11th Class Chemistry Hydrocarbons

  • question_answer 30) Addition of \[HBr\] to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

    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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