The total number of alkyl bromides (including stereoisomers) formed in the reaction ${\mathrm{Me}}_3\mathrm{C}-\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$ will be

$\left[\mathrm{P}\right]\stackrel{{\mathrm{Br}}_2}{⟶}{\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Br}}_2\frac{{\mathrm{NaNH}}_2}{{\mathrm{NH}}_3}\left[\mathrm{Q}\right]$

$\left[\mathrm{Q}\right]\frac{20\%{\mathrm{H}}_2{\mathrm{SO}}_4}{{\mathrm{Hg}}^{2+},\mathrm{\Delta }}\left[\mathrm{R}\right]\stackrel{\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}}{⟶}\left[\mathrm{S}\right]$

The species $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ respectively are:

The major products obtained in the following reaction is/are