Identifying an unknown C₄H₈O when one test is contaminated
A colourless liquid, boiling at 79–80 °C, is shown by combustion analysis to have molecular formula C₄H₈O. A student runs three tests in the same test tube: (i) 2,4-dinitrophenylhydrazine reagent gives an immediate orange precipitate; (ii) Tollens' reagent, warmed for 10 minutes, gives no silver mirror, but a faint grey film appears at the rim of the tube, which had earlier been used for a glucose test and rinsed only with tap water; (iii) iodine in sodium hydroxide gives a yellow precipitate with a characteristic sweet smell. The student labels the bottle 'butanal', arguing that a four-carbon carbonyl compound must be an aldehyde. A second student argues for but-3-en-2-ol, since that also gives a positive iodoform test. The teacher allows one further confirmatory experiment only, and asks the class to justify every observation before choosing it.
1. Assign a structure to the unknown, treating each observation explicitly, and state what happens to the 'butanal' and 'but-3-en-2-ol' proposals.
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Observation (i): a positive Brady's test means a carbonyl compound, aldehyde or ketone. It also rules out but-3-en-2-ol, which is an alcohol; its C=C does not give a 2,4-dinitrophenylhydrazone.
Observation (ii): the faint grey rim film is not a silver mirror. It is almost certainly silver left from the earlier glucose test, since the tube was only rinsed with tap water. Discounting it leaves a genuine negative Tollens result, which excludes an aldehyde and therefore excludes butanal.
Observation (iii): a positive iodoform test requires a CH₃CO– group or a CH₃CH(OH)– group. Of the remaining C₄H₈O carbonyls, butan-2-one, CH₃COCH₂CH₃, contains CH₃CO– and gives iodoform; 2-methylpropanal has no such unit and would also fail Tollens differently. Butanal, CH₃CH₂CH₂CHO, has no CH₃CO– unit and would not give iodoform.
So the consistent structure is butan-2-one. The boiling point (about 80 °C) fits. The student's reasoning failed because 'carbonyl plus four carbons' does not imply aldehyde; the ketone is the isomer that satisfies all three tests.
2. Which single confirmatory experiment should be done, and what result would settle the question?
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Repeat the Tollens test in a brand-new, never-used test tube, and run Fehling's or Schiff's test alongside it as an independent aldehyde probe. Aldehydes reduce Tollens' reagent to a true mirror and decolourise Schiff's reagent; ketones do not. If both tests are cleanly negative with freshly prepared reagents, the ketone assignment stands.
A melting-point check on a solid derivative is the stronger classical confirmation, because it is not vulnerable to dirty glassware. Prepare the 2,4-dinitrophenylhydrazone and, after recrystallisation, determine its melting point and compare with the literature value for butan-2-one 2,4-dinitrophenylhydrazone, not that of butanal. A sharp melting point matching the ketone derivative confirms the structure and simultaneously checks purity.
What should not be done is to accept the grey rim film as evidence, or to run more colour tests in the same contaminated tube; the glassware is the variable under suspicion, so the confirmatory test must eliminate it.
3. The student argues that because the iodoform test is positive, the compound 'must contain a methyl group'. Assess this statement.
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It is too broad to be useful. A positive iodoform test does indicate a particular methyl environment, but not any methyl group. The compound must have either CH₃CO– (a methyl ketone) or CH₃CH(OH)– (a methyl carbinol, such as ethanol or propan-2-ol), because the reaction proceeds through exhaustive iodination of that carbon followed by cleavage as CI₃⁻, which is protonated to iodoform, CHI₃.
Methyl groups elsewhere do nothing. Butan-2-one reacts because of CH₃COCH₂CH₃; 2-methylpropanal has two methyl groups and does not give iodoform; butanoic acid, with a terminal methyl, also does not. Equally, ethanol gives iodoform despite having no carbonyl at all, so the test alone cannot distinguish an alcohol from a methyl ketone.
The examinable judgement is that iodoform positivity narrows the structure to a CH₃CO– or CH₃CH(OH)– fragment, and the carbonyl evidence from the 2,4-DNP test is what decides which of those two the compound contains.