13
$\begingroup$

The susceptibility of a benzene ring to electrophilic attack depends on the type and number of groups bonded to the ring.

Activating groups donate/release electrons and increase the electron density in the benzene ring and so render it more vulnerable to electrophilic attack.

Deactivating groups do the opposite, withdrawing electrons and reducing electron density in the ring.

The phenyl group is said to be weakly activating, suggesting that it donates electrons to a benzene ring bonded to it.

biphenyl

One possible explanation I can think of is that phenyl group donates electrons through resonance with the other ring and it therefore increases electron density to a greater extent relative to hydrogen, which is the reference point for activating ability.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ "...the phenyl group (having the same structure) should have the exact same electronegativity as the benzene ring itself and hence have no activating or deactivating effect whatsoever." The activating/deactivating effects of substituents are not measured relative to the benzene ring, they are measured relative to hydrogen. $\endgroup$
    – orthocresol
    Commented Oct 12, 2015 at 14:32
  • $\begingroup$ It donates electron density to the ring via the resonance (or mesomeric) effect, which negates the electron withdrawal (through the inductive effect) to some extent. $\endgroup$
    – orthocresol
    Commented Oct 12, 2015 at 14:40

1 Answer 1

Reset to default
7
$\begingroup$

In the following diagram, which carbocation is more stable, A or B?

enter image description here

The answer is A because the positive charge on the positively charged carbon can be delocalized into the attached benzene ring (I've drawn one of the possible resonance structures to illustrate this).

The same thinking can be applied to your question. Look at resonance structures C and D (we can draw many more) for the attack of an electrophile on a benzene ring. If there is a second benzene attached and it is attached ortho or para to where the electrophile attacks, then the positive charge can be delocalized into the second benzene ring.

This additional delocalization will stabilize the system. Hence, the second benzene is an ortho-para activating substituent. However, since this stabilization involves electrons involved in pi bonds, the stabilization (activating effect) will be weaker than when a substituent with a "more readily available" lone pair of electrons (amino, hydroxyl, etc.) is involved.

Improve this answer
$\endgroup$
2
  • $\begingroup$ By your last sentence, do you mean an activating substituent on, say, the para position, would donate electrons into the ring and towards the cation/electrophile? Wouldn't that increase the strength of the stabilising effect? $\endgroup$
    – harry
    Commented May 1, 2021 at 6:55
  • 1
    $\begingroup$ @HarryHolmes In my last sentence I'm just saying that a lone pair on, say, a methoxy group is more available then the pi electrons in an aromatic ring. Hence the methoxy group would more strongly activate (and direct) than a phenyl ring. If the biphenyl has a p-methoxy substituent, and since the phenyl ring and methoxy don't activate the same available positions, the methoxy group will "win out" and control the location of the attacking electrophile. $\endgroup$
    – ron
    Commented May 2, 2021 at 15:20

Your Answer

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

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