Sn1 Stereochemistry
An optically active tertiary bromide is hydrolysed through a unimolecular pathway involving a planar carbocation; what stereochemical outcome is expected in the product?
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing halogens
- Topic
- sn1 stereochemistry
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
A racemic mixture with substantial loss of optical activity
Unimolecular nucleophilic substitution begins with slow ionisation of the carbon-halogen bond to give a carbocation that is sp2 hybridised and therefore trigonal planar at the reaction centre. Because this flat cation has two identical faces, the incoming water molecule can attack from either side with nearly equal probability, producing roughly equal amounts of the two enantiomers and hence a racemic mixture with large loss of optical activity. Complete retention is wrong because a planar intermediate erases the original three-dimensional arrangement. Complete inversion only is incorrect since that is the hallmark of concerted SN2 reactions, not of carbocation pathways. A single optically pure enantiomer of higher rotation is impossible because no new chiral information is introduced to favour one face. In practice slight steric shielding by the departing halide can cause a small excess of inversion, so real SN1 reactions often show partial rather than perfect racemisation, but the dominant outcome remains a near-equal enantiomer mixture. This behaviour is the diagnostic stereochemical signature that distinguishes a carbocation pathway from a concerted one. This is the classic NCERT explanation linking SN1 to racemisation. Sanity check: a symmetric planar cation attacked equally from both faces must give a near-50:50 enantiomer ratio, consistent with racemisation.
This medium difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of sn1 stereochemistry. It appeared in the 2025 exam.
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