Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two different hydrocarbons share the formula C six H twelve. Put each one through reductive ozonolysis, and each gives just one product. One product reduces Tollens' reagent, and the other does not. So which two hydrocarbons are they? You will crack it by the end of this video. Today we see how aldehydes and ketones are made, all on one map. It comes up often in JEE, so let's begin!

0:33The whole chapter on one map

So, here is the whole chapter. The box in the centre says aldehyde or ketone, and both are built on the carbonyl group, shown on screen: carbon joined to oxygen by a double bond, the two equal lines; the little angle shows the carbon's other two bonds. On the left, ten arrows point in, showing how to make aldehydes and ketones. On the right are their reactions, for the next chapter. With me? Now, how do we tell them apart? Look at C H three, C O, C H three. Which one is it? It is a ketone, yes, exactly! In an aldehyde, the carbonyl carbon carries at least one hydrogen, as in RCHO, okay? In a ketone it carries two carbon groups, RCOR prime, the small raised dash, the prime, meaning the second group may differ.

1:21Both, from alcohols

Right? Now, the left side. One with the small raised circle, the degree sign, means a primary alcohol, and two means secondary. With PCC, pyridinium chlorochromate, a primary alcohol gives an aldehyde. With C r O three, chromium trioxide, a secondary alcohol gives a ketone. Okay? Why not the usual acidified potassium dichromate for the aldehyde? Because it does not stop at the aldehyde! In water, the aldehyde goes on to the acid, unless we distil it out as it forms. On screen: a primary alcohol with K two C r two O seven, slash, H plus, gives the acid. The slash means with, and H plus means acid. So for an aldehyde, use PCC, clear? Now the second box on the map: pass the alcohol vapour over copper at five hundred and seventy three kelvin. Hydrogen is removed, and as the arrows show, a primary alcohol gives an aldehyde, and a secondary one a ketone. With me?

2:19Both, from alkenes and alkynes

Next, hydrocarbons. In reductive ozonolysis, an alkene is treated with ozone, O three, then with Z n slash H two O, zinc and water. The double bond is cut, and each end becomes a carbonyl group. Right? The equals sign in front marks the old double bond. An end C H two gives methanal, HCHO, while an end CHR gives an aldehyde, RCHO, and an end C R two gives a ketone, R two C O, okay? Keep this rule, because it solves our puzzle! Now alkynes. Add water, with mercuric sulphate, H g S O four, and dilute sulphuric acid. So, what does propyne give, propanal or propanone? It gives propanone, a ketone! Water adds the Markovnikov way, so the O H goes to the middle carbon, the more substituted one, and the enol formed turns into the ketone. Only ethyne gives an aldehyde, ethanal; every other alkyne gives a ketone. That is a common trap. Got it?

3:19Aldehydes only

Now, methods for aldehydes only. First, the Rosenmund reduction: an acid chloride, RCOC l, with hydrogen, over palladium on barium sulphate, P d slash B a S O four. Okay? Now, here is a question for you: why put the palladium on barium sulphate? Barium sulphate poisons the palladium, yes, exactly! So the reduction stops at the aldehyde, RCHO, and goes no further. Clear? For nitriles, RCN, there is the Stephen reaction: stannous chloride, S n C l two, with H C l, then hydrolysis with H three O plus, gives the aldehyde. Or use DIBAL H, diisobutyl aluminium hydride, then water. With me? For aromatic aldehydes, the Etard reaction. Ar C H three, where Ar means an aryl group, such as a benzene ring, with chromyl chloride, C r O two C l two, then H three O plus, gives Ar CHO, the aldehyde. Okay? And the Gattermann Koch reaction: an arene, Ar H, with carbon monoxide and H C l, over anhydrous aluminium chloride and cuprous chloride, also gives the aromatic aldehyde. Got it?

4:28Ketones only

Now, methods for ketones only. Take an acid chloride with a dialkylcadmium, R two C d. Two R prime COC l, plus R two C d, gives two R prime COR, the ketone, plus C d C l two. Okay? But a Grignard reagent, R M g X, also carries an R group. So why not use it directly? Because it does not stop! It attacks the ketone too, and goes on to a tertiary alcohol, three with the degree sign. The cadmium reagent is mild and leaves the ketone alone. That is the trap, right? Second, a nitrile with a Grignard reagent, then hydrolysis: RCN plus R prime M g X gives the ketone RCOR prime. Clear? And for aryl ketones, Friedel Crafts acylation: an arene, Ar H, with an acid chloride and anhydrous aluminium chloride, A l C l three. With me?

5:17One peek at the right side

Before the examples, one peek at the right side. Every aldehyde, aliphatic or aromatic, reduces Tollens' reagent and is itself oxidised to the acid, but simple ketones do not. Keep this for our puzzle, okay?

5:32Example 1: oxidising butanols and cyclohexanol

Now Example one, an easy one, from the concept Methods Common to Aldehydes and Ketones. Name the carbonyl compound formed from, first, butan one ol with PCC; second, butan two ol with C r O three; and third, cyclohexanol over copper at five hundred and seventy three kelvin. Pause the video and try it first. Done? Okay, let's check. Butan one ol has its O H on the end carbon, so it is primary, and PCC gives butanal. Butan two ol is secondary, so C r O three gives the ketone, butanone. Right? Now part c. Cyclohexanol, over hot copper: does it give an aldehyde or a ketone? It gives a ketone! The ring carbon bearing the O H is joined to two other carbons, so cyclohexanol is secondary, and we get cyclohexanone. So do not write an aldehyde here. Clear?

6:26Example 2: identify two C6H12 alkenes by ozonolysis

Now Example two, a harder one from the same concept: our opening puzzle! One hydrocarbon, C six H twelve, gives on reductive ozonolysis only one product, which does not reduce Tollens' reagent. Then a second isomer gives only one product, which does reduce it. Identify the two hydrocarbons. Pause the video and try it first. Done? Okay, let's check. Ozonolysis cuts the double bond into two halves. If there is only one product, what does that tell you about the halves? They must be the same, yes, exactly! The alkene is symmetrical, so its six carbons split three plus three, and each product has three carbons. Okay? First hydrocarbon: its product does not reduce Tollens' reagent, so it is a ketone with three carbons, propanone. Use the ozonolysis rule backwards: join two propanone halves by a double bond. C H three in brackets, with a small two, sits on each side of C double bond C: two three dimethylbut two ene. Right? Second hydrocarbon: its product reduces Tollens' reagent, so it is an aldehyde, propanal. Join two propanal halves the same way, and we get an ethyl group on each carbon of the double bond, which is hex three ene. Puzzle solved: same formula, but one gives a ketone and the other an aldehyde. Clear?

7:47Quick revision: routes to both, aldehydes only, ketones only

So, quick revision. Both come from alcohols, with PCC, C r O three or hot copper, and from ozonolysis and alkyne hydration. Aldehydes only: the Rosenmund, Stephen, DIBAL H, Etard and Gattermann Koch methods. Ketones only: R two C d, a nitrile with R M g X, and Friedel Crafts acylation. Now go and try the practice problems on this page. All the best!