![]() ![]() Therefore our analysis suggests that the para isomer will be the least deactivated and consequently react faster. 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. Clearly, a position that is doubly deactivated will react much slower than a position that is singly deactivated. In this case we find that each position is only deactivated by one of the nitro groups. We can analyze the para-dinitro isomer in a similar manner. One position is not deactivated since it is meta to both nitro groups. 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). I've carried out the first step of this exercise and show the results in the following figure. Next we can compare the molecules and see which one has the most activated positions. 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. And try not to provide an answer by directly copying from a source (especially large) thanks. I would like an explanation for each answer and especially the 1st one in detail. If my concept is wrong with respect to above context, please correct me.Īll of the remaining isomer pairs have same problem, but each is a bit different. 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). 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). 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. So which of these will reduce the electron density on benzene more. If the second nitro comes at C3 then the negative charge at C3 will be absorbed by nitro. So if another nitro comes at C4 then the positive charge at C4 will intensify as nitro draws more electron density. In the first pair of molecules, according to resonance the presence of one nitro will make C3 negative and C4 positive. In the following pairs of molecules, which is more reactive towards electrophilic substitution reaction?ġ) 1,4-dinitrobenzene or 1,3-dinitrobenzene (don't consider the ortho isomer)Ģ) benzene-1,3-diol or benzene-1,4-diol (don't consider the ortho isomer) ![]()
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