by-Step Solution:

1. **Identify the Reactants**:
We have the alkyne \( \text{CH}_3C \equiv CH \) and we are reacting it with one equivalent of DCl.

2. **Addition of DCl**:
When DCl is added to the alkyne, the triple bond reacts with DCl. The triple bond is nucleophilic, and it will attack the D atom from DCl, leading to the formation of a carbocation.

\[ \text{CH}_3C \equiv CH + DCl \rightarrow \text{CH}_3C^{+}D + Cl^{-} \]

3. **Formation of Carbocation**:
The addition of D leads to the formation of a carbocation. There are two possible carbocations that can form:
- \( \text{CH}_3C^{+}D \) (secondary carbocation)
- \( \text{CH}_2D^{+}CH_3 \) (primary carbocation)

The secondary carbocation \( \text{CH}_3C^{+}D \) is more stable due to hyperconjugation and inductive effects.

4. **Nucleophilic Attack by Cl**:
The chloride ion \( Cl^{-} \) will now attack the more stable carbocation \( \text{CH}_3C^{+}D \).

\[ \text{CH}_3C^{+}D + Cl^{-} \rightarrow \text{CH}_3CClD \]

5. **Final Product**:
The final product after the addition of DCl will be \( \text{CH}_3CClD \).

### Final Answer:
The major product of the reaction is \( \text{CH}_3CClD \).

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