Use these bond energies in kcal mol^-1: secondary C-H 98, H-Br 88, Br-Br 46, secondary C-Br 71. For the propagation steps of the radical bromination of propane at C2, (1) CH3CH2CH3 + Br^. -> CH3CH^.CH3 + HBr and (2) CH3CH^.CH3 + Br2 -> CH3CHBrCH3 + Br^., the values of Delta H for step (1) and step (2) and the overall propagation cycle are respectively

JEE Advanced levelSingle correct optionDifficulty 3 of 3g3149

Use these bond energies in $kcal mo l^{- 1}$ : secondary C-H 98, H-Br 88, Br-Br 46, secondary C-Br 71. For the propagation steps of the radical bromination of propane at C2, (1) $CH X_{3} CH X_{2} CH X_{3} + Br X^{∙} CH X_{3} CH X^{∙} CH X_{3} + HBr$ and (2) $CH X_{3} CH X^{∙} CH X_{3} + Br X_{2} CH X_{3} CHBrCH X_{3} + Br X^{∙}$ , the values of $Δ H$ for step (1) and step (2) and the overall propagation cycle are respectively

Step 1:

98 - 88 = + 10

(breaks C-H, forms H-Br). Step 2:

46 - 71 = - 25

(breaks Br-Br, forms C-Br). The sum is

+ 10 - 25 = - 15

kcal/mol. The slow endothermic first step is the selectivity-determining step.