Difficulty: Intermediate | Prerequisites: E2 elimination basics (see Part 1), alkene structure, carbocation stability, stereochemistry (R/S, E/Z)
This set of notes covers what happens to alkenes once they are formed: the addition reactions that break the pi bond and add new atoms or groups across the double bond. It also covers the reverse direction, acid-catalysed dehydration (E1), where an alcohol loses water to form an alkene. The exam tests your ability to predict products (including stereochemistry), draw reaction intermediates, write arrow-pushing mechanisms, and reason backwards from ozonolysis products to determine a starting material's structure. You will need to know which reactions are syn addition, which are anti, and which go through a carbocation intermediate.
Alkenes react with electrophiles (H+, Br2, BH3) by donating their pi electrons, forming intermediates like carbocations or cyclic bromonium ions that determine both the regiochemistry (where the atoms add) and the stereochemistry (which face they add to). Ozonolysis cleaves double bonds into carbonyls and can be worked backwards to deduce an unknown starting material. Acid-catalysed dehydration of alcohols is the reverse of addition: it goes through a carbocation intermediate, and the rate-determining step is the loss of water.
Ozonolysis
Cleavage of a carbon-carbon double bond using ozone (O3) followed by a reductive workup (Zn/H2O or dimethyl sulfide). Each carbon of the former double bond becomes a carbonyl (aldehyde or ketone). Think of it as a molecular scissors that cuts the double bond and caps each end with an oxygen.
Halohydrin
A compound bearing a halogen and a hydroxyl group on adjacent carbons, formed by treating an alkene with X2 in water. The reaction proceeds through a cyclic halonium ion, so the halogen and OH end up anti to each other.
Bromonium ion
A three-membered ring intermediate containing a positively charged bromine bridging two carbons. Formed when Br2 attacks an alkene. The ring prevents rotation and forces the second nucleophile to attack from the opposite face (anti addition). In simple terms, it is a temporary bridge of bromine that locks the stereochemistry.
Carbocation
A carbon atom bearing a positive charge, with only three bonds and an empty p orbital. Stability order: 3° > 2° > 1° > methyl. Carbocations are the key intermediates in Markovnikov addition of HX, E1 elimination, and acid-catalysed dehydration.
Hydroboration-oxidation
A two-step sequence: (1) BH3 adds across the double bond in a syn fashion, placing boron on the less-substituted carbon; (2) treatment with H2O2/NaOH replaces boron with OH, retaining configuration. The net result is anti-Markovnikov, syn addition of water.
Markovnikov's rule
In the addition of HX to an alkene, the hydrogen adds to the carbon with more hydrogens (the less-substituted carbon), and the halide adds to the more-substituted carbon. This happens because the more stable carbocation intermediate forms preferentially.
Anti-Markovnikov addition
The opposite regiochemistry to Markovnikov: the nucleophilic group (e.g. OH in hydroboration-oxidation) ends up on the less-substituted carbon. Occurs when the mechanism avoids a carbocation intermediate.
Syn addition
Both new groups add to the same face of the double bond. Examples: hydrogenation (H2/Pd), hydroboration.
Anti addition
The two new groups add to opposite faces of the double bond. Examples: bromination (Br2), halohydrin formation (X2/H2O).
E1 elimination
A two-step elimination: first the leaving group departs to form a carbocation, then a base removes a proton to form the alkene. Unimolecular: rate depends only on substrate concentration.
Dehydration
Loss of water from an alcohol to form an alkene, typically promoted by strong acid (H2SO4) and heat. The mechanism is E1 for secondary and tertiary alcohols.
The table below summarises the major alkene addition reactions tested on this exam. For each, know the reagents, the regiochemistry, the stereochemistry, and the intermediate.
Catalytic hydrogenation (H2/Pd or H2/Pt)
Adds H2 across the double bond.
Syn addition (both hydrogens delivered from the metal surface to the same face).
No regiochemistry issue (both atoms are H).
Product: the corresponding alkane with cis stereochemistry at the new C-H bonds if relevant.
Halogenation (Br2 or Cl2, no solvent nucleophile)
Adds two halogen atoms across the double bond.
Anti addition through a cyclic halonium ion (bromonium or chloronium).
If the alkene is symmetrical, one product. If unsymmetrical and the alkene has a plane of symmetry, a racemic mixture of enantiomers.
For cyclohexene + Br2: the product is trans-1,2-dibromocyclohexane (a racemic mix of the two enantiomers).
Halohydrin formation (X2/H2O)
Water acts as the nucleophile that opens the halonium ion.
Anti addition: halogen and OH end up on opposite faces.
Markovnikov-like regiochemistry for the OH: water attacks the more-substituted carbon of the halonium ion (the carbon better able to stabilise partial positive charge), so OH goes on the more-substituted carbon and X goes on the less-substituted carbon.
Acid-catalysed hydration (H3O+) and HX addition
Markovnikov addition: H adds to the less-substituted carbon, OH (or X) to the more-substituted carbon.
Goes through a carbocation intermediate.
Not stereospecific: the carbocation is planar, so nucleophilic attack occurs from both faces, giving a mixture of stereoisomers.
Hydroboration-oxidation (1. BH3; 2. H2O2/NaOH)
Anti-Markovnikov regiochemistry: OH ends up on the less-substituted carbon.
Syn addition: BH3 adds from one face in a concerted fashion, and the oxidation step retains configuration.
No carbocation intermediate.
Acid-catalysed dehydration (H2SO4, heat)
The reverse of hydration: converts an alcohol to an alkene.
E1 mechanism for 2° and 3° alcohols.
Zaitsev's rule applies: the more-substituted alkene is the major product.
Goes through a carbocation intermediate (after protonation and loss of water).
Ozonolysis (O3 followed by Zn/H2O or DMS) cleaves each C=C double bond and replaces it with C=O on both sides.
Rules for the carbonyl products:
A carbon that was part of the double bond and had two alkyl/aryl groups becomes a ketone (C=O with no H on the carbonyl carbon).
A carbon that was part of the double bond and had one H becomes an aldehyde (H-C=O).
A carbon that was part of the double bond and had two H's becomes formaldehyde (H2C=O).
Working backwards (retro-ozonolysis):
This is a common exam question type. You are given the carbonyl products and asked to determine the original alkene.
Take each carbonyl product and identify the C=O carbon.
Remove the oxygen and reconnect those carbons with a double bond.
If there are three carbonyl products, the starting material had two double bonds (ozone cleaves each one separately).
Worked example from the exam: three products are given (a diketone fragment, an aldehyde, and a ketoaldehyde). To reconstruct the starting material, connect the carbonyl carbons pairwise with double bonds. The number of C=O fragments tells you how many cuts ozone made, which tells you how many double bonds were present.
Key tip: count the total number of carbons across all products. That total must equal the number of carbons in the starting material. If there are three products from ozonolysis, the starting material has two double bonds (two cuts give three fragments).
Acid-catalysed dehydration of an alcohol follows an E1 pathway for secondary and tertiary substrates.
Step 1: Protonation of the hydroxyl group.
The oxygen lone pair acts as a base and picks up H+ from H2SO4 (or H3O+).
This converts the poor leaving group (OH-) into a good one (H2O).
Show the lone pair on oxygen attacking H+, forming an oxonium ion (protonated alcohol).
Step 2: Loss of water to form the carbocation (rate-determining step).
The C-O bond breaks heterolytically. Water departs with both electrons.
A carbocation forms on the carbon that was attached to the OH.
This is the slowest step because it involves breaking a bond and forming a high-energy intermediate.
Step 3: Deprotonation by a base to form the alkene.
A base (water, or the conjugate base of the acid) removes a beta-hydrogen.
The electrons from the C-H bond form the new C=C pi bond.
Zaitsev's rule applies: removal of the H that gives the more-substituted alkene is favoured.
Rate-determining step: Step 2 (loss of water to form the carbocation). This is always the answer when asked. The reasoning is that forming the carbocation requires the most energy; the other steps (protonation and deprotonation) are fast equilibria.
Carbocation rearrangements can occur between Steps 2 and 3 if a 1,2-hydride shift or 1,2-methyl shift produces a more stable carbocation. Always check for this.
Drawing intermediates correctly is worth significant marks. Here is what each one looks like and when it appears.
Carbocation intermediates
Formed in: HX addition to alkenes, E1 elimination, acid-catalysed dehydration, SN1 reactions.
Structure: the positively charged carbon is sp2-hybridised, trigonal planar, with an empty p orbital.
Draw it with three bonds, a positive charge on carbon, and (if asked) the empty p orbital.
Markovnikov's rule arises because the more-substituted (more stable) carbocation forms preferentially.
Example from the exam: HCl addition to a terminal alkene. The proton adds to the terminal carbon (less substituted), forming a secondary or tertiary carbocation on the internal carbon. Draw that carbocation as the intermediate.
Bromonium ion (and chloronium ion) intermediates
Formed in: Br2 addition to alkenes, halohydrin formation.
Structure: a three-membered ring with Br+ bridging the two carbons that were part of the double bond. Both carbons retain their four substituents (they are still sp3-like, not planar).
The positive charge is formally on the bromine, though the carbons bear partial positive charge.
Draw it as a triangle with Br at the apex and a + sign on Br.
Example from the exam: Br2 addition to a disubstituted alkene. Draw the bromonium ion with the bromine on top, bridging the two carbons, with a + on Br. The nucleophile (Br- or H2O) then attacks from the opposite face.
Protonated alcohol (oxonium ion)
Formed in: acid-catalysed dehydration (Step 1).
Structure: the oxygen has three bonds (two to H, one to C), a positive formal charge, and one lone pair remaining.
This is the intermediate formed just before water leaves in dehydration.
Draw the oxygen with three bonds, a + charge, and one lone pair.
Ozonolysis: alkene + O3, then Zn/H2O (or DMS) gives aldehydes and/or ketones. Each C=C becomes two C=O.
Hydroboration-oxidation: alkene + BH3, then H2O2/NaOH gives the anti-Markovnikov alcohol with syn addition.
Halohydrin: alkene + X2/H2O gives the halohydrin with anti addition and Markovnikov-like placement of OH on the more-substituted carbon.
Bromination: alkene + Br2 gives the vicinal dibromide with anti addition.
Hydrogenation: alkene + H2/Pd (or Pt) gives the alkane with syn addition.
Dehydration: alcohol + H2SO4/heat gives the Zaitsev alkene via E1.
Degree of unsaturation: (2C + 2 + N - H - X) / 2. Use this to determine how many rings or double bonds a molecular formula implies.
Students often assume all addition reactions are syn. They are not. Br2 addition and halohydrin formation are anti (via the bromonium/chloronium ion). Hydrogenation and hydroboration are syn.
Students forget that hydroboration-oxidation gives anti-Markovnikov regiochemistry. The OH ends up on the less-substituted carbon, which is the opposite of what acid-catalysed hydration gives.
Students mix up the intermediate for bromination (bromonium ion, a three-membered ring) with a carbocation (open, planar). The bromonium ion is what forces anti stereochemistry; a carbocation would give a mixture of syn and anti.
In ozonolysis problems, students sometimes reconnect the wrong carbonyl carbons. Remember: only the carbonyl carbon (the one bearing the C=O) was part of the original double bond. The rest of the fragment was attached to it. Count carbons carefully.
Product prediction questions (24 points on this exam) require you to know both regiochemistry and stereochemistry for every reaction. Drill these until automatic.
Arrow-pushing mechanisms for dehydration are a high-value question (8-12 points). You must show lone pairs, formal charges, and curved arrows for every step. The rate-determining step question always accompanies this.
Ozonolysis retrosynthesis (working backwards) appears as a 5-8 point question. Practice counting carbons and reconnecting fragments.
Drawing intermediates (bromonium ions, carbocations) is tested directly. Show the three-membered ring for bromonium, the positive charge in the correct place, and the sp2 geometry for carbocations.
Multi-step synthesis questions (6 points each) test whether you can chain reactions together, controlling regiochemistry and stereochemistry at each stage. Think backwards from the product.
True or False: Hydroboration-oxidation gives Markovnikov regiochemistry.
False. It gives anti-Markovnikov: OH on the less-substituted carbon.
True or False: The intermediate in Br2 addition to an alkene is a carbocation.
False. It is a bromonium ion (a three-membered ring with Br+).
Fill in the blank: The rate-determining step in acid-catalysed dehydration of a tertiary alcohol is ______.
Loss of water to form the carbocation.
True or False: Ozonolysis of a molecule with two double bonds gives two carbonyl products.
False. Two double bonds (two cuts) give three carbonyl fragments.
Fill in the blank: Br2 addition to an alkene gives ______ (syn/anti) addition.
Anti.
Q: Cyclohexene is treated with Br2. Draw the product and indicate the stereochemistry.
A: Trans-1,2-dibromocyclohexane (anti addition through a bromonium ion). The two bromines are on opposite faces of the ring. A racemic mixture of the two enantiomers is formed.
Q: 1-Methylcyclohexene is treated with Cl2/H2O. Where does the OH end up and where does the Cl end up?
A: OH goes on the more-substituted carbon (C1, the one bearing the methyl group) because water attacks the more electrophilic carbon of the chloronium ion. Cl goes on the less-substituted carbon (C2). They are anti to each other.
Q: A compound with molecular formula C7H13Cl is treated with NaOEt to give C7H12, which upon ozonolysis (O3, then Zn/H2O) gives a single product. What does the single ozonolysis product tell you about the alkene?
A: A single product from ozonolysis means the alkene is symmetrical about the double bond (both halves of the molecule are identical). The ozonolysis cuts the double bond and produces two identical fragments, which appear as one product.
Q: Write the arrow-pushing mechanism for the dehydration of 2-methyl-2-butanol with H2SO4 and heat. What is the rate-determining step?
A: Step 1: protonate the OH (lone pair on O attacks H+). Step 2 (rate-determining): water leaves, forming a tertiary carbocation on C2. Step 3: a base (water) removes a beta-H, electrons form the C=C. The major product is 2-methyl-2-butene (Zaitsev). The rate-determining step is Step 2, loss of water.
Q: 1-Methylcyclohexene is treated with (1) BH3, (2) H2O2/NaOH. What is the product and its stereochemistry?
A: Trans-2-methylcyclohexanol (with OH on C1, the less-substituted carbon of the original double bond, placed syn to the H that was delivered by BH3). Anti-Markovnikov regiochemistry, syn addition.
Alkene addition reactions are the reverse of elimination reactions covered in Part 1: understanding both directions helps you design multi-step syntheses. Carbocation chemistry connects to SN1 substitution and to rearrangements you will see again in electrophilic aromatic substitution. Ozonolysis reappears in the study of terpenes, fatty acids, and polymer chemistry, where determining double-bond positions in large molecules is a standard analytical technique.
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