Question

In the series of reactions: $ {{C}_{6}}{{H}_{5}}N{{H}_{2}}\xrightarrow[0.5\,{{\,}^{o}}C]{NaN{{O}_{2}}/HCl}X\xrightarrow[C{{H}_{2}}]{HN{{O}_{2}}}Y+{{N}_{2}}+HCl $ X and Y are, respectively:

• A. $ {{C}_{6}}{{H}_{5}}-N=N-{{C}_{6}}{{H}_{5}},{{C}_{6}}{{H}_{5}}N_{2}^{\oplus }C{{l}^{-}} $
• B. $ {{C}_{6}}{{H}_{5}}N_{2}^{\oplus }C{{l}^{-}},{{C}_{6}}{{H}_{5}}-N=N-{{C}_{6}}{{H}_{5}} $
• C. $ {{C}_{6}}{{H}_{5}}N_{2}^{\oplus }C{{l}^{-}},{{C}_{6}}{{H}_{5}}N{{O}_{2}} $
• D. $ {{C}_{6}}{{H}_{5}}N{{O}_{2}},{{C}_{6}}{{H}_{6}} $

Answer 1

Answer: (C)

Primary amine groups when reacts with $ \text{NaN}{{\text{O}}_{\text{2}}}\text{/HCl} $ or $ \text{HN}{{\text{O}}_{2}} $ at $ \text{0}{{\,}^{\text{o}}}\text{C,} $ it undergoes diazotisation and forms diazonium salt $ (-N=N-). $ So, the compound X will be $ {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}=\text{N}-\text{Cl} $ Benzene diazonium chloride) When benzene diazonium chloride reacts with $ \text{HN}{{\text{O}}_{\text{2}}} $ in presence of HCHO nitrobenzene is obtained. The reactions are as under: $ {{C}_{6}}{{H}_{5}}N{{H}_{2}}\xrightarrow[0-5{{\,}^{o}}C]{NaN{{O}_{2}}/HCl}\underset{Benzene\,diazonium\,chloride}{\mathop{{{C}_{6}}{{H}_{5}}N=NCl}}\,\, $ $ \xrightarrow[C{{H}_{2}}O]{HN{{O}_{2}}}{{C}_{6}}{{H}_{5}}N{{O}_{2}}+{{N}_{2}}+HCl $