Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle from Set three. Four different amines share one formula, C three H nine N, okay? Add nitrous acid to each, and all four react in some way. But how many of them give off bubbles of nitrogen gas? Hold that thought. Questions that mix ideas like this come up often in JEE Main. So, a quick fire round first, then two full problems. Let's go.

0:29Quick fire: count the carbons, find the N–H

First, butanoic acid is turned into its amide, butanamide, and then put through the Hoffmann bromamide reaction; each arrow means is turned into, and the question mark is the amine we want. The amide has a chain of four carbons, and the small numbers written low count the atoms. So the amine that comes out is butan one amine, right? That is the trap, so be careful. In the Hoffmann equation, the amide, RCONH two, where R is the carbon chain, reacts with bromine and four sodium hydroxide to give the amine, RNH two. The carbon of the C O group leaves as sodium carbonate, with sodium bromide and water. So the amine has one carbon fewer. With me? So the four carbons of butanamide become three, and the product is propan one amine, with N H two at the end of a three carbon chain. Got it? Next, here are three amines: ethanamine, N ethylethanamine and N, N diethylethanamine; the italic capital N means the ethyl group sits on nitrogen. Which of them cannot form hydrogen bonds between its own molecules? Amine molecules need an N H bond to hydrogen bond with one another; the dash is the bond. The small raised circle is read as degree, so one degree means primary. Primary and secondary amines have N H bonds, and a tertiary amine has none. Clear? So the answer is N, N diethylethanamine: C two H five in brackets, with a small three outside, means three ethyl groups on nitrogen, a tertiary amine. No N H bond, so no hydrogen bonds. Okay?

2:08Quick fire: basic strength and bromine water

Now let us look at basic strength. Arrange aniline, N methylaniline and N, N dimethylaniline in increasing basic strength, using the p K b values on screen; the small b written low stands for base. Remember, among the methylamines in water, the secondary amine is the strongest. So N methylaniline, the secondary one, is the strongest here too, right? It is not, so do not copy the aliphatic order; read the data. The smaller the p K b, the stronger the base: nine point three eight, nine point three zero, then eight point nine two. So aniline is weakest, then N methylaniline, then N, N dimethylaniline; the sign that opens to the right means less than. Clear? Now let us turn to the ring. With aniline itself, bromine water gives a white precipitate of two, four, six tribromoaniline, with bromine at both ortho positions and at the para position. So what does four nitroaniline give with bromine water? Three bromines again? Not three this time, and here is why. The N H two group still directs ortho and para. But in four nitroaniline, the para position already holds the nitro group, N O two. With me? So bromine goes only to the two ortho positions, two and six. The product is two, six dibromo four nitroaniline. Right?

3:32Example 1: which C3H9N amines give N2 with nitrous acid

Now, Example one, a medium problem from Set three, and it is our opening puzzle. How many of the isomeric amines C three H nine N liberate nitrogen with nitrous acid? Pause the video and try it first. Done? Okay, let's check. First, list the isomers class by class. Primary amines are counted by the alkyl groups, and there are two propyl groups, so we get two primary amines: propan one amine and propan two amine. With me? For secondary and tertiary amines, share the three carbons between the groups on nitrogen. Methyl and ethyl give N methylethanamine, the secondary one. Three methyl groups give N, N dimethylmethanamine, the tertiary one. So there are four isomers in all. Okay? All four react with nitrous acid, so all four give off nitrogen gas, N two, and the answer is four, right? Be careful here, because the answer is not four. Each primary aliphatic amine, RNH two, forms an unstable diazonium salt that loses nitrogen at once. So with nitrous acid, HNO two, it gives the alcohol, ROH, with nitrogen and water. Got it? The other two do react, but they give no gas. The secondary amine gives a yellow oily N nitrosoamine. And the tertiary aliphatic amine forms only a soluble nitrite salt. Clear? So nitrogen comes only from the two primary amines, and the answer is two. Puzzle solved: all four react, but only two fizz with nitrogen. With me?

5:12Example 2: pH of 0.05 M ethanamine solution

And now Example two, also from Set three. Find the p H of a zero point zero five molar solution of ethanamine; capital M means molar, moles per litre. K b, the base dissociation constant, is five point one times ten to the power minus four; the raised minus four is the power of ten. Pause the video and try it first. Done? Okay, let's check. We need the hydroxide ion concentration, written O H minus in square brackets, which mean concentration. Ethanamine gives hydroxide ions, so the hydroxide concentration is simply zero point zero five molar, right? That works only for a strong base. Ethanamine is a weak base: it takes a proton from water, giving the ethylammonium ion and hydroxide, but only partly. The two half arrows mean an equilibrium, and the raised plus and minus are the charges. Okay? So for a weak base of concentration c, when the degree of ionisation is small, the hydroxide concentration is the square root of K b times c; the tick shaped sign is the square root. Here, the concentration c is zero point zero five molar. Right? Now put in the numbers: K b times c equals five point one times ten to the power minus four, times zero point zero five, which gives two point five five times ten to the power minus five. Its square root is five point zero five times ten to the power minus three molar. Got it? Next, p O H is minus log of the hydroxide concentration: minus log of five point zero five times ten to the power minus three gives two point three zero. But this is p O H, not the p H we want, so do not stop here. With me? At two hundred and ninety eight kelvin, with capital K standing for kelvin, p H equals fourteen minus p O H, so fourteen minus two point three zero gives eleven point seven. That is basic, as an amine solution should be. Clear? Now let us check the condition for this shortcut. Alpha, the Greek letter for the degree of ionisation, is the hydroxide concentration over c, with the slash for over; it is about zero point one, and the wavy equals sign means approximately. Solving the full quadratic gives a p H of eleven point six eight, which still rounds to eleven point seven. Okay?

7:37Quick revision: Hoffmann, pKb, HNO2 test, weak base pH

So, quick revision. Hoffmann gives one carbon fewer, and tertiary amines have no N H bond, so they cannot hydrogen bond with each other. The smaller the p K b, the stronger the base, and N alkylanilines beat aniline. With nitrous acid, only primary aliphatic amines give off nitrogen. And for a weak base, the hydroxide concentration is the square root of K b times c when the ionisation is small; then p H is fourteen minus the p O H value. Now try the rest of Set three on this page. All the best!