'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.
A hydrocarbon ' P ' $\left(\mathrm{C}_4 \mathrm{H}_8\right)$ on reaction with HCl gives an optically active compound ' Q ' $\left(\mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}\right)$ which on reaction with one mole of ammonia gives compound ' $\mathrm{R}^{\prime}\left(\mathrm{C}_4 \mathrm{H}_{11} \mathrm{~N}\right)$. ' $\mathrm{R}^{\prime}$ on diazotization followed by hydrolysis gives ' S '. Identify $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and S .
JEE Main Subjects
Browse all chapters by subject







