Reactions of Aromatic Molecules
Nucleophilic Aromatic Substitution (NAS)
Last updated: October 15th, 2022 |
Nucleophilic Aromatic Substitution (NAS)
In many ways, nucleophilic aromatic substitution is the mirror opposite of electrophilic aromatic substitution.
- In Nucleophilic Aromatic Substitution, an electron-poor aromatic ring is attacked by a nucleophile, resulting in a substitution reaction
- The reaction proceeds through a negatively charged (carbanion) intermediate
- The reaction is accelerated by the presence of electron-withdrawing groups on the aromatic ring
- The placement of electron withdrawing groups ortho- or para- to the leaving group results in faster reactions than does the placement of electron withdrawing groups meta- to the leaving group
- Fluorine can act as a leaving group (!) in nucleophilic aromatic substitution reactions, since it is quite electron-withdrawing and C-F bond breakage is not the rate-limiting step.
Table of Contents
- Probably Not The Aromatic Substitution Reaction You Were Expecting
- Introducing….Nucleophilic Aromatic Substitution
- The Effect Of Substituents On The Ring
- The Effect Of The Leaving Group
- The Effect Of Substitution Pattern
- The “Meisenheimer” Intermediate Provides A Clue To The Mechanism of Nucleophilic Aromatic Substitution
- The Mechanism of Nucleophilic Aromatic Substitution
- Why Is The para- Isomer Faster Than The meta- Isomer ? It’s All About Stabilizing Negative Charge
- The Reaction Energy Diagram Of SNAr
- A Few Examples
- Summary: Nucleophilic Aromatic Substitution
- Quiz Yourself!
- (Advanced) References and Further Reading
1. Probably Not The Aromatic Substitution Reaction You Were Expecting
Let’s review electrophilic aromatic substitution (EAS). What have we learned?
- The aromatic ring acts as a nucleophile, and attacks an added electrophile E
- An electron-deficient carbocation intermediate is formed (the rate-determining step) which is then deprotonated to restore aromaticity
- electron-donating groups on the aromatic ring (such as OH, OCH3, and alkyl) make the reaction faster, since they help to stabilize the electron-poor carbocation intermediate
- Lewis acids can make electrophiles even more electron-poor (reactive), increasing the reaction rate. For example FeBr3 / Br2 allows bromination to occur at a useful rate on benzene, whereas Br2 by itself is slow).
Everything we’ve learned so far about substitution on aromatic rings would teach us that it proceeds much faster with methoxybenzene than with nitrobenzene, and much faster with an electrophile like Cl2 than with, say, an electron-rich nucleophile like NaOCH3.
Which brings us to the reaction below. The aromatic ring is electron-poor and we’re adding an electron-rich nucleophile.
What could happen here? Nothing, right?
“Nothing” is a good guess! Certainly, thinking of this as an electrophilic aromatic substitution, you’d be right in thinking that the answer to “what happens here?” is “jack squat”.
2. Introducing….Nucleophilic Aromatic Substitution
In fact, a substitution reaction does occur! (But, as you may suspect, this isn’t an electrophilic aromatic substitution reaction.)
In this substitution reaction the C-Cl bond breaks, and a C-O bond forms on the same carbon.
So while it is a substitution reaction, it has a few important differences:
- The species that attacks the ring is a nucleophile, not an electrophile
- The aromatic ring is electron-poor (electrophilic), not electron rich (nucleophilic)
- The “leaving group” is chlorine, not H+
- The position where the nucleophile attacks is determined by where the leaving group is, not by electronic and steric factors (i.e. no mix of ortho– and para- products as with electrophilic aromatic substitution).
In short, the roles of the aromatic ring and attacking species are reversed!
The attacking species (CH3O–) is the nucleophile, and the ring is the electrophile.
Since the nucleophile is the attacking species, this type of reaction has come to be known as nucleophilic aromatic substitution.
3. The Effect Of Substituents On The Ring
In nucleophilic aromatic substitution (NAS), all the trends you learned in electrophilic aromatic substitution operate, but in reverse.
The first trend to understand is that electron withdrawing groups (EWG’s) dramatically increase the rate of reaction, not decrease it.
From this, it follows that the more EWG’s there are, the faster the reaction.
For example, the rate of NAS for 2,4-dinitrophenyl chloride is about 105 times faster than for p-nitrophenyl chloride. [Note 1]
(I don’t have a rate constant for 2,4,6-trinitrophenyl chloride readily available, but it is orders of magnitude faster still).
4. The Effect Of The Leaving Group
One of the most eye-opening aspects of nucleophilic aromatic substitution is noting that fluorine is often used as a leaving group. This is seen in Sanger’s reagent for sequencing peptides, to take one example (more on that below).
After all, given the stern tones we instructors use in Org 1 on this subject, the words “FLUORINE IS NEVER A LEAVING GROUP IN SN2 AND SN1 REACTIONS” may as well have been carved on one of the stone tablets handed down to Moses on Mt. Sinai.
Here’s a thought: if even a “bad” leaving group like fluorine works in nucleophilic aromatic substitution, then surely a “better” leaving group like bromine or iodine would work even better. Right?
This is a good hunch. It is also wrong – which does not make it a dumb idea, only that organic chemistry is deep.
For one reaction studied [Note 3], F as the leaving group was observed to be 3300 times faster than iodine !
And between chlorine, bromine, and iodine, the difference was only by a factor of about 3.
So what could be different about nucleophilic aromatic substitution that makes the rate of reaction much less sensitive to the identity of the leaving group than the SN1 and SN2 reactions?
Well, for one thing, this would suggest that, unlike the SN1 and SN2 reactions, C-F bond cleavage does not occur in the rate-determining step. This information is helpful in coming up with a mechanism for the reaction.
5. The Effect Of Substitution Pattern
Unlike in electrophilic aromatic substitution, there are no “ortho-,para-” or “meta-” directors. The position of substitution is controlled by the placement of the leaving group.
However that isn’t to say that the rate of the reaction isn’t affected by the relative position of the leaving group and the electron-withdrawing group.
For example, nucleophilic aromatic substitution of p-nitrophenyl fluoride is orders of magnitude faster than m-nitrophenyl fluoride, even though the NO2 is closer to the leaving group and should presumably exert more of an inductive effect.
The ortho isomer is also faster than the meta by a large margin.
What’s going on?
6. The “Meisenheimer” Intermediate Provides A Clue To The Mechanism of Nucleophilic Aromatic Substitution
In the course of adding nucleophiles to various electron-poor aromatic molecules with a leaving group, intermediates have been isolated. One of the first was isolated in 1902 by Jacob Meisenheimer, and the general name “Meisenheimer complex” is given to these intermediates.
The intermediate is the (non-aromatic) addition product between the aromatic ring and the nucleophile. In the case below, the negative charge is delocalized to an oxygen on one of the nitro groups:
Meisenheimer intermediates can be isolated and characterized. However, if heated, the compound goes on to form the final nucleophilic aromatic substitution product.
This is very suggestive, to say the least.
7. The Mechanism of Nucleophilic Aromatic Substitution
Taking all of these observations into account we can now propose a mechanism for this reaction.
The first step is attack of the nucleophile on the electron-poor ring to generate a negatively charged intermediate (e.g. the “Meisenheimer” intermediate, above)
Since this disrupts the aromaticity of the ring, it’s also the rate-limiting step:
In electrophilic aromatic substitution (EAS) we saw that electron-rich substituents stabilized the electron-poor intermediate.
But in nucleophilic aromatic substitution (NAS) the tables are turned! Instead, the intermediate is electron-rich, and is stabilized by electron-withdrawing substituents, such as NO2.
The second step in nucleophilic aromatic substitution is expulsion of the leaving group:
8. Why Is The para- Isomer Faster Than The meta- Isomer ? It’s All About Stabilizing Negative Charge
This two-step mechanism where addition is the rate-determining step helps to explain our earlier puzzle of why the reaction with para-nitro is faster than the meta- isomer.
- Note how the anion in the para- intermediate can be delocalized to the oxygen on the nitro group, putting a negative charge on (more electronegative) oxygen.
- In the meta- intermediate, the negative charge cannot be delocalized to the nitro group, and is stuck on (less electronegative) carbon.
(This also explains why addition is fast for the ortho– isomer).
It also helps to explain why fluorine substituents increase the rate of nucleophilic aromatic substitution: the rate determining step is attack on the aromatic ring, not breaking the very strong C-F bond. The highly electronegative fluorine pulls electron density out of the ring, activating it towards attack.
So even though breaking a C-F bond is generally not energetically favorable, this is compensated by the fact that it restores aromaticity.
[UPDATE: In the comments, Matt brings up a very interesting recent study that suggests that some nucleophilic aromatic substitution reactions that don’t have F as a leaving group may proceed through a concerted mechanism. I would agree in considering this more advanced material is unlikely to be covered in most introductory courses, (except maybe Eugene’s) but very worthy of interest.]
9. The Reaction Energy Diagram Of SNAr
Putting it all together, we can sketch out a reaction-energy diagram for this mechanism that would look something like this:
(remember: transitition states are “peaks”, and intermediates are “valleys”. Intermediates can (at least theoretically) be isolated; transition states have partial bonds, only last a femtosecond, and can’t be isolated).
The nucleophile adds to the aromatic ring through transition state A (the rate limiting step) to give the negatively charged intermediate B, with a further input of energy (Ea) ascends to transition state C (loss of the leaving group, the fast step) and from there, the final product.
10. A Few Examples
Here’s three representative examples:
- The first is a straightforward nucleophilic aromatic substitution using an amine as a nucleophile.
- The second uses a stronger base (NaOH) to make a weaker base (the conjugate base of phenol) which attacks the electron-poor ring. (A variation of this reaction was used in a synthesis of the antibiotic vancomycin)
- The third example shows the N-terminus of a peptide reacting with 2,4-dinitrophenyl fluoride in a nucleophilic aromatic substitution reaction. Fred Sanger used this reagent to label the terminal residues in insulin, which led (after a lot of detective work Note 2], to the first reported sequence of a protein (and a Nobel Prize in chemistry in 1958).
11. Summary: Nucleophilic Aromatic Substitution
Well, now you’ve seen something we once said was impossible: fluorine as a leaving group. While we’re here, what other commandments from Org 1 can we possibly break?
How about putting a triple bond on an aromatic ring?
In the next post, we’ll cover a reaction that also qualifies as “nucleophilic aromatic substitution” although it goes through a completely different mechanism, involving this “triple bond”, above.
Next Post: Nucleophilic Aromatic Substitution (2): Arynes
Note 1. This is for the reaction below:
Note 2. Sanger’s detective work is detailed here.
Note 3. Link to the study is here.
(Advanced) References and Further Reading
This is on nucleophilic aromatic substitution, specifically the addition-elimination reaction.
- Ueber Reactionen aromatischer Nitrokörper
Liebig. Ann. Chem. 1902, 323 (2), 205-246
The addition intermediates from SNAr reactions can frequently be detected or isolated, and are called Meisenheimer complexes, after Jakob Meisenheimer, who first demonstrated their formation.
- The SN mechanism in aromatic compounds. Part VII
Peter Briner, Joseph Miller, M. Liveris, and (Miss)P. G. Lutz
J. Chem. Soc., 1954, 1265-1266
- The “Element Effect” as a Criterion of Mechanism in Activated Aromatic Nucleophilic Substitution Reactions
F. Bunnett, Edgar W. Garbisch Jr., and Kenneth M. Pruitt
Journal of the American Chemical Society 1957, 79 (2), 385-391
- Nucleophilic substitution. Linear free energy relations between reactivity and physical properties of leaving groups and substrates
Giuseppe Bartoli and Paolo Edgardo Todesco
Accounts of Chemical Research 1977, 10 (4), 125-132
These are mechanistic studies on the reaction, demonstrating that the order of reactivity is F > Cl > Br > I. In nucleophilic aromatic substitution, the formation of the addition intermediate is usually the rate-determining step so the ease of C-X bond breaking does not affect the rate.
- NUCLEOPHILIC AROMATIC SUBSTITUTION OF ARYL FLUORIDES BY SECONDARY NITRILES: PREPARATION OF 2-(2-METHOXYPHENYL)-2-METHYLPROPIONITRILE
Stéphane Caron, Jill M. Wojcik, and Enrique Vazquez
Org. Synth. 2002, 79, 209
A procedure in Organic Syntheses for an SNAr reaction, featuring nucleophilic displacement of an aryl fluoride.
- The terminal peptides of insulin
Biochemical Journal (1 January 1949) 45 (5): 563–574
One of this historically most significant examples of aromatic nucleophilic substitution is the reaction of amines with 2,4-dinitrofluorobenzene. This reaction was used by Frederick Sanger (who won 2 unshared Nobel Prizes in Chemistry) to develop a method for identification of the N-terminal amino acid in proteins.
- Concerted nucleophilic aromatic substitutions
Eugene E. Kwan, Yuwen Zeng, Harrison A. Besser & Eric N. Jacobsen
Nature Chemistry volume 10, pages 917–923 (2018)
This is at the cutting edge of chemistry – the authors use interesting NMR experiments to assert that SNAr reactions proceed through concerted rather than stepwise mechanism.
20 thoughts on “Nucleophilic Aromatic Substitution (NAS)”
IN the quiz question how do we decide which fluorine atom is going to get replaced by nucleophile??
Start by drawing the product of addition to the aromatic ring at each of the positions where there is a fluorine. Which anion will be more stable?
what will be the product in the following case?
2-bromo-4,6-dinitrophenol + methoxide ion –>?
You will put a CH3O in place of Br
can you explain quiz id 0543 for me ,please ?
only halogens act as leaving group or any other group can also act as LG ?
The best leaving group is F, but other groups can be leaving groups as well like Cl, Br, and OTs.
European Journal of Medicinal Chemistry, 2020, vol. 192
General procedure: To the solution of aromatic amine (0.7mmol) and the corresponding chlorinated 1,6-naphthyridine (0.7mmol) in isopropanol (10mL), HCl (20mmol%) was added drop-wise, and then heated to 90°C under nitrogen for 2h. The mixture was filtered, and the solid was dissolved in ethyl acetate. The solution was stired with K2CO3 (1mmol) at r.t. for 1h and filtered. The filtrate was concentrated in vacuum and purified by flash chromatography (CH2Cl2/MeOH=20:1) to yield the corresponding targets.
I want to ask: for some Heteroaromatic chloride, why can some acids (such as HCl, several drops) can promote SNAr reaction at high temperature. Thanks
What’s the specific reaction? What’s performing the substitution?
Can I ask for 3,4-dibromonitrobenzene reacting with excess pyrrolidine, only the Br para to NO2 will be substituted right?
Think about the negative charge that is formed through addition of pyrrolidine to the ring, and the resulting resonance forms. Attack at one position will be greatly favored over another. You have likely thought this through, judging from your answer, so I need not comment further.
So, you mean the negative charge will be more stable when it attacks at F than that at NO2? ?
Why the nucleophile does not attack on carbon at which NO2 is connected?
As a good nucleophile is not necessary and there will be more partial positive charge on NO2 ‘s carbon than that of F which attracts the nucleophile better?
If it attacks at that carbon then where does the negative charge go? :-?
Thankyou was very helpful
Great, thanks Pracheta.
You probably already read this, and I think it’s inappropriate to ever teach in Sophomore organic, because it contains too many exceptions on top of exceptions, but….
Recent research suggests many substrates react via a concerted mechanism. The computational evidence had suggested it for a while, and experimental research recently caught up. A better leaving group like bromine, and a weaker electron withdrawing group, tends to favor a concerted mechanism. The “original” reaction with a fluoride leaving group, and multiple nitro groups, is still step-wise.
I would like to read the original paper, but there’s a nice synopsis here:
It’s the same debate we can have over the mechanism of an alkyne hydrohalogenation–is it stepwise or concerted termolecular? I believe it is important to note to the students that there are some variations of the mechanism that go beyond the scope of the course, here we’ll only look at one “classic” mechanism. Otherwise students can get very confused and frustrated when they come across a different version or a deeper more involved discussion of the mechanism and feel like all they were taught is “incorrect” or start questioning what exactly they were taught and how much of it is “incorrect.”