Difficulty: Intermediate to Advanced | Prerequisites: Alkene/alkyne reactions and reagents (see Part 2 notes), carbocation and radical stability (see Part 1 notes), basic spectroscopy concepts (IR regions, mass spec molecular ion).
Mechanisms are the "why" behind the reactions you memorised in Part 2. Drawing a mechanism means showing each bond that forms and breaks, the intermediates that appear along the way, and the electron flow (curved arrows for ionic, fishhook arrows for radical) at every step. The exam also tests spectroscopy: given IR absorptions, a molecular ion from mass spec, and a ¹H NMR spectrum, you need to deduce a structure. Finally, a set of conceptual questions covers keto-enol tautomerism, allylic radical resonance, and alkene stability, all of which tie back to the stability principles from Part 1.
Br₂ addition to an alkene proceeds through a bromonium ion intermediate (the pi electrons attack Br₂, forming a three-membered ring, then bromide attacks from the back side for anti addition). Radical bromination uses fishhook (single-barbed) arrows to show one-electron movement during hydrogen abstraction. For spectroscopy, IR peaks near 3300 and 2200 cm⁻¹ signal a terminal alkyne; ¹H NMR splitting patterns and integration let you piece together a structure. Keto-enol tautomerism swaps the position of a proton between a carbonyl's alpha carbon and the oxygen.
Bromonium ion
A cyclic, three-membered intermediate formed in the first step of Br₂ addition to an alkene. The bromine bridges across both carbons of the former double bond, carrying a positive charge. Think of it as a bromine "hat" sitting on the double bond. Because it blocks one face, the nucleophile can only attack from the other side.
Curved arrow (double-barbed)
Represents the movement of a pair of electrons in an ionic mechanism. The arrow starts at the electron source (a bond or lone pair) and points to where the electrons go.
Fishhook arrow (single-barbed)
Represents the movement of a single electron in a radical mechanism. Used in homolytic bond cleavage and radical propagation steps.
Hydrogen abstraction
A propagation step in radical halogenation where a halogen radical (e.g. Br·) removes a hydrogen atom from a C-H bond, forming H-Br and a carbon radical. Two fishhook arrows show the one-electron movements.
Keto-enol tautomerism
An equilibrium between a ketone (or aldehyde) form and its enol form. The enol has a C=C double bond with an -OH on one of those carbons. The keto form is almost always favoured at equilibrium. In simple terms, a proton hops from the carbon next to the carbonyl onto the oxygen, converting C=O into C-OH and creating a new C=C in the process.
Enol
The tautomer of a carbonyl compound that features an -OH group on a carbon that is part of a C=C double bond. "En" for alkene, "-ol" for alcohol.
¹H NMR (proton nuclear magnetic resonance)
A spectroscopic technique that reveals the number of distinct hydrogen environments in a molecule, how many hydrogens are in each environment (integration), and what neighbouring hydrogens are present (splitting pattern).
Molecular ion (M⁺)
The peak in a mass spectrum corresponding to the intact molecule after losing one electron. Its m/z value equals the molecular weight of the compound.
This is a two-step ionic mechanism. You must show all intermediates and electron flow with curved (double-barbed) arrows.
Step 1: Formation of the bromonium ion
The pi electrons of the alkene act as the nucleophile and attack one bromine atom of Br₂.
Simultaneously, the Br-Br bond breaks heterolytically: the attacked bromine forms bonds to both carbons of the former double bond (creating the three-membered bromonium ring), and the other bromine leaves as Br⁻.
Arrow 1: from the C=C pi bond to one Br of Br₂.
Arrow 2: from the Br-Br bond to the departing Br (forming Br⁻).
Intermediate: a cyclic bromonium ion (three-membered ring with positive charge on Br) plus Br⁻.
Step 2: Nucleophilic attack by Br⁻
Br⁻ attacks one of the two carbons of the bromonium ion from the opposite face (anti attack, backside of the ring).
Arrow: from a lone pair on Br⁻ to one carbon of the bromonium ion.
The C-Br bond in the ring breaks, and both Br atoms end up on opposite faces of what was the double bond.
Product: a vicinal dibromide with anti stereochemistry (the two Br atoms are trans to each other).
Why anti addition?
The bromonium ion physically blocks one face of the molecule. The nucleophile (Br⁻) can only approach from the opposite face, forcing anti addition. This is the same reasoning that makes Br₂/H₂O give anti halohydrin products.
Radical reactions use fishhook (single-barbed) arrows. Each fishhook shows one electron moving.
The hydrogen abstraction step (propagation)
A bromine radical (Br·) approaches a C-H bond on butane (the exam example).
Fishhook 1: from the C-H bond to the hydrogen (H moves with one electron toward the Br radical).
Fishhook 2: from the Br· radical to the forming H-Br bond (the unpaired electron on Br pairs with the incoming electron from the C-H bond).
Products: a carbon radical (on butane, at the secondary position for selectivity reasons) and H-Br.
Why the secondary position?
Radical bromination is highly selective. The secondary C-H bond is preferentially abstracted because it produces the more stable secondary radical.
This selectivity is a direct consequence of the Hammond postulate: the transition state for a more exothermic (or less endothermic) step resembles the products less, but for bromination the selectivity is high because the Br· radical is relatively unreactive and the transition state is late (product-like), so stability differences in the radical product are fully reflected in the activation energy.
Full radical chain for reference
Initiation: Br₂ splits into 2 Br· (light or heat provides the energy for homolysis).
Propagation step 1: Br· + R-H gives R· + HBr (the hydrogen abstraction above).
Propagation step 2: R· + Br₂ gives R-Br + Br· (the carbon radical attacks Br₂).
Termination: any two radicals combine (Br· + Br·, R· + Br·, or R· + R·).
Problem 7 on the exam gives you spectroscopic data and asks you to deduce two structures. Here is the method, walked through with the exam's data.
Compound A clues
Molecular ion: m/z = 68. This is the molecular weight.
IR peaks at ~3300 cm⁻¹ and ~2200 cm⁻¹. The 3300 peak indicates a C-H stretch of a terminal alkyne (≡C-H). The 2200 peak indicates a C≡C stretch.
Conclusion: compound A is a terminal alkyne with MW 68.
Working backwards: C₅H₈ has MW = 68 and two degrees of unsaturation (one triple bond). With a methyl branch, the structure is 4-methylpent-2-yne (or equivalent).
The reaction: A treated with H₂O/HgSO₄/H₂SO₄
This is Markovnikov hydration of an alkyne, producing a ketone via enol tautomerisation.
Water adds across the triple bond with OH on the more substituted carbon, giving an enol that converts to the ketone.
Compound B clues
Molecular ion: m/z = 86. This is MW = 68 + 18 (water added).
¹H NMR interpretation:
1.2 ppm, 6H, doublet: six equivalent hydrogens split by one neighbour. This is two equivalent CH₃ groups attached to a CH (isopropyl group: (CH₃)₂CH-).
2.2 ppm, 3H, singlet: three hydrogens with no neighbours. This is a CH₃ group next to a carbonyl (CH₃-C=O), since the carbonyl carbon has no hydrogens.
2.7 ppm, 1H, septet: one hydrogen split by six neighbours. This is the CH of the isopropyl group, sitting between two CH₃ groups (6 equivalent neighbours) and next to the carbonyl.
Structure of B: 4-methylpentan-2-one (methyl isopropyl ketone). The NMR confirms three distinct hydrogen environments matching this structure.
Key IR frequencies to know for this exam
Region (cm⁻¹) | Assignment |
|---|---|
~3300 (sharp) | Terminal alkyne C-H stretch (≡C-H) |
~2100-2260 | C≡C stretch (alkyne) |
~1700-1750 | C=O stretch (carbonyl: ketone, aldehyde, acid) |
~1600-1680 | C=C stretch (alkene) |
~2850-3000 | C-H stretches (sp³, sp²) |
~3200-3600 (broad) | O-H stretch (alcohol, carboxylic acid) |
The exam shows a ketone and asks which of four structures is its enol form.
What the enol looks like
Take the ketone's C=O. Move a hydrogen from the alpha carbon (the carbon next to the carbonyl) onto the oxygen. This creates an -OH group on the oxygen and a new C=C double bond between the alpha carbon and the carbonyl carbon.
The enol has: an -OH group (where the C=O was) and a C=C double bond adjacent to it.
How to identify the correct enol
The carbon skeleton does not change. No bonds between carbon atoms are broken or formed.
The only change is: one alpha C-H breaks, the C=O becomes C-OH, and a new C=C forms between the alpha carbon and the former carbonyl carbon.
The exam answer is D: the structure that preserves the carbon skeleton and places the -OH and C=C in the correct positions.
Why it matters
Keto-enol tautomerism is the mechanism behind acid-catalysed alkyne hydration: the initial product is the enol, which tautomerises to the thermodynamically favoured ketone.
It also appears in alpha-halogenation and aldol reactions in Organic Chemistry II.
The exam gives an allylic radical and asks which structure is a valid resonance form.
Rules for drawing radical resonance
Only electrons move. No atoms change position.
The radical (unpaired electron) can be delocalised to another carbon through an adjacent pi bond, just as a charge is delocalised in resonance structures of ions.
Move the radical from its current carbon into the adjacent C=C bond, and shift the C=C bond one position over. The radical appears on the carbon at the far end of the former double bond.
Identifying the correct resonance form
The carbon framework stays identical. If an answer choice moves an atom to a different position, it is wrong.
The radical must end up on a carbon that was part of the original pi system.
The exam answer is D: the only structure where the unpaired electron has moved through the conjugated system to a new carbon position while keeping the carbon skeleton intact.
Real-world connection
Allylic radical resonance is why NBS bromination of a compound with a double bond gives products at both allylic positions. The radical intermediate is delocalised, and Br₂ can trap it at either end.
The exam asks you to rank three alkenes from least to most stable.
General rules for alkene stability
More substituted alkenes are more stable. Tetrasubstituted > trisubstituted > disubstituted > monosubstituted.
Trans alkenes are more stable than cis alkenes of the same substitution level, due to reduced steric strain.
Conjugation increases stability: a double bond adjacent to another double bond (conjugated diene) is more stable than an isolated double bond.
Exam ranking (from the key)
Three alkenes are compared. The answer is A < C < B (least to most stable).
A has the least substitution or the most steric strain (the exam structure shows it as less substituted or with ring strain).
B is the most substituted or most stabilised by hyperconjugation.
C falls in between.
How to approach these on the exam
Count the number of alkyl groups directly attached to the C=C carbons. More groups = more stable. If substitution is tied, check for trans vs cis geometry (trans wins) and conjugation (conjugated wins).
Students draw regular curved arrows in radical mechanisms. Radical steps require fishhook (single-barbed) arrows because only one electron moves at a time. Using double-barbed arrows in a radical mechanism is marked wrong.
When drawing the bromonium ion mechanism, students often skip the intermediate and jump straight to the product. The exam explicitly asks for all intermediates. Show the bromonium ion as a distinct species between the two steps.
Students confuse tautomers with resonance structures. Tautomers are different compounds (atoms move, specifically a hydrogen migrates) in equilibrium with each other. Resonance structures are different representations of the same compound (only electrons move, no atoms). The enol and keto forms are tautomers, not resonance structures.
For NMR, students often confuse splitting pattern with integration. The splitting tells you how many neighbours a hydrogen has (n+1 rule: n neighbours give n+1 lines). The integration tells you how many hydrogens are in that environment. These are separate pieces of information.
⚠️ The mechanism question (Problem 5) is worth 10 points and requires stepwise drawing with correct arrow types. Forgetting the bromonium ion intermediate, or using the wrong arrow type, costs significant marks.
⚠️ Spectroscopy (Problem 7) combines IR, MS, and NMR data. You need to use all three together. Start with the molecular formula (from MS), narrow down the functional group (from IR), then confirm the structure (from NMR).
⚠️ The miscellaneous multiple-choice (Problem 8) tests conceptual understanding: enol identification, radical resonance, and alkene stability. These are quick questions if you know the principles, but easy to get wrong if you are guessing.
⚠️ Degrees of unsaturation (index of hydrogen deficiency) is crucial for spectroscopy problems. For CₙH₂ₙ₊₂ (saturated), IHD = 0. Each ring or double bond adds 1. A triple bond adds 2.
True or false: the bromonium ion is drawn with a positive charge on bromine. ___
Fill in the blank: in radical mechanisms, electron flow is shown with ___ arrows.
True or false: an enol and a ketone are resonance structures of each other. ___
Fill in the blank: a doublet in ¹H NMR means the hydrogen has ___ neighbouring hydrogen(s).
True or false: a more substituted alkene is less stable than a less substituted one. ___
Answers: 1. True. 2. Fishhook (single-barbed). 3. False (they are tautomers). 4. One. 5. False (more substituted = more stable).
Q: Draw the stepwise mechanism for Br₂ addition to cyclohexene. Show all intermediates and use correct arrow notation.
A: Step 1: the pi electrons of cyclohexene attack Br₂ (curved arrow from C=C to Br). The Br-Br bond breaks (curved arrow from Br-Br to the departing Br). A cyclic bromonium ion forms on one face of the ring, with Br⁻ released. Step 2: Br⁻ attacks one carbon of the bromonium ion from the opposite face (curved arrow from Br⁻ to carbon). The product is trans-1,2-dibromocyclohexane.
Q: A compound has MW 68, with IR peaks at 3300 and 2200 cm⁻¹. What functional group is present?
A: A terminal alkyne. The 3300 cm⁻¹ peak is the ≡C-H stretch and the 2200 cm⁻¹ peak is the C≡C stretch.
Q: Compound B shows three ¹H NMR signals: 6H doublet at 1.2 ppm, 3H singlet at 2.2 ppm, and 1H septet at 2.7 ppm. Propose a structure.
A: 4-methylpentan-2-one (methyl isopropyl ketone). The 6H doublet is (CH₃)₂CH-, the 3H singlet is CH₃C=O, and the 1H septet is the CH between the two methyl groups and the carbonyl.
Q: What is the relationship between a ketone and its enol form: resonance structures or tautomers?
A: Tautomers. They are distinct structural isomers in equilibrium, involving the migration of a hydrogen atom. Resonance structures differ only in electron arrangement, not atom positions.
Q: An allylic radical has the unpaired electron on C1 of a three-carbon system with a double bond between C2 and C3. Where does the radical appear in the resonance form?
A: On C3. The unpaired electron delocalises through the pi bond: the radical moves from C1 to C3, and the double bond shifts from C2-C3 to C1-C2.
The bromonium ion mechanism taught here is the template for every electrophilic addition mechanism you will draw in this course and the next. Halohydrin formation, epoxide opening, and even some Diels-Alder problems follow the same logic of a bridged intermediate controlling stereochemistry.
Radical mechanisms reappear in polymer chemistry and in biological contexts (lipid peroxidation, antioxidant chemistry). The fishhook arrow notation and the initiation-propagation-termination framework carry through unchanged.
Spectroscopy only grows in importance. Organic Chemistry II adds ¹³C NMR, more complex ¹H NMR splitting, and expects you to combine all spectroscopic data to solve unknowns. Building fluency with IR functional group identification and the n+1 splitting rule now saves considerable time later.
Keto-enol tautomerism is the gateway to alpha-carbon chemistry in the second semester: enolate formation, aldol condensations, Claisen condensations, and alpha-halogenation all begin with this equilibrium.
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