Question

Two diastereomeric sets of enantiomers, A/B and C/D, exist for 3 -bromo-2-butanol. When enantiomer A or \(\mathrm{B}\) is treated with \(\mathrm{HBr}\), only racemic 2,3 -dibromobutane is formed; no meso isomer is formed. When enantiomer C or \(\mathrm{D}\) is treated with \(\mathrm{HBr}\), only meso 2,3 -dibromobutane is formed; no racemic 2,3 -dibromobutane is formed. Account for these observations.

Short answer

Answer: The stereochemistry of the starting enantiomers determines the spatial arrangement of bromine atoms during the reaction with HBr. For enantiomers A/B, bromide ion attacking carbon 2 results in racemic 2,3-dibromobutane without a plane of symmetry. For enantiomers C/D, bromide ion attacking carbon 2 leads to the formation of meso 2,3-dibromobutane with a plane of symmetry.

Step-by-step solution

Identify the groups in 3-bromo-2-butanol

3-bromo-2-butanol has a bromine atom on carbon 3 and a hydroxyl (OH) group on carbon 2. Since both of these groups are located on chiral centers, it leads to the formation of two diastereomeric sets of enantiomers, A/B and C/D.

Analyze the reaction mechanism with enantiomer A/B

When enantiomer A or B reacts with HBr, the hydroxyl group (\textrm{OH} ) on carbon 2 gets protonated to form a good leaving group (water). The resulting carbocation is resonance-stabilized, with a positive charge distributed between carbon 2 and carbon 3. If a bromide ion attacks carbon 2, racemic 2,3-dibromobutane is formed. No meso isomer is formed in this case since both bromines are placed in a way that doesn't yield a plane of symmetry in the molecule.

Analyze the reaction mechanism with enantiomer C/D

When enantiomer C or D reacts with HBr, the hydroxyl group (\textrm{OH} ) on carbon 2 gets protonated, similar to Step 2. However, bromide ion attack on carbon 2 leads to the formation of meso 2,3-dibromobutane due to a plane of symmetry in the molecule. In this case, no racemic product is formed as the reaction outcome depends on the spatial arrangement of the two bromine atoms.

Explain the observations

The difference in reaction outcomes between the two sets of enantiomers can be explained by their stereochemistry. In enantiomers A/B, the attack of bromide ion on carbon 2 leads to racemic 2,3-dibromobutane formation without a plane of symmetry. In enantiomers C/D, the attack of bromide ion on carbon 2 forms meso 2,3-dibromobutane with a plane of symmetry. Stereochemistry of the starting enantiomers determines the spatial arrangement of bromine atoms and the formation of racemic or meso products upon reaction with HBr.

Key concepts

Stereochemistry

Stereochemistry is a fundamental aspect of organic chemistry that deals with the three-dimensional arrangement of atoms within molecules. Understanding stereochemistry is essential when discussing diastereomers and enantiomers, two types of stereoisomers. Enantiomers are mirror images of each other and cannot be superimposed, like left and right hands. Diastereomers, on the other hand, are not mirror images and differ in the spatial arrangement of at least one (but not all) of their multiple chiral centers.

In the exercise, 3-bromo-2-butanol possesses chiral centers, which lead to the existence of multiple stereoisomers – two pairs of enantiomers that are also diastereomeric to each other (A/B and C/D). The stereochemistry of the starting material influences the outcome of reactions, including whether a racemic mixture or a meso compound is formed upon treatment with HBr.

Meso Compound Formation

Meso compounds are a special type of stereoisomer that have multiple chiral centers but are superimposable on their mirror image due to an internal plane of symmetry. As a result, these compounds are achiral. Despite having chiral centers, meso compounds do not exhibit optical activity because they are identical to their mirror images.

For instance, the reaction of enantiomer C or D of 3-bromo-2-butanol with HBr results exclusively in a meso compound, meso-2,3-dibromobutane. The formation of this meso compound means that after the reaction, there's a plane of symmetry that makes both halves of the molecule mirror images of each other, effectively canceling any optical activity.

Racemic Mixture

A racemic mixture contains equal amounts of two enantiomers of a chiral molecule, resulting in a solution that has no net optical activity. The presence of both enantiomers in equal quantity means that their optical activities cancel each other out.

In the case of the reaction of enantiomer A or B with HBr, a racemic mixture of 2,3-dibromobutane is produced. This outcome occurs because the spatial arrangement of the substituents in enantiomers A and B allows for the formation of both enantiomers of the product when a nucleophilic attack happens, resulting in no net chiral preference and thus no optical activity.

Reaction Mechanism

The reaction mechanism describes the step-by-step sequence of elementary reactions by which overall chemical change occurs. It includes details about how bonds are broken and formed, as well as how the reactants transform into the products.

In our exercise, when 3-bromo-2-butanol reacts with HBr, the -OH group is protonated, making water a good leaving group that departs to form a carbocation. The next step involves a nucleophilic attack by bromide ion. The nature of the carbocation intermediate and the point of attack by the nucleophile dictate whether a racemic mixture or a meso compound is formed. Enantiomers A and B go through a pathway leading to a racemic mixture, while C and D result in a meso compound. This is largely dependent on the initial stereochemistry of each enantiomer, as it influences the stability and geometry of the carbocation intermediate.