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A possible mechanism of the partial reduction of anthraquinone by SnCl2. This is the way that I think it proceedsI've heard of using $\ce{SnCl2}$ in the Stephen's reaction where you reduce a nitrile down to an aldehyde. I've come across a laboratory manual that discusses using tin in the partial reduction of anthraquinone1 so I guess it's possible but how would such a reaction work? Would the $\ce{SnCl2}$ act in a similar way to a dissolving metal reduction? That would create a secondary alcohol but I can't see how the alcohol can be further reduced to an alkyl.

Reference

  1. Joaquín Isac-García, José A. Dobado, Francisco G. Calvo-Flores, Henar Martínez-García, Chapter 11 - Microscale Experiments, Experimental Organic Chemistry, 2016, Pages 371-408, DOI: 10.1016/B978-0-12-803893-2.50011-5
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2 Answers 2

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Luca: Your mechanism looks good except you haven't resolved the reductive elimination in the last step. Assuming that the carbonyl oxygen is protonated as in 1a, then two events must occur in some order: protonation of the hydroxyl oxygen and a 2-electron reduction of the protonated carbonyl (2). Elimination of water from 2 affords the "hydroxyanthracene" that tautomerizes to anthrone 4. Alternatively, the carbonyl is not protonated but the hydroxyl is. A one-electron transfer to the carbonyl provides ketyl 1c, which, by the addition of a second electron and protonation, affords 3. As in the first mechanism, the order of events is not certain.

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According to this source Science Direct tin chloride will reduce anthraquinone to anthrone

Anthrone is a tricyclic aromatic ketone that can be used, for example for calorimetric determinations of carbohydrates in biological fluids. It can be synthesized by partial reduction of anthraquinone with several reagents, as sodium hydrogen sulfite, tin chloride, or tin.

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    $\begingroup$ Hello, I read the lab manual, what I'm curious about is how this reduction works mechanistically. Does it follow a dissolving metal style reduction, or does it proceed down a Stephen's reduction pathway, or something else. $\endgroup$
    – Luca
    Apr 16, 2022 at 18:04
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    $\begingroup$ I've added a mechanism of how I think the reaction proceeds. Thanks for the help! $\endgroup$
    – Luca
    Apr 16, 2022 at 18:34

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