Question

For the reaction, CH\(_3\)-CH=CH\(_2\) + HOCl \(\rightarrow\) A. The product A is

• A. CH\(_3\)-CHCl-CH\(_2\)OH
• B. CH\(_3\)-CH(OH)-CH\(_2\)Cl
• C. CH\(_3\)-CH\(_2\)-CH\(_2\)-COCl
• D. CH\(_3\)-C(Cl)(OH)-CH\(_3\)

Hint:

Markovnikov's rule: negative part adds to carbon with fewer hydrogens.

Answer 1

The Correct Option is B

To determine the product A of the reaction between propene (CH₃-CH=CH₂) and hypochlorous acid (HOCl), we should consider the steps involved in the reaction mechanism.

1. The reaction of an alkene with HOCl is similar to electrophilic addition reactions. Here, the alkene's double bond acts as a nucleophile, attacking the electrophilic center (Cl in HOCl), resulting in a chloronium ion intermediate.
2. Propene, being an unsymmetrical alkene, undergoes the addition of HOCl following the Markovnikov's rule. According to this rule, the hydroxyl group (OH) attaches to the more substituted carbon atom and the chlorine (Cl) to the less substituted carbon atom.
3. In propene, the \(\pi\)-bond electrons attack the chlorine, generating a chloronium ion. This intermediate can stabilize the positive charge better at the secondary (CH-CH) position because of more alkyl substitution, compared to the other carbon (CH₂).
4. Water attacks the more substituted carbon of the chloronium ion, opening the cycle and attaching the OH to that carbon.
5. This results in the formation of the product: CH₃-CH(OH)-CH₂Cl.

Thus, the correct product A formed in the reaction is CH₃-CH(OH)-CH₂Cl.