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Can anyone explain the reaction mechanism of the two reactions given below?

They are the conversion of nitrobenzene to aniline in catalytic ($\ce{H2/Pd}$ in ethanol) and acidic ($\ce{Sn/HCl}$ or $\ce{Fe/HCl}$) mediums respectively.

$$\ce{Ar-NO2 ->[\ce{H2/Pd}][\ce{EtOH}] Ar-NH2}$$
$$\ce{Ar-NO2 ->[\ce{Sn + HCl} \text{ or } \ce{Fe + HCl}] Ar-NH2}$$

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    $\begingroup$ What are your own thoughts on the matter? Do you have done any research or have any clue what might happen? $\endgroup$
    – Philipp
    Jul 30, 2014 at 16:22

1 Answer 1

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The first reaction is called a catalytic reduction. Here is a picture of how it works with an olefin like ethylene.

enter image description here

You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form. The same things happen with nitrobenzene. As the following diagram shows, the reduction passes through the nitroso and hydroxylamine intermediates.

enter image description here

The mechanism for the Sn\HCl reduction involves reductive electron transfer of the nitro group while the tin is being oxidized. This diagram shows the electron transfers involved

enter image description here

and here is the mechanistic scheme.

enter image description here

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  • $\begingroup$ How did you come to know the mechanism...? $\endgroup$
    – jNerd
    Jul 31, 2014 at 7:25
  • $\begingroup$ @user70856 The catalytic reduction I knew from some time ago. For the Sn\HCl reduction mechanism I looked on the internet. $\endgroup$
    – ron
    Jul 31, 2014 at 12:58
  • $\begingroup$ What would be the product if we use Sn/NaOH ? $\endgroup$
    – Jdeep
    Mar 26, 2021 at 18:45

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