1-methylcyclohexene on treatment with HCl gas gives:
(A) 1-chloro-2-methylcyclohexane
(B) 1-chloro-1-methylcyclohexane
(C) 1-chloro-cyclohexane
(D) None of these

Hint: In this question, 1- methylcyclohexene is an asymmetric alkene whereas HCl is a hydrogen halide. So, to solve this question we first need to know how the reaction between an asymmetric alkene and a hydrogen halide takes place.

Complete answer:
We know that hydrogen halides (HX) are protic acids. i.e., these acids produce protons as they ionize in solutions.
Now, when an asymmetric alkene reacts with a protic acid, the unsaturated carbon atom which has the most number of alkyl substituents will form a bond with the halide group whereas the unsaturated carbon atom which has the most number of hydrogen substituent will bind with the acidic hydrogen to form the major saturated product.
This phenomenon is known as Markovnikov's rule.
\[R-CH=C{{H}_{2}}+HX\to R-CH(X)-C{{H}_{3}}\text{ (major)}\]
So, according to this rule, the reaction between 1-methylcyclohexene and hydrochloric acid (HCl) will be as follows.

1-methylcyclohexene on treatment with HCl gas gives option (B) 1-chloro-1-methylcyclohexane

Additional information:
The speed of the reaction of the addition of hydrogen halide to alkene depends upon the hydrogen halide. The reaction rate increases as we go from HF to HI.
HF < HCl < HBr < HI
This is because, with the increase in the size of the halide atom, the bond strength of the hydrogen halide decreases, and the bonds break easily, which further helps in increasing the rate of the reaction.

Note:
It should be noted that when hydrogen bromide (HBr) is added to an asymmetrical alkene in the presence of an organic peroxide, the reaction follows the free radical mechanism, and the products are formed according to the anti-Markovnikov's rule which states that the protic acid is added to the more alkyl-substituted carbon atom.