Which of the statements below is correct?

a) The ortho and para positions are more active than the meta position.


b) The ortho and para positions are activated while the meta position is deactivated.

  • 1
    $\begingroup$ Neither of the statements is incorrect. Both statements are stating that the electrophilic substitution occurs at the ortho and para positions. The first statement just tells you where the reaction will occur, but does not provide any reason. The second provides a reason, but doesn't provide a cause. $\endgroup$ – LDC3 Jun 21 '15 at 19:44

As the figure below illustrates, substituents on aromatic rings that are electron donating like the methyl and t-butyl groups (and also the hydroxy group [e.g. phenol]) increase the reaction rate at all positions (ortho, meta and para) relative to benzene.

On the other hand, electron withdrawing substituents (the chlorine and ester carbonyl examples) decrease the reaction rate at all positions relative to benzene.

So for phenol, or any other electron donating substituent, while the meta position does react more slowly than the ortho and para positions, it still reacts faster than benzene. Therefore, it would be incorrect to say that the meta position is deactivated. Only when the rate at a given position is less than the rate for a single position in benzene could one say that the position is deactivated.

Statement "a" in your question is correct.

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(image source)

For future reference a note on terminology. The above figure tells us that nitration of toluene proceeds 24.5 times faster than the nitration of benzene; that is to say that the relative rate for nitration of toluene is 24.5 times greater than benzene. The meta position in toluene reacts 3 times faster than a single position in benzene, this is referred to as the partial rate factor.

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  • $\begingroup$ While the hydroxy group increases the electron density via mesomeric effects, wouldn't it decrease the electron density at meta position via inductive effects? $\endgroup$ – user8244 Jun 22 '15 at 2:24
  • $\begingroup$ Yes, but 1) resonance effects usually outweigh inductive effects in aromatic systems and 2) inductive effects drop off rapidly with distance. $\endgroup$ – ron Jun 22 '15 at 3:24
  • $\begingroup$ On drawing the resonance structures, isn't it only the ortho and para positions that get higher electron density? How does the meta position receive any resonance effect? $\endgroup$ – user8244 Jun 22 '15 at 5:47
  • 1
    $\begingroup$ That is a very good question, and also a very different question from your original question; you might want to ask it as a separate question. Basically, look at your resonance structures. Although the substituent does not interact with the meta position by resonance, it nonetheless stabilizes the overall meta intermediate because a substituent on a double bond stabilizes a double bond. Therefore, the meta intermediate is stabilized compared to the benzene intermediate and will react slightly faster than the benzene case. $\endgroup$ – ron Jun 22 '15 at 13:56

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