Difficulty: Intermediate | Prerequisites: Chapters 1–5 (bonding, structure, stereochemistry basics)

Big Picture: Alkenes are the first major reactive functional group in organic chemistry. The carbon-carbon double bond is electron-rich, making alkenes nucleophilic and susceptible to electrophilic addition reactions. Mastering alkene reactions gives you the toolkit to convert simple hydrocarbons into alcohols, haloalkanes, diols, epoxides, and more. If you can predict what happens when a reagent attacks a double bond, you can handle roughly a third of the reactions on any Orgo I exam.

TL;DR

Alkenes undergo electrophilic addition reactions because the pi bond is electron-rich. The reagent you choose determines the product (alcohol, haloalkane, diol, halohydrin, alkane, or ether) and the regiochemistry (Markovnikov vs anti-Markovnikov). Stereochemistry (syn vs anti addition) depends on the mechanism.

Key Terms

Electrophilic addition

A reaction in which an electrophile adds across a double bond, breaking the pi bond and forming two new sigma bonds. Think of it as: the electron-rich double bond donates electrons to an electron-poor species, and both ends of the former double bond pick up new groups.

Markovnikov's rule

In the addition of HX to an alkene, the hydrogen attaches to the carbon that already has more hydrogens, and the X attaches to the more substituted carbon. In simple terms, "the rich get richer" with hydrogen.

Anti-Markovnikov addition

The opposite of Markovnikov's rule: the hydrogen (or smaller group) ends up on the more substituted carbon, and the main group (e.g. OH or Br) goes to the less substituted carbon. This happens with hydroboration-oxidation and radical addition of HBr.

Syn addition

Both new groups add to the same face of the double bond. Think of it as both groups arriving from the same side of the molecule. Hydroboration is the classic example.

Anti addition

The two new groups add to opposite faces of the double bond. Halogenation (Br2 or Cl2) proceeds this way because the reaction goes through a cyclic halonium ion intermediate.

Carbocation

A positively charged carbon intermediate. Formed during Markovnikov additions (e.g. HX addition, acid-catalysed hydration). Stability order: 3° > 2° > 1° > methyl.

Halonium ion (bromonium or chloronium ion)

A three-membered ring intermediate with a positively charged halogen bridging two carbons. This intermediate blocks one face of the molecule, forcing anti addition.

Hydroboration-oxidation

A two-step reaction (1. BH3, 2. H2O2/NaOH) that converts an alkene to an alcohol with anti-Markovnikov regiochemistry and syn stereochemistry.

Oxymercuration-demercuration

A two-step reaction (1. Hg(OAc)2/H2O, 2. NaBH4) that gives Markovnikov alcohols without carbocation rearrangements.

Halohydrin

A molecule bearing both a halogen and an OH group on adjacent carbons, formed by treating an alkene with X2/H2O. The halogen and OH add anti to each other.

Vicinal diol (glycol)

A compound with OH groups on two adjacent carbons. Formed by OsO4/NaHSO3 (syn addition) or by acid-catalysed ring opening of an epoxide (anti addition).

Ozonolysis

Cleavage of a double bond using ozone (O3) followed by a reducing agent such as (CH3)2S. Breaks the alkene into two carbonyl compounds (aldehydes and/or ketones).

Core Content

Alkene to Alcohol Reactions

• Acid-catalysed hydration (H3O+/H2O) (Markovnikov, A)

  • Adds H and OH across the double bond

  • Markovnikov regiochemistry: OH goes to the more substituted carbon

  • Proceeds via carbocation intermediate, so rearrangements are possible

  • Not stereospecific

• Hydroboration-oxidation (1. BH3, 2. H2O2/NaOH) (anti-Markovnikov, C/D)

  • OH ends up on the less substituted carbon

  • Syn addition: both H and OH add from the same face

  • No carbocation intermediate, so no rearrangements

  • Concerted mechanism in the first step

• Oxymercuration-demercuration (1. Hg(OAc)2/H2O, 2. NaBH4) (Markovnikov, A)

  • Markovnikov regiochemistry, like acid-catalysed hydration

  • Key advantage: no carbocation rearrangements

  • Use this when you want Markovnikov alcohol without rearrangement risk

Alkene to Haloalkane Reactions

• Hydrohalogenation (HX) (Markovnikov, A)

  • Adds H and X across the double bond

  • Markovnikov: X goes to the more substituted carbon

  • Proceeds via carbocation, so rearrangements possible

• Radical addition of HBr (HBr, peroxides) (anti-Markovnikov, C)

  • Only works with HBr (not HCl or HI)

  • Anti-Markovnikov: Br goes to the less substituted carbon

  • Peroxides (ROOR) initiate the radical chain mechanism

  • Anti-Markovnikov selectivity because the more stable radical forms at the more substituted carbon

Alkene to Alkane

• Catalytic hydrogenation (H2/Pd, Pt, or Ni) (D)

  • Adds H2 across the double bond, reducing it to a single bond

  • Syn addition (both H atoms delivered from the catalyst surface)

  • Exothermic; heat of hydrogenation can be used to compare alkene stability

Alkene to Halohydrin

• X2/H2O (B)

  • Produces a halohydrin (adjacent X and OH)

  • Anti addition through a halonium ion intermediate

  • Markovnikov-like regiochemistry for the OH: water attacks the more substituted carbon of the halonium ion

Alkene to Vicinal Dihalide

• X2 (Br2 or Cl2) (B)

  • Adds two halogen atoms to adjacent carbons

  • Anti addition via halonium ion intermediate

  • Br2 decolourisation is a classic test for unsaturation

Alkene to Vicinal Diol

• OsO4, then NaHSO3 (syn addition, D)

  • Produces a cis-diol (both OH groups on the same face)

  • OsO4 is toxic and expensive; NaHSO3 is the reductant that frees the diol from the osmate ester

• Ozonolysis does not make diols. It cleaves the double bond entirely (see below).

Alkene to Aldehydes/Ketones (Cleavage)

• Ozonolysis (1. O3, 2. (CH3)2S)

  • Completely cleaves the C=C bond

  • Each carbon of the former double bond becomes a carbonyl (C=O)

  • Monosubstituted end gives an aldehyde; disubstituted end gives a ketone

  • Useful for working backwards: if you know the carbonyl products, you can deduce the original alkene

Alkene to Allylic Halide

• NBS (N-bromosuccinimide) (O)

  • Brominates at the allylic position (the carbon next to the double bond)

  • Radical mechanism, selective for the allylic C-H because the allylic radical is resonance-stabilised

  • The double bond itself remains intact

Alkene to Ether

• Not a direct alkene reaction on the roadmap; ethers are typically formed from haloalkanes via Williamson ether synthesis (see haloalkane notes)

Formulas and Diagrams

Markovnikov decision shortcut

Look at the alkene. Identify the more substituted carbon of the double bond. In Markovnikov addition, the electrophile (H) goes to the less substituted carbon, and the nucleophile (X, OH) goes to the more substituted carbon.

Degree of unsaturation (DoU) formula

DoU = (2C + 2 + N - H - X) / 2

Where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each degree of unsaturation indicates one ring or one double bond. A triple bond counts as two.

Key reagent-product summary table

Common Misconceptions

• Students often think Markovnikov's rule means "the halogen goes to the carbon with more hydrogens." It does not. The hydrogen goes to the carbon with more hydrogens; the halogen (or OH) goes to the more substituted carbon.

• Students confuse when to use HBr alone versus HBr with peroxides. Plain HBr gives Markovnikov product. HBr with peroxides gives anti-Markovnikov product. This radical pathway only works with HBr, not HCl or HI.

• Acid-catalysed hydration and oxymercuration-demercuration both give Markovnikov alcohols, but they are not interchangeable. If the substrate can rearrange (e.g. a secondary carbocation next to a tertiary centre), acid-catalysed hydration will give rearranged products. Oxymercuration will not.

• Students sometimes assume that "syn addition" and "Markovnikov" are linked. They are independent concepts. Syn/anti describes stereochemistry (which face). Markovnikov/anti-Markovnikov describes regiochemistry (which carbon).

Why It Matters / Exam Flags

⚠️ Predicting regiochemistry (Markovnikov vs anti-Markovnikov) is tested constantly. Know which reagents give which.

⚠️ Stereochemistry questions (syn vs anti) are common on exams. Hydroboration = syn. Halogenation = anti. OsO4 = syn.

⚠️ Carbocation rearrangements: if the question involves a substrate where a hydride or methyl shift could form a more stable carbocation, expect it to happen in acid-catalysed hydration and HX addition. It will not happen in hydroboration or oxymercuration.

⚠️ Ozonolysis is a favourite for "work backwards" problems: given two carbonyl products, draw the original alkene.

⚠️ The Br2 decolourisation test for alkenes appears regularly as a conceptual question.

Quick Self-Test

1. True or False: Hydroboration-oxidation gives Markovnikov alcohols.

1. Fill in the blank: In halogenation of an alkene with Br2, the stereochemistry is ______ addition because the mechanism proceeds through a ______ ion.

1. True or False: Adding HBr with peroxides to propene places the Br on carbon 1 (the terminal carbon).

1. Fill in the blank: Ozonolysis of an alkene followed by (CH3)2S produces ______ and/or ______.

1. True or False: Oxymercuration-demercuration is preferred over acid-catalysed hydration when carbocation rearrangement is a concern.

Answers: 1. False (anti-Markovnikov). 2. Anti; bromonium. 3. False (Br goes to C-1, which is the less substituted carbon, so this is true for anti-Markovnikov). 4. Aldehydes; ketones. 5. True.

Practice Q&A

Q: What reagents would you use to convert 1-butene to 1-butanol?

A: 1. BH3, 2. H2O2/NaOH (hydroboration-oxidation). This gives the anti-Markovnikov alcohol, placing OH on carbon 1.

Q: What reagents would you use to convert 1-butene to 2-butanol without risking rearrangement?

A: 1. Hg(OAc)2/H2O, 2. NaBH4 (oxymercuration-demercuration). This gives the Markovnikov alcohol without forming a free carbocation.

Q: You treat 2-methyl-2-butene with Br2. What is the product, and what is its stereochemistry?

A: 2,3-dibromo-2-methylbutane. The two Br atoms add anti to each other (trans relationship) because the reaction proceeds through a bromonium ion intermediate.

Q: You perform ozonolysis on 2-pentene with (CH3)2S workup. What products do you get?

A: Acetaldehyde (ethanal, CH3CHO) and propanal (CH3CH2CHO). Each carbon of the double bond becomes part of a new C=O.

Q: A student treats propene with HBr and obtains 1-bromopropane. What conditions were used?

A: HBr with peroxides (radical conditions). Without peroxides, Markovnikov addition would give 2-bromopropane instead.

Connections to Other Topics

Alkene reactions connect directly to alkyne chemistry (Chapter 7-8): alkynes undergo many of the same addition reactions, but can stop at the alkene stage or go further. Understanding Markovnikov regiochemistry here is essential for alkyne hydration to aldehydes/ketones.

The stereochemistry concepts (syn/anti addition) reappear in substitution and elimination reactions (Chapters 8-9), where SN2 inversion and E2 anti-periplanar geometry build on the same spatial reasoning.

Ozonolysis connects to carbonyl chemistry in later chapters: aldehydes and ketones are central to nucleophilic addition reactions you will encounter in Chapters 17-19.

Related Terms / Search Tags

Alkene reactions, electrophilic addition, Markovnikov rule, anti-Markovnikov, hydroboration oxidation, oxymercuration demercuration, acid catalysed hydration, hydrohalogenation, halogenation, halohydrin formation, ozonolysis, catalytic hydrogenation, syn addition, anti addition, bromonium ion, carbocation rearrangement, vicinal diol, NBS allylic bromination, organic chemistry chapter 6 7 8, Purdue organic chemistry, reaction roadmap