Difficulty: Intermediate to Advanced | Prerequisites: IUPAC nomenclature, functional group identification, carbocation/radical stability (see Part 1 notes), understanding of Markovnikov's rule.

Big Picture

This is the core of Organic Chemistry I: given a starting alkene or alkyne, pick the right reagent to get a specific product, or given a reagent, predict what forms. Every reagent has a personality: it controls regiochemistry (where atoms add), stereochemistry (which face of the molecule they add to), and the type of mechanism (ionic vs radical). The exam tests this from both directions, forward (reagents given, predict products) and backward (product given, identify reagents). You should already be comfortable with carbocation stability and Markovnikov's rule before tackling this material.

TL;DR

Each reagent combination gives a predictable product: HX adds Markovnikov, HBr with peroxides adds anti-Markovnikov, Br₂ alone adds anti across the double bond, Br₂/H₂O gives a bromohydrin, BH₃ then H₂O₂/NaOH gives anti-Markovnikov syn addition of OH, and ozonolysis (O₃ then Zn/H₂O) cleaves double bonds into carbonyls. Alkynes follow the same logic but can stop at one or two equivalents, and Lindlar catalyst vs Na/NH₃ controls whether you get a cis or trans alkene.

Key Terms

Markovnikov addition

In the addition of HX or H₂O across an alkene, the hydrogen adds to the less substituted carbon and the X (or OH) adds to the more substituted carbon. This follows because the more stable (more substituted) carbocation intermediate forms preferentially. In simple terms, "the rich get richer": the carbon that already has more hydrogens gets yet another one.

Anti-Markovnikov addition

The opposite regiochemistry: H goes to the more substituted carbon and Br (or OH) goes to the less substituted carbon. This occurs in radical additions (HBr/peroxides) and hydroboration-oxidation (BH₃ then H₂O₂/NaOH). Think of it as the exception that proves the rule: a different mechanism (radical or concerted) flips the usual selectivity.

Syn addition

Both new groups add to the same face of the double bond. Examples: hydroboration-oxidation, catalytic hydrogenation (H₂/Pd/C or H₂/Lindlar).

Anti addition

The two new groups add to opposite faces of the double bond. Examples: Br₂ addition (via a bromonium ion), Br₂/H₂O (halohydrin formation).

Bromonium ion

A three-membered ring intermediate formed when Br₂ approaches an alkene. The bromine bridges across both carbons of the former double bond, blocking one face and forcing the nucleophile to attack from the opposite side (anti addition).

Lindlar catalyst

A poisoned palladium catalyst (Pd/CaCO₃/quinoline) that reduces alkynes to cis-alkenes only. It stops at the alkene stage because the poison deactivates the catalyst enough to prevent further reduction.

Dissolving metal reduction

Na in liquid NH₃ reduces an alkyne to a trans-alkene. The mechanism involves radical anion intermediates that favour the more stable trans geometry.

Ozonolysis

Treatment of an alkene with O₃ followed by a reductive workup (Zn/H₂O or dimethyl sulfide) cleaves the double bond entirely, replacing each carbon of the C=C with a C=O (carbonyl). Internal double bonds give two ketones or aldehydes; terminal double bonds give formaldehyde as one product.

m-CPBA (meta-chloroperoxybenzoic acid)

A peracid that converts alkenes to epoxides via a concerted, stereospecific mechanism. The oxygen is delivered to one face of the double bond in a single step.

NBS (N-bromosuccinimide)

Used with peroxides or light for allylic bromination. It provides a low, steady concentration of Br₂, which favours substitution at the allylic position over addition to the double bond.

Core Content: Reagent Reference

The table below maps each reagent set to what it does, its regiochemistry, and its stereochemistry. This is the single most useful reference for Problems 2, 3, and 4 on the exam.

Core Content: Electrophilic Additions to Alkenes

HCl or HBr (1 eq, no peroxides) to a diene or alkene

• Follows Markovnikov's rule: H adds to the less substituted carbon, halide to the more substituted.

• With conjugated dienes at low temperature (-50 °C), 1,2-addition is the kinetic product. The exam shows HCl/1 eq at -50 °C giving Markovnikov addition across the more reactive double bond.

• The carbocation intermediate is key: the more stable cation forms, and the halide attacks it.

Br₂/H₂O (halohydrin formation)

• Bromonium ion forms first (anti addition). Water, the nucleophile in solution, attacks the more substituted carbon of the bromonium ion (Markovnikov-like for OH placement).

• Product: Br on the less substituted carbon, OH on the more substituted carbon, anti to each other.

• Produces a racemic mixture (+ enantiomer) when a new stereocentre forms.

BH₃ then H₂O₂/NaOH (hydroboration-oxidation)

• Anti-Markovnikov: OH ends up on the less substituted carbon.

• Syn addition: the H and OH are delivered to the same face of the double bond.

• With cyclic alkenes (e.g. cyclopentene), this gives cis product (OH and H on the same face of the ring), plus enantiomer.

H₂O/H₂SO₄ (acid-catalysed hydration)

• Markovnikov: OH on the more substituted carbon.

• Not stereospecific: the carbocation intermediate is planar and can be attacked from either face.

• Produces racemic mixture when a new stereocentre forms.

NBS/peroxides (allylic bromination)

• This is a substitution reaction, not an addition. It replaces an allylic C-H with C-Br via a radical mechanism.

• When there are two possible allylic positions, both products can form. The exam key shows two products from allylic bromination of a cyclic diene.

Core Content: Alkyne-Specific Reactions

Reduction of alkynes to alkenes

• H₂/Lindlar catalyst: reduces alkyne to cis-alkene (syn addition of H₂). The poisoned catalyst stops reduction at the alkene stage.

• Na/NH₃: reduces alkyne to trans-alkene (anti addition via radical anion intermediates). The mechanism favours the more stable trans geometry.

• H₂/Pd/C: full reduction of alkyne through to alkane. Cannot stop at the alkene stage.

Acetylide chemistry (NaNH₂ reactions)

• NaNH₂ deprotonates terminal alkynes (pKa ~25) to give acetylide anions (RC≡C⁻).

• The acetylide can then act as a nucleophile: with CH₃Br, it performs an Sₙ2 reaction to extend the carbon chain (alkylation).

• With H₂O as the second step, you simply re-protonate the acetylide, which in a synthetic sequence means you have performed an elimination to generate a terminal alkyne from a dihalide.

Alkyne hydration

• H₂O/HgSO₄/H₂SO₄: Markovnikov hydration of an alkyne. Water adds across the triple bond with OH going to the more substituted carbon, producing an enol intermediate that tautomerises to a ketone.

• For a terminal alkyne, the ketone has the carbonyl at C2 (the internal carbon). For a symmetrical internal alkyne, only one ketone is possible.

HBr additions to alkynes

• HBr, 1 eq: Markovnikov addition gives a vinyl halide.

• HBr, excess: two Markovnikov additions give a geminal dihalide (both Br on the same carbon, the more substituted one).

Halogenation of alkynes

• Br₂, 1 eq: adds across one pi bond of the triple bond to give a dibromo-alkene (anti addition).

• Br₂, 2 eq: adds across both pi bonds to give a tetrabromo-alkane.

Core Content: Oxidative and Radical Reactions

Ozonolysis (O₃, then Zn/H₂O)

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

• If a carbon had two alkyl groups, it becomes a ketone. If it had one alkyl group and one hydrogen, it becomes an aldehyde. If it had two hydrogens, it becomes formaldehyde.

• The Zn/H₂O reductive workup prevents over-oxidation to carboxylic acids.

Epoxidation (m-CPBA)

• Converts an alkene to an epoxide (a three-membered ring with oxygen) in a single concerted step.

• The oxygen is delivered to one face of the double bond (syn). With an asymmetric alkene, both enantiomers form (racemic mixture, shown as product + enantiomer).

• This is stereospecific: a cis-alkene gives a cis-epoxide, a trans-alkene gives a trans-epoxide.

Radical bromination with Br₂/light (or peroxides)

• A substitution reaction on alkanes or the saturated portion of an alkene-containing molecule.

• Highly selective for the most substituted C-H bond: tertiary > secondary > primary. This is because the more stable radical forms more readily.

• Used in synthesis sequences: brominate an alkane, then eliminate to form an alkene (using KOt-Bu or similar base), then functionalise the alkene further.

Multi-step synthesis example (from the exam)

• Cyclopentane to 1,2-dibromocyclopentane requires three steps:

  1. Br₂/light (or peroxides): radical bromination gives bromocyclopentane.

  1. KOt-Bu: E2 elimination gives cyclopentene.

  1. Br₂: anti addition across the double bond gives trans-1,2-dibromocyclopentane.

Common Misconceptions

• Students often confuse Markovnikov and anti-Markovnikov by thinking about which atom goes where, rather than which intermediate forms. The reliable method: draw the carbocation (or radical) intermediate and ask which one is more stable. The regiochemistry follows from there.

• Hydroboration-oxidation is frequently confused with acid-catalysed hydration. Both add water across a double bond, but they give opposite regiochemistry and different stereochemistry. BH₃/H₂O₂ = anti-Markovnikov, syn. H₂O/H₂SO₄ = Markovnikov, not stereospecific.

• Students often think NBS adds Br across a double bond. It does not. NBS with peroxides is an allylic substitution, not an addition. The double bond remains intact in the product.

• With alkynes, students forget that two equivalents of reagent can add. One equivalent of HBr to an alkyne gives a vinyl bromide; excess HBr gives a geminal dibromide. The exam tests whether you track how many equivalents are specified.

Why It Matters / Exam Flags

⚠️ The reagent-matching problem (Problem 2) is worth 20 points and requires you to identify the correct reagent from a list. Each reagent can only be used once, so if you misidentify one, it may cascade into other errors. Work through the most distinctive reactions first (ozonolysis, m-CPBA, Lindlar, Na/NH₃) before tackling the more ambiguous ones.

⚠️ Stereochemistry notation matters. If stereoisomers can form, you must indicate "+ enantiomer" or draw both. Forgetting to note this loses marks even if the main product is correct.

⚠️ Multi-step synthesis (Problem 6) is tested: you need to convert an alkane to a functionalised product using a sequence of reactions. The strategy is typically brominate → eliminate → add across the new double bond.

⚠️ Reaction scheme fill-in (Problem 3) tests both directions: given reagents predict the product, and given a product work backwards to identify the reagents or starting material.

Quick Self-Test

1. True or false: HBr with peroxides gives Markovnikov addition. ___

1. Fill in the blank: H₂ with Lindlar catalyst converts an alkyne to a ___ alkene.

1. True or false: Br₂/H₂O gives syn addition of Br and OH. ___

1. Fill in the blank: ozonolysis followed by Zn/H₂O cleaves a C=C and replaces it with two ___ groups.

1. True or false: NBS with peroxides adds Br across a double bond. ___

Answers: 1. False (anti-Markovnikov). 2. cis. 3. False (anti addition). 4. Carbonyl (C=O). 5. False (allylic substitution, double bond stays intact).

Practice Q&A

Q: An alkyne is treated with NaNH₂ followed by H₂O. What is the purpose of each step?

A: NaNH₂ deprotonates the terminal alkyne to form an acetylide anion. H₂O then re-protonates it. In a multi-step synthesis, this sequence is typically used after a double elimination from a dihalide to form the terminal alkyne.

Q: You need to convert cyclopentane into trans-1,2-dibromocyclopentane. Propose a synthesis.

A: Step 1: Br₂/light (radical bromination to give bromocyclopentane). Step 2: KOt-Bu (E2 elimination to give cyclopentene). Step 3: Br₂ (anti addition across the double bond to give trans-1,2-dibromocyclopentane).

Q: An alkene is treated with m-CPBA. What product forms, and what is the stereochemical outcome?

A: An epoxide forms via concerted syn delivery of oxygen. The stereochemistry of the alkene is preserved in the epoxide (cis-alkene gives cis-epoxide). Both enantiomers form (racemic).

Q: What is the difference in products when an internal alkyne is treated with H₂/Lindlar versus Na/NH₃?

A: H₂/Lindlar gives the cis-alkene (syn addition). Na/NH₃ gives the trans-alkene (anti reduction via radical anion intermediates).

Q: An alkene is treated with (1) O₃, then (2) Zn/H₂O, and one product is a ketone. What does that tell you about the starting alkene?

A: The carbon that became the ketone carbonyl had two alkyl substituents (no hydrogen) on it in the original alkene. If an aldehyde formed at the other end, that carbon had one hydrogen and one alkyl group.

Connections to Other Topics

These reactions are the building blocks for multi-step synthesis, which becomes the dominant question format in Organic Chemistry II. Every synthesis problem from here on is assembled from this reagent toolkit.

The stereochemistry rules (syn vs anti addition) connect directly to the stereochemistry unit earlier in the course. If you are shaky on R/S assignments, wedge/dash notation, or enantiomers vs diastereomers, review those notes: the reactions section assumes you can assign stereochemistry to products.

The mechanisms behind these reactions (covered in Doc 3) explain why each reagent gives its particular regio- and stereochemistry. Learning the mechanism is what lets you predict products for reactions you have not memorised.

Related Terms / Search Tags

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