A) Allyl bromide
B) n-propyl bromide
C) Isopropyl bromide
D) 3-bromo propane
Correct Answer: B
Solution :
The formation of n-propyl bromide in presence of peroxide can be explained as follows. Step-1: Peroxide undergo fission to give free radicals \[R-O-O-R\to 2-R-\dot{O}\] Step-2 : \[HBr\] combines with free radical to form bromine free radical \[R-\dot{O}+HBr\to R-OH+B\dot{r}\] Step-3 : \[B\dot{r}\] attacks the double bond of the alkene to form a more stable free radical \[C{{H}_{3}}CH=C{{H}_{2}}+B\dot{r}<\begin{matrix} \underset{\text{(more stable)}}{\mathop{C{{H}_{3}}-\dot{C}H-C{{H}_{2}}Br}}\, \\ {} \\ \underset{\text{(less stable)}}{\mathop{C{{H}_{3}}-\overset{Br\,}{\mathop{\overset{|}{\mathop{C}}\,H}}\,-\dot{C}{{H}_{2}}}}\, \\ \end{matrix}\] Step-4 : More stable free radical attacks the \[HBr\] \[C{{H}_{3}}-\dot{C}H-C{{H}_{2}}-Br+HBr\to \underset{\text{n-propyl bromide}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,+B\dot{r}\] Step-5 : \[B\dot{r}+B\dot{r}\to B{{r}_{2}}\]
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