IIT-JEE Questions (1997–2010): Acids

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. The phenoxide ion, from phenol, has more resonance structures than the benzoate ion, from benzoic acid. More structures should mean a more stable ion, and a stronger acid, right? But benzoic acid is the stronger acid! So how can that be? This was asked in IITJEE nineteen ninety seven. Today we crack it, and then we solve two more from this collection: an acidity order from two thousand and nine, and a labelled carbon from two thousand and eight. These ideas are worth knowing well, so let's begin.

0:39Why benzoic acid beats phenol

So, RCOOH is a carboxylic acid, with R standing for the rest of the molecule. With water, H two O, it forms the carboxylate ion, RCOO minus, and H three O plus. The small numbers written low count atoms, the raised signs are charges, and the double arrow means an equilibrium. Remember, the more stable this anion, the stronger the acid. Okay? Now, in a carboxylate ion like benzoate, the negative charge can sit on either oxygen. So there are two equivalent resonance structures, with the charge on oxygen in both. Keep that in mind, because the figure on screen is the phenoxide ion, and we compare the two next. Right? Now look at the figure, the phenoxide ion. The double-headed arrows mean resonance, and you can count four structures, against only two for benzoate. But in the three extra structures, where does the negative charge sit? On the ring carbons, yes, exactly! The red minus signs sit at the ortho, the para and the other ortho position. Only the first structure keeps the charge on oxygen. Clear? And carbon is less electronegative than oxygen, so these structures contribute very little, while both benzoate structures carry the charge on oxygen. That solves our puzzle: what matters is where the charge sits, not how many structures there are. With me? The numbers agree. The course writes p K a, a small p, a capital K and a small a written low, and a smaller p K a means a stronger acid. Carboxylic acids are at about four to five, and phenol at ten point zero. Got it?

2:21Groups on the ring: phenols and benzoic acids alike

Now, groups on the ring, and as the heading says, the rule is the same for phenols and for benzoic acids. If a group pulls electron density away from the ring, does the acid get stronger or weaker? It gets stronger, yes, exactly! An electron-withdrawing group, like nitro, N O two, or chlorine, C l, stabilises the anion, so the acid is stronger. Okay? And the opposite: an electron-releasing group, like methyl, C H three, or methoxy, OCH three, makes the anion less stable, so the acid is weaker. One caution: this rule is for groups at para, as in our examples. In a benzoic acid, a group at ortho is a special case, the ortho effect: almost any ortho group makes it stronger. Right?

3:08Heating a β-keto acid, then the iodoform test

Now the last idea, for the labelled carbon question. The map on screen sums up how carboxylic acids are made and how they react, and we will use one of its boxes soon. This acid is R, then C O, a keto group, then C H two and COOH, with dashes for single bonds. The carbon next to the carboxyl group is alpha, and the next is beta, the Greek letter on screen. The keto group is on the beta carbon, so it is a beta keto acid. Okay? So here is a question for you. What happens when a beta keto acid is heated gently? It loses carbon dioxide, yes! The carbon of the COOH group leaves as C O two, through a cyclic six-membered transition state, written T S on screen. Right? Next, the iodoform test: iodine, I two, plus sodium hydroxide, N a O H, and a positive test gives yellow iodoform, CHI three. With me? Now, ethanoic acid, C H three COOH in the badge, also has a C H three C O group. So does ethanoic acid give the iodoform test? No, it does not, and that is a classic trap! The C H three C O group must be attached to a carbon or a hydrogen, and in ethanoic acid it is attached to an O H group. Got it? The arrow means gives: in a methyl ketone, the C H three becomes iodoform, CHI three, and the rest becomes the salt of an acid with one carbon fewer. The sixth box on the left of the map says this too. Okay?

4:45Example 1: acidity order of phenols and benzoic acids

Time for Example one, from IITJEE two thousand and nine. Arrange in order of acidity: phenol, Roman one; four chlorophenol, Roman two; benzoic acid, Roman three; and four methylbenzoic acid, Roman four. The four in a name puts the group para to the main group. Pause the video and try it first. Done? Okay, let's check. Sort them into families first. Roman three and four are carboxylic acids, with a p K a of about four to five, and Roman one and two are phenols. So both acids come above both phenols. With me? Now, inside the acid family, four methylbenzoic acid has a methyl group at para. Which is stronger, benzoic acid or four methylbenzoic acid? It is benzoic acid, yes, exactly! Methyl releases electrons and weakens the acid. So Roman three is greater than Roman four, the greater than sign meaning more acidic. Right? Now the phenols. Chlorine withdraws electrons and stabilises the phenoxide ion, so Roman two is greater than Roman one. But here is the trap. Chlorine pulls electrons, while methyl pushes them. So should four chlorophenol beat four methylbenzoic acid? No, it should not, so be careful! Phenol starts far below, at a p K a of ten, against four to five for the acids, and a single chlorine cannot close that gap. So the order is Roman three, greater than four, greater than two, greater than one.

6:21Example 2: C-13 labelled beta-keto acid, heat, then iodoform

Clear? Now Example two, from IITJEE two thousand and eight. The acid is C six H five, the phenyl group, then C O, C H two and COOH, right? The raised thirteen marks that C H two carbon as carbon thirteen, a heavier isotope we can track. Heating gives E, and E, with iodine and N a O H, gives F and G, so identify all three. Pause the video and try it first. Done? Okay, let's check. The keto group is on the beta carbon, so this is a beta keto acid. Gentle heating removes the carbon of the COOH group as carbon dioxide. Clear? So, before we write E, think about this. Does our labelled carbon leave with the carbon dioxide? No, it stays behind, very good! The labelled C H two becomes C H three, so E is C six H five, C O, carbon thirteen H three: acetophenone, a methyl ketone. Its C H three C O group sits on a ring carbon, so it gives the iodoform test. So where does the labelled carbon go now? Into the iodoform, yes, exactly! The C H three becomes CHI three, so G is carbon thirteen iodoform, the yellow solid, carrying the label. Got it? And what about F, the other product? It is the salt of an acid with one carbon fewer: acetophenone has eight carbons, so we get sodium benzoate, C six H five COON a, with seven. Watch the trap: F has no label, because the labelled carbon left as iodoform. Right?

8:01Quick revision: acid vs phenol, substituents, iodoform

So, quick revision. Benzoate keeps the charge on oxygen in both its structures, so carboxylic acids beat phenols. Okay? Electron-withdrawing groups raise acidity, and electron-releasing groups lower it. Right? And a beta keto acid loses carbon dioxide on gentle heating, while C H three C O on carbon or hydrogen gives yellow iodoform. Now go through the other solved questions on this page, and try each one first. All the best!