Methods Common to Aldehydes and Ketones

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two hydrocarbons have the same formula, C six H twelve. We cut each one with ozone, and each gives only one product. But the first product does not reduce Tollens' reagent, and the second one does. Same formula, same reaction, so why the difference? We will solve it at the end. Today we learn the methods that make both aldehydes and ketones. They come up often in JEE, so let's begin.

0:32From alcohols: oxidation and dehydrogenation

So, road number one: oxidising an alcohol. The number one with a small circle, the degree sign, means a primary alcohol. Over the arrow are the reagents: PCC, pyridinium chlorochromate, or C r O three with no water, which is what anhydrous means. The small three written low counts atoms. And we get an aldehyde. Okay? But why such mild reagents? What happens if we use a strong oxidant in water instead? It does not stop at the aldehyde, exactly! Acidified potassium dichromate, K two C r two O seven, carries it on to the carboxylic acid. So with this reagent, the aldehyde must be distilled out as it forms, since it boils lower than the alcohol. Right? Now the two with the degree sign, a secondary alcohol. All three reagents work here: C r O three, PCC, or acidified dichromate, where slash H plus, the small plus written high, means with acid. Each one gives a ketone. Clear? Road number two is dehydrogenation, which means removing hydrogen. Pass the alcohol vapour over copper, C u, or silver, A g, at five hundred and seventy three kelvin, the capital K on screen. Primary gives an aldehyde, secondary a ketone. With me? And this is the better method for volatile aldehydes, the ones that evaporate easily. Got it?

1:59From alkenes: ozonolysis and the Wacker process

Now road number three: alkenes. First ozone, O three, then zinc and water, Z n slash H two O, or dimethyl sulphide instead. This is reductive ozonolysis. Okay? The rule is really simple. Split the carbon carbon double bond, the equals sign, and put an oxygen on each end. Right? So let me ask you this. One end of the double bond carries one hydrogen and one R group, where R stands for an alkyl group. After the split, is that end an aldehyde or a ketone? It becomes an aldehyde, yes, exactly! Equals C H two gives HCHO, methanal. Equals CHR gives RCHO, an aldehyde. And equals C R two gives R two C O, a ketone. Clear? And to make ethanal from ethene, we use the Wacker process. Two C H two double bond C H two, ethene, plus O two, over P d C l two and C u C l two in water, gives two C H three CHO, ethanal. Got it?

3:06From alkynes: hydration

Road number four: alkynes. H C, three lines, C H: the three lines are a triple bond, so this is ethyne. Add water with H g S O four and dilute sulphuric acid, H two S O four. We first get an enol, with an O H on a double bonded carbon, and it rearranges into ethanal. Okay? Now, a classic trap. Take propyne under the same conditions. Do we get propanal, or propanone? We get propanone, very good! Every other alkyne gives a ketone. For a terminal alkyne like propyne, that is because water adds in the Markovnikov way. Only ethyne gives an aldehyde. Right? Now the enol from propyne. The O H goes to the more substituted middle carbon: C H three, single bond, C, O H in brackets, double bond C H two. The small dash is the single bond, and the brackets mean the O H hangs on that carbon. It rearranges into C H three COCH three, propanone. So if an option says propanal, that is the trap. Clear?

4:11Two more: geminal dihalides and calcium salts

Road number five: geminal dihalides, with both halogen atoms on the same carbon. Hydrolysis turns RCHC l two into RCHO, an aldehyde, and R two C C l two into R two C O, a ketone. Okay? Road number six: dry distillation of calcium salts. The brackets with a small two outside mean two RCOO groups on one calcium, C a. On heating, we get RCOR, a ketone, plus C a C O three, calcium carbonate. With me? Here is a question for you. In calcium methanoate, each R is just a hydrogen atom. So what will it give on its own? It gives methanal, exactly! Put H for both R groups, and RCOR becomes HCOH, which is methanal. And a mixture of another calcium salt with calcium methanoate gives an aldehyde.

5:02Example 1: reductive ozonolysis products of three alkenes

Clear? Time for Example one. Name the products of reductive ozonolysis of part a, but two ene, part b, two methylpropene, and part c, two methylbut two ene. Pause the video and try it first. Done? Okay, let's check. But two ene is C H three C H, double bond, CHCH three. Split it, and each half gets an oxygen: C H three CHO, ethanal. Both halves are the same, so two moles of ethanal. Easy, isn't it? Now part b, two methylpropene. One carbon of the double bond carries two methyl groups, the other is a C H two. Which end gives the ketone? The end with two methyl groups, yes! It has no hydrogen, so it gives propanone, C H three in brackets, two, then C O, a ketone. The C H two end gives HCHO, methanal. Got it? And part c, two methylbut two ene, is a mix of the two. The end with two methyl groups gives propanone again, and the CHCH three end gives ethanal. Clear?

6:12Example 2: two C6H12 isomers, ozonolysis and Tollens test

Now Example two, a harder one: our puzzle from the start. One hydrocarbon, C six H twelve, gives only one product on reductive ozonolysis, and it does not reduce Tollens' reagent. Its isomer also gives only one product, but this one does reduce it. Identify the two hydrocarbons. Pause the video and try it first. Done? Okay, let's decode it. Ozonolysis splits the molecule into two halves. So what does only one product tell you? Both halves must be the same, exactly! Six carbons split equally gives three carbons in each half. Okay? Now the Tollens' clue. An aldehyde reduces Tollens' reagent; a ketone does not. So the first product is a ketone, and the second is an aldehyde. Right? Three carbons and a ketone: C H three in brackets, two, C O, propanone. Join two of them through their carbonyl carbons, a double bond in place of the two oxygens; the double arrow points to this alkene. C H three in brackets two, C, double bond, C, C H three in brackets two, which is two three dimethylbut two ene. With me? For the second, three carbons and an aldehyde: C H three C H two CHO, propanal. Joining two of them gives hex three ene, C H three C H two C H, double bond, CHCH two C H three. So the puzzle is solved: one alkene gives a ketone, the other an aldehyde. That's it!

7:45Quick revision: alcohols, alkenes, alkynes and other methods

So, quick revision. Alcohols: primary gives an aldehyde with PCC, secondary a ketone, or pass the vapour over copper at five hundred and seventy three kelvin. Alkenes: ozone, then zinc and water, and an oxygen on each end of the double bond. Alkynes: with H g S O four and dilute sulphuric acid, only ethyne gives an aldehyde. Also remember the Wacker process, geminal dihalides and calcium salts. Now try the four practice problems on this page. All the best!