Question

Two diastereomeric sets of enantiomers, \(\mathrm{A} / \mathrm{B}\) and \(\mathrm{CD}\), exist for 3 -bromo-2-butanol. When enantiomer \(\mathrm{A}\) or \(\mathrm{B}\) is treated with \(\mathrm{HBr}\), only racemic 2,3 -dibromobutane is formed; no meso isomer is formed. When enantiomer \(\mathrm{C}\) 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 treatment of enantiomers A/B with HBr results in racemic 2,3-dibromobutane due to the specific reaction mechanism and retention of configuration at the chiral C-2, while the treatment of enantiomers C/D with HBr results in meso 2,3-dibromobutane due to the retention of the configuration at C-2 but inversion at C-3, giving rise to a single meso isomer (having two chiral centers but still being achiral due to the presence of a mirror plane).

Step-by-step solution

Draw the structures of enantiomers A/B and C/D

To solve this problem, the first step is to draw the structures of the given sets of enantiomers A/B and C/D for 3-bromo-2-butanol. Note that they are diastereomers, which means that they have different configurations at one or more stereogenic centers but are not mirror images of each other.

Identify the chiral centers and stereoisomers

Identify the chiral center in each molecule. Chiral centers are carbon atoms with four different substituents attached to them. In this case, there is a chiral center at C-2 and C-3 of the 3-bromo-2-butanol for both sets of enantiomers.

React enantiomers A/B with HBr

Let's look at the reaction of enantiomers A/B when treated with HBr: 1. The nucleophilic hydroxyl group (\(\mathrm{OH}\)) of enantiomer A/B attacks the electrophilic bromine atom of HBr, leading to the formation of a bromonium ion. 2. The bromide ion (\(\mathrm{Br^{-}}\)) from HBr then attacks the carbon-3 of the bromonium ion from the opposite side. This creates 2,3-dibromobutane with inversion of configurations at the chiral C-3 center.

Examine the formation of racemic and meso 2,3-dibromobutane from enantiomer A or B

The reaction of enantiomer A/B with HBr results in racemic 2,3-dibromobutane due to the retention of the configuration at C-2 but inversion at C-3, thus giving rise to a racemic mixture instead of a meso isomer.

React enantiomers C/D with HBr

Now let's examine the reaction of enantiomers C/D when treated with HBr: 1. The nucleophilic hydroxyl group (\(\mathrm{OH}\)) of enantiomer C/D attacks the electrophilic bromine atom of HBr, leading to the formation of a bromonium ion. 2. The bromide ion (\(\mathrm{Br^{-}}\)) from HBr then attacks the carbon-3 of the bromonium ion from the opposite side. This creates 2,3-dibromobutane with inversion of configurations at the chiral C-3 center.

Examine the formation of meso 2,3-dibromobutane from enantiomer C or D

The reaction of enantiomer C/D with HBr results in meso 2,3-dibromobutane due to the retention of the configuration at C-2 but inversion at C-3, therefore giving rise to a single meso isomer (having two chiral centers but still being achiral due to the presence of a mirror plane). In conclusion, the treatment of enantiomers A/B with HBr results in racemic 2,3-dibromobutane due to the specific reaction mechanism and retention of configuration at the chiral C-2, while the treatment of enantiomers C/D with HBr results in meso 2,3-dibromobutane for the same reasons. This accounts for the observed outcomes of the reactions.

Key concepts

Diastereomers

Understanding diastereomers is crucial when delving into the stereochemistry of molecules. Diastereomers are compounds that share the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations (configurations) of their atoms in space, while not being mirror images of each other. This can happen when a molecule has multiple chiral centers; diastereomers will have at least one, but not all, of these centers inverted.

In the given textbook exercise, we come across two sets of diastereomeric enantiomers, labeled A/B and C/D for the compound 3-bromo-2-butanol. They must have different configurations at the chiral centers, which ultimately gives rise to differing chemical properties and reactions, such as their differing reactivity with hydrogen bromide (HBr) to form either racemic mixtures or meso compounds.

Chiral Centers

A chiral center, often a carbon atom, is one that is bonded to four different groups or atoms. Its presence is a key factor in the three-dimensional structure of molecules and thus their stereochemistry. This distinction creates non-superimposable mirror images, or enantiomers. An easy way to identify a chiral center is to look for a carbon with four unique substituents.

With reference to the exercise, 3-bromo-2-butanol features chiral centers at the carbon atoms C-2 and C-3. Each set of enantiomers has these chiral centers arranged differently, causing a varied response when these molecules interact with other chemical agents, such as HBr.

Racemic Mixture

A racemic mixture is a 50:50 blend of two enantiomers of a chiral molecule. Since these enantiomers have identical physical properties except for the direction they rotate plane-polarized light, the mixture shows no optical activity overall – it's optically inactive. This result occurs because the rotations caused by each enantiomer cancel each other out.

In considering the exercise scenario, when enantiomers A or B are reacted with HBr, a racemic mixture of 2,3-dibromobutane forms. This is due to the specific reaction mechanism and retention of configuration at C-2, while C-3 undergoes an inversion of configuration after the reaction.

Meso Compound

A meso compound is fascinating because it has chiral centers, yet it is achiral as a whole. This occurs due to an internal plane of symmetry that bisects the compound and makes one half of the molecule a mirror image of the other half. Because of this symmetry, meso compounds do not exhibit optical activity.

As per the provided solution, reacting enantiomers C or D with HBr uniquely results in a meso 2,3-dibromobutane. This specific outcome is tied to how the substituents on the chiral centers are arranged relative to one another, leading to an achiral compound despite the presence of chiral centers, because of that crucial mirror plane within the molecule.