11.6 Spectroscopic Structure Determination

Integrated Structure Determination

Quick Answer: When a PA-CAT item gives you a molecular formula or a mass-spec molecular ion plus an IR and an NMR, work in this order: (1) compute degree of unsaturation, (2) read IR for functional groups, (3) count 13C peaks for unique carbons, (4) read 1H shifts, integrals, and splitting for connectivity, (5) check that every datum fits one structure. Any mismatch means revise.

This is the integration node for the Bulletin organic Spectroscopy leaf. Chemistry is 16% (~38 items) with 54% Application-level, so a combined-spectroscopy item is plausible: a single compound probed by two or three techniques.

Degree of Unsaturation (DoU)

For a formula CcHhNnOoXx (X = halogen):

DoU = (2c + 2 + n − h − x) / 2 (oxygen ignored).

Each DoU unit is one ring or one π bond. A benzene ring consumes 4 DoU (one ring + three π bonds). A carbonyl consumes 1 DoU.

Worked Example — Unknown Compound X

Given: Mass spectrum shows M+ at m/z 122 (no significant M+2, so no Cl/Br). IR: strong, sharp peak at 1685 cm⁻¹; the very broad 2500–3300 cm⁻¹ envelope is absent, but there is a moderately broad band near 3300 cm⁻¹. 1H NMR (DMSO-d6): δ 6.9 (2H, d), 7.8 (2H, d), 9.8 (1H, s), 10.4 (1H, br s, exchangeable). 13C NMR: 5 signals, including one at 191 ppm.

Step 1 — Molecular formula and DoU. M+ = 122 with no M+2, so no chlorine or bromine. Two oxygens are likely given the carbonyl in the IR, and C7H6O2 has a molecular weight of 122.12 — the fit. (Check the near-miss: C8H6O2 weighs 134.13, not 122, so eight carbons is arithmetically impossible here. Always confirm the candidate formula's mass before building on it.) DoU = (2×7 + 2 − 6)/2 = (14 + 2 − 6)/2 = 10/2 = 5. Five DoU is exactly a benzene ring (4) plus one carbonyl (1), leaving no ring or π bond unaccounted for.

Step 2 — IR. A sharp peak at 1685 cm⁻¹ is a conjugated C=O stretch (lower than a saturated ketone ~1715 because conjugation weakens the bond). The absence of the very broad 2500–3300 cm⁻¹ envelope rules out a carboxylic acid; the moderately broad band near 3300 cm⁻¹ is instead a phenol or alcohol O–H. Learn that distinction — the two O–H shapes are different questions on the exam. The carbonyl could still be an aldehyde (which adds a C–H stretch near 2720/2820 cm⁻¹) or a ketone.

Step 3 — 13C NMR. Five unique carbons. A benzene ring alone has six carbons, so five total signals means the ring is symmetric: a para-disubstituted ring collapses its six carbons into four environments (C1, C2/C6, C3/C5, C4), leaving one signal for the single substituent carbon. The carbon at 191 ppm is a carbonyl; aldehydes typically fall at 190–200 ppm and ketones at 195–220, so an aldehyde is more likely.

Step 4 — 1H NMR. Two doublets of 2H each in the aromatic region (δ 6.9, 7.8) with mutual coupling are the signature of a para-disubstituted benzene (AA'BB' pattern). The singlet at δ 9.8 (1H) is the classic aldehyde proton, and the broad, exchangeable δ 10.4 (1H) is a phenol O–H — exchangeable protons are broad and their shift moves with solvent and concentration, so never use them for connectivity.

Step 5 — Assemble. The ring is para-disubstituted, one substituent is –CHO (it accounts for the 1H singlet at 9.8, the 13C at 191 ppm, and the only non-aromatic carbon). Count what is left: a para-disubstituted ring contributes C6H4 and the formyl group contributes CHO, giving C7H5O. To reach C7H6O2 the second substituent must be –OH, with no carbon of its own — which is exactly why the 13C shows five signals and not six. Compound X is 4-hydroxybenzaldehyde.

Cross-check every datum against that one structure: MW 122.12 ✓; DoU 5 = ring + C=O ✓; conjugated aldehyde C=O at 1685 cm⁻¹ ✓; phenol O–H near 3300 cm⁻¹, no acid envelope ✓; 5 13C signals with a 191 ppm carbonyl ✓; 6 total protons as 2 + 2 + 1 + 1 ✓. Every spectrum agrees, so the assignment stands. Contrast the near misses the exam will offer you: terephthalaldehyde (benzene-1,4-dicarbaldehyde) is C8H6O2 at MW 134 and its symmetry gives only four 13C signals; 4-methoxybenzaldehyde is C8H8O2 at MW 136 and would add a 3H methoxy singlet near δ 3.8; benzoic acid is C7H6O2 at the same MW 122 but shows the broad 2500–3300 acid envelope and a carbonyl near 167 ppm rather than 191. Same nominal mass does not mean same compound — the mass narrows the field, and the IR and NMR pick the winner.

Workflow Discipline

Common PA-CAT Traps

• A carbonyl in IR without an aldehyde C–H is a ketone, not an aldehyde.
• A broad O–H from 2500–3300 cm⁻¹ is a carboxylic acid; a sharp O–H near 3300 is a phenol or alcohol; an N–H near 3300 is an amine.
• 13C NMR with fewer peaks than the formula suggests symmetry—don't overcount carbons.
• A singlet integrating to 1H near 10 ppm is the aldehyde proton; a singlet near 2 ppm integrating to 3H is an acetyl CH3.