Question

Iodobenzene(C6H5I)is prepared from aniline(C6H5NH2)in a two step process as shown below:

i)C6H5NH2+HNO2+HCl——->(C6H5N2^+Cl^-)+2H20

ii)(C6H5N2^+Cl^-)+KI————>C6H5I+N2+KCl

In the actual preparation, 9.30g of aniline was converted to 16.32g of iodobenzene. The percentage yield of iodobenzene is:

a)8% b)50% c)75% d)80%

Answer 1

Heya!

Option A is correct !!

✍️C6H5NH2+HNO2+HCl->(C6H5N2^+Cl^-)+2H20

In the above reaction , the percentage yield is 50%

Thank you!❤