Preparation of Carboxylic Acids

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Chloromethane has one carbon atom, but ethanoic acid has two. So where does the extra carbon come from? We will answer that in the first example. Today we see the ways to prepare carboxylic acids, and two of them add exactly one carbon. These routes come up often in JEE, so let's begin.

0:25Oxidising primary alcohols and aldehydes

First, the figure. In the middle is our target, R dash COOH, where R is the rest of the molecule, the dash is a single bond, and COOH is the carboxyl group. The seven boxes on the left, under Preparation, are the ways to prepare it. The orange boxes on the right, under Reactions, are for the next concept. Okay? The first box: oxidise a primary alcohol, or an aldehyde. On screen, the small circle written high after the number one is a degree sign, and one degree means primary. The arrow means gives. So we get RCOOH, the acid. Right? Which oxidising agent? Potassium permanganate, K M n O four, where the small four written low means four oxygen atoms. It works in neutral, acidic or alkaline solution. Clear? Or acidified potassium dichromate, K two C r two O seven, with H plus after the slash. The slash means along with, and the small plus up top is a positive charge. Or C r O three in H two S O four, called the Jones reagent. Got it?

1:32Alkylbenzenes: the whole side chain goes

Now the second box: alkylbenzenes, a benzene ring with an alkyl side chain. We oxidise them with hot alkaline K M n O four, or chromic acid. In the figure, KOH with heat, then H three O plus, the hydronium ion, gives benzoic acid. With me? What if the side chain is long, say three or four carbons? Does only its end get oxidised, or something more? Yes, the whole side chain goes, whatever its length! The entire chain becomes one COOH group on the ring, so a ring with one such side chain gives benzoic acid. Right? But there is one condition. The benzylic carbon, the carbon joined directly to the ring, must carry at least one hydrogen, the H on screen. Clear? So now take a tert butyl group: one central carbon carrying three methyl groups, joined to the ring. Every side chain gets oxidised, right? So is this one oxidised too? If you said yes, careful, that is the trap! Its benzylic carbon carries three methyl groups and no hydrogen. So the tert butyl group survives. The word tert, in italics on screen, is short for tertiary. Got it?

2:45One carbon more: nitriles and Grignard reagents

Now, back to our puzzle: adding a carbon. Route three starts from an alkyl halide, R X, where X is a halogen. Potassium cyanide, KCN, written above the arrow, gives RCN, a nitrile. Okay? Then we hydrolyse the nitrile with H three O plus, an acid, or O H minus, an alkali, where the small minus up top is a negative charge, and heat. It goes first to RCONH two, an amide, then to RCOOH, the acid. Right? So, count the carbons. Compare R X at the start with RCOOH at the end. How many carbons has the chain gained? Exactly one, the carbon of the cyanide! So the acid has one carbon more than the halide, and the chain grows by one. Clear? Route four does the same job another way. R M g X, a Grignard reagent, adds to carbon dioxide, C O two, used as dry ice, giving RCOOM g X, okay? Then H three O plus gives RCOOH, again with one carbon more. With me?

3:49More routes to the acid

Back to the figure, box five. Acid chlorides, anhydrides and esters, the acid derivatives, can be hydrolysed back to the acid. Acid chlorides and anhydrides need just water. Right? Esters can be hydrolysed with acid, or, better, with alkali followed by acid. So the order is alkali first, then acid. Okay? In box six, a methyl ketone reacts with N a O X, where X is a halogen. This is the haloform reaction. And as the figure says, the acid has one carbon fewer than the ketone. Clear? In box seven, alkenes and alkynes with hot K M n O four are split apart at the multiple bond. This is oxidative cleavage. Got it? One more route, not in the figure, is hydrolysing one, one, one trihalides, with three halogen atoms on one carbon. C six H five C C l three gives C six H five COOH, benzoic acid. With me?

4:44Example 1: ethanoic acid from chloromethane, two routes

Now, remember our puzzle from the start? How is ethanoic acid obtained from chloromethane, by two different routes? Pause the video and try it first. Done? Okay, let's check. Chloromethane is C H three C l, one carbon. In route one, potassium cyanide gives C H three C N, ethanenitrile, and H three O plus with heat hydrolyses it to C H three COOH, ethanoic acid, with two carbons. Okay? In route two, magnesium in dry ether gives C H three M g C l, methylmagnesium chloride, our Grignard reagent. Then carbon dioxide, and then H three O plus, give C H three COOH, ethanoic acid again! Right? So, puzzle solved! The extra carbon came from C N minus, the cyanide ion, in route one, and from carbon dioxide in route two. One carbon in, two carbons out. See how neat that is?

5:45Example 2: nitrile or Grignard route, which and why

Now a harder one, JEE level. Which route, nitrile or Grignard, would you use to convert, in part A, two chloro two methylpropane into two, two dimethylpropanoic acid, and in part B, three bromopropan one ol into four hydroxybutanoic acid? And why? Pause the video and try it first. Done? Let's check part A first. The halide has four carbons, and the acid has five, so we need one carbon more. Easy, potassium cyanide, right? Will the nitrile route work here? Oh no, careful, that is the trap here! The halide is C H three in brackets, the small three outside meaning three methyl groups, then C, dash, C l, so the chlorine is on a tertiary carbon. With C N minus, a tertiary halide gives elimination, not the nitrile. Okay? So we use the Grignard route. The Grignard reagent, C H three in brackets three, C M g C l, first takes up carbon dioxide, and then H three O plus gives C H three in brackets three, CCOOH, two, two dimethylpropanoic acid. Right? Now, part B: three bromopropan one ol is HOCH two C H two C H two B r, three carbons, and the acid needs four. Look at the left end of the molecule. Can we make a Grignard reagent from this halide? No, we cannot, because there is an O H group! Remember, a Grignard reagent cannot be made in the presence of an O H group. So the Grignard route fails. Clear? So the nitrile route wins. Cyanide replaces the bromine, giving H O, C H two in brackets three, C N, okay? Then H three O plus hydrolyses it to H O, C H two in brackets three, COOH, four hydroxybutanoic acid. So part A takes the Grignard route, and part B the nitrile route. Got it?

7:45Quick revision: oxidation, nitrile and Grignard routes

So, quick revision. Oxidise primary alcohols, aldehydes and alkylbenzenes, where the benzylic carbon needs a hydrogen. For one carbon more, use the nitrile route, or a Grignard reagent with carbon dioxide. And a tertiary halide takes the Grignard route, while an O H group means the nitrile route. Now go and try the practice problems on this page. All the best!