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Organic Chemistry II Spectroscopy IR, NMR, MS cheat sheet - grade college

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This cheat sheet covers the core spectroscopy tools used in Organic Chemistry II: IR, NMR, and mass spectrometry. Students need these methods to identify functional groups, carbon frameworks, hydrogen environments, molecular masses, and likely molecular formulas. A compact reference helps connect spectral signals to structural evidence during homework, exams, and lab analysis.

The goal is to turn several spectra into one consistent structure rather than interpret each spectrum in isolation.

IR spectroscopy is most useful for recognizing key bonds and functional groups from diagnostic absorption ranges. 1H^1\mathrm{H} NMR and 13C^{13}\mathrm{C} NMR reveal chemical environments, splitting patterns, integration, and symmetry. Mass spectrometry gives molecular ion information, isotope patterns, and fragmentation clues.

Structure identification works best when you combine molecular formula, degrees of unsaturation, IR peaks, NMR data, and MS fragments into one evidence-based proposal.

Key Facts

  • The degree of unsaturation is DBE=2C+2+NHX2\mathrm{DBE} = \frac{2C + 2 + N - H - X}{2}, where XX counts halogens and oxygen is ignored.
  • A broad IR absorption near 3200 cm13200\text{ cm}^{-1} to 3600 cm13600\text{ cm}^{-1} often indicates an OH\mathrm{O-H} stretch, while a sharper peak near 3300 cm13300\text{ cm}^{-1} can indicate an NH\mathrm{N-H} or terminal alkyne CH\mathrm{C-H} stretch.
  • A strong IR absorption near 1650 cm11650\text{ cm}^{-1} to 1750 cm11750\text{ cm}^{-1} usually indicates a carbonyl C=O\mathrm{C=O} group.
  • In 1H^1\mathrm{H} NMR, the approximate splitting rule is n+1n + 1, meaning nn equivalent neighboring hydrogens split a signal into n+1n + 1 peaks.
  • In 1H^1\mathrm{H} NMR, integration is proportional to the number of hydrogens producing a signal, so a 3:2:13:2:1 integral ratio can represent 33, 22, and 11 hydrogens or any whole-number multiple.
  • Typical 1H^1\mathrm{H} NMR chemical shifts include alkyl protons at 0.8 ppm0.8\text{ ppm} to 2.0 ppm2.0\text{ ppm}, protons next to electronegative atoms at 3.0 ppm3.0\text{ ppm} to 4.5 ppm4.5\text{ ppm}, alkene protons at 4.5 ppm4.5\text{ ppm} to 6.5 ppm6.5\text{ ppm}, aromatic protons at 6.5 ppm6.5\text{ ppm} to 8.5 ppm8.5\text{ ppm}, and aldehyde protons near 9 ppm9\text{ ppm} to 10 ppm10\text{ ppm}.
  • In mass spectrometry, the molecular ion peak M+M^+ gives the molecular mass when it is visible, while the base peak is the tallest peak and is assigned 100%100\% relative intensity.
  • A chlorine-containing compound often shows an M:M+2M:M+2 isotope pattern of about 3:13:1, while a bromine-containing compound often shows an M:M+2M:M+2 pattern of about 1:11:1.

Vocabulary

Infrared spectroscopy
Infrared spectroscopy identifies bonds and functional groups by measuring the frequencies at which molecular bonds absorb IR radiation.
Chemical shift
Chemical shift is the position of an NMR signal in ppm\mathrm{ppm} relative to a reference standard and reflects the electronic environment of a nucleus.
Integration
Integration is the area under a 1H^1\mathrm{H} NMR signal and is proportional to the number of hydrogens in that environment.
Spin-spin splitting
Spin-spin splitting is the division of an NMR signal caused by neighboring nonequivalent nuclei, commonly estimated by the n+1n + 1 rule.
Molecular ion
The molecular ion is the ionized intact molecule, shown as M+M^+, and its m/zm/z value often equals the molecular mass.
Base peak
The base peak is the most intense peak in a mass spectrum and is assigned a relative abundance of 100%100\%.

Common Mistakes to Avoid

  • Using every IR peak as equally diagnostic is wrong because the fingerprint region below about 1500 cm11500\text{ cm}^{-1} is complex and is usually less useful than strong functional-group absorptions.
  • Assuming a missing M+M^+ peak means there is no molecular mass information is wrong because some compounds fragment easily, so isotope peaks and high-mass fragments may still provide clues.
  • Treating NMR integration as exact atom counts is wrong because integrations give ratios first and must be scaled to match the molecular formula.
  • Applying the n+1n + 1 rule to all NMR signals is wrong because equivalent neighbors do not split each other and exchangeable OH\mathrm{O-H} or NH\mathrm{N-H} protons may appear broad or unsplit.
  • Choosing a structure from one spectrum alone is wrong because IR, NMR, MS, and the molecular formula must all agree with the same proposed structure.

Practice Questions

  1. 1 A compound has formula C6H10O\mathrm{C_6H_{10}O} and a strong IR peak at 1715 cm11715\text{ cm}^{-1}. Calculate the degree of unsaturation and identify the likely functional group.
  2. 2 A 1H^1\mathrm{H} NMR signal integrates to 3H3\mathrm{H} and is split into a triplet by neighboring hydrogens. How many equivalent neighboring hydrogens caused the splitting?
  3. 3 A mass spectrum shows molecular ion peaks at m/z=122m/z = 122 and m/z=124m/z = 124 with nearly equal intensity. What halogen pattern is suggested, and why?
  4. 4 A proposed structure contains an alcohol, but the IR spectrum has no broad absorption near 3200 cm13200\text{ cm}^{-1} to 3600 cm13600\text{ cm}^{-1} and the 1H^1\mathrm{H} NMR has no exchangeable proton signal. Explain how this affects confidence in the structure.

Understanding Organic Chemistry II Spectroscopy IR, NMR, MS

Spectra are strongest when treated as constraints, not as pictures to memorize. Each observation removes some possible structures. A carbonyl signal narrows the functional group choices, but it does not by itself distinguish every kind of carbonyl compound.

The surrounding signals matter. For example, an alcohol O-H band often changes shape and position with concentration because hydrogen bonding changes.

Water left in a sample can create confusing broad absorption. IR peak intensity can vary too, so use the presence, shape, and location of several peaks rather than relying on one exact number.

In proton NMR, chemical shift comes from electron density around a hydrogen. Electrons shield a nucleus from the magnetic field. Nearby oxygen, nitrogen, halogens, pi bonds, and aromatic rings reduce shielding, moving a signal farther downfield.

Aromatic rings create a particularly important local magnetic effect. Their circulating electrons shift ring hydrogens into a characteristic region. This is why a proton's position in a molecule matters as much as the atom directly attached to it.

Exchangeable O-H and N-H protons can be broad, weak, or missing. They often fail to show normal splitting because they exchange rapidly with traces of water or other molecules.

Splitting gives information about neighboring hydrogens, but the simple neighbor rule has limits. It works best when neighboring protons are equivalent and have similar coupling strengths. A signal can become a complicated multiplet when it couples to different sets of nonequivalent neighbors.

Protons on the same carbon are not always identical. In a molecule with a chiral center or restricted shape, they may experience different environments and split each other. Students should first identify clean, simple patterns.

A three hydrogen triplet paired with a two hydrogen quartet often suggests an ethyl fragment. It remains a fragment, not proof of a complete structure.

Carbon NMR is useful for counting distinct carbon environments and recognizing symmetry. A molecule with many carbons may produce few signals if several carbons are equivalent. This prevents the mistake of assuming one signal means one carbon.

Standard carbon spectra do not give dependable peak heights for carbon counts, since different carbons relax at different rates. Attached proton methods can add useful detail.

They help separate carbons with no hydrogens from carbons bearing one, two, or three hydrogens. This can reveal whether a carbonyl carbon, a substituted aromatic carbon, or a branching point is present.

Mass spectra reflect the behavior of an ionized molecule, not a neatly broken structure diagram. Some molecules give a clear molecular ion, while others fragment so easily that this peak is weak or absent. Common fragments often form because they create a more stable positive ion.

Benzylic and allylic fragments are favored for this reason. Isotope patterns can provide an early warning that a halogen is present, before every NMR peak is assigned.

In lab work, impurities, solvent peaks, and imperfect integration are normal. Write down only conclusions supported by more than one piece of evidence, then test whether every atom, hydrogen count, and unsaturation requirement fits the proposed structure.