An iodoform test that is positive for four different compounds
An unknown compound Z has the molecular formula C₄H₈O and boils at 79–80 °C. A student carries out four tests and records the observations:
- Bromine water shaken with Z: the orange colour persists.
- Tollens' reagent warmed gently with Z: no silver mirror forms.
- Iodine in aqueous sodium hydroxide warmed with Z: a pale yellow precipitate forms.
- Sodium carbonate solution added to Z: no bubbles.
A second sample of Z is warmed with acidified potassium dichromate(VI): the orange solution stays orange.
The student's partner writes in the report: 'Z gives a positive iodoform test, so Z is ethanol.' The teacher marks this incorrect but comments that the iodoform observation really is ambiguous and that the partner has confused an ambiguous test with a decisive one. The teacher then asks for a structure for Z and for an outline of the mechanism of its reaction with HCN.
1. Deduce a structure for Z, showing what each test rules out and why the tests taken together are decisive.
Read the worked answer
Degrees of unsaturation = (2×4 + 2 − 8)/2 = 1. There is one ring or one double bond.
Bromine water is not decolourised, so there is no C=C; the unsaturation must be a carbonyl C=O. Bromine water would also oxidise an aldehyde, so this is further evidence against an aldehyde group.
No effervescence with sodium carbonate: no carboxylic acid (and the formula has only one oxygen, so no acid is possible anyway).
Tollens' reagent gives no silver mirror: not an aldehyde. With the carbonyl established, Z must be a ketone.
Iodoform positive: Z contains a CH₃CO– group, or a CH₃CH(OH)– group.
Dichromate(VI) remains orange: ketones resist oxidation under these conditions, whereas an alcohol or aldehyde would turn the solution green.
The only C₄H₈O ketone containing CH₃CO– is butanone, CH₃COCH₂CH₃. The boiling point of 79.6 °C matches the quoted 79–80 °C. Note that butan-2-ol and ethanol, which also give iodoform, would reduce dichromate and have different formulas - that is how the ambiguity is resolved.
2. Explain precisely why the iodoform result is ambiguous, and identify which single observation in the data set is most decisive in ruling out ethanol.
Read the worked answer
The iodoform reaction is given by any compound containing a CH₃CO– group or a CH₃CH(OH)– group, because the reagent oxidises the alcohol to the methyl ketone before cleaving the CH₃ group. Ethanal, ethanol, propan-2-ol, butan-2-ol and butanone all give a yellow precipitate of CHI₃. A positive result therefore identifies a structural fragment, not a compound, and the partner has treated a group test as a fingerprint.
The most decisive single observation is the molecular formula. Ethanol is C₂H₆O; Z is C₄H₈O with one degree of unsaturation, which cannot be satisfied by ethanol under any arrangement. If a formula were unavailable, the next most decisive observation would be the failure to reduce acidified dichromate(VI), since ethanol is readily oxidised (orange to green) whereas a ketone is not. The boiling point is close for both (ethanol 78 °C, butanone 80 °C) and is genuinely useless for deciding between them, so an answer that leans on boiling point alone is not creditworthy.
3. Outline the mechanism of the reaction of Z with HCN, name the organic product, and explain why the product is formed as a racemic mixture.
Read the worked answer
Z is butanone. The reaction is nucleophilic addition, base-catalysed (a trace of cyanide ion is generated from HCN plus hydroxide present in the reagent).
Steps, described in words:
- The carbonyl carbon is delta positive because oxygen is more electronegative, so the C=O carbon is attacked by the cyanide ion, :CN⁻. A curly arrow from the lone pair on carbon to the carbon atom, and a second arrow from the C=O pi bond to the oxygen, gives a tetrahedral alkoxide intermediate with O⁻ and a new C–CN bond.
- Protonation of the alkoxide oxygen by HCN (or by water) gives the neutral cyanohydrin, and regenerates CN⁻/OH⁻ catalyst.
Product: 2-hydroxy-2-methylbutanenitrile, CH₃CH₂C(OH)(CH₃)CN.
It is racemic because the carbonyl carbon is planar and sp² hybridised, so the cyanide ion can attack from either face with equal probability. When the intermediate is protonated, that carbon becomes tetrahedral and bears four different groups - CH₃, C₂H₅, OH and CN - so it is a stereogenic centre. Attack from one face gives one enantiomer, attack from the other gives the other, in equal amounts, so the product is optically inactive.