Question

Regarding E2 reactions, predict the stereochemistry of the bromoalkene that would result from the loss of one mol of `HBr` from meso- and from (±)-1,2-dibromo-1,2-diphenylethane. Which reaction should be slower? Why?

Answer 1

`"meso" → E ("slower"); "racemic" → Z ("faster")`

Explanation:

`"E2"` stereochemistry

In `"E2"` eliminations, the β-hydrogen must be antiperiplanar to the leaving group.

1,2-Dibromo-1,2-diphenylethane

The structure of 1,2-dibromo-1,2-diphenylethane is

E2 elimination of meso-1,2-dibromo-1,2-diphenylethane

The Newman projection of the meso isomer is

We know that the structure is meso because the order of groups `"H → Ph → Br"` is the same direction (clockwise) on each carbon atom. There must be an internal mirror plane.

In the transition state, the `"Ph"` on the back carbon is gauche to `"Ph"` and `"Br"` on the front carbon.

This gives a large activation energy and make for a slow reaction.

The product will have the two phenyl groups on the same side, giving an `E`-configuration.

(from chemwiki.ucdavis.edu)

E2 elimination of (±)-1,2-dibromo-1,2-diphenylethane

The Newman projection for one of the enantiomers is

We know that the structure is chiral because the order of groups `"H → Br → Ph"` is clockwise on one carbon atom and counterclockwise on the other. There can be no internal mirror plane.

In the transition state, the `"Br"` on the back carbon is gauche to `"Ph"` and `"Br"` on the front carbon.

The activation energy will be less, so this will be a faster reaction.

The product will have the two phenyl groups on opposite sides, giving a `Z`-configuration.

(from chemwiki.ucdavis.edu)