Question

Which of these reactants would be the most susceptible to electrophilic aromatic substitution (EAS)? `"PhN"("CH"_2)_3("CH"_3)_3^(+)` or `"PhN"("CH"_2)_3("CH"_3)_3^(+)`? Which would then yield the lowest, and highest, percentage of the meta isomer?

Answer 1

`"PhN"("CH"_2)_3("CH"_3)_3^(+)` would be most reactive towards EAS (most activating substituent), and `"PhN"("CH"_3)_3^(+)` would be the least reactive towards EAS (most deactivating substituent).

`"PhN"("CH"_2)_3("CH"_3)_3^(+)` would thus yield the lowest percentage of the meta isomer (ortho-para directing substituent), and `"PhN"("CH"_3)_3^(+)` would yield the highest (meta directing substituent).

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STRONGLY-ELECTRON-WITHDRAWING (DEACTIVATING) GROUPS ARE META DIRECTORS

The idea is that the trimethylammonium group (a positively-charged quaternary ammine) is a strongly-deactivating group, meaning it strongly withdraws electron density away from the aromatic ring, and is a meta director.

Try drawing the resonance structures of `"PhN"("CH"_3)_3^(+)`:

You should get partial positive charges at the ortho and para positions.

These `delta^+` repel incoming electrophiles (i.e. in electrophilic aromatic substitution) from binding at the ortho or para positions, and thus the meta position is favored.

As the length of the alkyl chain increases on the substituent, the substituent becomes more of an ortho-para directing group (it becomes a weakly-electron-donating group).

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Thus, for the first part of your question, `"PhN"("CH"_2)_3("CH"_3)_3^(+)` would be most reactive towards EAS (most activating substituent), and `"PhN"("CH"_3)_3^(+)` would be the least reactive towards EAS (most deactivating substituent).

For the second part of your question, `"PhN"("CH"_2)_3("CH"_3)_3^(+)` would thus yield the lowest percentage of the meta isomer (ortho-para directing substituent), and `"PhN"("CH"_3)_3^(+)` would yield the highest (meta directing substituent).