Fundamentholfundamenthol

Properties of Alkenes

ChemistryHydrocarbonsFor JEE aspirants

The chemical properties of alkenes are dominated by the bond, whose loosely held electrons make the double bond nucleophilic. Alkenes therefore undergo electrophilic addition with halogens, hydrogen halides, water and HOX, plus catalytic hydrogenation, polymerisation and, at high temperature, allylic substitution. Regiochemistry follows Markovnikov's rule in ionic conditions and reverses under peroxide (Kharasch) conditions. Physically, alkenes are non-polar, insoluble in water, and follow the same gas-liquid-solid pattern with chain length as alkanes.

Key Formulas - Quick Reference
  1. Halogenation: (vicinal dihalide, anti addition)
  2. Markovnikov addition:
  3. Peroxide effect:
  4. Acid hydration:
  5. Hydroboration-oxidation:
  6. Hydrogenation: , kJ/mol
  7. Halohydrin:

1. Physical Properties of Alkenes

Physical state of alkenes against chain length Alkenes with two to four carbon atoms are gases, those with five to seventeen carbons are liquids, and those with eighteen or more carbons are solids at room temperature. Alkenes are colourless, almost odourless, insoluble in water, less dense than water, and their boiling points rise with chain length and fall with branching. Physical state of straight-chain alkenes at 25 °C C2 to C4 gases C5 to C17 liquids C18 and above solids increasing number of carbon atoms → Colour and smell colourless, almost odourless Solubility insoluble in water, soluble in benzene and ether Density less than 1 g cm⁻³, so they float on water Boiling point rises with chain length, falls with branching
Figure 1: Alkenes follow the same gas to liquid to solid pattern with chain length as the alkanes, because both are held together only by weak van der Waals forces.
  • Physical state: C2 to C4 are gases, C5 to C17 are liquids, C18 and above are solids.
  • Polarity: almost non-polar. cis isomers have a small dipole moment; trans isomers are usually non-polar because the bond dipoles cancel.
  • Solubility: insoluble in water, freely soluble in benzene, ether and other organic solvents.
  • Boiling point: rises with chain length; branching lowers it by reducing surface contact.
  • Density: less than 1 g/cm3, so alkenes float on water.

2. Why Alkenes React: The Reactive Bond

The electrons sit farther from the nuclei than electrons and are held only loosely. That makes the double bond an electron-rich, nucleophilic site, attractive to any electron-poor species.

Sigma framework and pi bond of an alkene The sigma framework of an alkene is planar with bond angles of one hundred and twenty degrees around each sp2 hybridised carbon. The pi bond comes from sideways overlap of the two leftover p orbitals and places electron density above and below the molecular plane, where it is loosely held and available to electrophiles. Why the C=C bond is an electron-rich site Sigma framework The pi bond C C H H H H 120° each carbon is sp2 hybridised three sigma bonds, 120° apart all six atoms lie in one plane C C leftover p orbitals overlap sideways the pi cloud sits above and below the plane, far from the nuclei and loosely held, so the alkene is a nucleophile
Figure 2: The skeleton is planar and rigid; the cloud sits above and below it. Those exposed electrons are what every reagent in this chapter attacks.
General two step mechanism of electrophilic addition Step one: the pi electron pair of the alkene attacks the electrophile E plus, giving a planar sp2 carbocation. This step is slow and decides the rate. Step two: the nucleophile Nu minus adds rapidly to the positive carbon, giving the addition product E CH2 CH2 Nu. Typically electrophilic addition runs in two steps Step 1 · slow, rate deciding H2C CH2 + E+ E CH2 CH2 + carbocation: planar, sp², electron poor Step 2 · fast E CH2 CH2 + Nu− E CH2 CH2 Nu addition product
Figure 3: Step 1 is the slow step: the electrons attack and the carbocation forms. Step 2 is fast, the nucleophile simply traps it.
Addition beats substitution: Breaking one bond (251 kJ/mol) and forming two bonds ( kJ/mol) releases about 443 kJ/mol. A substitution would swap one bond for another of similar strength, so there is almost no energy gain. This is why the typical reaction of an alkene is electrophilic addition.
Energy profile of a two step electrophilic addition The reaction coordinate diagram shows two maxima separated by a shallow minimum. The first transition state is the highest point, so the first step is rate determining, and the minimum between them is the carbocation intermediate. A more stable carbocation lowers the first transition state and speeds the whole reaction up. Energy profile: why step 1 controls everything Potential energy reaction progress → alkene + E–Nu addition product TS 1 highest point TS 2 carbocation intermediate Ea of step 1 is the big one Whichever carbocation is more stable gives the lower TS 1, so that path wins.
Figure 4: Two humps, one well. The well is the carbocation; the taller first hump is why carbocation stability decides the major product.
Exam shortcut Step 1 is rate determining, so every question about rate, major product or regiochemistry in ionic addition is really asking one thing: which carbocation is more stable? Answer that and the rest follows.

3. What Makes One Alkene More Reactive Than Another

Effect of substituents on alkene reactivity Electron releasing groups raise the electron density of the pi bond and stabilise the carbocation, so electrophilic attack is faster. Electron withdrawing groups such as nitrile and nitro drain density from the double bond and slow the reaction down. The usual order of reactivity is vinyl ether, then alkyl substituted alkene, then ethene, then acrylonitrile. What makes one alkene react faster than another Electron releasing H CH3O H H pushes charge into the C=C FASTER pi density Plain alkene H H H H reference BASE pi density Electron withdrawing H N≡C H H drains charge from the C=C SLOWER pi density rate of electrophilic attack falls this way Typical order: vinyl ether > alkyl alkene > ethene > acrylonitrile
Figure 5: Reactivity order is ERG-CH=CH CH=CH EWG-CH=CH. Groups that push charge in help twice over: richer bond and a more stable cation.
  • Electron releasing groups (, ) raise the electron density, so electrophilic attack is faster.
  • They also stabilise the carbocation intermediate, lowering the activation barrier.
  • Electron withdrawing groups (, ) such as -CN and -NO2 deactivate the double bond.
  • Typical order: vinyl ether alkyl-substituted alkene ethene acrylonitrile.
Relative stability of carbocations Carbocation stability increases from methyl to primary to secondary to tertiary because each extra alkyl group donates electron density by the inductive effect and provides more alpha hydrogens for hyperconjugation. A tertiary carbocation with nine alpha hydrogens is the most stable of the four. Carbocation stability decides the major product CH3 + methyl 0 alpha H CH3CH2 + primary 2 alpha H (CH3)2CH + secondary 6 alpha H (CH3)3C + tertiary 9 alpha H stability → Each extra alkyl group adds +I donation and more alpha hydrogens for hyperconjugation, so the positive charge is spread further and the cation is stabilised.
Figure 6: Stability rises methyl. Almost every Markovnikov question reduces to picking the taller bar here.

4. Catalytic Hydrogenation

Hydrogen adds across the double bond over a metal catalyst to give the alkane. The uncatalysed reaction has too high a barrier to run at room temperature; the catalyst offers a lower-energy path by weakening the H-H bond on its surface.

Mechanism of catalytic hydrogenation of an alkene Hydrogen is adsorbed on a platinum, palladium or nickel surface and its bond is weakened. The alkene lies flat on the same surface, so both hydrogen atoms are delivered to the same face of the double bond. The addition is therefore syn and gives the cis product, releasing about one hundred and twenty kilojoules per mole. Catalytic hydrogenation: both H atoms from one face R H R H the alkene lies flat on the surface metal surface: Pt, Pd or Ni H H H–H bond weakened on the metal syn same face R R H H cis (syn) product both H atoms arrive on the same face, so they end up cis ΔH ≈ −120 kJ mol⁻¹ per double bond, so hydrogenation is strongly exothermic The heat released measures how unstable the alkene was to begin with. Rate falls as crowding rises: ethene > mono > di > tri > tetrasubstituted
Figure 7: Hydrogenation is a syn addition because the alkene sits flat on the metal and can only be reached from the exposed face. kJ mol.
Heat of hydrogenation and alkene stability All of these alkenes give the same kind of alkane, so the heat released measures how high in energy the alkene started. Ethene releases about one hundred and thirty seven kilojoules per mole while a tetrasubstituted alkene releases only about one hundred and ten, so more highly substituted alkenes are more stable. Heat of hydrogenation ranks alkene stability 137 kJ/mol CH2=CH2 ethene 126 kJ/mol RCH=CH2 mono 115 kJ/mol RCH=CHR di (trans) 112 kJ/mol R2C=CHR tri 110 kJ/mol R2C=CR2 tetra heat released → more alkyl groups → less heat released → the alkene was already more stable
Figure 8: Same product, different starting heights. Less heat released means a more stable alkene, which is why stability rises with substitution.

5. Addition of Halogens (Halogenation)

  • Works well with Cl2 and Br2. F2 is too violent and the iodine products decompose easily.
  • Solvents must be inert to halogens: CH2Cl2, CHCl3 or CCl4.
  • The addition is anti, because the intermediate is a bridged halonium ion opened by backside attack.
  • Decolourisation of bromine in CCl4 is the standard test for unsaturation.
Addition of bromine to an alkene through a bromonium ion Bromine adds to an alkene through a three membered bromonium ion rather than an open carbocation. Bromide ion then attacks the ring from the face opposite the bridging bromine, so the two bromine atoms finish on opposite faces and the product is the anti vicinal dibromide. Halogen addition goes through a bridged halonium ion H H H H Br2 CCl4 Br H H H H + bromonium ion blocks the top face Br- anti Br Br vicinal dibromide one Br front, one back Works with Cl2 and Br2. F2 is explosive; I2 products fall apart again Solvent inert to halogens: CH2Cl2, CHCl3 or CCl4 Stereochemistry anti, because the ring is opened from the back face Use in the lab red-brown Br2 fading is the standard test for unsaturation
Figure 9: The bridged ion is why halogen addition is anti, not syn. Br can only reach the face the bridge is not covering.
Bromine water test for unsaturation When bromine water is shaken with an alkene the red brown colour disappears because bromine adds across the double bond to give a colourless product. An alkane does not react in the dark and the colour stays. Alkynes, phenols and aldehydes also decolourise bromine water, so the test shows unsaturation rather than an alkene specifically. The bromine water test for unsaturation Alkene present Alkane only before red-brown shake after colourless Br2 adds across the C=C colour disappears at once before shake after no reaction in the dark colour stays Careful: alkynes, phenols and aldehydes also decolourise bromine water. The test proves unsaturation or easy oxidation, not an alkene specifically.
Figure 10: Decolourisation of Br water is the standard bench test. It confirms unsaturation, so always back it up with a second test.

6. Addition of HOX: Halohydrin Formation

Mechanism of halohydrin formation from an alkene Propene reacts with X2 in water. The halogen adds first to give a bridged halonium ion, which places more partial positive charge on the more substituted carbon. Water, present in large excess, attacks that carbon from the face opposite the halogen, and loss of a proton gives the halohydrin. The hydroxyl ends up on the more substituted carbon, the halogen on the less substituted one, and the two groups are anti to each other. Halohydrin formation: water opens the halonium ion propene halonium ion halohydrin H CH3 H H X2 in H2O X + CH3 H H H δ+ H2O water attacks the carbon that carries the larger δ+ anti – H+ OH X Intermediate bridged halonium ion, so no free carbocation is formed Stereochemistry anti: water opens the ring from the face opposite X Where OH goes the more substituted carbon, which carries more δ+ Where X goes the less substituted carbon Worked case propene + Cl2 in water gives 1-chloropropan-2-ol
Figure 11: Halohydrin formation is anti, and OH lands on the carbon better able to carry positive charge, so the regiochemistry looks Markovnikov.

7. Addition of Hydrogen Halides

Markovnikov's rule: When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part attaches to the carbon of the double bond bearing the least number of hydrogens; equivalently, the H+ adds to the carbon already carrying the most hydrogens.
Markovnikov addition of hydrogen bromide to propene Propene can be protonated at either carbon. Adding the proton to the terminal carbon gives a secondary carbocation, while adding it to the middle carbon would give a much less stable primary carbocation. The secondary route wins, so bromide finishes on the middle carbon and the major product is two bromopropane. Markovnikov's rule is really a carbocation rule H CH3 H H propene H–Br CH3–CH–CH3 + secondary more stable H goes to the CH2 end CH3–CH2–CH2 + primary not formed H goes to the CH end CH3–CHBr–CH3 2-bromopropane, the major product Rule of thumb: H+ adds to the carbon that already has more hydrogens. Real reason: that is the route that builds the more stable carbocation.
Figure 12: adds to the carbon already carrying more hydrogens, because that is the route that builds the more stable carbocation.
  • Reactivity of the acid: HI HBr HCl HF, matching the ease of releasing H+.
  • The reaction is not stereoselective, since the planar carbocation can be attacked from either face.
  • Carbocation rearrangement can occur, so watch for hydride and alkyl shifts.
  • The reaction is regioselective: two constitutional isomers are possible but one dominates.

The peroxide (Kharasch) effect

Radical chain mechanism of the peroxide effect with hydrogen bromide Peroxides break on heating to give alkoxy radicals, which take hydrogen from hydrogen bromide to give bromine radicals. The bromine radical adds to the less substituted carbon of the alkene because that gives the more stable secondary carbon radical. That radical then takes hydrogen from another molecule of hydrogen bromide, so bromine finishes on the less substituted carbon, opposite to Markovnikov's rule. The peroxide (Kharasch) effect: a radical chain 1 Initiation R–O–O–R heat 2 RO then Br from HBr 2 Propagation H CH3 H H Br CH3–CH–CH2Br secondary radical: the more stable one it then takes H from another HBr 3 Result CH3–CH2–CH2Br 1-bromopropane: Br on the LESS substituted carbon Only HBr does this. With HCl and HI one propagation step is endothermic, so no chain runs.
Figure 13: With peroxides, Br arrives first instead of , so the more stable radical forms and the product is anti-Markovnikov.
Markovnikov and anti Markovnikov addition of hydrogen bromide In the dark hydrogen bromide adds to propene by an ionic route and gives two bromopropane, the Markovnikov product. In the presence of peroxides or light the reaction switches to a radical chain and gives one bromopropane, the anti Markovnikov product. Both routes pick the more stable intermediate; only the species that adds first changes. Same alkene, same acid, opposite regiochemistry In the dark: ionic With peroxide: radical H CH3 H H H CH3 H H HBr dark HBr R2O2 or hν CH3–CHBr–CH3 2-bromopropane Markovnikov CH3–CH2–CH2Br 1-bromopropane anti-Markovnikov Both routes still pick the more stable intermediate. Only the first arrival changes: H+ in the ionic route, Br· in the radical route.
Figure 14: Same substrate, opposite regiochemistry. Only HBr shows this switch, and only when peroxide or light is present.
Reading the question: the words "peroxide", "R2O2", "sunlight" or "" with HBr means the radical, anti-Markovnikov route. With HCl or HI, the light makes no difference to the regiochemistry.

8. Addition of Water: Three Routes to an Alcohol

Three routes from an alkene to an alcohol Dilute sulfuric acid gives the Markovnikov alcohol but the carbocation may rearrange. Oxymercuration with mercuric acetate followed by sodium borohydride also gives the Markovnikov alcohol but goes through a bridged mercurinium ion so rearrangement is impossible. Borane followed by alkaline hydrogen peroxide gives the anti Markovnikov primary alcohol by a concerted syn addition. Three ways to add water across a double bond H CH3 H H propene dil. H2SO4 H2O CH3–CHOH–CH3 Markovnikov cheapest, but the carbocation can rearrange Hg(OAc)2/H2O then NaBH4 CH3–CHOH–CH3 Markovnikov bridged ion, so no rearrangement ever BH3·THF then H2O2/OH- CH3–CH2–CH2OH anti-Markovnikov concerted and syn, so no rearrangement
Figure 15: Pick the route by the alcohol you want. Only hydroboration-oxidation gives the anti-Markovnikov product.
Hydroboration oxidation of an alkene Borane adds across the double bond through a four centre transition state in which boron and hydrogen are delivered to the same face at the same moment. Boron attaches to the less hindered carbon because it is bulky and because the partial positive charge builds on the more substituted carbon. Oxidation with alkaline hydrogen peroxide replaces boron by hydroxyl with retention, giving the anti Markovnikov primary alcohol. Hydroboration: one step, syn and anti-Markovnikov Concerted four-centre transition state BH2 H CH3 H H H δ+ δ− B and H add across the same face at the same moment ‡ H2O2 OH− via the alkylborane OH propan-1-ol OH on the END carbon Where B goes to the less hindered, less substituted carbon Where H goes to the more substituted carbon After oxidation OH replaces B with retention, so OH ends up on the end carbon Stereochemistry syn: H and OH arrive on the same face Rearrangement impossible, because no carbocation is ever formed Exam signal see BH3 or B2H6 → expect the anti-Markovnikov alcohol
Figure 16: Boron goes to the less hindered carbon and H to the other, both from the same face, so the addition is syn and anti-Markovnikov.
Syllabus Acid-catalysed hydration and Markovnikov addition appear in NCERT and are examinable for JEE Main and NEET. Oxymercuration-demercuration and hydroboration-oxidation are JEE Advanced level; know the regiochemistry and stereochemistry even if the full mechanism is not asked.

9. Free Radical (Allylic) Substitution

Addition against allylic substitution in alkenes At low temperature chlorine adds across the double bond of propene to give one two dichloropropane. At six hundred degrees celsius, or with N bromosuccinimide, a chlorine radical instead removes the weak allylic hydrogen and the product is allyl chloride with the double bond still intact. The allylic carbon hydrogen bond is weakest because the radical left behind is resonance stabilised. Temperature decides: addition or allylic substitution Low temperature 600 °C, or NBS H CH3 H H H CH3 H H the C=C is attacked the allylic C–H breaks Cl2 Cl2 CH3–CHCl–CH2Cl 1,2-dichloropropane addition: C=C gone ClCH2–CH=CH2 allyl chloride substitution: C=C kept The allylic C–H is the weakest bond in the molecule: about 88 kcal/mol against 101, because the radical it leaves behind is resonance stabilised over three carbons. NBS works at room temperature by keeping the Br2 concentration permanently low.
Figure 17: Low temperature favours addition; high temperature or NBS favours allylic substitution, and the C=C survives.
Resonance stabilisation of the allyl radical The radical left after an allylic hydrogen is removed can be drawn in two equivalent resonance forms, so the unpaired electron is shared over both end carbons. That delocalisation makes the allylic carbon hydrogen bond about thirteen kilocalories per mole weaker than a normal carbon hydrogen bond, which is why a radical removes it first. Why the allylic C–H breaks first ↔ equal contributors the unpaired electron is shared over both end carbons delocalised allyl radical Allylic C–H ≈ 88 kcal/mol against a normal C–H ≈ 101 kcal/mol. Weaker bond + more stable radical = that is the hydrogen a radical takes.
Figure 18: Two equal resonance forms mean the allyl radical is spread over three carbons, so the allylic C-H is the weakest bond in the molecule.

10. Polymerisation

Alkenes are the standard monomers for addition polymers, also called chain-growth polymers, because monomer units simply add on to a growing chain with nothing else lost.

Addition polymerisation of an alkene In addition or chain growth polymerisation many alkene monomers join end to end to give one long chain and nothing else is produced. Because no small molecule is lost the polymer has the same empirical formula as the monomer, which is what distinguishes addition polymers from condensation polymers. Addition (chain-growth) polymerisation H H H H H H H H H H H H H H H H + … n monomers heat, P catalyst polymer repeating unit - - No small molecule is lost, so the polymer has the same empirical formula as the monomer. That is what separates addition (chain-growth) from condensation (step-growth) polymers, where water or HCl is expelled at every join.
Figure 19: Chain-growth polymerisation: monomers give one long chain, with no small molecule lost.
MonomerPolymerEveryday use
EthenePolytheneBags, bottles, insulation
PropenePolypropyleneRopes, containers
Chloroethene (vinyl chloride)PVCPipes, flooring
TetrafluoroetheneTeflonNon-stick coatings
IsopreneNatural rubber (all-cis)Tyres, elastic goods

11. Two More Reactions Worth Knowing

(a) Alkylation

Isobutylene and isobutane combine over H2SO4 to give 2,2,4-trimethylpentane, the compound that defines octane number 100 in petrol.

(b) Addition of NOCl (Tilden's reagent)

Nitrosyl chloride adds across the double bond, with Cl going to the carbon bearing fewer hydrogens, following the Markovnikov pattern.

12. Reagent to Product Summary

Summary map of alkene reactions and their products A single alkene gives ten different products depending on the reagent: an alkane with hydrogen over platinum, a vicinal dihalide with halogen, a halohydrin with halogen in water, a Markovnikov halide with hydrogen halide in the dark, an anti Markovnikov bromide with hydrogen bromide and peroxide, a Markovnikov alcohol with dilute acid, an anti Markovnikov alcohol by hydroboration oxidation, a vicinal diol with cold permanganate, carboxylic acids or ketones with hot permanganate, and aldehydes or ketones by ozonolysis. One alkene, ten reagents: the master map H H H H H2 / Pt alkane X2 vicinal dihalide X2 / H2O halohydrin HX (dark) Markovnikov halide HBr / R2O2 anti-Mark. bromide H2O / H+ Markovnikov alcohol BH3 then H2O2 anti-Mark. alcohol cold KMnO4 vicinal diol hot KMnO4 acids or ketones O3 then Zn/H2O aldehydes, ketones
Figure 20: Revision map. Read the reagent, read the product. Every JEE and NEET question on this chapter is somewhere on this page.
ReagentProductKey feature
H2 / Pt, Pd, NiAlkaneSyn addition
X2 (Cl2, Br2)Vicinal dihalideAnti addition
X2 in water (HOX)HalohydrinAnti, OH on more substituted C
HX (dark)Alkyl halideMarkovnikov, may rearrange
HBr with peroxideAlkyl bromideAnti-Markovnikov, radical
H2O / dil. H2SO4AlcoholMarkovnikov, may rearrange
Hg(OAc)2/H2O then NaBH4AlcoholMarkovnikov, no rearrangement
BH3 then H2O2/OH-AlcoholAnti-Markovnikov, syn
Cold dilute alkaline KMnO4Vicinal diol (glycol)Syn hydroxylation, Baeyer's test
Hot acidic KMnO4Acids, ketones, CO2Oxidative cleavage
O3 then Zn/H2OAldehydes and ketonesReductive ozonolysis
NBS or Cl2 at 600 °CAllylic halideSubstitution, C=C retained

Solved Examples

Solved Example 1
Predict the major product when 2-methylpropene reacts with HBr (a) in the dark and (b) in the presence of benzoyl peroxide.
Solution

(a) In the dark the route is ionic. H+ adds to the CH2 end, giving a tertiary carbocation (CH3)3C+. Bromide then attacks it. Major product: 2-bromo-2-methylpropane.

(b) With peroxide the route is radical. Br adds to the CH2 end, giving the tertiary carbon radical, which then takes H from HBr. Major product: 1-bromo-2-methylpropane.

Both steps pick the more stable intermediate; only the identity of the first-arriving species changes.

Solved Example 2
Identify the major products X and Y: CF3-CH=CH2 + HCl gives X, and CH3O-CH=CH2 + HCl gives Y.
Solution

For X, the CF3 group is strongly electron withdrawing, so it destabilises any adjacent positive charge. The proton adds to the carbon nearer CF3 so that the cation forms as far from CF3 as possible. X = CF3-CH2-CH2Cl, which looks anti-Markovnikov but is simply the more stable cation winning.

For Y, the CH3O group donates by resonance and stabilises a cation on the carbon next to it. Y = CH3O-CHCl-CH3, the normal Markovnikov product.

Solved Example 3
3,3-Dimethylbut-1-ene is hydrated in three ways: (i) dilute H2SO4, (ii) Hg(OAc)2 then NaBH4, (iii) BH3 then H2O2/OH-. Give the product in each case.
Solution

(i) Protonation gives a secondary cation which rearranges by a methyl shift to the tertiary cation. Water traps the rearranged ion, giving 2,3-dimethylbutan-2-ol.

(ii) The bridged mercurinium ion never becomes a free carbocation, so no rearrangement is possible. The Markovnikov alcohol 3,3-dimethylbutan-2-ol is obtained.

(iii) Boron attaches to the terminal carbon and oxidation replaces it with OH, giving the anti-Markovnikov primary alcohol 3,3-dimethylbutan-1-ol.

Solved Example 4
Why is tetrachloroethene, Cl2C=CCl2, unreactive towards Cl2, and why does adding AlCl3 make it react?
Solution

Four chlorine atoms withdraw electron density strongly by the effect, leaving the bond electron-poor. A neutral Cl2 molecule is too weak an electrophile to attack such a deactivated double bond.

AlCl3 is a Lewis acid. It polarises Cl2 to give an effective Cl+ species, a far stronger electrophile, which can now attack even this poor nucleophile.

Solved Example 5
Give the major product when propene reacts with (a) Cl2 at low temperature and (b) Cl2 at 600 °C.
Solution

(a) At low temperature the bond attacks Cl2 and the addition path runs, giving 1,2-dichloropropane.

(b) At 600 °C chlorine radicals form, and the weakest C-H bond is the allylic one because the resulting radical is resonance stabilised. The product is the substitution product 3-chloroprop-1-ene (allyl chloride), with the double bond intact.

Solved Example 6
An alkene C4H8 decolourises bromine water and, on treatment with HBr in the dark, gives a single monobromide. Identify the alkene.
Solution

Decolourising bromine water confirms a double bond. A single monobromide from HBr means both possible additions give the same compound.

But-2-ene, CH3CH=CHCH3, is symmetrical: whichever carbon is protonated, the product is 2-bromobutane.

Answer: but-2-ene (cis or trans). But-1-ene would give both 2-bromo and 1-bromobutane, and isobutylene would give mainly the tertiary bromide.

Solved Example 7
3-Methylbut-1-ene is treated with HCl and gives two products, 2-chloro-3-methylbutane and 2-chloro-2-methylbutane. Explain.
Solution

Protonation at the terminal CH2 gives the secondary cation (CH3)2CH-CH+-CH3. Chloride can trap it directly, giving 2-chloro-3-methylbutane.

That secondary cation can also undergo a 1,2-hydride shift from the neighbouring CH, producing the more stable tertiary cation (CH3)2C+-CH2CH3. Trapping that one gives 2-chloro-2-methylbutane.

Whenever a secondary cation sits next to a carbon that can hand over H or CH3 to make a tertiary cation, expect a rearranged product alongside the direct one.

Common Mistakes to Avoid

Watch out
  • Applying the peroxide effect to HCl or HI. Only HBr reverses; for the others the peroxide changes nothing.
  • Quoting Markovnikov's rule as a rule about hydrogens and stopping there. Always check which carbocation is more stable; with a strong electron withdrawing group the "Markovnikov" answer can look inverted.
  • Forgetting carbocation rearrangement in acid-catalysed hydration and in HX addition.
  • Expecting hydroboration to rearrange. It is concerted with no ionic intermediate, so it never does.
  • Calling halogen addition a syn addition. It is anti, because of the bridged halonium ion.
  • Mixing up cold dilute alkaline KMnO4 (gives the diol) with hot concentrated KMnO4 (cleaves the molecule).
  • Writing an allylic substitution product at low temperature. Substitution needs high temperature or NBS.
  • Assuming bromine water decolourisation proves an alkene specifically. Alkynes, phenols and aldehydes also decolourise it, so confirm with another test.

Frequently Asked Questions

Why do alkenes give addition reactions rather than substitution?

In an addition, one weak pi bond of about 251 kJ per mole is broken and two strong sigma bonds worth about 694 kJ per mole are formed, releasing roughly 443 kJ per mole. In a substitution a pi bond would be traded for a sigma bond of similar energy, so there is little driving force.

What is Markovnikov's rule and why does it work?

When an unsymmetrical reagent such as HX or water adds to an unsymmetrical alkene, the negative part goes to the carbon carrying fewer hydrogens. The real reason is carbocation stability: protonation happens at whichever carbon leaves the more stable cation behind, and tertiary beats secondary beats primary.

What is the peroxide or Kharasch effect?

In the presence of peroxides or light, HBr adds to an unsymmetrical alkene by a free radical chain instead of an ionic path. The bromine radical adds first and gives the more stable carbon radical, so bromine finishes on the less substituted carbon, opposite to Markovnikov's rule.

Why is the peroxide effect seen only with HBr?

Both chain propagation steps must be exothermic for the chain to run. With HCl the step where the carbon radical takes hydrogen is endothermic, and with HI the step where iodine adds to the alkene is endothermic. Only HBr has both steps favourable.

How do I choose between acid hydration, oxymercuration and hydroboration?

For the Markovnikov alcohol with no risk of rearrangement, use oxymercuration then NaBH4. For the anti-Markovnikov primary alcohol with syn stereochemistry, use BH3 then H2O2 with hydroxide. Dilute acid is the cheapest but may rearrange the carbocation.

Why does bromine water decolourise with alkenes?

The reddish brown bromine adds across the double bond to give a colourless vicinal dibromide or bromohydrin, so the colour disappears. Alkanes do not react at room temperature, which is why this is the standard test for unsaturation.

What is allylic substitution and when does it happen?

It is the replacement of a hydrogen on the carbon next to a double bond, leaving the double bond untouched. It needs a high temperature such as 600 degrees celsius with chlorine, or a reagent like N-bromosuccinimide that keeps the halogen concentration low. The allylic radical is resonance stabilised, which is why that C-H breaks first.

Why are more substituted alkenes slower to hydrogenate?

Hydrogenation happens on the surface of the catalyst, so the alkene must lie flat against it. Alkyl groups get in the way, and the more of them there are the harder adsorption becomes. So ethene reacts fastest and a tetrasubstituted alkene slowest.

What makes an alkene more reactive towards electrophiles?

Electron releasing groups such as alkyl, alkoxy or amino raise the electron density of the pi bond and stabilise the carbocation that forms, so they speed the reaction up. Electron withdrawing groups such as nitrile or nitro do the opposite.

Does hydroboration-oxidation ever rearrange the carbon skeleton?

No. Boron and hydrogen are delivered in one concerted step through a four centre transition state, so no carbocation is ever formed and there is nothing that could undergo a hydride or alkyl shift. This is exactly why it is chosen when the acid catalysed route would rearrange.

Previous year questions on Properties of Alkenes

4 questions from past papers, each with a step-by-step solution.

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