Carboxylic Acids

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Put a methyl group on the ring of benzoic acid, at para, across from the acid group. Methyl pushes electrons in, so the acid gets weaker. Now move that methyl group next door, to ortho. It should be weaker again, surely? But no, now it is stronger than benzoic acid itself! How can one group give opposite results? Soon you will know why. Acid strength comes up often in JEE, so let's begin.

0:33Stable carboxylate ion, stronger acid

So, a quick recall first. On screen is p K a: a small p, a capital K, and a small a written low. The smaller the p K a, the stronger the acid, and one unit of p K a is a factor of ten. Okay? When a carboxylic acid loses its proton, we get the carboxylate ion, RCOO minus, where R stands for the rest of the molecule and the minus sign written high is its charge. The more stable this ion, the stronger the acid. So electron-withdrawing groups, minus I, with I for inductive, stabilise it and strengthen the acid. With me? And electron-releasing groups, plus I, destabilise the ion and make the acid weaker. So tell me, with H, a methyl group or an ethyl group on COOH, which acid is the strongest? Methanoic acid, yes, exactly! H has no plus I effect, while alkyl groups are plus I, ethyl a little more than methyl. So HCOOH, three point seven five, beats C H three COOH, four point seven six, then propanoic acid, four point eight seven. Small numbers written low count atoms, brackets hold p K a values, and the sign between is greater than. Clear?

1:57How big is the effect? Number, strength, distance

Now, three things decide how big the effect is. First, the number of groups. One chlorine on ethanoic acid gives a p K a of two point eight seven. With two chlorines it is one point two six, and with three chlorines, C C l three COOH, it is zero point six four. Each chlorine adds its pull. Got it? Second, the strength of the group. For the halogens, the pull goes fluorine, greater than chlorine, greater than bromine, greater than iodine. So fluoroethanoic acid, two point five nine, is stronger than chloroethanoic acid, two point eight seven. Okay? And third, take chlorobutanoic acid. Carbon one is the COOH carbon, so carbon two is right next to it. Put one chlorine at carbon two, three or four. It is the same atom, so the same acidity, right? No, no, careful, that is the trap! The inductive effect fades with distance. So two-chloro is greater than three-chloro, greater than four-chlorobutanoic acid: the closer the chlorine, the stronger the acid. Clear?

3:05The carbon next to COOH, and benzoic acid

Now, the carbon next to COOH matters too. Three short lines mean a triple bond, two short lines a double bond, and each dash a single bond. So propynoic acid, with the triple bond, is strongest, then propenoic, then propanoic acid. With me? Why does this happen? The carbon attached to COOH is s p, then s p two, then s p three, the small numbers written high. More s character means a more electronegative carbon, so the s p carbon stabilises the carboxylate ion best. Right? Now, benzoic acid, C six H five COOH, has its acid group on an s p two carbon of the benzene ring. How does it compare with methanoic acid and ethanoic acid? It sits right in the middle, very good! At four point one nine, it beats ethanoic acid, four point seven six, as the s p two ring carbon withdraws electrons compared with methyl. But it is weaker than methanoic acid, three point seven five, as H has no plus I effect. Got it?

4:16Groups on the ring of benzoic acid

Now let us put groups on the ring. Minus R, with R for resonance, means pulling electrons through the pi system. Minus I and minus R groups, like N O two and C N, stabilise the carboxylate ion, so the acidity increases. Chlorine does the same, as its minus I pull wins over its weak electron release. Okay? Now flip it: put a methoxy group, OCH three, at para, straight across the ring from the COOH group. Is the acid stronger than benzoic acid, or weaker? It is weaker, yes, exactly! At para, OCH three and O H push electrons in by plus R, and C H three by its plus I effect. They destabilise the carboxylate ion, so the acidity decreases. Clear? And the numbers agree, where a small p in italics means para. Para-nitrobenzoic acid, three point four one, beats benzoic acid, four point one nine, which beats para-methoxybenzoic acid, four point four six. With me?

5:21The ortho effect

Now, back to our puzzle. Ortho means the position right next to COOH on the ring. Methyl at para weakens benzoic acid. So what does methyl at ortho do? Here is the surprise. With a group at ortho, written as a small o in italics, the acid is almost always stronger than benzoic acid. This is called the ortho effect. Okay? And it works whatever the group, even for C H three, which normally releases electrons.

5:52Example 1: chloroethanoic vs ethanoic acid from pKa

Got it? Alright, let us try example one. How many times stronger is chloroethanoic acid, with a p K a of two point eight seven, than ethanoic acid, with a p K a of four point seven six? Pause the video and try it first. Done? Okay, let's check. Chloroethanoic acid has the smaller p K a, so it is stronger. The difference is four point seven six minus two point eight seven, which equals one point eight nine. So it is one point eight nine times stronger, right? No, no, wait, that is a classic trap! One unit of p K a is a factor of ten, so the ratio is ten to the power one point eight nine, the small number written high being the power. Clear? And ten to the power one point eight nine is about seventy eight; the wavy equals sign means approximately. So a single chlorine makes the acid about seventy eight times stronger. Okay?

6:54Example 2: order of o-toluic, benzoic and p-toluic acid

Now example two, a step harder, and it is our puzzle. Toluic acids are methylbenzoic acids. Arrange benzoic acid, ortho-toluic acid and para-toluic acid in order of decreasing acidity. Pause the video and try it first. Done? Okay, let's check, starting with para-toluic acid. C H three at para releases electrons and destabilises the carboxylate ion, so it is weaker than benzoic acid. Right? Now for ortho-toluic acid, the ortho effect says it is stronger, but why? The methyl group sits right beside COOH, so the two are crowded. What does this crowding do to the COOH group? Very good, it pushes COOH aside! The methyl group twists it out of the plane of the ring. With me? Normally the ring releases electrons to COOH; once it is twisted, that release is cut off. So ortho-toluic acid becomes stronger than benzoic acid. Got it? So the order is ortho-toluic, three point nine one, greater than benzoic, four point one nine, greater than para-toluic, four point three four. Puzzle solved: the same methyl group, two positions, opposite results! See how simple that was?

8:18Quick revision: substituent effects and the ortho effect

So, quick revision. Minus I groups strengthen, plus I groups weaken, and more, stronger and closer groups act more. An s p carbon next to COOH helps most, and benzoic acid sits between methanoic and ethanoic acid. On the ring, minus I and minus R strengthen, plus R and plus I weaken, and an ortho group almost always strengthens. Now try the practice problems on this page. All the best!