Question

Cyanide ion acts as an ambident nucleophile. From which end does it act as a stronger nucleophile in an aqueous medium when it reacts with an alkyl halide?
A. It acts as a stronger nucleophile from the carbon end.
B. It acts as a stronger nucleophile from the nitrogen end.
C. It depends on the nature of the alkyl halide.
D. It has the same strength from both ends.

Answer 1

Hint: Think about what kind of reaction is taking place when the electron on the cyanide ion attacks the alkyl halide. Consider the type of bond that will be formed and which two atoms will be involved in the bond. A nucleophile is a species that is negatively charged.

Complete Solution :
An ambident nucleophile is a nucleophile that can share electrons from two or more than two sites in the ion to form a bond. In the cyanide ion, the 2 atoms that can share electrons and form bonds are $C$and $N$.

In this reaction, cyanide forms a bond with the alkyl group in the alkyl halide replacing the halogen atom present. For this type of bond, the cyanide ion has to attack from the carbon end as the $C-C$ bond is stronger than the $N-C$ bond.

Additional Information:
- Here, we can see that the negative charge is present on the carbon atom. If this charge moves to the nitrogen atom the group will attack from the nitrogen end and the compound formed will be called an isocyanide compound. Since, nitrogen is more electronegative than carbon, a negative charge on nitrogen will be more stable than a negative charge on carbon. Due to this, the cyanide group will act as a stronger nucleophile from the carbon end as it will try to get rid of the negative charge quickly.
So, the correct answer is “Option A”. It acts as a stronger nucleophile from the carbon end.’

Additional Information:
- The name of this reaction is the Kolbe nitrile synthesis, it is used to produce an alkyl nitrile using the corresponding alkyl halide and a metal cyanide like $KCN$. A similar reaction to form an aromatic nitrile is called the Rosenmund von Braun reaction.

Note: As nitrogen is more electronegative and may pull the electrons towards itself to create a more electronegative environment, it can easily attack the more electropositive sites during a reaction. But this is not the case here as the stability of the $C-C$ bond trumps the electronegative nature of nitrogen and its tendency to form bonds at electropositive sites. The molecule will become more stable if the cyanide ion attacks through the carbon end than the nitrogen end.