| Element | Percentage | Mole | Mole ratio |
| $C$ | $85.8$ | $\frac{85.8}{12}=7.15$ | $1$ |
| $H$ | $14.2$ | $\frac{14.2}{1}=14.2$ | $2$ |
Empirical formula $\left( CH _2\right)$
$14 \times n=84$
$n =6$
$\therefore$ Molecular formula $C _6 H _{12}$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{H_3}C - C - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}{\mkern 1mu} \mathop A\limits_{Major\,product} \, + \,\mathop B\limits_{Minor\,product} $
$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\text { (iii) } \mathrm{HBr}{(iv) \mathrm{Mg}, ether, then \mathrm{HCHO} / \mathrm{H}_3 \mathrm{O}^{+}}]{{(i)BH_3}{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2,{ }^{\text {(-) }} \mathrm{OH}}} \mathrm{A}$

ઉપરોક્ત પ્રકિયા માં નીપજ $(B)$ શું હશે ?