Spectroscopy (IR, NMR, MS)
The idea
Spectroscopy is how chemists read a molecule's structure without ever seeing it directly, by measuring how it interacts with energy. Three techniques carry most of the load. Infrared (IR) spectroscopy detects bond vibrations, so it flags functional groups: a strong band near 1700 cm⁻¹ shouts carbonyl, a broad band around 3300 cm⁻¹ signals O–H or N–H. Nuclear magnetic resonance (NMR) reports on the carbon–hydrogen framework: the number of signals counts chemically distinct hydrogens, chemical shift reflects their electronic environment, and integration gives their relative numbers. Mass spectrometry (MS) weighs the molecule and its fragments, fixing the molecular formula and mass.
The efficient workflow combines them with a single arithmetic anchor — the degree of unsaturation, (2C + 2 + N − H − X)/2 — computed from the formula MS provides. Each unit of unsaturation is one ring or one π bond, so the count tells you how much structural 'tension' must be accounted for before you draw anything. Then IR identifies the functional groups, and NMR maps the hydrogens onto a skeleton. The pieces constrain each other: a carbonyl in IR plus one degree of unsaturation means the molecule has no extra ring or double bond to hide.
The misconception to avoid is reading any single spectrum as a full answer. IR rarely pins an exact structure; NMR signal counts can mislead if symmetry makes distinct-looking hydrogens equivalent. Structure determination is a triangulation — let the molecular formula bound the possibilities, then use each technique to eliminate candidates until one survives.
Worked example
An unknown compound has molecular formula C₃H₆O. Its mass spectrum shows a molecular ion at m/z = 58, its IR spectrum has a strong absorption near 1715 cm⁻¹, and its ¹H NMR shows a single peak near δ 2.1. Deduce the structure.
- Confirm the formula against the mass: C₃H₆O weighs 3(12) + 6(1) + 16 = 58, matching the molecular ion at m/z = 58, so the formula is consistent.
- Compute the degree of unsaturation: (2×3 + 2 − 6)/2 = (8 − 6)/2 = 1, so the molecule contains exactly one ring or one double bond.
- Read the IR: a strong band near 1715 cm⁻¹ is the signature of a C=O stretch, which accounts for the single degree of unsaturation — there is no additional ring or alkene.
- Read the NMR: one signal means all six hydrogens are chemically equivalent. With a carbonyl already placed, two identical CH₃ groups flanking it fit perfectly; their position near δ 2.1 matches methyls next to a carbonyl.
- Assemble and verify: CH₃COCH₃ (acetone, propan-2-one) has formula C₃H₆O, one C=O, and six equivalent methyl hydrogens giving a single NMR peak — every clue is satisfied, so the unknown is acetone.
Answer. The compound is acetone (propan-2-one), CH₃COCH₃ — one carbonyl and two equivalent methyl groups consistent with all three spectra.
Check your understanding
- Why does the degree of unsaturation give you a structural budget before you interpret any peak, and how would a value of 4 immediately suggest an aromatic ring?
- How would the ¹H NMR of acetone differ from that of propanal, which has the same number of carbons but a different arrangement?
- Why can a single NMR signal hide more structural complexity than it seems to, and what role does molecular symmetry play?
- How would you use the relationship between IR bands and degrees of unsaturation to rule out a structure that otherwise fits the formula?
Build the foundations first
Spectroscopy (IR, NMR, MS) builds on these concepts. If any feel shaky, start there.