# Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution?

If I were to attempt to nitrate Pyridine (essentially benzene but with one carbon replaced with Nitrogen) would I get substitution at the Ortho-, Meta- or para position? Also why is this? I can only think of an inductive effect that the Nitrogen might have but I can't think of more factors and I also can't think whether it will be directing towards o-,m- or p- nor why. Can you help?

• Sorry, but this is rather common knowledge. The Wikipedia article on pyridine alone would have answered most of your questions and in nearly any general textbook on organic chemistry you would have found it as well. Your question triggered a nice answer by @ron but in general it would be nice if you'd do some research before asking a question here. We don't want to repeat what you can find in most textbooks or easily via google. Rather, we want to help with problems that can't be easily solved this way. – Philipp Dec 1 '14 at 23:35
• @Philipp I don't agree with you. Virtually all questions on this website can be solved by searching in textbooks/internet/scientific articles. If the question is not extremely trivial and not asked before and shows at least some kind of though process and think it is suitable and on topic here. – Jori Dec 2 '14 at 13:36
• @Jori Certainly, all questions on this website can be solved by searching in textbooks/internet/scientific article. But that was not my point. My point was that this question is concerned about the fact that a simple Wikipedia search would have solved the problem or a look in any general textbook on organic chemistry. I think this kind of research effort prior to asking a question here should be required by any OP because that's what we require for homework questions anyway. – Philipp Dec 2 '14 at 13:53
• @Philipp I agree that the OP should have put in more effort in this case. I only meant to say that the question is in principle not off topic or bad. – Jori Dec 2 '14 at 13:57
• @Jori I would agree with you that it is not off-topic, but it borders on a homework question. Also I would concede that it is not a particularly bad question but still I don't like such questions. It would be a perfectly acceptable question if the OP says that he looked it up at Wikipedia or some textbook and didn't understand this or that point about it. – Philipp Dec 2 '14 at 14:51

The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.

Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons. Below are the resonance structures we can draw for pyridine. They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects.

This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).

If we look at the resonance structures that we can draw for electrophilic attack at the ortho

meta

and para positions,

we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen. Therefor, meta-substitution is preferred in electrophilic attack on pyridine

So electrophilic attack on pyridine will produce meta-substituted pyridines. Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation. The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred. In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.

• Why are these resonance structures in pyridine even possible? The negative charge is surely occupying a p orbital on the Nitrogen but that is part of the delocalised system already – RobChem Dec 2 '14 at 13:18
• @user3764899 What do you mean? Why would these resonance structures not be possible? – Jori Dec 2 '14 at 13:45
• @ron Any particular reason for using sodium nitrate instead of nitric acid? I also wondered: this meta substitution is obviously kinetically controlled, but is it also thermodynamically more favorable? – Jori Dec 2 '14 at 13:53
• @user3764899 Any resonance structure you can draw is a "possible" contributor to the total picture of a molecule. The ones drawn above with a negative charge on nitrogen are significant contributors to the overall description of pyridine because they are placing the negative charge on a more electronegative element. These resonance structures tell us that in pyridine, the pi electron density is more concentrated in the region around the nitrogen - unlike benzene where the pi density is uniformly distributed. – ron Dec 2 '14 at 14:32
• @Jori Yields with nitric acid are very low. There are many "tricks" used in pyridine chemistry in order to get good yields (like the use of pyridine N-oxide). Sorry, don't know the relative thermodynamic stabilities of the products. – ron Dec 2 '14 at 14:45