Difficulty: Intermediate to Advanced | Prerequisites: Nomenclature and stereochemistry notes, reactions and mechanisms notes.
Spectroscopy is how chemists confirm what they have made. IR tells you which functional groups are present, mass spectrometry gives the molecular weight and isotope patterns, and ¹H NMR reveals the carbon-hydrogen framework of the molecule. Aromaticity is a special stability concept that determines reactivity for an entire class of ring compounds. Synthesis ties everything together: given a simple starting material, you must plan a sequence of reactions to reach a target product. This is the culmination of Organic Chemistry I.
Learn to read IR peaks (broad O-H, sharp C=O, C=C stretches), interpret mass spectra (molecular ion, isotope patterns for Cl and Br), and count/analyse ¹H NMR signals (chemical shift, integration, splitting). For aromaticity, apply Huckel's rule (4n+2 pi electrons in a planar, fully conjugated ring). For synthesis, work backwards from the product and chain known reactions together.
Infrared (IR) spectroscopy
Measures the absorption of infrared light by molecular bonds. Different functional groups absorb at characteristic frequencies (measured in cm⁻¹, called wavenumbers).
In simple terms, IR is like a fingerprint scanner for functional groups. A broad peak near 3300-3500 cm⁻¹ screams "O-H" or "N-H."
Mass spectrometry (MS)
Measures the mass-to-charge ratio (m/z) of a molecule and its fragments. The molecular ion peak (M⁺) gives the molecular weight.
Think of it as putting the molecule on a scale after knocking an electron off it.
Isotope pattern
The presence of certain elements produces distinctive patterns in the mass spectrum. Chlorine (³⁵Cl and ³⁷Cl) gives M and M+2 peaks in a 3:1 ratio. Bromine (⁷⁹Br and ⁸¹Br) gives M and M+2 peaks in a 1:1 ratio.
¹H NMR (proton nuclear magnetic resonance)
Measures the magnetic environment of hydrogen atoms in a molecule. Each chemically distinct set of hydrogens gives a separate signal.
In simple terms, NMR tells you how many different "types" of hydrogen are in your molecule, how many of each type there are (integration), and what is next door (splitting pattern).
Chemical shift (delta, ppm)
The position of an NMR signal on the x-axis. Electronegative groups pull signals downfield (higher ppm). Typical ranges: alkyl H = 0-2 ppm, allylic/benzylic = 1.5-2.5 ppm, H on C next to O or halogen = 3-4 ppm, vinyl H = 4.5-6.5 ppm, aromatic H = 6.5-8 ppm, aldehyde H = 9-10 ppm.
Splitting (multiplicity)
The n+1 rule: a signal is split into n+1 peaks, where n is the number of equivalent neighbouring hydrogens. A singlet means no neighbours, a doublet means one neighbour, a triplet means two neighbours, and so on.
Aromatic compound
A cyclic, planar, fully conjugated molecule with 4n+2 pi electrons (Huckel's rule). Aromatic compounds are unusually stable.
In simple terms, if you can count the pi electrons in a flat ring and the number fits the pattern 2, 6, 10, 14... it is aromatic.
Antiaromatic compound
A cyclic, planar, fully conjugated molecule with 4n pi electrons (4, 8, 12...). Antiaromatic compounds are unusually unstable and tend to distort out of planarity to avoid this.
Nonaromatic compound
A compound that does not meet all the criteria for aromaticity or antiaromaticity. It may be non-planar, or not fully conjugated, or not cyclic.
Retrosynthetic analysis
Working backwards from a target molecule to identify simpler precursors and the reactions that connect them. Each backwards step is marked with a "retrosynthetic arrow" (double-lined arrow, open at the tail).
Think of it as reverse engineering: you look at the product and ask, "What was the last reaction that made this?"
Region (cm⁻¹) | Bond / Group | Notes |
|---|---|---|
3200-3550 (broad) | O-H stretch | Alcohols, carboxylic acids (very broad for acids) |
3300 (sharp) | N-H stretch | Amines: primary gives two peaks, secondary gives one |
3300 (sharp) | ≡C-H stretch | Terminal alkyne |
~3100 | =C-H stretch | Alkene or aromatic C-H |
2850-3000 | C-H stretch (sp³) | Always present, not usually diagnostic |
2100-2260 | C≡C or C≡N | Triple bonds |
1630-1680 | C=C stretch | Alkenes |
1700-1750 | C=O stretch | Strong, sharp. Ketones, aldehydes, esters, carboxylic acids |
1550 | C=C aromatic | Aromatic ring stretches |
The molecular ion (M⁺) peak gives the molecular weight.
An odd molecular weight suggests an odd number of nitrogen atoms.
Isotope patterns are diagnostic: one Cl gives M and M+2 in a 3:1 ratio; one Br gives M and M+2 in a 1:1 ratio; two Cl atoms give M, M+2, M+4 in a 9:6:1 ratio.
The number of distinct ¹H NMR signals equals the number of sets of chemically non-equivalent hydrogens. Equivalent hydrogens (related by symmetry) give one signal.
Ethene (H₂C=CH₂): all four H are equivalent = 1 signal.
Propane: the two CH₃ groups are equivalent (2 signals total: CH₃ and CH₂).
A para-disubstituted benzene ring has two sets of aromatic H (2 signals from the ring, plus whatever the substituents contribute).
For the exam (Question 7a): count signals for each molecule by identifying all symmetry-equivalent sets of H.
Compound A: m/z = 84, IR peaks at 3100 cm⁻¹ and 1550 cm⁻¹, one degree of unsaturation.
3100 cm⁻¹ suggests =C-H or aromatic C-H. 1550 cm⁻¹ suggests aromatic ring stretches.
One degree of unsaturation with m/z = 84: possible formula C₆H₁₂ for a ring or alkene, but the IR says aromatic. Wait, a benzene ring alone is 4 degrees of unsaturation. With only 1 DoU and m/z = 84, this is likely a cyclohexene-type compound: C₆H₁₂ (DoU = 1). The 3100 and 1550 cm⁻¹ peaks could indicate C=C alkene character.
Reconsidering: m/z = 84 with one DoU. If all C and H: C₆H₁₂ = 84. That gives one ring or one double bond. The IR at 3100 (=C-H) and 1550 (C=C) points to a cyclic alkene or simply an alkene.
Compound B: Compound A + dilute H₂SO₄/H₂O → broad IR peak at 3400 cm⁻¹.
The broad 3400 cm⁻¹ peak indicates an -OH group. This is acid-catalysed hydration of the alkene (Markovnikov addition of water).
Compound B is an alcohol.
Compound C: Compound B + HCl → m/z 120 and 122 in a 3:1 ratio.
The 3:1 ratio of M and M+2 is the signature of one chlorine atom.
HCl converted the alcohol to an alkyl chloride.
The ¹H NMR shows: 6H singlet (two equivalent CH₃ groups, likely a gem-dimethyl or tert-butyl fragment), 2H triplet, 2H complex multiplet, 3H triplet. This is consistent with a structure like 2-chloro-2-methylpentane or a similar chloroalkane.
The key reasoning chain: alkene → alcohol (acid-catalysed hydration) → alkyl chloride (HCl, likely SN1 on a tertiary alcohol).
A compound is aromatic if it meets all four criteria:
Cyclic
Planar (or assumed planar, as the exam states)
Fully conjugated (every atom in the ring contributes a p orbital to the pi system)
Has 4n+2 pi electrons, where n = 0, 1, 2, 3... (giving 2, 6, 10, 14 pi electrons)
If criteria 1-3 are met but the pi electron count is 4n (4, 8, 12...), the compound is antiaromatic.
If any of criteria 1-3 is not met, the compound is nonaromatic.
Each C=C double bond in the ring contributes 2 pi electrons.
Each atom with a lone pair that is part of the conjugated system contributes 2 pi electrons (e.g. the nitrogen lone pair in pyrrole, the oxygen lone pair in furan).
A lone pair on an atom that already contributes to the pi system through a double bond does not add extra pi electrons (e.g. the nitrogen in pyridine uses its lone pair in the sigma system, not the pi system; its pi electrons come from the C=N double bond).
A positively charged carbon with an empty p orbital contributes 0 pi electrons but still participates in conjugation.
A negatively charged carbon with a lone pair in a p orbital contributes 2 pi electrons.
The practice exam shows four species and asks you to classify each as aromatic, antiaromatic, or nonaromatic.
Cyclopropenyl cation (3-membered ring, positive charge):
Cyclic, planar, fully conjugated, 2 pi electrons (4n+2 where n=0). Aromatic.
A five-membered ring with N and O (like an oxazole or isoxazole):
Count carefully. If the ring is fully conjugated and planar with 6 pi electrons, it is aromatic. The nitrogen and oxygen lone pairs that participate in the pi system must be counted.
A ring with an sp³ carbon or a non-conjugated atom:
The break in conjugation means the ring is nonaromatic, regardless of pi electron count.
Cyclooctatetraene or similar 4n system:
If planar and fully conjugated with 8 pi electrons (4n where n=2), it would be antiaromatic. In reality, cyclooctatetraene adopts a tub shape to avoid antiaromaticity, making it nonaromatic. But the exam says "assume completely conjugated rings are planar," so classify based on electron count if the ring is drawn as fully conjugated.
For each molecule, identify all sets of equivalent hydrogens:
Look for planes of symmetry and rotational symmetry.
Hydrogens related by a mirror plane or rotation are equivalent.
On a benzene ring: a monosubstituted ring has three types of aromatic H (ortho, meta, para). A 1,4-disubstituted ring (para) has two types if the substituents are different, one type if they are the same.
Compare the starting material and the target product. What bonds are new? What functional groups changed?
Work backwards (retrosynthetically): identify the last reaction that would give the target product.
Determine what intermediate is needed for that last reaction.
Continue working backwards until you reach the given starting material.
Write the synthesis in the forward direction, listing all reagents and isolable intermediates.
Target: A vicinal dichloride (Cl on two adjacent carbons) + enantiomer, from a terminal alkyne.
Retrosynthetic reasoning:
A vicinal dichloride with both Cl on adjacent carbons can come from addition of Cl₂ (or 2 HCl) across a double bond, or from addition of HCl across an alkyne.
The "+ enantiomer" label means racemic product, suggesting the reaction goes through a non-stereoselective step.
Starting from a terminal alkyne: first convert the triple bond to a double bond (e.g. H₂/Lindlar for cis, or Na/NH₃ for trans), then add HCl (Markovnikov, twice for 2 equiv HCl) or Cl₂.
Alternatively: treat the terminal alkyne with 2 equiv HCl directly, where the first addition follows Markovnikov (Cl to the more substituted carbon of the resulting vinyl chloride) and the second addition gives a geminal dichloride. But a vicinal product would require a different route.
The specific synthetic route depends on the exact structures drawn in the exam, but the key is recognising which reactions produce vicinal dihalides (Br₂ or Cl₂ addition to an alkene) vs geminal dihalides (2 equiv HX to an alkyne).
Target: A formate ester from cyclohexene.
Retrosynthetic reasoning:
The product is an ester: specifically a formate (HCOO-R). This means we need an alcohol intermediate that is then esterified, or we need an oxidative cleavage.
From cyclohexene: ozonolysis (O₃, then a workup) cleaves a double bond into two carbonyl groups. Cyclohexene would give a dialdehyde (a six-carbon chain with aldehydes at both ends).
Alternatively: epoxidation (mCPBA) followed by ring opening could give a diol, which could be further oxidised.
The exact route depends on the product structure, but the typical synthesis from a cyclic alkene to an open-chain ester involves ozonolysis or oxidative cleavage followed by an oxidation and esterification step.
Alkyne → cis alkene: H₂, Lindlar catalyst
Alkyne → trans alkene: Na, NH₃
Alkene → alcohol (Markovnikov): H₂O, H₂SO₄ or oxymercuration (Hg(OAc)₂, H₂O, then NaBH₄)
Alkene → alcohol (anti-Markovnikov): hydroboration-oxidation (BH₃, then H₂O₂/OH⁻)
Alkene → epoxide: mCPBA
Alkene → vicinal dihalide: Br₂ or Cl₂ (anti addition)
Alkene → vicinal diol: OsO₄ (cat.), NMO (syn addition) or KMnO₄ (cold, dilute)
Alcohol → alkyl halide: HBr, HCl, SOCl₂, PBr₃
Alcohol → tosylate: TsCl, pyridine
Alcohol → alkene: H₃PO₄ or H₂SO₄ (heat, E1)
Alkyl halide or tosylate + strong base → alkene (E2)
Alkyl halide or tosylate + nucleophile → substitution product (SN2)
Degree of unsaturation (DoU): (2C + 2 + N - H - X) / 2
Huckel's rule: 4n+2 pi electrons = aromatic (n = 0, 1, 2, 3...). So 2, 6, 10, 14 pi electrons are aromatic counts.
n+1 rule for NMR splitting: A hydrogen with n equivalent neighbours appears as an n+1 multiplet.
Isotope ratios in MS:
One ³⁵Cl/³⁷Cl: M : M+2 = 3 : 1
One ⁷⁹Br/⁸¹Br: M : M+2 = 1 : 1
Two Cl: M : M+2 : M+4 = 9 : 6 : 1
NMR spectroscopy is used daily in pharmaceutical labs to verify that a synthesised drug has the correct structure. MRI scanners in hospitals are based on the same nuclear magnetic resonance principles. IR spectroscopy is used in forensic science and environmental monitoring (e.g. detecting pollutants in water). Aromaticity governs the chemistry of benzene and its derivatives, which form the backbone of dyes, drugs, plastics, and explosives.
"A broad OH peak in IR means the compound is an alcohol." Carboxylic acids also show a very broad O-H stretch (even broader than alcohols, often spanning 2500-3300 cm⁻¹). Check for a C=O peak near 1710 cm⁻¹ to distinguish the two.
"The number of peaks in an NMR spectrum equals the number of carbons." No. The number of ¹H NMR signals equals the number of chemically distinct sets of protons. A highly symmetric molecule can have many carbons but few NMR signals.
"All cyclic conjugated systems are aromatic." Only if they also have 4n+2 pi electrons and are planar. Cyclobutadiene (4 pi electrons, planar) is antiaromatic. A non-planar ring is nonaromatic.
"Synthesis questions always have one correct answer." Multiple valid routes may exist. The exam asks for a reasonable synthesis with correct reagents, not the only possible one. Choose the route you are most confident in.
⚠️ The spectral analysis question (9 points) requires you to identify three compounds from IR, MS, and NMR data. Work through the clues systematically: molecular weight and DoU first, then functional groups from IR, then structural detail from NMR.
⚠️ The aromaticity sub-question asks you to classify species as aromatic, antiaromatic, or nonaromatic. Count pi electrons carefully. Lone pairs on heteroatoms are the most common source of error.
⚠️ The Diels-Alder sub-question gives you a product and asks for the diene and dienophile. Perform a retro Diels-Alder: the new sigma bonds in the product (the ones that close the ring) are the bonds that were formed, and breaking them gives you back the diene and dienophile.
⚠️ Synthesis questions (22 points) require all isolable intermediates to be drawn. Forgetting an intermediate costs points even if your reagents are correct.
True or false: A mass spectrum with M and M+2 in a 1:1 ratio indicates the presence of one bromine atom. (True)
Fill in the blank: A compound with 4n+2 pi electrons in a planar, fully conjugated ring is ______. (aromatic)
True or false: In ¹H NMR, a triplet indicates the hydrogen has three equivalent neighbours. (False, a triplet means two equivalent neighbours; n+1 rule.)
Fill in the blank: The IR absorption near 1700 cm⁻¹ is characteristic of a ______ bond. (C=O, carbonyl)
True or false: In a Diels-Alder reaction, the diene must be in the s-cis conformation. (True)
Q: A compound has a molecular ion at m/z = 78 and shows a strong IR absorption near 3300 cm⁻¹ (sharp) and 2100 cm⁻¹. What functional group is likely present?
A: The sharp 3300 cm⁻¹ peak combined with the 2100 cm⁻¹ absorption is characteristic of a terminal alkyne (≡C-H stretch and C≡C stretch). The molecular formula consistent with m/z = 78 and a terminal alkyne would be C₆H₆ (benzene is also 78, but benzene would show aromatic C-H near 3100 and no 2100 peak). A better fit is something like phenylacetylene-related, though the exact structure depends on the degree of unsaturation calculation.
Q: Classify the cyclopropenyl cation as aromatic, antiaromatic, or nonaromatic. Justify your answer.
A: The cyclopropenyl cation is a three-membered ring with two pi electrons (one C=C double bond; the positive carbon has an empty p orbital and contributes zero pi electrons but maintains conjugation). It is cyclic, planar, fully conjugated, and has 2 pi electrons (4n+2 where n = 0). It is aromatic.
Q: Cyclopentadienyl anion has a negative charge on one carbon of a five-membered ring. Is it aromatic?
A: Yes. The five carbons each contribute one p orbital to the pi system. Four carbons are involved in two C=C double bonds (4 pi electrons), and the carbanion contributes its lone pair (2 pi electrons), giving 6 pi electrons total. Cyclic, planar, fully conjugated, 4n+2 (n = 1). Aromatic.
Q: In a Diels-Alder reaction, the product is a cyclohexene with two -COOCH₃ groups. What are the diene and dienophile?
A: Perform a retro Diels-Alder. The two new sigma bonds in the cyclohexene ring (across from the double bond) are the bonds that formed. Breaking them gives the diene (a 1,3-butadiene derivative) and the dienophile (a double bond bearing the two -COOCH₃ groups, i.e. dimethyl fumarate or dimethyl maleate). The electron-withdrawing ester groups sit on the dienophile, which is consistent with the requirement for an electron-poor dienophile.
Q: Devise a synthesis of a formate ester from cyclohexene.
A: One route: (1) Ozonolysis of cyclohexene (O₃, then reductive workup with Me₂S or Zn) cleaves the double bond to give a six-carbon dialdehyde. (2) Oxidise one aldehyde to a carboxylic acid (e.g. with Jones reagent or KMnO₄) and esterify with methanol under acid catalysis, or use a Baeyer-Villiger-type approach. The exact route depends on the target structure. The key recognition is that breaking a ring double bond by ozonolysis opens the ring to a chain.
Q: How many ¹H NMR signals would you expect for para-bromochlorobenzene?
A: The ring has two pairs of equivalent hydrogens (the two H ortho to Br are equivalent to each other, and the two H ortho to Cl are equivalent to each other). That gives 2 signals from the aromatic ring, and no other hydrogens in the molecule. Total: 2 ¹H NMR signals.
Spectroscopy closes the loop on reactions: once you predict a product, IR and NMR are how you confirm you made the right thing in lab. Aromaticity connects back to hybridisation (every atom in an aromatic ring is sp²) and forward to electrophilic aromatic substitution in Organic Chemistry II. Synthesis planning draws on every reaction you have learnt, so it is the ultimate integration topic. If you can solve a multi-step synthesis problem, you have a working command of the entire CHEM 2301 syllabus.
IR spectroscopy, infrared, wavenumber, functional group identification, mass spectrometry, molecular ion, m/z, isotope pattern, chlorine 3:1, bromine 1:1, proton NMR, 1H NMR, chemical shift, splitting pattern, n+1 rule, integration, singlet, doublet, triplet, multiplet, aromaticity, Huckel rule, 4n+2, aromatic, antiaromatic, nonaromatic, cyclopropenyl cation, cyclopentadienyl anion, pi electrons, Diels-Alder, retro Diels-Alder, diene, dienophile, cycloaddition, synthesis, retrosynthetic analysis, multi-step synthesis, ozonolysis, hydroboration-oxidation, epoxidation, CHEM 2301, organic chemistry I, UMN, Salmon