β-Elimination and the Saytzeff ruleβ-विलोपन और सेज़ैफ़ नियमβ-Elimination and the Saytzeff rule

Class 12 Chemistry · Chapter 6 · interactive diagram

Parts and what they mean

• Haloalkane with a β-hydrogen (β-हाइड्रोजन वाला हैलोऐल्केन) — α-carbon carries X; the adjacent carbon is β. Only haloalkanes with a β-H can undergo this elimination.
• Base (e.g. OH⁻ from alcoholic KOH) (क्षारक (जैसे ऐल्कोहॉली KOH का OH⁻)) — Pulls off the β-hydrogen as a proton.
• Electron flow in elimination (विलोपन में इलेक्ट्रॉन प्रवाह) — Base lone pair → β-H; C–H bond pair → new C=C π bond; C–X bond pair → X (leaves as X⁻). Drawn for the common one-step pathway; NCERT gives the overall change.
• Alkene formed (बनी ऐल्कीन) — A C=C double bond forms between the α and β carbons.
• By-products (उप-उत्पाद) — The base picks up H (B–H, e.g. H₂O) and the halide ion leaves.
• 2-Bromopentane: two kinds of β-H (2-ब्रोमोपेन्टेन: दो प्रकार के β-H) — β-H on C-3 (the CH₂) and on C-1 (the CH₃) → two possible alkenes.
• Alcoholic KOH, heat (ऐल्कोहॉली KOH, गर्म) — Alcoholic (not aqueous) KOH favours elimination.
• Pent-2-ene — 81% (major) (पेन्ट-2-ईन — 81% (मुख्य)) — H removed from C-3; the C=C carries two alkyl groups → more substituted, preferred (NCERT).
• Pent-1-ene — 19% (minor) (पेन्ट-1-ईन — 19% (गौण)) — H removed from C-1; the C=C carries only one alkyl group.
• Saytzeff (Zaitsev) rule, 1875 (सेज़ैफ़ (ज़ैतसेव) नियम, 1875) — In dehydrohalogenation, the preferred product is the alkene with the greater number of alkyl groups on the doubly bonded carbons.

Original diagram drawn by Simple Learning Hub, based on the NCERT textbook. Report a mistake