12
$\begingroup$

If I have a strong deactivating group (like nitro) and a weak activating group (like methyl) attached to the benzene ring, which group's preference will direct the next substitution on the benzene ring? For example, if a chloro group is attached to meta-nitrotoluene, where will it substitute?

If I have both strong activating and also deactivating groups in the benzene ring, which position will be preferred for the next electrophilic substitution? For example, if I want to attach a chloro group to the meta-nitroaniline, where will it substitute?

Improve this question
$\endgroup$
8
  • $\begingroup$ Im not sure if I understood correctly, but I vaguely remember answering a similar question, not sure if it addreses your question but its found here: Rules for benzene halogenation $\endgroup$
    – xavier_fakerat
    Jun 8, 2017 at 7:59
  • $\begingroup$ if I want to attach a chlorine group to the meta-nitrotoluene, where will be chlorine attached ? @xavier_fakerat $\endgroup$
    – Hasin Punno
    Jun 8, 2017 at 8:06
  • $\begingroup$ For that you may refer to this site: Electrophilic Substitution of Disubstituted Benzene Rings but if you wish to have that summarised, I could make an answer for it. Hope it helps $\endgroup$
    – xavier_fakerat
    Jun 8, 2017 at 8:25
  • $\begingroup$ Thanks for the link. It helps a lot. But this link does not deal with strong deactivating and weak activating group combined at benzene ring like meta-nitrotoluene. Would you please clarify the question : if I want to attach a chlorine group to the meta-nitrotoluene, where will be chlorine attached ?" . Thanks in advance . @xavier_fakerat $\endgroup$
    – Hasin Punno
    Jun 8, 2017 at 8:45
  • $\begingroup$ To prevent a homework-answer scenario, I had to explain the concepts so you understand and be able to apply. Generally. I dislike this approach (question-direct answer) because it has long term negative consequences. So that is why I give Conceptual answers, once you get the concept you will apply it anywhere and still pass, suceed etc.. It may be a bit of content but trust me it helps :). $\endgroup$
    – xavier_fakerat
    Jun 8, 2017 at 10:43

1 Answer 1

Reset to default
12
$\begingroup$

Electrophilic Substitution of Disubstituted Benzene Rings

When a benzene ring has two substituents, each substituent exerts an effect on subsequent substitution reactions.

Electrophilic substitution of a disubstituted benzene ring is governed by the same resonance and inductive effects that affect monosubstituted rings. The only difference is that it’s necessary to consider the additive effects of two different groups.

  • If the directing effects of the two groups reinforce each other, the situation is straightforward. In p-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). A single product is thus formed on electrophilic substitution.

enter image description here

From above, the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). This is similar case with chlorine.

  • If the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed.

  • Further substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.

You may also wish to review this table showing positions where common substituents direct:

enter image description here

Basing on the above explanations, it is easy to predict the products of electrophilic substitution of disubstituted benzene ring, the only thing you have to know is the effect of each substituent whether it is activating or deactivating. In some cases, observe the positions of the substituents as point 3 mentions steric effects may come into play and substitution may not occur.

On the second note, concerning diazonium ion, I think you have to look at measured dipole moments, experimental data etc I doubt if there is a way of telling whether a substituent is strongly deactivating by merely looking at it

After explaining concepts now lets see what the answer for the question in comments will look like:

enter image description here

Reasons

  • the directing effects are opposing (nitro is meta while methyl is both ortho and para directing) but the methyl is more activating so we have a mixture:

    1. meta to nitro and methyl

    2. ortho to the methyl but para to the nitro

In this scenario since position 3 is not deactivated by nitro group hence there is also a possibility of a mixture here, also take note of how the nitro group is strongly deactivating at position 4 (more additive effect)

Hope this helps

References

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thanks. So between activating and deactivating groups, always activating groups are preferred (even the activating group is weak).right ? Except Stearic Hindrance effect. $\endgroup$
    – Hasin Punno
    Jun 8, 2017 at 11:00
  • $\begingroup$ I think this had to do with charge stabilisation. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization (e.g some activators like -OH), it will normally exert the product determining influence, even though it may be overall deactivating $\endgroup$
    – xavier_fakerat
    Jun 8, 2017 at 11:07
  • $\begingroup$ Thank you, The group which has lone pair electrons and if it can stabilize the adjacent atom, will be preferred. So if OH and nitro are present, OH is preferred. Right,sir ? $\endgroup$
    – Hasin Punno
    Jun 8, 2017 at 11:32
  • $\begingroup$ That only applies to case 2 :) otherwise you just look at all possible outcomes (the three points) $\endgroup$
    – xavier_fakerat
    Jun 8, 2017 at 11:39
  • $\begingroup$ And, for your last reaction, I want to know about meta-nitrotoluene but you explain about ortho-nitrotoluene.. meta-nitrotoluene has a problem as two group oppose. neither toluene or nitro has lone pair. So which group will be preferred ? Sir I understand many things but the answer of this question is still in confusion. :-( $\endgroup$
    – Hasin Punno
    Jun 8, 2017 at 11:40

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.