Question

Question. Section 17-12 showed how nucleophilic aromatic substitution can give aryl amines if there is a strong electron-withdrawing group ortho or para to the site of substitution. Consider the following example.

(a) Propose a mechanism for this reaction.
(b) We usually think of fluoride ion as a poor leaving group. Explain why this reaction readily displaces fluoride as the leaving group.
(c) Explain why this reaction stops with the desired product, rather than reacting with another dinitrofluorobenzene.

Short answer

(a)

(b) Formation of anionic-sigma complex “A” is the rate-determining step or slow step in nucleophilic aromatic substitution. The loss of fluoride ion occurs in a subsequent fast step in which aromaticity is restored and where the nature of leaving group does not affect the overall reaction rate. In the or mechanisms, however, the carbon-fluorine bond is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the rate.

(c) Amines can act as nucleophiles as long as the electron pair on the nitrogen is available for bonding. The initial reactant, methylamine, is a very reactive nucleophile. However, once the nitrogen is bonded to the benzene ring, the electron pair is delocalized onto the ring, especially with such strong electron-withdrawing groups like nitro group in the ortho and para positions. The electrons on nitrogen are no longer available for bonding so there is no danger of it acting as a nucleophile in another reaction.

Step-by-step solution

Step-1. Explanation of part (a):

Methylamine acts as a nucleophile and attacks at the ring at carbon-fluorine bond, which leads to formation of carbanion and this carbanion is resonance delocalized with nitro group as well as with the ring. In the resonating forms, positive charge also comes on nitrogen atom but this positive charge is stable as nitrogen’s octet is complete. Negative charge on completing delocalization enables the elimination of fluoride as leaving group and this fluoride acts as a base and abstracts the hydrogen from methylamine to neutralize positive charge on nitrogen thereby forming the required product.

Mechanism for the given reaction

Step-2. Explanation of part (b):

Formation of anionic-sigma complex “A” is the rate-determining step or slow step in nucleophilic aromatic substitution. The loss of fluoride ion occurs in a subsequent fast step in which aromaticity is restored and where the nature of leaving group does not affect the overall reaction rate. In the or mechanisms, however, the carbon-fluorine bond is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the rate.

Fluoride acts as a good leaving group as aromaticity of system gets restored upon its elimination

Step-3. Explanation of part (c):

Amines can act as nucleophiles as long as the electron pair on the nitrogen is available for bonding. The initial reactant, methylamine, is a very reactive nucleophile. However, once the nitrogen is bonded to the benzene ring, the electron pair is delocalized onto the ring, especially with such strong electron-withdrawing groups like nitro group in the ortho and para positions. The electrons on nitrogen are no longer available for bonding so there is no danger of it acting as a nucleophile in another reaction.