Acetanilide and p-Nitroacetanilide

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from the lab. You have aniline, and you want a nitro group on its ring, exactly opposite the nitrogen. The obvious move is to pour in nitric acid. But chemists do not do that! Instead, they first turn aniline into acetanilide, and only then do they nitrate it. So let us see both preparations, acetanilide and para nitroacetanilide; the small italic p in the title stands for para. Both are part of the JEE practical chemistry syllabus, so let's begin.

0:35The detour: through acetanilide

In this course figure, we need only the top row. On the left is aniline: the hexagon is a benzene ring, carrying an amino group, N H two. The arrows lead to acetanilide and then to para nitroacetanilide, with the reagents written on them. The reactions lower down belong to other lessons. So why not nitrate aniline directly? Here is the first reason. Nitric acid, HNO three, oxidises aniline to tarry products. The small three written low counts the oxygen atoms, okay? And the second reason. In the strong acid, we get the anilinium ion, which gives much meta product, not the para product we want. Right? So we acetylate first, and then nitrate, as the top row shows: the acetyl group goes on the nitrogen, and then the nitro group goes on the ring, opposite it. That position is para, as the label on the right of the top row says. With me? And at the very end, hot dilute acid or alkali removes the acetyl group, giving para nitroaniline, the compound from our puzzle. Clear?

1:40Acetanilide: acetylation of aniline

Now step one in detail. On the left, aniline plus acetic anhydride, where the brackets with a small two written low mean two acetyl groups, joined through an oxygen atom. After the arrow, we get acetanilide, with an acetyl group on the nitrogen, plus acetic acid. Okay? How does it happen? The nitrogen of aniline attacks the carbonyl carbon; the two little lines on screen are its double bond to oxygen. So it is a nucleophilic acyl substitution: aniline is the nucleophile, and acetate is the leaving group. Got it? Now, acetyl chloride can also acetylate aniline. So why does the lab use acetic anhydride, and not acetyl chloride? There are two reasons. Acetyl chloride reacts violently, and its H C l, hydrogen chloride, turns half the aniline into its hydrochloride, which does not react. The anhydride reacts smoothly, and as the equation above shows, its only by-product is acetic acid. Right?

2:44Acetanilide in the lab

Now the lab procedure. Aniline is warmed with acetic anhydride and glacial acetic acid, plus a pinch of zinc dust, which prevents oxidation of the aniline. With me? Next, the mixture is poured into ice-cold water. Why ice-cold water, do you think? Yes, exactly, it is because acetanilide is almost insoluble in cold water, so it separates as a solid, and we filter it off. The acetic acid stays dissolved, and so does any unused anhydride, once it is hydrolysed to acetic acid. Okay? Then we recrystallise it from hot water, because acetanilide is quite soluble in boiling water, but only slightly soluble in cold water. So most of it crystallises on cooling, and the soluble impurities stay behind. Clear? The result is white shining crystals, with a melting point of one hundred and fourteen degrees Celsius; the small raised circle on screen means degrees. Got it?

3:42p-Nitroacetanilide: nitration of acetanilide

Now step two, the nitration. Acetanilide plus nitric acid, with concentrated sulphuric acid, H two S O four, over the arrow; on screen, concentrated is shortened to its first four letters. In the product, the nitro group, O two N, sits at one end of the ring and the NHCOCH three group at the other; the short dashes are bonds. The ring is now C six H four, and water forms. Okay? In the lab, acetanilide is first dissolved in glacial acetic acid and concentrated sulphuric acid, and then cooled in ice. Right? Then the nitrating mixture, concentrated nitric acid plus concentrated sulphuric acid, is added drop by drop, below ten degrees Celsius. So why must it stay so cold? Very good if you thought of heat, because the reaction is strongly exothermic. Keeping it cold prevents dinitration, that is, a second nitro group going in. It also prevents hydrolysis of the amide, and stops more of the ortho isomer from forming. Clear?

4:48The electrophile, and why para

Now, the nitrating mixture has concentrated sulphuric acid in it. What is its role? What does it actually do? It protonates nitric acid and generates the electrophile; well done if you said that. Nitric acid plus two molecules of sulphuric acid gives N O two plus, the nitronium ion, plus H three O plus, and two hydrogen sulphate ions. The small raised signs are the charges. Got it? The nitronium ion is the electrophile. The group already on the ring, N H with its acetyl group, is ortho para directing: ortho is next to it, and para is opposite. The dash on screen is its bond to the ring. This group is bulky, so para is favoured. With me? The product is poured on crushed ice, and it holds a little of the minor ortho isomer, which is yellow. When we recrystallise from ethanol, the ortho isomer, being more soluble, stays in solution. Okay? And pure para nitroacetanilide forms pale yellow to colourless crystals, with a melting point of two hundred and fourteen degrees Celsius. Right?

5:53Example 1: acetanilide percentage yield from 4.65 g aniline

Time for Example one. Four point six five grams of aniline gives five point four grams of pure acetanilide. Find the percentage yield. The hint gives the molar masses, capital M: ninety three and one hundred and thirty five. Pause the video and try it first. Done? Okay, let's check. Small n, the moles of aniline, is four point six five over ninety three, which gives zero point zero five zero; m o l means moles. Now, some students simply take five point four over four point six five. Why is that wrong? Because those are grams of two different compounds; that ratio even comes out above a hundred percent. Compare moles instead. The equation is one to one, so at most zero point zero five zero moles of acetanilide can form. Times one hundred and thirty five, that is six point seven five grams, the theoretical yield. Clear? So the percentage yield is five point four over six point seven five, times one hundred, which gives eighty percent. Actual yield over theoretical yield, every time. Got it?

7:02Example 2: volume of acetic anhydride for 9.3 g of aniline

Now a harder one. What volume of acetic anhydride, density one point zero eight grams per millilitre, molar mass one hundred and two, is just enough to acetylate nine point three grams of aniline? The raised minus one in the unit means per. Pause the video and try it first. Done? Okay, let's check. Moles of aniline: nine point three over ninety three gives zero point one zero zero. Now, the anhydride carries two acetyl groups. So half as much, zero point zero five zero moles, is enough, right? That is not right, and this is the trap here. Look back at the equation: one anhydride acetylates one aniline, and the other acetyl group leaves as acetic acid. So we need zero point one zero zero moles, which times one hundred and two is ten point two grams. Okay? Last, from mass to volume. Capital V, the volume, is mass over density: ten point two over one point zero eight, which gives nine point four millilitres, m L on screen. The trap would give only half of that. With me?

8:09Quick revision: acetylation, nitration, para product, yields

So, quick revision. For acetanilide, acetylate aniline, pour into ice-cold water, and recrystallise from hot water. For para nitroacetanilide, add the nitrating mixture below ten degrees Celsius; the electrophile is the nitronium ion. Para is favoured, and the yellow ortho isomer stays in ethanol. For yields, compare moles, one anhydride per aniline. Now try the practice problems on this page. All the best!