Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Suppose you need iodobenzene, or fluorobenzene. Just halogenate benzene directly, right? But these two cannot be made conveniently by direct halogenation of benzene. Today we draw the complete map of one single salt, and questions on it come up often in JEE, okay? So tell me, how do chemists put iodine or fluorine on a benzene ring?

0:27Making the salt: diazotisation

So, the story starts on the left of the map. Aniline, C six H five N H two, reacts with sodium nitrite, N a N O two, and two H C l, hydrochloric acid. The small numbers written low are counts of atoms, okay? Now look at the arrow, which means gives, with the condition above it: two seventy three to two seventy eight kelvin, where capital K is kelvin. The product is benzenediazonium chloride, C six H five N two plus, C l minus; the raised signs are the charges on the two ions. Sodium chloride and water also form. This is diazotisation. Right? So why must we work this cold, at two seventy three to two seventy eight kelvin? What would happen if the solution warmed up? Yes, exactly, the salt is stable only in cold solution. Warm it, and it decomposes and loses nitrogen gas. So it is never stored; we make it and use it at once. Clear? Now one more question for you. Will any primary amine do the same, say ethanamine, an aliphatic amine? No, it will not, and this is a common trap. We need a primary aromatic amine, A r N H two, where A r stands for an aryl group, like phenyl. The ring stabilises the diazonium ion by resonance; an alkanediazonium ion has no such help and loses nitrogen at once. Got it?

1:53Replacement with copper: Sandmeyer and Gattermann

Now the right side of the map. The top eight arrows are replacement reactions: nitrogen gas, N two, leaves the ring, and a new group takes its place, okay? First come the copper salts. C u C l with hydrochloric acid gives the aryl chloride, A r C l; the slash just means with. Similarly, C u B r with H B r gives the aryl bromide. This is the Sandmeyer reaction. With me? Next comes the cyanide. C u C N with KCN gives the aryl nitrile, A r C N, and this is also a Sandmeyer reaction. Right? Now suppose we use copper powder with hydrochloric acid. We still get the aryl chloride. So tell me, is this also called the Sandmeyer reaction? Check the map before you answer. Careful now, because it has a different name. Copper powder with H C l or H B r is the Gattermann reaction: same products, different reagent. Students often mix these up, so remember: copper salt, Sandmeyer; copper powder, Gattermann. Clear?

3:02Replacement without copper

Now the reactions without copper. For iodine, simply warm the salt with aqueous potassium iodide, K I, and we get the aryl iodide. From our salt, that is iodobenzene, the first half of our puzzle. Got it? Next comes fluorine, the second half of our puzzle. Look at the map and tell me, which reagent brings in fluorine? Yes, exactly, it is fluoroboric acid, HBF four. It gives the diazonium fluoroborate, insoluble in water and stable at room temperature, so we filter it off and dry it. Heating the solid gives A r F, the aryl fluoride, and this is the Balz Schiemann reaction. Clear? Now let us warm the salt with water, H two O, from about two eighty three kelvin upward, as the map shows, and we get A r O H, phenol. That is exactly why we kept the salt cold. Right? Now here is a tricky one. Warm the salt with ethanol instead. Find ethanol on the map and tell me, what takes the place of the nitrogen on the ring? Yes, it is simply hydrogen. Hypophosphorous acid, H three P O two, or ethanol, gives A r H; from our salt, that is benzene itself. With me? The last replacement on the map gives a nitro group. Make the fluoroborate with HBF four, then heat it with sodium nitrite and copper to get A r N O two. So, eight replacement arrows, and in every one, nitrogen gas is lost. Okay?

4:35Coupling: nitrogen is kept

Now the last two arrows of our map, shown here in detail, where nitrogen is kept. The diazonium ion is a weak electrophile, so it attacks only strongly activated rings, like phenols and arylamines, usually at the para position. This is coupling. Right? With phenol, in weakly alkaline solution, p H nine to ten, we get para hydroxyazobenzene, which is orange; the italic p means para. In the figure, the O H minus above the arrow is the alkali, the same cold range, two seventy three to two seventy eight kelvin, is written below it, and the two rings are joined through two nitrogens with a double bond between them. Clear? Now replace phenol by aniline, in the second row of the figure. What product do you get, and what colour is it? Yes, exactly, you get para aminoazobenzene, and this one is yellow. But now the solution is weakly acidic, p H four to five, shown by the H plus above the arrow. Got it? And here is the reason for the colour. The N double bond N link, with dashes as bonds to the rings, joins both rings into one extended conjugated system. That is why azo compounds are coloured. With me?

5:48Example 1: products with CuBr, KI, warm water and H3PO2

Now Example one, an easy one from the section on replacement reactions in this chapter. Name the product of benzenediazonium chloride with, first, C u B r with H B r; second, potassium iodide; third, warm water; and fourth, hypophosphorous acid. Pause the video and try it first. Done? Okay, let's check. C u B r with H B r is the Sandmeyer reaction, so we get bromobenzene. Potassium iodide needs no copper, and gives iodobenzene, the compound from our puzzle. Right? Warm water gives phenol, which is why the salt is kept cold. And hypophosphorous acid replaces the nitrogen by hydrogen, so we get benzene back. Four reagents, four products, and nitrogen gas lost every time.

6:42Example 2: N2 volume at STP and organic product on warming

Clear? And now Example two, a little harder, from the section on diazotisation in this chapter. Read it with me: a solution containing zero point zero one mole of benzenediazonium chloride is warmed until decomposition is complete. What volume of gas at STP is evolved, and what organic product is left? Pause the video and try it first. Done? Okay, let us find the organic product first. The salt is a chloride, so warming it should give chlorobenzene, right? No, it does not give chlorobenzene, and that is exactly the trap of this question. Chlorobenzene needs copper, as in the Sandmeyer reaction. Here the salt is simply warmed in water, so we get phenol, C six H five O H, along with nitrogen gas and hydrogen chloride. Okay? Now let us find the gas. Look at the equation, and you see that one mole of the salt gives one mole of nitrogen gas, N two. So zero point zero one mole of salt gives zero point zero one mole of nitrogen. Got it? So what volume does zero point zero one mole of nitrogen occupy at STP? One mole of gas at STP occupies twenty two point four litres, or twenty two thousand four hundred millilitres. So the volume, capital V, is zero point zero one times twenty two thousand four hundred, which gives two hundred and twenty four millilitres of nitrogen, and phenol is left behind. Okay?

8:15Quick revision: diazotisation, replacement and coupling

So, quick revision. Diazotisation needs a primary aromatic amine, sodium nitrite and hydrochloric acid, at two seventy three to two seventy eight kelvin, and the salt is used at once. Replacement reactions lose nitrogen: Sandmeyer, Gattermann, potassium iodide, fluoroboric acid, water and hypophosphorous acid. Coupling keeps nitrogen and gives coloured azo compounds, orange with phenol and yellow with aniline. Now try the practice problems on this page. All the best!