3.3.8 Aldehydes and Ketones

The C=O bond is polar because oxygen is more electronegative than carbon, giving Cδ⁺=Oδ⁻. The electron-poor carbonyl carbon is therefore the site attacked by nucleophiles.

Marking the dipole before drawing a mechanism is a useful habit: it tells you why CN⁻ or H⁻ attacks the δ⁺ carbon; oxygen carries δ⁻.

Compare aldehydes with ketones: the aldehyde can be oxidised on to a carboxylic acid because the carbonyl carbon still has a hydrogen. The corresponding mild oxidation of a ketone would require C–C bond breaking.

That difference underlies the standard aldehyde/ketone oxidation tests. Use the structural reason as well as memorising the observation.

For the aldehyde/ketone tests, I’d learn the observation with the conclusion. Tollens’ reagent gives a silver mirror with an aldehyde and no corresponding reaction with a ketone under the test conditions.

Fehling’s solution gives the complementary observation: an aldehyde changes the blue Cu²⁺ mixture to a brick-red Cu₂O precipitate on warming, while a ketone gives no corresponding reaction. Learn the observation with the conclusion.

Aqueous NaBH₄ reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol. The C=O group becomes C–OH while the carbon skeleton remains unchanged.

You can predict the alcohol class directly from the starting carbonyl: terminal R–CHO gives R–CH₂OH; internal R₂C=O gives R₂CHOH.

When you reduce a carbonyl with NaBH₄, think addition of hydrogen across C=O. Aldehydes give primary alcohols and ketones give secondary alcohols.

NaBH₄ reduces aldehydes to primary alcohols and ketones to secondary alcohols by supplying hydride to the carbonyl carbon, followed by protonation of the oxygen-containing intermediate.

With KCN followed by dilute acid, CN⁻ attacks the δ⁺ carbonyl carbon and adds one carbon to the skeleton. Protonation of the O⁻ intermediate gives a hydroxynitrile containing both –OH and –C≡N.

Count the carbons before and after: cyanide addition lengthens the carbon chain by one because the carbon of CN⁻ becomes part of the product.

If nucleophilic attack on a planar carbonyl gives you a new chiral centre, remember that the nucleophile can attack from either face. Those two attack directions give the two enantiomers.

In an achiral reaction environment the two faces are equivalent, so the two enantiomers are formed in equal amounts and the product is racemic.

KCN and other cyanide reagents are severely toxic, so their use requires strict control of exposure and appropriate laboratory precautions. In particular, acidifying cyanide can release volatile HCN, so this is a genuine chemical hazard that applies outside the exam as well.

I would keep this safety point conceptually separate from the cyanide mechanism. Knowing why CN⁻ is a good nucleophile does not reduce the hazard of handling cyanide reagents.

For the overall reduction equation, [H] is acceptable shorthand for reducing equivalents: RCHO + 2[H] → RCH₂OH and R₂CO + 2[H] → R₂CHOH.

Use 2[H] because two hydrogen atoms are added overall across the carbonyl functionality. The shorthand is an equation device; the NaBH₄ mechanism itself uses hydride attack followed by protonation.

In the NaBH₄ mechanism, treat H⁻ as the nucleophile. Draw the first arrow from H⁻ to the carbonyl carbon and the second from the C=O π bond to oxygen, then protonate the alkoxide product.

The hydride arrow must start on the electron pair associated with H⁻ and end at carbon. Do not draw H⁻ attacking the δ⁻ oxygen.

The overall hydroxynitrile equation uses HCN because H and CN are added across the C=O bond overall. In the mechanism, however, CN⁻ is the attacking nucleophile and dilute acid supplies the proton in the later step.

Keep “overall reagent equation” and “mechanistic attacking species” separate; they legitimately use different species here.

For cyanide addition, start the curly arrow at CN⁻ and send it to the δ⁺ carbonyl carbon. Move the C=O π pair onto oxygen, then protonate the O⁻ intermediate with dilute acid to give the hydroxynitrile.

If CN⁻ attacks a planar carbonyl from either face and the product carbon ends up bonded to four different groups, you form opposite enantiomers. In an achiral environment the two faces are equally likely, so you get a racemic mixture.

Equal formation of the two mirror images gives a racemic mixture, so their optical rotations cancel overall.