Question

Which of the following alkyl halides exhibits racemisation in $ {{S}_{N}}1 $ reaction?
(A) $ {{\left( CH \right)}_{3}}CCl $
(B) $ C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl $
(C) $ C{{H}_{3}}C{{H}_{2}}Cl $ $ $
(D) $ {{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl $

Answer 1

Hint: We will be the concept of $ {{S}_{N}}1 $ reactions. This is a part of organic chemistry. Also, we need to know about the rate determining step. The alkyl halide forming the most stable carbocation shall form a racemic mixture.

Complete step by step answer
We already know that the $ {{S}_{N}}1 $ reaction - A Nucleophilic Substitution in which the Rate Determining Step involves 1 component. $ {{S}_{N}}1 $ reactions are unimolecular, proceeding through an intermediate carbocation. $ {{S}_{N}}1 $ reactions give racemization of stereochemistry at the reaction centre.
All the molecules undergoing $ {{S}_{N}}1 $ reaction show racemisation as there is formation of 2 types of product in equal amounts.
The reaction occurs only if the products formed are stable. The Reaction Rate Increases with Substitution at Carbon. Out of all these 4 alkyl halides, $ {{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl $ where, benzyl chloride has the strongest tendency to undergo SN​1 here thus leading to formation of racemic mixture as it contains benzylic carbocation which is most stable.
 So, we need to select the correct option.
The correct option is D.

Note
Some examples of sn1 reaction are: The hydrolysis of tert-butyl bromide with aqueous $ NaOH $ solution is an example of $ {{S}_{N}}1 $ reaction. The rate of the reaction depends on the concentration of tertiary butyl bromide but it is independent of the concentration of $ NaOH $ .Hence, the rate determining step only involves tert-butyl bromide.
 in a substitution reaction there really are two main factors that tell you whether it's $ {{S}_{N}}2 $ or $ {{S}_{N}}1 $ ,the leaving group propensity or the strength of an incoming nucleophile. Two molecules react, and one displaces a substituent on the other. The rate of reaction is determined by the rate determining step. While in some cases $ {{S}_{N}}1 $ will be faster while in some others $ {{S}_{N}}2 $ .