A compound $$'A'$$ when treated with \[HN{{O}_{3}}\] ( in presence of \[{{H}_{2}}S{{O}_{4}}\] ) gives compound $$'B',$$ which is then reduced with $$Sn$$ and $$HCl$$ to aniline ? The compound $$'A'$$ is
A.
toluene
B.
benzene
C.
ethane
D.
acetamide
Answer :
benzene
Solution :
\[A\xrightarrow{HN{{O}_{3}}/{{H}_{2}}S{{O}_{4}}}B\xrightarrow{Sn/HCl}{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]
This indicates that $$B$$ is \[{{C}_{6}}{{H}_{5}}N{{O}_{2}}\] and hence $$A$$ is \[{{C}_{6}}{{H}_{6}}\]