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

Options

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

Answer

To solve the given reaction step by step, let's analyze the process of the reaction involving the alkyne `CH3C≡CH` with DCl and DI.

### Step 1: Reaction with DCl
1. Start with the alkyne: `CH3C≡CH`.
2. When DCl (deuterium chloride) is added in one equivalent, the D (deuterium) and Cl will add across the triple bond.
3. The triple bond will convert to a double bond, and we will have:
- Chlorine (Cl) will attach to one of the carbons of the former triple bond.
- Deuterium (D) will attach to the other carbon.

The product after this step will be:
\[ CH_3C(Cl)=C(D) \]

### Step 2: Reaction with DI
1. Now, we take the product from Step 1, which is `CH3C(Cl)=C(D)`.
2. The next step involves the reaction with DI (deuterium iodide).
3. In this reaction, the iodine (I) will add to the carbon that already has chlorine (Cl), while deuterium (D) will add to the other carbon.

The final product after this step will be:
\[ CH_3C(Cl)(I)C(D) \]

### Final Product
The final major product of the reaction is:
\[ CH_3C(Cl)(I)C(D) \]

### Summary
The major product of the reaction `CH3C≡CH` with DCl followed by DI is `CH3C(Cl)(I)C(D)`.

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