Electrophilic Substitution

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Put a chlorine atom on a benzene ring. Chlorine pulls electrons away from the ring, so chlorobenzene reacts with electrophiles more slowly than benzene does. And yet, when a new group comes in, chlorine sends it to the carbons right next to it, or to the carbon opposite, as if it were pushing electrons into the ring. So is chlorine pulling electrons, or pushing them? It sounds like it cannot do both, right? Today, electrophilic substitution in haloarenes. It comes up often in JEE, so let's begin.

0:37The pull: the halogen slows the ring down

So, let us take the two effects one at a time, starting with the pull. Minus I in brackets stands for the inductive effect; the minus sign means the halogen withdraws electrons from the ring. Okay? Now, an electrophile loves electrons; it comes to the ring looking for them. So if the ring has fewer electrons, will the reaction be faster or slower? It gets slower, that is exactly right! Haloarenes react more slowly than benzene with electrophiles, and we call this deactivation. The halogen deactivates the ring. Clear?

1:14The push: resonance feeds ortho and para

Now let us look at the push. Chlorine, C l, has lone pairs, and by resonance, written plus R, one lone pair moves into the ring. Below, the course figure shows resonance in chlorobenzene: five structures, numbered one to five in Roman numerals, and the double headed arrows between them mean resonance. In structures two, three and four, chlorine has a double bond to the ring and a small plus charge, written high. With me? Now look at the small red minus signs in those three structures. Where do they sit, compared with chlorine? In two and four, they sit on the carbons right next to chlorine, the ortho positions. In three, on the carbon opposite, the para position. The meta carbons in between never get a minus sign. So there is more electron density at ortho and para than at meta: the small o, p and m on screen. The note under the structures says the same thing. Right? There is one more benefit. When the electrophile attacks at ortho or para, the intermediate it forms is stabilised by this resonance. Okay? So both reasons point the same way, and the halogen is ortho, para directing. Got it?

2:26Pull decides the speed, push decides the position

So now, back to our puzzle. The truth is, chlorine does both! The inductive pull lowers the electron density of the whole ring, so the reaction is slower. But the resonance push makes ortho and para richer than meta, so they are still the best places to attack. Here is a topper's line to remember: the pull decides the speed, and the push decides the position. Clear? Now look at these numbers: rates of nitration, with benzene taken as one. Toluene, which is methylbenzene, is about twenty five, and chlorobenzene is about zero point zero three; the wavy equals sign means approximately. With me? Methyl is ortho, para directing, and it activates the ring. Chlorine is also ortho, para directing, so it activates the ring too, right? No, no, wait, look at the numbers again! Toluene is faster than benzene, so methyl activates the ring. Chlorobenzene is far slower, so chlorine deactivates it, although both direct ortho and para. Students often assume every ortho, para director activates the ring, and halogens break that rule. Don't fall for it!

3:30Reactions of chlorobenzene: para is major

Now, the actual reactions of chlorobenzene. First, chlorination, with chlorine, C l two, the small two written low, and anhydrous F e C l three, ferric chloride; anhydrous means completely dry. The arrow means gives, and we get one, four dichlorobenzene, with the two chlorines at positions one and four, opposite each other. Okay? But wait, chlorobenzene has two ortho positions and only one para position. So the ortho product should be the major one, right? Oh no, careful here, because the para product is the major one. Para is far from chlorine, so it is less crowded, and it feels the inductive pull, minus I, more weakly than the ortho carbons right next door. And a small trick for names: the four in brackets means para, and the two means ortho. So the ortho product, one, two dichlorobenzene, also forms, but less. Clear? Next, nitration, with concentrated nitric acid, HNO three, plus concentrated sulphuric acid, H two S O four; conc. is short for concentrated. The major product is one chloro four nitrobenzene. Right? Sulphonation, with concentrated sulphuric acid and heat, gives four chlorobenzenesulphonic acid as the major product. With me? Now the Friedel Crafts reactions, with anhydrous aluminium chloride, A l C l three. Chloromethane, C H three C l, gives one chloro four methylbenzene. Okay? And ethanoyl chloride, C H three COC l, gives four chloroacetophenone. See the pattern here: every time, the major product has the new group para to chlorine, at position four. Got it?

5:13Example 1: major product of methylation of bromobenzene

Now let us try Example one. Name the major product of methylation of bromobenzene with C H three C l and A l C l three. This time the halogen is bromine. Pause the video and try it first. Done? Okay, let's check. Bromine is a halogen, so it deactivates the ring. So the methyl group goes to meta, right? That is the trap, so be careful, because deactivation only decides the speed. Bromine, B r, like chlorine, still pushes electrons to ortho and para by resonance, so it directs ortho and para. The push decides the position, remember? And between the two, para is the major product, so the methyl group, C H three, goes to position four. Right? So the answer is one bromo four methylbenzene, with the prefixes bromo and methyl in alphabetical order. Its common name is para bromotoluene; the small italic p stands for para. Clear?

6:15Example 2: 4- and 3-chlorobenzenesulphonic acid from benzene

Now a harder one, a synthesis question. Starting from benzene, how would you prepare, in part a, four chlorobenzenesulphonic acid, and in part b, three chlorobenzenesulphonic acid? The letters in brackets mark the parts. Pause the video and try it first. Done? Okay, let's check. Both products need one chlorination and one sulphonation. So the order of the two steps does not matter, right? No, no, wait, the order matters completely! The group already present on the ring decides where the second group goes. Okay? For part a, we want the two groups para to each other. Chlorine is an ortho, para director, so chlorine must go in first. Right? So benzene, C six H six, is chlorinated to chlorobenzene, C six H five C l; each arrow is one step. Then sulphonation puts the S O three H group para to chlorine, giving four chlorobenzenesulphonic acid as the major product. With me? Now for part b, the two groups must be meta to each other, and chlorine directs ortho and para, not meta. So which group must go in first? The sulphonic acid group, that is correct, because S O three H is a meta director. So we sulphonate first. Clear? Benzene is sulphonated to C six H five S O three H, benzenesulphonic acid. Then chlorination puts the chlorine meta to it, giving three chlorobenzenesulphonic acid. Same two steps, opposite order, and a different product!

7:52Quick revision: -I slows the ring, +R directs ortho, para

So, quick revision. First, the inductive pull deactivates the ring, so haloarenes react more slowly than benzene. Second, the resonance push makes halogens ortho, para directing, and the para product is the major one. Third, in a synthesis, the group already present decides where the next one goes, so plan the order. Now go and try the practice problems on this page. You are ready for them. All the best!