JEE Main and AIEEE Questions: Basicity
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Two past JEE Main questions on the basic strength of amines, solved step by step: why the smallest pKb means the strongest base, the methyl and ethyl orders in water versus the gas phase, why aniline and N-methylpyrrole are weak, and why the cage amine quinuclidine beats both triethylamine and diethylamine.
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Transcript
0:00Introduction
Hello everyone! Let us start with a small puzzle. Take triethylamine, a nitrogen holding three ethyl groups. Now join the far ends of those three chains to one extra carbon, and you get a cage called quinuclidine. Same nitrogen, three alkyl chains again. So, should the cage change the basic strength at all? It does, and enough to decide the answer of a JEE Main question! Today we solve two past JEE Main questions on basicity. Questions like these come up often, so let's begin.
0:34How we compare bases
So, what does basic strength mean? An amine, RNH two, where R is the carbon group, takes a proton from water and gives RNH three plus, the ammonium ion, and O H minus, hydroxide. The small numbers written low count atoms, the raised plus and minus are charges, and the two half arrows mean an equilibrium. Okay? Its equilibrium constant is K b, with a small b written low, for base. And p K b is minus log of K b; the small p in front simply means minus log. So a larger K b gives a smaller p K b. Right? Now, our first question asks for the smallest p K b. So tell me, does a smaller p K b mean a stronger base, or a weaker one? It means a stronger base, yes, exactly, because of that minus sign! Students flip this under exam pressure, so fix it now. Clear? And what makes one amine stronger than another? Anything that makes the lone pair on nitrogen more available, or that stabilises the ammonium ion formed. Both of today's questions turn on these two ideas. Got it?
1:43Alkylamines: gas phase versus water
Now, alkyl groups release electrons towards the nitrogen. This is the plus I effect, a plus sign and a capital I, for inductive. So every alkylamine is a stronger base than ammonia, N H three. Okay? In the gas phase, only this push acts. The little raised circle is read as degree: one degree is primary, with one alkyl group on nitrogen, two degree secondary, three degree tertiary. So tertiary beats secondary, which beats primary, which beats ammonia; each greater than sign means is a stronger base than. Right? But both of today's questions are about water. So in water, does the tertiary amine still win? It does not, and this is the heart of the chapter! In water, the ammonium ion is stabilised by H bonding, that is hydrogen bonding, through its N H bonds; the small dash is a bond. So more N H bonds help. And bulky groups hinder solvation, because water cannot get close. With me? So, for the methylamines in water, dimethylamine comes first, then methylamine, then trimethylamine, then ammonia; the small two or three after a bracket counts the methyl groups. Trimethylamine drops to third, because its ion has only one N H bond and is crowded. Clear? And the ethyl series in water is different: diethylamine first, then triethylamine, then ethanamine, then ammonia; C two H five is the ethyl group. In both series, the secondary amine is the strongest. Remember this line for Example two. Got it?
3:21When the lone pair is busy
Now the aromatic amines. In the figure, aniline, C six H five N H two, is a benzene ring carrying an N H two group. The two dots on the nitrogen are its lone pair, and the double headed arrows join resonance structures. In structures two to four, labelled with Roman numerals, the lone pair has moved into the ring, leaving a plus on the nitrogen and a minus at the ortho and para carbons. Okay? So the lone pair is spread over the ring. Then which is the stronger base, aniline or ammonia? It is ammonia that wins, yes, exactly! Aniline's lone pair is busy in the ring, so it is less available to a proton. Its p K b is nine point three eight, against four point seven five for ammonia, so aniline is a far weaker base. Right? Now pyrrole, a ring that contains the nitrogen itself. Its lone pair is part of the aromatic sextet, the six electrons that make the ring aromatic, so it is not free to take a proton. With me? So pyrrole is practically not a base; only very strong acids can protonate it. Keep pyrrole in mind, because it walks into Example two wearing a methyl group. Clear?
4:34Example 1: smallest pKb, methylamines vs aniline
Now Example one, from JEE Main twenty fourteen. Considering basic strength in aqueous solution, which of these four has the smallest p K b: dimethylamine, methylamine, trimethylamine, or aniline? Pause the video and try it first. Done? Okay, let's check. We fixed this at the start: the smallest p K b means the strongest base, and aqueous solution means water. Okay? First, throw out aniline, C six H five N H two. Its lone pair is delocalised into the ring, so it is the weakest of the four. Right? Now three methylamines are left. Trimethylamine has three methyl groups pushing electrons, the most push of all. So the answer is trimethylamine, right? That is the trap, so careful here: the push alone decides only the gas phase order. In water, dimethylamine comes first, then methylamine, then trimethylamine, because the trimethylammonium ion has one N H bond and is crowded. Clear? And the p K b values agree: three point two seven, three point three eight, four point two two, and nine point three eight, for dimethylamine, methylamine, trimethylamine and aniline, in that order. With me? So the answer is dimethylamine, the secondary amine. Smallest p K b, strongest base, and in water the secondary amine wins. Got it?
6:00Example 2: why quinuclidine is the strongest base
And now Example two, a harder one, from JEE Main twenty twenty two. Which is the strongest Bronsted base: triethylamine, diethylamine, N methylpyrrole, or quinuclidine? Quinuclidine's full name is in round brackets, one aza bicyclo, two two two, octane; the numbers in square brackets count the carbons in its three bridges. Pause the video and try it first. Done? Okay, let's check. Bronsted base means a proton acceptor. N methylpyrrole is pyrrole with a methyl on the nitrogen; the italic N shows where it sits. Its lone pair is still in the aromatic sextet, so it is practically not a base, and out it goes. Okay? Next, diethylamine against triethylamine. Remember the ethyl series in water: diethylamine is the stronger of the two. So diethylamine, the secondary amine, is our answer, right? Careful here, because we have not looked at the cage yet. First, see why triethylamine lost: its three ethyl groups move freely and crowd the nitrogen, so water cannot stabilise its ion well. Right? Now, quinuclidine is also a tertiary amine with three alkyl chains. So what is different about its chains? They are tied back, yes, exactly! In the cage, the three chains are pulled back into rings, away from the nitrogen. So the lone pair is exposed, free to take a proton, and nothing crowds it. With me? And its cation, the ion it forms with a proton, is easily solvated, because water can reach the nitrogen. So quinuclidine is a tertiary amine without the crowding. Clear? So the answer is quinuclidine. And that solves our puzzle: tying the chains back into a cage turns a crowded tertiary amine into the strongest base of the four. Got it?
7:59Quick revision: pKb, order in water, busy lone pairs
So, quick revision. The smallest p K b means the strongest base, and always check whether the question says gas phase or water. In water, the secondary amine leads both series, while aniline and pyrrole have their lone pairs busy in the ring. And when the chains are tied back, as in quinuclidine, the lone pair is exposed and the ion is easily solvated. Now go and try the other past JEE questions in this chapter. All the best!
More videos in Unit 18 · Organic Compounds Containing Nitrogen
- Overview
- Classification and Nomenclature
- Structure and Physical Properties
- Formulas, Isomer Counts and Analysis
- Past JEE Questions: Structure and Properties
- Reduction Methods
- From Alkyl Halides
- The Hoffmann Bromamide Degradation
- JEE Advanced and IIT-JEE Questions: Preparation
- AIEEE and Earlier JEE Main Questions: Preparation
- Aliphatic Amines
- Aromatic Amines and Other Nitrogen Bases
- Basicity Calculations and Separations
- JEE Advanced and IIT-JEE Questions: Basicity
- Alkylation and Acylation
- Carbylamine Reaction and Nitrous Acid
- Hinsberg's Test
- Ring Substitution in Aniline
- The Hoffmann Elimination
- JEE Advanced and IIT-JEE Questions (2001–2014): Reactions
- IIT-JEE Questions (1981–1999): Reactions
- AIEEE and Earlier JEE Main Questions: Reactions
- Overview
- Diazotisation
- Replacement of the Diazonium Group
- Coupling Reactions
- JEE Advanced and IIT-JEE Questions (2000–2016): Diazonium Salts
- IIT-JEE Questions (1990–1999): Diazonium Salts
- AIEEE and Earlier JEE Main Questions: Diazonium Salts
- Nitro Compounds
- Cyanides and Isocyanides
- Amides and Urea
- Conversions Through Nitrogen Compounds
- IIT-JEE Questions (1997–2009)
- IIT-JEE Questions (1986–1994)
- AIEEE and JEE Main Questions
- Overview
- Set 1
- Set 2
- Set 3
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks