oxidation of aldehydes and ketones

For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulphuric acid, you get propanone formed. Tollens’ reagent contains the diamminesilver(I) ion, Ag(NH3)2+. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate. To carry out the test, you add a few drops of the aldehyde or ketone to the freshly prepared reagent, and warm gently in a hot water bath for a few minutes.

Author: Stewart Hird

• Alternatively, you could write separate equations for the two stages of the reaction – the formation of ethanal and then its subsequent oxidation.
• You need to produce enough of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) to be able to test them.
• If nothing happens in the cold, the mixture is warmed gently for a couple of minutes – for example, in a beaker of hot water.
• The support material then goes on to say “The equations for their formation are not too difficult.” Does that mean that you have to know them?
• In testing, first you have to show that you have an aldehyde or ketone.

You need to use an excess of the oxidising agent and make sure that the aldehyde formed as the half-way product stays in the mixture. The full equation for this reaction is fairly complicated, and you need to understand about electron-half-equations in order to work it out. If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, CH3CHO. At the time of writing, no question involving these equations had been asked. Whether or not you find the equations difficult depends on how much time you have spent learning how to write equations from electron-half-equations. If you look at the equations on the page you have read, and find them scary, ignore them!

Reactions of Ions in Aqueous Solution

• If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, CH3CHO.
• In this case, there is no such hydrogen – and the reaction has nowhere further to go.
• The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde.

As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Methanal is such a powerful reducing agent that the copper(II) ions may be reduced to metallic copper – often seen as a very nice copper mirror on the tube. Because I am going to be using electron-half-equations quite a lot on the rest of this page, it would definitely be worth following this link if you aren’t happy about them. You will find details of these reactions further down the page. You are expected to know all of the above methods which can be used to distinguish between an aldehyde and a ketone! However, Tollens’ reagent is the most commonly used method, if trying to identify an unknown sample for example. That is particularly important with the ketone phenylethanone, or any other ketone where the carbonyl group is on a side chain attached to a benzene ring.

Why do aldehydes and ketones behave differently?

If you are studying a UK-based syllabus and haven’t got any of these things, follow this link to find out how to get them. You can draw simple structures to show the relationship between the primary alcohol and the aldehyde formed. Potassium manganate(VII) is a powerful enough oxidising agent to break carbon-carbon bonds in ketones, and so you wouldn’t get a reliable result. You will remember that the difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. These half-equations are then combined with the half-equations from whatever oxidising agent you are using.

Amino acids, Proteins & DNA

In testing, first you have to show that you have an aldehyde or ketone. Then you can do one of these tests to find out which you have got. On the other hand, ketones are resistant to oxidation by all of them. These half-equations are then combined with the half-equations from whatever oxidizing agent you are using.

The half-equation for the oxidation of the aldehyde obviously varies depending on whether you are doing the reaction under acidic or alkaline conditions. The problem is that what is important in using these reactions as tests is the colour change in the oxidising agent. In this particular reaction, you have to explain, for example, why the solution turns green. Any equation pin up casino that you write has got to show the production of the chromium(III) ions. Tertiary alcohols don’t have a hydrogen atom attached to that carbon.

Aldehydes & Ketones

A small amount of potassium dichromate(VI) solution is acidified with dilute sulphuric acid and a few drops of the aldehyde or ketone are added. If nothing happens in the cold, the mixture is warmed gently for a couple of minutes – for example, in a beaker of hot water. You need to produce enough of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) to be able to test them. There are various things which aldehydes do which ketones don’t. These include the reactions with Tollens’ reagent, Fehling’s solution and Benedict’s solution, and are covered on a separate page. You would then add a few drops of the alcohol to a test tube containing potassium dichromate(VI) solution acidified with dilute sulphuric acid.

Tertiary alcohols aren’t oxidised by acidified sodium or potassium dichromate(VI) solution. This statement is about distinguishing between aldehydes and ketones using various oxidising agents. In the case of a primary or secondary alcohol, the orange solution turns green. The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde.

Oxidation of Aldehydes

In my experience, these tests can be a bit of a bother to carry out and the results aren’t always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff’s reagent. Schiff’s reagent isn’t specifically mentioned by any of the UK-based syllabuses, but I have always used it. First you have to be sure that you have actually got an alcohol by testing for the -OH group. You would need to show that it was a neutral liquid, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes. Playing around with the reaction conditions makes no difference whatsoever to the product.

In turn the aldehyde is oxidised to the corresponding carboxylic acid. The alcohol is heated under reflux with an excess of the oxidising agent. When the reaction is complete, the carboxylic acid is distilled off. The syllabus asks that you should be able to work out whether you have an aldehyde or a ketone from the results of these tests. That doesn’t imply any need to know the equations of the reactions. Although you can oxidise aldehydes with potassium manganate(VII) solution, you wouldn’t choose to use it to distinguish between aldehydes and ketones.

Under acidic conditions, the aldehyde is oxidized to a carboxylic acid. Aldehydes reduce the complexed copper(II) ion to copper(I) oxide. Because the solution is alkaline, the aldehyde itself is oxidised to a salt of the corresponding carboxylic acid. In this case, there is no such hydrogen – and the reaction has nowhere further to go. If you need to work out the equations for these reactions, the only reliable way of building them is to use electron-half-equations.

Even if you don’t find them scary, I wouldn’t spend too much time on them. The support material then goes on to say “The equations for their formation are not too difficult.” Does that mean that you have to know them? In each of the following examples, we are assuming that you know that you have either an aldehyde or a ketone. There are lots of other things which could also give positive results.