I encountered the following question in an organic chemistry paper -

Multiple choice question on nucleophilic substitution through benzyne

The reaction follows benzyne mechanism. My answer was B.

However, the answer given in the test solutions was D. The mechanism given was as follows:

The mechanism for the nucleophilic substitution through benzyne

My question: The tert-butoxide being a bulky base is more likely to abstract the least hindered hydrogen, which is clearly present at the para-position to $\ce{-OCH3}$. Then the benzyne triple bond would form between the meta and para carbons, following which product (B) would be obtained. However, the given mechanism involves abstraction of the hydrogen from the ortho-position, which has two large groups on either side, and hence shouldn't occur.

Is my reasoning right? Are there any other factors affecting the reaction that decides which hydrogen will be removed?

I am also open to the possibility of the test solution's mechanism being wrong.


2 Answers 2


I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (like $\ce{OCH3}$ here), or accept a lone pair from the aromatic nucleus (like $\ce{NO2}$ would do) is not relevant in this case. However, inductive effects of the substituents in aromatic nucleus are still relevant (i.e. electron-withdrawing effects that occur through single $\sigma$-bonds).

However, $\ce{t-BuO-}$ is not as strong a base as $\ce{H2N-}$ in usual cases of benzyne reactions. Thus, it would abstract relatively most acidic available proton from to aromatic nucleus. Compared to para-position, ortho-position is more acidic due to proximity (ortho is one $\ce{C-C}$ bond away compared to para which is three $\ce{C-C}$ bonds away). Following is an example you can find in literature (Ref.1):

Nuc. Substitution


  1. José-Antonio García-López and Michael F. Greaney, "Synthesis of biaryls using aryne intermediates," Chem. Soc. Rev. 2016, 45(24), 6766-6798 (DOI: https://doi.org/10.1039/C6CS00220J).
  • 3
    $\begingroup$ Would $\ce{NH2-}$ deprotonate the para-position then? $\endgroup$
    – S R Maiti
    Commented May 1, 2021 at 18:31
  • 4
    $\begingroup$ To some extend yes, but ortho is preferred. If you look at experimental values with $\ce{PhLi}$, it preferably gives ortho intermediate. $\endgroup$ Commented May 1, 2021 at 18:34

Short answer: The $\ce{OMe}$ stabilizes the negative charge in the ortho position

From Master Organic Chemistry.com here see Case#2 below

So how do substituents on the ring affect addition to the triple bond?

Since the aromatic pi-system is at right angles to the triple bond, what’s NOT relevant is the ability to donate a lone pair to the ring (like $\ce{OCH3}$), or accept a lone pair from it (like $\ce{NO2}$).

However, inductive effects are still relevant (i.e. electron-withdrawing effects that occur through single bonds).

Why? Addition to the triple bond creates a negative charge on carbon, and electron-withdrawing groups stabilize negative charge: the closer, the better.

So a key principle in the addition of nucleophiles to arynes is that addition tends to happen so as to place the negative charge closer to an electron-withdrawing substituent.

Effect of EWG on nucleophilic aromatic substitution

What’s interesting is that $\ce{OCH3}$ behaves like an electron-withdrawing group in these examples, since the oxygen lone pairs can’t interact with the triple bond.

  • $\begingroup$ Your answer is most informative, but it mainly explains the point where nucleophile (NH2 in your case, OD in my question) attack after the benzyne triple bond is formed. Which will explain why the answer can be (A) and not (B), also can be (D) but not (C). $\endgroup$
    – TRC
    Commented May 1, 2021 at 13:51
  • $\begingroup$ However, I am ignorant as to whether a negative charge is formed during the benzyne formation. If it is, then of course for stability it will be at the ortho position, and your answer will be helpful in that regard also. Could you please elucidate if a negative charge temporarily appears during benzyne formation, on that carbon from where hydrogen is taken? Also, will the negative charge stability by -I of OCH3 group be sufficient to counter the steric hindrance experienced by the bulky tert-butoxide while abstracting the ortho hydrogen? $\endgroup$
    – TRC
    Commented May 1, 2021 at 13:54
  • 1
    $\begingroup$ A transient -ve charge is formed by deprotonation before elimination of the chloride ion. $\endgroup$
    – Waylander
    Commented May 1, 2021 at 14:24

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