Iodobenzene is prepared from aniline C6H5NH2 in a two step process as shown here- C6H5NH2 - HNO2 - HCl - C6H5N2- Cl- - 2H2O C6H5N2-Cl- - Kl - C6H5 I - N2 - KCl In an actual preparation- 9-30 g of aniline was converted to 12-32 g of iodobenzene- The percentage yield of iodobenzene is-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

Iodobenzene i...

Question

Iodobenzene is prepared from aniline $(C_{6} H_{5} N H_{2})$ in a two step process as shown here:
$C_{6} H_{5} N H_{2} + H N O_{2} + H C l \to C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O$
$C_{6} H_{5} N_{2}^{+} C l^{-} + K l \to C_{6} H_{5} I + N_{2} + K C l$
In an actual preparation, 9.30 g of aniline was converted to 12.32 g of iodobenzene. The percentage yield of iodobenzene is:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80%

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

(C_{6} H_{5} I)

(C_{6} H_{5} N H_{2})

C_{6} H_{5} N H_{2} + H N O_{2} + H C l \to C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O

C_{6} H_{5} N_{2}^{+} C l^{-} + K I \to C_{6} H_{5} I + N_{2} + K C l

9.30

16.32

C_{6} H_{5} N_{2}^{+} C l^{-} H_{3} P O_{2} - --- \to H_{2} O X + N_{2} ↑ + H C l + H_{3} P O_{3}

(i) C_{6} H_{6} + I C l A n h y d . A l C l_{3} - ------- \to iodobenzene

(i i) C_{6} H_{6} + I_{2} A n h y d . A l C l_{3} - ------- \to iodobenzene

Join BYJU'S Learning Program