Question

[
C_{6} H_{5} N H_{2}+H N O_{2}+H C l rightarrow C_{6} H_{5} N_{2}^{+} e l+2 H_{2}
]
( c_{6} H_{5} N_{2}+C l+H I quad longrightarrow quad C_{6} H_{5} I+N_{2}+H E X )
( C_{6} H_{5} N H_{2}+H N O_{2}+H I rightarrow C_{6} H_{5} I+N_{2}+2 H_{2} O )
( 93-139 m )
( 204 mathrm{gm} )
( 93 cdot 13 mathrm{gm} ) aniline reacts to give
( 204 mathrm{gm} )
Iodobenzene
93 gm aniline ( rightarrow 204 mathrm{gm} )
Iodo
benzene 9.3 gm anitine ( rightarrow x )
( x=20.4 quad ) gh
Iodobenzene
But got only ( 16 cdot 32 ) g Iodobersene
obtained
[
begin{array}{l}
=frac{16 cdot 32}{20 cdot 40} times 100
=80 %
end{array}
]

# lodobenzene (CHF) is prepared from aniline (CH NH) in a two step process as shown below CH NH2 + HNO2 + HCI → CHEN, CF + 2H,0 CH, N, CF + KI → CHI + N + KCI In an actual preparation 9.30 g of aniline was converted to 16.32 g of iodobenzene. The percentage yield of iodobenzene is : (A) 8 % (B) 50 % 2015 % (D) 80 %

Solution