Question

Predict the stereochemical results of the photobromination of \((R)-3\)-methylhexane. Is the product a racemic mixture? Is it a single enantiomer?

Short answer

The product of photobromination of \(R)-3\)-methylhexane is a racemic mixture.

Step-by-step solution

Understand Photobromination

Photobromination is a reaction where bromine is added to an organic compound through the absorption of light (usually UV). The reaction often proceeds via a radical mechanism.

Identify the Reaction Site

In \(R)-3\)-methylhexane, the photobromination will most likely occur at the tertiary carbon (the carbon atom to which the methyl group is attached) because tertiary radicals are more stable than secondary or primary radicals.

Radical Formation

The bromine radical abstracts a hydrogen atom from the tertiary carbon, resulting in the formation of a tertiary carbon radical at the 3-position.

Bromine Addition

A bromine molecule (\(Br_2\)) will react with the tertiary carbon radical, leading to the formation of \(R)-3\)-bromo-3-methylhexane. During this process, the bromine can attach from either the front or the back side of the planar radical.

Predict Stereochemical Outcome

Since the bromine can attach from either side of the planar radical, equal amounts of \(R\) and (S)-3-bromo-3-methylhexane will be produced. This results in a racemic mixture.

Conclusion

The product of photobromination of \(R)-3\)-methylhexane is a racemic mixture, containing equal amounts of \(R\) and (S)-3-bromo-3-methylhexane.

Key concepts

stereochemistry

In organic chemistry, stereochemistry refers to the spatial arrangement of atoms in molecules and how this affects their chemical behavior. It plays a crucial role in reactions such as photobromination. For instance, in the case of \((R)-3\)-methylhexane, the spatial configuration changes when a bromine atom attaches to the molecule. Understanding **stereochemistry** helps predict which enantiomers (mirror-image molecules) will form. This is crucial for determining the properties and reactions of the resulting compounds.

radical mechanism

A radical mechanism is a type of reaction pathway that involves free radicals, which are highly reactive species with unpaired electrons. In photobromination, UV light splits a bromine molecule into two bromine radicals.
These radicals then react with \(R)-3\text{-methylhexane}\) by abstracting a hydrogen atom from a tertiary carbon group, forming a carbon radical.
Understanding the **radical mechanism** is essential to anticipate the intermediate species that form during the reaction and how these intermediates go on to generate the final products.

racemic mixture

A racemic mixture consists of equal amounts of left- and right-handed enantiomers of a chiral molecule. In the photobromination of \((R)-3\)-methylhexane, bromine radicals can attach to the planar tertiary carbon radical from either the front or the back side.
This results in the formation of both \((R)\) and \((S)\) enantiomers of \(3\)-bromo-\(3\)-methylhexane in equal amounts. Hence, the final product is a racemic mixture.
Understanding this concept helps explain why a reaction doesn't always produce just one stereoisomer.

tertiary carbon radical

A tertiary carbon radical is a carbon atom bonded to three other carbon atoms and possessing an unpaired electron, which makes it highly reactive. These radicals are more stable compared to primary or secondary carbon radicals due to hyperconjugation and the inductive effect.
In the photobromination of \((R)-3\)-methylhexane, the bromine radical preferentially abstracts a hydrogen atom from the tertiary carbon.
This results in the formation of a stable tertiary carbon radical, which is key to understanding why the reaction yields specific products.