- AElectrophilic substitution reactions
- BNucleophilic substitution reactions
- ✓Electrophilic addition reactions
- DNucleophilic addition reactions
$C{H_3} - CH = C{H_2} + {H^ + } \xrightarrow{Slow} \mathop {C{H_3} - \mathop C\limits^ + H - C{H_3}}\limits_{{2^o}{\rm{ carbonium ion}}} $
$C{H_3} - \mathop {{\text{ }}C}\limits^ + H - C{H_3} + B{r^ - }\xrightarrow{{Fast}}$ $\mathop {\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}} \\
{|\,\,\,\,} \\
{\,Br\,\,\,}
\end{array}}\limits_{2 - Bromopropane} $
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Reason : Boron shows metallic nature.
Which combination gives the highest yield of $z,$ at equilibrium.

Given that $\left[ {\frac{{{{\left[ B \right]}_t}}}{{{{[C]}_t}}} = \frac{{16}}{9}} \right]$
$A\,\xrightarrow{{{K_1}\, = \,2\, \times \,{{10}^{^{ - 3}\,}}{S^{ - 1}}}}4B$
$A\to C$
$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$