Treatment of the following stereoisomer of 1-bromo-1,2-diphenylpropane with sodium ethoxide in ethanol gives a single stereoisomer of 1,2 -diphenylpropene. Predict whether the product has the \(E\) configuration or the \(Z\) configuration.

The substrate in this reaction is 1-bromo-1,2-diphenylpropane. Sodium ethoxide (NaOEt) is the base used to perform the elimination reaction, with ethanol as the solvent.

Sodium ethoxide is a strong base, and our substrate has a good leaving group (bromide ion). Thus, the reaction will proceed via an E2 mechanism (one step bimolecular elimination) rather than an E1 mechanism.

In an E2 reaction, it's important to note that the hydrogen that is being removed must be antiperiplanar (opposite direction) to the leaving group (bromide ion). This allows for the formation of the pi bond in the alkene product to occur in a concerted process known as syn-periplanar elimination. In an E2 mechanism, the most stable alkene is usually preferred, and the Zaitsev's rule states that the more substituted alkene (has more alkyl groups attached) is the major product.

Considering Zaitsev's rule, in this particular example, there is only one possible alkene product: 1,2-diphenylpropene. Now, we need to determine the configuration of the double bond, whether it's E (trans) or Z (cis). In this case, to determine the stereochemistry, we need to assign priority to the substituents using the Cahn-Ingold-Prelog (CIP) rules. On one end of the double bond, both carbons have phenyl groups attached, therefore they have equal priority. On the other end of the double bond, one carbon is bonded to a hydrogen, and the other to a phenyl group. The phenyl group has higher priority than hydrogen. Since the two carbons with the double bond have the high-priority groups on the same side, the configuration is predicted to be Z (cis). Therefore, the final product should have a Z configuration: Z-1,2-diphenylpropene.