IIT-JEE Questions (1990–1999): Diazonium Salts

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from IITJEE, nineteen ninety. You start with four nitroaniline: an amino group and a nitro group, para to each other on a benzene ring. And you must end with one, two, three tribromobenzene, three bromine atoms side by side, and nothing else on the ring. Both starting groups have vanished. So where did they go? The answer is the diazonium salt, which lets us swap an amino group for another group, or remove it completely. Today we work through the older IITJEE questions on diazonium salts, from nineteen ninety to ninety nine, and we will crack this puzzle at the end. So let's begin.

0:39The toolkit: make the salt, then swap the group

So, first the toolkit, and the course's map on the right shows all of it. On screen, A r N H two is an amino group on an aryl ring; A r means aryl, and the small two written low counts the hydrogens. Sodium nitrite and hydrochloric acid sit over the arrow, the slash meaning with, and the product is A r N two plus, C l minus, the diazonium salt, where the raised signs are the charges. Okay? This is diazotisation, and it needs a primary aromatic amine and the cold: two seventy three to two seventy eight kelvin, where capital K is kelvin. The salt is stable only in cold solution, so we use it at once. Right? Now the swaps, where nitrogen gas leaves the ring. Copper bromide, C u B r, gives the aryl bromide, and copper cyanide, C u C N, gives A r C N, the nitrile. Both are Sandmeyer reactions. Clear? And two swaps without copper. Warm water gives the phenol, A r O H, and hypophosphorous acid, H three P O two, replaces the group with plain hydrogen. Keep that last one in mind, because both of our hardest questions today need it. Got it?

1:49Coupling: the nitrogen stays

Now the second kind of reaction, coupling, in the figure on the right. Here the nitrogen stays. The diazonium ion is a weak electrophile, so it attacks only strongly activated rings, like phenol and aniline, usually at the para position. In the figure, the two rings end up joined through N double bond N, two nitrogens in the middle. With me? Here is IITJEE nineteen ninety eight. Benzenediazonium chloride, C six H five N two plus, C l minus, reacts with phenol in a weakly basic medium. What is formed? It is four hydroxyazobenzene, yes, exactly! The new N double bond N link goes para to the O H, at the four position, and you get the orange dye at the top of the figure, with O H minus, the base, over the arrow. The figure writes it as p hydroxyazobenzene, where the small p means para, the same thing as four. Okay? Now IITJEE nineteen ninety five. Two, four dinitroaniline is diazotised with sodium nitrite and H C l at five degrees Celsius, where the small circle means degrees, so it is still cold. Then anisole, which is methoxybenzene, is added. Right? But the course tells us that plain benzenediazonium chloride does not couple with anisole. So why does this diazonium ion manage it? Very good, it is because of the two nitro groups. They pull electrons away and make the diazonium ion a much stronger electrophile, strong enough to attack anisole. Clear? And it couples para to the methoxy group, as usual. The product is four methoxy two prime, four prime dinitroazobenzene; the little marks, called primes, put both nitro groups on the other ring. Got it?

3:34Topper's trick: use the amino group, then remove it

Now the topper's trick, from IITJEE nineteen ninety nine. Convert acetanilide into three, five diiodonitrobenzene in four steps. That means two iodine atoms, both meta to a nitro group. Okay? Step one is nitric acid with sulphuric acid. The acetamido group of acetanilide sends the nitro group para to itself. Right? Step two: heating with acid, H three O plus, hydrolyses the amide and frees the amine, giving four nitroaniline. With me? Now we iodinate. The amino group and the nitro group are para to each other. So where will the two iodine atoms go? Yes, exactly, they go ortho to the amino group, on both sides. The amino group directs ortho and para, and its para position is taken. Iodine monochloride, I C l, puts in the two iodines. Clear? Step four: diazotise at two seventy three kelvin, then H three P O two replaces the diazonium group with hydrogen. So the amino group put the iodines in place, and then it was taken away. That is the trick: use an amino group to direct, then remove it. Got it?

4:41Example 1: nitrobenzene to benzamide via benzonitrile

Time for Example one, from IITJEE nineteen ninety four. How is benzamide obtained from nitrobenzene? Pause the video and try it first. Done? Okay, let's check. First, tin and hydrochloric acid, S n and H C l, reduce the nitro group, giving aniline, C six H five N H two. Okay? Next, we diazotise with sodium nitrite and H C l in the cold, at two seventy three kelvin, to get benzenediazonium chloride. Right? Now, which group should replace the nitrogen, so that it can later become an amide? Yes, exactly, we bring in the cyano group. Copper cyanide, a Sandmeyer reaction, gives C six H five C N, benzonitrile. Clear? Now the last step. We just boil the nitrile with acid, right? No, no, careful, because that is exactly the trap in this question. Full hydrolysis goes all the way to the acid, benzoic acid. We need only partial hydrolysis. With me? So we use alkaline hydrogen peroxide, H two O two, or cold concentrated sulphuric acid, H two S O four, where conc. is short for concentrated. That stops at benzamide, C six H five CONH two. Got it?

5:59Example 2: 4-nitroaniline to 1,2,3-tribromobenzene

And now, our puzzle from the start, from IITJEE nineteen ninety. How is four nitroaniline converted into one, two, three tribromobenzene? Use the trick we just learnt. Pause the video and try it first. Done? Okay, let's check. First, bromine, B r two. The amino group directs ortho and para, and para is blocked by the nitro group, so two bromines go on both sides of the amino group. That gives two, six dibromo four nitroaniline. Okay? Now we need a third bromine, exactly in the middle, between these two. Which group can become that bromine? Yes, exactly, it is the amino group itself. Diazotise it, and treat it with copper bromide, a Sandmeyer reaction. Counting from the nitro group, we now have three, four, five tribromonitrobenzene. Right? So now we have three bromines in a row. But the nitro group is still there. How do we get rid of it? We reduce it first, with tin and hydrochloric acid, so the nitro group becomes an amino group. And an amino group, we already know how to remove. Clear? So we diazotise once more, and H three P O two puts hydrogen in its place, giving one, two, three tribromobenzene. And that solves our puzzle: the first amino group became the middle bromine, and the nitro group became an amino group and was removed. Got it?

7:23Quick revision: Sandmeyer, coupling, NH2 removal by H3PO2

So, quick revision. Diazotise the primary aromatic amine in the cold, then swap the group with copper bromide, copper cyanide, water or hypophosphorous acid. Coupling keeps the nitrogen and goes para to the O H, or to the methoxy group, OCH three, and nitro groups on the diazonium ion help it react. And the trick: use an amino group to direct, then remove it with H three P O two. Now try the other past questions on this page yourself. All the best!