Nitrobenzene is reduced to aniline either by catalytic hydrogenation, C6H5NO2 + 3H2 -> C6H5NH2 + 2H2O, or with tin and hydrochloric acid, 2C6H5NO2 + 3Sn + 12HCl -> 2C6H5NH2 + 3SnCl4 + 4H2O. Use C 12, H 1, N 14, O 16, Cl 35.5, Sn 119. The atom economies (catalytic; Sn/HCl) are

JEE Advanced levelSingle correct optionDifficulty 3 of 3ev211

Nitrobenzene is reduced to aniline either by catalytic hydrogenation, $C X_{6} H X_{5} NO X_{2} + 3 H X_{2} C X_{6} H X_{5} NH X_{2} + 2 H X_{2} O$ , or with tin and hydrochloric acid, $2 C X_{6} H X_{5} NO X_{2} + 3 Sn + 12 HCl 2 C X_{6} H X_{5} NH X_{2} + 3 SnCl X_{4} + 4 H X_{2} O$ . Use C 12, H 1, N 14, O 16, Cl 35.5, Sn 119. The atom economies (catalytic; Sn/HCl) are

Reactants of the catalytic route:

123 + 6 = 129

; product aniline

93

; atom economy

93/129 = 72.1%

. For the tin route reactants are

2 (123) + 3 (119) + 12 (36.5) = 246 + 357 + 438 = 1041

and product

2 (93) = 186

, giving

186/1041 = 17.9%

. The catalytic route generates far less waste.