Advanced Spectroscopy: Worked Problems

Introduction

The earlier pages of this unit treated each technique separately. Real problems rarely arrive labelled with the method you need; they arrive as a collection of spectra and a question. This page works through a set of integrated problems in the style of advanced examinations, showing how to set out calculations, how to use correlation data to assemble a structure and how to reconcile evidence from several spectroscopies. Try each problem before reading its solution.

Core explanation

Problem 1 — Assembling a molecule with 2D NMR. A compound C₄H₈O shows ¹H signals at 1.06 (t, 3H), 2.14 (s, 3H) and 2.44 (q, 2H) and ¹³C signals at 7.9, 29.4, 36.9 and 209 ppm.

- COSY: the 1.06 and 2.44 signals correlate, so they form an ethyl group, CH₃CH₂–. The 2.14 singlet has no COSY partner. - HSQC: 1.06 ↔ 7.9, 2.14 ↔ 29.4, 2.44 ↔ 36.9. The 209 ppm carbon has no attached proton, and its shift indicates a ketone C=O. - HMBC: the 2.14 methyl protons correlate with 209 (two bonds) and 36.9 (three bonds); the 2.44 CH₂ protons correlate with 209, 7.9 and 29.4.

The CH₃ singlet and the ethyl CH₂ must both be attached to the carbonyl carbon: the compound is butan-2-one, CH₃COCH₂CH₃. The HMBC link from the methyl singlet to the CH₂ carbon, across the carbonyl, is the decisive connection.

Problem 2 — An NOE distance. In a NOESY spectrum, a geminal CH₂ pair (fixed distance 1.78 Å) gives a cross-peak volume of 64 units. A second cross-peak, between protons on different rings, has a volume of 1.0 unit. Because the NOE scales as r⁻⁶ (in the initial-rate regime), r = 1.78 × (64/1.0)^(1/6) = 1.78 × 2.00 = 3.56 Å. A sixty-four-fold weaker peak means only a doubled distance.

Problem 3 — EPR of a radical. An X-band spectrometer at 9.500 GHz shows the centre of a radical signal at 339.0 mT. Then g = hν/(μ BB) = (6.626 × 10⁻³⁴ × 9.500 × 10⁹) ÷ (9.274 × 10⁻²⁴ × 0.3390) ≈ 2.002, close to the free-electron value 2.0023, as expected for an organic radical. The spectrum has four lines with intensities 1:3:3:1 spaced by 2.3 mT: coupling to three equivalent protons (2 × 3 × ½ + 1 = 4 lines, with Pascal's-triangle intensities). This is the methyl radical, ·CH₃.

Problem 4 — Mössbauer assignment. Two iron sites are fitted in a ⁵⁷Fe spectrum at 80 K (isomer shifts relative to α-Fe). Site A: δ = 1.15 mm/s, ΔE Q = 2.9 mm/s. Site B: δ = 0.47 mm/s, ΔE Q = 0.6 mm/s. Site A's large isomer shift and large quadrupole splitting fit high-spin Fe(II), whose extra d electron shields the s electrons and whose asymmetric t₂g⁴e g² configuration gives a large electric field gradient. Site B fits high-spin Fe(III), whose half-filled d⁵ shell is nearly spherical, so ΔE Q comes mainly from ligand asymmetry and is small.

Problem 5 — A Raman thermometer. For a band at 520 cm⁻¹, the Boltzmann factor at 298 K is exp(−1.4388 × 520 ÷ 298) = exp(−2.51) ≈ 0.081. If the measured, frequency-corrected anti-Stokes to Stokes ratio under the laser is 0.12, then 1.4388 × 520/T = ln(1/0.12) = 2.12, so T ≈ 353 K: the laser has heated the spot by about 55 K.

Formulae

r = r ref(V ref/V)^(1/6) for NOE distances; g = hν/(μ BB); number of hyperfine lines = 2nI + 1; I AS/I S ≈ exp(−hcΔν̃/kT), with hc/k = 1.4388 cm K.

Step-by-step reasoning

For any integrated problem:

1. List every piece of data and the method it came from. 2. Extract the direct deduction from each, with a calculation where needed. 3. Combine connectivity data first (COSY, HSQC, HMBC), then spatial data (NOE), then electronic data (EPR, Mössbauer). 4. Check that the final answer explains every signal, including the absent ones.

Visual explanation

Draw the butan-2-one skeleton and add arrows for each HMBC correlation: two arrows from the methyl singlet protons, three from the CH₂ protons. All arrows converge on or pass through the carbonyl carbon, visually showing that it is the hub joining the two fragments.

Real-world analogy

Solving an integrated spectroscopy problem is like completing a crossword. Each clue (spectrum) alone allows several answers, but the crossing letters (shared atoms and parameters) eliminate all but one consistent solution.

Real-world example

Natural-product chemists routinely assign new antibiotics from about 1 mg of material by combining HSQC, HMBC and NOESY data, exactly as in Problem 1 but on a scale of dozens of carbons. Stereochemistry is then fixed with NOE distances as in Problem 2, often supported by computed chemical shifts.

Why?

Why does a doubling of distance weaken an NOE by a factor of 64? The NOE arises from dipole–dipole cross-relaxation, whose rate depends on the square of the dipolar interaction, which itself falls as r⁻³. The product gives an r⁻⁶ dependence, and 2⁶ = 64.

Common misconception

"HMBC correlations always mean a three-bond relationship." HMBC shows both two- and three-bond couplings, occasionally four-bond ones in conjugated systems, and some three-bond correlations are missing when the dihedral angle makes ³J close to zero.

Worked example

Question: A nitroxide radical at 9.50 GHz gives three equal lines spaced by 1.6 mT. Explain the pattern, and predict the pattern if the ¹⁴N were replaced by ¹⁵N (I = ½).

Reasoning: ¹⁴N has I = 1, so 2 × 1 × 1 + 1 = 3 lines of equal intensity. ¹⁵N has I = ½, giving 2 lines. Its magnetogyric ratio is about 1.40 times larger in magnitude than that of ¹⁴N, so the splitting increases to roughly 2.2 mT.

Answer: Three equal lines from one ¹⁴N; with ¹⁵N, two lines about 2.2 mT apart.

Quick check

1. How many EPR lines, and in what intensity ratio, are expected for a radical coupled to two equivalent protons? Answer: Three lines, 2 × 2 × ½ + 1 = 3, with intensities 1:2:1.

Exam focus

Show every calculation with units, especially for g values, where mT must be converted to T. In structure problems state which correlation proves each bond. For Mössbauer, quote both δ and ΔE Q when assigning oxidation and spin state.

Advanced insight

NOE distances from strongly overlapped peaks or from flexible molecules are averages weighted towards the shortest distances, because ⟨r⁻⁶⟩ is dominated by close approaches. A molecule that spends 10% of its time with two protons at 2.5 Å can give a larger NOE than one fixed at 3.5 Å, so conformational ensembles, not single structures, are needed.

Summary

Integrated problems require extracting a single deduction from each spectrum and then combining them. COSY, HSQC and HMBC connect atoms, NOE intensities give distances through r⁻⁶, EPR g values and 2nI + 1 multiplets identify radicals, Mössbauer δ and ΔE Q assign oxidation and spin states, and anti-Stokes ratios measure temperature. A correct answer explains every signal.

Practice questions

1. An NOE of 8.0 units corresponds to 2.50 Å. What distance corresponds to an NOE of 1.0 unit? Answer: r = 2.50 × 8^(1/6) = 2.50 × 1.414 ≈ 3.54 Å. 2. A signal centred at 336.0 mT is recorded at 9.40 GHz. Calculate g. Answer: g = (6.626 × 10⁻³⁴ × 9.40 × 10⁹) ÷ (9.274 × 10⁻²⁴ × 0.3360) ≈ 2.00. 3. An iron site has δ = 0.45 mm/s and ΔE Q = 0.7 mm/s at 80 K. Assign its oxidation and spin state. Answer: High-spin Fe(III), because both the isomer shift and the quadrupole splitting are small. 4. How many EPR lines does the benzene radical anion show, and why? Answer: Seven lines, because the unpaired electron couples equally to six equivalent protons: 2 × 6 × ½ + 1 = 7.