In the following pairs of molecules, which is more reactive towards electrophilic substitution reaction?

1) 1,4-dinitrobenzene or 1,3-dinitrobenzene (don't consider the ortho isomer)

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2) benzene-1,3-diol or benzene-1,4-diol (don't consider the ortho isomer)

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3) 4-methylphenol or 3-methylphenol

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4) 4-nitrotoluene or 3-nitrotoluene

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I am having a conflict of concept.

In the first pair of molecules, according to resonance the presence of one nitro will make C3 negative and C4 positive. So if another nitro comes at C4 then the positive charge at C4 will intensify as nitro draws more electron density. If the second nitro comes at C3 then the negative charge at C3 will be absorbed by nitro. So which of these will reduce the electron density on benzene more.

In my opinion 1,3-dinitrobenzene will have less electron density on the benzene ring as (C3 is already negative due to nitro at C1) much of the electron density at C3 is absorbed. But in 1,4-dinitrobenzene the second nitro group is absorbing electron density from C4 (in which electron density is very less already due to presence of nitro at C1). So since the electron density is less at C4 then it won't be able to take much electron from ring (as its already positive and deficient of electron density).

If my concept is wrong with respect to above context, please correct me.

All of the remaining isomer pairs have same problem, but each is a bit different. I would like an explanation for each answer and especially the 1st one in detail. And try not to provide an answer by directly copying from a source (especially large) thanks.

  • $\begingroup$ Is "more reactive" equivalent to a lower activation energy? If so, you might want to draw the arenium ions (Wheland intermediates) for the the addition of an electrophile to the different isomers. $\endgroup$ Feb 27, 2016 at 17:23
  • $\begingroup$ @KlausWarzecha actually i mean which of the two isomers makes the benzene ring more activated.For example hydroxy group activates benzene lesser as compared to amino group.Similarly a methoxy goup is less activating as compared to hydroxy group.(actually i have read about activation energy in nucleophilic substitution reactions only).But i don't really like thermodynamics and all the heat and entopy.If you would please explain with respect to concentration of charges,resonance,inductive effect and destabilisation caused by charge concentration it would be great and easy enough to understand $\endgroup$ Feb 27, 2016 at 18:03
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    $\begingroup$ A nitro group doesn't lead to a resonance structure where there are negative charges on at meta carbons. They are simply more active than the ortho and para positions because they don't have a positive charge in any resonance structures. $\endgroup$ Feb 27, 2016 at 22:29
  • $\begingroup$ @DGS only if there is a negative charge at the carbon to which nitro group is attached then only it can show -R effect (if i am right) as in the case of para nitro phenol (where negative charge is at C4 and nitro group can take the electron pair in conjugation with carbon and nitrogen).how is then "A nitro group doesn't lead to a resonance structure where there are negative charges on at meta carbons" your statement correct.Sorry I didn't get what you explained after that.And why is it that you specifically mentioned meta.Thanks. $\endgroup$ Feb 28, 2016 at 2:46
  • $\begingroup$ I'm not sure what you're asking (if you're asking anything). $\endgroup$ Feb 28, 2016 at 2:58

2 Answers 2


We can answer your question by first looking at each individual, unsubstituted position in the benzene ring and determining the relative reactivity at each of these positions. Next we can compare the molecules and see which one has the most activated positions.

I've carried out the first step of this exercise and show the results in the following figure.

enter image description here

Looking at the top molecule, m-dinitrobenzene, we see that 3 positions are deactivated by the resonance effects of both nitro groups (that is, these positions are either ortho or para to each nitro group). One position is not deactivated since it is meta to both nitro groups. We can analyze the para-dinitro isomer in a similar manner. In this case we find that each position is only deactivated by one of the nitro groups. Clearly, a position that is doubly deactivated will react much slower than a position that is singly deactivated. Now, if we try to estimate the relative molecular reactivities by summing up and averaging the reactivities at each position, we see that the meta isomer is roughly deactivated by an average of (0+2+2+2/4) 1.5 at each position, whereas the para isomer is roughly deactivated by an average of (1+1+1+1/4) 1.0 at each position. Therefore our analysis suggests that the para isomer will be the least deactivated and consequently react faster.

In the dihydroxybenzene case, hydroxyl groups are ortho-para activating. Applying the same type of analysis leads us to conclude that the various positions in the meta isomer will, on average, be more reactive than the positions in the para isomer. Hence the meta isomer will react faster in an EAS reaction.

Similarly, in the cresol case both substituents are again ortho-para activating but with the hydroxyl group being more activating than the methyl substituent. In the meta isomer, 3 positions are doubly activated and one position is not activated (or deactivated). In the para isomer, each position is singly activated, and only 2 of these positions are activated by the strongly activating hydroxyl group. This would lead us to suspect that the meta isomer will react the fastest.

  • $\begingroup$ For the answer to the third question why is the meta carbon (with respect to methyl group) is not activated by the +I effect of methyl group? – $\endgroup$ Mar 1, 2016 at 12:57
  • $\begingroup$ Would you please provide the answer for fourth question.I will try to make explanation according to your answer. $\endgroup$ Mar 1, 2016 at 13:55
  • $\begingroup$ Inductive effects are transmitted over a fairly short range, the "meta" carbon is just to far away for the +I effect to be significant. $\endgroup$
    – ron
    Mar 1, 2016 at 14:09
  • $\begingroup$ For the fourth set of molecules the para isomer will react the fastest. It is the only isomer that has (2) carbons activated by the methyl group and not deactivated by the nitro group. Try to apply the above concepts and see if you can arrive at this answer. Let me know if you encounter a problem. $\endgroup$
    – ron
    Mar 1, 2016 at 14:12
  • $\begingroup$ but the figure shows para carbon is activated in the third question.para carbon is at a greater distance than meta then how can para carbon get activated by inductive effect if meta carbon doesn't get activated? $\endgroup$ Mar 1, 2016 at 14:27

Usually substitutents that activate electrophilic aromatic substitution like OH and CH3 are most effective at positions ortho and para to the substituent, so they direct the reaction to that site. You should be able to see that by drawing resonance structures. Deactivating substituents like NO2 are most deactivating at those same ortho and para sites, so they tend to favor the meta position for whatever reaction does occur.

Now you should be able to tell. Look for the molecule that has as many sites as possible in favorable positions, ortho/para to the activating groups but meta to the deactivating ones. For instance, in the first case the 1,3 di-nitro compound has a site meta to both deactivating NO2 group versus the 1,4 isomer having no such site.

  • $\begingroup$ so you are saying 1,3 dinitro benzene is less reactive towards electrophilic substitution than 1,4 dinitro benzene.Right?In 1,3dinitro benzene does the nitro group at C3 lead to resonance stabilisation (as C3 appears to have negative charge if draw the resonance structure of nitro benzene ) $\endgroup$ Feb 28, 2016 at 23:38
  • $\begingroup$ Look for a compound with more sites that are favorably licsted, for highly deactivating groups like NO2 the meta position is favored for any reaction that could occur. Which isomer has a site meta to both deactivating groups? $\endgroup$ Feb 29, 2016 at 0:28
  • $\begingroup$ it is 1,3 dinitro benzene.But if i am right at meta there is no resonance stabilization but rather only inductive stabilization (because,in resonance structure of nitro benzene we are not really having a pair of electron on meta carbon.right?Its just a comparitive negative charge,that is,we assigned negative charge for meta carbon on the fact that ortho and para carbon had a real positive charge i.e by forming a carbocation.But a meta,no carbanion is being formed) $\endgroup$ Feb 29, 2016 at 1:54
  • $\begingroup$ I am not sure where you get a carbenium from. I get carbocaions. In the nitro group you have two highly electronegative oxygen atoms and you can consider resonance structures with both oxygens negatively charged. Now you need two positive charges. One is on the nitrogen atom, the other ends up on the ring and goes to the ortho and para positions. Which deactivates them stringly and makes the nitro group meta directing. $\endgroup$ Feb 29, 2016 at 2:49
  • $\begingroup$ the carbanion which i mentioned is in the case of resonance structure by an electron donating group as in the case of phenol (at ortho and para).And what i meant is,such sort of a carbanion is not formed (at meta ) when an electron withdrawing group is present on a benzene ring .(the negative charge is not due to the formation/presence of carbanion,rather a comparitive negative charge) $\endgroup$ Feb 29, 2016 at 4:12

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