Nucleophilic addition to the carbonyl group

Attack on the carbon

The carbonyl carbon is flat and has a $\delta^+$ charge. Its pi bond is exposed above and below the plane. A nucleophile, a species with a lone pair to give, can approach from either side.

The reaction is nucleophilic addition. It has two steps.

1. The nucleophile uses its lone pair to bond to the carbonyl carbon. The pi bond of C=O moves onto oxygen. The carbon turns from $sp^2$ to $sp^3$, and the oxygen becomes an alkoxide ion, $\text{R}_{3}\text{C}\text{-}\text{O}^{-}$.
2. The alkoxide ion takes a proton from an acid, or from water, and becomes an alcohol group.

The carbon starts flat and ends tetrahedral. The new groups are close together, so crowded compounds find the reaction harder.

An acid can help. A proton on the oxygen makes the carbon more positive, and the attack is easier. This is acid catalysis.

Why aldehydes are more reactive than ketones

The order of reactivity towards nucleophiles is:

There are two reasons, and they work together.

• Electronic. An alkyl group pushes electrons towards the carbonyl carbon (the +I effect and hyperconjugation). A ketone has two of them and an aldehyde has one. So the ketone carbon is less positive.
• Steric. The nucleophile must reach the carbon, and the product has four groups round it. A ketone has two large groups, and an aldehyde has one large group and one small hydrogen. Methanal has two hydrogens.

Aromatic aldehydes are less reactive than aliphatic ones. The benzene ring shares electrons with the carbonyl group by resonance. This lowers the positive charge on the carbon.

A measure of the order is how far each compound adds water. The hydration constant $K$ is the ratio of hydrate to free carbonyl compound.

Methanal is almost wholly a hydrate in water. Ethanal is half hydrate. Propanone is only $0.14\ \%$ hydrate. Each extra alkyl group lowers $K$ by about a factor of a thousand.

Practise with new numbers

Nucleophiles that add

Hydrogen cyanide. HCN adds to give a cyanohydrin.

$\text{RCHO} + \text{HCN} \rightleftharpoons \text{RCH}(\text{OH})\text{CN}$

HCN is a weak acid and gives few cyanide ions. So a trace of base is added to make the reaction fast. The cyanide ion is the nucleophile. Aldehydes and most ketones react, and the product has one more carbon. Hydrolysis of the nitrile gives a 2-hydroxy acid. Ethanal gives 2-hydroxypropanoic acid, also called lactic acid.

Sodium hydrogensulphite. NaHSO$_3$ adds to give a white crystalline bisulphite compound.

$\text{RCHO} + \text{NaHSO}_{3} \rightleftharpoons \text{RCH}(\text{OH})\text{SO}_{3}\text{Na}$

Aldehydes, methyl ketones and small cyclic ketones react. Large ketones are too crowded. The addition is reversible, so dilute acid or base gives back the carbonyl compound. This separates an aldehyde from a mixture.

Alcohols. An aldehyde and an alcohol, with dry HCl gas, give a hemiacetal and then an acetal.

$\text{RCHO} + _{2}\text{R}'\text{OH} \xrightarrow{\text{dry}\ \text{HCl}} \text{RCH}(\text{OR}')_{2} + \text{H}_{2}\text{O}$

A ketone and ethylene glycol give a cyclic ketal. Acetals are stable to base but are hydrolysed by dilute acid. So a chemist uses an acetal to protect a carbonyl group while another part of the molecule reacts.

Ammonia derivatives. A compound $\text{H}_2\text{N-Z}$ adds to the carbonyl group. Then water is lost and a C=N bond forms.

$\text{C}{=}\text{O} + \text{H}_{2}\text{N}\text{-}\text{Z} \longrightarrow \text{C}{=}\text{N}\text{-}\text{Z} + \text{H}_{2}\text{O}$

• $\text{Z} = \text{OH}$, hydroxylamine: an oxime.
• $\text{Z} = \text{NH}_2$, hydrazine: a hydrazone.
• $\text{Z} = \text{NHC}_6\text{H}_5$, phenylhydrazine: a phenylhydrazone.
• $\text{Z} = \text{NHCONH}_2$, semicarbazide: a semicarbazone.
• 2,4-Dinitrophenylhydrazine gives an orange or red solid, the 2,4-DNP derivative. It is used to test for any aldehyde or ketone, and its melting point identifies the compound.

The medium is weakly acidic. In strong acid the amine takes a proton and cannot attack. In base the carbonyl is not helped by a proton.

Grignard reagents. The carbon of RMgX is a strong nucleophile. After water or dilute acid, methanal gives a primary alcohol, any other aldehyde gives a secondary alcohol, and a ketone gives a tertiary alcohol. The product has the carbons of both partners.

Try it in Python

The program turns the hydration constants into percentages of hydrate.

Common mistakes

• Starting with a proton. In most additions the nucleophile attacks first and the proton comes second.
• Thinking that all carbonyl compounds form bisulphite compounds. Large ketones do not.
• Strong acid for the ammonia derivatives. The amine is then protonated and unreactive.
• Forgetting that a Grignard reagent adds a carbon. The alcohol has the carbons of both the reagent and the carbonyl compound.
• Calling a hydrate a stable product. For most ketones it is only a trace.

Remember these

• Nucleophilic addition: attack on the carbon, then the proton. The carbon changes from $sp^2$ to $sp^3$.
• Reactivity: methanal, then other aldehydes, then ketones. Aromatic aldehydes are less reactive than aliphatic ones.
• HCN (trace of base): cyanohydrin. NaHSO$_3$: bisulphite compound. Alcohol and dry HCl: acetal.
• Ammonia derivatives give C=N compounds. 2,4-DNP gives an orange or red solid.

Check yourself

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

1. Which is more reactive towards HCN, propanone or ethanal?

2. Why is a trace of base added when HCN reacts with a carbonyl compound?

3. What does butanone give with hydroxylamine?

4. Why is an acetal used to protect an aldehyde group during a reaction with a base?

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 section on chemical reactions of aldehydes and ketones, nucleophilic addition.
• Clayden, Greeves and Warren, Organic Chemistry, the chapter on nucleophilic addition to the carbonyl group.