Organic Bases

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Trimethylamine has three methyl groups, and each one pushes electrons onto the nitrogen. So it should be the strongest base among the methylamines, right? In the gas phase, it is. But in water, it falls behind methylamine, which has only one methyl group! So how can that happen? Today we find out what makes an organic base strong or weak, and this comes up often in JEE, so let's begin.

0:31Measuring basic strength: Kb and pKb

First, let us measure the strength of a base. On screen, a base B takes a proton from water, H two O, giving B H plus and O H minus. The small numbers written low are subscripts, the signs written high are charges, and the two half arrows mean equilibrium. Okay? Its equilibrium constant is K b, with a small b written low, and p K b is minus log of K b. Now, a stronger base pushes this equilibrium to the right, so its K b is bigger. Is its p K b then bigger, or smaller? It is smaller, yes, exactly! Because of the minus sign, a bigger K b gives a smaller p K b. So, the smaller the p K b, the stronger the base. Right? For a base B and its conjugate acid, B H plus, at two hundred and ninety eight kelvin, p K a of B H plus, plus p K b of B, equals fourteen; the brackets name the species. With me? And a base is strong when two things hold: its lone pair is available, and its conjugate acid, B H plus, is stable. Clear?

1:41Aliphatic amines: gas phase versus water

Now, back to our puzzle. In the gas phase, with no water around, which is the strongest base: ammonia, or the amine with one, two or three methyl groups? It is trimethylamine, yes, exactly! In the gas phase only the plus I effect matters: alkyl groups push electrons onto nitrogen, so more alkyl groups make a stronger base. So tertiary, then secondary, then primary, then ammonia. The little circle up high means degree, and the greater than sign means stronger than. Got it? But in water, three factors compete: the plus I effect, the solvation of the ammonium ion by hydrogen bonding, and steric crowding. Okay? Water forms hydrogen bonds with the N H bonds of the ion; the short dash between N and H is that bond. So the primary ion, with the most N H bonds, is solvated best. Right? So in water, the methyl order is dimethylamine, methylamine, trimethylamine, then ammonia; the brackets group the methyl groups. Trimethylamine fell from first to third. Tell me the reason. Yes, exactly, it is solvation and crowding. Its ion has only one N H bond, so it is poorly solvated, and it is crowded. The p K b values agree: three point two seven, three point three eight, four point two two, and four point seven five. So our puzzle is solved. With me?

3:12Aromatic amines and amides

Now, aromatic amines. Aniline has a nitrogen lone pair, just like ammonia. So which is the stronger base, aniline or ammonia? It is ammonia, by far. Aniline's p K b is nine point three eight, so it is a much weaker base than ammonia. Its lone pair is delocalised into the benzene ring by resonance, so it is less available for a proton. Okay? But benzylamine, C six H five, C H two, N H two, has p K b four point seven zero. The C H two keeps the nitrogen off the ring, so it behaves like an alkylamine. Clear? Now, groups on the ring, say at the para position. Electron-releasing groups like OCH three, methoxy, and C H three, methyl, increase the basicity. Electron-withdrawing groups like N O two, nitro, C N, cyano, and the halogens decrease it. Right? Now a small test. In ortho toluidine, a methyl group sits next to the N H two group; the italic o means ortho. Methyl releases electrons, so it is stronger than aniline, right? Oh no, careful, that is the trap! By the ortho effect, ortho-substituted anilines are weaker bases than aniline, whatever the substituent, even methyl. The group at ortho hinders the solvation of the protonated ion. Got it? Last, amides are not basic in water. The nitrogen lone pair is delocalised onto the carbonyl oxygen: that is the O in C double bond O, where the double line on screen is the double bond. With me?

4:51Lone-pair orbital, pyrrole and guanidine

One more factor: the orbital holding the lone pair. The more s-character in it, the weaker the base; the double arrow means implies. So, which has more s-character, s p or s p three? It is s p, yes! An s p orbital is half s, while s p three is one quarter s; the three written high is a superscript. So amine, s p three, beats pyridine and imine, s p two, which beat nitrile, s p. Okay? Now, pyrrole is not basic at all. Its lone pair is part of the aromatic sextet, the six pi electrons of the ring, so it is not free to take a proton. Right? At the other end is guanidine: two N H two groups and a double bond to N H, all on one carbon; the double line is the double bond. It is a very strong base. Clear? The reason is that its cation has three equivalent resonance structures, so it is very stable. That is our second condition: a stable conjugate acid makes a strong base. Got it?

5:56Example 1: order of ethylamines and ammonia in water

Alright, example one. Arrange ethylamine, diethylamine, triethylamine and ammonia in order of decreasing basic strength in water. Pause the video and try it first. Done? Okay, let's check. Many students copy the methyl order: secondary, primary, tertiary. So diethylamine, ethylamine, triethylamine, then ammonia, right? No, no, wait, the ethyl order is different! Look at the p K b values; C two H five is the ethyl group. Diethylamine, three point zero zero. Triethylamine, three point two five. Ethylamine, three point two nine. Ammonia, four point seven five. Okay? The smallest p K b means the strongest base. So the answer is diethylamine, greater than triethylamine, greater than ethylamine, greater than ammonia. With ethyl groups, the tertiary amine comes second, not third, so learn both orders. Right?

7:01Example 2: basicity of o-, m- and p-nitroaniline

Now a harder one. Arrange ortho, meta and para nitroaniline in order of decreasing basic strength; the italic o, m and p stand for ortho, meta and para. Here is a hint: ask which effects of the nitro group act at each position. Pause the video and try it. Done? Let's check. The nitro group pulls electrons, and its inductive pull weakens with distance. Para is the farthest, so para nitroaniline is the strongest, right? Careful, that is the trap, so let us go position by position. At meta, only the minus I effect acts, the electron-withdrawing inductive effect. With me? At para, minus I acts, and minus R also acts: the nitro group pulls the lone pair through the ring by resonance. So para is weaker than meta. Okay? At ortho, minus I and minus R both act, and in addition there is the ortho effect we just learnt. So ortho is the weakest. Got it? So the answer is meta, greater than para, greater than ortho. Distance alone was not enough; resonance and the ortho effect decided it. Clear?

8:18Quick revision: pKb, gas vs water orders, ortho effect

So, quick revision. The smaller the p K b, the stronger the base, and p K a plus p K b equals fourteen at two hundred and ninety eight kelvin. In the gas phase, tertiary beats secondary beats primary; in water, methyl amines go secondary, primary, tertiary, and ethyl amines go secondary, tertiary, primary. And a delocalised lone pair, or more s-character, means a weaker base, and ortho anilines are weaker than aniline. Now try the practice problems on this page. All the best!