1. Identify the reaction type

The options indicate addition of $I$ and $Cl$ across a deuterated alkene. This corresponds to addition of iodine monochloride, $ICl$, to an alkene.

In $ICl$, chlorine is more electronegative, so the bond is polarized as:

$I^{\delta+}-Cl^{\delta-}$

Thus, the electrophile is effectively $I^+$.

2. Determine the alkene from the products

All options have the carbon skeleton:

$CH_3- C - C$

with two deuteriums distributed over the double-bond carbons. So the alkene must be:

$CH_3CD=CD_2$

Now we add $ICl$ across this double bond.

3. Mechanism of addition of $ICl$ to alkene

Addition proceeds through formation of an iodonium ion first, because $I^+$ attacks the double bond.

Then $Cl^-$ opens the halonium ion.

For such additions, the orientation is governed by the carbocation-like character in the bridged intermediate: the nucleophile $Cl^-$ attacks the more substituted carbon.

So effectively:

• $I$ goes first to the alkene,
• $Cl$ finally ends up on the more substituted carbon.

In the alkene

$CH_3CD=CD_2$

• Left alkene carbon ($CH_3CD=$) is more substituted.
• Right alkene carbon ($=CD_2$) is less substituted.

Therefore:

• $Cl$ goes to the left carbon,
• $I$ goes to the right carbon.

So the product should be:

$CH_3C(Cl)D-CHD_2(I)$

which can be written as:

$CH_3CD(Cl)CHD_2(I)$

Matching the options, this is Option B? Let us check carefully.

4. Check each option carefully

Starting alkene: $CH_3CD=CD_2$

After addition, each double-bond carbon gains one new substituent:

• Left carbon already has $CH_3$ and $D$, so it gets either $Cl$ or $I$ and no extra H/D.
• Right carbon already has $D_2$, so it gets either $Cl$ or $I$.

Thus the valid connectivity after addition must look like one of these two regioisomers:

$CH_3C(D)(Cl)-C(D_2)(I)$ nor $CH_3C(D)(I)-C(D_2)(Cl)$

Now inspect options:

• A: $CH_3CD_2CH(Cl)(I)$
  Impossible: left carbon has become $CD_2$, which would mean deuterium count changed incorrectly.

• B: $CH_3CD(Cl)CHD(I)$
  This has one $D$ on each carbon after addition, so total deuterium count is only 2, whereas alkene had 3 deuteriums. Impossible.

• C: $CH_3C(I)(Cl)CHD_2$
  This puts both $I$ and $Cl$ on the same carbon. Not possible for simple addition across a double bond.

• D: $CH_3CD(I)CHD(Cl)$
  Again only 2 deuteriums total, so impossible.

5. Conclusion from chemical reasoning

None of the options exactly matches the chemically correct product expected from addition of $ICl$ to $CH_3CD=CD_2$.

The correct major product should be:

$CH_3CD(Cl)-CD_2(I)$

or equivalently

$CH_3C(D)(Cl)CHD_2(I)$

Since this is not among the given options, the stored answer C is chemically inconsistent.

6. Comparison with stored answer

Stored correct answer: C

But option C is not a valid addition product, since it places both halogens on the same carbon.

Therefore, I disagree with the stored answer.