3.8: Nucleophilic Aromatic Substitution
After completing this section, you should be able to
- identify the conditions necessary for an aryl halide to undergo nucleophilic aromatic substitution, and give an example of such a reaction.
- write the detailed mechanism for a nucleophilic aromatic substitution reaction.
- compare the mechanism of a nucleophilic aromatic substitution reaction and the S N 1 and S N 2 mechanisms discussed earlier.
- identify the product formed when a given nucleophile reacts with a given aryl halide in a nucleophilic aromatic substitution reaction.
Although aromatic substitution reactions usually occur by an electrophilic mechanism, aryl halides that have electron-withdrawing substituents can also undergo a nucleophilic substitution reaction. For example, 2,4,6-trinitrochlorobenzene reacts with aqueous NaOH at room temperature to give 2,4,6-trinitrophenol. Here, the nucleophile OH – substitutes for Cl – .
Nucleophilic aromatic substitution is much less common than electrophilic substitution but nevertheless does have certain uses. One such use is the reaction of proteins with 2,4-dinitrofluorobenzene, known as Sanger’s reagent, to attach a “label” to the terminal NH 2 group of the amino acid at one end of the protein chain.
Although the reaction appears superficially similar to the S N 1 and S N 2 nucleophilic substitutions of alkyl halides discussed in the chapter on Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations, it must be different because aryl halides are inert to both S N 1 and S N 2 conditions. S N 1 reactions don’t occur with aryl halides because dissociation of the halide is energetically unfavorable, due to the instability of the potential aryl cation product. S N 2 reactions don’t occur with aryl halides because the halo-substituted carbon of the aromatic ring is sterically shielded from a backside approach. For a nucleophile to react with an aryl halide, it would have to approach directly through the aromatic ring and invert the stereochemistry of the aromatic ring carbon—a geometric impossibility.
Conditions for Nucleophilic Aromatic Substitution Reactions of Aryl Halides
The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides. The simple aryl halides generally are resistant to attack by nucleophiles in either S N 1 or S N 2 reactions. However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid
- when the aryl halide is activated by substitution with strongly electron-attracting groups such as NO 2 , and
- when very strongly basic nucleophilic reagents are used.
Addition-Elimination Mechanism
The nucleophilic substitutions on an aromatic ring proceed by the mechanism shown in Figure \(\PageIndex{1}\). The nucleophile first adds to the electron-deficient aryl halide, forming a resonance-stabilized, negatively charged intermediate called a Meisenheimer complex after its discoverer. Halide ion is then eliminated.
Nucleophilic aromatic substitution occurs only if the aromatic ring has an electron-withdrawing substituent in a position ortho or para to the leaving group to stabilize the anion intermediate through resonance (Figure \(\PageIndex{2}\)). A meta substituent offers no such resonance stabilization. Thus, p -chloronitrobenzene and o -chloronitrobenzene react with hydroxide ion at 130 °C to yield substitution products, but m -chloronitrobenzene is inert to OH – .
Propose a mechanism for the following reaction:
Strategy
when strongly electron-attracting groups are located on the ring at the ortho-para positions, the intermediate anion is stabilized by delocalization of electrons from the ring carbons to more favorable locations on the substituent groups. In this example, consider the displacement of bromine by methoxid (OCH 3 ) in the reaction of 4-bromonitrobenzene and methoxide ion.
The anionic intermediate formed by addition of methoxide ion to the aryl halide can be described by the valence-bond structures 5a-5d. Of these structures 5d is especially important because in it the charge is transferred from the ring carbons to the oxygen of the nitro substituent.
Solution
Electron-attracting groups in Meta Position
Substituents in the meta positions have much less effect on the reactivity of an aryl halide because delocalization of electrons to the substituent is not possible. No formulas can be written analogous to 5c and 5d in which the negative charges are both on atoms next to positive nitrogen, and
Note the differences between electrophilic and nucleophilic aromatic substitutions. Electrophilic substitutions are favored by electron-donating substituents, which stabilize a carbocation intermediate, while nucleophilic substitutions are favored by electron-withdrawing substituents, which stabilize a carbanion intermediate. Thus, the electron-withdrawing groups that deactivate rings for electrophilic substitution (nitro, carbonyl, cyano, and so forth) activate them for nucleophilic substitution. What’s more, these functional groups are meta directors in electrophilic substitution but are ortho–para directors in nucleophilic substitution. And finally, electrophilic substitutions replace hydrogen on the ring, while nucleophilic substitutions replace a leaving group, usually halide ion.
Exercises
Propose a mechanism for the following reaction:
- Answer
- The reaction starts when the phenol is deprotonated by KOH to give an anion that carries out a nucleophilic acyl substitution reaction on the 5-fluoro-2-nitroanisole.
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Make certain that you can define, and use in context, the key terms below.
- Meisenheimer complex
- nucleophilic aromatic substitution
A nucleophilic aromatic substitution reaction is a reaction in which one of the substituents in an aromatic ring is replaced by a nucleophile.
A Meisenheimer complex is a negatively charged intermediate formed by the attack of a nucleophile upon one of the aromatic-ring carbons during the course of a nucleophilic aromatic substitution reaction. A typical Meisenheimer complex is shown in the reaction scheme below. Notice how this particular complex can be formed from two different starting materials by using a different nucleophile in each case.