Difficulty: Intermediate–Advanced | Prerequisites: IUPAC nomenclature, Lewis structures, electronegativity, carbocation stability
Alkene addition reactions are the single largest topic on Dr. Salmon's Exam 4, spanning the 30-point contrasting reactions section, the 21-point synthesis section, and the 27-point mechanism section. If you know what each reagent does to a double bond, along with its regiochemistry and stereochemistry, most of this exam becomes a pattern-matching exercise.
Alkenes react by addition: something breaks the pi bond and two new sigma bonds form. The reagent determines what adds, which carbon it lands on (regiochemistry), and whether it comes from the same face or opposite faces of the double bond (stereochemistry). Learn the reagent, learn the pattern.
Electrophilic addition
The general mechanism for alkene reactions: an electrophile (electron-poor species) attacks the electron-rich pi bond, forming a carbocation intermediate (in most cases), which is then captured by a nucleophile. Think of it as the double bond acting as a nucleophile that donates its pi electrons to an approaching electrophile.
Markovnikov's rule
In the addition of HX to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogens (the less substituted carbon), and the X group goes to the more substituted carbon. In simple terms, "the rich get richer" in hydrogen count. The underlying reason is carbocation stability: the more substituted carbocation is more stable.
Anti-Markovnikov addition
The opposite regiochemistry: X ends up on the less substituted carbon. This happens in radical additions (HBr with peroxides, ROOR) and in hydroboration-oxidation.
Syn addition
Both new groups add to the same face of the double bond. Catalytic hydrogenation (H2/Pt or H2/Pd) is the classic example: both hydrogens come from the metal surface and land on the same face.
Anti addition
The two new groups add to opposite faces of the double bond. Halogenation (Br2) proceeds through a bromonium ion intermediate, forcing the second bromine to attack from the back side. Halohydrin formation (Br2/H2O) also gives anti addition.
Regioselectivity
Preference for one constitutional isomer over another. A regioselective reaction favours one positional isomer (Markovnikov product over anti-Markovnikov, for example).
Stereoselectivity
Preference for one stereoisomer over another. A stereospecific reaction gives exclusively one stereochemical outcome (syn or anti addition).
Carbocation
A carbon bearing a positive charge. Stability order: 3° > 2° > 1° > methyl. Carbocation stability determines Markovnikov regiochemistry and susceptibility to rearrangements.
Bromonium ion
A three-membered ring intermediate containing a positively charged bromine bridging two carbons. Formed during the addition of Br2 to an alkene. It prevents free rotation and forces anti addition because the nucleophile must attack from the opposite face.
Reagents: H2O, H2SO4 (acid catalyst), HgSO4 (mercury(II) sulfate catalyst for alkynes).
Regiochemistry: Markovnikov. The -OH ends up on the more substituted carbon.
Stereochemistry: not stereospecific (both syn and anti products form; the carbocation intermediate is planar, so nucleophilic attack occurs from either face).
For alkynes, this gives an enol intermediate that tautomerises to a ketone (Markovnikov product is the ketone, not the aldehyde).
Exam context: Question 3a (right arrow) tests this reaction on an alkyne.
Reagents: HCl or HBr.
Regiochemistry: Markovnikov. H adds to the less substituted carbon, X adds to the more substituted carbon.
Stereochemistry: not stereospecific.
For alkynes with 2 equivalents of HX, the first equivalent gives a vinyl halide (Markovnikov), and the second equivalent adds across the resulting alkene, again Markovnikov, placing both halogens on the same carbon (a geminal dihalide).
Exam context: Question 3a (left arrow) asks for the product of 2 equiv. HCl on an alkyne. Question 5a tests the mechanism of 2 equiv. HBr on an alkyne.
Reagents: HBr with a peroxide initiator (ROOR).
Regiochemistry: anti-Markovnikov. Br adds to the less substituted carbon.
Stereochemistry: not stereospecific (radical intermediates are planar).
Important: this only works with HBr, not HCl or HI. Peroxides reverse the regiochemistry by changing the mechanism from ionic to radical.
Exam context: Question 3e (left arrow).
Reagents: an alcohol (e.g. CH3OH) with an acid catalyst (H2SO4).
Product: an ether, formed by Markovnikov addition of the alcohol across the double bond.
Regiochemistry: Markovnikov. The -OR group goes to the more substituted carbon.
Stereochemistry: not stereospecific.
Exam context: Question 3e (right arrow).
Reagents: H2 gas with a metal catalyst (Pt, Pd, or Ni).
Product: an alkane (fully saturated). Both hydrogens add across the double bond.
Regiochemistry: not applicable (H adds to both carbons).
Stereochemistry: syn addition. Both hydrogens are delivered from the metal surface to the same face of the alkene.
Exam context: Question 2b tests knowing that H2/Pd gives syn addition. Question 3c (right arrow) tests the product.
Reagents: Br2 (sometimes written as Br2 in CCl4 or CH2Cl2).
Product: a vicinal dihalide (1,2-dibromide).
Regiochemistry: both carbons of the double bond get a bromine.
Stereochemistry: anti addition, via a bromonium ion intermediate. The two bromines end up on opposite faces.
For alkynes, 1 equiv. Br2 gives a dibromoalkene (anti addition giving the trans product); 2 equiv. Br2 gives a tetrabromoalkane.
Exam context: Question 3d (left arrow) tests 2 equiv. Br2 on an alkyne.
Reagents: Br2 in water (Br2/H2O).
Product: a halohydrin (bromine on one carbon, -OH on the adjacent carbon).
Regiochemistry: Markovnikov-like. Water (the nucleophile) attacks the more substituted carbon of the bromonium ion because that carbon bears more positive charge.
Stereochemistry: anti addition (bromonium ion intermediate forces back-side attack).
The reaction is both regiospecific and stereospecific.
Exam context: Question 2d asks about this. Question 3b (right arrow) tests the product.
Reagents: meta-chloroperoxybenzoic acid (mCPBA), a peroxyacid.
Product: an epoxide (a three-membered ring with an oxygen).
Regiochemistry: not applicable (oxygen bridges both carbons).
Stereochemistry: syn addition. The oxygen is delivered to one face of the alkene in a concerted mechanism. If the alkene is cis, the resulting epoxide retains cis geometry.
Exam context: Question 3c (left arrow).
Reagents: O3 (ozone), followed by a reductive workup such as dimethyl sulfide (S(CH3)2) or Zn/AcOH.
Product: the C=C double bond is cleaved completely. Each fragment becomes a carbonyl: a disubstituted alkene carbon becomes a ketone, a monosubstituted one becomes an aldehyde, and a terminal =CH2 becomes formaldehyde (CH2O).
This reaction is used diagnostically: if you know the carbonyl products, you can work backwards to deduce the original alkene structure.
Exam context: Question 3b (left arrow) and Question 6 (spectral analysis) both use ozonolysis.
Reagents: concentrated H2SO4 or H3PO4, with heat.
Product: an alkene (water is eliminated).
This is the reverse of acid-catalysed hydration. Le Chatelier's principle applies: removing water (or decreasing its concentration) pushes the equilibrium toward the alkene product.
Regiochemistry: Zaitsev's rule (the more substituted alkene is the major product).
Exam context: Question 2c asks what to do with water concentration to push equilibrium toward products (answer: decrease it).
Epoxidation with peroxyacids is used industrially to make epoxy resins (the coatings on aluminium cans, for instance). Catalytic hydrogenation is how vegetable oils are converted to margarine, with partial hydrogenation of the C=C double bonds in unsaturated fatty acids. Ozonolysis is used in analytical chemistry to determine the position of double bonds in unknown structures, which matters in natural product and pharmaceutical research.
Students often assume HBr/ROOR gives anti-Markovnikov addition for all hydrogen halides. It only works for HBr. HCl and HI do not undergo efficient radical chain reactions with peroxides.
Confusing "syn addition" with "Markovnikov." These describe different things. Syn/anti describes stereochemistry (same face vs opposite faces). Markovnikov/anti-Markovnikov describes regiochemistry (which carbon gets which group). A reaction can be Markovnikov and anti (halohydrin formation) or anti-Markovnikov and non-stereospecific (radical HBr).
Forgetting to write "+ enantiomer" when a racemic mixture forms. If a new stereocentre is created and the starting material has no existing chirality, both enantiomers form in equal amounts. Dr. Salmon's exam explicitly asks you to note this.
Treating ozonolysis as a simple addition. Ozonolysis cleaves the double bond entirely. The products are two separate carbonyl compounds, not one molecule with two new groups.
⚠️ Question 3 (Contrasting Reactions) is worth 30 points and tests ten separate reactions. Each box is 3 points. Getting the regiochemistry or stereochemistry wrong costs partial credit even if the connectivity is right.
⚠️ "If a mixture of enantiomers will be formed, draw one stereoisomer and write '+ enantiomer' in the box with your product." This instruction appears in bold on the exam. Forgetting "+ enantiomer" when a new stereocentre is created from an achiral starting material loses marks.
⚠️ "When multiple elimination products could potentially be formed, only draw the major isomer." For dehydration, this means the Zaitsev (more substituted) alkene.
⚠️ Question 5 (Mechanisms) is 27 points and requires curved arrows for every electron-movement step. Missing an arrow or skipping a resonance structure of a carbocation intermediate will cost points.
True or false: HBr with peroxides (ROOR) gives Markovnikov addition.
False. It gives anti-Markovnikov (radical mechanism).
Fill in the blank: Catalytic hydrogenation (H2/Pt) gives ______ addition.
Syn addition.
True or false: Halohydrin formation (Br2/H2O) is both regiospecific and stereospecific.
True.
Fill in the blank: Ozonolysis of a terminal alkene (R-CH=CH2) with reductive workup produces an aldehyde and ______.
Formaldehyde (CH2O).
True or false: To push the dehydration equilibrium toward the alkene product, you should increase the water concentration.
False. You should decrease the water concentration (remove water).
Q: An alkene is treated with Br2 in water. What is the product? Describe the regiochemistry and stereochemistry.
A: A bromohydrin. The bromine and hydroxyl are on adjacent carbons (vicinal). The -OH is on the more substituted carbon (Markovnikov-like, because water attacks the more substituted carbon of the bromonium ion). The stereochemistry is anti: Br and OH are on opposite faces of what was the double bond.
Q: You treat an alkene with mCPBA. What is the product and its stereochemistry?
A: An epoxide. The stereochemistry is syn: the oxygen is delivered to one face of the alkene in a concerted reaction. If the starting alkene is cis, the substituents remain cis in the epoxide.
Q: What products result from ozonolysis (O3, then S(CH3)2) of methylenecyclohexane (a cyclohexane ring with an exocyclic =CH2)?
A: The exocyclic double bond is cleaved. The =CH2 becomes formaldehyde (CH2O). The ring carbon that was part of the double bond becomes a ketone (cyclohexanone). Products: cyclohexanone + formaldehyde.
Q: Draw the mechanism for adding 2 equivalents of HBr to propyne. What is the final product?
A: Step 1, the pi electrons of the triple bond attack the H of HBr, following Markovnikov's rule, placing H on the terminal carbon (C1) and Br on C2. This gives 2-bromopropene (a vinyl bromide). Step 2, the pi electrons of the resulting alkene attack HBr again, and Markovnikov addition places the second Br on C2 (the carbon already bearing bromine, because that gives the more stable carbocation). Final product: 2,2-dibromopropane (a geminal dibromide).
Q: An alkene reacts with H2SO4 and CH3OH. What type of product forms and with what regiochemistry?
A: An ether. Methanol acts as the nucleophile, adding to the carbocation formed by protonation of the alkene. Regiochemistry is Markovnikov: the -OCH3 group ends up on the more substituted carbon.
Alkene addition reactions connect directly to alkyne chemistry: many of the same reagents work on triple bonds, sometimes with different stoichiometry (1 vs 2 equivalents). Carbocation stability, which you first learned with SN1 and E1 reactions, drives Markovnikov selectivity here. Stereochemistry concepts (R/S, syn/anti) from earlier chapters are applied every time a new stereocentre forms in an addition reaction.
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