My NCERT book(Class 12) said the following for reduction of nitrobenzene using $\ce{Sn/HCl}$ and $\ce{Fe/HCl}$ : enter image description here 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?

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    $\begingroup$ Probably preferred for cost reasons, iron being cheaper than tin $\endgroup$ – Waylander Mar 8 '19 at 10:14
  • $\begingroup$ @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? $\endgroup$ – Yusuf Hasan Mar 8 '19 at 10:37
  • $\begingroup$ I can see no problem with your. My guess is that the preference the authors express is for Fe/HCl over catalytic hydrogenation. $\endgroup$ – Waylander Mar 8 '19 at 10:39

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