Hydroxynitriles from Carbonyls Revision Notes | A Level AQA Chemistry

1. Why Carbonyl Compounds React with Cyanide

Picture the carbonyl group, C=O, at the heart of both aldehydes and ketones. The oxygen atom is considerably more electronegative than carbon, so it pulls electron density towards itself through the pi bond. This leaves the carbonyl carbon with a significant partial positive charge, $\delta^+$.

That electron-deficient carbon is a magnet for nucleophiles — species that carry a lone pair and are attracted to positive centres. The cyanide ion, $\text{CN}^-$, is an excellent nucleophile: it carries a lone pair on the carbon end of the ion and is small enough to approach the carbonyl carbon without too much steric hindrance.

The result of this attack is a class of compounds called hydroxynitriles (sometimes written as cyanohydrins). Understanding how and why they form — and what stereochemical consequences follow — is the focus of this lesson.

2. Overall Equations for Hydroxynitrile Formation

The reaction uses KCN (potassium cyanide) as the source of cyanide ions, followed by dilute acid to protonate the alkoxide intermediate. The net effect is the addition of HCN across the C=O bond. You need to be able to write the overall equation for both an aldehyde and a ketone.

For an aldehyde such as ethanal, the carbonyl carbon carries one hydrogen and one methyl group. After HCN addition, a new C–C bond forms and the carbon gains both an –OH and a –CN group, producing 2-hydroxypropanenitrile.

For a ketone such as propanone, both groups attached to the carbonyl carbon are carbon-containing. HCN addition gives 2-hydroxy-2-methylpropanenitrile. The general pattern is the same: HCN adds across C=O to give a product with –OH and –CN on the same carbon.

3. The Nucleophilic Addition Mechanism

The mechanism runs in two stages. In practice, KCN is used rather than HCN directly — KCN is a solid source of $\text{CN}^-$ ions in solution and is easier to handle than volatile HCN gas, though KCN itself is also highly toxic and must be treated with full safety precautions. The dilute acid added afterwards provides the proton needed to complete the reaction. The mechanism itself involves the cyanide ion acting as the nucleophile, not HCN.

In Stage 1, the $\text{CN}^-$ ion uses its lone pair (on the carbon end of the ion) to attack the $\delta^+$ carbonyl carbon. A full curly arrow — double-barbed, showing movement of an electron pair — goes from the lone pair on $\text{CN}^-$ to the carbonyl carbon. Simultaneously, a second curly arrow shows the C=O pi bond electrons moving onto the oxygen, giving it a negative charge. This produces an alkoxide intermediate, $\text{RR'C(O}^-\text{)(CN)}$.

In Stage 2, the alkoxide ion is protonated. The dilute acid provides $\text{H}^+$, and a curly arrow from the lone pair on $\text{O}^-$ points to $\text{H}^+$, forming the new O–H bond of the hydroxynitrile product. The overall result is addition of HCN across the C=O bond.

Worked example: Drawing the Nucleophilic Addition Mechanism for Ethanal + KCN/H+

Outline the mechanism for the reaction of ethanal with KCN followed by dilute acid to form 2-hydroxypropanenitrile. Show all curly arrows and intermediates.

1. In $\text{CH}_3\text{CHO}$, the carbonyl carbon carries $\delta^+$ due to the electronegativity of oxygen. Label it $\delta^+$ and the oxygen $\delta^-$. — This tells you where the nucleophile will attack. The nucleophile always attacks the electron-poor carbon, not the oxygen.
2. Draw a full curly arrow from the lone pair on the carbon end of $\text{CN}^-$ pointing to the carbonyl carbon. Draw a second curly arrow from the C=O pi bond to the oxygen atom. — The arrow must start from the lone pair on CN-, not from the nitrogen end. The pi bond breaks and both electrons go to oxygen, giving it a negative charge. The intermediate is $\text{CH}_3\text{CH(O}^-\text{)(CN)}$.
3. Draw $\text{CH}_3\text{CH(O}^-\text{)(CN)}$ with a negative charge on the oxygen. The carbon now has four different groups: $\text{CH}_3$, H, $\text{O}^-$, and $\text{CN}$. — This tetrahedral intermediate is the alkoxide ion. It is negatively charged and will be protonated in the next step.
4. Draw a curly arrow from the lone pair on $\text{O}^-$ to the $\text{H}^+$ provided by the dilute acid. This forms the O–H bond. — The dilute acid (e.g. dilute HCl or dilute H2SO4) provides the proton. The product is the neutral hydroxynitrile $\text{CH}_3\text{CH(OH)(CN)}$.
5. $\text{CH}_3\text{CH(OH)(CN)}$ — 2-hydroxypropanenitrile. The carbon bearing –OH and –CN is now a chiral centre. — Check: four different groups on C-2 are $\text{CH}_3$, H, OH, and CN. This chiral centre is why a mixture of enantiomers forms.

Answer: Stage 1: $\text{CN}^-$ attacks $\delta^+$ carbonyl carbon; C=O pi bond breaks, electrons move to O, forming alkoxide $\text{CH}_3\text{CH(O}^-\text{)(CN)}$. Stage 2: $\text{O}^-$ is protonated by $\text{H}^+$ from dilute acid to give $\text{CH}_3\text{CH(OH)(CN)}$, 2-hydroxypropanenitrile.

4. Enantiomers and the Chiral Centre

Here is where the chemistry becomes genuinely interesting. When $\text{CN}^-$ attacks the planar carbonyl carbon of ethanal, it can approach from either face of the trigonal planar C=O group — from above or from below the plane. Both faces are equally accessible because there is no existing chiral centre to block one approach over the other.

Attack from one face gives one enantiomer; attack from the other face gives the mirror-image enantiomer. Because both faces are equally likely to be attacked, the two enantiomers form in equal amounts. This equal mixture of two enantiomers is called a racemic mixture (or racemate).

The condition for enantiomer formation is the creation of a new chiral centre. After $\text{CN}^-$ addition to ethanal, C-2 of the product carries four different groups: $\text{CH}_3$, H, OH, and CN. Four different groups on one carbon = a chiral centre = two possible spatial arrangements = two enantiomers.

The same logic applies to unsymmetrical ketones. Propanone is symmetrical (both groups on the carbonyl carbon are CH3), so after CN addition the product carbon is attached to four substituents in total: CH3, CH3, OH, and CN. Because two of those substituents are identical methyl groups, there are only three distinct groups, so no chiral centre forms and no enantiomers result. Butanone, however, is unsymmetrical: the two groups on the carbonyl carbon differ, so the product carbon carries CH3, CH2CH3, OH, and CN — four different groups, giving a chiral centre and a racemic mixture.

5. Aldehydes vs Unsymmetrical Ketones: Comparing Reactivity and Stereochemistry

Both aldehydes and unsymmetrical ketones undergo nucleophilic addition with KCN/dilute acid, but there is a difference in reactivity worth noting. Aldehydes are generally more reactive than ketones towards nucleophilic addition. The carbonyl carbon in an aldehyde carries one hydrogen and one alkyl group, whereas in a ketone it carries two alkyl groups.

Alkyl groups are electron-donating: they push electron density towards the carbonyl carbon, reducing its $\delta^+$ character and making it less attractive to nucleophiles. Two alkyl groups in a ketone therefore reduce reactivity more than one alkyl group in an aldehyde. Additionally, two bulkier alkyl groups create more steric hindrance around the carbonyl carbon, physically obstructing the approach of the nucleophile.

The key AQA point is this: both aldehydes and unsymmetrical ketones produce a mixture of enantiomers because the nucleophile can attack either face of the planar carbonyl group with equal probability. Symmetrical ketones (like propanone) do not produce enantiomers because no chiral centre is created.

6. Hazards of Using KCN

KCN (potassium cyanide) is an extremely hazardous substance and AQA expects you to know its key risks. Cyanide ions are acutely toxic: they inhibit cytochrome c oxidase, an enzyme essential for cellular respiration in mitochondria, preventing cells from using oxygen. Even small doses can be fatal.

A critical hazard arises when KCN comes into contact with acid. The reaction produces hydrogen cyanide gas (HCN), which is volatile, highly toxic, and can be lethal at very low concentrations in air. This is why the reaction with dilute acid must be carried out in a fume cupboard, and why KCN must never be mixed with acid outside of controlled conditions.

• KCN is acutely toxic by ingestion, inhalation, and skin absorption — it must be handled with gloves, eye protection, and in a fume cupboard.
• Contact with dilute acid generates HCN gas, which is extremely toxic and has a characteristic bitter almond odour (though not everyone can detect it).
• In school and university laboratories, the reaction is often demonstrated using safer alternatives or very small quantities under strict supervision.
• Waste KCN solutions must be disposed of as hazardous chemical waste — never poured down the sink.
• Emergency procedures for cyanide exposure include moving the casualty to fresh air immediately and seeking urgent medical attention.

7. Exam Technique and Mark-Winning Guidance

Mechanism questions are among the most mark-rich in A-level organic chemistry, and they reward precision. Examiners award marks for correct curly arrow placement, correct intermediates, and correct products — not for vague descriptions. The guidance below consolidates the most common errors seen in this topic.