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) \[{{C}_{6}}{{H}_{5}}N{{O}_{2}},{{C}_{6}}{{H}_{6}}\]
Correct Answer: C
Solution :
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: \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\xrightarrow[0-5{{\,}^{o}}C]{NaN{{O}_{2}}/HCl}\underset{Benzene\,diazonium\,chloride}{\mathop{{{C}_{6}}{{H}_{5}}N=NCl}}\,\,\] \[\xrightarrow[C{{H}_{2}}O]{HN{{O}_{2}}}{{C}_{6}}{{H}_{5}}N{{O}_{2}}+{{N}_{2}}+HCl\]
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