The following reaction is specified in my notes; $$\ce{CH4 ->[Mo2O3, \Delta]HCHO}$$ The intermediate steps involve the dehydration of an alcohol to yield the final product.

Wouldn't this only work if there's acid in the reaction mixture? Or is it that water functions as a proton donor, inducing the protonation of the hydroxyl and causing the dehydration?

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    $\begingroup$ Is it not a reaction $\ce{CH4 + O2}$ ? $\endgroup$ – Maurice Apr 29 at 16:26
  • $\begingroup$ @Maurice: it was shown with $\ce{[O]}$ above the arrows in the reaction steps. $\endgroup$ – harry Apr 30 at 0:01
  • $\begingroup$ Sorry, There is no [O] above the arrow. There is only a molybdenum oxide $\endgroup$ – Maurice Apr 30 at 9:23
  • $\begingroup$ @Maurice: Yes, not in this simplified equation, but when you write it out, apparently you use nascent oxygen. It's the controlled oxidation of alkanes, to be specific. $\endgroup$ – harry Apr 30 at 9:31
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    $\begingroup$ The nature of the oxidizing agent should absolutely be given. Because it is the very heart of the problem. How is this "nascent oxygen" produced ? UV ? $\ce{H2O2}$ ? Spark ? $\endgroup$ – Maurice Apr 30 at 10:18

The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow. Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.

On industrial scale, formaldehyde is prepared from methanol by an heterogeneous redox-reaction. The oxygen needed may come from oxygen gas, or simply air. Thus, the last step known as Formox process is

$$ \ce{2 CH3OH + O2 -> 2 CH2O + 2 H2O} $$

(from wikipedia)

$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.

  • $\begingroup$ So oxidising alkanes up the oxidation ladder isn't a thing? $\endgroup$ – harry Apr 30 at 11:25
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    $\begingroup$ @HarryHolmes Compared to other compounds, saturated hydrocarbons / alkanes are not very much reactive. On occasion, you encounter them under the name «paraffin», derived from the Latin «parum» (little) and «afffinis» (related) to just describe their low reactivity in comparison to functionalized and unsaturated hydrocarbons. It is easier e.g., to chop long alkane chains into smaller alkenes (e.g., in a steam cracker) and then to oxidize these fragments (e.g., Wacker process with ethylene to yield acetaldehyde, $\ce{MeCHO}$). Formally, the remove of hydrogen from alkanes is an oxidation, too. $\endgroup$ – Buttonwood Apr 30 at 11:40

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