© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Desired Product
The product obtained from the addition of HCl is given as 1-chloro-1-methylcyclohexane. This means that, after the addition of HCl to the alkene, the cyclohexane ring carries both a chlorine and a methyl group on the same carbon (named the 1-position).
Step 2: Recall Markovnikov’s Rule
According to Markovnikov’s rule, when a hydrogen halide (such as HCl) adds to an alkene, the hydrogen ($H^+$) bonds to the carbon with more hydrogen atoms (i.e., the less substituted carbon), and the halide ($Cl^-$) attaches to the other carbon of the double bond (the more substituted carbon).
Step 3: Determine the Double-Bond Position Needed
To form 1-chloro-1-methylcyclohexane, the chlorine must go to the ring carbon that was more substituted in the alkene. Simultaneously, a methyl group must remain there. Hence, before reaction:
The ring carbon that ends up bearing both methyl and chlorine must have been one of the alkene carbons.
Among the alkene carbons, the carbon that will receive the chlorine (in Markovnikov fashion) is the one that is more substituted or can stabilize the carbocation better.
Step 4: Check the Provided Options
From the given images (A), (B), (C), and (D) (where (C) is (A) and (B)), options (A) and (B) have the double bond positioned in a way that on addition of HCl, the ring carbon which is substitued by the methyl group will get the chlorine (Markovnikov addition). Thus, both (A) and (B) can yield 1-chloro-1-methylcyclohexane.
Step 5: Conclude the Correct Answer
Hence, the alkenes in (A) and (B) will react with HCl according to Markovnikov’s rule to give 1-chloro-1-methylcyclohexane. Therefore, the correct answer is “(A) and (B).”
