Difficulty: Intermediate | Prerequisites: General chemistry, basic bonding and molecular structure.
Spectroscopy is how organic chemists figure out what a molecule looks like without being able to see it directly. Mass spectrometry tells you the molecular weight and how the molecule breaks apart. Infrared spectroscopy identifies which functional groups are present by measuring bond vibrations. Nuclear magnetic resonance reveals the carbon-hydrogen framework, including how many distinct proton environments exist and how they connect to each other. Together, these three techniques let you propose (or confirm) a structure from scratch. If you are coming in cold, you should already be comfortable with Lewis structures, electronegativity, and the basics of bonding.
Mass spec gives molecular weight and fragmentation clues. IR identifies functional groups through characteristic absorption frequencies. NMR maps the hydrogen (proton) environment of a molecule, telling you how many types of H are present, their electronic surroundings, and which hydrogens are neighbours.
Mass spectrometry (MS)
An analytical technique that ionises molecules with a high-energy electron beam, producing a radical cation, then separates the resulting ions by their mass-to-charge ratio (m/z) to determine molecular weight and structural fragments.
In simple terms, you blast a molecule with electrons, knock one off, and weigh the pieces.
Radical cation (M+.)
The species formed when a molecule loses one electron in the mass spectrometer. It carries both a positive charge and an unpaired electron.
Think of it as the molecule with one electron stolen, making it both charged and reactive.
Fragmentation
The process by which a radical cation breaks into a smaller cation and a radical. The mass spectrum records the cations (radicals are invisible to the detector).
In simple terms, the unstable radical cation falls apart, and you see the charged pieces.
Alpha cleavage (α-cleavage)
Breaking of the bond next to the atom bearing the radical/charge. Common in alcohols, amines, and carbonyls.
Think of it as cutting the bond right next to where the action is.
McLafferty rearrangement
A specific fragmentation in carbonyl compounds where a gamma hydrogen transfers to the carbonyl oxygen through a six-membered transition state, producing a neutral alkene and a radical cation enol fragment.
In simple terms, a hydrogen hops over to the oxygen through a convenient six-membered ring, and the molecule splits in two.
Infrared spectroscopy (IR)
A technique that measures the absorption of infrared light by molecular bonds. Different bonds vibrate at characteristic frequencies (reported in wavenumbers, cm⁻¹), allowing identification of functional groups.
Think of it as listening to which bonds are humming and at what pitch.
Hooke's Law (applied to IR)
The relationship v = (1/2π)√(k/μ), where v is the vibrational frequency, k is the force constant (bond strength), and μ is the reduced mass. Stronger bonds and lighter atoms vibrate at higher frequencies.
Wavenumber (cm⁻¹)
The unit used for the x-axis in IR spectroscopy. It equals 1/wavelength (in cm) and is directly proportional to frequency and energy.
Fingerprint region
The region below roughly 1500 cm⁻¹ in an IR spectrum. It is complex and unique to each molecule, like a fingerprint, but individual peaks are rarely assigned to specific vibrations.
Nuclear magnetic resonance (NMR)
A technique that exploits the magnetic properties of certain nuclei (most commonly ¹H) to reveal the number, type, and connectivity of atoms in a molecule. Proton precession frequencies shift depending on the local electronic environment.
Chemical shift (δ)
The position of an NMR signal on the spectrum, measured in parts per million (ppm) relative to a reference compound (TMS). Calculated as: δ = (Hz away from TMS) / (spectrometer frequency in MHz).
In simple terms, it tells you how shielded or exposed a proton is.
Shielded / deshielded
Shielded protons are surrounded by electron density and appear upfield (lower δ, closer to 0 ppm). Deshielded protons are near electronegative atoms or pi systems and appear downfield (higher δ, closer to 10 ppm).
Homotopic hydrogens
Hydrogens that are identical by symmetry and give the same NMR signal. Replacing either one with a test atom (e.g. deuterium) gives the same compound.
Diastereotopic hydrogens
Hydrogens on the same carbon that are not equivalent because replacing each one with a test atom gives diastereomers. They can show different chemical shifts.
Spin-spin splitting (coupling)
The splitting of an NMR peak into multiple lines caused by neighbouring, non-equivalent protons. The number of lines follows the n+1 rule, where n is the number of adjacent equivalent protons.
Coupling constant (J)
The distance (in Hz) between the lines of a split NMR peak. It is the same for both sets of coupled protons and is independent of spectrometer frequency. Typical values around 7 Hz for vicinal coupling. When a molecule cannot rotate around the C-C bond between two hydrogens, J can reach roughly 16 Hz.
Integration
The area under each NMR peak, which is proportional to the number of protons giving rise to that signal.
A high-energy electron beam strikes the sample molecule and knocks out one electron, forming a radical cation (M+.).
The radical cation is accelerated through a magnetic field. Its path curves depending on its mass-to-charge ratio (m/z), which is how the instrument sorts ions by weight.
The molecular ion peak (M+) gives the molecular weight directly.
The radical cation is unstable and breaks into a more stable cation plus a radical. Only the cation is detected.
Alkanes: CH₃-CH₃ → [CH₃CH₃]+. → CH₃. + CH₃⁺, or H. + CH₃CH₂⁺
Alcohols: Fragment by alpha-cleavage (breaking the bond next to the oxygen) or by dehydration (loss of water, m/z minus 18).
Alpha-cleavage produces a carbocation adjacent to the OH.
Dehydration gives a peak at M-18.
Amines: Fragment primarily by alpha-cleavage next to nitrogen. The nitrogen lone pair stabilises the resulting cation through resonance.
Carbonyl compounds: Undergo alpha-cleavage and, when a gamma hydrogen is available, the McLafferty rearrangement.
McLafferty: a six-membered transition state transfers a gamma-H to the carbonyl oxygen, producing a neutral alkene fragment and a radical cation enol fragment.
Stronger bond (higher force constant k) → higher frequency (higher wavenumber).
Greater mass difference between bonded atoms → higher frequency.
Larger dipole moment → larger (more intense) peak.
Region | Wavenumber (cm⁻¹) | What Absorbs Here |
|---|---|---|
O-H and N-H stretch | 3200 - 3600 | Broad for O-H (alcohols, carboxylic acids), sharper for N-H |
C-H stretch | 2800 - 3100 | sp³ C-H near 2900, sp² C-H above 3000, sp C-H near 3300 |
Triple bond region | 2000 - 2500 | C≡N (nitriles, sharp), C≡C (alkynes, sometimes weak) |
Carbonyl region | 1670 - 1780 | C=O stretch: acid chlorides ~1785, anhydrides ~1800+1750, esters ~1735 (aliphatic) / 1720 (aromatic), aldehydes ~1730 (aliphatic) / 1703 (conjugated), ketones ~1715 / 1680 (conjugated), amides ~1710 / 1680 (conjugated), carboxylic acids ~1710 |
C=C stretch | 1600 - 2000 | Alkenes, aromatic ring vibrations |
Fingerprint | below 1500 | Complex, unique to each compound |
Carbonyl region (1670-1780 cm⁻¹): Almost always the most diagnostic peak. Different carbonyl types absorb at slightly different positions.
Aromatic region (1600-2000 cm⁻¹): Overtone and combination bands can indicate an aromatic ring.
Triple bond region (2000-2500 cm⁻¹): Nitrile peaks are sharp and medium intensity. Alkyne C≡C can be weak or absent if symmetric.
C-H region (2500-3500 cm⁻¹): The position tells you hybridisation. Less s-character in the bond means lower frequency.
Scan left to right: O-H/N-H first, then C-H, then triple bonds, then C=O, then fingerprint. If you see a strong, sharp peak near 1700, you have a carbonyl. If you see a broad absorption around 3300, you likely have an O-H or N-H.
A strong external magnetic field causes protons to align with or against it. Radio-frequency energy flips protons between states, and the energy required depends on the local electronic environment.
The area (integration) under each signal is proportional to the number of protons producing that signal. A peak integrating for 6H next to a peak integrating for 2H tells you there are three times as many protons in the first environment.
Measured in parts per million (ppm) relative to tetramethylsilane (TMS), which is defined as 0 ppm.
Formula: δ = (Hz away from TMS) / (spectrometer frequency in MHz).
Upfield (low δ, near 0): shielded protons, surrounded by electron density.
Downfield (high δ, near 10): deshielded protons, near electronegative atoms, pi bonds, or aromatic rings.
Equivalent sets of protons give a single NMR signal. For example, the six methyl hydrogens in acetone are all equivalent.
Homotopic protons: interchangeable by a rotation. Always equivalent.
Diastereotopic protons: on the same carbon but in different environments (replacing each one gives diastereomers). These may show different chemical shifts.
To test equivalence, mentally replace each hydrogen with deuterium. If you get the same compound, they are homotopic. If you get enantiomers, they are enantiotopic (equivalent on a standard NMR). If you get diastereomers, they are diastereotopic (non-equivalent).
A proton's signal is split into multiple lines by non-equivalent protons on adjacent carbons.
n+1 rule: If a proton has n equivalent neighbours, its signal splits into n+1 lines.
0 neighbours → singlet (1 line)
1 neighbour → doublet (2 lines)
2 neighbours → triplet (3 lines)
3 neighbours → quartet (4 lines)
The spacing between the lines (in Hz) is the coupling constant, J.
J is measured in Hz and is independent of the spectrometer's operating frequency.
Typical vicinal (three-bond) coupling: approximately 7 Hz.
When the molecule cannot rotate freely around the C-C bond between two hydrogens (e.g. in a double bond or ring), J can be much larger, around 16 Hz for trans alkene protons.
Two mutually coupled protons always share the same J value.
Count the number of distinct signals (how many proton environments).
Read the integration for each signal (relative number of H in each environment).
Note the chemical shift (what kind of environment: alkyl, next to oxygen, aromatic, aldehyde).
Analyse the splitting pattern (how many neighbours).
Students often think the molecular ion peak is always the tallest peak in a mass spectrum. It is not. The base peak (tallest) is the most stable fragment, which may or may not be M+.
Students sometimes confuse wavenumber with wavelength. Higher wavenumber means higher energy and higher frequency, not longer wavelength.
A common error in NMR is counting the number of peaks rather than the number of distinct signals. Splitting can make one signal appear as multiple lines, but it is still one signal.
Students often assume all hydrogens on the same carbon are equivalent. Diastereotopic protons on the same carbon can be non-equivalent and show different chemical shifts.
Coupling constants (J values) do not change with spectrometer frequency. Chemical shifts in Hz do change, but J stays the same. This is why chemical shift is reported in ppm, not Hz.
⚠️ You will almost certainly be asked to identify a functional group from an IR spectrum. Know the carbonyl region cold: acid chloride (~1785), anhydride (two peaks ~1800 and ~1750), ester (~1735), aldehyde (~1730), ketone (~1715), amide (~1680-1710), carboxylic acid (~1710 + broad O-H).
⚠️ NMR problems are a staple. Be ready to determine the number of non-equivalent protons, predict splitting patterns using n+1, and assign chemical shifts to structural environments.
⚠️ Mass spec fragmentation: know that alcohols lose water (M-18), and carbonyls undergo both alpha-cleavage and McLafferty rearrangement.
⚠️ Distinguishing homotopic from diastereotopic hydrogens is a frequently tested concept. Use the substitution test.
⚠️ Coupling constants: know that J ~ 7 Hz is typical for vicinal coupling and J ~ 16 Hz is typical for trans alkene protons.
These three techniques are the everyday toolkit of any synthetic organic chemist. After running a reaction, the first thing you do is take an NMR to confirm your product formed. IR quickly tells you whether a key functional group (e.g. a carbonyl) is present or absent. Mass spec confirms molecular weight and can reveal unexpected side products. Forensic chemistry, pharmaceutical quality control, and environmental monitoring all rely heavily on these methods.
True or False: The molecular ion peak in a mass spectrum is always the tallest peak.
False. The base peak is the tallest. The molecular ion peak may be small or absent.
Fill in the blank: A stronger bond vibrates at a ______ frequency in IR.
Higher.
True or False: Two protons with the same coupling constant J must be coupled to each other.
True (assuming they are in the same molecule and J values match precisely).
Fill in the blank: A proton next to three equivalent neighbours appears as a ______ in the NMR.
Quartet (n+1 = 3+1 = 4 lines).
True or False: Chemical shift (in ppm) changes with spectrometer frequency.
False. Chemical shift in ppm is constant. The shift in Hz changes.
Q: An alcohol with molecular formula C₄H₁₀O shows a mass spectrum peak at m/z = 56. What fragment does this likely represent?
A: Loss of water from the molecular ion (M = 74, 74 - 18 = 56). This is the dehydration fragment, common for alcohols.
Q: You see a strong, sharp IR absorption at 1735 cm⁻¹ and no broad O-H stretch. What functional group is most likely present?
A: An ester. The carbonyl stretch at 1735 cm⁻¹ (aliphatic ester) without a broad O-H rules out carboxylic acid.
Q: A compound shows three signals in its ¹H NMR spectrum with integrations in a 3:2:1 ratio. The signal at δ 9.8 is a triplet. What structural feature does the downfield triplet suggest?
A: An aldehyde proton (δ ~ 9.8) coupled to adjacent CH₂ protons (n = 2, so the aldehyde H appears as a triplet).
Q: Two protons on the same carbon of a molecule containing a chiral centre give different chemical shifts. What type of protons are these?
A: Diastereotopic protons. Replacing each one with deuterium gives diastereomers, so they are non-equivalent.
Q: In the ¹H NMR of 1,1,2-trichloroethane (Cl₂CH-CH₂Cl), predict the splitting pattern for each set of protons.
A: The CHCl₂ proton has two neighbours (the CH₂Cl group), so it appears as a triplet. The CH₂Cl protons have one neighbour (the CHCl₂ proton), so they appear as a doublet.
Spectroscopy connects directly to every reaction you learn in organic chemistry: after any synthesis, these are the tools that confirm whether you made what you intended. IR carbonyl stretches will reappear when you study carboxylic acid derivatives (each type absorbs at a slightly different frequency). NMR becomes essential for distinguishing regiochemistry and stereochemistry in addition and substitution reactions.
Mass spectrometry, MS, radical cation, molecular ion, base peak, fragmentation, alpha cleavage, McLafferty rearrangement, infrared spectroscopy, IR, Hooke's Law, wavenumber, cm-1, functional group identification, carbonyl stretch, O-H stretch, N-H stretch, fingerprint region, NMR, nuclear magnetic resonance, proton NMR, 1H NMR, chemical shift, delta, ppm, TMS, tetramethylsilane, shielded, deshielded, upfield, downfield, integration, proton count, equivalent protons, homotopic, enantiotopic, diastereotopic, spin-spin splitting, coupling, coupling constant, J value, n+1 rule, singlet, doublet, triplet, quartet, CHEM 212, organic chemistry II spectroscopy