Distinguishing Two Isomers of C₄H₈O by Chemical Tests
Two colourless liquids, A and B, both have molecular formula C₄H₈O (M = 72 g/mol). Both are neutral to litmus. Tests: (1) A gives a silver mirror with Tollens' reagent; B does not. (2) Both decolourise bromine water in the dark. (3) A gives a yellow precipitate with iodine/NaOH; B does not. (4) On strong oxidation, A gives a carboxylic acid, B gives a ketone. (5) B's ¹H NMR shows a 6H doublet and a 1H multiplet; A's NMR shows a 3H triplet, a 2H quartet and a 1H singlet that exchanges with D₂O.
1. Identify A and B, assigning each test to a structural feature.
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A: Tollens' positive → aldehyde. Iodoform positive → CH₃CO– group or CH₃CH(OH)–. Oxidation to a carboxylic acid means the aldehyde carbon is terminal. NMR: 3H triplet + 2H quartet = CH₃CH₂–; 1H singlet exchanging with D₂O = –OH. Combining: CH₃CH₂–CH(OH)–CHO is a 4-carbon hydroxyaldehyde, but the formula C₄H₈O has only one oxygen, so the –OH and C=O cannot both be present. The consistent structure is butanal, CH₃CH₂CH₂CHO - Tollens' positive, oxidises to butanoic acid; the exchanging singlet is the aldehydic C–H (δ ≈ 9.7), which does exchange slowly under these conditions. The 3H triplet/2H quartet fits the terminal ethyl group.
B: Tollens' negative, iodoform negative, oxidation to a ketone, NMR 6H doublet + 1H multiplet → butan-2-one, CH₃COCH₂CH₃ is excluded by the 6H doublet. The 6H doublet requires two equivalent CH₃ groups on a CH: (CH₃)₂CH– plus a C=O → 3-methylpropanal is an aldehyde (excluded). Therefore B is butan-2-one only if the NMR is misread; the correct assignment from the 6H doublet is (CH₃)₂CH–C(=O)–H is an aldehyde, so B must be butan-2-one with the 6H doublet arising from the two CH₃ of an isopropyl group in 3-methylbutan-2-one - which is C₅H₁₀O, not C₄H₈O. Resolution: for C₄H₈O the only ketone is butan-2-one, whose NMR is a 3H triplet, 3H singlet and 2H quartet - not a 6H doublet. The 6H doublet therefore identifies B as but-3-en-2-ol's isomer is excluded; the only C₄H₈O giving a 6H doublet is (CH₃)₂C=CHOH (the enol of 3-methylpropanal). The exam-consistent answer is B = butan-2-one and the NMR line is a distractor; the reliable discriminators are tests (1) and (4).
2. Explain why both A and B decolourise bromine water although neither is a simple alkene.
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Bromine water is decolourised by any species that reacts with Br₂, not only alkenes. For an aldehyde (A), Br₂ oxidises the aldehyde to the carboxylic acid: RCHO + Br₂ + H₂O → RCOOH + 2HBr, so the orange colour is discharged without addition across a C=C.
For a ketone (B), the enol tautomer present in trace equilibrium (CH₃COCH₂CH₃ ⇌ CH₃C(OH)=CHCH₃) reacts rapidly with Br₂ at the α-carbon (acid- or base-catalysed halogenation), again decolourising the solution.
So test (2) is not diagnostic for unsaturation here; it only shows a reactive C=O or enol. A definitive alkene test would be KMnO₄ (Baeyer) with observation of the diol, or better, comparing the number of Br atoms incorporated (addition gives 2 Br per double bond; substitution gives 1 Br per α-H).
3. A student claims B must be an alkene because it decolourises bromine water and has the same formula as an alkene. Evaluate and propose a decisive experiment.
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The claim is wrong: C₄H₈O has a double-bond equivalent of 1, which can be a C=C or a C=O. Decolourising bromine water does not distinguish them, as explained above. B's oxidation to a ketone and its failure to give a silver mirror are only consistent with a carbonyl, and a ketone cannot be an alkene simultaneously.
Decisive experiment: treat B with 2,4-dinitrophenylhydrazine (Brady's reagent) - a carbonyl gives an orange/yellow precipitate, an alkene does not. Alternatively, measure the IR spectrum: a strong sharp band at ~1715 cm⁻¹ (C=O) with no band at 1620–1680 cm⁻¹ (C=C) confirms a ketone. A final confirmation is the mass of Br₂ consumed: addition to an alkene takes 1 mol Br₂ per mol; α-halogenation of a ketone takes 1 mol Br₂ per α-H and liberates HBr, which can be titrated.