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SAYTZEFF'S RULE for elimination reactions states that

Athe most substituted alkene (Zaitsev product) is the major product
Bthe LEAST substituted alkene (Hofmann product) is the major product
Celimination is always faster than substitution
Delimination only occurs at terminal carbons
Answer & Solution
Correct answer: A. the most substituted alkene (Zaitsev product) is the major product
1. NCERT §6.5 (Elimination reactions): when there's a CHOICE of which $\beta$-H to remove (giving different alkene products), the more SUBSTITUTED alkene predominates. 2. Why? More substituted alkenes are MORE STABLE (alkyl groups donate electrons by hyperconjugation, stabilising the double bond). 3. Example: 2-bromobutane $\mathrm{(CH_3)CHBr(CH_2CH_3)}$ + alcoholic KOH → mostly but-2-ene ($\mathrm{CH_3CH{=}CHCH_3}$, more substituted) rather than but-1-ene ($\mathrm{CH_2{=}CHCH_2CH_3}$, less substituted). 4. Named after Russian chemist Alexander Saytzeff (Zaitsev) who proposed it in 1875. 5. Option B describes the COUNTER rule (Hofmann's, which applies with bulky bases like $t$-butoxide). Option C is irrelevant. Option D is wrong physical chemistry. _Source: NCERT Class 12 Chemistry Part 2, Ch 6, §6.5 (Elimination — Saytzeff's rule), p. 13._
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