Aromatic Amines and Other Nitrogen Bases

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Pyridine and pyrrole are both rings with a nitrogen atom, and in both, the nitrogen carries a lone pair of electrons. So both should be bases, right? But pyridine is a base, and pyrrole is hardly basic at all. Same atom, same lone pair, and a different answer. So why is that? Hold on to that question, because by the end of this video, you will explain it yourself. Questions that ask you to arrange nitrogen compounds by basic strength come up often in JEE, so let's begin.

0:36Why aniline is a weak base

Let us begin with aniline, C six H five N H two. The small numbers written low are subscripts, and they count the atoms. Like every amine, aniline has a lone pair on N, the nitrogen, and a base uses that lone pair to take up a proton. Okay? But in aniline, the nitrogen is joined directly to a benzene ring. So tell me, where else can this lone pair go? Yes, exactly, it spreads into the ring. The lone pair is delocalised, and aniline has five resonance structures; the figure shows four, numbered one to four and joined by double headed arrows, and the fifth is simply the other Kekulé form of the first. In the first one, the two dots on the nitrogen are the lone pair. In the others, see the plus sign on the nitrogen, and the minus sign at the ortho and para positions. With me? Now add a proton. In the anilinium ion, C six H five N H three plus, the raised plus sign is the charge. The lone pair now holds the proton, so nothing is left to delocalise, and only the two Kekulé structures of the ring remain. Got it? So aniline, with five structures, is stabilised more than its ion, with only two. Taking a proton would throw away that extra stability, so protonation is unfavourable. And that is exactly why aniline is a weak base. Clear?

1:59Groups on the ring

Now let us put a group on the ring. Electron releasing groups, like OCH three, the methoxy group, and C H three, the methyl group, push electrons towards the nitrogen. The lone pair is more available, so the base is stronger. Okay? Now take the nitro group, which pulls electrons out of the ring. What will it do to the basic strength of aniline? It makes aniline a weaker base, correct. Electron withdrawing groups pull the lone pair away: N O two, the nitro group, C N, the cyano group, the halogens, S O three H, the sulphonic acid group, and COOH, the carboxylic acid group. Right? But for the nitro group, the position matters. From the ortho or para position, it pulls the lone pair by resonance, so it acts most strongly. From the meta position, only the inductive effect acts, so its effect is smaller. Clear? Now here is a trap. Methyl releases electrons. So two methylaniline, with the methyl group at the ortho position, must be a stronger base than aniline, right? No, careful, it is actually weaker than aniline, and this is called the ortho effect. Any group at the ortho position, whatever it is, makes aniline a weaker base, because it crowds the cation and hinders its solvation. So at ortho, do not trust the electron effect alone. With me?

3:25The order to remember

So here is the order to remember, strongest base first. The greater than sign means a stronger base, and the italic p stands for para. Para anisidine, that is four methoxyaniline, beats para toluidine, four methylaniline, then comes aniline, and then para chloroaniline. Okay? And it carries on with the three nitroanilines, where the italic m is meta and the italic o is ortho. Meta comes first, because from meta there is no resonance. Ortho is the weakest of all, with resonance and the ortho effect both acting. Right? There is one more point here. In N alkylanilines, the italic capital N tells you that the alkyl group sits on the nitrogen itself. These are slightly stronger bases than aniline. Got it? Now, benzylamine, C six H five C H two N H two. Its nitrogen sits on the C H two carbon, which is s p three; the small three is a superscript. So which is the stronger base, benzylamine or aniline? Benzylamine is much stronger, very good. Through that s p three carbon, the lone pair cannot reach the ring, so it is not delocalised, and benzylamine behaves as an alkylamine. Clear?

4:39Other nitrogen compounds

Now let us go beyond amines. In an amide, the nitrogen lone pair is delocalised onto the oxygen of the carbonyl group, so amides are practically neutral. And imides, like phthalimide, are even weakly acidic. Okay? Now, a nitrogen lone pair can sit in an s p three, an s p two, or an s p orbital. Which one gives the strongest base? The s p three lone pair of amines, correct. Then comes s p two, as in pyridine and imines, and the weakest is s p, as in nitriles. Right? The rule is simple: the more s character in the orbital holding the lone pair, the weaker the base. Even so, pyridine is still a base, with a p K b of eight point eight. In p K b, the small b is a subscript, and a larger p K b means a weaker base. Got it? And at the other end, amidines and guanidine are very strong bases. When they take a proton, the positive charge is shared by resonance over two or three nitrogen atoms, which makes the cation very stable. With me?

5:47Example 1: increasing basicity of para-substituted anilines

Now, here is our first example. Arrange in order of increasing basic strength: aniline, four nitroaniline, four methylaniline, and four methoxyaniline. Pause the video and try it first. Done? Okay, let's check. The four in each name puts the group at position four, which is para to the N H two group. Methoxy and methyl release electrons, and the nitro group withdraws them. And from our order, methoxy beats methyl. Right? Now careful, the question says increasing. So which compound do you write first? The weakest base comes first, so four nitroaniline leads. Then aniline, then four methylaniline, and four methoxyaniline is the strongest; the less than sign means a weaker base. Writing the strongest first is a common mistake, so read that word twice. Clear?

6:43Example 2: piperidine, pyridine, pyrrole in decreasing order

Now our second example takes us back to the opening puzzle. Arrange piperidine, pyridine and pyrrole in order of decreasing basic strength. Pause the video and try it first, and here is a hint: look at the nitrogen in each ring. Done? Let's check. Piperidine is a saturated ring, with only single bonds, so its nitrogen is s p three, just like an ordinary amine. Right? Pyridine is an aromatic ring, and its nitrogen is s p two. Its lone pair is free, so pyridine is a base, but weaker than piperidine, because of the greater s character. Okay? Now look at pyrrole: its nitrogen also has a lone pair. So why is pyrrole hardly basic at all? Because that lone pair is part of the aromatic sextet, the six electrons that make the ring aromatic: four from the two double bonds, and two from the nitrogen. To take a proton on that nitrogen, the ring would have to give up its aromatic character, so the lone pair is not available. That is our puzzle, solved! Got it? So the order is piperidine, then pyridine, then pyrrole, the weakest. See how one question checks two ideas, hybridisation and aromaticity? Clear?

8:02Quick revision: aniline, ring groups, hybridisation, pyrrole

So, quick revision. In aniline, the lone pair is delocalised into the ring, so aniline is a weak base. Electron releasing groups raise basicity, while electron withdrawing groups and any ortho group lower it. And for other nitrogen bases, s p three beats s p two, which beats s p, and a lone pair inside an aromatic sextet is practically not basic. Now go and try the practice problems on this page. All the best!