Reactions of Aldehydes and Ketones

Why do aldehydes and ketones react?

You should remember from Aldehydes and Ketones that they both contain the carbonyl functional group , $\mathrm{C}=\mathrm{O}$. This is a carbon atom joined to an oxygen atom by a double bond. Let's take a closer look at it.

If we compare the electronegativities of carbon and oxygen, we can see that oxygen is a lot more electronegative than carbon.

Electronegativity values of some common elements

As explored in Polarity, electronegativity is an atom's ability to attract a bonding pair of electrons towards itself. This means that the oxygen atom in the C = O bond pulls on the four bonding electrons that it shares with carbon quite strongly, tugging them closer. This makes the oxygen atom partially negatively charged and leaves the electron-deficient carbon atom partially positively charged. We can show this using the delta symbol , δ .

The carbonyl bond, shown here in methanal

The carbonyl bond is now polar. We can say that it has a permanent dipole moment. This makes it very attractive to certain molecules called nucleophiles, which we'll look at in just a second.

We can also see that aldehydes and ketones are unsaturated compounds - they contain a double bond. This means that they easily take part in addition reactions. If we put this all together, we can conclude that aldehydes and ketones often react with nucleophiles using nucleophilic addition. However, they also take part in other sorts of reactions such as oxidation and reduction, which we'll explore later.

Nucleophilic reactions of aldehydes and ketones

As mentioned above, aldehydes and ketones react with nucleophiles in nucleophilic addition reactions. Before we look at how, let's first remind ourselves of what a nucleophile is.

What is a nucleophile?

Oppositely charged particles attract one another, so nucleophiles are attracted to positively charged areas such as the carbon atom found in the carbonyl group. Some common examples include:

• The cyanide ion, \(CN^-\)
• The hydride ion, \(H^-\)
• Water \(H_2O\)
• Ammonia, \(NH_3\)

In this article we'll focus on reactions involving the first two molecules: the cyanide and hydride ions.

Reaction with hydrogen cyanide

Aldehydes and ketones react with hydrogen cyanide and dilute hydrochloric acid to form hydroxynitriles, which are also known as cyanohydrins. This is an important reaction because hydroxynitriles contain both a hydroxyl (\(-OH\)) and a nitrile ( $\mathrm{C}\equiv \mathrm{N}$) functional group, making them relatively reactive, meaning they are easily turned into other compounds. For example, we can use hydroxynitriles to make synthetic versions of amino acids.

Let's look at the general reaction between an aldehyde and cyanide ion.

The general reaction between an aldehyde and the cyanide ion

1. The negative cyanide ion is attracted to the partially positively charged carbon in the C=O bond. It forms a bond with this carbon and forces the C=O double bond to break, transferring one of the bonding pairs of electrons to the oxygen atom.
2. The oxygen atom is now a negatively charged ion with a lone pair of electrons. It reacts with a positive hydrogen ion from the solution to form a hydroxyl functional group, -OH.
3. The end product is a hydroxynitrile.

The reaction has the following general equation: \(RCHO + HCN \rightarrow RCH(OH)CN\)

To see how ketones react with cyanide ions, we simply replace the hydrogen atom bonded to the carbonyl group with another organic R group, as you can see in the following mechanism:

Nucleophilic addition of a ketone

Let's now apply that to a named aldehyde, ethanal.

Nucleophilic addition of ethanal using the cyanide ion

Because the product of this reaction contains two functional groups, both the nitrile group and the hydroxyl group, we use the prefix hydroxy and the suffix nitrile to name the molecule. The carbon atom attached to the nitrogen atom is always labeled carbon. 1. Here we can see that the alcohol functional group is joined to carbon 2, and so we call this molecule 2-hydroxypropanenitrile.

2-hydroxypropanenitrile

We can also react potassium cyanide with the ketone butanone, as shown below:

Nucleophilic addition of butanone using the cyanide ion

Look at the product. It still has a chain that is four carbon atoms long. But counting the carbon attached to the nitrogen as carbon 1, you can see that there is now both a methyl group and a hydroxyl group joined to carbon 2. We therefore name this molecule 2-hydroxy-2-methylbutanenitrile.

2-hydroxy-2-methylbutanenitrile

Isomeric products of nucleophilic addition

Let's look more closely at the previous example. If we react potassium cyanide with butanone, we can actually get two different products. This is because the cyanide ion can attack from either above or below the planar molecule. Both mechanisms are shown below for comparison.

Second mechanism for the nucleophilic addition of butanone

Second mechanism for the nucleophilic addition of butanone

Look at the two products. They are optical isomers of each other.

Optical isomers are also known as enantiomers. They are produced because the cyanide ion can attack from above or below the plane, as shown in the mechanisms above. The ion is equally as likely to attack from above as below, so the reaction produces a 50:50 mixture of the two isomers, known as a racemic mixture.

The two isomeric products of the reaction between butanone and the cyanide ion. They are optical isomers, known as enantiomers

Reduction of aldehydes and ketones

When you react aldehydes and ketones with a reducing agent which supplies the hydride ion $:{\mathrm{H}}^{-}$, you get an alcohol. This is also a type of nucleophilic addition reaction but is more often known as reduction.

Common reducing agents include sodium tetrahydridoborate (III) in an aqueous solution. That is a bit of a mouthful to remember but it is actually just the molecule ${\mathrm{NaBH}}_4$. It is also known as sodium borohydride.

Sodium borohydride

Reducing an aldehyde produces a primary alcohol and reducing a ketone produces a secondary alcohol. You might have realised that this is the opposite reaction to oxidisng alcohols, as explored in Oxidation of Alcohols. Oxidising a primary alcohol produces an aldehyde whilst oxidising a secondary alcohol produces a ketone.

Let's take a look at what happens when we reduce the aldehyde ethanal. It is very similar to the nucleophilic addition mechanism we looked at earlier.

The reduction of ethanal

1. The hydride ion acts as a nucleophile and attacks the partially positive carbon atom in the C=O bond. The C=O bond breaks and one of the bonding pairs of electrons is transferred to the oxygen atom, forming a negative oxygen ion with a spare pair of electrons.
2. The oxygen atom’s spare pair of electrons is attracted to a positive hydrogen ion from the solution and forms the hydroxyl functional group, -OH.
3. The product is a primary alcohol.

We can represent this reaction using [H] to show the reducing agent:

$$ RCHO + 2[H] \rightarrow RCH_2OH $$

The general equation for the reduction of a ketone is given below. Notice how it has two R groups attached to the $\mathrm{C}=\mathrm{O}$ carbon:

$$ RCOR' + 2[H] \rightarrow RCH(OH)R' $$

Isomeric products of reduction

Reducing a ketone can produce optical isomers in the same way as earlier in the article. However, reducing aldehydes does not produce optical isomers. This is because reduction of an aldehyde forms a primary alcohol. Primary alcohols have at least two hydrogen atoms bonded to the carbon containing the -OH group. To be a chiral centre, a molecule must have four different groups bonded to a carbon atom and here this clearly isn't the case. Therefore, primary alcohols don't show optical isomerism.

A primary alcohol

Oxidation of aldehydes

One way of creating aldehydes in the lab is by heating a primary alcohol with acidified potassium dichromate (VI). This oxidises the alcohol. If you distill the product off you get an aldehyde, but if you instead use reflux, the aldehyde will react further. It is oxidised again and forms a carboxylic acid.

For example, oxidising ethanal produces ethanoic acid.

$$ CH_3CHO + [O] \rightarrow CH_3COOH $$

Ketones don't react in this way - they can't be oxidised further because an oxidation reaction would involve breaking one of their strong CC bonds. This fact can be used to tell aldehydes and ketones apart with a number of different reagents.

Potassium dichromate (VI)

Oxidising an aldehyde with potassium dichromate (VI) turns the solution from orange to green. However, there will be no colour change with a ketone.

The reaction of acidified potassium dichromate (VI) with a ketone (left) and an aldehyde (right)

Tollens' reagent

Tollens' reagent is also known as the silver mirror test. It contains the diaminesilver (I) ion, $[\mathrm{Ag}{\left(\mathrm{NH}3\right)}_2{]}^{+}$. If you warm a ketone with Tollens' reagent, the solution will remain colourless, but adding an aldehyde will form a metallic silver mirror deposit on the sides of the test tube.

The reaction between Tollens' reagent and a ketone (left) and an aldehyde (right)

Fehling's test

Fehling's solution is made up of two other solutions called Fehling's A and Fehling's B. The solution is naturally blue. If you heat it with an aldehyde, you'll form a brick red precipitate. This tends to turn the whole solution red. However, warming with a ketone gives no colour change.

Reacting Fehling's solution with a ketone (left) and an aldehyde (right)

The following table should help summarise your knowledge of the different oxidation reactions you can use to tell aldehydes and ketones apart.

Testing for aldehydes and ketones

Distinguishing between molecules

Potassium dichromate is a much stronger oxidising agent than Tollens' reagent and Fehling's solution. It is also able to oxidise alcohols, as we mentioned earlier. This means it can help us distinguish between alcohols, aldehydes, and ketones.

Add three drops of an unknown aldehyde, ketone, or alcohol to a test tube that contains $2{\mathrm{cm}}^3$ acidified potassium dichromate (VI). Warm gently in a water bath and observe any changes.

• If the solution remains orange, your unknown compound is a ketone.
• If the solution turns green, it could be either an alcohol or an aldehyde.

Test your unknown compound further by adding three drops to $2{\mathrm{cm}}^3$ Tollens' reagent or Fehling's solution. In the presence of an aldehyde, Tollens' reagent will form a metallic silver mirror on the sides of the test tube and Fehling's solution will go from blue to brick red. However, these oxidising agents aren't strong enough to oxidise an alcohol so you won't see any change.