V asked in Science & MathematicsChemistry · 8 years ago

Strong Deactivator vs Weak activator on benzene ring - Electrophilic substitution?

I am currently working through 'Organic Chemistry II as a Second Language'. There is one problem I am unsure about (3.48)

We have a benzene ring with a weak activator & a strong deactivator that are META to each other. From what I understand, ortho-para directors always beat meta-directors. So given that, I would imagine that the overall activated positions would be: both ortho positions and para to the weak activator.

However, the answer key says only the two ortho positions to the weak activator are activated. Not the para position. This does not make any sense to me. Can someone please explain the electronics behind this? Perhaps it is a mistake?

1 Answer

  • 8 years ago
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    It's not a mistake, but it is a pretty tricky question. You really have to think about what a weak activator does: it activates via induction, not resonance. It puts a partial negative charge on the ortho positions, and because of the conjugation of the ring, also the para position. BUT the para position is so far away, that it's not as well activated (aka as much charge buildup).

    Without the addition of the meta-positioned strong deactivator, this isn't an issue and the substitution position would be determined by sterics and probability. However, you now have a positive charge being placed on the carbons that are ortho and para to the activator (and the deactivator as well, but it's easier to talk relative to just one substituent). So the para position is less activated than the ortho positions, and as such, the partial negative charge can't "overcome" the positive charge (or partial positive charge) from the deactivating group.

    The answer itself is really a lot of hand waving, though, and if you were asked this on a test, you could argue this answer OR the one you thought to be true. Hope I helped!

    Source(s): Organic chemist
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