3.3.15 NMR Spectroscopy

NMR signals correspond to chemically distinct environments, not to individual atoms. In ¹³C NMR, chemically equivalent carbons share a signal; in ¹H NMR, chemically equivalent protons share a signal.

I count environments before looking at detailed shift values. Symmetry can make several atoms equivalent, so the number of signals may be much smaller than the number of carbons or hydrogens in the formula; that signal count is already a strong structural constraint.

Know the comparison: at this level, ¹³C NMR gives simpler spectra than ¹H NMR. When you use a ¹³C spectrum, focus on how many different carbon environments are present and where their signals appear.

One ¹³C signal means one carbon environment, not necessarily one carbon atom. Symmetry can make several carbons equivalent and therefore give a single signal.

With an NMR spectrum, chemical shift δ is the horizontal position in ppm relative to TMS. Use the data-booklet range as evidence for the local environment; keep that separate from integration and splitting.

Peak position on the δ axis gives chemical shift. Peak area/integration and splitting carry different information, so do not use peak height as though it were a chemical-shift value.

Chemical shift depends on the molecular environment of the observed ¹H or ¹³C nucleus. Use the chemical-shift ranges in the data booklet as evidence for which environments are plausible.

Nearby electronegative atoms, functional groups and π systems can change δ. You do not need a detailed theory of the spectrometer here; use the measured chemical shift to help identify the local chemical environment.

The area under each ¹H-NMR signal is proportional to the number of equivalent protons producing it. Integration therefore gives a ratio between proton environments; it does not usually give an absolute proton count until you combine the ratio with the molecular formula or total hydrogen count.

Use integration alongside the molecular formula. If the simplified signal-area ratio accounts for only half the hydrogens present, scale every term by the same factor; chemical shift and splitting then tell you what those proton sets are attached to.

Use a solvent that does not add ordinary ¹H signals to the spectrum. Deuterated solvents replace most H with D; CCl₄ contains no hydrogen at all.

The purpose is to keep the solvent from swamping the spectrum with ordinary proton signals. Deuterium is ²H, so it is not detected in the same way in an ordinary ¹H-NMR spectrum.

For the NMR reference TMS, I’d remember two headline facts: Si(CH₃)₄ defines δ = 0 ppm, and all twelve protons are equivalent so it gives one sharp ¹H signal.

TMS is also chemically inert towards most samples, volatile and gives a signal away from most organic proton environments, which makes it a convenient reference in practice.

When a question asks why TMS is a good NMR reference, give the practical chemistry as well as δ = 0. It gives one sharp signal, is chemically inert towards most samples and is volatile enough to remove easily.

TMS is chemically inert towards most samples, volatile/easy to remove and gives one signal well away from most organic resonances. Those practical features make it a convenient internal reference.

For structure deduction, make every piece of NMR data constrain the same structure. ¹³C signal count gives carbon environments, chemical shifts suggest local environments, ¹H integration gives proton ratios and splitting tells you about neighbouring non-equivalent protons.

A proposed structure is acceptable only if it explains the whole spectrum. If one signal count, integration or splitting pattern does not fit, reject or modify the structure if any of those data fail to fit.

When you use integration, simplify the ratio first and then scale the whole ratio so the total matches the relevant number of hydrogens in the molecular formula. Do not scale one signal independently of the others.

For example, an integration ratio 2:3 with five relevant hydrogens already corresponds to 2 H and 3 H. If the same ratio had ten relevant hydrogens, scale both terms to 4 H and 6 H.

For the AQA n+1 cases, count the equivalent neighbouring protons on the adjacent carbon and add one. That gives the simple doublet, triplet and quartet patterns you need here.

A quick pattern check is n = 1 → doublet, n = 2 → triplet and n = 3 → quartet. Count equivalent neighbouring protons on the adjacent carbon in the simple aliphatic cases before applying the rule.