Question

The major product of the following reaction is:
$C{{H}_{3}}C\equiv CH\xrightarrow[(ii)DI]{(i)DCl(1eq)}$

A.$C{{H}_{3}}CD(Cl)CHD(I)$
B.$C{{H}_{3}}C{{D}_{2}}CH(Cl)(I)$
C.$C{{H}_{3}}CD(I)CHD(Cl)$
D.$C{{H}_{3}}C(I)(Cl)CH{{D}_{2}}$

Answer 1

Hint: To find out the appropriate product for this reaction, consider Markovnikov’s rule when the compound given is reacted with the halide. The D in the halide is an isotope of hydrogen and thus it will similarly act as a hydrogen atom does.

Complete step by step answer:
Given that, an alkyne compound is first being reacted with $DCl$ and then the product formed after this reaction is further reacted with $DI$. So, here we should know that chlorine and iodine are both halogen groups and the reactants given are halides. The D group in the halide is an isotope of hydrogen. So, it will similarly act just like hydrogen atoms do.
So, when alkynes react with halides, Markovnikov’s rule undergoes where a carbocation is formed on the carbon atom having more alkyl groups and the halide will get attached to this carbon breaking the triple bond while the isotope group or a hydrogen atom will get attached to the carbon containing less hydrogen atoms.
So, as per the given question the chlorine atom will get attached to the carbon containing more alkyl groups and the isotope atom will get attached to the carbon having less hydrogen atoms. And this reaction process can be shown as follows:
$C{{H}_{3}}CH\equiv CH\xrightarrow{DCl}C{{H}_{3}}-C(Cl)=CH-D$

Now, when further the product which is formed is reacted with $DI$, again Markovnikov’s rule undergoes and here the iodine atom will get attached to the carbon having more substituent groups and the isotope will be attached to the carbon having more hydrogen atoms. And the reaction is shown as follows:
$C{{H}_{3}}-C(Cl)=CH-D\xrightarrow{DI}C{{H}_{3}}C(Cl)(I)CH{{D}_{2}}$
So, the correct answer is “Option D”.

Note: It is important to note that, in Markovnikov’s rule when an asymmetric alkene or alkyne reacts with a hydrogen halide, the halide group gets attached to the carbon having more substituent or alkyl groups while it is opposite in case of anti-Markovnikov’s rule.