Consider the addition of bromine to Z- or E- 2-pentene:

A priori, we consider it possible that either diastereomer of the starting material should give all possible stereoisomers of the product. Since the product has two chiral centers, four stereoisomers are possible: two pairs of enantiomers, which are diastereomers of each other.

We know that diastereomers behave differently, so we also should expect a different result for each diastereomer of the starting material.

• Furthermore, since the diastereomeric products have different energies, we expect different proportions of them in the product.

• And, since the starting materials are achiral, we expect the enantiomeric products to be formed as racemic mixtures.

Here's the reaction for the E- diastereomer of the alkene.

Formation of the bromonium ion can occur from either the top or bottom face of the alkene.

• These are equivalent, since the molecular plane is a plane of symmetry.

• Attack on the left carbon of either bromonium ion gives a product that is the enantiomer of the product from attack on the right end

• Hence the two pathways for attack must have equal energies, and the products must be formed in equal amounts.

• We get the RS and SR enantiomers, but none of either RR or SS. The reaction is stereoselective.

If reaction occurs on the Z- alkene:

the same sequence of events occurs. Since the configuration of one end of the alkene was opposite, the configuration of one end of the dibromide product is opposite. We get the RR and SS enantiomers in equal quantities, but none of the RS or SR diastereomers. The reaction again is stereoselective.

Try this one yourself: Z- and E- 2-pentene reacting with OsO₄ to produce a diol. (Remember, this reaction is syn- stereoselective. Then go back and repeat both with Z- and E- 2-butene. You should be able to predict the outcome of all of these reactions before actually writing out the structures!