Welcome to the chemistry of the carbonyl group. In this topic, we will look at how the polar $C=O$ bond dictates the reactions of aldehydes and ketones, expanding on what you learned at AS-level.

What you'll learn:

• How to use chemical tests like Tollens' reagent and Fehling's solution to distinguish between aldehydes and ketones.
• How to write equations and draw mechanisms for the reduction of carbonyls to alcohols.
• How carbonyls react with $KCN$ to form hydroxynitriles, and the hazards involved.
• Why these addition reactions often produce a racemic mixture of optical isomers.

The carbonyl group

Both aldehydes and ketones contain the carbonyl group, $C=O$.

Oxygen is much more electronegative than carbon, which draws electron density away from the carbon atom. This leaves the carbon with a partial positive charge (${\delta }^{+}$) and the oxygen with a partial negative charge (${\delta }^{-}$). Because of this electron-deficient carbon, carbonyl compounds are prime targets for attack by nucleophiles (electron-pair donors).

Distinguishing aldehydes from ketones

Because aldehydes have a hydrogen atom attached to the carbonyl carbon ($RCHO$), they can be easily oxidised to form carboxylic acids ($RCOOH$). Ketones lack this hydrogen and cannot be oxidised under normal conditions.

We can use mild oxidising agents to chemically distinguish between the two. You need to know two specific chemical tests.

Tollens' reagent

Tollens' reagent is a colourless solution of ammoniacal silver nitrate, containing the complex ion $[Ag({NH}_3{)}_2{]}^{+}$.

• When warmed with an aldehyde, the aldehyde is oxidised to a carboxylic acid (or rather, a carboxylate ion in alkaline conditions). Simultaneously, the silver ions (${Ag}^{+}$) are reduced to solid silver atoms ($Ag$). This forms a distinctive "silver mirror" on the inside of the test tube.
• When warmed with a ketone, no reaction occurs and the solution remains colourless.

Fehling's solution

Fehling's solution is a blue alkaline solution containing copper(II) ions (${Cu}^{2+}$).

• When warmed with an aldehyde, the aldehyde is oxidised to a carboxylic acid, and the blue ${Cu}^{2+}$ ions are reduced to copper(I) oxide (${Cu}_{2}O$), which forms a brick-red precipitate.
• When warmed with a ketone, no reaction occurs and the solution remains blue.

Reduction of aldehydes and ketones

Just as primary and secondary alcohols can be oxidised to form aldehydes and ketones, the reverse process is also true: aldehydes and ketones can be reduced back to alcohols.

We use sodium tetrahydridoborate(III), ${NaBH}_4$, in an aqueous solution as the reducing agent.

• Aldehydes are reduced to primary alcohols.
• Ketones are reduced to secondary alcohols.

Writing the overall equations

When writing these equations, we use the symbol $[H]$ to represent the reducing agent. Because the reduction of a carbonyl group involves adding one hydrogen to the oxygen and one hydrogen to the carbon, you always need $2[H]$.

For the reduction of ethanal to ethanol:

$${CH}_{3}CHO+2[H]\to {CH}_3{CH}_{2}OH$$

For the reduction of propanone to propan-2-ol:

$${CH}_3{COCH}_3+2[H]\to {CH}_3{CH(OH)CH}_3$$

Mechanism of reduction

The reduction reaction proceeds via a nucleophilic addition mechanism. The ${NaBH}_4$ acts as a source of hydride ions, $H^{-}$. The hydride ion has a lone pair of electrons, making it a nucleophile.

Here is the sequence of events:

1. The lone pair on the hydride ion ($H^{-}$) attacks the ${\delta }^{+}$ carbon of the carbonyl group.
2. The carbon-oxygen double bond breaks, and the $\pi$ bonding pair of electrons moves fully onto the oxygen atom, forming an intermediate alkoxide ion with an $O^{-}$ charge.
3. The negatively charged oxygen donates a lone pair to a hydrogen ion ($H^{+}$) from the water solvent, forming the final alcohol.

Reaction with potassium cyanide

Another important nucleophilic addition reaction is the reaction of aldehydes and ketones with potassium cyanide ($KCN$) followed by dilute acid. This reaction produces hydroxynitriles — molecules containing both a hydroxyl ($-OH$) and a nitrile ($-CN$) group.

This reaction is exceptionally useful in organic synthesis because it increases the length of the carbon chain by one carbon atom.

Why KCN and not HCN?

Hydrogen cyanide ($HCN$) is a highly toxic gas that is difficult and dangerous to handle safely. Furthermore, it is a weak acid, so it only partially dissociates, meaning the concentration of the actual nucleophile (${CN}^{-}$) is very low.

Instead, we use a mixture of solid $KCN$ and dilute acid. $KCN$ is an ionic solid that fully dissociates in water to provide a high concentration of cyanide ions (${CN}^{-}$). The dilute acid then provides the $H^{+}$ ions needed for the final step of the mechanism.

Equations and mechanism

Although the reaction practically uses $KCN$ and dilute acid, the specification asks you to write overall equations using $HCN$ to represent the reagents adding across the double bond.

For the reaction of ethanal with $HCN$:

$${CH}_{3}CHO+HCN\to {CH}_{3}CH(OH)CN$$

The mechanism is another classic nucleophilic addition, mirroring the ${NaBH}_4$ reduction.

1. The cyanide ion (${CN}^{-}$) attacks the ${\delta }^{+}$ carbon.
2. The $C=O$ double bond breaks, pushing electrons onto the oxygen.
3. The $O^{-}$ intermediate picks up an $H^{+}$ ion from the dilute acid.

Optical isomerism from nucleophilic addition

When aldehydes and unsymmetrical ketones react with $KCN$ (followed by dilute acid), they produce molecules with a chiral centre — a carbon atom bonded to four different groups.

Because a chiral centre is formed, you might expect the product to be optically active (able to rotate plane-polarised light). However, the resulting mixture is almost always optically inactive because it forms a racemic mixture (a 50:50 mixture of two enantiomers).

Why does this happen? We must look at the geometry of the carbonyl group.

The atoms bonded directly to the carbonyl carbon are arranged in a planar (flat) triangle. When the nucleophile (${CN}^{-}$) approaches the planar $C=O$ bond, there is nothing blocking it from either side.

• There is a 50% chance the nucleophile attacks from above the plane.
• There is a 50% chance the nucleophile attacks from below the plane.

Because attack from either side is equally likely, equal amounts of the two enantiomers are formed. Their opposite effects on plane-polarised light cancel out perfectly, resulting in an optically inactive racemic mixture.