Question

In a nucleophilic substitution reaction:
$R - Br + C{l^ - }\xrightarrow{{DMF}}R - Cl + B{r^ - }$
Which one of the following undergoes complete inversion of configuration?
(A) ${C_6}{H_5}CH{C_6}{H_5}Br$
(B) ${C_6}{H_5}C{H_2}Br$
(C) ${C_6}{H_5}CH(C{H_3})Br$
(D) ${C_6}{H_5}CC{H_3}{C_6}{H_5}Br$

Answer 1

Hint: Here DMF stands for Dimethylformamide which is a polar aprotic solvent. DMF preferred the $S_N2$ mechanism of nucleophilic substitution reaction which is a one step reaction in which rate of reaction depends on both the reactants. This mechanism produces inversion configuration products.

Complete step-by-step answer:As per doing this, let we check all the options one by one to find the correct solution,

In both option (A) and (C) bromine is attached to the secondary carbon atom of the compound after which a substitution takes place as in the combination of $S_N1$ and $S_N2$ mechanism. As from which it is not possible configuration of complete inversion.

As for the compound present in option B, Bromine is attached to the primary carbon atom of the compound. Primary halides favours $S_N2$ mechanism due to low steric hindrance.

As talking about option (D) the Bromine atom is present to the tertiary carbon atom of the compound hence the substitution only takes place form the $S_N1$ mechanism hybridization.

As a result from the above conclusion we get that, ${C_6}{H_5}CH(C{H_3})Br$ is the correct answer because it undergoes the complete inversion of configuration by $S_N2$ mechanism.

Option ‘C’ is correct

Note:The inversion of configuration in $S_N2$ mechanism takes place because this is a one step reaction in which the leaving group and upcoming nucleophile are present in an intermediate form. So, the nucleophile gets attached to the opposite side from where the leaving group is leaving.