Question

Which compound undergoes nucleophilic substitution with \[NaCN\]at the faster rate?

Answer 1

Hint: In the given options, bromine is bonded which is a good leaving group. Thus, it leaves compounds easily and generates carbonation. This carbonation will get attacked by nucleophiles. And the rate of this nucleophile attack will be greater at the carbonation which is surrounded with less bulky groups (less steric hindrance).

Complete Step by Step Solution:
Reagent \[NaCN\] is a strong nucleophile. In a nucleophilic substitution reaction, this nucleophile substitutes the nucleophile present in the given compound. Bromine's whose electronegativity is quite high as compared to its substrate (C) is a good leaving group and tends to attain a negative charge after attracting bond electrons towards itself with an inductive effect. This is the reason bromine is a good nucleophile.

As bromine (nucleophile) is a good leaving group thus from all the compounds given in the option, bromine will leave taking bond electron with itself and generating carbonation such as





As a nucleophile leaves the compound, carbonation will form at which another nucleophile \[(C{{N}^{-}})\]will attack. Now nucleophiles attack fastly on that cation which is surrounded with a smaller number of groups (at the site of less steric hindrance) and show nucleophilic substitution reaction.

Compound A cation is less hindered thus, nucleophile firstly prefers to attack that compound. Further nucleophile attack on B compound as the bulky groups are far way the action. Compounds D and C are the compounds at which nucleophiles attack with difficulty.
Thus, the correct option is A.

Note: It is important to note that when a compound substrate is bonded with a strong base. Then this base tends to protonate through the hydrogen of carbon adjacent to the substrate and a new one more bond between substrate carbon and carbon to which protons get abstracted (alkane changes to alkene). This is an elimination reaction. If the substrate is bonded to the leaving group (strong nucleophile) thus a nucleophile takes the place of the leaving group and this is a substitution reaction.