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What is the reason behind the fact that methane on controlled oxidation gives methanol in presence of Cu catalysts whereas formaldehyde is obtained if we use molybdenum oxide ($\ce{Mo2O3}$)?

IUPAC says:

The term selectivity (S) is used to describe the relative rates of two or more competing reactions on a catalyst.

What is the reason behind favouring one products?

Would I be correct that in controlled oxidation of methane in presence of Cu catalyst, we get both methanol and formaldehyde?

Should I need to memorise what catalyst yield what products? Or there is any other way?

JM97
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  • I can't offer you an answer in as much detail as you'd probably like, but as you probably know, catalysts lower the activation energy by allowing a reaction to take place through an alternative mechanism. The Cu catalyst likely allows a mechanism of action that results in the formation of methanol preferred over the formation of methanal. I would assume that methanal could still be formed, but the methanol will form much faster in the presence of Cu and methanal formation would be minimal if present at all. However, I am only speculating based on what knowledge I do have of catalysts. – KeatonB Aug 21 '16 at 07:03
  • In short: yes, you have to memorize. Catalytic mechanisms can be pretty complicated, so in general you won't be able to predict what goes where. – Ivan Neretin Aug 25 '16 at 14:08

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