I know that zinc dust is used to obtain benzene from phenol. But can zinc get oxidised and remove OH group from a simple alcohol, for example ethanol?
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$\begingroup$ related chemistry.stackexchange.com/questions/87840/… chemistry.stackexchange.com/questions/81717/… $\endgroup$– MithoronCommented Nov 23 at 15:06
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$\begingroup$ chemistry.stackexchange.com/questions/60959/… $\endgroup$– MithoronCommented Nov 23 at 15:12
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No, not in any synthetically useful way
It has been observed here that the reaction of zinc powder with aliphatic alcohols at $\pu{330^\circ C}$ under solvothermal conditions produces $\ce{ZnO}$ nanoparticles. The reaction involves the cleavage of the $\ce{C-O}$ bond on the metal surface. The focus of this work appears to be the production of $\ce{ZnO}$ nanoparticles rather than the reduction of the alcohols.
$\ce{Zn}$ plus $\ce{NH4Cl}$ can reduce aldehydes and ketones to alcohols as described here but no alkanes are produced.
Standard Clemmensen ($\ce{Zn/HCl}$ or $\ce{Zn(Hg)/HCl}$) conditions reduce carbonyl compounds to alkanes passing through the intermediate alcohol, but the HCl is a key component here for protonation of the alcohol.
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$\begingroup$ How can zinc powder react with alcohol to produce $\ce{ZnO}$ without reducing the alcohol ? $\endgroup$– MauriceCommented Nov 23 at 16:34
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$\begingroup$ I can only see the abstract of the paper and no comment is made about the fate of the aliphatic groups - whether simple reduction occurs or a more complex process of oligermerisation $\endgroup$ Commented Nov 23 at 16:37
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$\begingroup$ Give me the reference of this article. I am pretty sure I can read the entire article, and not simply the abstract. $\endgroup$– MauriceCommented Nov 23 at 20:42
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$\begingroup$ link.springer.com/article/10.1007/s10876-007-0129-6#: Journal of Cluster Science vol 18 (2007) 660-670 $\endgroup$ Commented Nov 23 at 21:23
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$\begingroup$ Thank you for the reference in the Journal of Cluster Science. Reading it gives the equations for the reaction between zinc powder and alcohols. At $330$°C, the reaction is : $\ce{2 ROH + Zn -> R2 + ZnO + H2O}$. So methanol produces $\ce{R2}$ which is ethane, and ethanol produces $\ce{R2}$ which is n-butane $\endgroup$– MauriceCommented Nov 25 at 19:29