The reaction of \((\pm)-2,3\) -dibromobutane with ethoxide ion produces trans-2-bromo-2-butene while, under the same conditions, meso-2, 3 -dibromobutane produces the corresponding cis isomer. With the aid of three- dimensional representations, determine whether this is an anti (trans) elimination. Explain how it is possible for a "trans" elimination to produce both cis and trans product.

The reaction conditions given are as follows: 1. (±)-2,3-dibromobutane with ethoxide ion → trans-2-bromo-2-butene 2. meso-2,3-dibromobutane with ethoxide ion → cis-2-bromo-2-butene

To better understand the reaction mechanism, let's first visualize the three-dimensional structures of both (±)-2,3-dibromobutane and meso-2,3-dibromobutane. (±)-2,3-dibromobutane exists as a racemic mixture of R,R and S,S enantiomers. They both have a chiral carbon at the 2nd and 3rd carbon atoms. In the R,R enantiomer, the two bromine atoms are on the same side of the molecule (either both wedge or both dash), whereas in the S,S enantiomer they are on the opposite side. meso-2,3-dibromobutane is the diastereomer of (±)-2,3-dibromobutane, having one chiral carbon in R configuration and the other in the S configuration. These molecules have a plane of symmetry, and due to this symmetry, the two bromine atoms are also on the opposite side of the molecule.

The reaction of both dibromobutane isomers with ethoxide ions will undergo E2 (bimolecular elimination) reaction mechanism. In an E2 reaction, both the leaving group (bromide ion) and the hydrogen that is being eliminated are removed simultaneously in a single, concerted step leading to the formation of an alkene product. To determine if it's an anti (trans) elimination, we need to study the spatial alignment of the leaving group and the hydrogen atom being eliminated.

For (±)-2,3-dibromobutanes: In both enantiomers, the hydrogen atoms that are being eliminated are on the opposite side of the molecule as the corresponding leaving group (bromide ion). This can result in an anti (trans) elimination because the leaving groups and the hydrogen atoms being eliminated are aligned along the opposite sides of the molecule, resulting in the formation of the trans-2-bromo-2-butene product. For meso-2,3-dibromobutane: The two bromine atoms are on the opposite side of the molecule, and so are the hydrogen atoms that are being eliminated. In this case, an anti (trans) elimination is also possible since the leaving groups and the hydrogen atoms being eliminated are aligned along the opposite side of the molecule, resulting in the formation of cis-2-bromo-2-butene.

This problem demonstrates that an elimination reaction can be considered anti (trans) even if it results in different isomeric products, such as cis and trans. The key factor is that the relative spatial alignment of the leaving group and the hydrogen atom being eliminated is what dictates if the reaction proceeds via an anti (trans) elimination mechanism. In both reactions, the leaving group and the hydrogen atom being eliminated are aligned along the opposite side of the molecule, satisfying the condition for an anti (trans) elimination mechanism.