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I had a question in my textbook to convert propane to methane. The basic idea was chlorination in presence of sunlight. However, two products are obtained in this case - namely 2-chloropropane and 1-chloropropane. So it didn't work out at all.

I came up with another idea i.e. dehydrogenation, but it has not been taught to us at all. In addition the dehydrogenation reactions I searched on google were too complex. So, I hope there's another method for this.

Can anybody help me to sort this out?

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closed as off-topic by Mithoron, andselisk, Geoff Hutchison, Todd Minehardt, airhuff Jan 1 '18 at 20:11

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    $\begingroup$ Welcome to Chemistry.SE! Are you really meaning to convert propane, i.e. $\ce{CH3-CH2-CH3}$, methane, i.e. $\ce{CH4}$ and not maybe rather propane to propene, i.e. $\ce{CH3-CH=CH2}$? If this isn't a mistake I don't understand what you are trying to achieve with the chlorination or dehydrogenation. $\endgroup$ – Philipp Sep 15 '14 at 2:46
  • $\begingroup$ No the conversions may be lengthy I just could not determine what I need to do in the 1st step. $\endgroup$ – Dws_kool Sep 15 '14 at 2:48
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This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis

$\ce{C3H8 +O2 +H2O -> CO + H2 + CO2}$

This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed

$\ce{CO + 3 H2 -> CH4 + H2O}$

depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.

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enter image description here

Use: $$\begin{array}{} \ce{Br_2}/h\nu& \text{Bromination}\\ \ce{^{\large \circleddash} OEt/EtOH}& \text{Elimination: Dehydrohalogenation}\\ \ce{O_3/H_2O}& \text{Ozonolysis}\\ \ce{NaOH/CaO}/\Delta& \text{Decarboxylation}\\ \end{array}$$

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If you are trying to do a dehydrogenation, dehydrohalogenation of both your products would get you to propene, even though it takes two steps.

Good luck getting to methane. Of course, if you were in industry you would just heat the crap out of it and watch the money roll in... Well, maybe not with methane as the product...

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I wonder if it is not possible to convert prop to meth directly with a large excess of hydrogen $\ce{H2}$.

So $\ce{CH3-CH2-CH3 + 3H2 -> 3 CH4}$ where the propane/hydrogen ratio is something like 10/90 i.e. an over-excess of hydrogen under the influence of a heat source like a tungsten heated spiral ($1000{-}\pu{2500°C}$). Due to the heat, both dissociation and recombination should occur. Simple and direct.

I think that it should be possible.

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  • $\begingroup$ I like your thinking but I don’t think it is possible. $\endgroup$ – Jan Jan 1 '18 at 13:47

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