Track electrons, carbon atoms and charges through HCN addition, then distinguish the structural features detected by alkaline iodine.
Carbonyl polarity selects the attacking species
A nucleophile donates an electron pair to an electron-deficient centre. In C=O, oxygen withdraws electron density from carbon, so the carbon is δ+. Cyanide, CN⁻, attacks through its carbon atom, which supplies the bond-forming lone pair. The resulting new C–C bond extends the chain by one carbon.
HCN is used in the presence of KCN: KCN supplies enough CN⁻ to initiate nucleophilic attack, while HCN can supply a proton in the second step. Do not describe neutral HCN as the attacking nucleophile. HCN and soluble cyanides are highly toxic; mechanisms here explain chemistry rather than provide an unsupervised practical procedure.
Two steps with explicit arrow origins and charge bookkeeping
Step 1: put δ+ on the carbonyl C and δ− on O. Draw a full-headed curly arrow from the carbon lone pair of :C≡N⁻ to the carbonyl carbon. At the same time draw a curly arrow from the C=O π bond to O; carbon cannot retain five bonds. The intermediate is tetrahedral and contains a C–CN bond and O⁻. Its total charge is −1.
Step 2: draw a curly arrow from a lone pair on O⁻ to H of H–CN, and another from the H–CN bond back to the cyanide carbon. This forms OH and regenerates CN⁻. An acceptable simplified protonation depiction uses O⁻ to H⁺ when a proton source is indicated. A partial negative δ− on the intermediate oxygen is wrong: it has a full formal − charge.
For propanal the intermediate is CH₃CH₂CH(O⁻)CN and the product is CH₃CH₂CH(OH)CN, 2-hydroxybutanenitrile. The nitrile carbon is carbon 1 when naming the parent nitrile. Check four bonds at the former carbonyl carbon and three bonds between the C and N of CN.
Diagram placeholder
HCN addition: full curly-arrow mechanism
Labels to include:
- Propanal Cδ+=Oδ−
- Lone pair and − charge on cyanide carbon
- Arrow :CN⁻ carbon → carbonyl C; arrow C=O π bond → O
- Tetrahedral CH₃CH₂CH(O⁻)CN intermediate
- Arrow O⁻ lone pair → H of HCN; arrow H–CN bond → CN
- Neutral hydroxynitrile and regenerated CN⁻
The two arrows in the first step form a carbon–carbon bond while opening the carbonyl π bond. Proton transfer then gives OH. Every arrow represents movement of an electron pair; no arrow starts from a positive charge.
A planar starting centre can create a racemate
The geometry around the carbonyl carbon is trigonal planar. In an achiral environment, CN⁻ can attack either face. If the tetrahedral product carbon carries four different groups, these routes give equal amounts of two enantiomers, so the product mixture has no net optical rotation. It is the carbonyl region that is planar; an entire aldehyde molecule need not be planar.
Propanal gives a product centre bonded to H, OH, CN and CH₂CH₃, so it forms a racemate under the simple conditions above. Propanone gives (CH₃)₂C(OH)CN: there are two identical CH₃ groups and hence no chiral centre. Planar attack by itself does not guarantee optical isomerism.
As a synthetic connection, hydrolysing the nitrile converts CN to COOH and can give a hydroxy acid while retaining the new carbon. Always follow the added cyanide carbon; deleting it when drawing the acid undoes the chain extension.
Alkaline iodine detects a particular arrangement
Warm iodine with aqueous sodium hydroxide and the compound. A positive iodoform reaction produces a yellow precipitate of triiodomethane, CHI₃. Among aldehydes and ketones, the reacting arrangement is CH₃–C(=O)–: methyl ketones react, and ethanal is the only simple aldehyde that does. Do not infer that every ketone reacts.
The methyl group next to C=O is ultimately removed as CHI₃, while the remaining acyl portion becomes a carboxylate. For butan-2-one the other organic product is propanoate, not ethanoate. Counting the starting four carbons gives three in propanoate and one in CHI₃.
The test also works with ethanol and alcohols containing CH₃CH(OH)– because iodine in alkali first oxidises them to ethanal or a methyl ketone. Propan-1-ol does not have the required oxidisable pattern. Therefore a positive result supports that structural motif or its precursor; it does not by itself prove a carbonyl was initially present.
| Compound | Result | Structural reason |
|---|---|---|
| Ethanal, CH₃CHO | Positive | CH₃CO– with H as the other substituent |
| Pentan-2-one, CH₃COCH₂CH₂CH₃ | Positive | Methyl ketone |
| Pentan-3-one, CH₃CH₂COCH₂CH₃ | Negative | Neither carbon next to C=O is a methyl group directly bonded to it |
| Butan-2-ol, CH₃CH(OH)CH₂CH₃ | Positive | Oxidises to a methyl ketone |
Worked deduction from a short evidence chain
An original unknown has formula C₄H₈O. It gives a DNPH precipitate, no Tollens’ silver mirror and a positive alkaline-iodine result. DNPH supports aldehyde/ketone; Tollens’ favours ketone; iodoform requires CH₃CO–. The compatible acyclic structure is CH₃COCH₂CH₃, butan-2-one.
Do not jump from “yellow precipitate” to a named molecule. Start with what each reagent tests, impose the molecular formula, and draw a structure satisfying all constraints. If more than one candidate survives, request or interpret further evidence such as NMR rather than inventing certainty.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Where must the first curly arrow begin in cyanide addition to propanone?Show answer
At the lone pair on the carbon atom of CN⁻, and it ends at the carbonyl carbon. A second arrow takes the C=O π-bond pair to oxygen. An arrow from nitrogen would imply different connectivity.
Q2. Name and write the hydroxynitrile formed from ethanal.Show answer
CH₃CH(OH)CN, 2-hydroxypropanenitrile. The CN carbon is included in the three-carbon parent chain; the original ethanal contains only two carbons.
Q3. Why does propanal produce a racemate but propanone produce an achiral hydroxynitrile?Show answer
Both carbonyl centres are planar and can be attacked from either face. Propanal’s product has four different groups at the new tetrahedral centre; propanone’s has two identical methyl groups. Only the first gives two enantiomers.
Q4. Which give a positive iodoform test: pentan-2-one, pentan-3-one, ethanol and propan-1-ol? Explain.Show answer
Pentan-2-one has CH₃CO– and ethanol oxidises to ethanal, so both react. Pentan-3-one has ethyl groups on both sides of C=O, and propan-1-ol oxidises to propanal; neither gives the required methyl-carbonyl pattern.
Q5. Write the two carbon-containing products of the iodoform reaction of pentan-2-one.Show answer
CHI₃ and CH₃CH₂CH₂COO⁻, butanoate, in alkaline solution. The original five carbons divide into one in CHI₃ and four in the carboxylate. Writing butanoic acid would omit the alkaline conditions.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification, Issue 3 — Topic 17, printed pp.38–39; Year 13 content building on 8CH0/9CH0 Topic 6.
- Chemrevise: UK Edexcel Organic Chemistry II — Guide pp.8–10; secondary coverage reference. Explanations, worked data and questions are original Finesse material.
- Pearson 9CH0/02 June 2023 mark scheme — Q7(b)(i–ii), PDF pp.24–25. Cyanide mechanism and racemate explanation.
- Pearson 9CH0/02 June 2023 examiner report — Q7(b)(i–ii), printed/PDF pp.39–42. Local carbonyl planarity, cyanide attack and charge notation.
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