A2 · Practice questions · Carbon-13 NMR spectroscopy

Map the skeleton.

Six original Cambridge-style questions on carbon-13 NMR and combined spectroscopy.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Environments, ranges, then combine.

Peaks = carbon environments; use symmetry.
Ranges: alkyl 5–55, C–O 50–90, C=C and aromatic 110–160, C=O 160–220.
No splitting, and heights do not count carbons.
Combine: MS for M_r, IR for groups, NMR for the skeleton.

Analysis

[2 marks]

State the number of peaks in the ¹³C NMR spectra of (a) propan-1-ol and (b) propan-2-ol, and explain the difference.

(a) 3 peaks: all three carbons are in different environments. ✓
(b) 2 peaks: the molecule is symmetrical, so the two CH₃ carbons are equivalent. ✓

Analysis

[3 marks]

Predict the ¹³C NMR spectrum of ethyl ethanoate, CH₃COOCH₂CH₃, giving the approximate shift of each peak.

Four carbon environments, so four peaks. ✓
C=O of the ester about 171 (160–185); O–CH₂ about 60 (50–90). ✓
Two alkyl CH₃ carbons at about 21 and 14 (5–55). ✓

Explain

[2 marks]

Give two differences between a routine ¹³C NMR spectrum and a high-resolution ¹H NMR spectrum of the same compound.

¹³C peaks are single lines with no splitting; ¹H peaks are split by the n+1 rule. ✓
¹³C peak heights do not show the number of atoms, while ¹H peak areas give the ratio of H atoms (also, the ¹³C shift range is much wider, about 0–220 ppm against 0–13). ✓

Analysis

[3 marks]

Compound Z, C₃H₆O, shows: IR strong absorption at 1715 cm⁻¹ and no broad band at 3200–3600 cm⁻¹; ¹³C NMR peaks at 206 and 30; ¹H NMR one singlet. Deduce the structure of Z.

IR: C=O present and no O–H, so Z is an aldehyde or ketone. ✓
Two carbon environments with the C=O at 206 (ketone range); one proton environment with no neighbouring H. ✓
Z is propanone, CH₃COCH₃. ✓

Calculation

[2 marks]

In the mass spectrum of a hydrocarbon the M⁺ peak has height 24.0 and the M+1 peak has height 1.06. Calculate the number of carbon atoms in the molecule.

n = 100 × (M+1 height) ÷ (1.1 × M⁺ height). ✓
n = 100 × 1.06 ÷ (1.1 × 24.0) = 4.02, so 4 carbon atoms. ✓

Analysis

[3 marks]

Pentan-2-one, CH₃COCH₂CH₂CH₃, and pentan-3-one, CH₃CH₂COCH₂CH₃, give almost identical IR spectra. Explain how ¹³C NMR distinguishes them.

Both are ketones, so both absorb near 1715 cm⁻¹ in IR and both have a C=O peak at 190–220 in ¹³C NMR. ✓
Pentan-2-one has no symmetry: five carbon environments, so five peaks. ✓
Pentan-3-one is symmetrical about the C=O: the two CH₂ carbons are equivalent and so are the two CH₃ carbons, giving only three peaks. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.