Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two routes in this chapter do the same job: each turns an alkyl halide into an acid with one carbon more, one through cyanide, one through a Grignard reagent. So why learn two routes for one job? Keep that question in mind; by the end, you will know. This chapter comes up often in JEE, so let's begin!

0:26The whole chapter on one map

So, here is the whole chapter on one page. In the centre sits the carboxylic acid, RCOOH, where R is an alkyl or aryl group and the dash is a bond. Seven arrows point in from the left, the ways to make it, and ten point out to the right, its reactions. With me?

0:43Making acids by oxidation

First, top left: a primary alcohol, written as one with a small raised circle, the degree sign, or an aldehyde, is oxidised by K M n O four, or by K two C r two O seven with H plus. The small low numbers count atoms, the slash means with, and the raised plus is a charge. Okay? Next box: an alkylbenzene, a benzene ring carrying an alkyl group, heated with K M n O four and KOH, then H three O plus, dilute acid. So take ethylbenzene, with two carbons in its side chain. What acid do you get? Benzoic acid, yes, exactly! The arrow means becomes: the whole side chain, whatever its length, becomes one COOH group. But the benzylic carbon, the one joined to the ring, needs at least one hydrogen. Right? And the bottom box: an alkene or alkyne, heated with K M n O four, is split at the multiple bond, which we call cleavage, and acids are formed; but an alkene carbon carrying no hydrogen gives a ketone instead.

1:48Adding one carbon, and other routes in

Clear? Now the routes that change the chain. An alkyl halide, R X, with X a halogen, reacts with potassium cyanide, KCN, to give the nitrile, RCN, and heating with H three O plus turns it into the acid; each reagent sits above its arrow. Got it? The second route: the Grignard reagent, R M g X, with carbon dioxide, C O two, as dry ice, then dilute acid. Now count carbons. Compared with the halide, how many more carbons does the acid have, in each route? One more in both, exactly! The new carbon comes from cyanide in one route, and from carbon dioxide in the other. Keep this pair in mind for Example two. Okay? Next, simple hydrolysis: an acid chloride or an anhydride reacts with water, H two O, and gives back the acid. An ester needs help: dilute acid, H plus, or alkali, O H minus, followed by acid. With me? And the last line: a methyl ketone with N a O X, a sodium hypohalite. This is the haloform reaction, and the acid has one carbon fewer than the ketone.

2:55Reactions: salts, esters, acid chlorides

Right? Now the right side. The acid forms a salt with sodium, sodium hydroxide, N a O H, or sodium hydrogencarbonate, N a HCO three, and with the hydrogencarbonate, carbon dioxide fizzes out. Phenol is acidic too. Does phenol release carbon dioxide from sodium hydrogencarbonate? No, it does not, because phenols are weaker acids; they give no carbon dioxide, except some nitrophenols with several nitro groups. So this fizz tells an acid from a phenol. Clear? Next, an alcohol, R prime O H, with concentrated sulphuric acid, H two S O four, gives an ester; the small raised dash is called prime. And note the word reversible: this is an equilibrium. Got it? With P C l five, P C l three, or thionyl chloride, SOC l two, the acid gives an acid chloride. Remember this arrow for Example one. Okay? Then ammonia, N H three, followed by heat, gives the amide; and heating with P two O five gives the anhydride. With me?

3:57Reduction, decarboxylation, HVZ and the ring

Next, reduction. Sodium borohydride, N a B H four, reduces aldehydes and ketones. So will it reduce a carboxylic acid to an alcohol? No, it will not, and that is a classic trap! We need lithium aluminium hydride, L i A l H four, or diborane, B two H six, to get the primary alcohol. Right? Heating the sodium salt with soda lime, N a O H with C a O, removes carbon dioxide and gives the hydrocarbon, R H, an alkane when R is alkyl; this is decarboxylation. Kolbe electrolysis of the salt joins two R groups instead. Clear? Then the Hell-Volhard-Zelinsky reaction, HVZ for short. X two, chlorine or bromine, with a little red phosphorus, puts a halogen on the alpha carbon, the one next to COOH; alpha is the Greek letter. So the acid needs an alpha hydrogen. Got it? And last, an aromatic acid, A r COOH, where A r is an aryl group. The ring is substituted at the meta position, and there is no Friedel-Crafts reaction. With me?

5:05Example 1: ethanoic acid with LiAlH4, PCl5, then alc. KOH

Time for Example one, asked in JEE Main twenty fourteen, from the concept AIEEE and Earlier JEE Main Questions. Ethanoic acid with L i A l H four gives A, then P C l five gives B, and alcoholic KOH gives the final product C, okay? Pause the video and try it first. Done? Okay, let's check. From our map, lithium aluminium hydride reduces the acid to a primary alcohol. So compound A is ethanol, C H three C H two O H, with me? Now P C l five. On our map, P C l five turns an acid into an acid chloride. So B must be ethanoyl chloride, right? No, no, wait, that is the trap! Compound A is an alcohol now, not an acid. P C l five turns an alcohol into an alkyl chloride, so B is chloroethane, C H three C H two C l. Always check what the reagent acts on. Clear? Finally, alcoholic KOH removes H C l from chloroethane, an elimination. So C is ethene, C H two double bond C H two; the equals sign inside the formula is the double bond.

6:19Example 2: nitrile or Grignard route to the acid, and why

Got it? Now Example two, a harder one, from the concept Preparation of Carboxylic Acids. It is our opening puzzle! Which route, nitrile or Grignard, would you use to convert two chloro two methylpropane into two two dimethylpropanoic acid, and three bromopropan one ol into four hydroxybutanoic acid? And why? Pause the video and try it first. Done? Okay, let's check. In the first part, C H three in brackets, with a small three outside, means three methyl groups, all on the carbon carrying C l, so the halide is tertiary, and the acid has COOH where the chlorine was. So, can cyanide replace the chlorine on a tertiary carbon? No, it cannot, because cyanide with a tertiary halide gives elimination instead. So the nitrile route fails here, and we use the Grignard route. Okay? In the second part, the chain is H O, three C H two groups, then B r, and the acid simply ends in COOH instead. Notice the O H group. Can we make a Grignard reagent from this bromide? No, we cannot, and that is the second trap! You cannot make a Grignard reagent while an O H group is present, so the nitrile route wins here. Same job, but the structure decides the route. And that is our puzzle, solved!

7:42Quick revision: making acids, key reactions, LiAlH4, HVZ

So, quick revision. Make acids by oxidation, by hydrolysis, or by adding one carbon through KCN or a Grignard reagent with carbon dioxide. Acids give salts, esters, acid chlorides and amides, and lithium aluminium hydride, not sodium borohydride, reduces them. Soda lime removes carbon dioxide, and HVZ needs an alpha hydrogen. Now try the practice problems in the next concepts. All the best!