Dehydration of alcohols: mechanism and conditionsऐल्कोहॉल का निर्जलीकरण: क्रियाविधि और परिस्थितियाँDehydration of alcohols: mechanism and conditions
Class 12 Chemistry · Chapter 7 · interactive diagram
Parts and what they mean
- Proton from the acid (H₂SO₄ or H₃PO₄) (अम्ल (H₂SO₄ या H₃PO₄) से प्रोटॉन) — Protic acid used as catalyst; anhydrous ZnCl₂ or alumina also work.
- Step 1 arrow: lone pair of O → H⁺ (पद 1 तीर: O का एकाकी युग्म → H⁺) — Oxygen uses a lone pair to bind the proton (fast, reversible).
- Protonated alcohol (ethyloxonium ion) (प्रोटॉनित ऐल्कोहॉल (एथिलऑक्सोनियम आयन)) — Now –OH₂⁺ is a good leaving group (leaves as water).
- Step 2 arrow: C–O bond pair moves onto O (पद 2 तीर: C–O बंध युग्म O पर जाता है) — Water leaves taking both electrons; this is the slowest (rate-determining) step.
- Carbocation CH₃CH₂⁺ (कार्बधनायन CH₃CH₂⁺) — Tertiary carbocations are more stable and form more easily — so 3° alcohols dehydrate most easily.
- H on the carbon next to C⁺ (C⁺ के पास वाले कार्बन पर H) — This H is lost as H⁺.
- Step 3 arrow: C–H bond pair forms the C=C π bond (पद 3 तीर: C–H बंध युग्म C=C π बंध बनाता है) — The acid used in step 1 is released in step 3; ethene is removed as it forms to drive the equilibrium right.
- Ethene (alkene product) (एथीन (ऐल्कीन उत्पाद)) — Loss of one water molecule overall.
- Primary: harsh conditions (प्राथमिक: कठोर परिस्थितियाँ) — Ethanol needs conc. H₂SO₄ at 443 K.
- Secondary: milder (द्वितीयक: मृदु) — Propan-2-ol with 85% H₃PO₄ at 440 K gives propene.
- Tertiary: mildest (तृतीयक: सबसे मृदु) — 2-Methylpropan-2-ol with 20% H₃PO₄ at 358 K gives 2-methylpropene.
Original diagram drawn by Simple Learning Hub, based on the NCERT textbook. Report a mistake