Chapter 1: Problem 112
Iodobenzene is prepared from aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) in a two-step process as shown here: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HNO}_{2}+\mathrm{HCl} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+2 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+\mathrm{KI} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}+\mathrm{N}_{2}+\mathrm{KCl}\) In an actual preparation, \(9.30 \mathrm{~g}\) of aniline was converted to \(16.32 \mathrm{~g}\) of iodobenzene. The percentage yield of iodobenzene is \((\mathrm{I}=127)\) (a) 8\% (b) \(50 \%\) (c) \(75 \%\) (d) \(80 \%\)
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