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Elimination Reactions and Saytzeff Rule

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
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
Then show how to use it
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.
3
Add one concrete 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.
4
Avoid this incomplete answer
Choosing but-1-ene as major from 2-bromobutane is wrong because it is less substituted than but-2-ene.

Definition

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.

Memory hook

Alcoholic KOH removes HX; Saytzeff selects the alkene with the richer double bond.

Examples and method

Worked example

For CH3-CHBr-CH2-CH3, the alpha carbon bears Br. Beta-hydrogens are on carbon 1 and carbon 3. Removing H from carbon 3 gives but-2-ene; removing H from carbon 1 gives but-1-ene. But-2-ene is more substituted, so it is major.

Method to apply

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.

Diagram support

A beta-hydrogen marking diagram can help students see which hydrogens may be removed, but the rule can be applied from structures.

How CBSE asks it

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

Avoid common mistakes

Common confusion

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

Common wrong answer

Choosing but-1-ene as major from 2-bromobutane is wrong because it is less substituted than but-2-ene.

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.

Quick check

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

But-2-ene is the major product according to Saytzeff rule.

Answer writing and exam use

1-mark answer

2-mark answer

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. 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. 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-mark answer

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. 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. For CH3-CHBr-CH2-CH3, the alpha carbon bears Br. Beta-hydrogens are on carbon 1 and carbon 3. Removing H from carbon 3 gives but-2-ene; removing H from carbon 1 gives but-1-ene. But-2-ene is more substituted, so it is major. Exams ask for major product prediction, reagent distinction between alcoholic and aqueous KOH, and explanation of Saytzeff orientation. Choosing but-1-ene as major from 2-bromobutane is wrong because it is less substituted than but-2-ene.

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