The major product of the following reaction is:
CH3C≡CH + DCl (1 equiv) → ?

The reaction of an alkyne with DCl follows Markovnikov's rule. The deuterium (D) will add to the carbon atom with more hydrogen atoms, while the chlorine (Cl) will add to the carbon atom with fewer hydrogen atoms. Since only one equivalent of DCl is used, the reaction will stop at the formation of a vinyl halide.

Step 1: The deuterium adds to one of the carbons of the triple bond, forming a vinyl cation intermediate.

Step 2: The chloride ion attacks the vinyl cation, yielding the final product.

Therefore, the major product is CH3CD=CHD(Cl).

However, none of the options match this exactly. The options seem to imply a subsequent reaction, perhaps with DI. Let's consider the reaction with DI. If DI is added, the reaction would proceed similarly, but this time an iodine atom would add to the carbon with fewer deuteriums/hydrogens.

Given the options provided, and assuming a secondary reaction with DI, the correct answer is CH3CD(Cl)CHD(I). The initial Markovnikov addition of DCl leads to the formation of a vinyl chloride, and subsequent reaction with DI leads to the addition of an iodine atom according to Markovnikov's rule. This results in the formation of CH3CD(Cl)CHD(I).