Lesson 10 · Alkene Reactions
Alkene chapters look like a wall of reagents, but almost everything here is one idea repeated: the electron-rich pi bond attacks an electrophile, and two groups add across the double bond. Learn the mechanism behind each pattern and the reagent chart stops being a memorization exercise.
Rule to remember
The pi bond is the nucleophile. Ask three questions of every reaction: what adds, where does it go, and what is the 3D result.
Learning goals
Key terms (click to reveal)
Part 01
A C=C double bond is two different bonds stacked together. The sigma bond is strong, buried between the nuclei, and survives almost every reaction in this lesson. The pi bond is weaker, sits above and below the plane of the molecule, and holds an exposed, electron-rich cloud. That cloud is the reactive part.
Because the pi electrons are loosely held and easy to reach, the alkene behaves as a nucleophile. It donates its pi electrons to electrophiles: H⁺ from a strong acid, Br₂, borane, a metal surface loaded with H₂. That is why nearly every reaction here belongs to one family, electrophilic addition: the pi bond breaks, and two new sigma bonds form, one on each of the old alkene carbons.
The sigma bond stays. The pi cloud attacks the electrophile.
Big picture
One reaction family, ten variations. Every entry in this lesson except ozonolysis keeps the carbon skeleton intact and simply adds two groups across the old double bond.
Part 02
Every alkene reaction on an exam can be broken down with the same three questions. Get in the habit of answering all three every single time, even when the question only asks for the product.
Question 1
What gets added?
Identify the two pieces that end up on the old alkene carbons. H and Br? H and OH? Two OH groups? The reagents tell you.
Question 2
Where does each group go?
Regiochemistry. If the two carbons of the alkene are different, which one gets which group? This is the Markovnikov question, and the mechanism decides it.
Question 3
What is the 3D result?
Stereochemistry. Same face (syn), opposite faces (anti), or a mixture? The intermediate decides: concerted means syn, bridged ion means anti, free carbocation means no control.
Part 03
This is the whole lesson in one place. Do not memorize it row by row yet. Read the rest of the lesson first, then come back: every token in the Regio and Stereo columns should feel like a conclusion, not a fact.
| Reaction | Reagents | What adds | Regio | Stereo | Rearranges? |
|---|---|---|---|---|---|
| Hydrohalogenation | HCl, HBr, or HI | H + X | Markovnikov | Not stereospecific | Possible |
| Acid-catalyzed hydration | H₂O, H₂SO₄ (cat.) | H + OH | Markovnikov | Not stereospecific | Possible |
| Oxymercuration-demercuration | 1. Hg(OAc)₂, H₂O 2. NaBH₄ | H + OH | Markovnikov | Not stereospecific | No |
| Hydroboration-oxidation | 1. BH₃·THF 2. H₂O₂, NaOH | H + OH | Anti-Markovnikov | Syn | No |
| Halogenation | Br₂ or Cl₂ (in CH₂Cl₂) | X + X | n/a | Anti | No |
| Halohydrin formation | Br₂, H₂O | X + OH | OH on more sub. C | Anti | No |
| Hydrogenation | H₂, Pd/C (or Pt, Ni) | H + H | n/a | Syn | No |
| Epoxidation | mCPBA | O (epoxide ring) | n/a | Stereospecific | No |
| Syn dihydroxylation | OsO₄ (cat.) or cold dilute KMnO₄ | OH + OH | n/a | Syn | No |
| Ozonolysis | 1. O₃ 2. Me₂S (or Zn, H₂O) | Cleaves C=C into two C=O | n/a | n/a | No |
Part 04
Take propene plus HBr. The pi bond attacks H⁺ first. Protonation can happen at either alkene carbon, but it happens in the direction that puts the positive charge on the more stable carbocation: 3° beats 2° beats 1°. For propene, protonating the CH₂ end gives a 2° cation instead of a 1° cation, so that is the path the reaction takes. Br⁻ then attacks the cation.
Protonate toward the more stable carbocation, then the nucleophile attacks. Markovnikov falls out automatically.
Notice what this one mechanism explains at once. Markovnikov regiochemistry is not a separate rule: it is carbocation stability. The lack of stereocontrol makes sense too, because the flat carbocation can be attacked from either face. And because a free carbocation exists, hydride and methyl shifts can occur whenever a more stable cation is one shift away. Acid-catalyzed hydration is the same mechanism with water as the nucleophile, so it inherits all three traits.
Hydroboration is the opposite personality. BH₃ adds H and B to the alkene in a single concerted step: no intermediate, no free carbocation, no chance to rearrange. Boron, the bulkier and more electrophilic partner, ends up on the less substituted carbon, and because both pieces come from the same BH₃ molecule, H and B land on the same face. The H₂O₂/NaOH oxidation then swaps boron for OH with retention at that carbon.
One concerted step: B on the less substituted carbon, H and B on the same face, nothing free to rearrange.
So the three-part exam description, anti-Markovnikov plus syn plus no rearrangement, is really one sentence about the mechanism: everything happens at once, with boron choosing the roomier carbon. That description is far more durable than memorizing the reagent string.
When Br₂ approaches an alkene, the pi bond attacks one bromine and kicks the other out as Br⁻. But instead of a normal carbocation, the attached bromine bridges both carbons in a three-membered bromonium ion. That bridge physically blocks one face of the molecule, so the incoming nucleophile has to attack from the opposite face. Anti addition is a geometric consequence, not a rule to memorize.
The bromine bridge blocks one face. Backside attack forces anti addition.
Run the same reaction in water and water outcompetes Br⁻ as the nucleophile. Water opens the bromonium ion at the more substituted carbon, because that carbon carries more partial positive charge in the bridge. The result is a halohydrin: Br on one carbon, OH on the more substituted one, still anti. Same intermediate, different nucleophile.
The remaining reactions are all deliveries to a single face. Hydrogenation happens on a metal surface: the alkene lies flat on the catalyst and both hydrogens are delivered from below, so the addition is syn. mCPBA transfers its oxygen to one face of the alkene in a single concerted step, making an epoxide and preserving the alkene geometry: a cis alkene gives the cis epoxide. OsO₄ (or cold dilute KMnO₄) adds both oxygens through a five-membered cyclic intermediate on one face, giving the syn diol.
Ozonolysis is the outlier of the whole lesson. O₃ does not add across the double bond, it cleaves it entirely: both the pi bond and the sigma bond break, and each old alkene carbon becomes a carbonyl carbon. With a Me₂S or Zn workup, the fragments are aldehydes and ketones. Exams love running ozonolysis backwards: given the fragments, rebuild the alkene by joining the two carbonyl carbons.
Ozonolysis snaps the C=C completely. Each alkene carbon becomes a C=O.
Part 05
Exams rarely test one reaction in isolation. They test whether you can pick the right member of a family. Two families matter most in this chapter.
H₂O, H₂SO₄
Acid hydration
1. Hg(OAc)₂, H₂O 2. NaBH₄
Oxymercuration
1. BH₃·THF 2. H₂O₂, NaOH
Hydroboration
Same net change, H + OH across the double bond, but three different answers to the three questions. If a question says "Markovnikov alcohol, no rearrangement," it is fishing for oxymercuration. If it says "OH on the terminal carbon," it is fishing for hydroboration.
Br₂, CH₂Cl₂
Halogenation
Br₂, H₂O
Halohydrin
mCPBA
Epoxidation
The solvent is doing real work in this family. Br₂ alone gives the dibromide; add water and the water intercepts the bromonium ion to give the halohydrin. Watch the conditions line of every question, not just the reagent.
Part 06
Every alkene problem should run through the same pipeline. Slow at first, automatic by exam day:
Worked example: 1-methylcyclohexene plus BH₃·THF, then H₂O₂/NaOH. Reagent: hydroboration-oxidation. Mechanism: concerted, no carbocation. Regio: OH goes to the less substituted alkene carbon, C2. Stereo: H and OH add syn, so they end up cis on the ring. Product: trans-2-methylcyclohexan-1-ol drawn with the OH and the new H on the same face, no rearranged skeletons anywhere.
Reagent, mechanism, regio, stereo, product. Answer all four before drawing anything.
Exam habit
A reagent chart without mechanisms produces memorized answers that collapse under pressure. If you can name the intermediate, the regio and stereo columns fill themselves in.
Watch out
Forgetting rearrangements
HX addition and acid hydration go through free carbocations. If a 3° cation is one hydride or methyl shift away, the exam expects the rearranged product. Oxymercuration and hydroboration never rearrange.
Mixing up the two Markovnikov hydrations
H₃O⁺ and oxymercuration both give the Markovnikov alcohol, but only acid hydration can rearrange. When the substrate has a quaternary carbon next to the alkene, the reagent choice changes the answer.
Assigning stereochemistry where there is none
Drawing wedges and dashes on a carbocation product implies stereocontrol that does not exist. HX and acid hydration on a simple alkene give a mixture; only claim syn or anti when the mechanism earns it.
Putting OH on the wrong carbon in halohydrins
In Br₂/H₂O, water opens the bromonium ion at the more substituted carbon, so OH goes there and Br goes to the less substituted carbon. Students often reverse this.
Confusing cold and hot KMnO₄
Cold dilute KMnO₄ gives the syn diol, matching OsO₄. Hot concentrated KMnO₄ cleaves the alkene oxidatively instead. The temperature is part of the reagent.
Reading ozonolysis in only one direction
Exams give the fragments and ask for the alkene as often as the reverse. Practice stitching two carbonyl carbons back together into a C=C.
Checkpoint
2-methyl-2-butene reacts with HBr. Which carbon gets the Br, and why?
Br ends up on C2, the more substituted alkene carbon.
Protonation happens first, and it happens in the direction that forms the more stable carbocation. Protonating C3 puts the positive charge on the 3° carbon C2. Br then attacks that carbocation. Markovnikov regiochemistry is just carbocation stability in disguise.
You need to convert 1-methylcyclohexene into a trans product with Br on both former alkene carbons. Which reagent, and why does it give trans?
Br₂ in CH₂Cl₂. The bromonium ion forces anti addition, which shows up as trans on the ring.
The first Br bridges both carbons as a bromonium ion, blocking one face. Br⁻ can only attack from the opposite face, so the two bromines end up on opposite sides of the ring.
You want to hydrate 1-hexene so the OH lands on C1. Which reagent set do you pick, and what would H₃O⁺ give instead?
Hydroboration-oxidation: 1. BH₃·THF, then 2. H₂O₂, NaOH. Acid hydration would put OH on C2 instead.
Boron adds to the less substituted carbon for steric and electronic reasons, and oxidation swaps B for OH with retention. That gives the anti-Markovnikov alcohol. Acid hydration goes through the more stable 2° carbocation at C2, so the OH would land there.
3,3-dimethyl-1-butene reacts with dilute H₂SO₄ in water. Why is the major product NOT 3,3-dimethyl-2-butanol?
The 2° carbocation rearranges by a methyl shift into a 3° carbocation, so the major product is 2,3-dimethyl-2-butanol.
Acid hydration forms a free carbocation at C2. A methyl group on the neighboring quaternary carbon shifts over, converting the 2° cation into a 3° cation before water attacks. Any free-carbocation mechanism next to a quaternary carbon is a rearrangement trap on exams.
An unknown alkene is treated with O₃ and then Me₂S, giving acetone and butanal. What was the starting alkene?
2-methyl-2-hexene.
Ozonolysis works in reverse for structure determination: rejoin the two carbonyl carbons with a double bond. The acetone carbonyl carbon carries two methyls, and the butanal carbonyl carbon carries a propyl chain and an H. Stitch them together at the C=O carbons and you get 2-methyl-2-hexene.
Practice workbook
A full set of product prediction, reagent selection, and retro-ozonolysis questions, organized by the reagent, mechanism, regio, stereo workflow.
Lesson summary