Consider the following sequence of reactions.

Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce

$\_\_\_\_$ mg of $B$.

(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{Br}: 80$ )

'A' is a neutral organic compound (M. F : $\mathrm{C}_8 \mathrm{H}_9 \mathrm{ON}$ ). On treatment with aqueous $\mathrm{Br}_2 / \mathrm{HO}^{(-)}$, ' A ' forms a compound ' B ' which is soluble in dilute acid. ' B ' on treatment with aqueous $\mathrm{NaNO}_2 / \mathrm{HCl}\left(0-5^{\circ} \mathrm{C}\right)$ produces a compound ' C ' which on treatment with $\mathrm{CuCN} / \mathrm{NaCN}$ produces ' D '. Hydrolysis of ' D ' produces ' E ' which is also obtainable from the hydrolysis of ' A '. ' E ' on treatment with acidified $\mathrm{KMnO}_4$ produces ' F '. ' F ' contains two different types of hydrogen atoms. The structure of ' A ' is

Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

An organic compound $(\mathrm{P})$ on treatment with aqueous ammonia under hot condition forms compound $(\mathrm{Q})$ which on heating with $\mathrm{Br}_2$ and KOH forms compound $(\mathrm{R})$ having molecular formula $\mathrm{C}_6 \mathrm{H}_7 \mathrm{~N}$. Names of P, Q and R respectively are.