See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 elimination of HBr requires removal of a beta-hydrogen by a base. The ease of elimination depends on the acidity of the beta-hydrogen (i.e., how readily it can be removed) and the stability of the transition state. Electron-withdrawing groups (EWGs) on the carbon bearing the beta-hydrogen increase the acidity of that hydrogen, making it easier to remove and thus facilitating elimination. Conversely, electron-donating groups or the absence of activating groups make the beta-hydrogen less acidic and less susceptible to elimination. Reasoning: - Option (a): Br-CH2-CH2-NO2. The nitro group (-NO2) is a very strong EWG. It strongly activates the beta-hydrogens (on the CH2 adjacent to NO2) toward removal by a base, making this compound highly susceptible to elimination. - Option (b): Br-CH2-CH2-CH3. The propyl group (-CH3) is electron-donating (alkyl group). There is no EWG to activate the beta-hydrogens. The beta-hydrogens are least acidic among all four options, making this compound least susceptible to elimination of HBr. - Option (c): Br-CH2-CH2-CN. The cyano group (-CN) is a strong EWG, activating beta-hydrogens toward elimination, making it more susceptible than (b). - Option (d): Br-CH2-CH2-CO2Et. The ester group (-CO2Et) is a moderate EWG, also activating beta-hydrogens toward elimination, making it more susceptible than (b). Among all options, only (b) has an electron-donating alkyl substituent, which does not activate the beta-hydrogens. All other options have EWGs that facilitate base-mediated elimination. Therefore, (b) Br-CH2-CH2-CH3 is least susceptible to elimination of hydrogen bromide. Therefore, the correct answer is B.