MCQ
$HCl$ does not show peroxide effect since
$C{{H}_{3}}-CH=C{{H}_{2}}+\overset{\centerdot }{\mathop{X}}\,\to C{{H}_{3}}-\overset{\centerdot }{\mathop{C}}\,H-C{{H}_{2}}X$ $(I)$
$C{{H}_{3}}-\overset{\centerdot \,\,\,\,}{\mathop{CH}}\,-C{{H}_{2}}X+H-X\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}X+\overset{\centerdot }{\mathop{X}}\,$ $(II)$
$C{{H}_{3}}-CH=C{{H}_{2}}+\overset{\centerdot }{\mathop{X}}\,\to C{{H}_{3}}-\overset{\centerdot }{\mathop{C}}\,H-C{{H}_{2}}X$ $(I)$
$C{{H}_{3}}-\overset{\centerdot \,\,\,\,}{\mathop{CH}}\,-C{{H}_{2}}X+H-X\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}X+\overset{\centerdot }{\mathop{X}}\,$ $(II)$
- A$(I)$ and $(II)$ both steps are endothermic
- ✓$(I)$ is exothermic and $(II)$ is endothermic
- C$(I)$ is endothermic and $(II)$ is exothermic
- D$I$ and $(II)$ both are exothermic steps
