Question

In an ${{S}_{N}}1$ reaction on chiral centres, there is:
A.100% retention
B. 100% inversion
C. 100% racemization
D. inversion more than retention leading to initial racemization

Answer 1

The carbocation formed in the first step is $s{{p}^{2}}$ hybridised and planar. It has vacant unhybridized p orbital perpendicular to this plane, so attack of nucleophile can take place from either side of the plane.

Complete Solution :
- In order to answer our question, we need to learn about the ${{S}_{N}}1$ reaction. Nucleophilic substitution unimolecular or ${{S}_{N}}1$ is a two step unimolecular reaction. The first step is the slow ionisation of substrate and is the rate-determining step. The second step is the rapid reaction between the carbocation (formed in the first step) and the nucleophile. ${{S}_{N}}1$ reactions generally proceed in polar protic solvents such as ${{H}_{2}}O$.
- The energy required to break C—X bond in step 1, is compensated to a large extent by the solvation of ionic intermediates formed in this step. Since the first step is the rate-determining step, the rate of reaction depends upon the concentration of alkyl halide only and is independent of the concentration of nucleophile. Further, greater the stability of carbocation, it will be easy for the formation from alkyl halide and faster will be the rate of reaction. Thus reactivity order of alkyl halides towards ${{S}_{N}}1$ reaction is: tertiary halide> secondary halide > primary halide > methyl halide.

- Tertiary Alkyl halides undergo nucleophilic substitution mainly by ${{S}_{N}}1$ mechanism whereas secondary Alkyl halides undergo partly by ${{S}_{N}}2$ and partly by ${{S}_{N}}1$. If the solvent is nonpolar, the ${{S}_{N}}1$ mechanism stops working. Primary Alkyl and methyl halides undergo nucleophilic substitution mainly by ${{S}_{N}}2$. Hence, both retention and inversion occurs, however, there is more retention than inversion.
So, the correct answer is “Option D”.

Note: It is to be noted that for the same alkyl group, the reactivity order of alkyl halides is same towards ${{S}_{N}}1$ as well as ${{S}_{N}}2$ i.e., R- > R-Br > R-CI > R-F. Allyl and benzyl halides follow ${{S}_{N}}1$ mechanism because the carbocation formed in each case is stabilised by resonance.