Question

Consider the following reactions:
(i)${{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}Br\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}O{{C}_{2}}{{H}_{5}}+HBr$
(ii) ${{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}Br\xrightarrow{{{C}_{2}}{{H}_{5}}{{O}^{-}}}{{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}O{{C}_{2}}{{H}_{5}}+B{{r}^{-}}$
The mechanism of reactions(i) and (ii) are respectively:
(a) ${{S}_{N}}1$ and ${{S}_{N}}2$
(b) ${{S}_{N}}1$and${{S}_{N}}1$
(c) ${{S}_{N}}2$and ${{S}_{N}}2$
(d) ${{S}_{N}}2$and ${{S}_{N}}1$

Answer 1

The ${{S}_{N}}1$ and ${{S}_{N}}2$ both are nucleophilic substitutions reactions i.e. involves the nucleophile and the, substitution of functional group in the molecule with the nucleophile.
In ${{S}_{N}}1$ reactions, first there is formation of carbocation and then attack of nucleophile but in ${{S}_{N}}2$reactions, the nucleophile attack and the departure of the leaving group occurs simultaneously. Now, identify the reactions.

Complete step by step solution:
- All the reactions occurring in organic chemistry follow a particular reaction mechanism i.e., it follows a step by step sequence in which the reaction takes place and the way reactants follow and join together after the formation of either carbocation or through the transition states and how they are transformed into the products.
- Similarly, in the same way, ${{S}_{N}}1$ and ${{S}_{N}}2$ are two such reaction mechanisms. Here, in this S is for substitution i.e. those chemical reactions in which one functional group is replaced by the other and N stands for the nucleophilic because the nucleophile ( nucleus loving) is being substituted with the other one.
- ${{S}_{N}}1$ reactions occur in two steps . In the first step the formation of carbocation takes and is the slowest step and is the rate determining step and in the second step, nucleophile attacks the (which can donate electrons) carbocation and is a very fast reaction. So , thus, in these reactions rate of reaction depends only on the slowest rate determining step i.e. the alkyl halide and thus, are unimolecular reactions.
- The reaction: ${{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}Br\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}O{{C}_{2}}{{H}_{5}}+HBr$
follows the ${{S}_{N}}1$ mechanism because in this the alcohol is weak nucleophile and first results in the formation of carbocation and then forms the final product.
- On the other hand, in ${{S}_{N}}2$ reactions, the reaction occurs in a single step through the formation of the transition state and the reactions depend on both the reactants involved and thus, are bimolecular reactions. In this, the incoming nucleophile attacks the molecule from the back side i.e. the opposite side from which the leaving group leaves.
- The reaction: ${{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}Br\xrightarrow{{{C}_{2}}{{H}_{5}}{{O}^{-}}}{{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}O{{C}_{2}}{{H}_{5}}+B{{r}^{-}}$
Follows the ${{S}_{N}}2$mechanism because in this the ${{C}_{2}}{{H}_{5}}{{O}^{-}}$ is strong nucleophile and quickly the attacks the molecule from the back side as the leaving group is leaving from the front side.
So, from this, it is clear that the reaction (i) follows the ${{S}_{N}}1$ mechanism and the reaction (ii) follows the ${{S}_{N}}2$ mechanism.

Hence, option (a) is correct.

Note: in ${{S}_{N}}1$, the nucleophile attacks only after the leaving group leaves the molecule and is thus a two-step reaction. On the other hand, in ${{S}_{N}}2$, the nucleophile attacks the molecule from the back side of the molecule as the leaving group is leaving from the front side and is thus, is a single step reaction and is, therefore also called as the concerted mechanism.