Question

Reaction of HBr with \((R)-3\) -methylhexan-3-ol leads to racemic 3 -bromo3-methylhexane. Explain.

Short answer

The reaction forms a racemic mixture because a carbocation intermediate allows Br- to attack from either side, resulting in both enantiomers.

Step-by-step solution

Understand the Reactants

We begin by identifying the reactants involved in the chemical reaction. We have (R)-3-methylhexan-3-ol, which is an alcohol with a chiral center at carbon 3. The reagent is HBr, which will act as both a strong acid and a nucleophile.

Recognize the Reaction Type

The reaction between (R)-3-methylhexan-3-ol and HBr is a nucleophilic substitution reaction. The hydroxyl (-OH) group of the alcohol will be replaced by the bromide ion (Br-).

Formation of a Carbocation

When HBr reacts with the alcohol, the hydroxyl group is protonated, making it a better leaving group. This results in the formation of water and an unstable carbocation at the tertiary carbon (carbon 3).

Carbocation Intermediate Inversion

The tertiary carbocation can rearrange, but in this molecule, it remains as is due to stability. The planar nature of the carbocation means the incoming bromide ion can attack from either side, leading to different stereochemical outcomes.

Formation of Racemic Mixture

Since the nucleophilic attack by Br- can occur from either side of the planar carbocation, both the (R) and (S) enantiomers of 3-bromo-3-methylhexane are formed in equal amounts, resulting in a racemic mixture.

Key concepts

Carbocation Intermediate

In the nucleophilic substitution reaction, the formation of a carbocation is a key step. When the alcohol (R)-3-methylhexan-3-ol reacts with HBr, the hydroxyl group gets protonated, facilitating its departure as water. What is left is a positively charged carbocation.
This carbocation is particularly important because it determines how the rest of the reaction will proceed. It occurs at the tertiary carbon, which is quite stable despite its positive charge. The stability is crucial as it allows the reaction to proceed before the carbocation rearranges, which is common in less stable carbocations. The planar shape of the carbocation means it can be attacked by a nucleophile from either side. This characteristic is what ultimately leads to the formation of a racemic mixture.

Racemic Mixture

The concept of a racemic mixture arises from the nature of nucleophilic attack on a carbocation. When the bromide ion (Br-) attacks the carbocation formed during the reaction, it can approach from either side due to the planar geometry. This reactivity results in the formation of two enantiomers: (R) and (S).
Since both enantiomers are formed in equal amounts, the product is a racemic mixture. Racemic mixtures are significant in chemistry as they demonstrate the non-selective nature of certain reactions. In this reaction, the equal probability of an attack from either side of the carbocation underscores this concept. A racemic mixture is optically inactive, meaning it does not rotate plane-polarized light, which is a crucial indicator in identifying its formation.

Chiral Center

Chirality is an important aspect of many organic molecules. A chiral center is a carbon atom that is bonded to four different atoms or groups, giving rise to non-superimposable mirror images called enantiomers. In this exercise, (R)-3-methylhexan-3-ol has a chiral center at the third carbon.
This chiral center starts in a specific configuration—(R) in this case. However, the reaction with HBr leads to the disruption of this configuration due to the formation of a carbocation. The flat structure of the carbocation means that the resulting product can have either (R) or (S) configuration. This is why a racemic mixture is formed. Understanding chiral centers helps to appreciate the complexity and specificity of stereochemistry in chemical reactions.