Ring Substitution in Aniline

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Usually, a benzene ring needs a catalyst to take even one bromine atom. But add bromine water to aniline at room temperature, with no catalyst, and a white solid comes out at once, with three bromine atoms on the ring! And the amino group directs ortho and para, yet nitrating aniline directly gives almost half meta product. Strange, isn't it? One idea explains both, and it comes up often in JEE questions. So, let's begin.

0:33Why the amino group activates the ring

Look at this figure from Chapter three. Structure one is the usual structure of aniline: the two dots on the nitrogen are its lone pair, and in N H two, the small two written low means two hydrogens. In structures two, three and four, that lone pair has become a double bond between the nitrogen and the ring, leaving a plus charge on the nitrogen and pushing a minus charge, the small red dash, onto a ring carbon. The double headed arrows join these resonance structures, numbered in Roman numerals. Okay? Now look at the labels under the structures, and tell me. Which ring carbons get that minus charge? Yes, exactly, they are ortho and para. Structures two and four put it on the ortho carbons, next to the nitrogen, and structure three on the para carbon. So these positions become rich in electrons, as the note under the figure says. With me? So an electrophile, hungry for electrons, goes straight there, and fast. The N H two group is very strongly activating, and ortho, para directing. That is the key idea for this whole lesson.

1:41Bromination: it does not stop at one

Now, bromination. We add bromine water to aniline at room temperature; the plus sign means we mix them. No catalyst is needed; the ring is that active. Clear? So tell me, those three bromine atoms from our puzzle, where on the ring do they go? Correct, one at each active position: both ortho carbons and the para carbon. You get a white precipitate of two, four, six tribromoaniline; these numbers are the ring positions of the bromines, counting from the carbon with the nitrogen. So the first puzzle is solved. Got it?

2:19Controlling it: acetylate first

But what if we want only one bromine, at the para position? Then we calm the ring by acetylating the amino group first, with ethanoic anhydride, turning aniline into acetanilide; the arrow means gives. Okay? Now think about this. Acetanilide still has a nitrogen with a lone pair, right next to the ring. So why should it be any less reactive? Because that lone pair now has a second partner, the carbonyl group, the carbon oxygen double bond written C equals O on screen. So less of it reaches the ring, and the ring is only moderately activated. Very good if you saw that. Now brominate with B r two, bromine, in ethanoic acid, and we get mainly para bromoacetanilide; the italic p in front stands for para. Right? Finally, hydrolyse with acid or alkali to remove the acetyl group, and out comes para bromoaniline. Protect, react, then remove the protection.

3:20Nitration: the meta surprise

Clear? Now, nitration. Done directly, it gives tarry oxidation products. But the bigger surprise is this. N H two directs ortho and para, so how much meta product do you expect? Almost none, right? That is the trap, because the actual mixture is fifty one percent para, forty seven percent meta, and only two percent ortho. Almost half of it is meta! That is our second puzzle. Okay? Here is the reason. Nitration uses a strongly acidic mixture, and aniline is a base, so most of it takes a proton and becomes the anilinium ion, C six H five N H three plus; the small raised plus is its charge. Got it? Now the nitrogen has no lone pair to give, and its plus charge pulls electrons out of the ring. So the N H three plus group is deactivating and meta directing, hence the meta product. Right? So, for para nitroaniline in good yield, use the same trick. Acetylate, nitrate at two hundred and eighty eight kelvin, the capital K on screen, then hydrolyse.

4:27Sulphonation, and no Friedel–Crafts

Clear? Next, sulphonation. With concentrated sulphuric acid, H two S O four, aniline first forms a salt, anilinium hydrogensulphate; conc on screen is short for concentrated. Okay? Heat this salt at four hundred and fifty three to four hundred and seventy three kelvin, the dash meaning to, and it rearranges to para aminobenzenesulphonic acid, which we call sulphanilic acid. With me? Sulphanilic acid melts very high and is insoluble in organic solvents, because it exists as a zwitterion, a dipolar ion: plus on the H three N end, minus on the S O three end, with the C six H four ring between; the dashes are bonds. So it is held in an ionic lattice, like a salt. Right? And now the last reaction, Friedel Crafts. It needs aluminium chloride, A l C l three, a Lewis acid. But aniline is a Lewis base. So what happens when they meet? They form a salt, exactly. Aniline gives its lone pair to aluminium chloride, so the nitrogen, the capital N on screen, becomes positive and deactivates the ring. So, aniline gives no Friedel Crafts reaction. Got it?

5:43Example 1: mass of tribromoaniline from 0.93 g aniline

Now, example one. Find the mass of two, four, six tribromoaniline formed from zero point nine three grams of aniline, the small g, and excess bromine water. The B r equals eighty in brackets is the atomic mass of bromine. Pause the video and try it first. Done? Okay, let's check. Aniline has a molar mass of ninety three, the capital M on screen. So zero point nine three over ninety three gives zero point zero one mole; mol is short for mole. One aniline gives one tribromoaniline, so that is zero point zero one mole of product. Okay? Now the molar mass of the product. Aniline is ninety three, and three bromines add two hundred and forty, so three hundred and thirty three, right? Careful, that is a trap, because each bromine replaces a ring hydrogen, so three hydrogens leave. The formula is C six H two B r three N H two. Carbon gives seventy two, hydrogen four, nitrogen fourteen, and bromine two hundred and forty, a total of three hundred and thirty grams per mole; the raised minus one means per mole. With me? So the mass is zero point zero one times three hundred and thirty, which gives three point three zero grams. Very good if you avoided the trap!

7:06Example 2: order of reactivity towards bromination

Now a harder one, using every idea from today. Arrange these in order of decreasing reactivity towards bromination: aniline, acetanilide, benzene and anilinium chloride. Pause the video and try it first. Done? Let's go. Aniline is the fastest, with its lone pair free. Acetanilide is next, with its lone pair shared with the C equals O group. And benzene, with no lone pair to give, is slower still. The greater than sign means more reactive than. Okay? Now the tricky one. Anilinium chloride still has a nitrogen on the ring, so surely it beats plain benzene, right? It does not, and that is the trap. Its nitrogen is N H three plus, with no lone pair, and that plus charge pulls electrons out of the ring. So it is deactivated, the slowest of all. Clear? So the final order is aniline, then acetanilide, then benzene, then anilinium chloride. The same lone pair explains the whole order. Got it?

8:12Quick revision: lone pair, acetylation, anilinium ion

So, quick revision. First, the free lone pair makes aniline very fast and ortho, para directing, so bromine water gives the tribromo product at once. Second, acetylate first to control it, for para bromo and para nitro anilines. And third, when the nitrogen turns positive, in strong acid or with aluminium chloride, you get much meta product, and no Friedel Crafts. Now go and try the practice problems on this page. All the best!