Question

This concerns Marc Loudon's Organic Chemistry 6th Edition's 5.2 page 187.

Consider 2-methyl-1-propene and its halohydrin formation mechanism. Usually, there will be formation of bromonium, where I assume there is no way that solvent stabilization will create carboncation.

Q:
Why does not hydronium attack C1 from opposite side of bromide of bromonium ion to open the ring? Consider swapping methyl group at C2 position by t-butyl group. I would expect significant steric hindrance from 2 t-butyl groups and this would force ring opening attack at C1 position as attacking happens at opposite side of bromide. If I remember correctly from Sunderberg's Advanced Organic Chemistry reading on bromonium cation, there is possibility of steric hindrance happening.

Answer 1

The formation of halohydrins from alkenes involves the creation of a bromonium ion intermediate, and the nucleophile usually attacks the less substituted carbon due to less steric hindrance. In the case with significant steric hindrance at one carbon (e.g., due to a t-butyl group), the nucleophile may attack the other carbon, but the overall reaction is influenced by several factors.

Step-by-step explanation:

The question is centered around the mechanism of halohydrin formation from alkenes and the factors that influence the site of nucleophilic attack in the intermediate bromonium ion. In the case of 2-methyl-1-propene reacting with a halogen like bromine, a bromonium ion intermediate is formed.

The question asks why a hydronium ion (H3O+) doesn't attack the more substituted carbon (C1), particularly when steric hindrance is considered by replacing the methyl group at C2 with a t-butyl group.

Generally, in halohydrin formation, the nucleophile (which can be water in this case) attacks the more accessible carbon atom, which is typically the less substituted one, because it offers less steric hindrance. In the case where a t-butyl group is present, the steric hindrance at C2 becomes significant, possibly prompting an attack at C1.

However, the actual outcome of such a reaction can still be influenced by other factors such as the nature of the solvent, the overall stability of the carbocation that might form as an intermediate, and the conditions under which the reaction is carried out.

It is important to note that while in SN1 reactions, solvolysis can lead to carbocation formation with subsequent attack by nucleophiles on either side, the reaction involving alkenes and halogens to form halohydrins typically proceeds through the more stable bromonium ion intermediate without the formation of a free carbocation.