- A$6$
- B$0$
- C$2$
- ✓$4$
The product formed has two chirality centres, therefore, four stereoisomers $\left(2^{2}\right)$ will exist.
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C{H_2} - C = CH - C{H_3}}
\end{array}$ $\xrightarrow[{Peroxide}]{{HBr}}$ $\mathop {\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
|
\end{array}} \\
{C{H_3} - C{H_2} - {}_*CH - \mathop {CH}\limits^{*\,} - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br}
\end{array}}\limits_{Major\,\,product} $
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| Ltst $I$ | List $II$ |
| $A$ $XeF _4$ | $I$ See-saw |
| $B$ $SF _4$ | $II$ Square planar |
| $C$ $NH _4^{+}$ | $III$ Bent T-shaped |
| $D$ $BrF _3$ | $IV$ Tetrahedral |
Choose the correct answer from the options given below: