Alkene preparation methods
An alkene's calling card is its carbon-carbon double bond, , and the double bond is a site of reactivity, not something a molecule keeps by accident. So the standard way to make an alkene is to start from a saturated or functionalised precursor and pull two atoms or groups off adjacent carbons, letting the leftover electrons snap together into a bond. This article collects the three routes an ISC Class 11 student must know cold: dehydrohalogenation of an alkyl halide, dehydration of an alcohol, and partial hydrogenation of an alkyne. All three are elimination-flavoured, all three need a named reagent and named conditions to work, and two of them are governed by the same rule — Saytzeff's rule — for choosing which alkene wins when more than one is possible.
Alkenes are built by eliminating two atoms or groups from adjacent carbons (alkyl halide with alcoholic KOH, or alcohol with an acid/alumina dehydrating agent) or by stopping hydrogen addition to an alkyne halfway with Lindlar's catalyst, and where a choice of product exists, the more substituted alkene forms preferentially by Saytzeff's rule.
The core idea
Dehydrohalogenation of alkyl halides. Heat an alkyl halide with alcoholic KOH — potassium hydroxide dissolved in ethanol, deliberately not water — and one molecule of HX leaves: an H from one carbon, the halogen X from the carbon next door.
where: X is a halogen (Cl, Br or I) and marks that heating is required. Swap in aqueous KOH and the reaction changes character entirely — the hydroxide ion now attacks the carbon bearing the halogen and substitutes for it, giving an alcohol instead of an alkene. So "alcoholic vs aqueous" is not a minor detail; it decides which product family you get.
Dehydration of alcohols. Heat an alcohol with a dehydrating agent — concentrated , or vapour passed over hot alumina, — and a molecule of water leaves: the –OH from one carbon, an –H from the carbon next door.
where: the dehydrating agent is essential — writing this as if water simply falls off on heating, with no acid or alumina named, is the single commonest slip students make on this reaction.
Partial hydrogenation of alkynes. An ordinary catalyst (Pd, Pt or Ni) reduces a triple bond all the way to a single bond. To stop cleanly at the alkene stage, the catalyst is deliberately weakened — palladium supported on barium sulfate or calcium carbonate and "poisoned" with quinoline, known as Lindlar's catalyst. This gives predominantly the cis-alkene.
where: the poisoning is what makes the catalyst selective — it slows hydrogen addition enough that the reaction halts after one addition instead of racing on to the alkane. Across all three, the pattern is the same: pull two atoms/groups off adjacent carbons (true elimination), or interrupt an addition to a triple bond partway through (controlled/partial addition). Either way, the product is a molecule with one fewer -bond pair and one new bond.
Why it works
An elimination reaction is, at its heart, a trade: the molecule gives up two single bonds (a C-H and a C-X, or a C-H and a C-OH) and forms one new pi bond plus a small stable byproduct (HX, or H2O). That trade is thermodynamically favourable because HX and H2O are themselves very stable small molecules, so the leaving groups are happy to go, and the two electrons left behind on the adjacent carbons pair up into the new pi bond rather than sitting unused. Alcoholic KOH and hot dehydrating agents both supply the two things this trade needs: a base or acid environment that helps detach the leaving group, and enough heat to drive off the small molecule (HX or H2O) so the reaction proceeds forward instead of reversing. Saytzeff's rule follows from a separate piece of reasoning: not all alkenes are equally stable. An alkene with more alkyl groups attached directly to its double-bond carbons is more stable than one with fewer, because alkyl groups release electron density toward the double bond and support it (the same idea used qualitatively at this level, without invoking hyperconjugation by name). When an elimination has a choice of which adjacent C-H to remove, the transition state that leads to the more substituted alkene is lower in energy, so that pathway is favoured and its product dominates. This is why but-2-ene (internal, disubstituted-on-each-carbon) outcompetes but-1-ene (terminal, less substituted) when both are accessible from the same alkyl halide.
Worked example
Problem: 2-Bromobutane, , is heated with alcoholic KOH. Two alkenes are possible. Identify both, name the major product by Saytzeff's rule, and state what changes if aqueous KOH is used instead.
Step 1. Locate the halogen and the adjacent carbons available for elimination. The Br sits on C2. Elimination needs an H from a carbon next to C2, so either C1 (a group) or C3 (a group) can lose its H.
Step 2. Write the product from removing an H on C1.
Step 3. Write the product from removing an H on C3.
Step 4. Compare substitution to apply Saytzeff's rule. In but-2-ene, each double-bond carbon carries one alkyl group and one H; in but-1-ene, one double-bond carbon carries two H atoms and no alkyl group. But-2-ene is the more substituted, hence more stable, alkene, so by Saytzeff's rule it is the major product; but-1-ene is the minor product.
Step 5. Note what changes with aqueous KOH. Swapping alcoholic KOH for aqueous KOH switches the reaction from elimination to substitution — the hydroxide displaces Br directly on C2 — giving butan-2-ol instead of either alkene. Answer: but-2-ene is the major product and but-1-ene the minor product with alcoholic KOH; aqueous KOH instead gives the substitution product, butan-2-ol, and no alkene.
Common traps
- Using aqueous KOH and expecting an alkene — aqueous KOH favours substitution (an alcohol), not elimination; alcoholic KOH is required for dehydrohalogenation.
- Writing alcohol dehydration with no reagent shown, as if water leaves spontaneously — conc. or hot plus heat must always be stated.
- Confusing Lindlar's catalyst (poisoned Pd, stops at the cis-alkene) with an ordinary Pd/Pt/Ni catalyst, which reduces an alkyne all the way to the alkane.
- Applying Saytzeff's rule mechanically even on a symmetric substrate (like 2-bromopropane) where both elimination directions give the identical product, so there is nothing to choose between.
Ask the tutor: if predicting the major vs minor alkene product for an unsymmetrical substrate feels tricky, ask the AI tutor for a few practice eliminations with step-by-step reasoning.
Why does alcoholic KOH give an alkene but aqueous KOH gives an alcohol from the same alkyl halide?
The solvent changes what the hydroxide ion does. In ethanol, KOH behaves as a base that pulls off an H from a carbon next to the halogen, triggering elimination of HX and forming the double bond. In water, the same hydroxide ion instead behaves as a nucleophile that attacks the carbon bearing the halogen directly, displacing it and forming an alcohol by substitution. Same ion, different job, because the solvent changes which pathway is easier. At this level you are only expected to know and apply the outcome — alcoholic gives alkene, aqueous gives alcohol — not the full mechanistic reasoning, which arrives in Class 12.
Why is Saytzeff's rule stated as more substituted rather than just bigger alkene?
Because size alone is not the deciding factor — the number of alkyl groups directly attached to the two double-bond carbons is. An alkyl group next to a double bond donates electron density that stabilises it, so a double bond flanked by two alkyl carbons (like in but-2-ene) is more stable than one flanked by only one alkyl carbon and a bare CH2 (like in but-1-ene), even though both molecules have the same molecular formula, . Saytzeff's rule simply says the elimination pathway leading to the more stable, more substituted alkene is favoured, so it forms as the major product.
Does Lindlar's catalyst always give exactly cis and nothing else?
Lindlar's catalyst gives predominantly the cis-alkene, which is what you should write in an exam answer, but "predominantly" is doing real work in that sentence — trace amounts of the trans-isomer and even some fully reduced alkane can form as minor byproducts under real laboratory conditions. The syllabus point is the qualitative one: a poisoned Pd catalyst is selective for stopping at the alkene stage and for delivering the two hydrogen atoms to the same face of the triple bond, which is why the major product is cis. You are not expected to quantify the minor products at this level.