Electrophilic Substitution

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. On paper, benzene has three double bonds, and an alkene simply adds bromine to give a dibromide, right? But benzene reacts with bromine only when F e B r three is present, and even then it gives bromobenzene, not a dibromide. So why does benzene refuse to add? Hold on to that. Electrophilic substitution comes up often in JEE, so let's begin.

0:30The mechanism in three steps

So, let us build the mechanism with the figure on the right. Step one: make the electrophile, capital E with a small plus written up high, a superscript that means a positive charge. We make it using a strong acid or a Lewis acid. Okay? Now, look at the benzene ring in the figure. Where will E plus find electrons? Yes, exactly, in the pi electrons of the ring! Pi is the Greek letter you see in step two. They bond to E plus, and this attack, the arrow marked slow, is the slow step. Right? We get the arenium ion, also called the sigma complex; sigma is the Greek letter in brackets. The carbon holding both H and E becomes s p three, with the small three written up high. The double-headed arrows join three resonance structures, and the red plus signs show the charge shared by the two ortho carbons and the para carbon. So the ring is no longer aromatic. Clear? That costs the ring its special stability. So tell me, how can it become aromatic again? Very good, it throws out a proton, H plus, from the s p three carbon! That is the arrow marked fast, with minus H plus written under it. The aromatic ring is back, with E where a hydrogen was. That is substitution, not addition. Got it?

1:53Five reactions, five electrophiles

Now, five reactions, each with its own electrophile. First, nitration, with concentrated HNO three and H two S O four, nitric and sulphuric acids; conc. means concentrated, and the small numbers written low are subscripts. The temperature is three hundred and twenty three to three hundred and thirty three kelvin, written with capital K, and the electrophile is N O two plus, the nitronium ion. The arrow leads to nitrobenzene, C six H five N O two. With me? Second, halogenation, with C l two or B r two and anhydrous F e C l three, F e B r three or A l C l three; anhyd. means anhydrous, completely dry. The electrophile is C l plus or B r plus, and the arrow gives C six H five X, with X standing for the halogen. Okay? Third, sulphonation, with fuming sulphuric acid, called oleum. Careful here: the electrophile is S O three, with no plus sign at all. The product is C six H five S O three H, benzenesulphonic acid. Right? Fourth, alkylation, with R C l, where R is an alkyl group, and anhydrous A l C l three. Chlorine gave C l plus. So what electrophile will R C l give? Yes, exactly, R plus, a carbocation! And the product is C six H five R, an alkylbenzene. Clear? Fifth, acylation, with RCOC l, or the anhydride, RCO in round brackets with a small two, then O, and anhydrous A l C l three. The electrophile is RCO plus, the acylium ion, giving C six H five COR, a ketone. Alkylation and acylation are the Friedel-Crafts reactions, okay?

3:37How each electrophile is made

Alright, where do these electrophiles come from? For nitration, HNO three plus two H two S O four gives N O two plus, H three O plus, and two HSO four minus. The big two in front counts molecules. Right? Now let me ask you this. In this reaction, is nitric acid acting as an acid, or as a base? If you said acid, oh no, careful, that is exactly the trap many students fall into! Nitric acid acts as a base here, because sulphuric acid is the stronger acid and gives it a proton. Clear? For halogenation and alkylation, the Lewis acid does the job. C l two plus F e C l three gives C l plus and F e C l four, in square brackets, with a minus charge outside. In the same way, R C l plus A l C l three gives R plus and A l C l four minus. The Lewis acid pulls the chloride away. With me? And sulphonation is reversible. Heat the sulphonic acid with dilute acid or steam, and the S O three H group comes off again. Got it?

4:42Friedel–Crafts: the traps

Now, the Friedel-Crafts traps. Take benzene with one chloropropane, a three carbon chain with chlorine at the end, and A l C l three. That gives propylbenzene, right? No, no, careful, that is exactly the trap! The primary carbocation rearranges by a hydride shift to the isopropyl cation, so you get mainly isopropylbenzene. With me? Here is the second problem: the alkylbenzene is more reactive than benzene, because the alkyl group activates the ring. So it keeps reacting, giving polyalkylation. Okay? Acylation behaves much better. The acylium ion does not rearrange, and the acyl group deactivates the ring, so the product reacts no further. Right? Now here is the topper's trick. Reduce that ketone with zinc amalgam, that is zinc mixed with mercury, written Z n dash H g, and H C l. You get a straight-chain alkylbenzene, with no rearrangement. Clear? And neither reaction works on a strongly deactivated ring like nitrobenzene, or on aniline, which forms a salt with A l C l three. Got it?

5:49Example 1: why benzene gives bromobenzene, not a dibromide

Now, remember our puzzle? Let's solve it. Benzene reacts with bromine only in the presence of F e B r three, and gives bromobenzene, not a dibromide. Explain both facts. Pause the video, and try it first. Done? Okay, let's check. For the first fact, F e B r three, the Lewis acid, polarises B r two into a strong electrophile. That is why it is needed. Right? Now for the second fact. The ring attacks, and we reach the arenium ion, as in the figure. Should it now add B r minus, like an alkene would, or lose H plus? Adding B r minus would give a product that is not aromatic, so the ring would never get its special stability back. That does not happen. With me? Losing H plus restores aromaticity, so we get C six H five B r, bromobenzene, a substitution product. Puzzle solved: benzene would rather keep its aromatic ring. Clear?

6:51Example 2: propylbenzene from benzene without rearrangement

Now a harder one, in two steps. How would you prepare propylbenzene from benzene, without rearrangement? Remember the traps. Pause the video, and try it first. Done? Okay, let's check. One chloropropane, C H three C H two C H two C l, with A l C l three, is tempting. But we just saw that it gives mainly isopropylbenzene. Right? So we acylate instead. Benzene plus propanoyl chloride, C H three C H two COC l, with A l C l three written over the arrow, gives the ketone C six H five COCH two C H three, plus H C l. The acylium ion does not rearrange. Okay? But we want propylbenzene, not a ketone. So what is the next step? Yes, exactly, reduce the ketone! With zinc amalgam and H C l over the arrow, the C O group becomes C H two, and we get propylbenzene, with a straight chain. See how neatly we avoided the trap?

7:57Quick revision: three steps, electrophiles, Friedel-Crafts

So, quick revision. Three steps: make E plus, a slow attack giving the arenium ion, and a fast loss of H plus. The electrophiles are N O two plus, C l plus or B r plus, S O three, R plus, and RCO plus. And alkylation rearranges and repeats, while acylation does neither, then we reduce. You have done really well today. Now try the practice problems on this page. All the best!