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This cheat sheet covers the main tools chemists use to identify unknown organic compounds: IR spectroscopy, NMR spectroscopy, mass spectrometry, and UV-Vis spectroscopy. Students need these methods to connect molecular structure with experimental evidence. It is designed as a quick reference for recognizing functional groups, interpreting spectra, and combining clues from multiple instruments.

IR spectroscopy identifies bonds by their absorption frequencies, while NMR shows the chemical environments of hydrogen or carbon atoms. Mass spectrometry gives molecular mass and fragmentation clues, and UV-Vis measures electronic transitions in conjugated systems. The most important strategy is to use each spectrum for the information it gives best, then check that all evidence supports one structure.

Key Facts

  • IR absorption wavenumber is related to wavelength by ν~=1λ\tilde{\nu} = \frac{1}{\lambda}, where ν~\tilde{\nu} is usually measured in cm1\text{cm}^{-1}.
  • Broad O-H\text{O-H} stretching usually appears near 3200 to 3600cm13200\text{ to }3600\,\text{cm}^{-1}, while carboxylic acid O-H\text{O-H} can be very broad near 2500 to 3300cm12500\text{ to }3300\,\text{cm}^{-1}.
  • A strong carbonyl C=O\text{C=O} absorption usually appears near 1650 to 1750cm11650\text{ to }1750\,\text{cm}^{-1}, making it one of the most useful IR signals.
  • In proton NMR, chemical shift is calculated by δ=Δνν0×106\delta = \frac{\Delta \nu}{\nu_0} \times 10^6 and is reported in parts per million, or ppm\text{ppm}.
  • The n+1n+1 rule predicts that a proton signal with nn equivalent neighboring protons is split into n+1n+1 peaks.
  • NMR integration gives the relative number of protons producing a signal, so an integration ratio of 3:2:13:2:1 represents relative proton counts.
  • In mass spectrometry, the molecular ion peak M+\text{M}^{+\cdot} gives the molecular mass when the molecule remains intact after losing one electron.
  • In UV-Vis spectroscopy, absorbance follows Beer-Lambert law, A=εbcA = \varepsilon bc, where AA is absorbance, ε\varepsilon is molar absorptivity, bb is path length, and cc is concentration.

Vocabulary

Wavenumber
Wavenumber is the number of wave cycles per centimeter and is commonly used as the x-axis unit in IR spectra.
Chemical Shift
Chemical shift is the position of an NMR signal relative to a reference standard, measured in ppm\text{ppm}.
Integration
Integration is the area under an NMR signal and shows the relative number of nuclei producing that signal.
Molecular Ion
The molecular ion is the charged form of the entire molecule in mass spectrometry, often written as M+\text{M}^{+\cdot}.
Base Peak
The base peak is the tallest peak in a mass spectrum and is assigned a relative intensity of 100%100\%.
Chromophore
A chromophore is a part of a molecule that absorbs UV or visible light because of electronic transitions.

Common Mistakes to Avoid

  • Treating every IR peak as equally important is wrong because the diagnostic region and strong functional group absorptions usually give the clearest structural clues.
  • Confusing NMR integration with peak height is wrong because integration depends on signal area, not how tall the peak appears on the spectrum.
  • Applying the n+1n+1 rule to non-equivalent neighboring protons is wrong because splitting only comes from nearby equivalent protons under typical introductory conditions.
  • Assuming the base peak is always the molecular ion is wrong because the base peak is simply the most abundant fragment, while M+\text{M}^{+\cdot} may be weak or absent.
  • Using one spectrum alone to prove a full structure is risky because IR, NMR, mass spectrometry, and UV-Vis each provide different types of evidence.

Practice Questions

  1. 1 An IR spectrum shows a strong absorption at 1715cm11715\,\text{cm}^{-1} and no broad peak above 3000cm13000\,\text{cm}^{-1}. What functional group is strongly suggested?
  2. 2 A proton NMR signal has 22 equivalent neighboring protons. How many peaks should appear according to the n+1n+1 rule?
  3. 3 A UV-Vis sample has ε=15000L mol1cm1\varepsilon = 15000\,\text{L mol}^{-1}\text{cm}^{-1}, b=1.00cmb = 1.00\,\text{cm}, and c=2.0×105mol L1c = 2.0 \times 10^{-5}\,\text{mol L}^{-1}. Calculate AA using A=εbcA = \varepsilon bc.
  4. 4 Why is it useful to combine IR, NMR, and mass spectrometry instead of relying on only one spectrum when identifying an unknown compound?