Question
When 2-butanol is heated with concentrated H₂SO₄ at 170°C, what is the major product? Explain why the more substituted alkene is preferred (Zaitsev’s rule). Also state the order of ease of dehydration: 1° > 2° > 3° or 3° > 2° > 1°?
(JEE Main 2023, similar pattern)
Solution — Step by Step
The major product is but-2-ene (more substituted alkene), not but-1-ene.
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Protonation: The -OH group gets protonated by H₂SO₄ to form a good leaving group ().
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Loss of water: Water leaves to form a secondary carbocation ().
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Proton loss: A proton is lost from an adjacent carbon. Two options exist — loss from C3 gives but-2-ene, loss from C1 gives but-1-ene.
Zaitsev’s rule: In elimination reactions, the major product is the more substituted (more stable) alkene — the one formed by removing a proton from the carbon with fewer hydrogens.
But-2-ene is a disubstituted alkene (2 alkyl groups on C=C). But-1-ene is monosubstituted. Hyperconjugation from adjacent C-H bonds stabilises the more substituted alkene, making it thermodynamically more stable and kinetically preferred in E1 reactions.
This follows carbocation stability. Tertiary alcohols form stable 3° carbocations easily, so they dehydrate fastest — sometimes even at temperatures as low as 85°C with dilute acid. Primary alcohols need harsh conditions (conc. H₂SO₄, 170°C) because 1° carbocations are too unstable; they often undergo E2 instead.
Why This Works
The E1 mechanism for dehydration proceeds through a carbocation intermediate. The stability of this intermediate controls the reaction rate — hence the 3° > 2° > 1° order.
Once the carbocation forms, it loses a proton to give the alkene. The proton that’s lost determines which alkene forms. The more substituted alkene has more hyperconjugative stabilisation (more -hydrogens on neighbouring carbons donating electron density into the empty p-orbital of the bond), making it the lower-energy product.
In but-2-ene, the double bond has alkyl groups on both sides, maximising hyperconjugation. This translates to real energy differences: trans-but-2-ene is about 7 kJ/mol more stable than but-1-ene.
Alternative Method
For quick product prediction: identify all possible elimination products by removing -OH and an H from adjacent carbons. The product with the most substituted double bond wins.
Zaitsev’s rule applies to E1 and most E2 reactions. The exception is when a bulky base like (CH₃)₃CO⁻ (tert-butoxide) is used — then the Hofmann product (less substituted alkene) becomes major because the base can’t reach the more sterically hindered proton. JEE loves this exception.
Common Mistake
Students sometimes confuse the conditions for dehydration vs substitution. With concentrated H₂SO₄ at 170°C, you get elimination (alkene). At 140°C, you get substitution to form ethers (for primary alcohols). The temperature difference matters — higher temperature favours elimination because it has higher activation energy and is entropically favoured (more molecules in the product).