Source: CHM 255 PSO Exam 2 Practice Problems
Alkenes react with electrophiles because of their electron-rich pi bond. The key exam content is knowing which reagent gives which product, whether the addition follows Markovnikov or anti-Markovnikov regiochemistry, and whether the stereochemistry is syn or anti. Master the six core reactions below and you cover most of this exam.
Difficulty: Intermediate to Advanced
Prerequisites: Acid-base chemistry in organic molecules (see companion notes), Lewis structures, functional group recognition, basic stereochemistry (R/S configuration, cis/trans).
This is the heart of CHM 255 Exam 2. Alkene addition reactions show up as product prediction, reagent identification, mechanism drawing, and stereochemistry labelling. Every reaction follows a pattern: something attacks the pi bond, a new sigma bond forms, and the regiochemistry and stereochemistry depend on the mechanism. If you missed the acid-base material, go back to that first, because proton transfers initiate several of these reactions.
Markovnikov addition (Markovnikov's rule)
In the addition of HX or H2O to an alkene, the hydrogen attaches to the carbon with more hydrogens (less substituted), and the X or OH attaches to the more substituted carbon. Think of it as: "the rich get richer" in terms of substituents.
Anti-Markovnikov addition
The opposite regiochemistry: the hydrogen ends up on the more substituted carbon and the functional group (OH, H) on the less substituted carbon. Hydroboration-oxidation is the classic example.
Regioselectivity
Which carbon of the double bond the new group attaches to. Markovnikov vs anti-Markovnikov is a question of regioselectivity.
Stereoselectivity
Whether the addition occurs from the same face (syn) or opposite faces (anti) of the double bond.
Syn addition
Both new groups add to the same face of the double bond. Hydroboration-oxidation and catalytic hydrogenation are syn additions. OsO4 dihydroxylation is also syn.
Anti addition
The two new groups add to opposite faces of the double bond. Bromination (Br2) and halohydrin formation (Br2/H2O) are anti additions because they proceed through a cyclic bromonium ion intermediate.
Carbocation
A positively charged carbon intermediate. Formed during acid-catalysed hydration and HBr addition. Its stability order: 3° > 2° > 1° > methyl.
Carbocation rearrangement
A 1,2-hydride shift or 1,2-methyl shift that converts a less stable carbocation into a more stable one. This can change the position of the functional group in the product.
Bromonium ion
A three-membered ring intermediate containing a positively charged bromine bridging two carbons. Forms during Br2 addition to an alkene. Its cyclic structure forces anti addition because the nucleophile must attack from the opposite face.
Meso compound
A molecule with stereocentres that is nevertheless achiral because it has an internal plane of symmetry. Can result from anti addition of Br2 to certain symmetric alkenes.
Enantiomers
Non-superimposable mirror-image stereoisomers. Many alkene addition products form as racemic mixtures (equal amounts of both enantiomers).
Diastereomers
Stereoisomers that are not mirror images of each other. Differ in configuration at one or more (but not all) stereocentres.
Regiochemistry: Markovnikov. OH ends up on the more substituted carbon.
Stereochemistry: Not controlled (carbocation intermediate is planar, so attack occurs from both faces). Racemic mixture if a stereocentre forms.
Mechanism: Protonation of the alkene forms the more stable carbocation. Water attacks the carbocation. Deprotonation gives the alcohol.
Watch for: Carbocation rearrangements. If the initial carbocation can rearrange to a more stable one via a 1,2-hydride or 1,2-methyl shift, the product will reflect the rearranged skeleton.
Regiochemistry: Markovnikov. Br ends up on the more substituted carbon.
Stereochemistry: Not controlled (racemic at a new stereocentre).
Mechanism: Protonation of the alkene gives the more stable carbocation. Bromide ion attacks.
Carbocation rearrangement: If a secondary carbocation can shift to a tertiary one, the product will show the rearranged skeleton. Alkenes adjacent to a quaternary carbon centre (e.g. 3,3-dimethyl-1-butene) are classic rearrangement candidates.
Regiochemistry: Not applicable (both atoms added are bromine).
Stereochemistry: Anti addition. The reaction proceeds through a bromonium ion, forcing the second Br to attack from the opposite face.
Products: A vicinal dibromide. If the alkene is symmetric, the product may be a meso compound. If asymmetric, expect a racemic mixture of enantiomers (or diastereomers, depending on the substrate).
Regiochemistry: Markovnikov-like. Water (the nucleophile) attacks the more substituted carbon of the bromonium ion (the carbon better able to bear partial positive charge).
Stereochemistry: Anti addition (through the bromonium ion). OH and Br end up on opposite faces.
Product: A bromohydrin (OH and Br on adjacent carbons, anti to each other).
Regiochemistry: Anti-Markovnikov. OH ends up on the less substituted carbon.
Stereochemistry: Syn addition. Boron and hydrogen add to the same face in the hydroboration step, and the oxidation step replaces B with OH with retention of configuration.
Product: An alcohol on the less substituted carbon with syn stereochemistry.
This is one of the most important contrasts on the exam: same functional group (alcohol) as acid-catalysed hydration, but opposite regiochemistry and controlled stereochemistry.
Occurs in reactions that proceed through carbocation intermediates (acid-catalysed hydration, HBr addition). Does not occur in concerted or bridged-ion mechanisms (Br2 addition, hydroboration).
A 1,2-hydride shift moves H with its bonding electrons from an adjacent carbon to the carbocation centre.
A 1,2-methyl shift moves a methyl group with its bonding electrons.
Both convert a less stable carbocation to a more stable one (e.g. 2° to 3°).
To predict whether rearrangement will occur: draw the initial carbocation, then check whether moving an H or a methyl from an adjacent carbon would produce a more stable carbocation.
Product: A 1,2-diol (glycol), with both OH groups on adjacent carbons.
Stereochemistry: Syn addition. Both OH groups add to the same face of the double bond.
The OsO4 forms a cyclic osmate ester intermediate across both carbons, which is then cleaved by the reducing agent (NaHSO3) to give the diol.
For a trisubstituted or tetrasubstituted alkene, this can create two new stereocentres with a specific relative configuration.
Product: The double bond is cleaved entirely. Each carbon of the former double bond becomes a carbonyl (C=O).
If the carbon had one hydrogen attached, it becomes an aldehyde. If it had no hydrogens (fully substituted), it becomes a ketone.
Retro-ozonolysis (working backwards from products): Given the carbonyl products, join the C=O carbons together and replace the two oxygens with a double bond to reconstruct the original alkene.
The worksheet tests this in both directions: predicting products from a known alkene, and deducing the starting alkene from given carbonyl fragments.
Product: An alkane (complete reduction of the double bond).
Stereochemistry: Syn addition. Both hydrogens add to the same face of the double bond.
If the starting material has two double bonds, excess H2 reduces both. With controlled conditions (1 equiv. H2, Lindlar catalyst for alkynes), partial reduction is possible.
The product's stereochemistry matters: on a ring system, syn addition of H2 gives cis-fused products at the new stereocentres.
This is a high-yield exam topic (Q9 on the worksheet). For any given alkene, you should be able to predict the alcohol product from each of these four reagent sets:
H2SO4, H2O: Markovnikov alcohol, no stereocontrol, carbocation rearrangement possible.
1. BH3; 2. H2O2, NaOH: Anti-Markovnikov alcohol, syn addition, no rearrangement.
Br2, H2O: Markovnikov-like bromohydrin (not a simple alcohol, but OH is on the more substituted carbon), anti addition.
1. OsO4; 2. NaHSO3, H2O: 1,2-diol (both carbons get OH), syn addition.
Given a starting alkene and several possible products, work backwards:
An alcohol on the more substituted carbon = acid-catalysed hydration (Markovnikov).
An alcohol on the less substituted carbon = hydroboration-oxidation (anti-Markovnikov).
A vicinal dibromide = Br2 (not applicable for M/AM labelling).
Ozonolysis products (aldehydes/ketones) = O3 then DMS (not applicable for M/AM labelling).
A bromohydrin = Br2, H2O (Markovnikov-like, though some instructors label this NA).
Students often assume all additions to alkenes follow Markovnikov's rule. Hydroboration-oxidation is anti-Markovnikov. Know which reactions go which way.
Students confuse syn and anti addition. A useful mnemonic: reactions with cyclic intermediates (bromonium ion, osmate ester) lock the stereochemistry. Bromonium = anti. Osmate ester = syn. Hydroboration = syn (concerted, four-centre transition state).
Students forget to check for carbocation rearrangement in acid-catalysed hydration and HBr addition. If the initial carbocation is secondary and an adjacent carbon is tertiary, a shift will occur.
Students label bromination as Markovnikov or anti-Markovnikov. When both atoms being added are the same (Br2), regioselectivity labels do not apply.
Students sometimes draw the product of OsO4 dihydroxylation as anti (OH groups on opposite faces). OsO4 gives syn diols. If you want anti diols, you would use a different reagent (epoxidation followed by acid-catalysed ring opening, which is not on this exam).
⚠️ The "four ways to make alcohols from alkenes" question (Q9) is a near-certainty. Be able to draw all four products from a single alkene, with correct regiochemistry and stereochemistry.
⚠️ Mechanism drawing for bromination (Q5) is commonly tested. You must show the bromonium ion intermediate and the anti attack by bromide.
⚠️ Carbocation rearrangement questions (Q11) test whether you can spot a substrate where the initial carbocation is secondary but a shift would produce a tertiary carbocation.
⚠️ Stereochemistry labelling (achiral, enantiomers, diastereomers, meso) comes up with bromination and dihydroxylation products. Practise drawing both enantiomers and checking for meso symmetry.
⚠️ Retro-ozonolysis (Q7, Q12b) requires you to work backwards from carbonyl products to the original alkene. Join the carbonyl carbons with a double bond.
⚠️ Q10 and Q12 test your ability to match reagents to products. For each transformation, ask: What is the regiochemistry? What is the stereochemistry? Does the mechanism involve a carbocation?
True or false: Hydroboration-oxidation gives a Markovnikov alcohol. (False. It gives anti-Markovnikov.)
Fill in the blank: Br2 addition to an alkene proceeds through a ______ ion intermediate. (bromonium)
True or false: Catalytic hydrogenation (H2/Pd) is an anti addition. (False. It is syn addition.)
Fill in the blank: OsO4 dihydroxylation gives a ______ diol. (syn / cis)
True or false: Carbocation rearrangement can occur during hydroboration-oxidation. (False. No carbocation intermediate is formed.)
Q: What is the major product of treating 1-methylcyclohexene with H2SO4 and H2O?
A: 1-Methylcyclohexanol. The OH adds to the more substituted carbon (Markovnikov). No stereocontrol (racemic if applicable).
Q: What is the major product of treating 1-methylcyclohexene with BH3 followed by H2O2/NaOH?
A: trans-2-Methylcyclohexanol. The OH adds to the less substituted carbon (anti-Markovnikov) with syn addition.
Q: Br2 is added to cis-2-butene. What are the products and their stereochemical relationship?
A: Anti addition of two bromines across the double bond. Because cis-2-butene is symmetric, the product is the meso-2,3-dibromobutane (an internal plane of symmetry, achiral despite two stereocentres).
Q: An alkene undergoes ozonolysis to give formaldehyde (CH2O) and acetone (CH3COCH3). What was the starting alkene?
A: 2-Methylpropene (isobutylene). Join the carbonyl carbons with a double bond: CH2=C(CH3)2.
Q: Which of the following alkenes is most likely to undergo carbocation rearrangement upon addition of HBr: 2-methylpropene, 3,3-dimethyl-1-butene, or cyclohexene?
A: 3,3-Dimethyl-1-butene. The initial carbocation is secondary, but a 1,2-methyl shift from the adjacent quaternary carbon gives a tertiary carbocation. The other two do not have a rearrangement pathway to a more stable carbocation.
Q: A disubstituted cyclopentene with a methyl group is treated with H2/Pd. What is the stereochemistry of the product?
A: Syn addition of H2 means both hydrogens add to the same face of the ring, giving a cis relationship between the newly added H atoms. The methyl group's orientation relative to the new H atoms depends on which face the H2 delivers to. A racemic mixture of cis products is expected if the starting material is symmetric.
Alkene addition reactions connect back to acid-base chemistry (the first step of acid-catalysed hydration is a proton transfer) and forward to alkyne reactions (covered in the next set of notes), which follow similar addition patterns but with additional considerations for partial vs complete addition.
Stereochemistry concepts from this section (syn, anti, meso, enantiomers, diastereomers) carry through to substitution and elimination reactions later in the course.
Carbocation rearrangements reappear in SN1 reactions and Friedel-Crafts alkylation.
Tags: alkene reactions, electrophilic addition, Markovnikov, anti-Markovnikov, hydroboration-oxidation, acid-catalysed hydration, bromination, halohydrin, bromonium ion, syn addition, anti addition, carbocation rearrangement, 1,2-hydride shift, 1,2-methyl shift, OsO4 dihydroxylation, ozonolysis, catalytic hydrogenation, stereochemistry, meso compound, enantiomers, diastereomers, CHM 255, Purdue organic chemistry