Cyanohydrin Formation

Introduction

A cyanohydrin forms when a suitable aldehyde or ketone adds the elements of hydrogen cyanide across its C=O bond. The product carbon bears both –OH and –CN. This reaction matters because it creates a new carbon–carbon bond: the carbon of CN⁻ becomes attached to the former carbonyl carbon. Study the bonding and mechanism conceptually; cyanide compounds require specialist safety controls in practice.

Core explanation

For an aldehyde R–CHO, the formal product is R–CH(OH)–CN. For a ketone R–CO–R′, it is R–C(OH)(CN)–R′. The carbon skeleton therefore gains one carbon atom. The nitrile group is not bonded through nitrogen: the carbon end of CN⁻ is the nucleophilic atom that attacks the electrophilic carbonyl carbon. Showing an N–C bond to the carbonyl carbon would give the wrong connectivity.

The first conceptual step is nucleophilic addition. Electron density at the carbon end of cyanide attacks carbonyl carbon; the C=O π electrons move to oxygen. A tetrahedral alkoxide intermediate now has a newly formed C–C bond. The second step is proton transfer to oxygen, giving the hydroxyl group. The overall addition is reversible for many substrates, so product yield depends on structure and conditions. Writing an equilibrium arrow is often more honest than implying every carbonyl converts completely.

Why can this addition be useful in synthesis? A carbonyl already marks a predictable electrophilic site. Installing –CN there adds a carbon-bearing functional group that can be transformed in later synthetic planning, while the carbonyl oxygen is retained as OH. The reaction thus changes both carbon count and functional-group pattern. In a structure problem, count carbons before and after: propanone has three carbons, whereas its cyanohydrin has four.

Aldehydes tend to be less crowded and often more receptive to nucleophilic addition than comparable ketones. However, a ketone such as propanone can form a cyanohydrin, and equilibrium effects matter. Bulky substituents around C=O can hinder attack and favour starting carbonyl. The lesson is a structural prediction, not a claim that all carbonyls give the same isolated yield.

The stereochemistry also needs care. The carbonyl carbon begins approximately trigonal planar and becomes tetrahedral. If its four attached groups are different in the product, it becomes a stereogenic centre. Attack from either face of an achiral planar substrate can give enantiomers in the absence of a chiral influence. Ethanal cyanohydrin, CH₃–CH(OH)–CN, has such a centre; propanone cyanohydrin, (CH₃)₂C(OH)CN, does not because two methyl groups are identical.

Chemically, cyanide and hydrogen cyanide are acutely hazardous. A teaching diagram is enough to understand carbonyl addition and carbon-count changes; it is not an instruction for handling these substances. In an exam, prioritise electron flow, product connectivity, equilibrium and stereochemical consequences.

Step-by-step reasoning

1. Find the C=O carbon and mark it electrophilic. 2. Draw attack from the carbon end of CN⁻ to that carbon. 3. Move C=O π electrons onto oxygen to form an alkoxide. 4. Protonate oxygen in the conceptual mechanism. 5. Verify that OH and CN attach to the same former carbonyl carbon.

Visual explanation

Draw a planar R–C(=O)–R′ triangle and then a tetrahedral R–C(OH)(CN)–R′ centre. Highlight the new bond from carbonyl carbon to the carbon in CN, and show oxygen changing from double-bonded to OH.

Real-world analogy

Imagine adding one new carriage to a train at a marked coupling point. The carbonyl marks the coupling point; cyanide contributes the new carbon carriage, while the existing oxygen stays with the train in a changed role.

Real-world example

In a synthetic route diagram, the conversion of an aldehyde into a cyanohydrin is a recognizable one-carbon extension. A chemist can identify that extension on paper without relying on reagent preparation or laboratory handling details.

Why?

Why is this described as carbon–carbon bond formation rather than simple oxygen protonation? Protonation completes the OH group, but the skeleton-changing event is the bond made between cyanide carbon and carbonyl carbon.

Common misconception

“CN⁻ attaches through nitrogen because N is more electronegative.” The standard cyanohydrin connectivity places the carbon of CN at the former carbonyl carbon. Determine bonds from the product formula, not electronegativity alone.

Worked example

Predict the constitutional product of ethanal CH₃CHO with conceptual cyanide addition followed by protonation. Attach the carbon of CN to the former aldehyde carbonyl carbon; convert C=O to C–OH. The product is CH₃CH(OH)CN, containing three carbons rather than the starting aldehyde's two. That central carbon has H, CH₃, OH and CN, so it can be stereogenic.

Quick check

1. What atom of cyanide bonds to the carbonyl carbon? Answer: The carbon atom of CN⁻, creating a new C–C bond; the nitrile nitrogen remains terminal.

Exam focus

Show the new C–C bond, OH and CN on the same carbon; count the extra carbon. Qualify product formation as reversible when discussing yield.

Advanced insight

The OpenStax treatment at https://openstax.org/books/organic-chemistry/pages/19-6-nucleophilic-addition-of-hcn-cyanohydrin-formation presents this as a nucleophilic carbonyl addition. Product chirality depends on whether the tetrahedral carbon has four distinct substituents, not merely on whether the starting carbonyl was planar.

Summary

Cyanohydrin formation adds cyanide carbon to carbonyl carbon and protonates the oxygen-derived alkoxide. The product carries CN and OH on the former carbonyl carbon, often extending the skeleton by one carbon. Substrate structure influences equilibrium and possible chirality.

Practice questions

1. Give the structural formula for propanone cyanohydrin. Answer: (CH₃)₂C(OH)CN; the former ketone carbonyl carbon bears OH and CN. 2. Does that product have a stereogenic central carbon? Answer: No. Two identical CH₃ groups are attached to that carbon. 3. What happens to the carbonyl π bond in the first step? Answer: Its electrons move onto oxygen as the new C–C bond forms. 4. Why can yield not be predicted from the product drawing alone? Answer: Cyanohydrin formation can be reversible, and substrate structure and conditions determine the equilibrium composition.