JEE Advanced and IIT-JEE Questions: Basicity

Transcript

0:00Introduction

Hello everyone! Let us start with two molecules that look almost like twins. Acetamide, and an amidine, which is the same molecule with its oxygen swapped for an N H group. But one of them is practically neutral, and the other is the strongest base in its question! How can one swapped atom do that? The IITJEE paper of two thousand and one asked exactly this, and questions on basic strength come up often. So, let's begin.

0:30Idea 1: two things decide basic strength

So, let us start with idea one. Almost every basicity question comes down to two things. First, how free the lone pair on nitrogen is to take a proton: the more available, the stronger the base. The double arrow on screen means implies. Okay? Second, how well the cation formed is stabilised: the better it is stabilised, the stronger the base. Right? Now, alkyl groups push electrons towards nitrogen: the plus I effect, a plus sign and a capital I, for inductive. In the gas phase it acts alone, so tertiary beats secondary, which beats primary, which beats ammonia, N H three; the small numbers written low count atoms. The little raised circle is read as degree, and each greater than sign means is a stronger base than. Clear? Now let me ask you this. In water, is trimethylamine still the strongest? It has three methyl groups pushing, so it should win, right? It does not, and the IITJEE paper of nineteen ninety eight asked exactly why. In water, the cation is stabilised by hydrogen bonding, H bonding on screen, through its N H bonds; the small dash is a bond. The dimethylammonium ion has two N H bonds and is less crowded, so water stabilises it better than the trimethylammonium ion, with only one. With me? So in water the order is dimethylamine, then methylamine, then trimethylamine, then ammonia; the small number after a bracket counts the groups inside it. Right? And JEE Advanced twenty fourteen tested this too: when methylamine, a primary amine, beats trimethylamine, a tertiary amine, in water, hydrogen bonding plays the central role. Got it?

2:11Idea 2: a benzene ring takes the lone pair

Now idea two, a benzene ring on the nitrogen. Look at the figure. The Roman numerals under the figure number the structures. In aniline, structure one, the two dots on nitrogen are its lone pair. In structures two, three and four, that pair has moved into the ring, leaving a plus charge on nitrogen and a minus charge on an ortho or para carbon. The double headed arrows join resonance structures. Okay? Now add a proton. In the anilinium ion, C six H five N H three plus, where the small raised plus is its positive charge, the lone pair holds that proton, so nothing is left to spread into the ring. Right? So tell me, which one is stabilised more by resonance, aniline or the anilinium ion? Aniline is stabilised more, exactly, so taking a proton costs it that extra stability. So aniline is a weak base: its p K b, with a small b written low, is nine point three eight, against three point three eight for methylamine. Bigger p K b, weaker base. Clear? That answers the IITJEE question of nineteen eighty two: cyclohexylamine is a stronger base than aniline, because its saturated ring keeps the lone pair localised on nitrogen, and releases electrons too. Got it? Now a trap from nineteen ninety three. Someone draws the anilinium ion like structure two, with a double bond from nitrogen to the ring, the C equals N on screen. Is that acceptable? The answer is no, it is not. In the anilinium ion, nitrogen already has four bonds, three to hydrogen and one to the ring. So a double bond would make five bonds and ten valence electrons, as the screen says, and nitrogen can never have that. With me?

3:57Idea 3: groups on the ring

Now idea three, groups on the ring. Electron releasing groups, like methyl, C H three, or methoxy, OCH three, placed para to the N H two group, make the base stronger. Okay? So what will a nitro group, N O two, do? It pulls electrons away from the ring. It makes the base weaker, yes, exactly! So do C N, the cyano group, and the halogens. Right? And the position matters. From ortho or para, the nitro group pulls the lone pair by resonance as well as by induction; from meta, by induction only. Clear? So meta nitroaniline is a stronger base than para nitroaniline; the small italic m and p stand for meta and para. Got it?

4:41Example 1: strongest base, benzylamine vs aniline (2000)

Time for Example one, an easy one from the IITJEE paper of two thousand. Which is the strongest base: aniline, C six H five N H two; para nitroaniline; meta nitroaniline; or benzylamine, C six H five C H two N H two? Pause the video and try it first. Done? Okay, let's check. Both nitroanilines carry the nitro group, which pulls the lone pair away, so both are weaker than aniline. They are out. Okay? So it is aniline against benzylamine. Both have a benzene ring, so both must be weak bases, right? Careful now, because that is exactly the trap. In aniline, nitrogen sits directly on the ring, so its lone pair is delocalised into it, as we saw in idea two. With me? In benzylamine, a C H two group sits in between. That carbon is s p three, the s p with a raised three, so the lone pair cannot reach the ring. It stays on nitrogen, just like in an alkylamine. Clear? So the strongest base is benzylamine, C six H five C H two N H two. The ring only matters when it touches the nitrogen. Got it?

5:55Example 2: amidine, amines, amide in basicity order (2001)

Now Example two, the harder one, from IITJEE two thousand and one, our puzzle from the start. Arrange four compounds in order of basicity. Compound one is the amidine: a carbon with a methyl group, an N H joined by a double bond, the equals sign in brackets, and an N H two group. Compound two is ethylamine, compound three is dimethylamine, and compound four is acetamide. Pause the video and try it first. Done? Okay, let's check. Acetamide, C H three CONH two, has an N H two group too. So it is a base like ethylamine, right? The answer is no, and that is the trap. In the amide, the lone pair on nitrogen is pulled onto the carbonyl oxygen, the C double bond O on screen. Amides are practically neutral, so compound four is the weakest. Okay? Now the amidine takes a proton. Where does the plus charge go? It is shared by both nitrogen atoms, very good! The cation is C H three C with two N H two groups, the two outside the bracket, and resonance spreads its plus charge over both nitrogens. So the cation is well stabilised, and compound one is the strongest base. Right? Now the two simple amines. Dimethylamine, the secondary amine, is the stronger of the two. The p K b values agree: three point two seven for dimethylamine is less than three point two nine for ethylamine, so it is just slightly stronger. Clear? So the order is one, then three, then two, then four: the amidine first and the amide last. Swap one oxygen for N H, and the weakest becomes the strongest. Our puzzle is solved! Got it?

7:41Quick revision: lone pair, H-bonding in water, ring effects

So, quick revision. The base is stronger when its lone pair is available and its cation is well stabilised. In water, hydrogen bonding and crowding decide, so dimethylamine beats trimethylamine. And a ring on nitrogen, a nitro group or a carbonyl group takes the lone pair and weakens the base, while the amidine's shared charge makes it strong. Now work through the other past questions on this page yourself. All the best!