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?

#"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)