Question

What stereochemistry do you expect for the alkene obtained by E2 elimination of \((1 R, 2 R)-1,2\) -dibromo- 1,2 -diphenylethane? Draw a Newman projection of the reacting conformation.

Short answer

The alkene formed will be trans-stilbene through E2 elimination.

Step-by-step solution

Understand the Reactant

The reactant, \( (1R, 2R)-1,2\)-dibromo-1,2-diphenylethane, is a dibromo compound with phenyl groups (\(Ph\)) on each carbon. The bromine atoms and the phenyl groups are on the first and second carbon, and we need to determine the stereochemistry upon elimination of HBr to form an alkene.

Find Suitable Conformation for E2 Elimination

For E2 elimination to occur, we need an anti-periplanar geometry, meaning the hydrogen and bromine must be 180 degrees apart. The Newman projection can help visualize this. Put the molecule into a conformation where the Br and a neighboring H (on the adjacent carbon) are anti-periplanar.

Draw the Newman Projection

Visualize along the C1-C2 bond. The front carbon (C1) will show one bromine atom and a phenyl group. Similarly, the back carbon (C2) will have another bromine atom and phenyl group. Identify and draw the hydrogen that is antiperiplanar to one bromine. This form will lead directly to the required stereochemical result after elimination.

Determine the Product of E2 Elimination

With elimination, the Br and H being anti-periplanar will leave, which will form a double bond between C1 and C2. The phenyl groups, due to this anti arrangement, will be on opposite sides of the formed double bond, resulting in a trans-alkene configuration.

Key concepts

Stereochemistry of Alkenes

Alkenes are hydrocarbons containing at least one carbon-carbon double bond, and the stereochemistry of alkenes is a key aspect in understanding their structure and reactivity. In E2 elimination reactions, the configuration of products is greatly impacted by the starting stereochemistry. When discussing stereochemistry, alkenes can be classified within two major categories:

• Cis (Z) - where substituents on the double-bonded carbons are on the same side.
• Trans (E) - where substituents are on opposite sides.

In the case of the E2 elimination of (1R, 2R)-1,2-dibromo-1,2-diphenylethane, we produce a trans-alkene (E). This is dictated by the anti-periplanar arrangement required in E2 mechanisms. Keeping these stereochemical outcomes in mind helps predict the final structure of the product.

Newman Projection

A Newman projection is a method used to visually analyze and represent the conformation of a molecule. This tool is incredibly useful in understanding stereochemical aspects of chemical reactions, particularly the E2 elimination. To draw a Newman projection:

• Look along the axis of a bond, typically between two carbon atoms, and represent the front atom as a dot.
• The back atom appears as a circle around the dot.
• Draw substituents coming off each atom, where they usually appear as sticks projecting outward.

For the given compound (1R, 2R)-1,2-dibromo-1,2-diphenylethane, viewing along the C1-C2 bond allows us to see the bromine and hydrogen substituents in their critical orientation for the anti-periplanar configuration. Properly employing a Newman projection helps ensure accurate visualization and understanding of molecular interactions necessary for the E2 reaction.

Anti-Periplanar Geometry

Anti-periplanar geometry is pivotal in reactions, specifically E2 eliminations, as it requires specific spatial relationships between atoms and substituents. This geometry translates to a situation where atoms or groups are positioned on opposite sides but in the same plane. For E2 reactions:

• It involves a hydrogen ( H) and a leaving group (such as bromine, Br), which should be 180 degrees apart when viewed in a Newman projection.
• This arrangement is essential as it allows for optimal overlap of atomic orbitals and facilitates a concerted mechanism where the hydrogen is removed, and the leaving group exits simultaneously.

With anti-periplanar orientation, (1R, 2R)-1,2-dibromo-1,2-diphenylethane undergoes efficient elimination forming a trans-alkene. Understanding this geometry is crucial because it dictates the stereochemical outcome of these reactions, ensuring the formation of the most stable alkene.