Comparing the reactivity towards aromatic electrophilic substitution reaction

Question

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)

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

3) 4-methylphenol or 3-methylphenol

4) 4-nitrotoluene or 3-nitrotoluene

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.

Answer 1

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.

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.

Answer 2

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.