Answer:
\[\underset{\text{Propene}}{\mathop{{{H}_{3}}C-CH=C{{H}_{2}}+HBr}}\,\xrightarrow{Peroxide}\underset{\text{n-Propyl}\,\text{bromide}}{\mathop{{{H}_{3}}C-C{{H}_{2}}C{{H}_{2}}Br}}\,\]
Peroxide effect is effective
only in the case of \[HBr\]and not seen in the case of \[HCl\] and HI. This is
due to the following reasons:
(i) H?Cl bond (103 kcal/mol) is
stronger than H?Br bond(87 kcal/mol).
H?Cl bond is not decomposed by
the peroxide free radical, whereas the\[H-I\] bond is weaker (71 kcal/mol).
(ii) Iodine free radical \[({{I}^{^{\bullet
}}})\] formed as \[H-I\]bond is weaker but iodine free radicals readily
combine with each other to form iodine molecules rather attacking the double
bond.
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