NEET 2021 Chemistry Question 68 Solution

When 2-bromopentane is heated with alcoholic potassium hydroxide ($alc.\; KOH$), base-induced $\beta$-elimination of $HBr$ occurs across the $C_\alpha-C_\beta$ bond:

- Pathway A (Loss of $\beta_1$-H): Yields pent-1-ene ($CH_3CH_2CH_2CH=CH_2$), which has only 1 alkyl substituent and 2 hyperconjugative $\alpha$-hydrogens (19% yield, minor). - Pathway B (Loss of $\beta_2$-H): Yields pent-2-ene ($CH_3CH_2CH=CHCH_3$), which has 2 alkyl substituents and 5 hyperconjugative $\alpha$-hydrogens (81% yield, major).

Alexander Zaitsev (Saytzeff) formulated in 1875 that: *"In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms."* This is Saytzeff's Rule, corresponding to Option (1).