Question

Which of the following is an example of \[{S_N}2\] reaction?
A. \[C{H_3}Br + O{H^ - } \to C{H_3}OH + B{r^ - }\]
B. \[{(C{H_3})_2}CHBr + O{H^ - } \to {(C{H_{3}})_2}CHOH + B{r^ - }\]
C. \[CH_{3}^{}C{H_2}OH\xrightarrow{{{H_2}O}}C{H_2} = C{H_2}\]
D. \[{(C{H_3})_3}C - Br + O{H^ - } \to {(C{H_3})_3}COH + B{r^ - }\]

Answer 1

Hint: \[{S_N}2\] means nucleophilic substitution reaction of second order. It is a bimolecular reaction taking place in a single step. \[{S_N}2\] reaction requires a reactant having covalent character in presence of polar aprotic solvent.

Complete step-by-step answer:\[{S_N}2\] has the following characteristics using which we can identify whether a reaction is following \[{S_N}2\] mechanism or not.
No intermediate is formed during \[{S_N}2\] reaction. It takes place in a single step, proceeding through formation of a transition state.
More the steric hindrance in the reactant that is more the bulky group in the reactant, less is the rate of \[{S_N}2\] mechanism.
The rate of \[{S_N}2\] is $1^\circ > 2^\circ > 3^\circ $carbon
Solvents like ${H_2}O$, $KOH$(aq) are not preferred. Rather best \[{S_N}2\] solvents include acetone, DMF, DMSO

OPTION D: In\[{(C{H_3})_3}C - Br\], the leaving group is connected to a $3^\circ $carbon atom. The three methyl groups have increased the bulkiness of the reactant hence the incoming nucleophile experiences hindrance, hence the rate of \[{S_N}2\]reaction decreases. It is favoured by \[{S_N}1\] because of the formation of extremely $3^\circ $stable carbocation.

Option C: It is an example of elimination reaction rather than substitution. $OH$ group gets converted into a very good leaving group \[{H_2}O\] by protonation and the positive charge eliminates another proton(deprotonation) to undergo elimination.

Option B: In\[{(C{H_3})_2}CHBr\], the leaving group is connected to a $2^\circ $carbon atom which is not much favoured by both \[{S_N}1\] and \[{S_N}2\]. Hence, the choice is of solvent that determines the mechanism path to be followed.

Option A: \[C{H_3}Br\] undergoes \[{S_N}2\] mechanism because leaving group is attached to $1^\circ $ carbon atom and a single methyl does not offer hinderance to incoming nucleophile. Moreover, \[{S_N}1\] can’t be favoured because of unstable carbocation formation.

Option ‘A’ is correct

Note: The rate of reaction in \[{S_N}2\] is dependent on the power of both substrate and nucleophile. Hence, this shows that nucleophile has an impact on the rate of reaction. Thus, strong nucleophiles (generally charged) are used. This is not the case in \[{S_N}1\].