Topic 15 · page 28 of 28

Putting the data together

Use accurate mass for the molecular formula, carbon-13 NMR for the number of carbon environments and the functional groups, and proton NMR for the environments, ratio and neighbours. Then assemble a structure that fits every piece.

16 min read15.17(i)15.19(i)15.20(iv)

Read first: Accurate relative molecular mass, Carbon-13 NMR, Predicting and deducing from proton NMR

What each technique tells you 15.17(i)

A four-step flow chart. Step 1, accurate mass: the molecular ion peak gives the molecular formula. Step 2, carbon-13 NMR: the number of peaks is the number of carbon environments and the shifts show C=O, C–O and C–C. Step 3, proton NMR: shift gives the environment, area gives how many H and splitting gives the neighbours. Step 4, assemble and check: build the structure and test it against every piece of data.
Each technique answers a different question. The structure must satisfy all of them.
Technique Question it answers
accurate mass spectrum what is the molecular formula?
carbon-13 NMR how many carbon environments, and which functional groups (C=O, C–O)?
proton NMR, shift what is each hydrogen next to?
proton NMR, area how many hydrogens in each environment?
proton NMR, splitting which groups are next to each other?

MethodDeducing a structure from spectra

  1. Find the molecular formula from the accurate mass.
  2. Count the carbon-13 peaks and use the shifts to find the groups, especially the C=O and any C–O.
  3. Use the proton NMR shifts and areas to list the pieces (such as CH₃CO–, –OCH₂–, CH₃CH₂–).
  4. Use the splitting to say which pieces are joined.
  5. Check: predict the spectra of your structure and compare.

Worked example, an ester 15.19(i)

Worked exampleIdentify compound P

The molecular ion peak of P is at m/z = 88.0522. The carbon-13 NMR spectrum has four peaks at δ = 171, 60, 21 and 14. The proton NMR spectrum has three peaks:

δ / ppm Area Splitting
4.1 2 quartet
2.0 3 singlet
1.2 3 triplet
  1. Formula. The calculated masses are C₃H₄O₃ 88.0159, C₄H₈O₂ 88.0522 and C₅H₁₂O 88.0885. The measured value matches C₄H₈O₂.
  2. Carbon-13. Four peaks mean four carbon environments. δ = 171 is a C=O of an acid or ester (160–185). δ = 60 is a C–O (50–90). δ = 21 and 14 are alkyl carbons (5–55). The compound is an ester, since it has a C=O (160–185) and a C–O but no peak above 190.
  3. Proton NMR. The area ratio is 2 : 3 : 3, which totals 8 H, as in the formula. The quartet (2 H, δ 4.1) is an OCH₂, next to a CH₃. The triplet (3 H, δ 1.2) is that CH₃, so these form –OCH₂CH₃. The singlet (3 H, δ 2.0) is CH₃CO–, which has no neighbouring hydrogens.
  4. Assemble. CH₃CO– joined to –OCH₂CH₃ gives CH₃COOCH₂CH₃.
  5. P is ethyl ethanoate.

The carbon-13 spectrum alone does not decide between ethyl ethanoate and its isomer methyl propanoate, because both have four carbon environments including a C=O and a C–O. The proton NMR does: methyl propanoate has a singlet at about 3.7 for OCH₃ and no singlet at 2.0.

Worked example, a ketone 15.20(iv)

Worked exampleIdentify compound Q

Q has the molecular formula C₃H₆O. Its carbon-13 NMR spectrum has two peaks at δ = 206 and 30. Its proton NMR spectrum has a single peak, a singlet at δ = 2.1.

  1. The peak at 206 is in the 190–220 range: C=O of an aldehyde or ketone.
  2. Two carbon environments, but three carbons in the formula, so two carbons are equivalent: two CH₃ groups on a central carbon.
  3. A singlet for 6 H has no neighbouring hydrogens, consistent with two CH₃ groups attached to C=O (δ 2.1 is in the 2.0–2.7 range).
  4. Q is propanone, CH₃COCH₃. Propanal would give three carbon-13 peaks and a proton peak above 9.4 for the CHO hydrogen.

Check yourself

  1. An unknown has a molecular ion peak at 58.0417. Using the accurate masses C₃H₆O = 58.0417, C₂H₂O₂ = 58.0054 and C₂H₆N₂ = 58.0530, decide the molecular formula.
  2. A compound of formula C₃H₈O has three carbon-13 peaks at δ = 64, 26 and 10. Which of propan-1-ol, propan-2-ol and methoxyethane (CH₃OCH₂CH₃) fit, and which proton NMR feature would then decide?
  3. Why does carbon-13 NMR alone not tell ethyl ethanoate from methyl propanoate? State what does.
  4. A compound C₄H₈O has carbon-13 peaks at δ = 209, 37, 29 and 7, and a proton NMR with a singlet (3 H) at 2.1, a quartet (2 H) at 2.4 and a triplet (3 H) at 1.0. Deduce the structure and say why it is not butanal.
  5. State which technique you would use to (a) find the molecular formula, (b) find the number of carbon environments, and (c) show that a molecule contains a CH₂CH₃ group.

Answers cover these until you have tried the questions

  1. The measured 58.0417 matches only C₃H₆O.
  2. Propan-2-ol has only 2 carbon environments, so it is ruled out. Propan-1-ol (3 peaks: CH₃, CH₂, CH₂OH) and methoxyethane (3 peaks: CH₃, CH₂, CH₃O) both have three. The shifts help: 64 is the only peak in the C–O range (50–90), so one carbon is bonded to oxygen, which fits propan-1-ol. Methoxyethane has two carbons bonded to O and would show two peaks above 50. The proton spectrum would also decide: propan-1-ol has four peaks (3 : 2 : 2 : 1), methoxyethane three (3 : 2 : 3) including a singlet of area 3.
  3. Both esters have four carbon environments, including a C=O (about 171 to 174) and a C–O, so carbon-13 NMR gives four peaks and the same functional groups for each. Proton NMR tells them apart, because ethyl ethanoate has a singlet at about 2.0 and a quartet at about 4.1, and methyl propanoate has a singlet at about 3.7 and a quartet at about 2.3.
  4. Butanone, CH₃COCH₂CH₃. The C=O at 209 is a ketone. The singlet (3 H, 2.1) is CH₃CO–, the quartet (2 H) and triplet (3 H) form –CH₂CH₃. It is not butanal because there is no proton peak at 9.4–10.0 (no CHO hydrogen), and no triplet near 9.8.
  5. (a) Accurate mass spectrometry. (b) Carbon-13 NMR: count the peaks. (c) Proton NMR: a triplet (3 H) and a quartet (2 H) together.