MCQ
In a reaction, if half of the double bond is broken and two new bonds are formed, this is a case of
- AElimination
- ✓Addition
- CDisplacement
- DRearrangement
✓
Answer
Correct option: B.
Addition
b
(b) e.g. $C{H_2} = C{H_2} + B{r_2} \to \begin{array}{*{20}{c}}
{C{H_2} - C{H_2}} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,Br\,\,\,\,\,}
\end{array}$
(b) e.g. $C{H_2} = C{H_2} + B{r_2} \to \begin{array}{*{20}{c}}
{C{H_2} - C{H_2}} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,Br\,\,\,\,\,}
\end{array}$
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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