Reactions of Phenol

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Add bromine water to phenol, and a white precipitate forms: three bromine atoms have gone onto the ring. Now use bromine in carbon disulphide, at two hundred and seventy three kelvin, and mainly one bromine atom goes in. Same phenol, same bromine, so why three in one case and only one in the other? Today we study the reactions of phenol: first its O H group, then its ring. These reactions come up often in JEE, so let's begin.

0:35The OH group: esters

Let's start with the O H group. You know that an alcohol and a carboxylic acid give an ester. So phenol with a carboxylic acid should give an ester just as easily, right? Oh no, careful, phenol does not react easily with the acids themselves, as the round brackets say. So for phenol, we make esters with acid chlorides or with acid anhydrides. Okay? Phenol, C six H five O H, plus ethanoic anhydride, C H three C O in brackets with a small two, then O; the small numbers written low are subscripts. The arrow means gives: the ester phenyl ethanoate, C six H five OCOCH three, plus C H three COOH, which is ethanoic acid. With me? On salicylic acid, the same reaction gives aspirin. Acetylation puts the acetyl group, C H three C O, onto the phenolic O H group, and the name of aspirin, in brackets, is two acetoxybenzoic acid. Got it?

1:34More reactions of the OH group

Next, ethers: sodium phenoxide, the sodium salt of phenol, plus an alkyl halide gives an ether. In brackets is its name, the Williamson synthesis. Clear? Now heat phenol with zinc dust, and the O H group is removed, giving benzene. And neutral ferric chloride, F e C l three, gives a violet colour with phenols, and also with enols, as the brackets say. Right?

1:58The ring: strongly activated

Now let's turn to the ring. The O H group strongly activates the benzene ring towards electrophiles, and directs them to the ortho and para positions. Okay? And the phenoxide ion, C six H five O minus, with the small raised minus as its charge, is even more reactive than phenol. Remember this ion; it will solve our puzzle. With me? So, nitration first: dilute nitric acid, HNO three, at two hundred and ninety eight kelvin, written with a capital K, gives ortho and para nitrophenol; the italic o and p stand for ortho and para. Right? Now make the nitric acid concentrated, and the strongly activated ring takes three nitro groups, at both ortho positions and the para position: two four six trinitrophenol. Its common name, in brackets, is picric acid. Clear? And for a better yield of picric acid, phenol is first sulphonated to phenol two four disulphonic acid, and that is then nitrated. Got it?

2:56Bromination: the solvent decides

Now, back to our puzzle. Take bromine, B r two, in a solvent of low polarity, like carbon disulphide, C S two, or chloroform, CHC l three, at two hundred and seventy three kelvin. This gives monobromophenols, mainly the para isomer. Okay? But with bromine water, a white precipitate of two four six tribromophenol forms. So why does water make such a big difference? Think about the ion we just met. The phenoxide ion, yes, exactly! Water ionises phenol to the very reactive phenoxide ion, and it also polarises the bromine. In carbon disulphide neither happens, so only one bromine goes in, and that solves our puzzle. Right?

3:38Kolbe's reaction and the Reimer–Tiemann reaction

Now two named reactions, starting with Kolbe's reaction: sodium phenoxide with carbon dioxide, C O two, at four hundred kelvin and four to seven atmospheres, and then acid, gives salicylic acid. Clear? Now tell me, why do we use sodium phenoxide here, and not phenol itself? Because carbon dioxide is only a weak electrophile, yes, exactly! It needs the phenoxide ion, which is much more reactive than phenol. Okay? Next, the Reimer Tiemann reaction: phenol with chloroform and aqueous sodium hydroxide, N a O H, at three hundred and forty kelvin, and then acid, gives salicylaldehyde. With me? The electrophile here is dichlorocarbene, C C l two, and the two dots on its carbon are a lone pair of electrons. And note the twist: with carbon tetrachloride, C C l four, in place of chloroform, the product is salicylic acid. Got it?

4:33More reactions to know

Now a few more, quickly: chromic acid, sodium dichromate, N a two C r two O seven, with sulphuric acid, H two S O four, the slash meaning together, oxidises phenol to benzoquinone. Okay? Hydrogen, H two, with nickel, N i, at four hundred and thirty three kelvin, gives cyclohexanol. And with methanal, phenol gives bakelite. Right? Next, coupling: a benzenediazonium salt in weakly alkaline solution gives an orange azo dye with phenol, para hydroxyazobenzene. Clear? And with phthalic anhydride and concentrated sulphuric acid, phenol gives phenolphthalein, but only when the para position of the phenol is free. With me?

5:15Example 1: tribromophenol mass from 0.47 g phenol

Now Example one, a numerical from the course, using the bromine water reaction from our puzzle. Find the mass of two four six tribromophenol, with molar mass, capital M, three hundred and thirty one, precipitated from zero point four seven grams of phenol with excess bromine water. Pause the video and try it first. Done? Okay, let's check. The molar mass of phenol, C six H five O H, is seventy two plus six plus sixteen, which is ninety four. Zero point four seven over ninety four gives zero point zero zero five moles; the slash means divided by. Now, three bromine atoms go onto each ring, so the moles of product are three times this, right? No, no, wait, that is the trap! One mole of phenol gives one mole of tribromophenol, because the three bromine atoms are already counted in the three hundred and thirty one. So we get zero point zero zero five moles of product as well. Clear? So the mass is zero point zero zero five times three hundred and thirty one, which is one point six five five, or about one point six six grams; the wavy equals sign means approximately. Got it?

6:26Example 2: phenol to anisole, salicylic acid, ester, quinone

Now Example two, the harder one, built from today's reactions. How would you convert phenol into, part A, anisole; part B, two hydroxybenzoic acid; part C, phenyl ethanoate; and part D, para benzoquinone? Pause the video and try it first. Done? Okay, let's check. Part A, anisole, is C six H five OCH three, an ether. So use sodium hydroxide first, to make sodium phenoxide, and then methyl iodide, C H three I, which is the Williamson method. Right? Part B, two hydroxybenzoic acid, is salicylic acid, so this is Kolbe's reaction: sodium hydroxide, then carbon dioxide at four hundred kelvin under pressure, then acid. With me? Part C, phenyl ethanoate, is an ester. So we just heat phenol with ethanoic acid, right? Oh no, careful, remember the brackets from the start: phenol does not react easily with acids. So we use ethanoic anhydride, written C H three C O in brackets with a small two, then O, and not the acid itself. Okay? And for part D, para benzoquinone, we oxidise phenol with chromic acid, that is, sodium dichromate with sulphuric acid. Every one came from today's list. Got it?

7:45Quick revision: OH group, ring, Kolbe, Reimer Tiemann

So, quick revision. The O H group gives esters with acid chlorides or anhydrides, ethers by the Williamson method, and a violet colour with neutral ferric chloride. The ring is activated at ortho and para, the phenoxide ion even more, so bromine water gives tribromophenol. Kolbe's reaction gives salicylic acid, and Reimer Tiemann gives salicylaldehyde, or the acid with carbon tetrachloride. Now go and try the practice problems on this page. All the best!