See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 elimination rate depends on the accessibility of the beta-hydrogen atoms and the stability of the transition state. In E2 reactions, the rate is governed by the number of available beta-hydrogens and the ease with which the base can abstract them, as well as steric factors around the carbon bearing the leaving group. Structure identification: (a) 2-bromo-2-methylpropane (tertiary alkyl bromide): CH3-C(Br)-CH3 with an additional CH3, i.e., (CH3)3CBr — actually from the image it is CH3-C(Br)-CH3 with two CH3 groups, making it a tertiary bromide (tert-butyl bromide). It has 9 beta-hydrogens but is highly sterically hindered around the central carbon. (b) 2-bromopropane (secondary alkyl bromide): CH3-CH(Br)-CH3. It has 6 beta-hydrogens and moderate steric hindrance. (c) 1-bromopropane... wait — CH3-CH-Br appears to be 1-bromoethane or ethyl bromide based on the image (CH3-CH2Br, a primary bromide). It has fewer beta-hydrogens and least steric hindrance. Step-by-step reasoning: 1. E2 reaction rate increases with increasing substitution of the substrate because: more beta-hydrogens are available, and the transition state is stabilized by hyperconjugation/alkene stability. 2. Tertiary > Secondary > Primary in E2 reactivity under normal (non-bulky base) conditions because the greater number of beta-H atoms and more stable alkene product formed. 3. Therefore: (a) tertiary > (b) secondary > (c) primary, giving the order a > b > c. Why other options fail: - Option (a) c > b > a: This is the SN2 order, not E2. - Option (c) b > a > c: Incorrect; tertiary should be faster than secondary in E2. - Option (d) c > a > b: Incorrect ordering entirely. Therefore, the correct answer is B.