Nucleophilic Aromatic Substitution Via Arynes

by James

Description: Treatment of an aromatic containing a good leaving group with a strong base will lead to formation of an “aryne”. Attack of the aryne with a nucleophile.


Notes: The leaving group (LG) is a typical good leaving group like Br, I, Cl, OTs, or OMs. The choice of strong base is usually an amide (NaNH2). The solvent for NaNH2 is generally liquid ammonia (NH3) so this is often the nucleophile that ends up attacking the aryne intermediate.



Notes: Example 1 shows a typical example where NH2 has replaced Br. Example 2 is very interesting: note that two products are formed (meta and para) for reasons explained below. The same is true in the third example.

Mechanism: First, it is worth making a note about the structure of arynes. The “triple bond” shown is not a true triple bond. In an aryne, there are two sp2 orbitals in the plane of the C-H bonds – one contains a lone pair (and bears a negative charge) and the other orbital is empty.


The reaction begins by deprotonation of C-H by the strong base NaNH2, resulting in the cleavage of the C-Br bond (Step 1, arrows A, B and C). This results in formation of the aryne. Attack on the aryne by the nucleophile (Step 2, arrows D and E) results in negatively charged carbon, which is then protonated (Step 3, arrows F and G) and then the positively charged nitrogen is deprotonated (Step 4, arrows H and I).
Note that it’s also possible to show steps 3 and 4 in various different ways (e.g. using R-NH3+ as the acid in step 3 and NH3 as the base in step 4, or even through an intramolecular “proton transfer” reaction).  The key is merely to show formation of C-H and breakage of N-H.

One interesting variation arises when a substituted starting material is used, as in this example. Here, deprotonation gives the aryne as before. Now, two pathways are possible. Path A (attack on C-4) leads to formation of the “para” product, whereas path B (attack on C-3) results in formation of the “meta” product. These two products are formed in roughly equal amounts.



{ 1 comment… read it below or add one }

Rich Aversa

In the mechanism, the last entry in the “bond broken” list may require some attention.


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