Aldehydes and ketones

In a nutshell

Aldehydes and ketones both contain the carbonyl group (C=O). The carbon of that group carries a δ+ charge, so these compounds are attacked by nucleophiles.

This subtopic is really three things: how to tell an aldehyde from a ketone (Fehling's and Tollens'), the two nucleophilic addition reactions you must mechanism (reduction with NaBH₄, and KCN then dilute acid to make hydroxynitriles), and why that second reaction gives a mixture of enantiomers.

Assumed knowledge: Alcohols, Optical isomerism (3.3.7).

Core content

The carbonyl group and why nucleophiles attack it

An aldehyde has the carbonyl carbon at the end of the chain, bonded to at least one H (written -CHO, name ends in -al, e.g. ethanal CH₃CHO).

A ketone has the carbonyl carbon joined to two other carbon atoms (written -CO-, name ends in -one, e.g. propanone CH₃COCH₃).

Oxygen is more electronegative than carbon, so the C=O bond is polar: the carbon is δ+ and the oxygen is δ-.

That electron-deficient δ+ carbon is what a nucleophile (a species with a lone pair of electrons that it can donate) is attracted to. This is why the carbonyl group reacts by nucleophilic addition: a nucleophile adds on and the C=O double bond opens up.

Still don't get it? · what "nucleophilic addition" means

Think of the carbonyl carbon as the positive end of a magnet and a nucleophile as the negative end: opposites attract, so the nucleophile is pulled straight onto that carbon.

Now build it up slowly. The C=O is a double bond, which is two pairs of shared electrons. Oxygen hogs them, so the carbon is left a little short of electrons (δ+). A nucleophile always carries a spare lone pair, so it moves that lone pair in to form a new bond to the δ+ carbon. But carbon can only have four bonds, so as the new bond forms, one of the two C=O pairs has to get out of the way, and it shifts entirely onto the oxygen, leaving the oxygen with a negative charge.

In exam language: the nucleophile uses its lone pair to attack the δ+ carbon, the C=O bond breaks heterolytically with both electrons going to the oxygen, and because a group has been added across the double bond the reaction is called nucleophilic addition.

Oxidation: aldehydes to carboxylic acids

Aldehydes are readily oxidised to carboxylic acids. Ketones are not oxidised (there is no H on the carbonyl carbon to remove).

Using an oxidising agent, written [O]:

CH₃CHO + [O] → CH₃COOH

With acidified potassium dichromate(VI) the dichromate ion is reduced, so the colour change is orange to green (Cr₂O₇^2- to Cr³⁺). A ketone leaves it orange.

This difference in behaviour (easily oxidised or not) is exactly what the two confirmatory tests below exploit.

Distinguishing aldehydes from ketones: Fehling's and Tollens'

Both tests oxidise an aldehyde and are reduced themselves (giving a visible change); a ketone gives no reaction. This is a test-tube skill (warm gently in a water bath).

In each test the aldehyde is oxidised to a carboxylic acid (as a carboxylate salt under these conditions) while the metal ion is reduced:

• Tollens': silver(I) is reduced to silver, [Ag(NH₃)₂]⁺ + e^- → Ag + 2NH₃
• Fehling's: copper(II) is reduced to copper(I), forming red Cu₂O.

Acidified potassium dichromate(VI) (orange to green with an aldehyde) is sometimes used too, but it is not specific: it also oxidises primary and secondary alcohols. Fehling's and Tollens' are the clean carbonyl tests.

Reduction: NaBH₄ makes alcohols (nucleophilic addition)

Sodium tetrahydridoborate(III), NaBH₄, in aqueous solution reduces carbonyls to alcohols at room temperature. It is the source of the nucleophile, the hydride ion, H^-.

• an aldehyde is reduced to a primary alcohol
• a ketone is reduced to a secondary alcohol

Write the overall equations with [H] as the reductant (two [H] are added):

CH₃CHO + 2[H] → CH₃CH₂OH

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

The mechanism is nucleophilic addition. The H^- lone pair attacks the δ+ carbon; the C=O electrons shift onto the oxygen to give a negatively charged alkoxide intermediate; that oxygen is then protonated by a water molecule to give the -OH group.

Nucleophilic addition of KCN then dilute acid: hydroxynitriles

Carbonyls react with potassium cyanide (KCN) followed by dilute acid to form hydroxynitriles (molecules with both an -OH and a -CN group). KCN supplies the nucleophile, the cyanide ion :CN^-.

Write the overall equation using HCN:

CH₃CH₂CHO + HCN → CH₃CH₂CH(OH)CN

The mechanism is again nucleophilic addition: :CN^- attacks the δ+ carbon (the curly arrow starts at the lone pair on the carbon of the cyanide ion), the C=O opens onto the oxygen to give the alkoxide, and dilute acid then protonates that oxygen to the -OH.

Naming the product trips people up. The nitrile carbon is carbon 1 and counts as part of the main chain:

• from ethanal: CH₃CH(OH)CN is 2-hydroxypropanenitrile
• from propanal: CH₃CH₂CH(OH)CN is 2-hydroxybutanenitrile
• from propanone: (CH₃)₂C(OH)CN is 2-hydroxy-2-methylpropanenitrile

Why a mixture of enantiomers forms

Aldehydes and unsymmetrical ketones produce a mixture of enantiomers in this reaction.

The reason is the shape of the starting material. The carbonyl group is planar, so :CN^- can attack from either side of that plane with equal probability.

In the product the old carbonyl carbon is now bonded to four different groups (a chiral centre), so the two routes give the two enantiomers (non-superimposable mirror images) in equal amounts, which is a racemic mixture. A racemate is optically inactive, because the two enantiomers rotate plane-polarised light equally in opposite directions, so the effects cancel.

A symmetrical ketone does not do this. Propanone gives (CH₃)₂C(OH)CN, whose central carbon carries two identical CH₃ groups, so there is no chiral centre and only one (achiral) product forms.

Still don't get it? · why a flat molecule gives a 50:50 mix of mirror images

Imagine dropping a coin flat on a table and sticking a blob of putty onto its face. You could have stuck it on the heads side or the tails side, and because the coin was lying flat, both were equally easy. You end up with a jar of coins, half blobbed on one face and half on the other.

The carbonyl is that flat coin. Its carbon is trigonal planar, so the incoming :CN- has a completely free choice of which face to hit, top or bottom, with no preference. Hitting the top builds the four groups around the carbon one way; hitting the bottom builds the same four groups the mirror-image way. Nothing favours one face, so you get equal numbers of each.

In exam language: the planar carbonyl group is attacked from either side with equal probability, giving equal amounts of the two enantiomers, i.e. a racemic mixture. Because equal and opposite rotations of plane-polarised light cancel, the mixture is optically inactive.

The hazards of KCN

Potassium cyanide is very toxic (poisonous). The danger is the cyanide ion.

With acid, KCN releases hydrogen cyanide, HCN, a toxic gas, so the reaction must be carried out in a fume cupboard, avoiding skin contact and inhalation.

KCN is used rather than HCN itself because HCN is a volatile, toxic gas that is hard to contain, and because HCN is a weak acid that is only partially ionised, so it gives a low concentration of CN^-. KCN is fully ionised, giving a higher [CN^-] and a faster reaction.

Worked examples

1. Writing the reduction equation with [H]. Reduce butanone with NaBH₄; give the overall equation.

• Butanone is a ketone, so it is reduced to a secondary alcohol, butan-2-ol.
• Add 2[H] across the C=O:

$CH_3COCH_2CH_3 + 2[H] \rightarrow CH_3CH(OH)CH_2CH_3$

Where marks are lost: leaving out the [H] (the equation will not balance), or writing an aldehyde as R-COH. The aldehyde group is always written -CHO.

2. Making and naming a hydroxynitrile. Give the equation and product name for propanal with KCN then dilute acid.

• Nucleophilic addition of HCN across the carbonyl:

$CH_3CH_2CHO + HCN \rightarrow CH_3CH_2CH(OH)CN$

• Count carbons with the nitrile carbon as carbon 1: CN (C1), CH(OH) (C2), CH₂ (C3), CH₃ (C4) is a four-carbon chain with -OH on C2.
• Product: 2-hydroxybutanenitrile.

Where marks are lost: calling it "propanenitrile" (forgetting the nitrile carbon counts), or shortening "butanenitrile" to "butanitrile" (rejected).

3. Model answer: "Explain why the reaction of ethanal with KCN followed by dilute acid gives an optically inactive product." (4 marks)

1. The product 2-hydroxypropanenitrile has a carbon bonded to four different groups, a chiral centre, so it exists as two enantiomers.
2. The carbonyl group in ethanal is planar, so the :CN^- can attack from either side with equal probability.
3. This produces equal amounts of the two enantiomers, a racemic mixture.
4. The two enantiomers rotate plane-polarised light equally in opposite directions, so the effects cancel and the mixture is optically inactive.

Where marks are lost: writing about the optical activity of ethanal itself instead of the product (no marks without reference to the product), and saying the light is "bent" or that the molecules rotate, rather than plane-polarised light being rotated in opposite directions.

Common exam mistakes

• Writing an aldehyde as R-COH (e.g. CH₃CH₂COH); the aldehyde group must be written -CHO.
• Naming the hydroxynitrile from the wrong chain length. The nitrile carbon is carbon 1, so ethanal gives 2-hydroxypropanenitrile, not ethanenitrile. "Butanitrile" is rejected; it must be "butanenitrile".
• Starting the cyanide curly arrow from the nitrogen. It must start from the lone pair on the carbon of :CN^-, and the new bond is from the carbonyl carbon to that carbon.
• Marking the carbonyl carbon with a full positive charge; it is δ+, not +. Also forgetting the negative charge on the nucleophile and on the alkoxide intermediate.
• Breaking the C=O bond before the nucleophile has attacked. The attack and the shift of the C=O electrons happen together, with the electrons going onto the oxygen.
• Showing BH₄^- as the attacking species in the NaBH₄ mechanism. The nucleophile must be shown as H^-.
• Claiming both aldehydes and symmetrical ketones give a racemate. A symmetrical ketone (e.g. propanone) gives a product with no chiral centre, so a single achiral product forms.
• Saying "planar molecule" or "planar bond" when explaining enantiomer formation. The mark needs planar carbonyl group.
• Using Fehling's or Tollens' on the wrong compound, writing an incomplete Tollens' reagent (silver nitrate alone, or [Ag(NH₃)₂]⁺ without its charge), or recording "red solution" for Fehling's (it must be a red precipitate).
• Explaining why HCN reacts slowly by confusing the acid dissociation constant with a rate constant, instead of linking a weak acid to a low [CN^-].

Key definitions

• Nucleophile - a species that has a lone pair of electrons which it can donate (to an electron-deficient carbon atom).
• Nucleophilic addition - a reaction in which a nucleophile adds across the polar C=O double bond of a carbonyl compound.
• Hydroxynitrile - a compound containing both a hydroxyl (-OH) group and a nitrile (-CN) group on the same carbon skeleton.
• Enantiomers - non-superimposable mirror images (of each other).
• Racemic mixture (racemate) - an equal mixture (50:50) of two enantiomers, which is optically inactive because their effects on plane-polarised light cancel.

Specification

• I can state that aldehydes are readily oxidised to carboxylic acids, and ketones are not.
• I can describe and give the results of the chemical tests (Fehling's solution and Tollens' reagent) used to distinguish between aldehydes and ketones.
• I can state that aldehydes are reduced to primary alcohols and ketones to secondary alcohols using NaBH₄ in aqueous solution, and that this is nucleophilic addition.
• I can write overall equations for these reduction reactions using [H] as the reductant.
• I can outline the nucleophilic addition mechanism for reduction with NaBH₄, showing the nucleophile as H^-.
• I can write overall equations for the formation of hydroxynitriles using HCN.
• I can outline the nucleophilic addition mechanism for the reaction with KCN followed by dilute acid.
• I can explain why the reaction of aldehydes and unsymmetrical ketones with KCN followed by dilute acid can produce a mixture of enantiomers.
• I can state the hazards of using KCN.

Related notes

• Alcohols
• Organic analysis
• Optical isomerism (3.3.7)
• Carboxylic acids and derivatives (3.3.9)