Spectroscopy Practice Questions

Spectroscopy problems reward a systematic approach more than memory. The same logic works every time: get the molecular formula, count the degrees of unsaturation, identify functional groups from IR, then build the carbon skeleton from NMR, and finally check every piece of data. These questions start with single techniques and finish with combined problems of the kind set in advanced exams.

Useful background: mass spectrometry, infrared spectroscopy, NMR explained and how chemists identify an unknown compound.

Reference data used in the answers

Infrared (approximate, cm⁻¹): O–H alcohol 3200–3550 (broad); O–H carboxylic acid 2500–3300 (very broad); N–H 3300–3500; C–H 2850–3100; C=O 1680–1750; C=C 1620–1680; C–O 1000–1300.

¹H NMR chemical shifts (approximate, ppm): R–CH₃ 0.9; R–CH₂–R 1.3; CH₃–C=O 2.0–2.6; CH₃–O or –CH₂–O 3.3–4.2; benzene ring H 6.5–8.0; aldehyde CHO 9–10; carboxylic acid COOH 10–12.

¹³C NMR (approximate, ppm): alkyl C 10–40; C–O 50–70; C=C and aromatic 110–160; C=O in acids and esters 160–185; C=O in aldehydes and ketones 190–220.

Degree of unsaturation (DBE): (2C + 2 + N − H − X) ÷ 2

Part A: Mass spectrometry

1. A compound’s mass spectrum shows peaks at m/z 78 and 80, in a 3 : 1 height ratio, with M⁺ at 78. What does this suggest?

Answer: One chlorine atom. Chlorine-35 and chlorine-37 occur in roughly a 3 : 1 ratio, so molecules containing one Cl give M⁺ and M+2 peaks in that ratio. A mass of 78 with one Cl fits C₃H₇Cl (chloropropane).

2. A compound with M⁺ at 58 gives a strong peak at m/z 43. Suggest the fragment lost and the fragment detected.

Answer: 58 − 43 = 15, the loss of a CH₃ group. The m/z 43 fragment is likely CH₃CO⁺ (or C₃H₇⁺). Together with M = 58, this fits propanone, CH₃COCH₃, which loses CH₃ to give CH₃CO⁺.

Part B: Infrared

3. A compound C₃H₆O shows a strong absorption at 1715 cm⁻¹ and no absorption between 3200 and 3600 cm⁻¹. It gives no reaction with Tollens’ reagent. Identify it.

Answer: 1715 cm⁻¹ is C=O; no O–H band rules out alcohols and acids. A C₃H₆O carbonyl compound is propanal or propanone. No reaction with Tollens’ reagent rules out the aldehyde. The compound is propanone, CH₃COCH₃.

4. How would the IR spectra of ethanol, ethanoic acid and ethyl ethanoate differ?

Answer:

• Ethanol: broad O–H at 3200–3550 cm⁻¹; no C=O.
• Ethanoic acid: very broad O–H at 2500–3300 cm⁻¹ (overlapping the C–H region) and C=O near 1710 cm⁻¹.
• Ethyl ethanoate: C=O near 1740 cm⁻¹ and strong C–O near 1240 cm⁻¹, but no O–H.

Part C: ¹³C NMR

5. How many peaks would you expect in the ¹³C NMR spectra of (a) propan-1-ol, (b) propan-2-ol, (c) propanone?

Answer: (a) CH₃CH₂CH₂OH: three different carbon environments → 3 peaks. (b) (CH₃)₂CHOH: the two CH₃ carbons are equivalent → 2 peaks. (c) (CH₃)₂CO: the two CH₃ carbons are equivalent → 2 peaks, one near 30 ppm (CH₃) and one near 205 ppm (C=O).

6. An isomer of C₄H₁₀O shows only two peaks in its ¹³C NMR spectrum. Suggest its structure.

Answer: Only two carbon environments among four carbons means high symmetry. Candidates: 2-methylpropan-2-ol, (CH₃)₃COH (three equivalent CH₃ carbons + one central C), or ethoxyethane, CH₃CH₂OCH₂CH₃ (two CH₃ equivalent, two CH₂ equivalent). IR would distinguish them: the alcohol has a broad O–H band, the ether doesn’t.

Part D: ¹H NMR

7. Predict the ¹H NMR spectrum of ethanol, CH₃CH₂OH (ignore coupling to the OH).

Answer: Three environments:

• CH₃: about 1.2 ppm, 3H, triplet (2 neighbouring H on CH₂)
• CH₂: about 3.7 ppm, 2H, quartet (3 neighbouring H on CH₃); shifted downfield by the O
• OH: variable (often 2–5 ppm), 1H, usually a singlet because the OH proton exchanges rapidly

8. A compound shows a single peak in its ¹H NMR spectrum at 2.2 ppm. Its formula is C₃H₆O. Identify it.

Answer: One environment for all six H atoms, near a C=O (2.2 ppm): propanone, CH₃COCH₃. Propanal would show three signals, including an aldehyde H near 9.8 ppm.

9. What would adding D₂O do to the ¹H NMR spectrum of ethanol, and why is it useful?

Answer: The OH proton exchanges with deuterium (O–H → O–D). Deuterium doesn’t appear in ¹H NMR, so the OH peak disappears. This confirms which peak belongs to an OH or NH group.

Part E: Combined problems

10. Compound A, C₄H₈O₂:

• IR: very broad absorption 2500–3300 cm⁻¹; strong absorption at 1710 cm⁻¹
• ¹H NMR: 11.8 ppm (1H, singlet); 2.6 ppm (1H, septet); 1.2 ppm (6H, doublet)

Identify A.

Answer:

• DBE = (8 + 2 − 8) ÷ 2 = 1 → one C=O.
• IR: very broad O–H plus C=O → carboxylic acid. Confirmed by the 11.8 ppm singlet (COOH).
• A 6H doublet + 1H septet is the classic isopropyl pattern, (CH₃)₂CH–: the six equivalent CH₃ hydrogens are split by one CH neighbour (doublet); the CH is split by six neighbours (6 + 1 = septet).
• A = 2-methylpropanoic acid, (CH₃)₂CHCOOH. The CH at 2.6 ppm is next to the C=O. ✓

11. Compound B, C₈H₁₀:

• ¹H NMR: 7.2 ppm (5H, multiplet); 2.6 ppm (2H, quartet); 1.2 ppm (3H, triplet)
• ¹³C NMR: 6 peaks

Identify B.

Answer:

• DBE = (16 + 2 − 10) ÷ 2 = 4 → a benzene ring likely (ring + 3 C=C).
• 5H at 7.2 ppm → a monosubstituted benzene ring, C₆H₅–.
• 2H quartet + 3H triplet → an ethyl group; the CH₂ at 2.6 ppm is attached to the ring.
• B = ethylbenzene, C₆H₅CH₂CH₃.
• ¹³C check: ring carbons have 4 environments (the attached carbon, the two ortho, the two meta, and the para), plus CH₂ and CH₃ = 6 peaks ✓. The dimethylbenzene isomers (xylenes), also C₈H₁₀, would show only 4 aromatic H and a 6H singlet for the two CH₃ groups.

12. Compound C, C₅H₁₀O:

• Mass spectrum: M⁺ at 86; strong peaks at 57 and 29
• IR: strong absorption at 1715 cm⁻¹; no O–H
• ¹H NMR: 2.4 ppm (4H, quartet); 1.1 ppm (6H, triplet)
• ¹³C NMR: 3 peaks

Identify C.

Answer:

• DBE = (10 + 2 − 10) ÷ 2 = 1 → one C=O (IR confirms; ketone or aldehyde).
• No aldehyde H (9–10 ppm) → ketone.
• 4H quartet + 6H triplet → two equivalent ethyl groups.
• C = pentan-3-one, CH₃CH₂COCH₂CH₃.
• Mass spectrum: loss of C₂H₅ (29) from 86 gives CH₃CH₂CO⁺ at 57 ✓; C₂H₅⁺ at 29 ✓.
• ¹³C: CH₃, CH₂ and C=O → 3 peaks ✓. (Pentan-2-one, also C₅H₁₀O, would show a 3H singlet near 2.1 ppm and 5 carbon peaks.)

A checklist for combined problems

1. Formula → DBE. Is there a ring or C=O?
2. IR: O–H? C=O? Both? Neither?
3. ¹³C: how many environments? Is there a C=O peak above 160 ppm?
4. ¹H: count signals, use integration ratios, use shifts, use splitting (n + 1).
5. Look for patterns: ethyl (quartet + triplet), isopropyl (septet + 6H doublet), tert-butyl (9H singlet), monosubstituted benzene (5H near 7.2 ppm).
6. Assemble a structure, then check every piece of data, including the mass spectrum fragments.
7. Consider isomers and state which data rules them out.

Key takeaways

• Start with the molecular formula and DBE before looking at any spectrum.
• IR finds functional groups; the absence of a band can be just as informative as its presence.
• ¹³C NMR counts carbon environments; symmetry reduces the count.
• ¹H NMR gives environments, ratios, shift and neighbours; learn the common splitting patterns.
• Always confirm the final structure against all the data and rule out isomers.