Question

We saw in Section 8-7 that bromohydrins are converted into epoxides when treated with base. Propose a mechanism, using curved arrows to show the electron flow.

Short answer

The mechanism for the conversion of bromohydrins to epoxides on treating with base is,

Step-by-step solution

Halohydrin to epoxide

In converting halohydrin to epoxide, the base deprotonates the alcohol and forms an alkoxide intermediate. Further by S_N2 reaction, the nucleophile (alkoxide) attacks the electrophilic C and displaces the leaving group (halide ion).

The mechanism for the conversion of bromohydrins to epoxides on treating with a base

When a starting material, bromohydrin, is rotated by 180^o, the C-C bonds are rotated so that the –OH and –Br bonds are 180^o apart. Now, bromohydrin is treated with a base, and proton removal occurs.

After that, the S_N2 displacement -Br^-of -O⁺ by occurs, which leads to the formation of epoxide ring and ion. This reaction is an intramolecular S_N2 displacement.

The mechanism for the conversion of bromohydrins to epoxides