Three NMR Signals and a Formula: Ethyl Acetate or Methyl Propanoate?
A colourless neutral liquid X, molecular formula C4H8O2, boils at 77 °C, is insoluble in water, miscible with ether, and gives no effervescence with aqueous sodium hydrogen carbonate. Its IR spectrum shows a strong sharp band at 1740 cm^-1, no absorption between 2500 and 3300 cm^-1, and nothing in the 3300–3600 cm^-1 region. Its proton NMR in CDCl3 with TMS shows exactly three signals: a triplet at δ 1.21 (3H), a singlet at δ 2.05 (3H) and a quartet at δ 4.12 (2H). No signal appears downfield of δ 5. Two structures are proposed in the laboratory: methyl propanoate, CH3CH2COOCH3, and ethyl acetate, CH3COOCH2CH3. A student argues that since the two are isomers with the same functional group, the IR cannot separate them and the three NMR signals could be assigned to either structure.
1. Decide between the two structures. Assign every NMR signal, giving the multiplicity, the approximate chemical shift expected for that type of proton, and the reason each signal appears where it does.
Read the worked answer
Structure: ethyl acetate, CH3COOCH2CH3.
The logic rests on which protons sit near oxygen and which sit near the carbonyl, because those two deshielding effects differ by about 2 ppm.
Ethyl acetate:
- δ 2.05, singlet, 3H → CH3 directly bonded to C=O. Carbonyl anisotropy deshields it to ~2.0 ppm, and because no proton is on the adjacent atom (the neighbouring atom is the ester oxygen), there is no coupling, so it is a singlet.
- δ 4.12, quartet, 2H → the OCH2 of the ethoxy group. Attachment to oxygen pushes it to ~4.1 ppm; it is coupled to the adjacent CH3 (3 equivalent neighbours) so n + 1 = 4 lines.
- δ 1.21, triplet, 3H → the terminal CH3 of the ethoxy group, coupled to the adjacent CH2, n + 1 = 3 lines.
Methyl propanoate would give a different fingerprint: CH3CH2 triplet near δ 1.15 (3H), CH2 quartet near δ 2.3 (2H, this CH2 is α to the carbonyl), and OCH3 singlet near δ 3.7 (3H).
The decisive mismatches are the quartet at 4.12 (a CH2 attached to oxygen, not the 2.3 ppm CH2 of methyl propanoate) and the singlet at 2.05 (an acyl CH3, not an OCH3 at 3.7). The student is right that IR alone barely separates the two isomers - both are saturated esters with a C=O near 1735–1745 cm^-1 - but IR plus NMR together are conclusive.
2. Compute the double-bond equivalent, list the functional-group types still compatible with it, and explain whether the absence of a signal beyond δ 5 is, on its own, safe evidence against a carboxylic acid.
Read the worked answer
DBE = (2C + 2 − H)/2 = (8 + 2 − 8)/2 = 1. So X has exactly one ring or one π bond.
Compatible types. With two oxygens and one degree of unsaturation, the plausible families are: a saturated acyclic ester (one C=O); a saturated acyclic carboxylic acid; a hydroxy ketone or hydroxy aldehyde; or a cyclic ether/cyclic ester (lactone) with no C=O only if a ring is present - but a ring plus a C=O would need DBE = 2, so a lactone is excluded here.
Is 'no signal beyond δ 5' sufficient? No, and this is the trap. An O–H proton can be broad, concentration-dependent, and sometimes invisible or exchange-broadened in CDCl3, so the absence of a downfield peak is weak evidence on its own. A carboxylic acid would also fail the sodium hydrogen carbonate test only if it were insoluble or hindered, and butanoic acid is in fact water-soluble and does fizz. What actually excludes the acid is the combination of three independent facts: (i) no O–H stretch anywhere in 2500–3600 cm^-1, (ii) no CO2 with NaHCO3, and (iii) only three proton environments spanning 8 H with an OCH2 pattern at 4.12 and an acyl CH3 singlet at 2.05.
Could it be a hydroxy ketone? CH3COCH2CH2OH would show a broad O–H stretch near 3400 cm^-1, a CH2–O triplet near 3.6 ppm, and more than three environments. It is ruled out by the same three pieces of evidence plus the 1740 cm^-1 band, which is ester-region rather than the ~1715 cm^-1 of a simple ketone.
3. X is boiled with excess aqueous NaOH and the mixture is worked up. Identify the products, and describe two simple tests - one that confirms a product of X and one that would fail if X had been methyl propanoate.
Read the worked answer
Hydrolysis. CH3COOCH2CH3 + NaOH → CH3COONa + CH3CH2OH. Base hydrolysis of an ester is effectively irreversible because the carboxylate is resonance-stabilised and cannot be attacked by the alcohol.
Test 1 - confirming the carboxylate. Acidify a portion of the hydrolysate with dilute H2SO4 and add solid NaHCO3: brisk effervescence of CO2 confirms a carboxylic acid (acetic acid, bp 118 °C, pungent). Alternatively, add a few drops of neutral FeCl3 to the sodium acetate solution - a deep red complex of ferric acetate forms. Note the weakness of the first test: propanoic acid would also fizz, so it identifies the functional group, not the chain.
Test 2 - the discriminating iodoform test. Distil or extract the alcohol fraction, add iodine and NaOH, and warm. Ethanol is oxidised under the conditions to acetaldehyde, which has the CH3–CH(OH)– / CH3–C(=O)– unit, so a yellow precipitate of CHI3 with its characteristic antiseptic smell appears. Methanol - the alcohol that methyl propanoate would have released - has no such unit and gives no precipitate.
Judgement point. The iodoform test is the one that actually decides between the two candidate esters; the NaHCO3 test only confirms that an ester, not an ether or a ketone, was hydrolysed. In an exam, cite the test that discriminates, not just any test that is positive.