2014年7月30日 · 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. and here is the mechanistic scheme.

2014年8月28日 · I am trying to understand this mechanism for nitrobenzene reduction with $\ce{Sn/HCl}$. I do not understand why it implies that you pull off the $\ce{C-H}$ hydrogen instead of the $\ce{O-H}$ hydrogen in the methanol.

2018年8月2日 · Hydrogenation of aromatic nitro groups (over, for example, $\ce{Pd/C}$) usually results in reduction to the corresponding aniline and is probably easier to carry out and work up as compared to $\ce{Sn/HCl}$. Wikipedia has a list of several conditions for reduction of $\ce{ArNO2}$ to $\ce{ArNH2}$.

2022年9月29日 · The std electrode potential for Sn/Sn(II) is -0.14, the std electrode potential for 2H+/H2 is 0 (by definition) so the reaction of Sn + 2H+ to Sn(II) + H2 is energetically favourable. The std electrode potential of Sn(II)/Sn(IV) is +0.15 so H+ going to H2 is not a strong enough oxidant to oxidise Sn(II) to Sn(IV).

\begin{aligned}\ce{ HCl + SnCl2 &-> SnCl\\ HCl + KMnO4 &-> MnCl2 + KCl + H2O }\end{aligned} If we look at the oxidation states of our reactants, and products, $\ce{H}$ has +1, $\ce{Cl}$ -1, and for the products $\ce{Sn}$ +2, $\ce{Cl}$ -1 for the first reaction. The first reaction gains two electrons and thus is reduced.

2019年3月8日 · 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 …

2019年12月4日 · $\ce{HCl}$ is added to the nitrile group to give the imidoyl chloride salt first, which is then reduced by the tin(II) chloride. The initial product is the complex with $\ce{SnCl4}$ (which is often crystalline) that is then hydrolyzed to the corresponding aldehyde (see the insert in …

2018年10月10日 · For tin/HCl the reduction proceeds as a series of one-electron transfers from metal surface mixed with protonation. It seems that the lowest free orbital in the compound belongs mostly to nitro group (which makes sense as it has a strong π-withdrawing effect, stronger than carbonyl group), so it is reduced to a nitroso group and then further ...

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2020年3月8日 · Also, in case of 1,3- dinitrobenzene, $\ce{Sn + HCl}$ converts both the $\ce{-NO2}$ groups to $\ce{-NH2}$ groups. Now if we use $\ce{NH4HS}$ or some other nitrogen and sulphur containing ammonia like compounds, we get reduction again, but only for one group out of the two. So my question is, how do we : Decide that there is going to be reduction