Chapter 12, Problem 12.64P

Interpretation Introduction

Interpretation:

Two different ways to convert $\text{2-methylbut-2-ene}$ to $\text{3-bromo-2-methylbutan-2-ol}$ are to be shown.

Concept introduction:

When alkene reacts with molecular bromine in carbon tetrachloride as a solvent, vicinal dibromide is formed. But, when the same reaction is carried out in the water as the solvent, addition of Br and OH across the double bond takes place. In this reaction, the first step is the formation of the bromonium ion intermediate. The bromonium ion is then attacked by water rather than the bromide ion because the concentration of water is much higher.

The attack by water takes place at the more substituted carbon, opening up the bromonium and forming a protonated alcohol. The uncharged alcohol is formed by deprotonation.

Alkene, on treatment with peracids, forms an epoxide. The epoxide adds bromine at the less substituted carbon with ring-opening on treatment with NaBr under neutral conditions. Acid workup gives the corresponding alcohol.