12.6 Structure Elucidation: NMR Spectroscopy, Combined Spectral Problems, and Qualitative Organic Tests

12.6 Structure Elucidation: NMR Spectroscopy, Combined Spectral Problems, and Qualitative Organic Tests

Quick Answer: Organic area G (10 items) tests identification of simple organic compounds by chemical tests and spectroscopy, and NMR also appears in the Analytical instrumental area. Combine a molecular formula (degrees of unsaturation), IR bands (Section 7.2), mass spectra (Section 7.4), and proton and carbon NMR into one consistent structure. Then confirm functional groups with classical tests.

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NMR Principles

Nuclei with spin $I = \tfrac{1}{2}$ ($^1H$, $^{13}C$, $^{19}F$, $^{31}P$) align with or against an applied field $B_0$ and absorb radio-frequency energy at the Larmor frequency $\nu = \gamma B_0 / 2\pi$. A 9.4 T magnet makes protons resonate at about 400 MHz.

Electrons around a nucleus oppose $B_0$ (shielding), so chemically different protons resonate at slightly different frequencies. The chemical shift is independent of the spectrometer:

$\delta = \frac{\nu_{sample} - \nu_{TMS}}{\nu_{spectrometer}} \times 10^6\ \text{ppm}$

Tetramethylsilane (TMS) is the reference at 0 ppm. On a 400 MHz instrument, 1 ppm is 400 Hz. Electronegative atoms and $\pi$-system ring currents deshield protons and move signals downfield (to higher $\delta$).

Integration: peak areas are proportional to the number of protons, so they give ratios, not absolute counts.

Spin-Spin Coupling

A proton with $n$ equivalent neighbouring protons (usually on adjacent carbons) splits into $n + 1$ lines with Pascal's-triangle intensities. Equivalent protons do not split each other.

The coupling constant $J$ (Hz) does not depend on field strength. Typical values: free-rotating vicinal $H{-}C{-}C{-}H$ 6-8 Hz; trans alkene 12-18 Hz; cis alkene 6-12 Hz; ortho aromatic 7-9 Hz. O-H and N-H protons exchange rapidly, so they usually appear as broad singlets and vanish after shaking with $D_2O$.

Carbon-13 NMR

$^{13}C$ is only 1.1% abundant, and spectra are normally proton-decoupled, so each unique carbon gives one line.

DEPT spectra sort carbons by attached hydrogens. In DEPT-135, CH and $CH_3$ peaks point up, $CH_2$ peaks point down, and quaternary carbons are absent. DEPT-90 shows only CH carbons.

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Solving a Combined Spectral Problem

1. Degrees of unsaturation (rings plus $\pi$ bonds): $\text{DBE} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1$. Four or more often signals a benzene ring.
2. IR: look for $C{=}O$ near 1700-1750 $cm^{-1}$, broad $O{-}H$ near 3300 (or 2500-3300 for acids), $N{-}H$, $C{\equiv}N$ near 2250, and aromatic overtones.
3. MS: the molecular ion gives $M$; an $M+2$ peak of one-third intensity means Cl, and an equal $M+2$ means Br.
4. Proton NMR: count signal sets, integrate, read splitting, and place groups by chemical shift.
5. Check: the proposed structure must explain every peak.

Worked example: $C_4H_8O_2$; IR strong 1740 $cm^{-1}$, no O-H; $^1H$ NMR $\delta$ 4.12 (q, 2H), 2.04 (s, 3H), 1.26 (t, 3H).

• DBE $= 4 - 4 + 1 = 1$: one $C{=}O$, and the 1740 band suggests an ester.
• The quartet and triplet form an ethyl group; the $CH_2$ at 4.12 ppm sits on oxygen.
• The 3H singlet at 2.04 ppm is a methyl group next to $C{=}O$.
• The structure is ethyl acetate, $CH_3COOCH_2CH_3$. Its isomer methyl propanoate would show a 3H singlet near 3.7 ppm ($OCH_3$) and a quartet near 2.3 ppm.

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Classical Qualitative Organic Tests

Section 12.3 explains the carbonyl test mechanisms, and Section 12.2 covers the Lucas test.