Set 2

Transcript

0:00Introduction

Hello everyone! Let us begin with a puzzle from Set two. The N H two group of aniline sends new groups to the ortho and para positions. So when we nitrate aniline, we expect almost no meta product. Yet forty seven percent of the product comes out meta, along with a lot of tar! How can that happen? It sounds strange, doesn't it? Today you will see why, and how to make clean para nitroaniline. Problems like these come up often in JEE Main, so let's begin.

0:33Amines by reducing nitro compounds

First, a quick recap. To make an amine, we can reduce a nitro compound. RNO two goes to RNH two: R is the carbon group, the small two written low counts the atoms, and the arrow means gives. We use tin, iron or zinc with hydrochloric acid, written S n slash H C l and so on; the slash means together with. Okay? Or we use hydrogen gas, H two, over nickel, palladium or platinum, the metals N i, P d and P t. That is catalytic hydrogenation. Right? And remember this number: six hydrogen atoms are used per nitro group. With me? Now look at the figure: nitrobenzene, C six H five N O two, sits on the left. Does every reducing agent turn it into aniline? No, it does not, because the medium decides the product. With a metal and acid, such as tin or iron with H C l, or with hydrogen over palladium, we get aniline, the top box. But in a neutral medium, zinc with ammonium chloride, N H four C l, we get N phenylhydroxylamine; the italic capital N in front means the phenyl group sits on nitrogen. Clear? In an alkaline medium, zinc with sodium hydroxide, N a O H, gives hydrazobenzene, where two N H groups are joined by a single bond. And lithium aluminium hydride, L i A l H four, gives azobenzene, with a double bond between the nitrogens. So, for an amine, use a metal with acid, or catalytic hydrogenation. Got it?

2:10Nitrating aniline: why acetylate first

Now, the ring of aniline. In the figure, the two dots on nitrogen are its lone pair, and the double-headed arrows link the resonance structures shown. Nitrogen gets a plus sign, and the minus signs, the negative charges, land on the ortho and para positions. So the N H two group strongly activates the ring and directs ortho and para. Okay? So let us nitrate aniline directly, with nitric acid and sulphuric acid. We should get only ortho and para products, right? Sadly it is not that simple, because two things go wrong. First, direct nitration gives tarry oxidation products, a messy tar. With me? Second, the products come out as para fifty one percent, meta forty seven percent, and ortho just two percent. That is our puzzle from the start, forty seven percent meta! Right? So where does all that meta product come from? Here is a hint: aniline is a base, and the nitrating mixture is strongly acidic. Yes, exactly, aniline is largely protonated. It becomes the anilinium ion, C six H five N H three plus; the small plus written high is its positive charge. And the anilinium ion is meta-directing, so a big share of the product goes meta. Clear? So here is the topper's way: acetylate first. Acetylation turns the N H two into NHCOCH three, and aniline becomes acetanilide. Capital M is the molar mass, and each such change raises it by forty two. Got it? Now, here is why acetylation helps. In acetanilide, the lone pair of nitrogen is shared with the C O group; the equals sign is the double bond between carbon and oxygen. Right? So the ring is only moderately activated, and the reaction stays under control. Okay?

4:03Example 1: mass of ethanamine from 7.5 g of nitroethane

Now, Example one. Find the mass of ethanamine obtained by reducing seven point five grams of nitroethane completely; the small g is grams. Pause the video and try it first. Done? Okay, let's check. Nitroethane, C H three C H two N O two, is reduced to ethanamine, C H three C H two N H two. Each molecule keeps its one nitrogen atom, so one mole gives one mole; m o l is short for mole. Right? Now the molar mass, capital M, of nitroethane, C two H five N O two. Two carbons give twenty four, five hydrogens give five, one nitrogen gives fourteen, and two oxygens give thirty two. That adds up to seventy five grams per mole; the minus one written high on mol means per mole. With me? So the number of moles, small n, is seven point five over seventy five, which gives zero point one mole. Okay? Now, a trap. Reduction adds hydrogen, six atoms per nitro group. So the amine must weigh more than seven point five grams, right? No, it is lighter, and here is why. Ethanamine is C two H five N H two: twenty four, plus seven for the hydrogens, plus fourteen, gives forty five grams per mole. Compare the formulas: the two oxygens, worth thirty two, are gone, and only two hydrogens come onto nitrogen. Clear? So the mass is zero point one times forty five, which gives four point five grams of ethanamine. It is simple once you count moles. Got it?

5:42Example 2: aniline to 4-nitroaniline via acetanilide

Now a harder one, Example two. Aniline is treated in turn with ethanoyl chloride, then with nitric acid and sulphuric acid at two eighty eight kelvin, and then with hot dilute acid. Give the name and molar mass of the main product. HNO three slash H two S O four is the nitrating mixture, and capital K is kelvin. Pause the video and try it first. Done? Okay, let's check. In step one, ethanoyl chloride acetylates the amino group, so aniline becomes acetanilide, C six H five NHCOCH three. We do this to tame the ring, exactly as we just saw. Right? Now step two, the nitration. Where does the nitro group go this time: ortho, meta or para? It goes mainly to the para position, yes. The ring is only moderately activated now, so the reaction is under control, and we get para nitroacetanilide; the italic p stands for para. Okay? So the answer is para nitroacetanilide, right? That is the trap, because the question has a third step. Hot dilute acid hydrolyses the amide, so the NHCOCH three group goes back to N H two. The acetyl group was only a guard: on in step one, off in step three. With me? So the main product is four nitroaniline, which is para nitroaniline. On screen, O two N, the nitro group, and N H two sit at opposite ends of the ring, C six H four, and the dashes are the bonds. And that is the clean route to para nitroaniline we promised at the start. Clear? Now the molar mass of C six H six N two O two. Six carbons give seventy two, six hydrogens give six, two nitrogens give twenty eight, and two oxygens give thirty two. The total is one hundred and thirty eight grams per mole. Got it? And here is a quick check. Aniline is ninety three; swap one hydrogen for a nitro group of forty six. Ninety three minus one plus forty six gives one hundred and thirty eight. The forty two from acetylation was undone in step three, so it does not count.

8:01Quick revision: nitro reduction, anilinium ion, acetylation

So, quick revision. Reduce a nitro compound with a metal and acid, or with hydrogen over nickel, and one mole gives one mole of amine. Direct nitration of aniline gives tar, and forty seven percent meta, through the anilinium ion. So acetylate, nitrate and hydrolyse to get four nitroaniline, one hundred and thirty eight grams per mole. Now try all fifteen problems of Set two in twenty five minutes, then check the solutions. All the best!