Source: Brown, Iverson, Anslyn, Foote, Organic Chemistry 8th ed., Chapters 21 (Part I), 12, 13, 14, 20.3
Difficulty: Intermediate | Prerequisites: CHM 25500 (functional groups, bonding, stereochemistry, reaction mechanisms through SN1/SN2/E1/E2)
Tags: aromatic chemistry, aromaticity, Hückel's rule, benzene, spectroscopy, mass spectrometry, infrared spectroscopy, IR, NMR, nuclear magnetic resonance, proton NMR, carbon-13 NMR, UV-Vis spectroscopy, fragmentation pattern, chemical shift, coupling constant, degree of unsaturation
This module opens CHM 25600 by pairing two large ideas: what makes a molecule aromatic, and how chemists figure out what an unknown molecule looks like. Aromaticity underpins much of the reactivity you will study for the rest of the semester (electrophilic aromatic substitution, nucleophilic aromatic substitution, amine chemistry). Spectroscopy, meanwhile, is the toolkit you use to confirm the structure of anything you make or isolate in the lab. If CHM 25500 taught you how to push electrons, this module teaches you how to read the evidence that tells you whether your mechanism was right. You should already be comfortable drawing Lewis structures, understanding electronegativity, and recognising common functional groups.
Aromatic compounds are unusually stable cyclic, planar, fully conjugated systems that follow Hückel's rule (4n + 2 pi electrons). The four spectroscopic methods covered here (MS, IR, NMR, UV-Vis) each reveal different structural information, and exam problems typically require you to combine data from all four to deduce an unknown structure.
Aromaticity
The special thermodynamic stability associated with cyclic, planar, fully conjugated systems that contain 4n + 2 pi electrons. In simple terms, aromatic rings are far more stable than you would predict from their number of double bonds alone.
Hückel's Rule
A molecule is aromatic if it is cyclic, planar, fully conjugated, and contains 4n + 2 pi electrons (where n = 0, 1, 2, 3...). Think of it as the "magic number" test: 2, 6, 10, 14 pi electrons pass; 4, 8, 12 do not.
Antiaromatic
A cyclic, planar, fully conjugated system with 4n pi electrons. These molecules are less stable than the equivalent open-chain conjugated system. In simple terms, antiaromaticity is the penalty you pay when the electron count is wrong for the ring geometry.
Degree of Unsaturation (Index of Hydrogen Deficiency, IHD)
The number of rings plus double bonds in a molecular formula. Calculated as (2C + 2 + N − H − X) / 2 for a formula C_cH_hN_nO_oX_x. Think of it as counting how many pairs of hydrogens are "missing" relative to the fully saturated version of that formula.
Mass Spectrum (MS)
A plot of ion abundance vs. mass-to-charge ratio (m/z). The molecular ion peak (M⁺) gives the molecular weight; fragmentation peaks reveal structural pieces. Think of it as weighing the molecule and then seeing which pieces break off.
Base Peak
The tallest peak in a mass spectrum, assigned a relative abundance of 100%. It is the most stable fragment ion formed, not necessarily the molecular ion.
Fragmentation
The breaking of bonds in the molecular ion to produce smaller cation and radical pieces. Common patterns include loss of 15 (CH₃), 18 (H₂O), 28 (CO or C₂H₄), 29 (CHO), 31 (OCH₃), and 45 (OC₂H₅).
Infrared Spectroscopy (IR)
Measures absorption of infrared light by molecular vibrations (stretching and bending of bonds). Each functional group absorbs at a characteristic frequency (reported in wavenumbers, cm⁻¹). Think of it as a fingerprint for which functional groups are present.
Wavenumber (cm⁻¹)
The unit used for IR absorption frequencies. Higher wavenumber means higher energy vibration, which corresponds to stronger or lighter bonds.
Chemical Shift (δ, ppm)
The position of an NMR signal relative to a reference compound (TMS, tetramethylsilane, at δ = 0). Reflects the electronic environment around the nucleus. In simple terms, the more deshielded a proton is (closer to electronegative atoms or pi systems), the further downfield (higher ppm) it appears.
Shielding and Deshielding
Shielded nuclei are surrounded by higher electron density and resonate upfield (lower ppm). Deshielded nuclei have lower electron density and resonate downfield (higher ppm).
Spin-Spin Coupling (Splitting)
Neighbouring non-equivalent hydrogens split an NMR signal into multiple peaks. The n + 1 rule says that n equivalent neighbours produce n + 1 peaks (doublet, triplet, quartet, etc.).
Coupling Constant (J, Hz)
The distance in Hz between the peaks of a split signal. Mutually coupled protons share the same J value, which helps you identify which signals are neighbours.
Integration
The area under an NMR peak, proportional to the number of protons producing that signal. Read as a ratio, not an absolute count.
Carbon-13 NMR (¹³C NMR)
Each chemically distinct carbon gives one peak. No splitting in broadband-decoupled spectra. Typical range 0–220 ppm.
DEPT (Distortionless Enhancement by Polarisation Transfer)
A ¹³C NMR technique that distinguishes CH₃, CH₂, CH, and quaternary C.
UV-Vis Spectroscopy
Measures absorption of ultraviolet and visible light, which promotes electrons from bonding/non-bonding orbitals to antibonding orbitals (pi to pi* and n to pi* transitions). Reports lambda max (λ_max) in nm. Think of it as detecting how extended the conjugation in a molecule is.
Chromophore
The part of a molecule responsible for UV-Vis absorption, typically a conjugated pi system.
Bathochromic Shift (Red Shift)
A shift of λ_max to longer wavelength, usually caused by extending conjugation.
Benzene is the archetype: six carbons, six pi electrons, planar, cyclic, every atom sp² hybridised
Criteria for aromaticity (all four must be met):
Cyclic
Planar (all atoms in the ring can achieve a coplanar geometry)
Fully conjugated (every atom in the ring has a p orbital available for overlap)
4n + 2 pi electrons (Hückel's rule)
Common aromatic systems beyond benzene:
Cyclopentadienyl anion (6 pi electrons, the lone pair on the sp² carbon enters the pi system)
Cycloheptatrienyl cation (tropylium, 6 pi electrons)
Pyridine, pyrrole, furan, thiophene (heteroaromatic rings)
Naphthalene (10 pi electrons), anthracene (14 pi electrons)
Antiaromatic systems (4n pi electrons, cyclic, planar, conjugated):
Cyclobutadiene (4 pi electrons), planar cyclopropenyl anion (4 pi electrons)
These systems distort or react to avoid antiaromaticity
Non-aromatic systems fail one or more criteria:
Cyclooctatetraene is tub-shaped (not planar), so it is non-aromatic despite having a continuous ring of p orbitals
The molecular ion (M⁺) peak gives the molecular weight
Even M⁺ means an even number of nitrogen atoms (or zero)
Odd M⁺ means an odd number of nitrogen atoms (nitrogen rule)
High-resolution MS can give the exact molecular formula
Isotope patterns:
Chlorine: M and M+2 in roughly 3:1 ratio
Bromine: M and M+2 in roughly 1:1 ratio
Common neutral losses to memorise:
15: CH₃ (methyl)
17: OH
18: H₂O (alcohols)
28: CO (aldehydes, ketones) or C₂H₄ (ethylene, from ethyl groups)
29: CHO (aldehydes)
31: OCH₃ (methyl ethers)
45: OEt (ethyl ethers)
46: NO₂ or C₂H₅OH
McLafferty rearrangement: a carbonyl compound with a gamma hydrogen can transfer that hydrogen to the carbonyl oxygen via a six-membered transition state, losing a neutral alkene
Key absorptions to know:
O–H stretch: broad, 3200–3600 cm⁻¹ (alcohols and carboxylic acids; acids are broader and extend to ~2500 cm⁻¹)
N–H stretch: 3300–3500 cm⁻¹ (primary amines show two peaks; secondary amines show one)
C–H stretch: ~3000 cm⁻¹ (sp³ C–H just below 3000; sp² and sp C–H just above 3000)
C≡C or C≡N stretch: 2100–2260 cm⁻¹ (triple bond region)
C=O stretch: 1650–1800 cm⁻¹ (the single most diagnostic region in IR)
Esters: ~1735–1750 cm⁻¹
Aldehydes: ~1720–1740 cm⁻¹
Ketones: ~1705–1720 cm⁻¹
Carboxylic acids: ~1710–1715 cm⁻¹
Amides: ~1630–1690 cm⁻¹
Anhydrides: two C=O peaks
C=C stretch: 1600–1680 cm⁻¹
Aromatic C=C: ~1450–1600 cm⁻¹ (often two or three peaks)
The fingerprint region (below ~1500 cm⁻¹) is complex but useful for matching known compounds
Chemical shift ranges (approximate, memorise these):
0.9–1.0 ppm: R–CH₃
1.2–1.4 ppm: R–CH₂–R
1.5–2.0 ppm: allylic C–H
2.0–2.5 ppm: C–H next to C=O
3.3–4.0 ppm: C–H next to O or N (e.g. ethers, alcohols)
4.5–6.5 ppm: vinyl (alkene) C–H
6.5–8.5 ppm: aromatic C–H
9.0–10.0 ppm: aldehyde C–H
10–12 ppm: carboxylic acid O–H
Splitting patterns (n + 1 rule):
Singlet: 0 neighbours
Doublet: 1 neighbour
Triplet: 2 equivalent neighbours
Quartet: 3 equivalent neighbours
Integration gives the ratio of protons in each signal
Equivalent protons do not split each other
Coupling constants (J values):
Geminal (2-bond): typically 0–3 Hz (often not observed in equivalent CH₂)
Vicinal (3-bond): 6–8 Hz (free rotation), can vary with dihedral angle
Trans alkene: 12–18 Hz
Cis alkene: 6–12 Hz
Aromatic ortho: 6–10 Hz
Aromatic meta: 1–3 Hz
Aromatic para: 0–1 Hz
Each chemically distinct carbon gives one line (in broadband-decoupled spectra)
Chemical shift ranges:
0–50 ppm: sp³ carbons (alkyl)
50–90 ppm: C bonded to O or N
100–150 ppm: sp² carbons (alkene, aromatic)
150–220 ppm: carbonyl carbons (C=O)
DEPT experiments distinguish CH₃, CH₂, CH, and quaternary C
Symmetry reduces the number of peaks: benzene gives one ¹³C peak
Molecules with extended conjugation absorb at longer wavelengths (lower energy)
Isolated double bonds absorb below 200 nm (not typically observed with standard instruments)
Conjugated dienes absorb around 215–250 nm
Each additional conjugated double bond shifts λ_max roughly 30–40 nm to longer wavelength
Aromatic rings: benzene absorbs at ~254 nm; substituents shift this
Woodward's rules (Woodward-Fieser rules) allow prediction of λ_max for conjugated dienes and enones by adding increments for substituents, ring residues, and solvent
Systematic approach:
From the molecular formula, calculate the degree of unsaturation
From MS, confirm molecular weight and look for heteroatom clues (nitrogen rule, isotope patterns)
From IR, identify functional groups present (O–H, N–H, C=O, C≡C, etc.)
From ¹H NMR, determine the number of proton environments, their integration ratios, splitting patterns, and chemical shifts
From ¹³C NMR and DEPT, count distinct carbons and classify them
From UV-Vis, assess conjugation
Propose a structure consistent with all data, then verify each piece of data against your proposal
Degree of unsaturation = (2C + 2 + N − H − X) / 2
where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Oxygen and sulfur do not appear in the formula.
A degree of unsaturation of 4 strongly suggests a benzene ring (three double bonds plus one ring).
Wavenumber to wavelength: ṽ (cm⁻¹) = 1 / λ (cm) = 10⁴ / λ (μm)
NMR splitting: number of peaks = n + 1, where n is the number of equivalent neighbouring protons
Beer-Lambert Law (UV-Vis): A = εlc, where A = absorbance, ε = molar absorptivity, l = path length (cm), c = concentration (mol/L)
IR spectroscopy is used routinely by forensic chemists to identify unknown powders and by environmental scientists to monitor atmospheric pollutants such as CO₂ and methane. NMR is the basis of MRI (magnetic resonance imaging) in medicine, using the same nuclear spin physics but applied to hydrogen atoms in water and fat inside the body.
Students often think any cyclic conjugated molecule is aromatic. It is not: the molecule must also be planar and satisfy 4n + 2 pi electrons. Cyclooctatetraene is conjugated and cyclic but adopts a tub shape to avoid being antiaromatic.
Students frequently confuse the molecular ion peak with the base peak. The base peak is simply the tallest peak (most abundant ion) and may or may not be M⁺.
A common mistake is adding oxygen into the degree of unsaturation formula. Oxygen does not change the hydrogen count of a saturated molecule, so it drops out.
Students sometimes assume that all protons on the same carbon are equivalent. They are not always: diastereotopic protons on a CH₂ next to a stereocentre are non-equivalent and can show different chemical shifts and complex splitting.
⚠️ Degree of unsaturation is nearly always the first step on a spectral identification problem. Get this wrong and everything downstream is off.
⚠️ The carbonyl C=O stretch (1650–1800 cm⁻¹) is the single most reliable IR absorption. If it is absent, the molecule has no carbonyl.
⚠️ Hückel's rule questions often appear as "is this molecule aromatic, antiaromatic, or non-aromatic?" You must check all four criteria, not just the electron count.
⚠️ NMR splitting problems are very common on exams. Practise assigning splitting patterns and integration ratios for simple molecules like ethanol, diethyl ether, and ethyl acetate.
⚠️ Molecular model kits are permitted on exams. Use them for stereochemistry and planarity questions.
True or false: A molecule with 4n pi electrons in a cyclic, planar, fully conjugated system is aromatic.
Fill in the blank: The degree of unsaturation of C₈H₆O is ____.
True or false: A broad O–H stretch around 2500–3300 cm⁻¹ in an IR spectrum is characteristic of a carboxylic acid.
Fill in the blank: A proton with two equivalent neighbours appears as a ____ in the ¹H NMR spectrum.
True or false: In ¹³C NMR, the number of peaks equals the number of carbon atoms in the molecular formula.
Answers: 1. False (that is antiaromatic). 2. 6 (consistent with a benzene ring plus two additional degrees). 3. True. 4. Triplet. 5. False (it equals the number of chemically distinct carbons; symmetry can reduce the count).
Q: A compound with molecular formula C₉H₁₀O shows a strong IR absorption at 1715 cm⁻¹ and no broad O–H stretch. Its ¹H NMR shows signals at δ 7.2–7.4 (5H, multiplet), δ 3.6 (2H, singlet), and δ 2.1 (3H, singlet). Propose a structure.
A: The degree of unsaturation is 5 (consistent with a phenyl ring plus one C=O). IR at 1715 cm⁻¹ indicates a ketone. The 5H aromatic multiplet is a monosubstituted benzene. A 2H singlet at 3.6 ppm suggests a CH₂ between the ring and the carbonyl. A 3H singlet at 2.1 ppm suggests a methyl ketone. The structure is phenylacetone (1-phenyl-2-propanone, C₆H₅CH₂COCH₃).
Q: Cyclopentadienyl anion is aromatic. Explain why, including a count of pi electrons.
A: The cyclopentadienyl anion is a five-membered ring where every carbon is sp² hybridised. It has two C=C double bonds contributing 4 pi electrons plus the lone pair on the sp² carbanion contributing 2 more, giving 6 pi electrons total. With n = 1, 4(1) + 2 = 6, satisfying Hückel's rule. It is also cyclic, planar, and fully conjugated, meeting all four criteria.
Q: In a mass spectrum, a compound shows M⁺ at m/z = 88 and a significant fragment at m/z = 60. What neutral fragment was lost, and what does this suggest?
A: The loss is 88 − 60 = 28, which corresponds to CO or C₂H₄. If the compound contains a carbonyl group (confirmed by IR), loss of 28 as CO is likely. If no carbonyl is present, loss of ethylene via a McLafferty-type rearrangement or simple bond cleavage should be considered.
Q: How do you distinguish between a primary amine and a secondary amine using IR spectroscopy?
A: Primary amines (–NH₂) show two N–H stretching peaks near 3300–3500 cm⁻¹ (symmetric and asymmetric stretch). Secondary amines (–NH–) show only one N–H stretching peak in the same region.
Q: A ¹H NMR signal appears as a quartet with a coupling constant of 7.1 Hz. A nearby signal appears as a triplet, also with J = 7.1 Hz. What structural fragment does this indicate?
A: An ethyl group (–CH₂CH₃). The quartet arises from the CH₂ (split by 3 equivalent CH₃ neighbours), and the triplet arises from the CH₃ (split by 2 equivalent CH₂ neighbours). The matching J value confirms they are coupled to each other.
This module feeds directly into Module 2, where you will see carbonyl-group chemistry (aldehydes, ketones, carboxylic acids). The IR and NMR skills learned here will be essential for identifying products of those reactions. Aromaticity returns in Module 3 for electrophilic and nucleophilic aromatic substitution, so a solid grasp of Hückel's rule and ring stability now will save significant revision later. Spectroscopy is also the primary evidence tool in lab courses and in pharmaceutical/analytical chemistry careers.
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