Organic Chemistry · Alkenes: Reactions and Synthesis
Preparing Alkenes: A Preview of Elimination Reactions
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Chapter 7 treated alkenes as starting materials; this chapter flips the perspective. Alkenes are often the products we want, and the standard way to make them is to remove two atoms or groups from adjacent carbons — an elimination, the microscopic reverse of an addition. Instead of adding a reagent across a π bond, we remove a small molecule (H–X, H–OH, or X2) from a saturated precursor, creating the double bond.
This topic previews the three classic routes: dehydrohalogenation Elimination of H and X from an alkyl halide using a strong base. Full entry → of alkyl halides (removing H and X), dehydration Elimination of water from an alcohol using acid and heat. Full entry → of alcohols (removing H and OH), and dehalogenation Elimination of two halogens from a vicinal dihalide using Zn. Full entry → of vicinal dihalides (removing X2). Each has its own reagents, conditions, and regiochemistry; each is treated mechanistically in Chapter 11. The unifying idea is Zaitsev's rule Elimination favors the more substituted alkene. Full entry →: when an elimination can form two alkenes, the more substituted (more stable) alkene is the major product. Mastering this preview makes the mechanism chapters far easier.
Why this matters
Elimination is how the lab and the chemical industry convert cheap saturated compounds into the reactive alkene intermediates everything else is built from. Industrially, alcohol dehydration is a route to ethylene and propylene (Chapter 7, Topic 1). In the lab, dehydrohalogenation converts an alkyl halide (Chapter 10) into a specific alkene, and it competes with substitution whenever a base meets an alkyl halide — understanding elimination is essential for predicting what a reaction will actually give. In biology, elimination forms double bonds in fatty acid and terpene metabolism. On exams, Zaitsev's rule and the conditions/reagents of each route are near-guaranteed questions that resurface in every synthesis problem.
The college version
Core Concepts
Dehydrohalogenation of alkyl halides
An alkyl halide (R–CH2–CHX–R') treated with a strong base — typically potassium hydroxide in ethanol (KOH/EtOH) or sodium ethoxide — loses H–X to give an alkene:
R-CH2-CHX-R' + KOH EtOH, Δ⟶ R-CH=CH-R' + KX + H2O
The base abstracts a proton from the carbon adjacent to the halogen-bearing carbon (the β-carbon), the electron pair forms the C=C, and the halide leaves — all in one concerted step in the common E2 mechanism Concerted elimination: base, H, and leaving group in one step. Full entry → (Chapter 11). The leaving group and proton must be appropriately aligned (anti-periplanar geometry in acyclic systems), which matters in cyclohexane chemistry (Chapter 11, Topic 9). Strong, bulky bases favor elimination over substitution; the classic choice is KOH in ethanol.
Dehydration of alcohols
An alcohol loses water when heated with a strong acid — concentrated sulfuric or phosphoric acid — typically at 100–200 °C:
R-CH2-CH(OH)-R' H2SO4, Δ⟶ R-CH=CH-R' + H2O
The mechanism (Chapter 11, Topic 10) is the reverse of acid-catalyzed hydration: acid protonates the OH, water leaves to form a carbocation (E1 pathway for tertiary/secondary alcohols), and a proton is lost from an adjacent carbon to make the alkene. Because a carbocation is involved, rearrangements (Chapter 7, Topic 11) are possible, and tertiary alcohols dehydrate most readily. The reaction is reversible; removing water or distilling the alkene drives it forward. Water is the only byproduct, which is why dehydration is an attractive "green" route when the alcohol is bio-derived.
Dehalogenation of vicinal dihalides
A vicinal dihalide (two halogens on adjacent carbons) reacts with zinc metal to give an alkene, eliminating the halogens as ZnX2:
R-CHX-CHX-R' + Zn → R-CH=CH-R' + ZnX2
This is a reductive elimination: the metal supplies electrons and the halogens leave as halide ions. It is useful when the alkene is best made from a dihalide (e.g., from halogenation of another alkene, Chapter 8, Topic 2) and handy for converting 1,2-dihalides back to alkenes.
Zaitsev's rule: which alkene forms?
When elimination can remove a proton from either of two β-carbons, two alkene isomers are possible. Zaitsev's rule (also spelled Saytzeff) states that the major product is the more substituted alkene — more alkyl groups on the double-bond carbons. This is a thermodynamic preference: the more substituted alkene is more stable (Chapter 7, Topic 6), and both E1 and E2 deliver it preferentially. Example: 2-bromobutane with KOH/EtOH gives predominantly 2-butene (disubstituted) rather than 1-butene (monosubstituted). (The Hofmann elimination with bulky bases, Chapter 11, is the notable exception — it favors the less substituted alkene for steric reasons.)
Elimination versus substitution: the competition
Alkyl halides can react with nucleophiles/bases in two ways: substitution (the nucleophile replaces the halogen) or elimination (the base removes H–X). The balance depends on the base, substrate, and conditions: strong, bulky bases and high temperatures favor elimination; good nucleophiles and polar aprotic solvents favor substitution. The competition is treated systematically in Chapter 11; the preview point is that KOH/EtOH and heat are deliberately chosen to bias toward elimination when an alkene is the goal.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Elimination | substitution | Elimination forms a π bond by removing H–X; substitution replaces X with a nucleophile. Both can occur with the same alkyl halide + base. |
| Dehydration | dehydrogenation | Dehydration removes H2O (from an alcohol, acid-catalyzed); dehydrogenation removes H2 (from an alkane, catalytic). |
| Zaitsev product | Hofmann product | Zaitsev: more substituted alkene (normal bases); Hofmann: less substituted alkene (bulky base, steric control) — Chapter 11. |
| β-hydrogen | α-hydrogen | The β-H is on the carbon next to the leaving group; it is the one removed in elimination. |
| E2 = always Zaitsev | E2 = always anti | E2 requires anti-periplanar H/leaving-group alignment (stereoelectronic), but which alkene forms is governed by Zaitsev regiochemistry. |
| Dehydration of 3° alcohol | E2 dehydrohalogenation | Tertiary alcohol dehydration is E1-like (carbocation), so rearrangements are possible; E2 dehydrohalogenation is concerted and usually rearrangement-free. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine two train cars joined by a link, each carrying a passenger (an H and an X). An elimination is like a conductor pulling both passengers off at the same time: the two cars snap together with a tight double connection — the alkene — and the removed passengers walk away as a little molecule (HX or water). If there are two ways to pull the passengers off, the conductor chooses the one that connects the cars where they already have the most neighbors — the "more substituted" alkene.
Worked example
Example 1 — Predict the major alkene from 2-bromobutane with KOH/EtOH. The substrate is CH3–CHBr–CH2–CH3. Elimination can remove H from C1 (giving 1-butene, monosubstituted) or from C3 (giving 2-butene, CH3–CH=CH–CH3, disubstituted). By Zaitsev's rule the more substituted alkene is major:
CH3CHBrCH2CH3 KOH/EtOH⟶ CH3CH=CHCH3 (major: 2-butene)
1-Butene forms as the minor product. Both are reasonable elimination products; the stability difference (disubstituted > monosubstituted, Chapter 7 Topic 6) decides the ratio.
Example 2 — Dehydration of 2-methyl-2-butanol. The alcohol (CH3)2C(OH)–CH2–CH3 loses water with H2SO4 and heat. Which alkene is major? The carbocation formed is tertiary: (CH3)2C⁺–CH2–CH3. Proton loss can occur from a methyl group (giving 2-methyl-1-butene, CH2=C(CH3)–CH2–CH3, disubstituted) or from the CH2 (giving 2-methyl-2-butene, (CH3)2C=CH–CH3, trisubstituted). Zaitsev predicts the trisubstituted alkene:
(CH3)2C(OH)CH2CH3 H2SO4, Δ⟶ (CH3)2C=CHCH3 (major: 2-methyl-2-butene)
Example 3 — Stoichiometry of dehydration. How much water is produced when 50.0 g of 2-methyl-2-butanol (molar mass 88.15 g/mol) is dehydrated completely? The reaction gives 1 mol water per mol alcohol. Write the relation, then substitute:
n(alcohol) = 50.0 g88.15 g mol-1 = 0.567 mol
m(H2O) = 0.567 mol × 18.02 g mol-1 = 10.2 g
So 10.2 g of water is eliminated, and the remainder of the mass appears in the alkene products (about 39.8 g combined).
Example 4 — Dehalogenation of a vicinal dihalide. 1,2-Dibromocyclohexane (a vicinal dibromide) is treated with zinc dust. Write the outcome. Answer: Elimination of the two bromines gives cyclohexene:
C6H10Br2 + Zn → C6H10 + ZnBr2
The zinc removes both bromides as ZnBr2, and the ring carbons form the double bond.
Key takeaways
- Elimination = removal of two atoms/groups from adjacent carbons to form a π bond; the reverse of addition.
- Three routes: dehydrohalogenation (alkyl halide + strong base, e.g., KOH/EtOH), dehydration (alcohol + H2SO4/H3PO4, heat), dehalogenation (vicinal dihalide + Zn).
- Zaitsev's rule: the major alkene is the more substituted one (more stable).
- Dehydration goes through a carbocation (E1-like) for tertiary/secondary alcohols; rearrangements are possible.
- Dehydrohalogenation is typically concerted (E2), requiring anti-periplanar alignment of H and leaving group.
- Elimination competes with substitution; strong bulky bases + heat favor elimination.
- 2-Bromobutane + KOH/EtOH → mainly 2-butene (disubstituted), not 1-butene.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Define elimination and give the three standard routes for preparing alkenes.
Show answer
An elimination removes two atoms/groups from adjacent carbons to form a π bond. Routes: dehydrohalogenation (alkyl halide + strong base), dehydration (alcohol + acid/heat), dehalogenation (vicinal dihalide + Zn).
State Zaitsev's rule and explain the physical reason behind it.
Show answer
The major product of an elimination is the more substituted alkene. Reason: the more substituted alkene is more stable (hyperconjugation, Chapter 7 Topic 6), and the transition state for its formation is lower in energy, so it forms faster.
What is the major alkene from 2-chloropentane with KOH/EtOH? (Consider 1-pentene, 2-pentene.)
Show answer
2-Pentene (CH3–CH=CH–CH2–CH3) is major — it is disubstituted; 1-pentene is monosubstituted and minor.
Why is dehydration of a tertiary alcohol faster than dehydration of a primary alcohol?
Show answer
Tertiary alcohols form a stable tertiary carbocation in the rate-determining ionization step; primary alcohols would have to form a very unstable primary cation, so their dehydration requires harsher conditions and follows a different (often concerted) pathway.
What reagent converts a vicinal dihalide back into an alkene?
Show answer
Zinc metal (Zn), which removes the two halogens as ZnX2.
Why does strong, bulky base plus heat favor elimination over substitution?
Show answer
A strong bulky base abstracts the β-proton efficiently (favored for elimination), and heat supplies the energy for the bond-breaking process; substitution requires the nucleophile to attack the hindered carbon, which bulky bases avoid.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- elimination reaction
- A reaction that removes two atoms/groups from adjacent atoms, forming a multiple bond.
- dehydrohalogenation
- Elimination of H and X from an alkyl halide using a strong base.
- dehydration
- Elimination of water from an alcohol using acid and heat.
- dehalogenation
- Elimination of two halogens from a vicinal dihalide using Zn.
- Zaitsev's rule
- Elimination favors the more substituted alkene.
- β-carbon
- The carbon adjacent to the carbon bearing the leaving group.
- E2 mechanism
- Concerted elimination: base, H, and leaving group in one step.
- E1 mechanism
- Stepwise elimination: leaving group departs first to form a carbocation.
Sources & references
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