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Can we actually control catalytic hydrogenation by introducing specific molar quantities of hydrogen gas? My prof insists on no. No we cannot use a certain equivalency of hydrogen gas to stop at a desired product.

Yet in the book I keep seeing examples of specific molar quantities of hydrogen gas being used ...

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Using the equipment that I have and working on the scale that I tend to work on (<1 mol typically), I can't deliver only one equivalent of hydrogen gas. We can measure the amount of hydrogen consumed as the reaction proceeds. The problem you have shown could be interpreted that way, i.e. 1 mol is consumed but more may be present.

On industrial scale it may be possible to deliver a more accurate amount of hydrogen. I'll leave it to others that might have more experience on large scale to confirm or deny.

There are alternative sources of hydrogen that can more easily be delivered accurately on small scale. Either ammonium formate or cyclohexadiene can be used to generate hydrogen in the presence of palladium. The former liberates ammonia and carbon dioxide as byproducts, while the latter forms benzene as byproduct.

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  • $\begingroup$ Why can't you deliver only one equivalent of H2 gas? Can't you just put in 1 mole of H2 gas into the reaction chamber? $\endgroup$ – Dissenter Nov 17 '14 at 15:27
  • $\begingroup$ @Dissenter Delivering a gas to a reaction is not as simple as a solid or liquid. You can just weigh out a solid. You can measure the volume of a liquid or weigh it out. Usually solids and liquids are dissolved in the solvent, and we know that the measured quantities are available to react. $\endgroup$ – jerepierre Nov 17 '14 at 16:23
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    $\begingroup$ @Dissenter The gas must be delivered out of a gas cylinder. For simple hydrogenation reactions, we can just fill a balloon and attach it to the reaction vessel. Some of the hydrogen will dissolve and be available to react, but the rest will remain in the atmosphere. As hydrogen is consumed, more will dissolve, but eventually we'll get to a point where the pressure is low enough there won't be an appreciable amount in solution. I suppose we could account for that, but I've never seen anyone do that in practice, again with the caveat that I don't work on production scale. $\endgroup$ – jerepierre Nov 17 '14 at 16:26

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