Question

When propene reacts with $HCl$ in presence of peroxide, the product is:
(A) 1- chloropropane
(B) 1,1- dichloropropane
(C) 2- chloropropane
(D) 1,2- dichloropropane

Answer 1

Hint: The question has specifically mentioned peroxide. That means there is a process without the presence of peroxide. Maybe the results of both these processes will be opposite.

Complete answer:
The following diagram depicts what will happen when peroxide is absent in the reaction mixture:
\[C{{H}_{3}}-CH=C{{H}_{2}}+HCl\to C{{H}_{3}}-CHCl-C{{H}_{3}}\]
This reaction follows Markovnicov’s rule. In simpler terms this rule states- the addition of a protic acid HX to an asymmetric alkene leads to the attachment of the conjugate base (${{X}^{-}}$ ) at the carbon with least hydrogen substituents.
The above reaction follows an intermediate state where the double bond breaks to form a carbocation and carbanion. As a secondary carbocation is more stable than a primary one, that is where the conjugate base (${{X}^{-}}$ ) attaches itself. The intermediate state is shown below:
\[C{{H}_{3}}-C{{H}^{+}}-C{{H}_{2}}^{-}\]
The reaction does not follow this route when peroxide is present in the reaction medium. It follows the anti-Markovnikov's rule, which as its name suggests, is the opposite to the above reaction. The reaction in this case is the following:
\[C{{H}_{3}}-CH=C{{H}_{2}}+HCl\xrightarrow{{{H}_{2}}{{O}_{2}}}C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Cl\]
As you can see, the $-C{{l}^{-}}$ group is attached to the last carbon which is a primary carbocation. The intermediate stage in this case looks like:
\[C{{H}_{3}}-C{{H}^{-}}-C{{H}_{2}}^{+}\]
This phenomenon is known as the peroxide effect.
From the above discussion it is concluded that the answer to this question is (A) 1-chloropropane.

Note:
It should be kept in mind that the Marcovnicov’s rule is applicable to any asymmetric alkene or alkyne, but the presence of peroxide does not always guarantee the “anti” process. There are specific conditions that have to be met first.