Aldol and Cannizzaro reactions, and the benzene ring

The acidic hydrogen

The hydrogens on the carbon next to the carbonyl group are the alpha hydrogens. They are far more acidic than hydrogens on an alkane: the pKa is about 17 to 20, against about 50. When one leaves as a proton, the negative charge is shared between carbon and oxygen. This ion is the enolate ion.

$\text{CH}_{3}\text{CHO} + \text{OH}^{-} \rightleftharpoons \text{CH}_{2}{=}\text{CH}\text{-}\text{O}^{-} + \text{H}_{2}\text{O}$

The enolate is a strong nucleophile at carbon. The base only makes a little of it at a time. But the little is enough.

At pH $13$, propanone has one enolate for about two million molecules. An ester has one in $10^{12}$. The lower the pKa, the more enolate.

Practise with new numbers

The aldol reaction

Take dilute sodium hydroxide and ethanal. The steps are:

1. Hydroxide takes an alpha hydrogen, and an enolate forms.
2. The enolate carbon attacks the carbonyl carbon of a second ethanal.
3. The alkoxide takes a proton from water. The product is 3-hydroxybutanal, which has an aldehyde group and an alcohol group. This is why the reaction is called aldol.
4. On heating, water leaves from the carbon with the OH and the carbon with the alpha hydrogen. This gives but-2-enal, a conjugated aldehyde.

$_{2}\text{CH}_{3}\text{CHO} \xrightarrow{\text{dil.}\ \text{NaOH}} \text{CH}_{3}\text{CH}(\text{OH})\text{CH}_{2}\text{CHO} \xrightarrow{\text{heat}} \text{CH}_{3}\text{CH}{=}\text{CHCHO} + \text{H}_{2}\text{O}$

Two molecules of ethanal join to make one of four carbons. A ketone behaves in the same way. Propanone gives 4-hydroxy-4-methylpentan-2-one, and then 4-methylpent-3-en-2-one.

For the aldol reaction the compound must have at least one alpha hydrogen.

Crossed aldol reaction. Two different carbonyl compounds, with alpha hydrogens, give four products. Each can be the enolate, and each can be attacked. A mixture is not useful. The reaction is useful when one partner has no alpha hydrogen. Benzaldehyde is such a partner. With ethanal in base it gives 3-phenylprop-2-enal, which is cinnamaldehyde. With propanone it gives 4-phenylbut-3-en-2-one. With acetophenone it gives chalcone.

The Cannizzaro reaction

An aldehyde with no alpha hydrogen, treated with concentrated alkali, goes through the Cannizzaro reaction. One molecule is oxidised to an acid salt and another is reduced to an alcohol.

$_{2}\text{C}_{6}\text{H}_{5}\text{CHO} \xrightarrow{\text{conc.}\ \text{NaOH}} \text{C}_{6}\text{H}_{5}\text{CH}_{2}\text{OH} + \text{C}_{6}\text{H}_{5}\text{COONa}$

Hydroxide adds to one aldehyde first. The anion that forms then passes a hydride ion to a second aldehyde. This hydride transfer is the slow step. Methanal and 2,2-dimethylpropanal also do it.

In a crossed Cannizzaro of methanal with benzaldehyde, methanal is the better target for hydroxide, so it becomes methanoate. The benzaldehyde receives the hydride, and becomes benzyl alcohol.

Aldol needs a dilute base and an alpha hydrogen. Cannizzaro needs a concentrated base and no alpha hydrogen.

The benzene ring

The carbonyl group takes electrons from the ring, so benzaldehyde and acetophenone are less reactive than benzene. The group is deactivating and meta-directing.

$\text{C}_{6}\text{H}_{5}\text{CHO} \xrightarrow{\text{conc.}\ \text{HNO}_{3}\ \text{conc.}\ \text{H}_{2}\text{SO}_{4}} _{3}\text{-}\text{O}_{2}\text{N}\text{-}\text{C}_{6}\text{H}_{4}\text{-}\text{CHO}$

Nitration of benzaldehyde gives 3-nitrobenzaldehyde. Bromine with aluminium chloride and acetophenone give 3-bromoacetophenone. Friedel-Crafts reactions do not work on rings with such a group.

Phosphorus pentachloride replaces the oxygen of a carbonyl group with two chlorine atoms: benzaldehyde gives benzal chloride, $\text{C}_{6}\text{H}_{5}\text{CHCl}_{2}$.

Try it in Python

The program works out how many carbonyl molecules there are for each enolate ion in a basic solution, for three compounds with different alpha hydrogens.

Common mistakes

• Giving an aldol product to benzaldehyde. It has no alpha hydrogen.
• Using concentrated base for the aldol. The aldol needs dilute base.
• Expecting Cannizzaro from ethanal. It has an alpha hydrogen and does aldol.
• Writing an ortho or para product for the nitration of benzaldehyde. The group is meta-directing.
• Forgetting the loss of water on heating. The aldol gives the unsaturated aldehyde only when heated.

Remember these

• Alpha hydrogens are acidic (pKa about 17 to 20). The enolate attacks another carbonyl group.
• Aldol: dilute NaOH, at least one alpha hydrogen, then heat to lose water.
• Cannizzaro: concentrated NaOH, no alpha hydrogen. Half the product is the alcohol and half is the acid salt.
• The carbonyl group is meta-directing and deactivating.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which of benzaldehyde, ethanal and methanal can undergo the aldol reaction?

2. What does methanal give with concentrated NaOH?

3. Write the product of nitrating benzaldehyde.

4. How many products can a mixture of ethanal and propanal give in dilute NaOH?

References

• OpenStax, Chemistry 2e, Chapter 20, section 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters.
• NCERT, Chemistry Part II for Class XII, Chapter 8, the sections on aldol condensation, Cannizzaro reaction and electrophilic substitution.
• Clayden, Greeves and Warren, Organic Chemistry, the chapters on enolates and the aldol reaction.