# Preference for tin or iron in the reduction of nitrobenzene

My NCERT book(Class 12) said the following for reduction of nitrobenzene using $$\ce{Sn/HCl}$$ and $$\ce{Fe/HCl}$$ : Now, I have been following this mechanism for this particular reduction, and it seems to me that the metal is getting oxidized to +2 state wherever it is shown that 2 electrons are being transferred. If that is the case, then:

•Why is $$\ce{Fe + HCl}$$ being preferred?

•Because I looked it up on Wikipedia and it says that $$\ce{SnCl2}$$ also gets hydrolysed to liberate $$\ce{HCl}$$ as follows: $$\ce{SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)}$$

Or have I interpreted the mechanism incorrectly?

• Probably preferred for cost reasons, iron being cheaper than tin – Waylander Mar 8 '19 at 10:14
• @Waylander But the way the book has stated it,it seems that this regeneration of HCl takes place only with Fe. So is there a problem with the yield in case of Sn+HCl? – Yusuf Hasan Mar 8 '19 at 10:37
• I can see no problem with your. My guess is that the preference the authors express is for Fe/HCl over catalytic hydrogenation. – Waylander Mar 8 '19 at 10:39

The book's reasoning is correct. Fe/HCl gives $$\ce{FeCl2}$$ on reaction and it gets hydrolyzed by steam vapors to produces more hydrochloric acid and hydrogen to push the reaction forward thus making the reaction self-sustaining to produce more aniline from nitrobenzene.
$$\ce{Fe + 2HCl ->[160°C,CH3OH] FeCl2 + H2}$$
$$\ce{3FeCl2 + 4H2O(steam) ->[Hydrolysis] Fe3O4 + 6HCl + H2 }$$
whereas $$\ce{SnCl2}$$ gives $$\ce{Sn(OH)Cl}$$ and 1 mole of hydrogen chloride which is insufficient and it gets exhausted after sometime.