Question

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 %

IIT/JEE

Chemistry

Solution

168

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[ 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} ]