Difficulty: Intermediate-Advanced | Prerequisites: Nomenclature and stability of alkenes/alkynes, basic arrow-pushing
This exam tests your ability to predict the products of alkene and alkyne reactions, paying close attention to regiochemistry (where groups add) and stereochemistry (syn vs anti addition, racemic mixtures). The core reactions include dissolving metal reduction, Lindlar hydrogenation, halogenation, acid-catalysed hydration, radical addition of HBr, epoxidation with mCPBA, ozonolysis, and catalytic hydrogenation. Each reagent set gives a specific product with specific stereochemistry, and confusing them is the fastest way to lose marks.
Markovnikov addition
In the addition of HX to an alkene, the hydrogen adds to the less substituted carbon and the X (halide) adds to the more substituted carbon. This occurs because the reaction proceeds through the more stable (more substituted) carbocation intermediate. In simple terms, "the rich get richer" : the carbon with more hydrogens gets the new hydrogen.
Anti-Markovnikov addition
The opposite regiochemistry: H adds to the more substituted carbon, and the other group (e.g. Br) adds to the less substituted carbon. This happens when the mechanism goes through a radical intermediate rather than a carbocation, typically triggered by peroxides (ROOR).
Syn addition
Both new groups add to the same face of the double bond. Examples: catalytic hydrogenation (H₂/Pt or H₂/Pd), hydroboration-oxidation, Lindlar hydrogenation of alkynes.
Anti addition
The two new groups add to opposite faces of the double bond. Examples: halogenation with Br₂ or Cl₂ (via a cyclic halonium ion intermediate), epoxidation followed by ring opening.
Lindlar catalyst
A "poisoned" palladium catalyst (Pd/CaCO₃ with lead acetate and quinoline) that reduces an alkyne to a cis (Z) alkene and stops there. It cannot reduce the alkene further. Think of it as a gentle, selective hydrogenation.
Dissolving metal reduction (Na/NH₃)
Sodium metal in liquid ammonia reduces an internal alkyne to a trans (E) alkene. The mechanism involves radical anion intermediates that favour the more stable trans geometry.
mCPBA (meta-chloroperoxybenzoic acid)
A peracid that converts an alkene into an epoxide (a three-membered ring with an oxygen). The addition is syn and concerted, meaning both C-O bonds form on the same face simultaneously.
Ozonolysis
Treatment of an alkene with ozone (O₃) followed by a workup reagent. O₃ then H₂O (or DMS or Zn) gives a reductive workup, cleaving the double bond into two carbonyl compounds (aldehydes and/or ketones). O₃ then H₂O₂ gives an oxidative workup, producing carboxylic acids instead of aldehydes.
Radical chain mechanism
A three-phase mechanism: initiation (homolytic cleavage of the peroxide O-O bond to generate radicals), propagation (radical addition to the alkene, then abstraction of H from HBr), and termination (any two radicals combine).
Halonium ion (bromonium or chloronium ion)
A three-membered ring intermediate formed when Br₂ or Cl₂ adds to an alkene. The halogen bridges both carbons, blocking one face and forcing the second halogen to attack from the opposite face (anti addition).
Converts an internal alkyne to a trans (E) alkene.
Mechanism goes through a vinyl radical anion. The trans product is favoured because the bulky groups prefer to be on opposite sides during the electron-addition steps.
Does not work on terminal alkynes the same way (the acidic terminal H complicates things).
Converts an alkyne to a cis (Z) alkene.
Syn addition of H₂ across the triple bond, stopping at the alkene stage because the poisoned catalyst cannot reduce further.
Lindlar = Pd/CaCO₃/Pb(OAc)₂/quinoline.
One equivalent of Br₂ or Cl₂ adds across the triple bond to give a dihaloalkene (anti addition, giving the trans dihaloalkene).
Two equivalents give a tetrahalide (all four halogens added).
The exam specifically tested 2 equiv. Cl₂ addition to a terminal alkene (Question 3a), yielding a vicinal dichloride and then a geminal/vicinal tetrachloride.
NaNH₂ is a strong base that deprotonates terminal alkynes to form acetylide anions (RC≡C⁻).
With 2 equivalents of NaNH₂ on a vicinal dihalide, double elimination occurs: first elimination gives a vinyl halide, second elimination gives an alkyne.
This is a key synthetic tool for making alkynes from dihalides (tested in Question 6).
Adds water across a double bond following Markovnikov regiochemistry: OH goes to the more substituted carbon.
Mechanism: protonation of the alkene to form the more stable carbocation, then nucleophilic attack by water, then deprotonation.
Produces an alcohol. If the carbocation intermediate is achiral, expect a racemic mixture (the nucleophile can attack from either face of the planar carbocation).
Adds HBr with anti-Markovnikov regiochemistry: Br goes to the less substituted carbon.
Mechanism: radical chain. Initiation produces radicals from the peroxide. In propagation, a bromine radical adds to the less substituted end of the alkene (forming the more stable, more substituted carbon radical), then that radical abstracts H from HBr.
Only works with HBr and peroxides. HCl/ROOR and HI/ROOR do not give anti-Markovnikov products (the thermodynamics do not favour radical propagation with those hydrogen halides).
Stereochemistry: the radical intermediate is planar (sp²), so attack on either face is equally likely. Expect a racemic mixture at any new stereocentre.
Br₂ adds across the double bond with anti stereochemistry via a cyclic bromonium ion intermediate.
The product is a vicinal dibromide with the two Br atoms on opposite faces.
If the starting alkene is symmetric, the product may be a meso compound. If asymmetric, expect a racemic pair of enantiomers (+ enantiomer).
mCPBA delivers an oxygen atom to the double bond in a single concerted step, forming an epoxide (oxirane).
Syn addition: the oxygen bridges from one face. The stereochemistry of the starting alkene is preserved in the epoxide.
A cis alkene gives a cis-substituted epoxide, a trans alkene gives a trans-substituted epoxide.
If the alkene face is equally accessible from both sides, expect a racemic mixture of epoxides (+ enantiomer).
Adds H₂ across the double bond with syn stereochemistry.
Both hydrogens add to the same face of the alkene because the reaction occurs on the metal surface.
Converts an alkene to an alkane. No regiochemistry to worry about (both atoms added are H).
Pt and Pd are not poisoned, so they will reduce alkynes all the way to alkanes (unlike Lindlar, which stops at the alkene).
Cleaves the C=C double bond completely, replacing it with two C=O bonds.
Reductive workup (O₃ then H₂O, or O₃ then DMS/Zn): terminal =CH₂ gives formaldehyde (H₂C=O); internal =CHR gives an aldehyde (RCHO); =CR₂ gives a ketone (R₂C=O).
Oxidative workup (O₃ then H₂O₂): aldehydes are further oxidised to carboxylic acids.
This reaction is useful in synthesis for "cutting" molecules and is useful analytically for determining the position of a double bond.
Students frequently confuse Markovnikov and anti-Markovnikov regiochemistry. Remember: Markovnikov is the default for ionic (polar) additions like H₂O/H₂SO₄ and HBr without peroxides. Anti-Markovnikov only occurs with HBr in the presence of peroxides (radical mechanism).
A common mistake is assuming that HCl/ROOR or HI/ROOR also give anti-Markovnikov products. They do not. Only HBr works with peroxides for anti-Markovnikov addition.
Students often forget to write "+ enantiomer" when a racemic mixture forms. If a reaction creates a new stereocentre through a planar intermediate (carbocation or radical), both enantiomers form in equal amounts.
Another frequent error is confusing syn and anti addition. Hydrogenation and Lindlar reduction are syn. Halogenation (Br₂, Cl₂) is anti. mCPBA epoxidation is syn (it delivers the O from one face).
On ozonolysis questions, students sometimes forget that a reductive workup (H₂O or DMS) gives aldehydes and ketones, while an oxidative workup (H₂O₂) converts aldehydes further to carboxylic acids. The exam specified O₃ then H₂O, which is a reductive workup.
⚠️ Question 3 (Contrasting Reactions) is worth 30 points, nearly a third of the exam. It presents pairs of reactions on the same starting material with different reagents. You must predict the correct product for each, with correct regiochemistry and stereochemistry.
⚠️ The exam pairs reactions that look similar but give different products. For example, H₂O/H₂SO₄ on an alkene (Markovnikov alcohol) is contrasted with HBr/ROOR on the same alkene (anti-Markovnikov bromide). Know the differences cold.
⚠️ Question 4 (Synthesis, 20 points) requires you to chain multiple reactions together to convert a starting material into a target. You need to know which reagents achieve each transformation and show all isolable intermediates.
⚠️ Always indicate stereochemistry in your products. If a racemic mixture forms, draw one enantiomer and write "+ enantiomer." If a meso compound forms, draw it. If a specific diastereomer forms (e.g. anti addition), show it with wedges and dashes.
True or false: HBr/ROOR gives Markovnikov addition. (False, it gives anti-Markovnikov)
Fill in the blank: Lindlar hydrogenation of an internal alkyne gives a ____ alkene. (cis / Z)
True or false: Br₂ addition to an alkene proceeds through a bromonium ion and gives anti addition. (True)
Fill in the blank: Na/NH₃ reduction of an internal alkyne gives a ____ alkene. (trans / E)
True or false: Ozonolysis with H₂O workup converts a terminal =CH₂ to a carboxylic acid. (False, reductive workup gives formaldehyde; oxidative workup with H₂O₂ would give a carboxylic acid)
Q: 1-Butene is treated with H₂O/H₂SO₄. What is the major product?
A: 2-Butanol (Markovnikov addition: the OH ends up on the more substituted carbon, C2). The mechanism proceeds through a secondary carbocation at C2.
Q: 1-Butene is treated with HBr/ROOR. What is the major product?
A: 1-Bromobutane (anti-Markovnikov addition: Br ends up on the less substituted carbon, C1). The radical forms at C2 (more stable secondary radical), and Br is already on C1 from the initial radical addition step.
Q: An internal alkyne is treated with Na/NH₃, then with Br₂. What products form and what is the stereochemistry?
A: First step gives the trans (E) alkene. Second step (Br₂ addition) gives a vicinal dibromide with anti stereochemistry via the bromonium ion. The overall product is the anti addition product of Br₂ across a trans alkene. Identify the specific stereocentres to determine whether you get a racemic mixture or a meso compound.
Q: Propose a synthesis of a trans alkene from a terminal alkyne (propyne to trans-2-pentene, for example).
A: First, deprotonate the terminal alkyne with NaNH₂ to form the acetylide anion. Then, alkylate with an appropriate alkyl halide (ethyl bromide for this case) to extend the chain and form an internal alkyne (2-pentyne). Finally, reduce with Na/NH₃ to give trans-2-pentene.
Q: Cyclohexene is treated with mCPBA. Describe the product.
A: The product is 1,2-epoxycyclohexane (cyclohexene oxide). The oxygen is delivered from one face in a syn, concerted fashion. Since cyclohexene is symmetric, both faces are equally accessible, so you get a racemic mixture of enantiomeric epoxides (+ enantiomer).
Q: 2-Methyl-2-butene is treated with O₃ then H₂O. What products form?
A: Ozonolysis cleaves the double bond. 2-Methyl-2-butene has the double bond between C2 and C3. Cleavage gives acetone (from the C2 side, which bears two methyl groups and becomes a ketone) and acetaldehyde (from the C3 side, which bears one methyl and one H, becoming an aldehyde).
Every reaction here builds on the stability and nomenclature concepts from the first set of notes. Markovnikov regiochemistry exists because the more substituted carbocation is more stable. Anti-Markovnikov radical addition exists because the more substituted radical is more stable. The mechanisms tested in Question 5 of the exam (radical chain for HBr/ROOR, electrophilic addition for HCl) require comfortable arrow-pushing through these intermediates.
The spectral analysis question (Question 6) ties reactions to structure determination: you must recognise that treating an alkene with Cl₂ gives a vicinal dichloride, and that double elimination with NaNH₂ converts that dichloride into an alkyne.
Alkene reactions, alkyne reactions, Markovnikov, anti-Markovnikov, radical addition, HBr ROOR, acid-catalysed hydration, halogenation, bromonium ion, chloronium ion, mCPBA, epoxidation, ozonolysis, Lindlar catalyst, dissolving metal reduction, Na NH3, syn addition, anti addition, catalytic hydrogenation, NaNH2, acetylide anion, synthesis, organic chemistry I, CHEM 2301, Exam 4, stereochemistry, regiochemistry, racemic mixture, enantiomer