Question

obtained by chlorination of n-butane, will be

• A. meso form
• B. racemic mixture
• C. d-form
• D. I-form

Answer 1

The Correct Option is B

To determine the product obtained by chlorination of n-butane, let us explore the chemical process and reasoning involved:

Understanding the Reaction

When n-butane undergoes chlorination, a free radical substitution reaction occurs. In this reaction, chlorine atoms are substituted into the hydrocarbon chain, primarily at the secondary carbon position because it is more stable than the primary carbon. This results in the formation of different chlorinated isomers.

Formation of Isomers

The two primary isomers formed during the chlorination of n-butane are:

• 1-Chlorobutane
• 2-Chlorobutane

Analysis of Isomers

1-Chlorobutane is a straightforward linear compound without any chirality, while 2-Chlorobutane features a chiral center. This chiral center is responsible for the formation of optical isomers.

2-Chlorobutane can exist as two enantiomers: (R)-2-Chlorobutane and (S)-2-Chlorobutane. These are non-superimposable mirror images due to the presence of a chiral carbon center.

Resultant Mixture: Racemic Mixture

Since the chlorination of n-butane leads to equal formation of both enantiomers of 2-chlorobutane, the final product is a racemic mixture. A racemic mixture contains equal amounts of both enantiomers, resulting in no net optical activity as the effects of the two enantiomers cancel each other out.

Why Other Options Are Incorrect

• Meso Form: Meso compounds have multiple chiral centers and an internal plane of symmetry, leading to optical inactivity. 2-Chlorobutane does not fit this description.
• d-form and l-form: These terms denote specific optical activities due to one enantiomer predominating, which does not happen in this case of equal enantiomers.

Conclusion

The chlorination of n-butane predominantly yields a racemic mixture of enantiomers for 2-chlorobutane, thus the correct answer is racemic mixture.