Aldehydes and ketones contain a polar carbonyl group (\(\text{C=O}\)). Due to the higher electronegativity of oxygen, the double bond is polarised, leaving the carbonyl carbon electron-deficient. This makes it an ideal target for attack by nucleophiles. The reaction of carbonyls with hydrogen cyanide (\(\text{HCN}\)) is a key nucleophilic addition reaction, extending the carbon chain and yielding a class of compounds called hydroxynitriles.
🔑 Key Principle
Nucleophilic addition of \(\text{HCN}\) to aldehydes and ketones increases the carbon chain length by one carbon. Because \(\text{HCN}\) is a weak acid that dissociates poorly, the reaction requires a catalyst such as potassium cyanide (\(\text{KCN}\)) to provide a high concentration of the nucleophile, \(\text{CN}^-\). If the starting carbonyl is asymmetrical, the addition produces a racemic mixture due to equal probability of attack from either side of the planar group.
1. Reaction Conditions and the Role of the KCN Catalyst
The overall reaction between a carbonyl compound and hydrogen cyanide produces a hydroxynitrile (sometimes called a cyanohydrin):
\[ \text{RCHO} + \text{HCN} \xrightarrow{\text{KCN}} \text{RCH(OH)CN} \]
\[ \text{RCOR}' + \text{HCN} \xrightarrow{\text{KCN}} \text{RCR}'\text{(OH)CN} \]
Hydrogen cyanide, \(\text{HCN}\), is a weak acid in aqueous solution and only partially dissociates:
\[ \text{HCN(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{CN}^-(\text{aq}) \]
Because the concentration of the \(\text{CN}^-\) nucleophile is extremely low in pure \(\text{HCN}\), the reaction is painfully slow on its own. To overcome this, potassium cyanide (\(\text{KCN}\)) is added as a catalyst. \(\text{KCN}\) is a soluble salt that dissociates completely in water, supplying a high concentration of \(\text{CN}^-\) ions to initiate the reaction:
\[ \text{KCN(aq)} \rightarrow \text{K}^+(\text{aq}) + \text{CN}^-(\text{aq}) \]
An organic compound containing both a nitrile functional group (\(\text{-C}\equiv\text{N}\)) and a hydroxyl functional group (\(\text{-OH}\)) bonded to the same carbon atom.
An ionic salt used in the addition of HCN to carbonyls. It dissociates completely to provide a high concentration of cyanide nucleophiles, dramatically accelerating the reaction rate.
2. The Two-Step Mechanism
The addition of \(\text{HCN}\) to a carbonyl compound occurs via a two-step nucleophilic addition pathway:
- Step 1: Nucleophilic Attack. The lone pair of electrons on the carbon atom of the cyanide ion (\(\text{CN}^-\)) attacks the electron-deficient carbonyl carbon (\(\text{C}^{\delta+}\)). As the new C-C bond forms, the \(\text{C=O}\) \(\pi\) bond breaks, transferring both electrons to the oxygen atom. This forms a negatively charged alkoxide intermediate.
- Step 2: Protonation. The lone pair on the intermediate's oxygen atom (\(\text{O}^-\)) attacks the hydrogen atom of an \(\text{HCN}\) molecule. This protonates the intermediate to form the stable hydroxyl group (\(\text{-OH}\)) and regenerates the \(\text{CN}^-\) catalyst.
Pay close attention to where your curly arrows start and end:
1. The arrow in step 1 must start from the lone pair or the negative charge on the carbon of the cyanide ion (\(\text{⁻:CN}\)). Starting the arrow from the nitrogen atom is incorrect and will not score.
2. In step 2, the arrow must start from the lone pair on the intermediate's negatively charged oxygen (\(\text{O}^-\)) and point to the hydrogen atom of \(\text{HCN}\).
3. Don't forget to draw the final arrow showing the \(\text{H-CN}\) covalent bond breaking, transferring both electrons to the carbon atom to reform the \(\text{CN}^-\) catalyst.
3. Stereochemical Outcome
If the starting reactant is an aldehyde (except methanal) or an unsymmetrical ketone, the carbonyl carbon is bonded to two different groups. When the \(\text{CN}^-\) group adds, it becomes bonded to four different groups (\(\text{-R}\), \(\text{-R}'\), \(\text{-OH}\), and \(\text{-CN}\)), creating a chiral centre.
Because the carbonyl group has a flat, trigonal planar geometry, the nucleophile has an equal (50%) probability of attacking from either the top face or the bottom face. This results in the formation of an equimolar (50:50) mixture of the two mirror-image enantiomers. This is a racemic mixture, which is optically inactive because the equal and opposite rotations of plane-polarised light cancel each other out.
4. Synthetic Importance of Hydroxynitriles
The addition of \(\text{CN}^-\) is highly valued in synthetic organic chemistry for two main reasons:
- Chain Extension: It increases the length of the carbon backbone by one carbon atom. This is one of the few reactions available at A-Level (along with reacting halogenoalkanes with \(\text{KCN}\) and Friedel-Crafts alkylation/acylation) that can build larger organic molecules.
- Versatile Derivatives: The nitrile group (\(\text{-CN}\)) in the resulting hydroxynitrile can be easily converted into other functional groups:
- Hydrolysis: Heating with dilute acid hydrolyses the \(\text{-CN}\) group to a carboxylic acid (\(\text{-COOH}\)), yielding a 2-hydroxycarboxylic acid (such as lactic acid).
- Reduction: Reacting with hydrogen in the presence of a nickel catalyst reduces the \(\text{-CN}\) group to a primary amine (\(\text{-CH}_2\text{NH}_2\)), producing a 2-hydroxyamine.
Draw the structure of the intermediate alkoxide ion and the final product. Explain if the product is optically active.
Solution:
Step 1: Intermediate Structure
The cyanide ion attacks the carbonyl carbon, and the double bond breaks to form an oxygen anion. The central carbon is now tetrahedral, bonded to \(\text{-CH}_3\), \(\text{-CH}_2\text{CH}_3\), \(\text{-CN}\), and \(\text{-O}^-\):
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CH₃ - C - CH₂CH₃
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CN
Step 2: Final Product Structure and Optical Activity
Protonation of the oxygen anion gives 2-hydroxy-2-methylbutanenitrile:
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CH₃ - C* - CH₂CH₃
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CN
Optical Activity Explanation:
- The product contains a chiral centre at carbon-2 (bonded to \(\text{-CH}_3\), \(\text{-CH}_2\text{CH}_3\), \(\text{-OH}\), and \(\text{-CN}\)).
- The carbonyl group in the reactant butanone is trigonal planar.
- The cyanide nucleophile is equally likely to attack the planar carbonyl carbon from either side.
- This produces a 50:50 racemic mixture of the two enantiomers.
- Therefore, the product is optically inactive because the equal and opposite rotations of plane-polarised light cancel each other out.
Solution:
There are two reasons: electronic and steric.
- Electronic (Inductive) Effect: Alkyl groups are electron-donating. Ketones have two electron-donating alkyl groups attached to the carbonyl carbon, whereas aldehydes only have one. These alkyl groups release electron density towards the carbon atom, reducing its partial positive charge (\(\delta+\)). Because the carbonyl carbon in a ketone is less electron-deficient, it is less attractive to nucleophiles.
- Steric Effect: Ketones have two relatively bulky alkyl groups surrounding the carbonyl carbon, whereas aldehydes have one alkyl group and one tiny hydrogen atom. The bulky groups in ketones crowd the reaction site, making it harder for the incoming nucleophile to approach and attack. This is called steric hindrance.