Cyanohydrin Formation

Introduction

Aldehydes and relatively unhindered ketones can add HCN across their carbonyl group to form a cyanohydrin. The product has an OH group and a nitrile group, –C≡N, attached to the same former carbonyl carbon. The useful carbon–carbon bond is made by cyanide ion attacking at its carbon end; the oxygen is protonated afterward. The process is reversible, and a small amount of base can accelerate it by generating nucleophilic CN⁻ from HCN.

Core explanation

The carbonyl carbon is partially positive because oxygen draws electron density through the C=O bond. Cyanide ion has a nucleophilic carbon end that can donate its electron pair to that carbon. Draw one arrow from cyanide carbon to carbonyl carbon and another from the C=O pi bond to oxygen. The immediate product is a tetrahedral alkoxide in which the former carbonyl carbon bears the original groups, a new C–C≡N substituent and O⁻. HCN or another proton donor then protonates O⁻, forming the cyanohydrin and regenerating CN⁻ in a simple base-catalysed cycle.

For an aldehyde R–CHO, the product is R–CH(OH)–C≡N. For a ketone R–CO–R′, it is R–C(OH)(R′)–C≡N. The –OH and –CN are geminal , attached to the same carbon, unlike a vicinal halohydrin where OH and halogen are on neighbouring carbons. The cyanide carbon is a new atom in the organic skeleton, so the product has one more carbon than the starting aldehyde or ketone.

The addition is reversible. Pure HCN supplies relatively little CN⁻, so the reaction can be slow; base increases the concentration of the carbon nucleophile. Conditions also influence equilibrium. Very crowded ketones are less favourable because cyanide has difficulty approaching the carbonyl carbon and the tetrahedral product is sterically crowded. Aldehydes commonly react more readily than comparable ketones for the same steric and electronic reasons discussed in general carbonyl addition.

Cyanohydrin formation is valuable because the nitrile group can be transformed. Hydrolysis can eventually convert –C≡N into a carboxylic acid group, while suitable reduction can convert it to a primary amine side chain. Thus cyanide addition is not merely a way to add OH; it installs a versatile one-carbon extension. In a retrosynthetic plan, the carbonyl carbon becomes the OH-bearing centre and the new nitrile carbon supplies the extension.

If the former carbonyl carbon becomes stereogenic, cyanide can attack the two faces of a planar aldehyde or ketone. In an achiral environment, an enantiomeric mixture may form. Benzaldehyde, for example, gives mandelonitrile, whose OH-bearing carbon is chiral; facial control requires a chiral influence. Do not assume that a product drawing without wedges specifies one pure stereoisomer.

The reaction is often schematised as HCN addition, but the mechanistic attacking species is CN⁻. Drawing neutral HCN attacking through carbon without first accounting for cyanide generation can obscure why base accelerates the reaction. HCN and cyanide are highly toxic in real settings, so actual operations require controlled professional procedures; the course focus is electron flow and product structure.

Step-by-step reasoning

Identify the carbonyl carbon and list its original substituents. Attach the carbon end of CN⁻ to that carbon and move the C=O pi pair to oxygen. Draw the tetrahedral O⁻ intermediate, then transfer a proton from HCN or another donor to oxygen. Check that CN and OH are on the same carbon, that the nitrile retains its C≡N triple bond and that the organic skeleton gained one carbon.

Visual explanation

Draw a planar aldehyde with O above the carbonyl carbon. A CN⁻ arrow approaches from one face, with its carbon end leading; a second arrow moves the C=O pi pair to O. The intermediate has O⁻ and C–CN on the same central carbon. A final HCN-to-O proton transfer makes OH while another CN⁻ appears in the catalytic scheme. Draw both attack faces if a chiral centre forms.

Real-world analogy

A flat junction receives a one-carbon extension at its exposed centre, while an upper connector temporarily carries negative charge and is later capped with hydrogen. The extension is cyanide's carbon, the junction is the carbonyl carbon and the cap is protonation of oxygen. The analogy helps keep CN and OH on the same carbon rather than drifting onto adjacent positions.

Real-world example

Benzaldehyde can add HCN to form mandelonitrile, a cyanohydrin. The new nitrile carbon can later be transformed into a carboxylic-acid carbon, giving access to an alpha-hydroxy acid skeleton. The synthesis uses carbonyl addition to create a new C–C bond, and the stereochemistry of the OH-bearing carbon may need control if one enantiomer is desired.

Why?

Why does a little base speed the reaction? HCN is a weak acid, and base converts some of it to CN⁻, the strong carbon nucleophile that actually attacks the electrophilic carbonyl carbon. After CN⁻ addition, HCN protonates the alkoxide and can regenerate CN⁻. The overall addition may be favourable, but the availability of the active nucleophile controls the initial rate.

Common misconception

"The nitrogen end of cyanide bonds to the carbonyl carbon in cyanohydrin formation." The standard product has a new carbon–carbon bond: carbonyl C–C≡N. The nitrile nitrogen stays at the far end of the triple bond. Draw CN⁻ with its carbon attack point and verify the product contains one additional carbon atom.

Worked example

Question: Predict the organic product when ethanal reacts with HCN in the presence of a small amount of base.

Reasoning: CN⁻ attacks ethanal's carbonyl carbon through its carbon end. The C=O pair moves to O, giving CH₃–CH(O⁻)–CN. Proton transfer from HCN converts O⁻ to OH.

Answer: CH₃–CH(OH)–C≡N, 2-hydroxypropanenitrile, forms as a cyanohydrin; an achiral medium can give an enantiomeric mixture.

Quick check

1. Are the OH and CN groups in a cyanohydrin on the same carbon or neighbouring carbons? Answer: On the same former carbonyl carbon, so they are geminal substituents.

Exam focus

Draw carbon-end cyanide attack and simultaneous C=O pi movement, then a separate protonation. Count the extra carbon from CN and keep C≡N intact. State that aldehydes and unhindered ketones react more readily than crowded ketones and that cyanohydrin formation is reversible under suitable conditions.

Advanced insight

Because the addition is reversible, product yield depends on both reaction barriers and equilibrium composition. A cyanohydrin's nitrile can be hydrolysed or reduced in later steps, making the initial addition a strategic one-carbon homologation. Chiral catalysts or enzymes can favour one carbonyl face, converting an otherwise racemic cyanohydrin formation into an enantioselective reaction.

Summary

Cyanohydrin formation joins the carbon end of CN⁻ to an aldehyde or ketone carbonyl carbon. The C=O pi pair moves to oxygen, and protonation yields OH and CN on the same carbon. Base helps generate CN⁻ from HCN, and the addition can reverse. The product gains one carbon and provides a nitrile handle for later chemistry; steric crowding and facial attack affect yield and stereochemistry.

Practice questions

1. Which end of CN⁻ attacks in standard cyanohydrin formation? Answer: The carbon end, making a new carbon–carbon bond to the former carbonyl carbon. 2. What is the immediate charged intermediate after cyanide adds to an aldehyde? Answer: A tetrahedral alkoxide with O⁻ on the former carbonyl oxygen. 3. Why are very crowded ketones poor cyanohydrin substrates? Answer: Steric hindrance raises the barrier for cyanide approach and can disfavour the crowded tetrahedral adduct. 4. What useful group can the new nitrile be converted into by hydrolysis? Answer: A carboxylic acid group, retaining the carbon introduced by cyanide.