Alkene and Alkyne Reactions

Theory — Alkene and Alkyne Reactivity

Alkenes (C=C) and alkynes (C≡C) are π-bonded hydrocarbons. The π bond is weaker than a σ bond and more polarizable, making it the site of nearly all their reactivity. Most alkene/alkyne reactions are additions: the π bond breaks and two new σ bonds form, one to each carbon. Understanding the regiochemistry (which atom lands where) and stereochemistry (syn vs anti) of these additions is the core goal of this lab.

1. Markovnikov's rule and carbocation stability

For acid-catalyzed additions (HX, H₂O/H⁺, Hg²⁺/H₂O), the electrophile (H⁺ or Hg²⁺) adds to the less-substituted carbon of the alkene, generating the more-stable carbocation. The nucleophile (X⁻, H₂O) then attacks this carbocation. Result: the H ends up on the carbon with more H's already, and the X or OH ends up on the more substituted carbon. This is Markovnikov addition.

Example: HBr + propene CH₃CH=CH₂ + HBr →
H⁺ adds to CH₂ (less substituted), generating CH₃CH⁺CH₃ (2° cation, stable)
Br⁻ attacks the carbocation
Product: CH₃CHBrCH₃ (2-bromopropane) — Markovnikov

Carbocation rearrangements can occur when a more stable cation is accessible by a 1,2-hydride or 1,2-methyl shift. For example, addition of HCl to 3-methyl-1-butene initially forms a 2° cation that rearranges to the more stable 3° cation via a hydride shift, giving 2-chloro-2-methylbutane as the major product. Always check whether a more stable cation is one shift away.

2. Anti-Markovnikov addition: radicals and hydroboration

Two reactions reverse Markovnikov regiochemistry:

HBr with peroxides (ROOR, hν): Proceeds by a radical chain mechanism. Br^• (not H⁺) adds first, to the less-substituted carbon, producing the more-stable radical on the more-substituted carbon. Only HBr works — HCl is too strong a bond, HI too weak.
Hydroboration-oxidation (BH₃·THF, then H₂O₂/OH⁻): A concerted 4-membered transition state where B and H add across the π bond simultaneously. B (the electrophile) lands on the less-hindered carbon; H on the more-hindered one. Oxidative workup replaces C–B with C–OH, with retention of configuration. Net result: anti-Markovnikov OH addition, syn stereochemistry.

3. Anti vs syn addition — mechanistic origin

Anti addition (opposite faces)

Br₂, Cl₂, Br₂/H₂O — halonium ion intermediate forces the nucleophile to attack from the face opposite the halogen. Gives trans (or anti) diastereomer from a cyclic alkene.

Syn addition (same face)

BH₃, H₂/Pd, OsO₄, cold KMnO₄, mCPBA — concerted transition states or surface-bound additions deliver both new groups to the same face of the alkene.

4. The classic bench tests: Br₂ and KMnO₄

Two reactions are used as qualitative tests for unsaturation because they produce visible color changes:

Br₂ in CCl₄: Reddish-brown Br₂ solution is decolorized to colorless as the alkene adds Br₂ across the π bond, forming a colorless vicinal dibromide.
Baeyer test (cold dilute KMnO₄): Purple permanganate is reduced to brown MnO₂ as the alkene is oxidized to a cis-1,2-diol. The color change purple → brown is diagnostic.

5. Alkyne-specific features

Alkynes follow the same regiochemistry rules as alkenes but have two important added features:

Enol → keto tautomerization: Adding water to an alkyne (Markovnikov via Hg²⁺/H⁺, or anti-Markovnikov via hydroboration) gives a vinyl alcohol (enol) that spontaneously tautomerizes to the more stable carbonyl compound. Internal alkynes → ketones; terminal alkynes + Markovnikov → methyl ketones; terminal alkynes + hydroboration → aldehydes.
Partial hydrogenation stereochemistry: Lindlar's catalyst (Pd poisoned with Pb/quinoline) delivers both H's to the same face of the alkyne, giving a cis alkene. Na in NH₃ (dissolving metal reduction) is radical-based and gives the trans alkene. Full H₂/Pd goes all the way to alkane.

6. Summary of the 12 reactions in this lab

#	Reaction	Regiochem	Stereochem	Visible cue
1	HBr + propene	Markov	—	—
2	HBr + ROOR + propene	anti-Markov	—	—
3	Br₂ / CCl₄ + cyclohexene	—	anti	orange → colorless
4	Br₂ / H₂O + propene	Mark-OH	anti	orange → pale
5	BH₃; H₂O₂/OH⁻ + propene	anti-Markov	syn	bubbles
6	H₂O / H₂SO₄ + 2-methylpropene	Markov	—	—
7	cold dilute KMnO₄ + cyclohexene	—	syn diol	purple → brown
8	O₃; Zn/H₂O + 2-methyl-2-butene	—	cleavage	—
9	HBr (1 eq) + 1-butyne	Markov	trans vinyl	—
10	Hg²⁺/H₂O/H⁺ + 1-pentyne	Mark (enol→ketone)	—	—
11	BH₃; H₂O₂ + 1-hexyne	a-M (enol→aldehyde)	—	bubbles
12	H₂ / Lindlar + 2-butyne	—	syn (cis)	—

Instructions

This lab's Simulation section has four sub-sections (tabbed). Complete them in order.

Section I — Reaction Bench (12 reactions). Pick a reaction from the list. Click Add Reagent to perform the transformation — watch the flask contents change color and observations appear in the notebook. Then identify the correct product from three options. Correct answers show a green checkmark; incorrect answers show why and let you retry.

Section II — Regiochemistry Sorter (12 cards). Classify each reaction by its regio/stereochemistry — Markovnikov, anti-Markovnikov, syn, or anti addition. Drag each card into the correct bin.

Section III — Mechanism Animations (3 key mechanisms). Watch step-by-step arrow-pushing animations of the three most important mechanisms: halonium ion (bromination), carbocation (Markovnikov addition), and concerted hydroboration.

Section IV — Retrosynthesis Challenge (6 targets). For each target compound, pick the correct reagent from four options. Tests your forward understanding in reverse.

Prepare your lab report. Use the Example Report as your template for write-up.

Safety note: Several reagents would be hazardous in a real lab — Br₂ is corrosive, O₃ is toxic, BH₃ is pyrophoric, Hg²⁺ is toxic. In this virtual lab you're safe, but learn to recognize the hazards for when you encounter them in person.

Simulation

The simulation has four parts. Move through them in order using the tabs below.

Select a Reaction

▼ Fume Hood — Reaction in progress

Notebook — Observations Select a reaction to begin.

Predict the Product

Which structure is the correct product of this reaction? Consider regiochemistry AND stereochemistry.

Select a reaction

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Starting Material

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Reagent(s)

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Conditions

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What to watch for

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Drag each reaction card into the bin that matches its primary classification. Each reaction has one correct bin based on its dominant teaching point (regiochemistry for acyclic additions; stereochemistry for cyclic additions; "cleavage" for ozonolysis).

Reactions to classify

Markovnikov

Anti-Markovnikov

Syn addition

Anti addition

Oxidative cleavage

Score: 0 / 12 · Drag cards from the left into a bin.

Three key mechanisms with arrow-pushing animations. Use the buttons to switch mechanisms; use Next Step → to walk through each one.

Step 0 of 4

For each target product, choose the correct reagent. This tests your forward understanding in reverse — given the product, work back to the reagent that made it.

Score: 0 / 6

Team Questions

Discuss these questions with your team before starting the Reaction Bench. Each question targets a key concept that the simulation will reinforce.

Question 1

Carbocation stability. When HBr adds to 3-methyl-1-butene, the major product is 2-bromo-2-methylbutane, NOT 2-bromo-3-methylbutane. Draw both possible carbocations from H⁺ addition. Which is more stable, and why? What 1,2-shift occurs to convert one to the other?

Discuss: Identify the initial 2° carbocation, then recognize that a 1,2-hydride shift converts it to the 3° carbocation. The relative stability is 3° > 2° > 1°. Br⁻ traps the more stable cation, giving the rearranged product.

Question 2

The Br₂ test for unsaturation. Why does adding Br₂ in CCl₄ to an alkene cause the orange color to fade, while adding Br₂ to an alkane (e.g., cyclohexane) does NOT cause the color to fade in the dark? What would happen with cyclohexane under UV light?

Discuss: Alkenes react with Br₂ via electrophilic addition (no light needed). Alkanes are unreactive with Br₂ in the dark because they lack the π electrons. Under UV light, alkanes undergo radical halogenation (substitution, not addition), and the color also fades — but this gives an alkyl halide + HBr, not a 1,2-dibromide.

Question 3

Why does Br₂/H₂O give a halohydrin instead of a dibromide? When the bromonium ion forms, both Br⁻ and H₂O are present. Why does H₂O usually win? And why does H₂O attack the more-substituted carbon?

Discuss: Water is present in vast excess (it's the solvent) compared to the trace Br⁻ released by each addition. Water attacks the more-substituted C of the bromonium because that carbon bears more of the positive charge — it's better at stabilizing the partial bond breaking. So OH ends up on the more-substituted C (Markovnikov-OH), Br on the less-substituted C, with overall anti stereochemistry from the bromonium.

Question 4

Hydroboration regioselectivity. BH₃ doesn't generate a carbocation, yet it gives "anti-Markovnikov" addition. What controls the regioselectivity? Why does B end up on the LESS-substituted carbon?

Discuss: Two factors. Steric: B is the bigger atom; less steric strain when it lands on the less-hindered (less-substituted) carbon. Electronic: in the partial 4-membered TS, the carbon that bears more of the developing positive charge is the more-substituted one (Markovnikov sense), and H (the H of BH₃, slightly hydridic, δ⁻) goes there. So B goes to the less-substituted C, H to the more-substituted C. Net OH after oxidation = anti-Markovnikov.

Question 5

Keto-enol tautomerization. Hg²⁺/H₂O hydration of 1-pentyne gives the enol HOC(=CH-)..., which tautomerizes to 2-pentanone. Why is the keto form (ketone) more stable than the enol form here? What about cases where the enol is actually preferred?

Discuss: The keto form has a stronger C=O π bond (~178 kcal/mol) than the C=C of the enol (~65 kcal/mol). The keto also has a stronger C-H bond replacing the weaker O-H of the enol. So keto is favored for simple ketones/aldehydes. Enols ARE preferred when stabilized — for example, phenol (the enol of cyclohexa-2,4-dienone) is overwhelmingly enol due to aromatic stabilization. β-diketones (like 2,4-pentanedione) also have substantial enol content due to intramolecular H-bonding and conjugation.

Question 6

Lindlar vs Na/NH₃. Both Lindlar's catalyst and dissolving-metal reduction (Na/NH₃(l)) reduce alkynes to alkenes. Why does Lindlar give the cis isomer while Na/NH₃ gives trans?

Discuss: Lindlar is heterogeneous catalysis — both H atoms are delivered from the metal SURFACE, on the same face of the alkyne (syn addition) → cis alkene. Na/NH₃ is single-electron transfer through a vinyl radical → vinyl anion. The vinyl anion adopts the less-strained trans geometry (sp² lone pair anti to the larger group) before it picks up a proton → trans alkene.

Example Report

An example student lab report demonstrating the format and depth of analysis expected. Use this as a template for your own write-up.

Alkene and Alkyne Reactions: Predicting Regiochemistry and Stereochemistry

Submitted by: [Student Name]

Course: Organic Chemistry · Section: 221-A · Date: April 24, 2026

Abstract

This lab examined twelve canonical addition reactions of alkenes and alkynes to verify rules for regiochemistry (Markovnikov vs anti-Markovnikov) and stereochemistry (syn vs anti). Each reaction was simulated with the appropriate reagent system, and the product was identified from a list of plausible alternatives. Mechanistic explanations were applied to rationalize each observation, with three reactions (halonium ion bromination, carbocation HX addition, concerted hydroboration) examined in step-by-step mechanistic detail.

Introduction

Alkenes and alkynes have π bonds that act as nucleophiles toward electrophiles. The selectivity of additions to these π systems is controlled by two principles: (1) the electrophile attacks the carbon that produces the more-stable cationic (or radical, or partial-cation) intermediate (Markovnikov), and (2) the geometry of the intermediate determines whether the second group adds to the same face (syn) or opposite face (anti). Special cases include radical-mediated HBr addition (which reverses regiochemistry), and concerted reactions like hydroboration (which place B on the less-substituted carbon). For alkynes, an additional layer of chemistry applies: enol intermediates from hydration tautomerize to the more-stable carbonyl, and the choice of catalyst (Lindlar vs Na/NH₃) determines cis vs trans alkene products.

Results

All twelve reactions were performed in the virtual fume hood, observations recorded, and products correctly identified on the first or second attempt for each. Two reactions produced visible color changes diagnostic of unsaturation: Br₂ in CCl₄ on cyclohexene (orange → colorless) and cold dilute KMnO₄ on cyclohexene (purple → brown). Three reactions involving aqueous H₂O₂ workup produced visible bubbling (hydroboration of propene and 1-hexyne).

Rxn #	Substrate	Reagent	Product	Classification
1	propene	HBr	2-bromopropane	Markovnikov
2	propene	HBr, ROOR, hν	1-bromopropane	anti-Markovnikov (radical)
3	cyclohexene	Br₂ / CCl₄	trans-1,2-dibromocyclohexane	anti (halonium)
4	propene	Br₂ / H₂O	1-bromo-2-propanol	Markovnikov-OH, anti
5	propene	BH₃; H₂O₂/OH⁻	1-propanol	anti-Markovnikov, syn
6	2-methylpropene	H₂O / H₂SO₄	2-methyl-2-propanol	Markovnikov (3° cation)
7	cyclohexene	cold KMnO₄	cis-1,2-cyclohexanediol	syn dihydroxylation
8	2-methyl-2-butene	O₃; Zn/H₂O	acetone + acetaldehyde	oxidative cleavage
9	1-butyne	HBr (1 eq)	2-bromo-1-butene	Markovnikov (vinyl)
10	1-pentyne	Hg²⁺/H₂O/H⁺	2-pentanone	Mark. → methyl ketone
11	1-hexyne	BH₃; H₂O₂	hexanal	anti-Mark. → aldehyde
12	2-butyne	H₂ / Lindlar	cis-2-butene	syn (cis alkene)

Discussion

Regiochemistry trends: The Markovnikov vs anti-Markovnikov dichotomy is fundamentally about WHICH intermediate (cation, radical, or partial-cation) forms during the rate-determining step. Acid-catalyzed and ionic additions proceed through the more-stable cation (Markovnikov). Radical chains and concerted hydroboration access the less-substituted carbon for the new sigma bond to "B" or "Br•", since these mechanisms place the developing positive charge (or unpaired electron) on the more-substituted carbon during the TS.

Stereochemistry trends: Anti addition arises from a bridged intermediate (halonium, oxonium) that blocks one face of the alkene. Syn addition arises from concerted (hydroboration, dihydroxylation) or surface-bound (catalytic hydrogenation) mechanisms where both new bonds form to the same face simultaneously. For acyclic alkenes, this distinction matters only when the product is stereogenic; for cyclic alkenes, it cleanly produces cis vs trans diastereomers.

Alkyne tautomerization: The most surprising results for an organic novice are the alkyne hydration outcomes. Hg²⁺/H₂O on 1-pentyne gives 2-pentanone (a methyl ketone, not an enol or alcohol) — because the Markovnikov enol tautomerizes spontaneously to the more-stable ketone. Hydroboration on 1-hexyne gives hexanal (an aldehyde), because the anti-Markovnikov enol tautomerizes to the aldehyde. These two reactions thus give COMPLEMENTARY carbonyl products from terminal alkynes — Hg²⁺ for ketones, BH₃ for aldehydes.

Mechanistic insights: The three mechanism animations clarified the origin of each addition's selectivity. The halonium ion mechanism explained why Br₂ addition is strictly anti. The carbocation mechanism explained why HX addition is Markovnikov but susceptible to rearrangement. The hydroboration TS explained why B goes to the less-hindered carbon and why no rearrangement is possible (no carbocation forms).

Conclusions

All twelve reactions confirmed the predicted product based on the regio/stereochemistry rules. The reactions can be organized into a small number of mechanistic categories: ionic Markovnikov, radical anti-Markovnikov, halonium-mediated anti addition, concerted syn addition, and tautomerization-coupled alkyne reactions. Mastery of these five categories enables prediction of products for nearly any reagent system encountered with alkenes or alkynes.

References

1. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed., Oxford University Press, 2012, Ch. 19–20.
2. Smith, M. B. March's Advanced Organic Chemistry, 8th ed., Wiley, 2020, Ch. 15.
3. Brown, H. C. Hydroboration; Benjamin: New York, 1962.

Practice Questions

Test your understanding. Try each one before peeking at the hint.

Practice 1

Predict the major product: 2-methyl-2-butene + HCl. Will rearrangement occur? Why or why not?

Hint: H⁺ adds to give the most stable cation directly — already 3° (the central carbon). No rearrangement possible because we're already at the most stable cation. Product: 2-chloro-2-methylbutane.

Practice 2

Reagent selection: Starting from cyclohexene, propose a sequence to make trans-1,2-cyclohexanediol (trans diol).

Hint: KMnO₄ and OsO₄ both give CIS (syn) diols. To get TRANS, use a two-step sequence: (1) mCPBA forms an epoxide with retention of alkene geometry; (2) acidic aqueous workup (H₃O⁺) opens the epoxide with anti stereochemistry → trans diol.

Practice 3

Mechanism: Why doesn't HBr add to 1-butene with REARRANGEMENT to give 2-bromo-2-butene? (Hint: there's a 2° to 2° "shift" that wouldn't help.)

Hint: The initial cation from 1-butene + H⁺ is the 2° cation at C2 (CH₃-CH₂-CH⁺-CH₃, sec-butyl cation). A hydride shift from C3 to C2 would give... another 2° cation (just the mirror image). Carbocation rearrangements only occur if they go to a MORE stable cation. 2° → 2° gains nothing, so no rearrangement.

Practice 4

Synthesis design: How would you make pentanal from 1-pentyne? How would you make 2-pentanone from the same starting material?

Hint: Pentanal (anti-Markovnikov, aldehyde) → use disiamylborane then H₂O₂/NaOH. 2-pentanone (Markovnikov, methyl ketone) → use Hg²⁺/H₂O/H₂SO₄. Both work on the same alkyne; the choice of reagent system determines which carbonyl is formed.

Practice 5

Identify reactions by observation: A student adds reagent X to two different unknowns. With cyclohexene, a deep purple color fades to brown. With cyclohexane, no color change. What reagent is X? What is the product with cyclohexene? What does this tell you about the difference between alkenes and alkanes?

Hint: The purple-to-brown color change is the Baeyer test for unsaturation: cold dilute KMnO₄. Cyclohexene → cis-1,2-cyclohexanediol (with brown MnO₂ precipitate from reduction of Mn(VII) to Mn(IV)). Cyclohexane lacks π electrons, so it doesn't react. This is a key qualitative test for unsaturation, alongside Br₂ decolorization.

Practice 6

Stereochemistry challenge: Cyclohexene reacts with Br₂ to give racemic trans-1,2-dibromocyclohexane. WHY racemic? (Two stereocenters, one product — but what about (R,R) vs (S,S)?)

Hint: The bromonium ion can form on either face of the planar alkene with equal probability. If it forms on the top face, Br⁻ attacks from below → one enantiomer (e.g., R,R). If it forms on the bottom face, Br⁻ attacks from above → the other enantiomer (S,S). Equal populations of both = racemic mixture of the trans diastereomer. The (R,S) meso form is NOT trans — it would be cis, which doesn't form here.