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Elimination Reactions and Saytzeff Rule

Elimination reaction of a haloalkane removes hydrogen halide from adjacent carbon atoms to form an alkene. Saytzeff rule states that the more substituted alkene is usually the major product in dehydrohalogenation.

In one line: Alcoholic KOH removes HX; Saytzeff selects the alkene with the richer double bond.

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Understand it

  • When a haloalkane is heated with alcoholic KOH or a strong base, a beta-hydrogen and halide are removed to form a carbon-carbon double bond.
  • E1 involves carbocation formation and is favoured by more substituted carbocations.
  • E2 is concerted and needs a suitable beta-hydrogen arrangement.
  • Product prediction depends on which beta-hydrogen is removed.

Example

2-bromobutane on heating with alcoholic KOH gives but-2-ene as the major product and but-1-ene as a minor product because but-2-ene is more substituted.

Rule to remember

Dehydrohalogenation pattern: R-CHX-CH2-R' + alcoholic KOH, heat → alkene + KX + H2O.

Saytzeff rule: the alkene with more alkyl groups attached to double-bonded carbons is generally major.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

For CH3-CHBr-CH2-CH3, the alpha carbon bears Br.

  1. 1Beta-hydrogens are on carbon 1 and carbon 3.
  2. 2Removing H from carbon 3 gives but-2-ene; removing H from carbon 1 gives but-1-ene.
  3. 3But-2-ene is more substituted, so it is major.

Method you can reuse

Find the carbon bearing halogen, identify adjacent beta-carbons with hydrogen, draw possible alkenes, compare substitution around the double bond, and choose the more substituted alkene as major.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

What is the major product when 2-bromobutane is heated with alcoholic KOH?

Think first, then show answer
But-2-ene is the major product according to Saytzeff rule.

Common wrong answer

Choosing but-1-ene as major from 2-bromobutane.

Why it loses marks: It is less substituted than but-2-ene.

Don't mix this up

Students often use aqueous KOH instead of alcoholic KOH. Aqueous KOH favours substitution to alcohol, while alcoholic KOH favours elimination to alkene.

Exam tip

For product prediction, locate beta-carbons first. Then form all possible alkenes and choose the more substituted alkene as the major product unless special conditions are stated.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Exams ask for major product prediction, reagent distinction between alcoholic and aqueous KOH, and explanation of Saytzeff orientation.
  1. 1

    Enough for 1 mark

    Elimination reaction of a haloalkane removes hydrogen halide from adjacent carbon atoms to form an alkene. Saytzeff rule states that the more substituted alkene is usually the major product in dehydrohalogenation.

  2. 2

    Add for 2 marks

    Example: 2-bromobutane on heating with alcoholic KOH gives but-2-ene as the major product and but-1-ene as a minor product because but-2-ene is more substituted.

  3. 3

    Add for 3 marks

    When a haloalkane is heated with alcoholic KOH or a strong base, a beta-hydrogen and halide are removed to form a carbon-carbon double bond.

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