Spectroscopy Practice: 6 Full Problems with Detailed
Solutions
These problems are designed to mirror exam-style structure-identification questions using combinations of
MS, IR, ¹H NMR (and occasionally UV/Raman/EPR). For each problem you'll find the spectral data
followed by a detailed step-by-step solution and final structure. Use the workflow: (1) molecular weight &
isotopes (MS), (2) functional groups (IR), (3) connectivity & environments (¹H NMR), (4) cross-check &
propose structure.
Problem 1 — Methyl benzoate (MS + IR + ¹H NMR)
Spectral data:
Spectral data (Unknown A):
- MS: molecular ion peak M■ at m/z = 136 (strong base peak at m/z 77).
- IR (cm■¹): strong absorption at 1715; aromatic C–H region peaks near 3100; no broad O–H.
- ¹H NMR (400 MHz, CDCl■): δ 7.95–7.30 (5H, multiplet), 3.92 (3H, singlet).
Detailed solution:
Step 1 — MS: M■ = 136 → likely molecular formula C■H■O■ (exact mass ≈ 136.05). The base peak at
m/z 77 is characteristic of the phenyl cation (C■H■■), common in aromatic fragmentation.
Step 2 — IR: Strong band at ~1715 cm■¹ indicates a carbonyl group (C=O). Absence of a broad
3200–3600 cm■¹ band rules out a free O–H (no carboxylic acid or alcohol OH). A carbonyl plus no OH
suggests ketone, ester, or aldehyde; but aldehydes show C–H ~2720–2820 cm■¹ and a distinctive ¹H
NMR aldehyde proton near 9–10 ppm (not observed).
Step 3 — ¹H NMR: A 3H singlet at δ 3.92 is typical of a methyl group attached to an oxygen (–OCH■, a
methyl ester). The aromatic region integrates to 5H, consistent with an unsubstituted phenyl ring
(monosubstituted benzene).
Step 4 — Assemble: Monosubstituted aromatic ring (C■H■–), an ester (–COOCH■) gives methyl
benzoate (C■H■O■). The MS fragment m/z 77 corresponds to the phenyl cation (C■H■■), common for
benzene-containing compounds.
Final answer: **Methyl benzoate** (structure: C■H■–COOCH■).
Problem 2 — p-Bromotoluene (MS isotope pattern + ¹H NMR)
Spectral data:
Spectral data (Unknown B):
- MS: two molecular-ion peaks at m/z 157 and m/z 159 of approximately equal intensity (M and M+2 ~1:1);
base peak at m/z 91.
- IR: no strong C=O band; aromatic C–H present.
- ¹H NMR (400 MHz, CDCl■): δ 7.20 (2H, doublet, J = 8.5 Hz), 7.02 (2H, doublet, J = 8.5 Hz), 2.34 (3H,
singlet).
Detailed solution:
Step 1 — MS: M and M+2 peaks of equal intensity point strongly to one bromine atom (■■Br and ■¹Br
give ~1:1). The molecular ion around m/z 157 suggests formula near C■H■Br (exact mass ~157). The
base peak at m/z 91 is the tropylium ion (C■H■■), common in benzyl-type fragments.
Step 2 — IR: No carbonyl; aromatic remains.
Step 3 — ¹H NMR: Two aromatic doublets (2H each) with the same coupling constant (J = 8.5 Hz) indicate
a para-disubstituted benzene (two sets of equivalent protons). A singlet at 2.34 integrating to 3H is a
methyl group not split by neighboring protons → likely a benzylic methyl (–CH■ on the ring).
Step 4 — Assemble: A para-disubstituted benzene with a methyl group (–CH■) and one bromine atom fits
p-bromotoluene (4-bromotoluene). The tropylium ion at m/z 91 is consistent with benzyl fragments.
Final answer: **p-Bromotoluene (4-bromotoluene)**.
Problem 3 — Benzaldehyde (IR + MS + ¹H NMR)
Spectral data:
Spectral data (Unknown C):
- MS: M■ at m/z = 106.
- IR: strong C=O absorption at 1705 cm■¹ and a weak band near 2720 cm■¹ (aldehyde C–H overtone).
- ¹H NMR (400 MHz, CDCl■): δ 9.98 (1H, singlet), 8.05–7.44 (5H, multiplet).
Detailed solution:
Step 1 — MS: M = 106 → consistent with C■H■O (benzaldehyde formula weight 106.12).
Step 2 — IR: Carbonyl at ~1705 cm■¹ and the weak band near 2720 cm■¹ are classic indicators of an
aldehyde (the C–H stretch of the aldehyde often appears as a weak band around 2720–2820 cm■¹).
Step 3 — ¹H NMR: A singlet at δ 9.98 (1H) is diagnostic for an aldehydic proton. The aromatic multiplet
integrating to 5H shows an unsubstituted phenyl ring (monosubstituted benzene).
Step 4 — Assemble: Monosubstituted benzene with an aldehyde group → benzaldehyde (C■H■–CHO).
Final answer: **Benzaldehyde**.
Problem 4 — UV-Vis & Raman (conjugation identification)
Spectral data:
Two unknowns (D and E):
- D: UV-Vis λmax = 200 nm, low molar absorptivity; Raman: weak C=C signal.
- E: UV-Vis λmax = 290 nm, strong molar absorptivity; Raman: strong band at 1600 cm■¹ (C=C stretch),
IR shows no strong polar functional groups.
Question: Which compound is more conjugated? Which is likely to be an aromatic system vs a simple
alkene? Explain.
Detailed solution:
Interpretation — UV-Vis: A larger λmax (bathochromic shift toward longer wavelengths) and higher molar
absorptivity indicate a more extensive π-conjugation or an aromatic chromophore. Compound E (λmax =
290 nm, strong ε) absorbs at a longer wavelength than D (200 nm).
Raman & IR: A strong Raman band at ~1600 cm■¹ corresponds to C=C or aromatic ring stretches. The
absence of polar functional-group absorptions in IR plus a strong Raman C=C band supports a nonpolar
conjugated/aromatic system.
Conclusion: **E is more conjugated** and likely contains an aromatic system (or an extended conjugated π
system) while **D** is likely a simple isolated alkene or non-conjugated chromophore. The spectral
evidence (longer λmax and strong Raman C=C) points to aromaticity/conjugation for E.
Problem 5 — EPR conceptual (free radicals)
Spectral data:
Question: What types of chemical species produce EPR (ESR) signals? What information does hyperfine
splitting in an EPR spectrum provide? Give one short example of an organic radical and how EPR helps
characterize it.
Detailed solution:
Which species produce EPR signals? EPR detects species with one or more unpaired electrons: organic
radicals (e.g., R·), transition-metal complexes with unpaired d-electrons, radical anions/cations, and some
defects in solids.
Hyperfine splitting: When an unpaired electron interacts (couples) with nearby magnetic nuclei (commonly
¹H, ¹■N, etc.), the EPR signal is split into multiple components. The pattern and spacing of hyperfine lines
report how many equivalent nuclei couple to the electron and the coupling constants (A values) give
information on the electron density at those nuclei. This helps identify which atoms or groups are close to
the unpaired electron and the delocalization of the radical.
Example: The nitroxide radical (R■N–O·■) often shows a three-line EPR spectrum from coupling to a
single ¹■N nucleus (I = 1), with further splitting from nearby protons. The relative intensities and coupling
constants allow assignment of whether the unpaired electron is localized on N/O or delocalized into the π
system.
In short: EPR → detects unpaired electrons; hyperfine structure → tells you what nuclei are interacting
with that electron and how the spin is delocalized.
Problem 6 — Benzyl acetate (MS + IR + ¹H NMR)
Spectral data:
Spectral data (Unknown F):
- MS: molecular ion M■ at m/z = 150.
- IR: strong ester C=O band at 1740 cm■¹; C–O stretch in the 1200–1300 cm■¹ region; aromatic C–H
present.
- ¹H NMR (400 MHz, CDCl■): δ 7.35 (5H, multiplet), 4.95 (2H, singlet), 2.08 (3H, singlet).
Detailed solution:
Step 1 — MS: M = 150 suggests a formula near C■H■■O■ (benzyl acetate is C■H■■O■, MW =
150.18).
Step 2 — IR: Strong band at 1740 cm■¹ is characteristic of an ester carbonyl (esters absorb at slightly
higher frequency than ketones). C–O stretches around 1200–1300 cm■¹ further support an ester
functional group.
Step 3 — ¹H NMR: An aromatic multiplet integrating to 5H indicates an unsubstituted phenyl ring
(C■H■–). A singlet at δ 4.95 integrating to 2H suggests a benzylic CH■ (–CH■–O–) adjacent to oxygen
but not split (no neighboring hydrogens). A 3H singlet at δ 2.08 corresponds to a methyl group attached to
the carbonyl (the acetate CH■, –COCH■).
Step 4 — Assemble: A benzyl group (Ph–CH■–O–) bound to an acetyl (–COCH■) fits **benzyl acetate**
(Ph–CH■–O–CO–CH■). The masses, IR, and NMR match.
Final answer: **Benzyl acetate**.