This covers the reactions you need for Exam 4: electrophilic additions to alkenes (HBr, Br₂, H₂O/H₂SO₄, halohydrin, hydroboration-oxidation, epoxidation), radical additions (HBr/ROOR), catalytic hydrogenation (H₂/Pd/C, Lindlar), dissolving metal reduction (Na/NH₃), and ozonolysis. For each, know the mechanism type, regiochemistry, stereochemistry, and major product.
Difficulty: Intermediate to Advanced | Prerequisites: Alkene/alkyne structure and nomenclature (see companion study notes), carbocation stability, hybridisation, curved-arrow formalism, radical basics.
After learning what alkenes and alkynes are, the course moves to what they do. This section covers every major addition reaction to C=C and C≡C bonds that appears on Exam 4. Each reaction is defined by four things: the reagents, the mechanism (ionic electrophilic, radical, or concerted), the regiochemistry (Markovnikov or anti-Markovnikov), and the stereochemistry (syn, anti, or non-specific). If you can identify those four elements for each reaction, you can predict the major product of nearly any problem on the exam. The mechanism questions (16 points) require you to draw curved arrows showing electron movement through every intermediate.
Electrophilic addition
The general mechanism by which an electrophile (an electron-poor species) reacts with the pi electrons of an alkene or alkyne. The pi bond acts as the nucleophile. Think of it as the double bond "donating" its electrons to the electrophile.
Markovnikov's rule
In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens, and the X group attaches to the carbon with fewer hydrogens. In simple terms, the electrophile (H⁺) goes to the less substituted carbon, forming the more stable carbocation intermediate.
Anti-Markovnikov addition
The opposite regiochemistry: the X group ends up on the less substituted carbon. This occurs in radical addition of HBr (with peroxides) and in hydroboration-oxidation.
Bromonium ion
A three-membered ring intermediate formed when Br₂ reacts with an alkene. One bromine bridges across both carbons of the former double bond, carrying a positive charge. This intermediate forces the second nucleophile to attack from the opposite face (anti addition).
Halohydrin formation
Addition of X₂ (Br₂ or Cl₂) in the presence of water. Water acts as the nucleophile that opens the halonium ion. The product is a halohydrin (one carbon bears the halogen, the adjacent carbon bears -OH). Regiochemistry: water attacks the more substituted carbon of the halonium ion. Stereochemistry: anti.
Hydroboration-oxidation
A two-step sequence: (1) BH₃·THF adds across the alkene in a concerted, syn fashion, placing boron on the less substituted carbon; (2) H₂O₂/NaOH replaces boron with -OH (with retention of configuration). Net result: anti-Markovnikov, syn addition of water.
Acid-catalysed hydration
H₂O/H₂SO₄ adds water across an alkene via a carbocation intermediate. Markovnikov regiochemistry (OH ends up on the more substituted carbon). Not stereospecific because the carbocation is planar.
Epoxidation
Treatment of an alkene with a peroxyacid (e.g. mCPBA or CH₃COOOH) gives an epoxide (a three-membered ring with an oxygen). The reaction is concerted and syn: both C-O bonds form on the same face.
Radical chain reaction
A reaction proceeding through radical intermediates, with initiation (homolytic cleavage to generate radicals), propagation (radicals reacting to form product and regenerate a new radical), and termination (radical-radical coupling) steps.
HBr/ROOR (radical addition)
Peroxides (ROOR) cause HBr addition to proceed via a radical mechanism rather than the usual ionic one. This reverses the regiochemistry to anti-Markovnikov (Br ends up on the less substituted carbon). Only works with HBr, not HCl or HI.
Catalytic hydrogenation (H₂/Pd/C)
Addition of H₂ across a double bond (or triple bond) using a metal catalyst. Syn addition. With excess H₂ and Pd/C, both alkenes and alkynes are fully reduced to alkanes.
Lindlar catalyst
A "poisoned" palladium catalyst (Pd on CaCO₃, treated with lead acetate and quinoline) that reduces an alkyne to a cis (Z) alkene only, stopping at the alkene stage.
Dissolving metal reduction (Na/NH₃)
Reduction of an internal alkyne to a trans (E) alkene using sodium metal in liquid ammonia. The mechanism involves sequential electron and proton transfers.
Ozonolysis
Cleavage of a C=C double bond using ozone (O₃), followed by a reductive workup (Zn/H₂O or dimethyl sulfide). Each carbon of the former double bond becomes a carbonyl (C=O). If the carbon had one hydrogen, an aldehyde forms; if it had no hydrogens, a ketone forms.
Step 1: The pi electrons of the alkene attack one bromine of Br₂, displacing Br⁻ and forming a cyclic bromonium ion intermediate. Draw the curved arrow from the C=C pi bond to Br.
Step 2: Br⁻ (the nucleophile) attacks the more substituted carbon of the bromonium ion from the back side (anti to the bromine bridge). This opens the ring and gives a vicinal dibromide with anti stereochemistry.
The exam showed 2-methylpropene + Br₂ giving a vicinal dibromide. Both bromines end up anti to each other.
Initiation:
Step 1: Heat or light causes homolytic cleavage of the O-O bond in ROOR, generating two RO· radicals (use fishhook arrows).
Step 2: RO· abstracts a hydrogen from HBr, generating ROH and a Br· radical.
Propagation:
Step 1: Br· adds to the less substituted carbon of the alkene (to form the more stable, more substituted radical intermediate). Draw the fishhook arrow from Br· to one carbon of the C=C, and the other fishhook from the pi bond to the other carbon.
Step 2: The carbon radical abstracts H from another HBr molecule, forming the anti-Markovnikov product and regenerating Br· to continue the chain.
Termination steps exist but the exam explicitly says not to draw them.
Norbornene + Cl₂/H₂O (halohydrin formation)
Chloronium ion forms on the more accessible face of norbornene.
Water opens the chloronium ion at the more substituted position (anti to Cl).
Product: a trans-chlorohydrin (Cl and OH anti to each other).
Alkene + HBr/ROOR (radical addition)
Anti-Markovnikov regiochemistry: Br ends up on the less substituted carbon.
Not stereospecific (mixture of syn and anti).
Internal alkyne + Na/NH₃ (dissolving metal reduction)
Converts an internal alkyne to a trans (E) alkene.
The mechanism involves radical anion intermediates, and the thermodynamic trans product forms preferentially.
Cyclopentene + BH₃·THF, then H₂O₂/NaOH (hydroboration-oxidation)
Syn addition of H and OH across the double bond.
Anti-Markovnikov regiochemistry: OH goes on the less substituted carbon.
For cyclopentene (symmetric), this gives a cis arrangement of H and OH on the same face.
2-Methylbut-2-ene + H₂O/H₂SO₄ (acid-catalysed hydration)
Markovnikov addition: the proton goes to the less substituted end, forming the more stable 3° carbocation, then water attacks.
Product: a tertiary alcohol.
Not stereospecific.
Diene + CH₃COOOH (epoxidation with a peroxyacid)
The peroxyacid reacts with the more electron-rich (more substituted) double bond preferentially.
Concerted syn delivery of the oxygen to form an epoxide.
If two double bonds are present, the more substituted one epoxidises first.
Diene/enyne + H₂ (excess) / Pd/C (catalytic hydrogenation)
Excess H₂ with Pd/C reduces all pi bonds (both double and triple bonds) to give the fully saturated alkane.
Syn addition, but since the product is an alkane, stereochemistry is often moot.
Alkyne + HBr (1 eq, ionic, no peroxides)
Markovnikov addition: H to the terminal carbon, Br to the internal carbon.
Product: a vinyl bromide (if starting from a terminal alkyne) with Markovnikov regiochemistry.
With one equivalent, the reaction usually stops at the vinyl halide stage.
Internal alkyne + H₂/Lindlar catalyst
Partial reduction of the triple bond to a cis (Z) alkene.
Lindlar catalyst is "poisoned" and cannot reduce past the alkene stage.
Alkene + O₃, then Zn/H₂O (ozonolysis)
The double bond is cleaved completely. Each carbon of the former C=C becomes a C=O.
A carbon that bore one H gives an aldehyde (RCHO). A carbon that bore no H (two R groups) gives a ketone (R₂C=O).
For the specific exam substrate (a trisubstituted alkene with a methyl and another substituent), identify each fragment by counting what was on each side of the double bond.
The exam showed a transformation of 2-methylpentane to a tertiary alkyl bromide and asked for the reagents.
This is a radical halogenation with selectivity at the 3° C-H bond: HBr/hv or Br₂/hv (radical conditions) would brominate selectively at the most substituted position.
Students often assume that HBr with peroxides gives anti addition. It does not. Radical addition of HBr/ROOR is anti-Markovnikov in regiochemistry, but it is not stereospecific (it gives a mixture of syn and anti products).
Hydroboration-oxidation is frequently confused with acid-catalysed hydration. Both add water across an alkene, but they give opposite regiochemistry. Hydroboration-oxidation is anti-Markovnikov (OH on the less substituted carbon); acid-catalysed hydration is Markovnikov (OH on the more substituted carbon).
Students forget that Lindlar reduction gives the cis (Z) alkene while dissolving metal reduction (Na/NH₃) gives the trans (E) alkene. These are tested side by side.
Ozonolysis is sometimes confused with epoxidation. Ozone cleaves the double bond entirely (giving two carbonyl fragments), while a peroxyacid inserts an oxygen to form an epoxide without breaking the carbon-carbon bond.
When drawing radical mechanisms, students often use double-headed (ionic) curved arrows instead of single-headed fishhook arrows. Radicals involve one-electron movement, so every arrow in initiation and propagation should be a fishhook.
⚠️ Mechanism drawing is worth 16 points (the single biggest question type). You must show every intermediate, use the correct arrow type (double-headed for ionic, fishhook for radical), and show formal charges.
⚠️ Product prediction across 10 different reactions is worth 30 points. The fastest way to prepare is to build a mental table: for each reagent set, recall the mechanism type, regiochemistry, and stereochemistry.
⚠️ Hydroboration-oxidation vs acid-catalysed hydration is a classic exam comparison. Know which is Markovnikov and which is anti-Markovnikov, and which is syn vs non-stereospecific.
⚠️ Lindlar vs Na/NH₃ reduction of alkynes appears on nearly every CHEM 2301 exam. Cis (Lindlar) vs trans (dissolving metal) is a two-point giveaway if you know it.
⚠️ Ozonolysis product identification requires you to mentally "cut" the double bond and cap each side with =O. Practise this on several substrates.
⚠️ Radical addition only reverses regiochemistry for HBr (not HCl or HI). Peroxides have no effect on HCl or HI additions.
Fill in the blank: HBr/ROOR gives ___ regiochemistry via a ___ mechanism. (anti-Markovnikov, radical)
True or false: Hydroboration-oxidation gives Markovnikov addition of water. (False, it is anti-Markovnikov.)
Fill in the blank: Lindlar catalyst reduces an internal alkyne to a ___ alkene, while Na/NH₃ gives a ___ alkene. (cis/Z, trans/E)
True or false: In the bromonium ion mechanism, the nucleophile attacks the same face as the bromine. (False, it attacks the opposite face, giving anti addition.)
Fill in the blank: Ozonolysis of a trisubstituted alkene followed by Zn/H₂O gives one ___ and one ___. (ketone, aldehyde)
Q: What is the major product of cyclopentene + BH₃·THF followed by H₂O₂/NaOH?
A: Cyclopentanol, with the OH delivered syn and to the less substituted carbon. Since cyclopentene is symmetric, the product is simply cis-cyclopentanol (H and OH on the same face).
Q: An alkene is treated with Br₂ in water. What type of product forms, and what is its stereochemistry?
A: A bromohydrin (one carbon bears Br, the adjacent carbon bears OH). The stereochemistry is anti because the reaction proceeds through a bromonium ion intermediate. Water attacks the more substituted carbon from the opposite face.
Q: Draw the initiation and propagation steps for the reaction of 2-methylpropene with HBr in the presence of ROOR.
A: Initiation: ROOR cleaves homolytically to give 2 RO·. Then RO· abstracts H from HBr to give ROH + Br·. Propagation: Br· adds to the less substituted carbon (CH₂ end) of 2-methylpropene, forming a 3° radical on the more substituted carbon. That radical then abstracts H from HBr, giving the anti-Markovnikov product (1-bromo-2-methylpropane) and regenerating Br·.
Q: What are the products of ozonolysis (O₃, then Zn/H₂O) of 2-methylbut-2-ene?
A: The double bond between C2 and C3 is cleaved. C2 (bearing two carbon substituents and no H) becomes a ketone: acetone (propan-2-one). C3 (bearing one carbon substituent and one H) becomes an aldehyde: acetaldehyde (ethanal).
Q: How would you convert an internal alkyne to a trans alkene? To a cis alkene?
A: Trans: dissolving metal reduction with Na in liquid NH₃. Cis: catalytic hydrogenation with H₂ and Lindlar catalyst.
Q: Provide the reagents to convert 2-methylpentane into a tertiary alkyl bromide.
A: Br₂ with heat or light (hv). Radical bromination is highly selective for the 3° C-H bond.
Every reaction here builds on carbocation stability and hybridisation concepts from the companion notes. The synthesis question on Exam 4 (covered in the next study notes doc) chains several of these individual reactions together, so you need to be able to recall each one quickly. Epoxidation connects forward to ring-opening reactions covered later in the course, and ozonolysis is a key tool in retrosynthetic analysis for identifying alkene starting materials.
electrophilic addition mechanism, Markovnikov rule, anti-Markovnikov, radical addition HBr peroxides, bromonium ion, halohydrin formation, chlorohydrin, hydroboration oxidation BH3 THF, acid catalysed hydration, epoxidation mCPBA peroxyacid, ozonolysis O3 Zn H2O, Lindlar catalyst cis alkene, dissolving metal reduction Na NH3 trans alkene, catalytic hydrogenation Pd/C, syn addition, anti addition, alkene reactions, alkyne reactions, CHEM 2301 exam 4, Novikov, University of Minnesota, organic chemistry I