Difficulty: Intermediate–Advanced | Prerequisites: Part 1 of these notes (sections 8.1–8.7), carbocation stability, stereochemistry (R/S, E/Z), and nucleophile/electrophile concepts.
Part 1 covered the "Markovnikov family" of additions (HX, H₂O) and hydrogenation. This second half introduces the reactions that go the other way or do something different entirely. Hydroboration-oxidation gives anti-Markovnikov alcohols. Halogen addition and epoxidation are stereospecific for anti and syn addition respectively. Ozonolysis cleaves the double bond altogether. Together with Part 1, these reactions form the complete toolkit for transforming alkenes, and retrosynthetic analysis (8.14) is where you learn to use that toolkit backwards to plan a synthesis.
Hydroboration-oxidation adds H and OH across a double bond with anti-Markovnikov regiochemistry and syn stereochemistry. Halogen addition (Br₂, Cl₂) goes through a cyclic halonium ion and gives anti addition. Epoxidation with a peroxy acid is a concerted syn addition that produces a three-membered oxygen ring. Ozonolysis cleaves the double bond completely, giving two carbonyl compounds. Retrosynthesis ties all of these reactions together by working backwards from a target molecule to choose the right alkene and reagent.
Hydroboration
Addition of a boron hydride (BH₃ or R₂BH) across a carbon–carbon pi bond, placing boron on the less substituted carbon and hydrogen on the more substituted carbon. Think of it as the "anti-Markovnikov" partner to acid-catalysed hydration.
Organoborane
The intermediate formed after hydroboration: a molecule with a carbon–boron bond. It is not the final product; it still needs oxidation.
Oxidation (of an organoborane)
Treatment of the organoborane with hydrogen peroxide (H₂O₂) and aqueous base (NaOH) replaces the C–B bond with a C–OH bond, yielding the alcohol.
Anti-Markovnikov addition
A regiochemical outcome where the nucleophilic portion (e.g. OH) ends up on the less substituted carbon of the former double bond. Hydroboration-oxidation is the classic example.
Halonium ion
A three-membered cyclic intermediate containing a positively charged halogen atom bridging two carbons. In simple terms, the halogen forms a tiny ring that locks the two carbons in place, forcing the second halide to attack from the opposite face.
Vicinal dihalide
A compound with two halogen atoms on adjacent (vicinal) carbons. The product of halogen addition to an alkene.
Halohydrin
A compound with a halogen and a hydroxyl group on adjacent carbons.
Epoxide (oxirane)
A three-membered ring containing one oxygen atom. Formed by treating an alkene with a peroxy acid.
Peroxy acid (peracid)
An organic acid with an extra oxygen (RCOOOH). The oxygen donor in epoxidation. A common example is mCPBA (meta-chloroperoxybenzoic acid).
Ozonolysis
Cleavage of a carbon–carbon double bond by ozone (O₃), followed by a workup step, to give two carbonyl compounds (aldehydes and/or ketones).
Ozonide
The unstable cyclic intermediate formed when ozone reacts with an alkene. It is not isolated; it decomposes during the aqueous workup.
Enantioselective addition
An addition reaction that preferentially forms one enantiomer over the other, typically achieved using a chiral catalyst or enzyme.
Retrosynthetic analysis
Working backwards from a target molecule to identify viable starting materials and reactions. Drawn with a retrosynthetic arrow (⇒) pointing from product to precursor.
Two-step sequence: hydroboration (adds BH₃ across the double bond), then oxidation (replaces C–B with C–OH).
Overall result: addition of H and OH across the double bond.
Regiochemistry: anti-Markovnikov
Boron bonds to the less substituted carbon of the double bond. After oxidation, the OH ends up on the less substituted carbon.
This is the opposite of acid-catalysed hydration, which places OH on the more substituted carbon.
Stereochemistry: syn addition
Both the B–H addition and the subsequent oxidation proceed with retention of configuration. The net result is syn addition of H and OH.
Example: hydroboration-oxidation of 1-methylcyclopentene gives trans-2-methylcyclopentanol (H and OH added to the same face, then the ring geometry makes them trans to the methyl group).
Hydroboration step
Step 1: BH₃ approaches the alkene. Electrons flow from the pi orbital of the alkene into the empty 2p orbital on boron, forming a pi complex.
Step 2: The pi complex rearranges in a concerted fashion: the B–H bond breaks, boron bonds to the less substituted carbon, and hydrogen migrates to the more substituted carbon. Both new bonds form on the same face (syn).
Because no carbocation intermediate forms, there is no opportunity for rearrangement. This is a key contrast with HX and H₂O additions.
Oxidation step
Step 1: Hydrogen peroxide is converted to its anion (HOO⁻) in basic solution.
Step 2: The hydroperoxide anion attacks boron as a nucleophile, forming an oxygen–boron bond.
Step 3: Carbon migrates from boron to oxygen in a 1,2-shift, displacing hydroxide. This step retains the stereochemistry at carbon.
Step 4: Hydrolysis of the resulting alkoxyborane cleaves the B–O bond, yielding the alcohol.
The key thing to remember: hydroboration gives anti-Markovnikov, syn addition. No carbocation, no rearrangement.
Halogens in diatomic form (Br₂, Cl₂) add across the double bond to give a vicinal dihalide (two halogens on adjacent carbons).
More substituted double bonds react faster (relative rates for bromine addition: ethylene 1.0, propene 61, 2-methylpropene 5400, 2,3-dimethyl-2-butene 920,000).
Mechanism (two steps, via halonium ion)
Step 1 (slow, rate-determining): The alkene's pi electrons attack one atom of X₂. A cyclic halonium ion forms (three-membered ring with a positively charged halogen bridging both carbons), and X⁻ is released.
The halonium ion is more stable than an open carbocation because the octet rule is satisfied for both carbons and the halogen.
Step 2 (fast): The halide ion (nucleophile) attacks one of the carbons of the halonium ion from the side opposite the bridging halogen (backside attack, like SN2). This forces anti addition.
Stereochemistry: anti addition
Anti addition of Br₂ to (E)-2-butene gives meso-2,3-dibromobutane.
Anti addition of Br₂ to (Z)-2-butene gives a racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane.
Two factors to keep track of: the E or Z configuration of the starting alkene, and the anti stereochemistry of the addition.
Halohydrins
When halogen addition is carried out in water, the solvent (water) can act as the nucleophile instead of the halide ion, giving a halohydrin (halogen and OH on adjacent carbons) with anti stereochemistry.
An epoxide is a three-membered ring containing one oxygen atom (also called an oxirane).
Epoxidation is carried out by reacting an alkene with a peroxy acid (RCOOOH). The byproduct is a carboxylic acid.
Mechanism (concerted)
All the bonding changes occur in a single transition state: the oxygen is transferred from the peroxy acid to the less crowded face of the alkene, forming the epoxide ring and the carboxylic acid simultaneously.
Because the mechanism is concerted, there is no intermediate. This means no carbocation, no rearrangement.
Stereochemistry: syn addition
The oxygen is delivered to one face of the double bond in a single step, so the stereochemistry of the starting alkene is preserved in the epoxide.
A cis alkene gives a cis epoxide; a trans alkene gives a trans epoxide.
Reactivity trend
More substituted alkenes react faster (same trend as all the other electrophilic additions). Relative rates with peroxyacetic acid: ethylene 1.0, propene 22, 2-methylpropene 484, 2,3-dimethyl-2-butene 6526.
Ozone (O₃) is a powerful electrophile that reacts with alkenes to cleave the double bond entirely.
The initial product is an ozonide (a cyclic peroxide), which is unstable and is decomposed during workup.
Products
Hydrolysis of the ozonide with water gives two carbonyl compounds plus hydrogen peroxide.
What you get depends on the substitution of the original double bond carbons:
A carbon bearing two hydrogens gives formaldehyde (H₂C=O).
A carbon bearing one hydrogen and one alkyl group gives an aldehyde (RCHO).
A carbon bearing two alkyl groups gives a ketone (R₂C=O).
You work this out for each carbon of the former double bond independently, so one alkene always yields exactly two carbonyl fragments.
Practical use
Ozonolysis is often used as a diagnostic tool: if you know the carbonyl products, you can work backwards to deduce the structure of the original alkene.
Enzymes can catalyse hydration reactions enantioselectively, converting achiral alkenes into a single enantiomer of a chiral alcohol.
This is possible because the enzyme active site is itself chiral, so it distinguishes between the two enantiotopic faces of the double bond.
In the lab, chiral catalysts (such as Sharpless epoxidation catalysts, covered in later courses) achieve similar selectivity.
Key principle: an optically active product can only form from an optically inactive starting material if at least one chiral agent (reagent, catalyst, or enzyme) is involved.
Retrosynthesis means starting from the target molecule and working backwards to identify what starting materials and reagents would produce it.
The retrosynthetic arrow (⇒) means "can be made from."
For this chapter, the key question is: which alkene, combined with which reagent, gives the target product?
Strategy
Look at the target's functional group and its position.
OH on the more substituted carbon? Acid-catalysed hydration (Markovnikov).
OH on the less substituted carbon? Hydroboration-oxidation (anti-Markovnikov).
Halide on the more substituted carbon? HX addition (Markovnikov).
Two halogens on adjacent carbons (anti)? X₂ addition.
Epoxide? Peroxy acid.
Then identify which alkene precursor has the double bond in the right position.
Lecture worked examples are the best resource for practising this skill.
Hydroboration: C=C + BH₃ → organoborane (B on less substituted C, H on more substituted C)
Oxidation: organoborane + 3 H₂O₂ + HO⁻ → alcohol + 2 ROH + B(OH)₄⁻
Halogen addition: C=C + X₂ → vicinal dihalide (anti addition via halonium ion)
Epoxidation: C=C + RCOOOH → epoxide + RCOOH (syn, concerted)
Ozonolysis: C=C + O₃ → ozonide; then H₂O → two carbonyl compounds + H₂O₂
Ozonolysis product guide: –CH₂ → formaldehyde, –CHR → aldehyde, –CR₂ → ketone
Epoxides are important building blocks in the chemical and pharmaceutical industries because the strained three-membered ring is highly reactive and can be opened by a variety of nucleophiles to install two functional groups at once. Ethylene oxide (the simplest epoxide) is produced on a massive industrial scale and used to make antifreeze, polyester, and surfactants. Ozonolysis is used in structure determination: by identifying the carbonyl fragments, chemists can work out where a double bond was located in a complex natural product.
Students frequently confuse Markovnikov and anti-Markovnikov addition. A useful mnemonic: acid-catalysed hydration = Markovnikov (OH on the more substituted carbon); hydroboration-oxidation = anti-Markovnikov (OH on the less substituted carbon). Both give alcohols, but at different positions.
A common error is applying Markovnikov's rule to halogen addition (Br₂). Markovnikov's rule applies only when the two ends of the adding reagent are different (H vs. X, H vs. OH). Both atoms in Br₂ are the same, so regiochemistry is not the relevant question; stereochemistry (anti addition) is.
Students sometimes draw a carbocation intermediate for epoxidation. There is none: the mechanism is concerted. If you drew a carbocation, you have the wrong mechanism.
In ozonolysis, students occasionally forget to split the molecule at the double bond and instead try to add O₃ across the bond. Ozonolysis cleaves the bond completely.
⚠️ "Give the reagents to convert alkene X into alcohol Y" is a classic exam question. You must decide between acid-catalysed hydration (Markovnikov) and hydroboration-oxidation (anti-Markovnikov) based on where the OH ends up.
⚠️ Predicting the stereochemistry of Br₂ addition to E and Z alkenes (meso vs. racemic products) is a frequent exam item. Draw the halonium ion, then show backside attack from both possible carbons.
⚠️ Ozonolysis problems can run forwards ("what are the products?") or backwards ("what was the starting alkene?"). Both directions are tested. Practise reassembling alkenes from pairs of carbonyl fragments.
⚠️ Retrosynthesis questions tie together every reaction in the chapter. Be ready to choose the correct reaction and draw the mechanism to justify your answer.
True or False: Hydroboration-oxidation places the OH group on the more substituted carbon. Answer: False. It places OH on the less substituted carbon (anti-Markovnikov).
Fill in the blank: Halogen addition to an alkene proceeds through a cyclic ______ ion intermediate. Answer: halonium (or bromonium for Br₂)
True or False: Epoxidation of a cis alkene gives a trans epoxide. Answer: False. The cis geometry is retained (syn addition).
Fill in the blank: Ozonolysis of 2-butene gives two molecules of ______. Answer: acetaldehyde (CH₃CHO)
True or False: The hydroboration step involves a carbocation intermediate. Answer: False. It is concerted, with no carbocation.
Q: What reagents would you use to convert 1-methylcyclohexene into trans-2-methylcyclohexanol? Explain your choice.
A: Hydroboration-oxidation (BH₃, then H₂O₂/NaOH). This places the OH on the less substituted carbon (anti-Markovnikov) via syn addition. Because the H and OH are delivered to the same face of the ring, the product is the trans alcohol (OH and methyl end up on opposite sides).
Q: Predict the products of ozonolysis of 2-methyl-2-butene, followed by aqueous workup.
A: Acetone (CH₃COCH₃) and acetaldehyde (CH₃CHO). One carbon of the double bond bears two alkyl groups (gives a ketone) and the other bears one alkyl group and one hydrogen (gives an aldehyde).
Q: When Br₂ is added to (Z)-2-butene, what is the stereochemistry of the product? Is it optically active?
A: Anti addition to the Z alkene gives a racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane. The mixture is optically inactive overall because the two enantiomers are present in equal amounts, but each individual molecule is chiral.
Q: A student treats 1-butene with a peroxy acid and obtains an epoxide. What is the product, and what is the stereochemistry of the reaction?
A: 1,2-Epoxybutane (butylene oxide). The reaction is a concerted syn addition, so the geometry of the starting alkene is preserved. Since 1-butene has no E/Z isomerism, the stereochemical point here is simply that both C–O bonds form on the same face.
Q: You are given two carbonyl fragments from an ozonolysis experiment: formaldehyde and cyclohexanone. What was the starting alkene?
A: Methylenecyclohexane (a cyclohexane ring with an exocyclic =CH₂ group). One carbon of the double bond was attached to two hydrogens (giving formaldehyde) and the other was part of the ring bearing two alkyl connections (giving cyclohexanone).
Hydroboration-oxidation complements acid-catalysed hydration (Part 1, section 8.6): together they let you place an OH on either carbon of an unsymmetrical double bond. The halonium ion mechanism in halogen addition connects to epoxide ring-opening reactions you will meet in Chapter 9, because both involve nucleophilic attack on a strained three-membered ring from the backside. Ozonolysis links forward to carbonyl chemistry (aldehydes and ketones, Ch. 17–19), where you will study the reactivity of the very fragments this reaction produces. Retrosynthetic thinking introduced here becomes central to the multi-step synthesis problems that dominate later organic chemistry courses.
Hydroboration-oxidation, BH3 borane, anti-Markovnikov addition, syn addition, organoborane, halogen addition to alkenes, bromine addition, chlorine addition, halonium ion, bromonium ion, cyclic intermediate, vicinal dihalide, anti addition, halohydrin, epoxidation, epoxide, oxirane, peroxy acid, peracid, mCPBA, concerted mechanism, ozonolysis, ozone, ozonide, carbonyl products, aldehyde, ketone, formaldehyde, enantioselective addition, chiral catalyst, enzyme catalysis, retrosynthetic analysis, retrosynthesis, retrosynthetic arrow, synthesis planning, Organic Chemistry I, OChem Chapter 8