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Why are halogens ortho- para- directors?

The previous post in this series tried to show that the key to understanding why a substituent is an ortho-, para- director or meta– director lies in understanding how it influences the stability of the ortho-, para- and meta- carbocation intermediates.

In ortho- and para- addition, there’s a resonance form where the carbocation ends up directly bonded to the substituent.

In meta– addition, the carbocation ends up on the carbon adjacent to the carbon bonded to the substituent.

All activating groups are ortho-, para– directors. These groups can donate electron-density to an adjacent carbocation through inductive effects (a.k.a. “sigma-donation”, as with alkyl groups) and/or pi-donation, where donation of a lone pair from an attached oxygen or nitrogen provides a key resonance form where all carbons have a full octet. Carbocations, being electron-poor, are stabilized by electron-rich neighbors.

Most deactivating groups are meta- directors. They withdraw electron-density through an adjacent carbocation through being “sigma-acceptors” (such as the electron-withdrawing CF3 group, or the ammonium [–NR3+] group) and/or “pi acceptors”, such as nitro, carbonyl, or sulfonyl groups. Carbocations are destabilized by electron-poor neighbors.

Which brings us to the peculiar case of halogens.

Halogens are deactivating substituents, which is to say that the rate of electrophilic aromatic substitution is lowered when a halogen replaces hydrogen (H) as a substituent. [See this earlier post on “activating vs. deactivating substituents“] . This reflects their high electronegativity, withdrawing electron density from the ring. [Note]

At first glance, this might seem to preclude them from being ortho-para directors.  But lo,  they are!

How can we rationalize this observation?

Recall that  “activating” vs. “deactivating” just compares how well a substituent stabilizes a carbocation relative to hydrogen.

That’s not the right comparison here. Just like the old joke goes, it’s not about outrunning the bear – it’s about outrunning the other guy.  

The key for a substituent being an ortho-, para- versus meta- director is the stability of the ortho- and para– carbocation intermediates versus the meta- carbocation intermediate.

We can rationalize the ortho-, para- directing ability of halogens by noting that these atoms have attached lone pairs, and can (albeit poorly) act as pi-donors. This results in a resonance form where carbon has a full octet.

Note that I didn’t say “predict” – I said “rationalize” : – ) .  Rationalization involves looking backward from a result and trying to understand why something might have happened.  There are several variables at work here that tug in opposite directions, and predicting the magnitude of these individual effects in the absence of a strong computational model is a fool’s errand. That’s why we run experiments!

From these experiments, it seems that a carbocation intermediate which has a pi-donor is more important toward determining whether it is an ortho-, para- director than whether it is a strong electron withdrawing group.

Just for fun, we could take this a bit further.

Are there any other deactivating ortho-, para- directors?

Yes. NO.


Yes, NO. Nitroso.

Knowing what we now know about halogens, what predictions would you make for the nitroso group, a group that is somewhat electron withdrawing, but also bears a lone pair on the nitrogen.


The yields aren’t great, but there you go.

[I can’t find any papers detailing how deactivating the nitroso group is – if anyone could share a source of experimental data, I’d welcome it. ]

[Note 1] It is interesting to note, however, that despite having the highest electronegativity, fluorine is actually the most activating of the halogens (the other halogens are relatively similar in their deactivating powers). This can be attributed to the better orbital overlap of the fluorine sp3 orbitals with the 2p orbitals of the pi system. [For similar reasons, BF3 is a worse Lewis acid than BCl3 and BBr3 ,  since the fluorine orbitals overlap much better with the empty boron 2p orbital].



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