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Organic Chemistry 1 · Substitution and Elimination

Preparation of Alkenes via Elimination

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Alkenes are prepared by elimination: E2 removes HX from an alkyl halide with a strong base, while E1 acid-catalyzed removes water from an alcohol. The predicts the more substituted alkene as the major product; the predicts the less substituted alkene with bulky bases or poor leaving groups. Carbocation rearrangements occur in E1 and can change the product skeleton.

Why this matters

Dehydration of alcohols is a route to small alkenes used industrially, and elimination chemistry underlies double-bond formation in pharmaceutical synthesis. Choosing E2 versus E1 conditions determines which alkene isomer forms — and therefore the three-dimensional shape and activity of a downstream drug fragment — so reading a synthetic route correctly depends on these regiochemical rules.

The college version

1. Dehydrohalogenation (E2)

An alkyl halide reacts with a strong base (for example, OH⁻ or OR⁻) to eliminate HX and form an alkene in one concerted step. The base removes a β-hydrogen while the halide leaves; the rate depends on both the substrate and the base. The hydrogen and the leaving group must be anti-periplanar.

2. Dehydration of Alcohols (E1, acid-catalyzed)

An alcohol is protonated by acid, converting the poor leaving group −OH into the good leaving group water. Water departs to give a carbocation, which then loses a β-hydrogen to form the alkene. This is E1: two steps, a carbocation intermediate, and a rate that depends only on the alcohol.

3. Zaitsev, Hofmann, and Rearrangements

Zaitsev's rule: the major product is the more substituted (more stable) alkene. Hofmann's rule: with bulky bases or poor leaving groups, the less substituted alkene can predominate. In E1, carbocations rearrange (hydride or alkyl shifts) to more stable carbocations before deprotonation, changing the product skeleton.

How it works

  1. Choose the starting material: an alkyl halide (for E2) or an alcohol (for E1).
  2. Select conditions: a strong base for E2; acid plus heat for E1.
  3. Identify all β-hydrogens and the possible alkene products.
  4. Apply Zaitsev (or Hofmann for bulky bases) and watch for E1 rearrangements.
  5. State the major product and the minor byproducts.

Common confusions

Do not confuseWithDifference
E2E1E2 is concerted with a strong base; E1 is stepwise via a carbocation
ZaitsevHofmannZaitsev = more substituted; Hofmann = less substituted
DehydrationDehydrohalogenationAn alcohol loses water; a halide loses HX
E1 rearrangementE2 rearrangementE1 rearranges; E2 does not
Zaitsev ruleA mechanistic certaintyIt is a predictive shortcut, not the mechanism itself

Memory aids

"Zaitsev Zaps in the most substituents; Hofmann Hesitates with bulky bases." More substitution = Zaitsev.

Quick review

Topic Recap

Alkenes are prepared by E2 dehydrohalogenation of alkyl halides (strong base, concerted) and by E1 acid-catalyzed dehydration of alcohols (carbocation intermediate). Zaitsev predicts the more substituted alkene; Hofmann predicts the less substituted one with bulky bases. E1 carbocations rearrange. Choosing the substrate and conditions is the core of alkene .

Knowledge Check

  1. What conditions favor E2 over E1?
  2. Predict the major product of dehydrating 2-methylbutan-2-ol.
  3. Why must an alcohol be protonated before dehydration?
  4. Does E2 or E1 involve a ?
  5. When does the Hofmann product predominate?

Answers and Rationales

  1. A strong base (for example, OH⁻ or OR⁻) in a polar aprotic solvent.
  2. 2-methylbut-2-ene — the Zaitsev, more substituted alkene.
  3. −OH is a poor leaving group; protonation converts it to water, a good leaving group.
  4. E1 — the carbocation intermediate can rearrange; E2 is concerted and cannot.
  5. With bulky bases or poor leaving groups, where steric/electronic factors favor the less substituted alkene.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of an alkene as a double bond you build by removing two small "blocks" — a hydrogen and a leaving group (a halogen or an OH) — from adjacent carbons. In E2, a strong base grabs the hydrogen and the leaving group departs in one smooth, single motion. In E1, the leaving group leaves first, creating a charged, unstable carbocation that only later loses a hydrogen.

A comparison: E2 is like two people exiting through the same door at the same moment (concerted); E1 is like one person leaving first and the second following later (stepwise).

Where it stops being exact: the "remove two blocks" picture hides the ordering and geometry. In E2 it is concerted, and the departing hydrogen and leaving group must be anti-periplanar. In E1 it is stepwise, and the carbocation intermediate can rearrange to a more stable carbocation before losing a proton — a possibility the simple picture omits.

Simple Example

Dehydrohalogenation of 2-bromobutane with a strong base gives but-2-ene (Zaitsev, the more substituted alkene) as the major product and but-1-ene as the minor product.

Worked example

E2 dehydrohalogenation of 2-bromobutane.

  1. A double-headed curved arrow shows the base's lone pair moving to a β-hydrogen, forming a new O–H bond. Simultaneously, the C–H σ-bond electrons move (second curved arrow) to form the C=C π bond, and the C–Br σ-bond electrons leave with bromine as Br⁻ (third curved arrow).
  2. All bond changes happen in one concerted step. The H and Br must be anti-periplanar.
  3. Two β-hydrogens are available: removal at C1 gives but-1-ene (monosubstituted); removal at C3 gives but-2-ene (disubstituted).
  4. The more substituted but-2-ene is the Zaitsev major product.

E1 dehydration of 3-methylbutan-2-ol.

  1. Protonate the OH group to give −OH2⁺, a good leaving group.
  2. Water departs, leaving a secondary carbocation.
  3. A hydride (or methyl) shift converts the secondary carbocation into a more stable tertiary carbocation (rearrangement).
  4. Deprotonation at a β-carbon yields the more substituted alkene, 2-methylbut-2-ene (Zaitsev), with a rearranged skeleton.

Key takeaways

  • High yield: Zaitsev = the more substituted (more stable) alkene is major.
  • High yield: E2 is concerted; E1 proceeds through a carbocation.
  • High yield: Alcohols need acid protonation to turn OH into a good leaving group (water).
  • High yield: E1 carbocations rearrange; E2 does not rearrange.
  • Bulky bases (for example, tert-butoxide) give the Hofmann (less substituted) product.
  • Dehydration of secondary and tertiary alcohols favors E1; primary alcohols tend toward E2-like pathways.
  • Zaitsev and Hofmann are prediction tools — verify them with mechanism analysis.

Keep learning

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Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Describe E2 dehydrohalogenation of alkyl halides and E1 acid-catalyzed dehydration of alcohols.
  • Apply the Zaitsev and Hofmann rules to predict major elimination products.
  • Predict carbocation rearrangements in E1 dehydration and explain their effect on the product.
  • Plan simple syntheses of alkenes by choosing an appropriate substrate and reaction conditions.

Key vocabulary

Dehydrohalogenation
E2 removal of HX from an alkyl halide
Dehydration
Acid-catalyzed loss of water from an alcohol (E1)
Zaitsev rule
More substituted alkene is major
Hofmann rule
Less substituted alkene favored (bulky base/poor LG)
Carbocation rearrangement
Hydride/alkyl shift to a more stable cation
Substrate dependence
Substitution level determines E2 vs E1 behavior
Product distribution
Mixture of alkene isomers formed
Synthesis planning
Choosing substrate and conditions for a target alkene

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