The Unknown With Formula C4H8O2
An undergraduate practical in Girona issues an unknown colourless liquid with molecular formula C4H8O2 (Mr 88.11) and boiling point 163 °C; the student must decide whether it is a carboxylic acid or an ester before the session ends. Available reagents are 0.100 M NaOH, phenolphthalein, aqueous NaHCO3 and an IR spectrometer. The student weighs 0.220 g of the unknown, dissolves it in ethanol–water and titrates: 25.0 mL of NaOH is needed to the phenolphthalein endpoint. Adding a few drops of the unknown to NaHCO3 solution produces brisk effervescence. The IR spectrum shows a strong band at 1710 cm⁻¹ and a broad absorption around 3000 cm⁻¹. The printed ¹H NMR spectrum shows δ 0.95 (3H, triplet), 1.70 (2H, sextet), 2.35 (2H, triplet) and 11.8 (1H, broad). A classmate insists the compound is ethyl acetate 'because esters also have the formula C4H8O2'.
1. Use the titration and the qualitative tests to decide whether the unknown is an acid or an ester, and identify its structure.
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Moles of NaOH used: n = 0.100 mol L⁻¹ × 0.0250 L = 2.50 × 10⁻³ mol. If the compound reacts 1:1 with NaOH, its molar mass is 0.220 g/2.50 × 10⁻³ mol = 88.0 g mol⁻¹, which matches C4H8O2 (88.11) to within experimental error. The clean, stoichiometric consumption of base at room temperature is characteristic of a carboxylic acid being deprotonated, not of an ester, which would require heating with excess base to saponify.
The NaHCO3 test confirms it: only a carboxylic acid is acidic enough (pKa ≈ 5) to liberate CO2 from hydrogencarbonate; esters and alcohols do not. The IR band at 1710 cm⁻¹ is a carboxylic acid C=O (esters usually absorb near 1735–1740 cm⁻¹), and the broad 3000 cm⁻¹ band is the hydrogen-bonded O–H of the acid, which esters lack.
NMR: three aliphatic signals in a 3:2:2 pattern plus a broad exchangeable proton at δ 11.8. The triplet at 0.95 (CH3) coupled to a sextet at 1.70 (CH2), which is coupled to a triplet at 2.35 (CH2 next to C=O), gives CH3–CH2–CH2–COOH. The compound is butanoic acid, consistent with its boiling point of 163 °C (ethyl acetate boils at 77 °C).
2. The titration alone gives the molar mass but not the full structure. Explain why, and show how the NMR distinguishes butanoic acid from its acid isomer.
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Both butanoic acid and 2-methylpropanoic acid have the same formula, the same molar mass (88.11 g mol⁻¹) and one carboxyl group, so both consume exactly 1 mol of NaOH per mole and both give the same 25.0 mL endpoint for 0.220 g. The titration therefore fixes the family (monocarboxylic acid) and the molecular mass, but not the carbon skeleton.
NMR distinguishes them by signal count and splitting:
- Butanoic acid, CH3CH2CH2COOH: four signals - triplet at 0.95 (3H), sextet at 1.70 (2H), triplet at 2.35 (2H), broad COOH at 10–12 (1H). This matches the printout exactly.
- 2-Methylpropanoic acid, (CH3)2CHCOOH: three signals - a doublet near 1.2 integrating to 6H (two equivalent methyls split by one CH), a multiplet near 2.6 integrating to 1H, and the COOH proton. A 3:1:1 pattern with a six-proton doublet is not observed here.
The observed 3:2:2 aliphatic pattern with two triplets and one sextet can only arise from a straight three-carbon chain attached to the carboxyl group, so the structure is confirmed as butanoic acid. Note that the titration data would also be satisfied by a mixture of the two acids, which is why the spectroscopic evidence, not the endpoint alone, settles the structure.
3. Evaluate the classmate's claim that the compound is ethyl acetate. How many esters share the formula C4H8O2, and what would each test have shown for ethyl acetate?
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The claim is wrong on every test. Ethyl acetate, CH3COOCH2CH3, has the same formula and molar mass, so it would give the same 88 g mol⁻¹ if it reacted - but it does not react under these conditions. Hydrolysis of an ester needs heating with excess aqueous base; a cold titration to phenolphthalein with one equivalent would not go to completion, and the 1:1 stoichiometry obtained would be coincidental. Ethyl acetate also gives no CO2 with NaHCO3, because it has no acidic proton.
Spectroscopically: ethyl acetate's IR carbonyl is at about 1735–1740 cm⁻¹ with no broad O–H at 3000 cm⁻¹; its ¹H NMR shows a triplet near 1.25 (CH3), a quartet near 4.1 (OCH2, strongly deshielded by oxygen) and a singlet near 2.05 (CH3CO), i.e. a quartet, not the observed sextet, and no signal near 11.8. Its boiling point (77 °C) is also far below the measured 163 °C.
Isomer count: the esters with C4H8O2 are methyl propanoate, ethyl acetate, propyl methanoate and isopropyl methanoate - four structural isomers, plus the two carboxylic acids (butanoic and 2-methylpropanoic). A positive NaHCO3 test, a 1:1 base titration and the acid O–H band eliminate all four esters at once, so the classmate's 'same formula' argument confuses molecular formula with functional group.