Coupling Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Benzene is colourless, and pure aniline is colourless too. But join two benzene rings through two nitrogen atoms, and suddenly you get a bright orange or yellow compound. And the methyl orange indicator you use in titrations is made in exactly this way. So where does all that colour come from? Today we will answer that with coupling reactions of diazonium salts, and questions on them come up often in JEE, okay? So, let us begin with the reaction itself.

0:35Coupling: the nitrogen stays

So, here is our diazonium ion, written A r N two plus. Here, A r stands for an aryl group like phenyl, the small two is the number of nitrogen atoms, and the raised plus is the charge. It is an electrophile, so it wants electrons, but it is a weak one. In the figure on the right, you see it as benzenediazonium chloride, with the chloride ion, C l minus, beside it, reacting with phenol on top and with aniline below. Okay? Now let me ask you this. This weak electrophile meets a plain benzene ring. Will it react? No, it will not, and that is the whole point. It is too weak for an ordinary ring, so it attacks only strongly activated rings, rings rich in electrons. Clear? And the strongly activated rings are phenols, and arylamines like aniline. The attack happens usually at the para position, opposite the O H or the N H two group; the italic word on screen is para. Right? The product is an azo compound: A r, N double bond N, A r prime. The dashes are bonds to the two rings, and the prime mark means a second aryl group, which can be different. And notice, both nitrogen atoms stay in the product, so the nitrogen is retained. Got it?

1:53Phenol or aniline: the pH decides

Now the conditions, and questions often test these. Phenol couples in weakly alkaline solution, p H nine to ten; in the figure, the O H minus above the arrow is the alkali, and the two seventy three to two seventy eight kelvin below it tells you the reaction is kept cold. The product is para hydroxyazobenzene, where the italic p means para, and it is orange. Okay? The reason for the alkali is simple: the base removes the proton from phenol and gives the phenoxide ion, C six H five O minus. The phenoxide ion is much more reactive than phenol itself, so it is the real reactive species. Clear? Now let us look at aniline. It couples in weakly acidic solution, p H four to five, shown by the H plus above the arrow. The product is para aminoazobenzene, and this one is yellow. Okay? So if a little acid helps with aniline, a strongly acidic solution should be even better, right? That is the trap, so be careful here. In strong acid, aniline becomes the anilinium ion, C six H five N H three plus, and its N H three plus group deactivates the ring. So enough of the aniline must stay as the free amine, and the solution must not be too acidic. With me? And alkali is no good for aniline either, because alkali converts the diazonium ion itself into diazotate, which is unreactive. So p H four to five is acidic enough to protect the diazonium ion, but not so acidic that most of the aniline is protonated. Got it?

3:29Para blocked, and the azo dye test

Now, what if the para position is already taken? Look at four methylphenol: the methyl group sits at the para position, opposite the O H group. So where can the diazonium ion attack now? Yes, exactly, it goes ortho, the position right next to the O H group. So the rule is simple: para first, and if para is occupied, then ortho. Okay? Now the azo dye test. We diazotise the amine at two seventy three to two seventy eight kelvin, and then add an alkaline solution of two naphthol. Clear? So tell me, which amines will give a dye in this test? Only primary aromatic amines, like aniline, because only they give an aryl diazonium salt stable enough in the cold to couple. It couples with two naphthol at C one, the carbon next to the one carrying the O H group, and gives an orange red dye. So this is the test for primary aromatic amines. With me?

4:29Why azo compounds are coloured

Now, back to our puzzle about colour. Look at the N double bond N link in the figure; the dashes are the bonds to the two rings. Benzene alone is colourless, so why does this link give colour? Because the two rings and the N double bond N link form one long, extended conjugated system. That system absorbs light in the visible region, so we see colour: orange, yellow, or orange red. Clear? Now, the methyl orange from our puzzle. Its first part is sulphanilic acid, which is diazotised to give a diazonium salt. Right? That salt is then coupled with N N dimethylaniline; the two italic capital letters N tell you that both methyl groups sit on the nitrogen. So the indicator in your titration flask is simply an azo dye. Got it?

5:19Example 1: which compounds couple with a diazonium salt

Now, Example one, a quick check of the rule. Which of these couple with benzenediazonium chloride: phenol, toluene, N N dimethylaniline, nitrobenzene, and two naphthol? Pause the video and try it first. Done? Okay, let's check. Start with toluene. It has a methyl group, which activates the ring, so toluene couples too, right? That is the trap, so be careful. The methyl group, C H three, does activate the ring, but not strongly enough for our weak electrophile. So toluene does not couple. Okay? Nitrobenzene is even easier. The nitro group, N O two, deactivates the ring strongly, so it certainly does not react with this weak electrophile. Right? Now N N dimethylaniline. It is a tertiary amine, and tertiary amines do not give the azo dye test. So it cannot couple, can it? It does couple, and many students lose marks here. It is not the amine being diazotised; it is the partner ring, and its ring is strongly activated, just as in methyl orange. Phenol and two naphthol are strongly activated too. So the answer is phenol, N N dimethylaniline and two naphthol. Clear?

6:38Example 2: why 4-nitrobenzenediazonium couples with anisole

And now Example two, a harder one. Four nitrobenzenediazonium chloride couples even with anisole, but plain benzenediazonium chloride does not. Can you explain why? Pause the video and try it first. Done? Okay, let's check. We just said that a nitro group deactivates a ring. So the nitro compound should react less, not more, right? Now look carefully at where the nitro group is. It sits on the diazonium ring, the electrophile, and not on anisole, the ring being attacked. That changes everything. With me? The nitro group withdraws electrons, so it pulls electron density away from the diazonium group, N two plus. That makes the diazonium group even hungrier for electrons. Okay? So the four nitrobenzenediazonium ion is a stronger electrophile. It is strong enough to attack anisole, which the plain benzenediazonium ion cannot do. Right? And here is the general rule: electron withdrawing groups on the diazonium ring make it a stronger electrophile. So a nitro group slows down the ring that is attacked, but it helps the ion that attacks. Clear?

7:51Quick revision: para attack, pH 9-10 phenol, pH 4-5 aniline

So, quick revision. The diazonium ion is a weak electrophile: it attacks only strongly activated rings, usually at para, and the nitrogen is kept. Phenol couples at p H nine to ten and gives orange; aniline couples at p H four to five and gives yellow; and alkaline two naphthol gives the orange red dye. And electron withdrawing groups on the diazonium ring make it a stronger electrophile. Now go and try the practice problems on this page. You will find them easy. All the best!