3
$\begingroup$

In phenol, the OH group has -I effect due to electronegative oxygen. So, OH group should deactivate the benzene ring towards electrophilic substitution.

[P.S.: I am not talking about the o,p-directing ability of OH, which appears due to lone pair. Also please do not suggest me the question at the link Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?, because that is not what I am looking for.]

$\endgroup$

1 Answer 1

9
$\begingroup$

The oxygen in the $\ce{-OH}$ group in phenol has unpaired electrons that are capable of being donated to the ring. Although oxygen is electronegative and does exhibit some I- effects, the electron sharing that oxygen is capable of (shown by when drawing the resonance structures) has a stronger effect and results in $\ce{-OH}$ being a good ring activator. I've included the resonance structures below.

Phenol resonance

$\endgroup$
4
  • $\begingroup$ in such case, the halogens should also work as activating groups, but they are deactivating. How do you explain it then??? $\endgroup$ Dec 10, 2015 at 15:01
  • 3
    $\begingroup$ Halogens are weekly deactivating, there are two competing mechanisms: inductive (deactivating), resonance (activating), overall = weakly deactivating $\endgroup$
    – Jeanno
    Dec 10, 2015 at 15:46
  • 1
    $\begingroup$ Halogens are deactivating groups because they are much so electronegative (even more than oxygen). Because of this, electrophilic addition reactions with halobenzenes will be considerably slower than with phenols. They are, however, ortho/para directions like oxygen because, like oxygen, they have unbonded pairs of electrons that get donated to the ring, forming similar resonance structures. $\endgroup$
    – ringo
    Dec 10, 2015 at 17:59
  • 1
    $\begingroup$ You could even improve this answer with orbitals; the hydroxy group introduces a seventh p-orbital into the pi system and two more electrons; the latter make for the stronger activation. $\endgroup$
    – Jan
    Dec 11, 2015 at 2:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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