Reactions Involving α-Hydrogen

CHEReactions of Aldehydes & Ketones

The acidity of the $\alpha\text{-}H$ (next to $\text{C=O}$ ) drives the aldol condensation, while carbonyls lacking $\alpha\text{-}H$ undergo the Cannizzaro reaction instead. This subtopic tests both reactions, their products, and which compound follows which path.

Aldol condensation of ethanal

2CH_3CHO \xrightarrow{\text{dil. }NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta,\ -H_2O} CH_3CH{=}CHCHO

aldol (3-hydroxybutanal) dehydrates to but-2-enal; needs an

\alpha\text{-}H

Cannizzaro reaction

2HCHO \xrightarrow{\text{conc. }NaOH} CH_3OH + HCOONa

\alpha\text{-}H

: one molecule reduced to alcohol, one oxidised to carboxylate

Aldol condensation: a carbonyl with at least one $\alpha\text{-}H$ reacts with dilute NaOH; two ethanal molecules give 3-hydroxybutanal (aldol), which on heating loses water to give but-2-enal.
Requirement: aldol needs an $\alpha\text{-}H$ . Ethanal and propanal undergo aldol; HCHO, benzaldehyde and 2,2-dimethylpropanal $(CH_3)_3CCHO$ have no $\alpha\text{-}H$ and cannot.
Cannizzaro reaction: an aldehyde with NO $\alpha\text{-}H$ disproportionates in conc. alkali - one molecule is oxidised to the carboxylate (salt) and another reduced to the $1^\circ$ alcohol.
HCHO undergoes Cannizzaro (no $\alpha\text{-}H$ ) giving methanol + sodium formate; benzaldehyde gives benzyl alcohol + sodium benzoate; $(CH_3)_3CCHO$ also gives Cannizzaro.
Decision rule: $\alpha\text{-}H$ present $\rightarrow$ aldol; $\alpha\text{-}H$ absent $\rightarrow$ Cannizzaro.
Why the $\alpha\text{-}H$ is acidic: removing it gives a resonance-stabilised enolate (negative charge delocalised onto the carbonyl O), which is the nucleophile that attacks the second carbonyl in aldol.
Aldol vs aldol condensation: the cold, dilute reaction stops at the $\beta$ -hydroxy carbonyl (the aldol); heating drives dehydration to the conjugated $\alpha,\beta$ -unsaturated carbonyl - the "condensation" is this loss of water.
Ketones also do aldol if they have $\alpha\text{-}H$ : propanone gives 4-hydroxy-4-methylpentan-2-one ("diacetone alcohol"), though the equilibrium is less favourable than for aldehydes.
Crossed (mixed) aldol uses two different carbonyls; it is synthetically useful only when ONE partner has no $\alpha\text{-}H$ (e.g. benzaldehyde + ethanal) so it acts solely as the electrophile, avoiding a four-product mixture.
Cannizzaro is a disproportionation/redox in one substrate: the same aldehyde is simultaneously oxidised (to acid salt) and reduced (to alcohol); a hydride ion is transferred from one molecule to the other.
Crossed Cannizzaro with $HCHO$ : formaldehyde is preferentially oxidised (it is the best hydride donor), so the OTHER aldehyde is reduced to its alcohol - a way to selectively make a $1^\circ$ alcohol.
Watch what counts as an $\alpha\text{-}H$ : it must be on the carbon directly attached to $\text{C=O}$ ; $(CH_3)_3CCHO$ has no such H (the carbonyl carbon's neighbour is quaternary), so it gives Cannizzaro, not aldol.

Where the marks go

Counting H on the wrong carbon: $\alpha\text{-}H$ is on the carbon adjacent to $\text{C=O}$ ; benzaldehyde and $(CH_3)_3CCHO$ have NONE and so cannot do aldol.
Forgetting the dehydration step - if the question says "condensation" or "on heating", the answer is the $\alpha,\beta$ -unsaturated product (but-2-enal), not the $\beta$ -hydroxy aldol.
Using dilute base for Cannizzaro or concentrated base for aldol - aldol needs DILUTE NaOH, Cannizzaro needs CONCENTRATED alkali.
Naming both Cannizzaro products as neutral molecules - one product is the carboxylate SALT (e.g. sodium formate), formed because the medium is strongly basic.
Saying an aldehyde must do only one of the two - some (e.g. ethanal) do aldol; some (e.g. HCHO) do Cannizzaro; classify by the presence/absence of $\alpha\text{-}H$ .

How the board asks it

Predict the productaldol condensation and Cannizzaro reaction
Write the structures of the organic products and the equations when (i) two molecules of ethanal are warmed with dilute $NaOH$ and the product is then heated, and (ii) $HCHO$ is treated with concentrated $NaOH$ .
Give reasonspresence/absence of $\alpha$ -H and the resonance-stabilised enolate
Account for the fact that $(CH_3)_3CCHO$ and benzaldehyde undergo the Cannizzaro reaction, whereas ethanal and propanal undergo the aldol condensation.
Distinguishthe $\alpha$ -H decision rule (aldol vs Cannizzaro)
How will you distinguish between ethanal and benzaldehyde by their behaviour with dilute and with concentrated $NaOH$ ?
Conversioncrossed Cannizzaro and crossed aldol
How will you convert benzaldehyde into benzyl alcohol, and benzaldehyde into cinnamaldehyde ( $3$ -phenylprop-2-enal)?
Identify / classifyCannizzaro products as identifiers
An aldehyde $A$ of molecular formula $C_7H_6O$ has no $\alpha$ -H and, on warming with concentrated $NaOH$ , gives a primary alcohol $B$ and the sodium salt of an acid $C$ . Identify $A$ , $B$ and $C$ and name the reaction.

Practise this topic

PreviousNucleophilic Addition & Reactivity NextOxidation & Distinguishing Tests

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.