Consider the following SN2 reaction of optically pure
1-chloro-3-ethylcyclopentane (X).H3CH2C–(cyclopentane ring)–Cl + OH− → YX
Q.22 The structure of Y in the above reaction is
(A) H3CH2C–(cyclopentane)–H / OH
(B) H3CH2C–(cyclopentane)–OH / H
(C) (cyclopentane)–OH with inverted configuration
(D) (cyclopentane)–OH with retained configuration
Introduction
Questions based on SN2 reactions and stereochemistry are very important
in competitive examinations such as JEE Advanced and NEET. These problems test understanding of
backside attack, Walden inversion, and three-dimensional molecular structures.
In this article, we analyze the SN2 reaction of optically pure
1-chloro-3-ethylcyclopentane with hydroxide ion and determine the correct structure
of the product with clear reasoning.
Key Concepts Used
- SN2 reactions occur in a single step
- Nucleophile attacks from the backside
- Results in inversion of configuration (Walden inversion)
- Optically pure reactant gives optically pure product with opposite configuration
Step-by-Step Stereochemical Analysis
Step 1: Identify the Chiral Center
The carbon bearing the Cl atom is a chiral center because it is attached to four
different groups. Therefore, stereochemistry must be carefully considered.
Step 2: Nature of the Nucleophile
OH− is a strong nucleophile and favors the SN2 mechanism.
Step 3: Apply SN2 Rule
The nucleophile attacks from the side opposite to the leaving group (Cl), causing
inversion of configuration at the chiral carbon.
Correct Answer
Option (A) correctly represents the product.
OH replaces Cl with complete inversion of configuration, while the relative positions of
hydrogen and the ethyl group are maintained correctly.
Quick Exam Tip
For SN2 reactions:
- Always expect inversion of configuration
- No racemization or retention
- Optically pure reactant → optically pure product
Conclusion
The SN2 reaction of optically pure 1-chloro-3-ethylcyclopentane proceeds through
backside attack by OH−, leading to complete inversion of configuration.
Careful tracking of stereochemistry allows us to confidently identify the correct product.
Final Answer: Option (A)
