Question

Two diastereomeric sets of enantiomers, \(A / \mathrm{B}\) and \(\mathrm{C} / \mathrm{D}\), exist for S-bromo-2-butanol. CC(O)C(C)Br CC(O)C(C)Br CC(O)C(C)Br \(A\) \(\mathrm{B}\) 4 1) When enantiomer \(A\) or \(\mathrm{B}\) is treated with HBr, only racemic 2,3 -dibromohutane is formed; ao meso isomer is formed. When enantiomer \(\mathrm{C}\) or \(\mathrm{D}\) is treated with HBr, only meso 2,3-dihromobutane is formed; no racenaic 2,3-dibroenobutane is formed. Ancount for these oboervarions.

Short answer

Question: Explain why the reaction of enantiomers A/B with HBr results in racemic 2,3-dibromobutane, while the reaction of enantiomers C/D with HBr results in meso 2,3-dibromobutane. Answer: The reaction of enantiomers A/B with HBr results in racemic 2,3-dibromobutane because the inversion of configuration at C3 creates two new stereocenters with opposite configurations, giving equal amounts of both product enantiomers. In contrast, the reaction of enantiomers C/D with HBr results in meso 2,3-dibromobutane because when the configuration at C3 is inverted, both stereocenters have the same configuration, giving a molecule with internal symmetry and no net optical rotation.

Step-by-step solution

Analyze the structure of the starting enantiomers

Each set of enantiomers (A/B and C/D) has an S-bromo-2-butanol structure. They are diastereomers, meaning they have different configurations at one or more stereocenters. In this case, there are two stereocenters: the carbon bearing the bromine (C2) and the one bearing the hydroxyl group (C3). Assign the R,S configurations to each enantiomer for further analysis.

Understand the reaction mechanism with HBr

The reaction of 2-bromobutanol with HBr proceeds via an S_N2 mechanism, meaning that the bromide anion (\mathrm{Br}^-) will perform a nucleophilic attack on the carbon with the hydroxyl group (C3). Simultaneously, the hydroxyl group (-OH) is released from the molecule. This results in a substitution of -OH by a -Br group at C3. The stereochemistry of this reaction is important, as an S_N2 reaction causes an inversion of configuration at the attacked carbon atom.

Derive the product stereochemistry for each enantiomer set

For each enantiomer (A, B, C, and D) and their respective mechanisms, determine the stereochemistry of the 2,3-dibromobutane products obtained after the reaction with HBr. Consider that the configuration at C3 will be inverted and that the configuration at C2 remains the same.

Account for the observations for the reaction of A/B enantiomers with HBr

For enantiomers A and B, reacting with HBr leads to racemic 2,3-dibromobutane. This suggests that both product enantiomers are formed in equal amounts and that there is no meso compound, which would exhibit internal symmetry and no net optical rotation. In this case, C3 inversion results in the formation of two new stereocenters with configurations opposite to those of the starting materials. Each enantiomer gives either R or S configuration at both C2 and C3.

Account for the observations for the reaction of C/D enantiomers with HBr

When enantiomers C and D react with HBr, it is observed that only meso 2,3-dibromobutane is formed. This suggests that the stereochemistry of the products has internal symmetry, leading to no net optical rotation. For these enantiomers, when the configuration at C3 is inverted, both stereocenters have the same configuration (either R or S). Meso 2,3-dibromobutane has the same configuration at both C2 and C3, and a plane of symmetry exists between these two stereocenters, making the molecule achiral. By analyzing the stereochemistry of the starting enantiomers and the resulting products, we can account for the formation of either racemic or meso 2,3-dibromobutane products, depending on whether the starting enantiomers are A/B or C/D.

Key concepts

S_N2 Reaction Mechanism

Understanding the S_N2 reaction mechanism is crucial when studying organic chemistry, especially for interpreting the outcomes of reactions involving chiral centers. In an S_N2 reaction, the 'S_N' stands for nucleophilic substitution, and the '2' indicates that the rate-determining step involves two species: the nucleophile and the electrophile.

The signature feature of an S_N2 reaction is the backside attack by the nucleophile. This attack occurs on the electrophilic carbon atom that is bonded to the leaving group. As the nucleophile approaches, it forms a bond with the carbon, simultaneously displacing the leaving group. This concerted mechanism—one step with bond-making and bond-breaking events happening at once—results in the inversion of the configuration at the chiral center, which is called 'Walden inversion'.

An everyday example that mirrors an S_N2 process is opening a door (the electrophile) by pushing the handle (the nucleophile) where your hand replaces the original handle position (the leaving group). The inverted stereochemistry following an S_N2 reaction is vital to its outcome, particularly in the creation of diastereomers, racemic mixtures, or meso compounds from chiral reactants.

Diastereomers

Diastereomers are a type of stereoisomer found in organic molecules with two or more stereogenic (chiral) centers. Unlike enantiomers, which are non-superimposable mirror images of each other, diastereomers are not mirror images and differ in configuration at one or more (but not all) of the chiral centers.

Imagine that your hands represent molecules with fingers as substituents. If all fingers (substituents) are arranged differently between your left and right hand, you have enantiomers, but if only some fingers differ, this represents diastereomers. They have distinct physical properties, like melting points and solubilities, and often react differently in chemical reactions, unlike enantiomers, which usually behave identically in an achiral environment.

In our textbook problem, the S-bromo-2-butanol and its respective products have multiple stereocenters. This complexity contributes to variations in the molecules' three-dimensional orientations, leading to different compounds with unique properties—diastereomers.

Racemic Mixture

A racemic mixture, also simply called a racemate, is an equimolar mixture of two enantiomers of a chiral molecule. These mixtures do not exhibit optical activity, which means they do not rotate plane-polarized light because the rotations caused by each enantiomer cancel each other out.

Think of it as a dance floor with pairs of dancers spinning in opposite directions. Each pair represents an enantiomer pair, and as they spin equally but oppositely, the overall movement on the dance floor appears to be stationary. This is analogous to a racemic mixture's effect on polarized light—no net rotation occurs.

In chemical reactions like the one depicted in the exercise, if both enantiomers are formed in equal amounts (like A and B giving rise to enantiomers of 2,3-dibromobutane), a racemic mixture is the result. This condition is common when a reaction, like an S_N2 reaction, proceeds through a mechanism that does not favor one enantiomer over the other.

Meso Compound

A meso compound is a unique type of stereoisomer. Despite having multiple stereocenters, it is achiral because it has an internal plane of symmetry. This symmetry means that the molecule is superimposable on its mirror image, a property not held by chiral molecules.

Think of a meso compound as a symmetrical building with two identical wings. If you draw a line down the center of the building, both sides are mirror images, creating no difference overall, regardless of your perspective. In the context of the textbook problem, the meso compound is 2,3-dibromobutane, resulting from enantiomers C and D after the S_N2 reaction with HBr. Instead of yielding a racemic mixture, the reactions lead to a meso product, which is achiral and therefore does not contribute to optical activity.

Meso compounds play an important role in organic synthesis and stereochemistry because they offer a pathway to synthesize achiral compounds from chiral starting materials without requiring a separate resolution step to separate enantiomers.