Source: Exam 4 Key, University of Minnesota Twin Cities
Difficulty: Intermediate | Prerequisites: IUPAC nomenclature, carbocation stability, stereochemistry basics (R/S, E/Z) Tags: electrophilic addition, Markovnikov, anti-Markovnikov, hydration, halogenation, hydroboration-oxidation, radical addition, stereochemistry, syn addition, anti addition, alkyne reactions, dissolving metal reduction, Lindlar catalyst
Alkenes and alkynes undergo addition reactions that break pi bonds and form new sigma bonds. The regiochemistry (where groups add) follows Markovnikov's or anti-Markovnikov's rule depending on the reagent. The stereochemistry (syn or anti addition) depends on the mechanism. Knowing which reagent gives which combination of regiochemistry + stereochemistry is the core skill tested.
Electrophilic addition
A reaction in which an electrophile adds to a pi bond, forming two new sigma bonds. The pi electrons attack the electrophile in the first step. In simple terms, the electron-rich double bond grabs an electron-poor species, opening the double bond so two new groups can attach.
Markovnikov's rule
In the addition of HX to an unsymmetrical alkene, the hydrogen adds to the less substituted carbon and the halide to the more substituted carbon. This is because the more stable (more substituted) carbocation intermediate forms preferentially. Think of it as "the rich get richer": the carbon with more hydrogens gets the new hydrogen.
Anti-Markovnikov addition
The opposite regiochemistry: the hydrogen ends up on the more substituted carbon and the other group on the less substituted carbon. Achieved with radical conditions (HBr + peroxides) or hydroboration-oxidation (BH₃ then H₂O₂/NaOH).
Syn addition
Both new groups add to the same face of the double bond. Examples: hydrogenation (H₂/Pd or H₂/Pt), hydroboration-oxidation, epoxidation. Think of it as both groups landing on the same side of the flat alkene.
Anti addition
The two new groups add to opposite faces of the double bond. Examples: halogenation (Br₂ or Cl₂, via a cyclic halonium ion intermediate), halohydrin formation. Think of it as the groups landing on opposite sides.
Halonium ion (bromonium ion, chloronium ion)
A three-membered ring intermediate formed when a halogen (Br₂ or Cl₂) reacts with an alkene. The ring blocks one face of the double bond, forcing the nucleophile to attack from the opposite face (anti addition).
Hydroboration-oxidation
A two-step sequence: (1) BH₃ adds across the double bond in a syn, anti-Markovnikov fashion; (2) H₂O₂/NaOH oxidises the C-B bond to C-OH with retention of configuration. Net result: syn, anti-Markovnikov addition of water.
Carbocation rearrangement
A 1,2-hydride shift or 1,2-methyl shift that converts a less stable carbocation to a more stable one. Always check for rearrangements when a carbocation intermediate is involved (acid-catalysed hydration, HX addition). In simple terms, the positive charge "moves" to a neighbouring carbon if doing so makes it more stable.
Ozonolysis
Cleavage of a C=C double bond using ozone (O₃), followed by a reductive workup (Zn/H₂O or DMS). Each carbon of the former double bond becomes a carbonyl (aldehyde or ketone, depending on substitution).
Dissolving metal reduction
Reduction of an alkyne to a trans-alkene using an alkali metal (Li or Na) in liquid ammonia (NH₃). Produces the E (trans) alkene via an anti addition of two hydrogens.
Lindlar catalyst
A poisoned palladium catalyst (Pd on CaCO₃ with lead acetate and quinoline) that reduces an alkyne to a cis-alkene (Z). It stops at the alkene stage because the catalyst is deactivated ("poisoned") for further reduction. Think of it as the gentle catalyst that only half-reduces the triple bond and delivers syn hydrogenation.
Mechanism: protonation of the alkene forms a carbocation (Markovnikov), then water attacks the carbocation, and deprotonation gives the alcohol.
Regiochemistry: Markovnikov (OH ends up on the more substituted carbon).
Stereochemistry: not controlled (the carbocation is planar, so nucleophilic attack occurs from both faces).
Rearrangements: yes. Because a carbocation intermediate forms, always check for 1,2-hydride or 1,2-methyl shifts to a more stable carbocation.
Exam example: hydration of a trisubstituted alkene with H₂SO₄/H₂O gave both a rearranged product (major, from a hydride shift to a more stable 3° carbocation) and a non-rearranged product.
Mechanism: protonation of the alkene forms a carbocation (Markovnikov), then halide attacks.
Regiochemistry: Markovnikov (halide on the more substituted carbon).
Stereochemistry: mixture of syn and anti (planar carbocation).
Rearrangements: possible (carbocation intermediate).
Excess HCl with an alkyne: two sequential additions give a geminal dihalide (both halogens on the same carbon, the more substituted one per Markovnikov's rule applied twice).
Mechanism: the pi bond attacks the halogen to form a cyclic halonium ion, then the halide ion attacks the more substituted carbon from the opposite face.
Regiochemistry: both halogens add (anti-periplanar opening of the halonium ion).
Stereochemistry: anti addition. If the alkene is cyclic or has defined geometry, this produces trans-dihalide products.
Stereoisomers: a racemic mixture of enantiomers typically forms (attack can occur on either face of the alkene to start, giving two enantiomeric halonium ions).
Mechanism: same halonium ion intermediate as halogenation, but water (the solvent) acts as the nucleophile instead of halide.
Regiochemistry: water attacks the more substituted carbon of the halonium ion (Markovnikov-like for the OH).
Stereochemistry: anti addition (nucleophile attacks opposite the halonium bridge).
Product: a halohydrin (adjacent C-X and C-OH, trans to each other).
Mechanism: a concerted [2+1] cycloaddition where the peroxyacid delivers an oxygen atom to the alkene.
Stereochemistry: syn addition. The geometry of the alkene is preserved in the epoxide. A cis-alkene gives a cis-epoxide; a trans-alkene gives a trans-epoxide.
If the alkene is achiral and internal, a racemic mixture of enantiomeric epoxides forms.
Mechanism: peroxide-initiated radical chain. Br· adds to the less substituted carbon first (forming the more stable radical on the more substituted carbon), then H· adds.
Regiochemistry: anti-Markovnikov (Br ends up on the less substituted carbon).
Stereochemistry: mixture of syn and anti (radicals are planar, attack from both faces).
Important: only works with HBr. HCl and HI do not undergo useful radical addition to alkenes.
Mechanism: BH₃ adds across the double bond in a concerted, four-centred transition state. The boron attaches to the less substituted carbon (anti-Markovnikov). Oxidation with H₂O₂/NaOH replaces B with OH, retaining the stereochemistry.
Regiochemistry: anti-Markovnikov (OH on the less substituted carbon).
Stereochemistry: syn addition. Both the B (later replaced by OH) and the H add to the same face.
No rearrangements: there is no carbocation intermediate, so no 1,2-shifts.
Both hydrogens add to the same face of the alkene (syn addition) because the reaction occurs on the metal surface.
Completely reduces alkenes to alkanes. Also reduces alkynes to alkanes (goes all the way through).
Stereochemistry: syn. On a cyclic alkene, both H atoms end up cis.
Not an addition reaction, but a substitution at the most stable radical position.
Br₂ with light (or heat) on an alkane or at an allylic position proceeds through a radical mechanism.
Selectivity: bromine is highly selective for the most substituted C-H (3° > 2° > 1°) because the more stable radical intermediate has a lower-energy transition state.
Stereochemistry: if bromine replaces a hydrogen on a stereogenic centre, a racemic mixture forms (planar radical intermediate).
One equivalent of Br₂ adds across a triple bond to give a trans-dibromoalkene (anti addition via a bromonium-like intermediate).
Two equivalents give a tetrahalide.
Reduces an internal alkyne to a trans (E) alkene.
Mechanism: stepwise addition of electrons and protons. The geometry is anti because the vinyl radical/anion intermediates prefer the trans configuration to minimise steric strain.
This is the way to make a trans-alkene from an alkyne.
Reduces an alkyne to a cis (Z) alkene.
The poisoned palladium catalyst stops reduction at the alkene stage.
Stereochemistry: syn addition of H₂ gives the Z-alkene.
This is the way to make a cis-alkene from an alkyne.
NaNH₂ deprotonates the terminal alkyne (pKa ~25) to give an acetylide anion.
The acetylide is a strong nucleophile/base and undergoes SN2 with a primary alkyl halide (CH₃I, etc.).
This extends the carbon chain by forming a new C-C bond.
Acid-catalysed (H₂O, H₂SO₄, HgSO₄): Markovnikov addition of water gives an enol, which tautomerises to a ketone (or aldehyde from a terminal alkyne, though Markovnikov gives a methyl ketone).
The mercury(II) sulfate catalyst is required specifically for alkyne hydration; it is not used for alkene hydration.
Alkene ozonolysis (O₃, then Zn/H₂O or DMS): cleaves the double bond completely. Each carbon becomes a carbonyl. Terminal alkene carbons become aldehydes; internal ones become ketones.
Useful for structure determination: if you know the ozonolysis products, you can work backwards to deduce the alkene structure.
Students assume anti-Markovnikov addition happens with any HX + peroxides. Only HBr undergoes radical addition. HCl and HI do not.
Students forget to check for carbocation rearrangements in acid-catalysed hydration and HX addition. Any time a carbocation forms, a shift to a more stable carbocation is possible.
Students confuse syn and anti addition. A useful mnemonic: hydroboration and hydrogenation are both "H" reactions and both syn. Halogenation through a halonium ion is anti.
Students think dissolving metal reduction gives the cis-alkene. It gives the trans (E) alkene. Lindlar gives cis (Z). Mix these up and the synthesis falls apart.
Students forget that ozonolysis products are carbonyls, not alcohols. Each side of the broken double bond becomes an aldehyde (if terminal) or a ketone (if internal).
⚠️ Predicting major products is the most heavily tested skill. For each reagent set, you must know: regiochemistry (Markovnikov or anti-Markovnikov), stereochemistry (syn, anti, or no preference), and whether rearrangements are possible.
⚠️ Stereochemistry notation matters. If stereoisomers form, you must indicate them ("+enantiomer" or draw both). Losing marks for omitting "+enantiomer" is common.
⚠️ Multi-step synthesis questions require you to chain reactions together. Think backwards from the target: what is the last step, and what starting material does it need?
⚠️ Mechanism questions want every arrow, every intermediate, and correct electron flow direction. Forgetting to show the halonium ion in halogenation or the oxymercuration intermediate in hydration costs full marks.
⚠️ The Lindlar vs dissolving metal reduction distinction is tested almost every exam. Lindlar = cis (Z). Li/NH₃ = trans (E).
True or False: HBr with peroxides adds Br to the more substituted carbon. False. Anti-Markovnikov: Br goes to the less substituted carbon.
Fill in the blank: Halogenation of an alkene with Br₂ proceeds through a ______ intermediate and gives ______ addition. Bromonium ion; anti.
True or False: Hydroboration-oxidation gives Markovnikov addition of water. False. It gives anti-Markovnikov, syn addition.
Fill in the blank: To convert an internal alkyne to a trans-alkene, use ______ in ______. Li (or Na); liquid NH₃.
True or False: Acid-catalysed hydration of an alkene can produce rearranged products. True. The carbocation intermediate can undergo 1,2-shifts.
Q: Draw the expected product(s) when a trisubstituted alkene is treated with H₂SO₄/H₂O. Should you expect rearrangement?
A: Markovnikov hydration gives the alcohol on the more substituted carbon. Yes, always check for rearrangement: if a 1,2-hydride or methyl shift converts the initial carbocation to a more stable one, the rearranged product forms alongside (or instead of) the direct product.
Q: A cyclohexene is treated with Cl₂. What is the stereochemistry of the product?
A: Anti addition through a chloronium ion gives trans-1,2-dichlorocyclohexane (+ enantiomer, as a racemic mixture).
Q: You need to place an OH on the less substituted carbon of an alkene with syn stereochemistry. Which reagent set do you use?
A: Hydroboration-oxidation (1. BH₃, 2. H₂O₂/NaOH). It delivers anti-Markovnikov regiochemistry with syn addition.
Q: What product forms when an alkyne is treated with 1. NaNH₂, 2. CH₃I, then H₂ with Lindlar catalyst?
A: Step 1-2: the terminal alkyne is deprotonated and alkylated, adding a methyl group to give an internal alkyne. Step 3: Lindlar hydrogenation gives the cis (Z) alkene.
Q: An alkene is treated with Cl₂/H₂O. Draw the mechanism showing the intermediate and the final product.
A: The alkene attacks Cl₂ to form a chloronium ion (three-membered ring with Cl bridging the two carbons). Water, acting as the nucleophile, opens the ring by attacking the more substituted carbon from the opposite face (anti). Deprotonation gives a chlorohydrin with trans geometry between the Cl and OH.
Q: Predict the product of radical polymerisation of methyl acrylate (CH₂=CH-COOMe) with peroxides and heat.
A: The radical adds to the less substituted end (the CH₂ end) of each monomer. The polymer backbone has the repeating pattern -(CH₂-CH(COOMe))n-, with the ester groups alternating on one side. This is the head-to-tail polymer.
Electrophilic addition connects directly back to carbocation stability and Hammond's postulate from the nomenclature/intermediates notes. Every Markovnikov addition goes through the more stable carbocation.
Stereochemistry of addition reactions feeds into the stereochemistry chapter: you need R/S assignments to fully describe the products of anti addition to cyclic alkenes, and E/Z to describe alkyne reduction products.
Synthesis problems chain these reactions together. A common pattern: start with an alkyne (easy to manipulate regiochemistry), reduce to the alkene you want (Lindlar for cis, dissolving metal for trans), then functionalise the alkene.
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