Answer:
Addition of \[HBr\] to propene is an
ionicelectrophilic addition reaction, in which the electrophile in the step
1 is\[HBr\].
Step 1.\[HBr\to
\underset{Attacking\,electrophile}{\mathop{{{H}^{+}}}}\,+_{\centerdot
}^{\centerdot }B{{r}^{-}}\]
Step 2. The proton \[({{H}^{+}})\]
attacks the \[\pi \]-bond to give a stable carbocation.
Step 3. The
nucleophile bromide ion attacks the more stable \[2{}^\circ \]carbocation to
give 2-bromopropane(major product).
\[C{{H}_{3}}-\overset{+}{\mathop{CH}}\,-C{{H}_{3}}+_{\centerdot
}^{\centerdot }B{{r}^{-}}\xrightarrow{Fast}C{{H}_{3}}-\underset{\underset{\underset{2-Bromopropane}{\mathop{\text{Br}}}\,}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}\]
In presence of benzoylperoxide, the reaction is
freeradical electrophilic addition reaction, in which the electrophile here
is \[B{{r}^{\centerdot }}\] free radical which is obtained by the action of
benzoylperoxide on\[HBr\].
Step 1. Peroxide
undergoes fission to give free
radicals (Initiation).
\[{{C}_{6}}{{H}_{5}}CO-O-O-CO{{C}_{6}}{{H}_{5}}--\xrightarrow[(Homolytic\,Fission)]{Heat}\] \[\underset{\begin{smallmatrix}
Benzoyl\,radical \\
(unstable)
\end{smallmatrix}}{\mathop{2{{C}_{6}}{{H}_{5}}CO{{O}^{\centerdot
}}}}\,\to \underset{Phenyl\,free\,readical}{\mathop{[2\overset{\centerdot
}{\mathop{{{C}_{6}}{{H}_{5}}}}\,+2C{{O}_{2}}]}}\,\]
Step 2.\[~HBr\]
combines with free radical to formbromine free radical.
Step 3. Br attacks the
double bond of alkene to
form a more stable free radical (Propagation).
Step 4. More stable
free radical attacks the \[HBr\](Termination).
\[\overset{\centerdot }{\mathop{Br}}\,+\overset{\centerdot
}{\mathop{Br}}\,\to B{{r}_{2}}\]
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