Elimination Reactions: Forming Double Bonds
Learn elimination, the reverse of addition: dehydrohalogenation of 2-bromopropane with KOH, dehydration of alcohols, and Zaitsev's rule for the major alkene.
Quick answer
In an elimination reaction, atoms are removed from adjacent carbons of a molecule and a carbon–carbon double bond forms. It is the reverse of addition: where addition consumes a pi bond, elimination creates one, typically by expelling a small molecule such as HBr or H2O.
Example: 2-bromopropane reacts with potassium hydroxide to give propene: CH3CHBrCH3 + KOH → CH3CH=CH2 + KBr + H2O. A hydrogen and a bromine from neighboring carbons are eliminated together.
What you will learn
- Define elimination and recognize it as the reverse of addition.
- Describe dehydrohalogenation of haloalkanes with KOH.
- Explain acid-catalyzed dehydration of alcohols.
- Apply Zaitsev's rule to predict the major alkene product.
What elimination means
In an elimination reaction, small atoms or groups are removed from two adjacent carbons, and a double bond forms between those carbons. The general pattern is X–C–C–Y → C=C + XY.
Elimination is the exact reverse of addition: addition breaks a pi bond to form two sigma bonds, while elimination breaks two sigma bonds to form a pi bond. The same molecules can often interconvert, with conditions deciding the direction.
The two classic eliminations are dehydrohalogenation (removing H and a halogen from a haloalkane) and dehydration (removing H and –OH, as water, from an alcohol).
Dehydrohalogenation
A haloalkane heated with potassium hydroxide loses hydrogen and halogen from neighboring carbons:
CH3CHBrCH3 + KOH → CH3CH=CH2 + KBr + H2O
The H comes from a carbon adjacent to the one bearing bromine; the two leave together as HBr (captured by KOH as KBr and water), and the double bond forms between the two carbons.
Chemists specify the base carefully: KOH in ethanol favors elimination, while aqueous KOH favors substitution to the alcohol. Same reagents, different conditions, different reaction family — a reminder that conditions are part of the reaction.
Dehydration of alcohols
Alcohols eliminate water when heated with concentrated sulfuric acid, which acts as a dehydrating agent:
CH3CH2OH → CH2=CH2 + H2O
The acid first protonates the –OH group, converting it into H2O+ — essentially a water molecule ready to leave. Water departs, and a neighboring hydrogen is removed, forming the double bond. (This is a conceptual description of the chemistry, not a procedure to attempt: concentrated sulfuric acid is severely corrosive.)
Dehydration is a major industrial route to alkenes — including ethene from bio-ethanol — because alcohols are readily available from fermentation and other sources.
Zaitsev's rule
When elimination can produce two different alkenes, Zaitsev's rule predicts the major product: the more substituted alkene predominates — the one with more alkyl groups attached to the double-bond carbons.
For 2-bromobutane (CH3CHBrCH2CH3), removing H from carbon 1 gives but-1-ene, while removing H from carbon 3 gives but-2-ene. But-2-ene, with two alkyl substituents on the double bond versus one, is the major product.
The reason is stability: alkyl groups donate electron density that stabilizes the double bond, so more substituted alkenes are lower in energy. As with Markovnikov's rule, the observed selectivity reflects the relative stability of the species along the reaction pathway.
Elimination versus substitution
Haloalkanes sit at a crossroads: with base they can undergo substitution (giving an alcohol) or elimination (giving an alkene). The conditions tip the balance.
Strong, bulky bases and high temperatures favor elimination — the bulky base struggles to attack carbon directly but readily plucks an exposed hydrogen. Smaller nucleophiles and lower temperatures favor substitution.
This competition is a central theme in organic problem-solving. When you see a haloalkane plus base, always ask which pathway the conditions favor — and remember that mixtures are common when the balance is close.
Why elimination matters
Elimination completes the reaction network: substitution installs functional groups on alkanes, and elimination converts those functionalized molecules into alkenes — which then undergo addition to give new functional groups.
Follow the loop: alkane → (substitution) → haloalkane → (elimination) → alkene → (addition) → alcohol or dihalide. Each step is a standard transformation, and chaining them is the essence of synthesis.
Mastering elimination therefore does double duty: it is both a useful reaction in its own right and the conceptual mirror of addition. Whenever you understand one direction thoroughly, the reverse comes nearly free.
Common mistakes
They are distinct reaction types with different bond changes: substitution swaps one group for another on a single carbon, while elimination removes groups from two adjacent carbons and creates a double bond between them. The product types — substituted alkane versus alkene — are completely different.
The opposite: Zaitsev's rule favors the more substituted alkene — the one with more carbon groups (and therefore fewer hydrogens) on the double-bond carbons. Greater substitution stabilizes the double bond, so that product forms preferentially.
Alcohols need an acid catalyst, such as concentrated sulfuric acid, to dehydrate at a useful rate. The acid protonates the –OH group, turning it into water — a far better leaving group — which is what makes elimination possible.
Real-world applications
Dehydration of ethanol is an industrial route to ethene, a cornerstone chemical feedstock.
Elimination steps in synthesis build the double bonds found in fragrances, vitamins, and pharmaceuticals.
Catalytic cracking in refineries uses elimination-type chemistry to convert heavy fractions into valuable alkenes.
Understanding elimination versus substitution lets chemists steer reactions toward the desired product family.
Frequently asked questions
How is elimination the reverse of addition?
Addition takes an alkene plus XY and makes X–C–C–Y; elimination takes X–C–C–Y and regenerates the alkene plus XY. The same bond changes run backward: addition breaks the pi bond to form two sigma bonds, while elimination breaks two sigma bonds to form the pi bond.
What does Zaitsev's rule predict for 2-bromobutane?
Two alkenes are possible: but-1-ene (double bond at the end) and but-2-ene (double bond in the middle). Zaitsev's rule predicts but-2-ene as the major product because its double bond carries two alkyl substituents versus one — it is the more substituted, more stable alkene.
Why is concentrated sulfuric acid used to dehydrate alcohols?
The acid plays two roles: it protonates the –OH group to make water, an excellent leaving group, and it strongly absorbs the water produced, driving the equilibrium toward the alkene. Without the acid, the –OH group is a poor leaving group and elimination is extremely slow.
Can elimination happen more than once?
Yes, if the product still has removable neighbors. A dihaloalkane can eliminate twice to form an alkyne, and repeated dehydrations build conjugated systems. Each elimination needs a hydrogen on a carbon adjacent to the leaving group — without that, the reaction cannot proceed.