Reactions with Metals; Ring and Side-Chain Halogen

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. I have two bottles. Each holds a compound with the formula C seven H seven C l, and each has a benzene ring and one chlorine atom. I treat both with alcoholic silver nitrate. One bottle gives a white precipitate, and the other gives nothing at all. Same atoms, same formula, so why such a difference? Hold that thought, because by the end of this video you will solve it in ten seconds. Today we also see how sodium, copper and magnesium react with aryl halides. Questions like these do come up in JEE, so let's begin.

0:41Sodium joins halides: Wurtz–Fittig and Fittig

So, first, sodium. On screen, A r stands for an aryl group, a benzene ring such as phenyl, R for an alkyl group, and X for a halogen, with each small dash a bond. In the Wurtz Fittig reaction, an aryl halide, A r X, and an alkyl halide, R X, react with two sodium atoms, N a, in dry ether, the condition written on the arrow. We get A r joined to R, an alkylarene, plus two N a X, the sodium halide. Okay? Now let me ask you this. What if we take only the aryl halide with sodium, and no alkyl halide at all? Yes, the aryl groups join each other! This is the Fittig reaction: two A r X plus two sodium, in dry ether, give A r joined to A r, a biaryl, with two rings joined directly. Got it?

1:32Copper and magnesium: Ullmann and Grignard

Now copper, in the Ullmann reaction. Two aryl iodides, A r I, where I is iodine, are heated with two copper atoms, C u, and heat is the condition on the arrow. Again A r joins A r to give a biaryl, and the copper takes the iodine as C u I, cuprous iodide. With me? Next, magnesium makes a Grignard reagent. Bromobenzene, C six H five B r, with magnesium, M g, in dry ether, gives C six H five M g B r, phenylmagnesium bromide. The small numbers written low count the atoms, right? Now tell me this. If I take chlorobenzene instead, will the same dry ether do the job? No, it will not, and this is a common trap. Chlorobenzene needs THF, short for tetrahydrofuran, a solvent with a higher boiling point. So remember the pair: bromobenzene in dry ether, chlorobenzene in THF, okay?

2:26Ring halogen versus side-chain halogen

Alright, now the second big idea. In chlorobenzene, C six H five C l, the chlorine sits directly on the benzene ring. Clear? Now benzyl chloride, C six H five, C H two, C l. One extra carbon, the C H two, sits between the ring and the chlorine. So this chlorine is on an s p three carbon, written s p with a small raised three, just like in an ordinary alkyl halide. Right? So which of these two will behave like an alkyl halide? Benzyl chloride, of course, and a very reactive one! Chlorobenzene does not react under the same conditions. Just one carbon, and the whole behaviour changes. Got it?

3:07Telling them apart: the silver nitrate test

So how do we tell them apart in the lab? Boil each with aqueous sodium hydroxide, N a O H, then acidify with dilute nitric acid, HNO three, and add silver nitrate, A g N O three. Which one gives a white precipitate? Benzyl chloride, yes! Its chlorine comes off during the boiling, so we get a white precipitate of silver chloride, A g C l. Chlorobenzene gives none, because its ring chlorine never came off. Okay? The same rule holds with other reagents. With potassium cyanide, KCN, ammonia, N H three, or sodium alkoxide, N a O R, only the side-chain halogen, X, is replaced. With me?

3:49Oxidation: the side chain goes, the ring halogen stays

One more tool, oxidation. Hot potassium permanganate, K M n O four, converts any side chain with a benzylic hydrogen into COOH, the carboxylic acid group, where the arrow means gives. And a benzylic hydrogen is a hydrogen on the carbon attached directly to the ring, okay? Now, what about a chlorine sitting on the ring itself? Does it also get oxidised away? No, the ring halogen survives the oxidation. So the acid still carries its ring chlorine, and that tells you where the chlorine was. Keep this in mind for our second example. Clear?

4:26Example 1: by-products of the Wurtz-Fittig toluene synthesis

Now let's try one. Chlorobenzene and chloromethane are treated with sodium in dry ether. Toluene forms, but which by-products come along with it? Pause the video and try it first. Done? Okay, let's check. First the cross product: the phenyl group of chlorobenzene joins the methyl group of chloromethane, giving C six H five, C H three, which is toluene. That is the Wurtz Fittig product we want, okay? But sodium does not choose partners. So tell me, can two chlorobenzene molecules join each other? Yes, exactly! That is just the Fittig reaction, and it gives C six H five, C six H five, which is biphenyl. Right? In the same way, two chloromethane molecules join to give C H three, C H three, which is ethane. That is the Wurtz reaction from the last chapter, where two different halides gave a mixture of three products. Clear? So the by-products are biphenyl and ethane. Many students write only toluene and stop there, so always think of all three pairs, okay?

5:28Example 2: identify two C7H7Cl isomers by AgNO3 and KMnO4

Now the harder one, and it is our puzzle from the start. Compound A, C seven H seven C l, gives a white precipitate with alcoholic silver nitrate, and is oxidised to benzoic acid. Its isomer, compound B, gives no precipitate, and is oxidised to an acid, C seven H five C l O two, that melts higher than its two position isomers. Identify both compounds. Pause the video and try it first. Done? Okay, let's check. Compound A gives a white precipitate. So is its chlorine on the ring, or on the side chain? On the side chain, yes, because a white precipitate means a reactive side-chain chlorine. The double arrow on screen means implies. With me? Seven carbons, and the ring has six, so one carbon sits outside the ring, carrying the chlorine. So compound A is C six H five, C H two, C l, benzyl chloride. Hot permanganate turns that side chain into COOH, giving benzoic acid. Everything matches, right? Now look at compound B, which gives no precipitate, so its chlorine is on the ring. Then the seventh carbon must be a methyl group, C H three, also on the ring. Clear? But here is the catch. The chlorine and the methyl group can be ortho, meta or para to each other. So which one is compound B? Look at the acid. Oxidation turns the methyl group into COOH, and the ring chlorine survives, giving C l, C six H four, COOH, a chlorobenzoic acid. It melts higher than its two position isomers, and that suggests the para isomer. Got it? So compound B is four chlorotoluene. And there you go, mystery solved! Those were our two bottles from the start: benzyl chloride gives the precipitate, and four chlorotoluene does not. See how one carbon changes everything?

7:24Quick revision: Na, Cu, Mg reactions, ring vs side-chain Cl

So, quick revision. Sodium in dry ether gives A r R by Wurtz Fittig, and A r A r by Fittig. Copper gives the Ullmann reaction, and for the Grignard reagent, bromobenzene goes in ether, but chlorobenzene in THF, okay? And a side-chain halogen reacts, while a ring halogen stays, even with hot permanganate. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!