Chirality And Stereoisomerism
Consider 2-chlorobutane prepared as a single batch by free-radical chlorination of butane; how many optically active stereoisomers can this particular structure exhibit?
Select the correct option:
Solution
Two, a pair of non-superimposable mirror-image enantiomers
Optical activity requires at least one chiral carbon, a carbon bonded to four different groups, so that the molecule and its mirror image cannot be superimposed. In 2-chlorobutane the second carbon carries a hydrogen, a chlorine, a methyl group and an ethyl group, four distinct substituents, making C2 a single stereocentre. A molecule with one stereocentre exists as exactly two enantiomers, a left- and a right-handed form, both optically active. The achiral option is wrong because C2 clearly bears four different groups and is a genuine chiral centre. The four-isomer option is incorrect because there is only one chiral carbon, not two, so 2^2 does not apply. The three-isomer meso option is wrong because a meso compound needs two stereocentres with an internal mirror plane, which this molecule lacks. The two enantiomers rotate plane-polarised light by equal magnitudes in opposite directions, so a racemic batch produced by random free-radical chlorination is optically inactive overall even though each individual molecule is chiral. Separating the mixture into its pure enantiomers, called resolution, would be required to observe net rotation. This follows the NCERT treatment of optical isomerism in haloalkanes. Sanity check: one stereocentre gives 2^1 = 2 enantiomers, matching two optically active forms.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing halogens
- Topic
- chirality and stereoisomerism
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Two, a pair of non-superimposable mirror-image enantiomers
Optical activity requires at least one chiral carbon, a carbon bonded to four different groups, so that the molecule and its mirror image cannot be superimposed. In 2-chlorobutane the second carbon carries a hydrogen, a chlorine, a methyl group and an ethyl group, four distinct substituents, making C2 a single stereocentre. A molecule with one stereocentre exists as exactly two enantiomers, a left- and a right-handed form, both optically active. The achiral option is wrong because C2 clearly bears four different groups and is a genuine chiral centre. The four-isomer option is incorrect because there is only one chiral carbon, not two, so 2^2 does not apply. The three-isomer meso option is wrong because a meso compound needs two stereocentres with an internal mirror plane, which this molecule lacks. The two enantiomers rotate plane-polarised light by equal magnitudes in opposite directions, so a racemic batch produced by random free-radical chlorination is optically inactive overall even though each individual molecule is chiral. Separating the mixture into its pure enantiomers, called resolution, would be required to observe net rotation. This follows the NCERT treatment of optical isomerism in haloalkanes. Sanity check: one stereocentre gives 2^1 = 2 enantiomers, matching two optically active forms.
This hard difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of chirality and stereoisomerism. It appeared in the 2025 exam.
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