Proton removal during benzyne formation in substituted haloarenes

Question

I encountered the following question in an organic chemistry paper -

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:

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.

Answer 1

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):

References:

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).

Answer 2

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.

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.