Question

Reaction of meso-1, 2 -dibromo \(-1,2\) -diphenylethane (stilbene dibromide) with alc. \(\mathrm{KOH}\) (or \(\left.\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OK} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) gives (a) cis \(-1\) -bromo \(-1,2\) -diphenyl ethene (b) trans-1-bromo-1, 2-diphenyl ethene (c) \(1: 1\) mixture of (a) and (b) (d) \(1: 2\) mixture of (a) and (b)

Short answer

The major products formed in the reaction are a 1:1 mixture of cis-1-bromo-1,2-diphenyl ethene and trans-1-bromo-1,2-diphenyl ethene.

Step-by-step solution

Write down the reactants

Write meso-1,2-dibromo-1,2-diphenylethane (stilbene dibromide) reacting with an alcoholic KOH or \(\left. C_{2} H_{5} OK / C_{2} H_{5} OH\right)\). It will look like this: meso-1,2-dibromo-1,2-diphenylethane + \(\left. C_{2} H_{5} OK / C_{2} H_{5} OH\right)\)

Identify the functional groups in the reactants

In the meso-1,2-dibromo-1,2-diphenylethane molecule we have a dihalide (two bromine atoms) and two phenyl groups. The alcoholic KOH or \(\left. C_{2} H_{5} OK / C_{2} H_{5} OH\right)\) is a strong base.

Determine the reaction mechanism

Since we have a dihalide reacting with a strong base, this reaction will follow the E2 (Elimination Bimolecular) mechanism, where the base abstracts a proton from the substrate and the halides leave the substrate to form a new double bond.

Perform the E2 elimination in meso-1,2-dibromo-1,2-diphenylethane

During the E2 elimination, the alcoholic KOH (or \(\left. C_{2} H_{5} OK / C_{2} H_{5} OH\right)\) will act as a strong base, abstracting the proton from the alpha-carbon (the carbon adjacent to the carbon attached to the two bromine atoms), leading to the formation of a double bond and the elimination of one bromine atom: meso-1-bromo-1,2-diphenylethane Since the remaining molecule has symmetry, any cis or trans isomers formed will be in 1:1 ratio.

Identify the correct answer

In conclusion, after eliminating one bromine atom, the reaction will yield a 1:1 mixture of cis-1-bromo-1,2-diphenyl ethene and trans-1-bromo-1,2-diphenyl ethene. So, the correct option is: (c) 1:1 mixture of (a) and (b)

Key concepts

Meso Compounds

A meso compound is an achiral molecule that contains multiple stereocenters. Despite having stereocenters, meso compounds have an internal plane of symmetry. This symmetry allows them to be superimposable on their mirror image, making them achiral. In the context of the reaction we're looking at, meso-1,2-dibromo-1,2-diphenylethane is a meso compound. It possesses symmetrical features, with two bromine atoms and two phenyl groups arranged in a way that reflects symmetry across a central plane.

• Key point: Meso compounds are a curious case where symmetry negates the chirality typically found in molecules with stereocenters.

Understanding meso compounds is important for predicting outcomes in reactions, particularly when considering stereochemistry.

Dihalides

Dihalides are organic compounds containing two halogen atoms. In our exercise, we focus on meso-1,2-dibromo-1,2-diphenylethane, which has two bromine atoms attached to adjacent carbons. These compounds are important in elimination reactions, particularly E2 mechanisms. The presence of two halides makes dihalides excellent candidates for reactions where one or both halogens can be removed, leading to the formation of alkenes.

• The reactivity is influenced by the position of the halogens and the type of base present.
• In elimination reactions, halogens serve as good leaving groups.

Dihalides often participate in reactions that form alkenes, as seen with their reaction with strong bases like alcoholic KOH in this exercise.

Stereochemistry

Stereochemistry involves the study of the spatial arrangement of atoms within molecules. It's crucial in organic chemistry to understand how different isomers of a compound can have vastly different properties. In our specific scenario, the reaction of meso-1,2-dibromo-1,2-diphenylethane with a strong base results in the assessment of stereochemistry to determine the ratio of the resulting products.

• When meso compounds react, it is possible to form different geometric isomers (cis and trans in this case).
• The symmetrical nature of meso compounds can lead to equal quantities of these isomers.

In this exercise, recognizing the symmetry in the reactant helps predict a 1:1 ratio of cis and trans products, due to the equal chance of forming each isomer.

Strong Bases in Organic Reactions

Strong bases are vital in many organic reactions, particularly elimination reactions. They can remove a proton from a substrate, leading to the formation of double bonds. In the exercise, alcoholic KOH acts as a strong base, facilitating the E2 elimination reaction of meso-1,2-dibromo-1,2-diphenylethane.

• Strong bases like \( ext{KOH}\) or \( ext{C}_2 ext{H}_5 ext{OK}\) are key in E2 reactions where they deprotonate the molecule, promoting double bond formation.
• The presence of a strong base also ensures that the reaction proceeds through a concerted process where proton removal and halide departure occur simultaneously.

Understanding how strong bases influence the reactivity of molecules helps in predicting the products and their stereochemical outcomes.