Four Unlabelled Bottles and Two Overconfident Claims
Four unlabelled bottles from a school store cupboard contain butanal, butan-2-one, phenol and ethanoic acid. A student is allowed four reagents: 2,4-dinitrophenylhydrazine (2,4-DNP), Tollens' reagent, aqueous NaHCO3 and neutral FeCl3 solution, plus alkaline iodine solution borrowed from the next bench for one confirmatory test. Bottle Q gives a yellow-orange precipitate with 2,4-DNP, no silver mirror with Tollens', and a yellow precipitate with alkaline iodine. Bottle R decolourises bromine water and gives a violet colour with FeCl3, but shows no effervescence with NaHCO3. A second student claims Q must be butanal because 'only aldehydes give the iodoform test', and that R must be a carboxylic acid because 'only acids react with bromine water'.
1. Identify Q and R, giving the full reasoning, and state precisely what is wrong with each of the second student's two claims.
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Q: the 2,4-DNP precipitate shows a carbonyl compound (aldehyde or ketone). No silver mirror with Tollens' rules out an aldehyde, so Q is a ketone. The iodoform test is positive only when a CH3CO– group (or CH3CH(OH)–) is present; the only methyl ketone available is butan-2-one, CH3COCH2CH3. Butanal, CH3CH2CH2CHO, has no CH3CO– group and gives no iodoform.
R: violet colour with FeCl3 is the classic phenol test (formation of a coloured iron–phenoxide complex), and bromine water is decolourised because phenol undergoes electrophilic substitution to give 2,4,6-tribromophenol, a white precipitate. No effervescence with NaHCO3 confirms it is not a carboxylic acid.
Student's claim 1 is false: iodoform is given by methyl ketones and by alcohols with the CH3CH(OH)– unit (ethanol, propan-2-ol), not only by aldehydes - only acetaldehyde among simple aldehydes responds. Claim 2 is false: bromine water is decolourised by phenols and by alkenes/alkynes as well as by some easily oxidised compounds, so it is not an acid test.
2. Using only the four permitted reagents, describe the minimum sequence that would distinguish butanal from butan-2-one, stating the observation for each.
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Both compounds give an orange-yellow precipitate with 2,4-DNP (carbonyl group present), neither dissolves with effervescence in NaHCO3, and neither gives a violet colour with FeCl3. So three of the four reagents give indistinguishable results and cannot separate them.
The decisive test is Tollens' reagent, warmed gently: butanal gives a bright silver mirror (it is oxidised to butanoate, reducing Ag⁺ to Ag), while butan-2-one gives no mirror because a ketone cannot be oxidised under these conditions without breaking a carbon chain.
If the borrowed alkaline iodine is available, it gives the reverse pattern and confirms the conclusion: butan-2-one gives a yellow precipitate of iodoform from its CH3CO– group, butanal does not. A good answer states explicitly that 2,4-DNP only classifies the compound as a carbonyl compound and that the discriminating reagent is Tollens' (or iodoform), not the one that reacts with both.
3. Phenol dissolves in NaOH but not in NaHCO3, whereas ethanoic acid dissolves in both. Explain using pKa values, and predict whether 2,4,6-trinitrophenol (picric acid) would dissolve in NaHCO3.
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An acid–base reaction goes in the direction that produces the weaker acid. Relevant pKa values: ethanoic acid 4.76, H2CO3 (first) 6.35, phenol 10.0, water 15.7.
With NaOH, the base is OH⁻, whose conjugate acid is water (pKa 15.7). Both phenol (10.0) and ethanoic acid (4.76) are far stronger acids than water, so both are deprotonated and dissolve.
With NaHCO3, the base is HCO3⁻, conjugate acid H2CO3 (pKa 6.35). Only an acid with pKa well below 6.35 can be deprotonated: ethanoic acid (4.76) yes, releasing CO2; phenol (10.0) no, so it stays undissolved. The phenoxide ion is stabilised only by delocalisation into the ring, not by a strongly electron-withdrawing carboxylate-type group, so phenol is too weak an acid.
Picric acid has pKa ≈ 0.4: three nitro groups withdraw electron density and stabilise the phenoxide by resonance and induction, making it far stronger than H2CO3. It should therefore dissolve in NaHCO3 with effervescence.