A) Elimination
B) Addition
C) Displacement
D) Rearrangement
Correct Answer: B
Solution :
e.g. \[C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\to \underset{\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}-C{{H}_{2}}}}\,\] Half of the double bond is broken. It means \[\pi \]bond is broken while sigma bond is retained also two new \[C-Br\] bonds are formed.
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