Question

$Ph-N{H}_2\underset{0\text{-}{5}^{\circ }\mathrm{C}}{\overset{\mathrm{NaN}{\mathrm{O}}_2/HCl}{\to }}\underset{\text{dil.HCl}}{\overset{\mathrm{ph}\text{-}\mathrm{N}{\mathrm{H}}_2}{\to }}(X)$
Find the sum of number of nitrogen atoms present in $(X)$ and the total number of stereoisomers of $(X)$ formed

Answer 1

Answer: 5

$ph-N{H}_2\underset{0\text{-}{5}^{\circ }\mathrm{C}}{\overset{\mathrm{NaN}{\mathrm{O}}_2/HCl}{\to }}ph-\stackrel{+}{N_2}\stackrel{-}{\mathrm{Cl}}$

No. of N-atoms in X is 3
No. of stereoisomers = 2 (syn/anti)