2
$\begingroup$

Something's wrong with how I think of this:

methoxyl has the partial negative charge on oxygen which means it's activating. However, halogens are more electronegative yet they're considered deactivating. Why?

On my lecture notes, I have one section that says "if Q is an electron withdrawing group then attack on the ring is slowed because this leads to additional positive charge on the ring".

While another section says "all meta directing groups have either a partial or full positive charge on the atom directly attached to the aromatic ring."

$\endgroup$
  • $\begingroup$ Methoxy not methoxyl $\endgroup$ – orthocresol Jan 28 '16 at 23:03
  • 2
    $\begingroup$ methoxyl has the partial negative charge on oxygen which means it's activating. Where did you get this idea from? Having a partial negative charge on oxygen has nothing to do with methoxy groups being activating. $\endgroup$ – bon Jan 29 '16 at 10:34
0
$\begingroup$

Halogens are considered deactivating groups due to the inductive effects of their overwhelming electronegativity withdrawing electrons away from the ring, thus slowing the overall reaction. However, halogens are also able to donate one of their lone pars to the ring through resonance which is characteristic of activating groups. Thus the observance of ortho and para directing. Ultimately, halogens inductive effects will trump resonance due to their electronegativity.

I forgot to add, that while oxygen can donate a lone pair through resonance it's electronegativity is not of the same magnitude as that of a halogen. So the alkoxy group would lead to a more electron rich aromatic system, thus it is a strong activator.

$\endgroup$
  • 2
    $\begingroup$ Of the halogens, only fluorine is more electronegative than oxygen. $\endgroup$ – ron Jan 28 '16 at 3:19
2
$\begingroup$

Because fluorine has such a high electronegativity, its inductive withdrawing effect is much larger than any resonance electron-donating effect.

The other halogens have poor orbital overlap with the pi system of an aromatic ring, which precludes any appreciable resonance electron-donation (this is why you almost never see a C-X double bond). Therefore, their inductive capabilities lead to a net deactivation of the aromatic ring.

This is why an alkoxy group is more activating than a halogen. Oxygen does have a higher electronegativity than does chlorine, bromine or iodine. However, oxygen's orbital overlap with carbon allows it to more readily donate into the ring's pi system. The larger halogens aren't able to do this (to an appreciable extent).

$\endgroup$
-1
$\begingroup$

All the good points been drawn, the electron withdrawing ability of oxygen is reduced by the I+ effect of the alky group of the alkoxide, thus all the more favouring electron donation to the aromatic ring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.