Question

predict the major product.

a.

b.

c.

d.

Explanation:

Step1: Formation of bromonium ion

When the alkene reacts with \(Br_2\) in \(H_2O\), a bromonium ion intermediate is formed. The bromine atom (from \(Br_2\)) attacks the double - bond, creating a three - membered ring with a positive charge on bromine.

Step2: Nucleophilic attack by water

Water (\(H_2O\)) acts as a nucleophile. It attacks the more substituted carbon of the bromonium ion. In this case, the carbon attached to the methyl group (more substituted) is attacked. The attack of water follows the anti - addition rule (because of the bromonium ion intermediate). The oxygen of water donates a pair of electrons to the carbon, breaking one of the carbon - bromine bonds in the bromonium ion.

Step3: Deprotonation

After the nucleophilic attack, a proton (\(H^+\)) is removed from the oxonium ion (formed from the water nucleophile) to give the final product. The bromine atom remains attached to the less substituted carbon (relative to the site of water attack) in an anti - configuration.

The product has the \(Br\) and \(OH\) groups in an anti - relationship. Also, considering the steric and electronic effects (more substituted carbon is more electrophilic in the bromonium ion for the nucleophilic attack of water), the correct structure is the one where \(OH\) is on the more substituted carbon (attached to methyl) and \(Br\) is on the adjacent carbon in an anti - orientation.

Answer:

A. Option Text