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NEET CHEMISTRYAlcohols, Phenols and EthersMedium

Question

The 2-chlorobutane obtained by chlorination of n-butane will be:

A

Meso form

B

Racemic mixture

C

d-form

D

l-form

Step-by-Step Solution

During the free radical chlorination of n-butane, substitution at the secondary carbon (C2) yields 2-chlorobutane, which contains a chiral centre. The reaction proceeds via the formation of a secondary free radical intermediate (sec-butyl free radical). This free radical is essentially planar (sp2sp^2 hybridized). The subsequent attack by a chlorine molecule/radical can occur from either face of the planar intermediate with equal probability. As a result, an equimolar (50:50) mixture of both enantiomers (d- and l-forms) is formed, which is optically inactive and is known as a racemic mixture.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Alcohols, Phenols and Ethers. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYAlcohols, Phenols and Etherschlorobutaneobtainedchlorinationnbutane

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