Fundamentholfundamenthol

Preparation of Alkenes

ChemistryHydrocarbonsFor JEE aspirants

Alkenes are prepared mainly by elimination reactions, in which two groups on adjacent carbons leave and a bond forms in their place. The standard routes are dehydrohalogenation of alkyl halides with alcoholic KOH, dehydration of alcohols with concentrated H2SO4, dehalogenation of vicinal dihalides with zinc, and partial reduction of alkynes. Which alkene dominates is decided by Saytzeff's rule for small bases and by Hofmann's rule for bulky ones. The Wittig reaction and Kolbe electrolysis add two non-elimination routes.

Key Formulas - Quick Reference
  1. Dehydrohalogenation:
  2. Dehydration:
  3. Dehalogenation:
  4. Syn reduction:
  5. Anti reduction:
  6. E2 rate law: ; E1 rate law:
  7. Ease of elimination: and

1. Overview: Every Route Is an Elimination

To make a C=C you must remove two atoms or groups from adjacent carbons. The name of the method just records what gets removed.

MethodWhat is removedTypical reagent
DehydrohalogenationH and XAlcoholic KOH, NaNH2, t-BuOK
DehydrationH and OHConc. H2SO4 at 160 °C, Al2O3
DehalogenationX and XZn dust in acetic acid, NaI in acetone
Partial reductionAdds H and H to an alkyneH2/Lindlar, Na in liquid NH3
Thermal eliminationH and OCOR, or H and NR2OHeat alone
Hofmann eliminationH and NR3Heat the quaternary ammonium hydroxide
Wittig reactionReplaces C=O by C=CPh3P=CR2 ylide
Kolbe electrolysisTwo CO2 groupsElectrolyse the dicarboxylate salt
Every preparation of an alkene is an elimination Each route removes two groups from adjacent carbons, a beta hydrogen and a leaving group, and the electrons left behind form the new carbon carbon double bond. Every preparation is an elimination C C H X β-H and leaving group on adjacent carbons – HX base or heat C C R R R R the π bond appears between them Dehydrohalogenation H + X Dehydration H + OH Dehalogenation X + X Hofmann elimination H + NR3 only the pair being removed changes; the outcome is always a new C=C
Figure 1. Every method on this page removes two pieces from adjacent carbons. Only the pair being removed changes; the product is always a new C=C.

2. Dehydrohalogenation of Alkyl Halides

A strong base pulls a hydrogen off the carbon next to the one carrying the halogen, and the halide leaves. Because the hydrogen comes from the carbon, this is called -elimination.

Reagents: hot alcoholic KOH (that is EtO- in EtOH), NaNH2, or potassium tert-butoxide in t-BuOH.

The E2 mechanism

E2 happens in a single concerted step. The base removes the -H while the C-X bond breaks and the bond forms at the same instant. Because both the substrate and the base appear in the rate-determining step, the reaction is second order.

E2 elimination in one step with anti-periplanar H and Br E2 dehydrohalogenation drawn in the anti-periplanar conformation: the beta hydrogen points up and the bromine points down, both in the plane of the page, 180 degrees apart. In one concerted step the base removes the beta hydrogen, the carbon hydrogen bond pair becomes the new pi bond and bromide leaves. The two groups drawn on wedges, R1 and R3, end up cis in the alkene, so E2 is stereospecific. A Newman projection along the beta to alpha bond shows the H and Br 180 degrees apart. The rate equals k times the concentration of alkyl halide times the concentration of base, second order. E2: one concerted step, with H and Br anti-periplanar The base pulls off the β-H while Br− leaves from the opposite side C C H Br R1 R3 R2 R4 β α B − one step no intermediate C C R1 R3 R2 R4 + BH + Br− H up, Br down, both in the page: 180° apart the two wedged groups end up cis Looking along Cβ to Cα (Newman) H R1 R2 Br R3 R4 180° Why one step, and why anti rate = k[alkyl halide][base] second order, one step needs a strong base: alc. KOH or C2H5O−Na+ C-H and C-Br must line up anti (180°) so the new π bond can form as Br− leaves Groups on the same side of the H-Br line end up cis: E2 is stereospecific
Figure 2. E2 is one concerted step that needs the -H and the leaving group anti-periplanar ( apart), so groups on the same side end up cis. Rate [alkyl halide][base].
  • The -H and the leaving group must be anti-periplanar, so E2 is a stereospecific anti elimination.
  • Ease of dehydrohalogenation: alkyl halide.
  • Rate with the halogen: R-I R-Br R-Cl R-F, matching the ease of C-X bond breaking.
  • Greater conjugation in the product alkene means greater stability, so the elimination is faster.

Saytzeff's rule

When more than one -H is available, the major product is the more substituted, more stable alkene.

Newman projection showing the anti-periplanar requirement of E2 Two Newman projections viewed along the carbon carbon bond. Left: the beta hydrogen on the front carbon and the bromine on the rear carbon are anti, one hundred and eighty degrees apart, so their orbitals lie in one plane and overlap to form the new pi bond; E2 elimination occurs. Right: the hydrogen and bromine are gauche, sixty degrees apart, the orbitals are not aligned and E2 cannot occur from this conformation. Why anti-periplanar: the Newman view β-H and Br anti β-H and Br gauche H R R Br R R 180° good overlap: E2 ✓ H R R Br R R 60° poor overlap: no E2 ✗ anti: C-H and C-Br lie in one plane, so their orbitals form the π bond in a ring, if no β-H can sit anti to the leaving group, E2 fails
Figure 3. Sighting along the C-C bond shows the E2 requirement: the -H and Br must be anti-periplanar () so their orbitals line up with the forming bond; in the gauche () form they cannot.
Saytzeff versus Hofmann elimination of 2-bromo-2-methylbutane 2-bromo-2-methylbutane drawn with its hydrogens shown. The alpha carbon carries bromine. The inner beta hydrogens sit on the CH2 group and are crowded; the outer beta hydrogens sit on the CH3 group and are exposed. Small ethoxide removes an inner beta hydrogen and gives 2-methylbut-2-ene, a trisubstituted alkene, the Saytzeff product. Bulky tert-butoxide reaches an outer beta hydrogen far more easily and gives 2-methylbut-1-ene, a disubstituted alkene, the Hofmann product. Same substrate, two bases: which β-H is removed? H3C C H H C CH3 Br C H H H β α β 2-bromo-2-methylbutane inner β-H: crowded on the CH2 group outer β-H: exposed on the CH3 group EtO− small base O− C CH3 CH3 CH3 bulky base (t-BuO−) EtO− / EtOH, heat t-BuO− K+ / t-BuOH, heat Saytzeff product (major) C C H3C H CH3 CH3 1 2 3 2-methylbut-2-ene trisubstituted, about 70% Hofmann product (major) C C H3C H5C2 H H 1 2 2-methylbut-1-ene disubstituted, about 73% small base → more stable alkene; bulky base → most exposed β-H
Figure 4. Ethoxide is small enough to reach the crowded inner -H, so it gives the more substituted, more stable alkene (Saytzeff). Bulky tert-butoxide reaches the exposed -H of the group far more easily, so the less substituted alkene wins (Hofmann).

The Hofmann exception: bulky bases

A large base cannot squeeze in to reach the internal, crowded -H. It takes the exposed hydrogen of a terminal methyl instead, and the less substituted alkene becomes the major product.

Why a bulky base gives the less substituted (Hofmann) alkene E2 elimination of 2-bromo-2-methylbutane with potassium tert-butoxide. The inner beta hydrogen on the CH2 group sits between two methyl groups, so the bulky tert-butoxide ion clashes with them and cannot reach it. The outer beta hydrogen on the terminal methyl group is exposed, so the base removes it and 2-methylbut-1-ene, the Hofmann alkene, is the major product (about 70 percent). A small base such as ethoxide gives mainly 2-methylbut-2-ene, the Saytzeff alkene. Access, not stability, decides 2-bromo-2-methylbutane has two kinds of β-H. Which one can (CH3)3CO− reach? Inner β-H (on CH2): hard to reach the base's CH3 groups clash with the substrate's Outer β-H (on CH3): easy to reach nothing in the way: this H is removed C Br CH3 C H H H3C C H H H α β β C CH3 CH3 CH3 O − ✗ C Br CH3 C H H H3C C H H H α β β C CH3 CH3 CH3 O − E2 (−HBr), minor path E2 (−HBr), main path H3C CH C CH3 CH3 2-methylbut-2-ene trisubstituted, more stable: Saytzeff alkene minor, about 30% H3C CH2 C CH2 CH3 2-methylbut-1-ene disubstituted, less stable: Hofmann alkene major, about 70% Bulky bases: (CH3)3CO−K+, LDA, (C2H5)3N Small base (C2H5O−): Saytzeff alkene major. Bulky base: Hofmann alkene major.
Figure 5. With a bulky base such as , access, not alkene stability, decides: the crowded inner -H is hard to reach, so the exposed -H of the group is removed and the Hofmann alkene is major.
Endocyclic versus exocyclic: With 1-bromo-1-methylcyclohexane, losing a proton from a ring CH2 gives 1-methylcyclohexene, where the double bond is inside the ring (endocyclic) and trisubstituted. Losing one from the methyl gives methylenecyclohexane, an exocyclic and only disubstituted alkene. Saytzeff favours the endocyclic product.

3. Dehydration of Alcohols

Heating an alcohol with a strong acid removes water and gives an alkene. Since the OH- ion is a poor leaving group, the acid first protonates it to H2O+, which leaves readily.

E1 mechanism for the acid-catalysed dehydration of an alcohol Dehydration of tert-butyl alcohol (2-methylpropan-2-ol) with 20 percent phosphoric acid at 358 K. Step 1, fast and reversible: a lone pair on oxygen takes a proton, giving the protonated alcohol (oxonium ion). Step 2, slow and rate determining: the carbon-oxygen bond breaks, water leaves and a planar tertiary carbocation forms. Step 3, fast: water removes a beta hydrogen and the carbon-hydrogen bond pair becomes the carbon-carbon double bond of 2-methylpropene, regenerating the acid as hydronium ion. Tertiary alcohols dehydrate most easily (20 percent phosphoric acid, 358 K), secondary need 85 percent phosphoric acid at 440 K and primary need concentrated sulfuric acid at 443 K. E1 dehydration of an alcohol 2-methylpropan-2-ol (tert-butyl alcohol), 20% H3PO4, 358 K Step 1: protonation (fast, reversible) C H3C CH3 CH3 O H + H+ fast C H3C CH3 CH3 O H H + tert-butyl alcohol protonated alcohol (oxonium ion) Step 2: water leaves (slow, rate-determining) C H3C CH3 CH3 O H H + slow RDS C H3C CH3 CH3 + + H2O 3° carbocation (planar, sp²) Step 3: water removes a β-H (fast) C H3C CH3 C H H H + β H2O fast C H3C CH3 CH2 + H3O+ 2-methylpropene acid regenerated Step 2 decides the rate: the more stable the carbocation, the milder the conditions 3° alcohol 20% H3PO4, 358 K > 2° alcohol 85% H3PO4, 440 K > 1° alcohol conc. H2SO4, 443 K ease of dehydration: 3° > 2° > 1° (same order as carbocation stability)
Figure 6. Dehydration by the E1 path: protonation, slow loss of water to give a carbocation, then loss of a -H. Because step 2 makes the carbocation, ease of dehydration follows .
ReagentConditionsNote
Conc. H2SO4160 °CMost common laboratory choice
H3PO4HeatLess charring than H2SO4
P2O5HeatPowerful dehydrating agent
Al2O3350 °C, vapour phaseHeterogeneous, industrial

Watch for rearrangement

The carbocation intermediate is free to rearrange. A hydride or alkyl shift that converts a secondary cation into a tertiary one happens fast, and the alkene then forms from the rearranged ion.

Carbocation rearrangement by a 1,2-methyl shift during dehydration Dehydration of 3,3-dimethylbutan-2-ol. Loss of water gives a secondary carbocation on carbon 2, whose empty p orbital lines up with a carbon 3 to methyl bond. The methyl group slides to carbon 2 with its bonding pair through a bridged transition state, so the positive charge moves to carbon 3 and becomes a more stable tertiary carbocation. Loss of a proton then gives 2,3-dimethylbut-2-ene, the tetrasubstituted alkene, as the major product. Step 1: water leaves, giving a 2° carbocation H3C CH OH C CH3 CH3 CH3 2 3 3,3-dimethylbutan-2-ol conc. H2SO4, Δ − H2O H3C CH C CH3 CH3 CH3 + 2 3 2° carbocation Step 2: 1,2-methyl shift, seen in 3D H3C H CH3 CH3 CH3 + C C 2 3 empty p orbital H3C H CH3 CH3 CH3 C C δ+ δ+ ‡ H3C H3C H CH3 CH3 + C C 2 3 empty p orbital 2° C+, less stable CH3 bridges C-2 and C-3 3° C+, more stable Step 3: loss of H+ gives the alkene H3C C CH3 H C CH3 CH3 + 2 3 3° carbocation − H+ C C H3C H3C CH3 CH3 2,3-dimethylbut-2-ene major product (tetrasubstituted) The CH3 moves with its bond pair, so the + charge moves the other way: 2° → 3°
Figure 7. The group (violet) moves from C-3 to C-2 with its bonding pair into the empty orbital, so the positive charge moves from C-2 to C-3 and the cation becomes a more stable cation. Losing then gives the tetrasubstituted alkene as the major product.

4. Dehalogenation of Vicinal Dihalides

A vicinal dihalide has the two halogens on adjacent carbons; a geminal dihalide has both on the same carbon. Only the vicinal type gives an alkene on dehalogenation.

E1 compared with E2 E1 goes through a carbocation in two steps with a first order rate law and can rearrange, while E2 is a single concerted step with a second order rate law that demands anti-periplanar geometry. E1 compared with E2 Feature E1 E2 Steps two (cation forms first) one, concerted Rate law k [substrate] k [substrate][base] Base needed weak base is enough strong base required Substrate 3° > 2° > 1° 3° > 2° > 1° Rearrangement possible never Geometry no special requirement anti-periplanar both give the Saytzeff alkene unless the base is bulky
Figure 8. The comparison worth memorising. The single most useful row is the last one: only E2 imposes a geometric requirement.
Anti dehalogenation of 2,3-dibromobutane is stereospecific Dehalogenation of a vicinal dibromide with sodium iodide in acetone or zinc dust. Iodide attacks one bromine, the carbon bromine bond pair becomes the new pi bond and the second bromide leaves from the opposite side, so the two bromines are removed anti. Meso-2,3-dibromobutane (2R,3S), drawn with both bromines anti, has its methyl groups on opposite sides and gives trans-but-2-ene. The (2R,3R) isomer, one of the racemic pair, has both methyl groups on the same side and gives cis-but-2-ene. Gem-dihalides do not react this way. Dehalogenation of a vicinal dibromide is anti and stereospecific NaI in acetone, or Zn dust, removes both Br atoms; they must be anti and on adjacent carbons meso-2,3-dibromobutane (2R,3S) C C Br Br H3C H H CH3 I − NaI, acetone or Zn dust C C H3C H H CH3 trans-but-2-ene CH3 groups were on opposite sides (2R,3R)-2,3-dibromobutane, one of the (±) pair C C Br Br H3C H CH3 H NaI, acetone or Zn dust C C H3C H CH3 H cis-but-2-ene CH3 groups were both wedged Wedge groups end up on the top side: the two Br leave anti, so each isomer gives one alkene By-products: IBr + NaBr with iodide (IBr + I− then gives I2 + Br−); ZnBr2 with zinc gem-Dihalides (both Br on one carbon) do not give alkenes this way
Figure 9. The two Br atoms leave anti, so meso-2,3-dibromobutane gives trans-but-2-ene and the isomer gives cis-but-2-ene: the reaction is stereospecific.
  • Reagents: Zn dust in acetic acid or ethanol, or NaI in acetone.
  • Both are E2-type, stereospecific anti eliminations.
  • Since vic-dihalides are themselves made from alkenes plus Br2, this route is mostly used to purify or protect a double bond rather than to create a new one.

5. Partial Reduction of Alkynes

Adding just one molecule of hydrogen across a triple bond gives an alkene. The trick is to stop there, and the choice of reagent also fixes the geometry of the product.

(a) Syn addition: cis alkene

Lindlar hydrogenation: syn addition gives the cis alkene Side view of Lindlar hydrogenation of an internal alkyne on palladium supported on calcium carbonate and poisoned with lead acetate and quinoline. Panel 1: hydrogen splits into hydrogen atoms held on the metal surface. Panel 2: the alkyne lies flat on the surface and both hydrogen atoms add to the two carbons from below, the same face, which is syn addition. Panel 3: the cis alkene leaves the surface, and the poison sites stop a second addition of hydrogen, so the reaction stops at the alkene. Nickel boride (P-2) also gives the cis alkene, sodium or lithium in liquid ammonia gives the trans alkene, and unpoisoned palladium on carbon reduces all the way to the alkane. Lindlar catalyst: both H atoms add from the metal side, so the alkene is cis Pd on CaCO3, poisoned with lead acetate and quinoline 1 H2 splits on Pd the H-H bond breaks on the metal Pd Pd Pd Pd Pd Pd Pd Pd H H H H H atoms held on the surface 2 Alkyne lies flat both H atoms come from below Pd Pd Pd Pd Pd Pd Pd Pd R C C R H H same face for both: syn addition 3 cis-Alkene leaves poison sites stop further H2 addition Pd Pd Pd Pd Pd Pd Pd Pd C C R R H H dark squares: Pb and quinoline H2, Lindlar (Pd, poisoned): cis-alkene, syn H2, Ni2B (P-2 catalyst): cis-alkene, syn Na or Li, liq. NH3: trans-alkene, anti H2, Pd/C with no poison: all the way to alkane Both H atoms reach the carbons from the same face, so they end up cis
Figure 10. Both H atoms are delivered from the metal surface to the same face of the alkyne, so with Lindlar's catalyst gives the cis-alkene (syn addition); the poison stops the reduction at the alkene.

(b) Anti addition: trans alkene

Sodium in liquid ammonia reduces an internal alkyne to the trans alkene Dissolving metal reduction of an internal alkyne with sodium or lithium in liquid ammonia, in four steps. Step 1: sodium gives one electron, forming a radical anion in which the two R groups sit trans, with the unpaired electron and the lone pair in sp2 orbitals on opposite carbons. Step 2: ammonia gives a proton to the anionic carbon, forming a trans vinyl radical. Step 3: a second electron from sodium gives the more stable trans vinyl anion. Step 4: ammonia protonates it, giving the trans alkene. The two hydrogens are added anti. Lindlar hydrogenation instead gives the cis alkene by syn addition, and a terminal alkyne is mostly deprotonated to sodium acetylide. Na in liquid NH3: anti addition gives the trans alkene Two electrons (from Na) and two protons (from NH3) add one at a time R C C R − C C R R C C R H R 1 + e− from Na 2 + H+ from NH3 internal alkyne radical anion: trans set here vinyl radical 3 + e− from Na − C C R H R 4 + H+ from NH3 C C R H H R vinyl anion (trans, more stable) trans-alkene Why trans? In the radical anion the two R groups sit on opposite sides, as far apart as possible. Each H+ adds where the electrons were, so the shape never flips. H2, Lindlar's catalyst: cis-alkene (syn) Na or Li, liq. NH3: trans-alkene (anti) Overall: alkyne + 2Na + 2NH3 → trans-alkene + 2NaNH2 A terminal alkyne is mostly deprotonated to the sodium acetylide instead
Figure 11. Na or Li in liquid adds , , , in turn. The R groups are trans from the first intermediate onwards, so the two H atoms end up anti and the product is the trans alkene.
ReagentMode of additionProduct geometry
H2, Lindlar catalyst (Pd/CaCO3, quinoline)Syncis-alkene
H2, P-2 catalyst (Ni2B)Syncis-alkene
Na or Li in liquid NH3Antitrans-alkene
H2, Pd or Pt or Ni (no poison)Syn, twiceAlkane, not alkene

6. Thermal and Pyrolytic Eliminations

Some eliminations need no base at all, only heat. They pass through a cyclic transition state in which the departing hydrogen and the leaving group must be on the same face, so these are syn eliminations and they follow the Hofmann pattern.

Ester pyrolysis: syn elimination through a six-membered cyclic transition state An alkyl acetate is heated to about 500 degrees Celsius without any base. In a six-membered cyclic transition state the carbonyl oxygen takes the beta hydrogen, the carbon hydrogen bond pair becomes the new carbon carbon double bond, and the alpha carbon to oxygen bond pair becomes the new carbon oxygen double bond, giving an alkene and acetic acid in one concerted step. Newman projections along the beta to alpha bond show that the hydrogen and the ester oxygen are syn periplanar (0 degrees) in ester pyrolysis, whereas in E2 the hydrogen and the bromine are anti periplanar (180 degrees). Xanthate (Chugaev) and amine oxide (Cope) eliminations are also syn. Ester pyrolysis (Ei): a six-membered cyclic transition state Heat alone, no base: the C=O oxygen pulls off the β-H itself H C C O C O R H H H CH3 heat ≈ 500 °C H C C O C O R H H H CH3 β α ‡ R CH CH2 H3C C O H O + alkene acetic acid alkyl acetate all bonds change at once Look down the Cβ-Cα bond Ei (ester pyrolysis): syn, 0° H and OCOCH3 on the same side E2 (base): anti, 180° H and Br on opposite sides OCOCH3 H H H R H Br H H H R H Other syn (Ei) eliminations: xanthates (Chugaev, 6-membered ring) and amine oxides (Cope, 5-membered ring) Syn elimination by heat alone; E2 needs a base and an anti H
Figure 12. Ester pyrolysis is a syn (Ei) elimination: in one concerted step the -H moves to the carbonyl oxygen through a six-membered ring, so the H and the group must leave from the same side, unlike the anti arrangement of E2.
BEYOND NCERT Ester pyrolysis, the Cope elimination of amine oxides, and the Hofmann elimination are standard in JEE Advanced-level organic chemistry but are not part of the NCERT hydrocarbons chapter. Learn the syn geometry and the Hofmann orientation; detailed mechanisms are rarely asked in JEE Main or NEET.
Hofmann elimination of a quaternary ammonium hydroxide gives the less substituted alkene Hofmann elimination. Step 1: an amine is treated with excess methyl iodide (exhaustive methylation) to give a quaternary ammonium iodide, and moist silver oxide swaps the iodide for hydroxide while silver iodide precipitates. Step 2: on heating, (butan-2-yl)trimethylammonium hydroxide undergoes E2 elimination. Hydroxide removes a primary beta hydrogen from the methyl end, the C-H electrons form the double bond and trimethylamine leaves. But-1-ene, the less substituted alkene, is the major Hofmann product; but-2-ene, the Saytzeff product, is minor. The bulky trimethylammonium group and the carbanion-like transition state explain the preference. Step 1: make the quaternary ammonium hydroxide RNH2 excess methyl iodide (exhaustive methylation) RN+(CH3)3 I− moist silver oxide (AgI precipitates) RN+(CH3)3 OH− Step 2: heat (E2), shown for R = butan-2-yl H H N(CH3)3 + HO − 1° β-H 2° β-H (butan-2-yl)trimethylammonium hydroxide Δ 1° H lost 2° H lost but-1-ene (major) Hofmann product but-2-ene (minor) Saytzeff product both routes also release trimethylamine and water WHY THE LESS SUBSTITUTED ALKENE? 1. Size: the bulky trimethylammonium group shields the inner 2° β-H, so hydroxide removes an exposed 1° β-H from the methyl end. 2. Acidity: the C-H bond breaks ahead of the C-N bond (carbanion-like E2), and a 1° β-H is the most acidic because its carbon carries the fewest alkyl groups. Hofmann rule: the least substituted alkene wins (the opposite of Saytzeff)
Figure 13. Heating a quaternary ammonium hydroxide, , removes a -hydrogen by E2. The bulky leaving group steers hydroxide to the exposed -H, so but-1-ene, the least substituted alkene (Hofmann product), is major.

7. The Wittig Reaction

A phosphorus ylide, made from an alkyl halide and triphenylphosphine followed by a strong base, converts an aldehyde or ketone directly into an alkene.

Wittig reaction mechanism: carbonyl to alkene through an oxaphosphetane Wittig reaction mechanism. A phosphorus ylide, triphenylphosphonium ylide, adds to the carbonyl group of an aldehyde or ketone: the negative ylide carbon bonds to the carbonyl carbon while the carbonyl oxygen bonds to phosphorus, giving a four-membered oxaphosphetane ring. The ring splits into an alkene and triphenylphosphine oxide, so the new carbon-carbon double bond forms exactly where the carbon-oxygen double bond was. Side panel: the ylide is made from triphenylphosphine and an alkyl halide, then a strong base such as butyllithium or sodium hydride. Step 1: [2+2] addition across C=O R R Ph3P C + − phosphorus ylide R′ R′ O C aldehyde or ketone R R R′ R′ Ph3P O C C oxaphosphetane (four-membered ring) Step 2: ring splits R R R′ R′ C C alkene + Ph3P O triphenylphosphine oxide strong P=O bond drives the reaction HOW THE YLIDE IS MADE Ph3P + R2CHX → Ph3P+CHR2 X− then a strong base (n-BuLi or NaH) removes the H on that carbon, giving the ylide The ylide carbon replaces the O: the C=C forms exactly where the C=O was
Figure 14. In the Wittig reaction the ylide carbon bonds to the carbonyl carbon and O bonds to P, closing a four-membered oxaphosphetane ring. The ring then splits into and the alkene, so the new sits exactly where the was.
Quick way to write the product: remove H and X from the -carbon of the alkyl halide, remove O from the carbonyl carbon, and join those two carbons with a double bond.

8. Kolbe Electrolytic Synthesis

Electrolysing the potassium salt of a dicarboxylic acid such as succinic acid discharges both carboxylate groups at the anode and leaves an alkene behind.

Kolbe electrolytic synthesis of an alkene Electrolysis of an aqueous solution of potassium succinate gives ethene and carbon dioxide at the anode, and hydrogen with potassium hydroxide at the cathode. Kolbe electrolytic synthesis potassium succinate in water KOOC-CH2-CH2-COOK anode (+) cathode (−) CH2=CH2 + 2CO2 oxidation at the anode H2 + 2KOH reduction at the cathode the alkene comes from the anode; the salt of a dicarboxylic acid is essential
Figure 15. Electrolysis of potassium succinate gives ethene and CO at the anode, H and KOH at the cathode.

9. Choosing the Right Method

If you wantUseBecause
The more substituted alkeneAlcoholic KOH, or acid dehydrationSaytzeff control
The less substituted alkenet-BuOK, or Hofmann eliminationSteric control
A cis alkeneH2 with Lindlar catalystSyn addition
A trans alkeneNa in liquid NH3Anti addition
The double bond at an exact positionWittig reactionNo carbocation, no rearrangement
To avoid rearrangementDehydrohalogenation by E2, not dehydrationNo free carbocation forms
Choosing a preparation route to an alkene Each starting material has its own reagent: alkyl halides use alcoholic potassium hydroxide, alcohols use concentrated sulfuric acid, vicinal dihalides use zinc, alkynes use Lindlar or sodium in ammonia, and carbonyls use a Wittig ylide. Choosing a route to an alkene Alkyl halide alc. KOH, Δ dehydrohalogenation Alcohol conc. H2SO4, 160 °C dehydration Vicinal dihalide Zn dust dehalogenation Alkyne (cis) H2 / Lindlar partial reduction Alkyne (trans) Na / liq. NH3 dissolving metal Carbonyl Ph3P=CR2 Wittig need a specific geometry? choose the alkyne route Lindlar gives cis, Na in liquid ammonia gives trans
Figure 16. Pick the row that matches your starting material. If the question fixes the geometry of the product, the alkyne routes are the only ones that give you that control.

Solved Examples

Solved Example 1
Which alkyl halide would give each of the following as the only alkene on treatment with alcoholic KOH? (i) (CH3)2C=CH2, (ii) CH3CH2CH2CH=CH2
Solution:

(i) The alkene has four carbons arranged as isobutylene, so the halide must be 2-chloro-2-methylpropane, (CH3)3C-Cl. Every -H is on an equivalent methyl group, so only one alkene is possible.

(ii) The alkene is pent-1-ene, so the halide is 1-chloropentane, CH3CH2CH2CH2CH2Cl. A primary halide with -H on only one carbon can give only this alkene.

Solved Example 2
Predict the major product when 2-bromo-2-methylbutane is treated with (a) sodium ethoxide in ethanol and (b) potassium tert-butoxide in t-butanol.
Solution:

(a) Ethoxide is small, so Saytzeff control applies. The major product is 2-methylbut-2-ene, the trisubstituted alkene.

(b) tert-Butoxide is bulky, so it abstracts the accessible terminal hydrogen. The major product is 2-methylbut-1-ene, the Hofmann product.

Both reactions run by E2; only the base size differs.

Solved Example 3
Compound A (C7H15Br) is not a primary alkyl bromide. It yields a single alkene B on heating with sodium ethoxide in ethanol. Hydrogenation of B gives 2,4-dimethylpentane. Identify A and B.
Solution:

Hydrogenation gives 2,4-dimethylpentane, so B has the same carbon skeleton: (CH3)2CH-CH2-CH(CH3)2 as the backbone.

Only one alkene forms, which requires all -hydrogens to be equivalent. Putting Br on C3, the central carbon, gives 3-bromo-2,4-dimethylpentane, a secondary bromide with two equivalent isopropyl groups on either side.

A = 3-bromo-2,4-dimethylpentane, B = 2,4-dimethylpent-2-ene.

Solved Example 4
How would you convert hex-3-yne into (a) cis-hex-3-ene and (b) trans-hex-3-ene?
Solution:

(a) Treat with H2 over Lindlar's catalyst. Both hydrogens are delivered from the catalyst surface in a syn fashion, so both ethyl groups end up on the same side: cis-hex-3-ene.

(b) Treat with Na in liquid NH3. The reduction passes through a trans vinyl anion, so the two ethyl groups end up on opposite sides: trans-hex-3-ene.

Solved Example 5
Explain why dehydration of 3,3-dimethylbutan-2-ol gives 2,3-dimethylbut-2-ene as the major product rather than 3,3-dimethylbut-1-ene.
Solution:

Protonation and loss of water give a secondary carbocation at C2. A methyl group migrates from the neighbouring quaternary carbon, a 1,2-methyl shift, producing a much more stable tertiary carbocation.

Loss of a -H from this rearranged cation gives 2,3-dimethylbut-2-ene, a tetrasubstituted and highly stable alkene. The unrearranged path would give only a disubstituted alkene, so it is minor.

Solved Example 6
What alkenes can form when HBr is lost from 2-bromo-2-methylbutane, and which one dominates with alcoholic KOH?
Solution:

Two different carbons carry hydrogens: the CH2 of the ethyl group and the two methyls on C2.

Removing an H from the CH2 gives 2-methylbut-2-ene (trisubstituted). Removing an H from a methyl gives 2-methylbut-1-ene (disubstituted).

With the small ethoxide base, Saytzeff control makes 2-methylbut-2-ene the major product.

Common Mistakes to Avoid

Watch out
  • Using aqueous KOH for elimination. Aqueous conditions give substitution and an alcohol; elimination needs alcoholic KOH.
  • Removing the hydrogen from the same carbon as the halogen. Elimination is always , never .
  • Applying Saytzeff's rule when the base is bulky. With t-BuOK, expect the Hofmann product.
  • Forgetting carbocation rearrangement in acid dehydration. Always check whether a hydride or methyl shift gives a more stable cation.
  • Trying to dehalogenate a geminal dihalide to an alkene. Only vicinal dihalides work.
  • Using Na in liquid NH3 on a terminal alkyne. It only forms the sodium acetylide salt; no reduction occurs.
  • Forgetting that ester pyrolysis and the Cope elimination are syn, while E2 and dehalogenation are anti.
  • Assuming hydrogenation of an alkyne stops at the alkene without a poisoned catalyst. With plain Pd or Pt the reaction runs on to the alkane.

Frequently Asked Questions

What is the best laboratory method to prepare an alkene?

Dehydrohalogenation of an alkyl halide with hot alcoholic KOH is the standard laboratory route, because the conditions are mild and the alkyl halide is easy to make. Dehydration of alcohols is preferred industrially since alcohols are cheaper, but it risks carbocation rearrangement.

Why is alcoholic KOH used for elimination but aqueous KOH for substitution?

In alcohol the base exists largely as ethoxide, a strong base that pulls off a beta hydrogen and drives E2 elimination. In water the hydroxide is heavily solvated and behaves more as a nucleophile, so substitution takes over and an alcohol is formed instead.

What is Saytzeff's rule?

Saytzeff's rule says that in an elimination the major alkene is the more highly substituted one, because it is the more stable product and the transition state leading to it is lower in energy. A small base such as ethoxide follows this rule.

When is the Hofmann product formed instead?

When the base is bulky, as with potassium tert-butoxide or di-isopropylamine, or when the leaving group is bulky, as in the quaternary ammonium hydroxide of the Hofmann elimination. Steric access then decides which beta hydrogen is removed, and the less substituted alkene dominates.

How do you get a cis alkene rather than a trans alkene from an alkyne?

Use hydrogen with Lindlar's catalyst, which is palladium on calcium carbonate poisoned with quinoline, or with the P-2 nickel boride catalyst. Both hydrogens add from the catalyst surface, that is a syn addition, so the cis alkene results. For the trans alkene use sodium or lithium in liquid ammonia.

Why does sodium in liquid ammonia fail with terminal alkynes?

A terminal alkyne has an acidic C-H, so the sodium simply deprotonates it to form the sodium acetylide salt rather than reducing the triple bond. Only internal alkynes are reduced to trans alkenes under these conditions.

Why does dehydration of alcohols sometimes give an unexpected alkene?

Dehydration proceeds through a carbocation, and a less stable carbocation can rearrange by a hydride or alkyl shift into a more stable one. The alkene then forms from the rearranged ion, so the double bond appears at an unexpected position.

Which eliminations are syn and which are anti?

Dehydrohalogenation by E2 and dehalogenation of vicinal dihalides are anti eliminations, requiring the two departing groups to be 180 degrees apart. Ester pyrolysis and the Cope elimination of amine oxides are syn eliminations, going through a cyclic transition state.

What makes the Wittig reaction useful?

It puts the double bond exactly where the carbonyl oxygen was, with no ambiguity about position and no carbocation rearrangement. That certainty of placement is something the elimination routes cannot guarantee.

Previous year questions on Preparation of Alkenes

1 question from past papers, each with a step-by-step solution.

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