Laurelium
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Beginner question related to patent reading
Hello kind stranger,
I recently came across a patent which expressed the various chemical ratios in ‘parts’. Normally I would assume them to be moles although the
patent specifies that “...all parts and percentages used in the examples are calculated on a weight basis.”. I’m simply unsure whether I should
assume that ‘weight basis’ means grams instead of moles.
The patent used: 12.5 parts Nitroalkene, 53 parts Absolute Ethanol, 1 part Glacial acetic acid and 2.5 parts of 5% Pd/C.
The molar ratio given below uses 6 moles of said nitro compound, hence I divided each part by “2.0834” (ie.: 53 / 2.0834 = 25.44 moles) since 12.5
/ 2.0834 give 6 moles.
Every grams of 5% catalyst was replaced by 0.5g of 10% since the metal loading is the same.
Ratio using Moles:
979g Nitroalkene
1172g Anhydrous ethanol
64g 10% Pd/C
28.824g Glacial acetic acid
Ratio using grams: (Also uses 6 moles but each part gets divided by 0.012768131 since 12.5 / 979 gives such number.)
979g Nitroalkene
4151g Anhydrous ethanol
98g 10% Pd/C
78.32g Glacial acetic acid
~~
1Kg Nitroalkene
4.24Kg Anhydrous Ethanol
100g 10% Pd/C
80g Glacial acetic acid
~~
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averageaussie
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not too sure on this, but I want to give it a shot anyways
I think what it means is the ratio for the reaction is 12.5:53:1:2.5 by weight.
so, for example, you would need 12.5 grams of Nitroalkene, 53 grams of absolute ethanol, 1 gram glacial acetic acid, 2.5 grams Pd/C
please don't try anything before confirming the actual amounts of each required from someone with more experience, don't want you to waste any
expensive chemicals.
Also, can you by any chance link the patent you are following?
[Edited on 10-5-2023 by averageaussie]
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B(a)P
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Agreed with averageaussie.
So for your last list of reagents that you say is weight based, it should be
1Kg Nitroalkene
4.24Kg Anhydrous Ethanol
200g 10% Pd/C
80g Glacial acetic acid
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Laurelium
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Thank you both for confirming, it does seem much more likely to be by weight.
B(a)P: Do you mean 200g of 5% Pd/C hence 100g of 10% Pd/C ? ~~~ 12.5 / 1000g = "0.0125" as our magic number to divide the parts by. (Since 0.0125 *
1000g gives 12.5 parts)
Hence:
53 / 0.0125 = 4240g
2.5 / 0.0125 = 200g of 5% or 100g of 10%
1 / 0.0125 = 80g
Sorry if I'm being redundant, I just don't see my mistake. To go a step further in proving my ratio, in this case 1 part is equal to 80g, which if we
multiply by 53 gives 4240g.
Attachment: Reduction of Arylnitroalkenes.pdf (458kB)
This file has been downloaded 100 times
[Edited on 11-5-2023 by Laurelium]
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B(a)P
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Quote: Originally posted by Laurelium  |
B(a)P: Do you mean 200g of 5% Pd/C hence 100g of 10% Pd/C ? ~~~ 12.5 / 1000g = "0.0125" as our magic number to divide the parts by. (Since 0.0125 *
1000g gives 12.5 parts)
Hence:
53 / 0.0125 = 4240g
2.5 / 0.0125 = 200g of 5% or 100g of 10%
1 / 0.0125 = 80g
Sorry if I'm being redundant, I just don't see my mistake. To go a step further in proving my ratio, in this case 1 part is equal to 80g, which if we
multiply by 53 gives 4240g.
[Edited on 11-5-2023 by Laurelium] |
Apologies, you are correct regarding the mass of Pd/C. I missed that you had changed the percentage from what was in the patent.
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Pumukli
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10 g 5% PdC is not always equivalent to 5g 10% PdC in a real world flask! PdC is more than Pd + C mixed intimately together!
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Laurelium
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Pumukli: Could you provide a more detailed explanation to your example for why it couldn't be feasible. If it's due to the surface area, wouldn't a
longer reaction time or a pre-reduction of the catalyst be sufficient in providing the same results.
Additionally, would I be wrong to assume that such a scenario is usually only applicable to small scale procedures ? (I was unfamiliar to that
possibility, thank you for mentioning it.)
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averageaussie
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I have no clue about this, but I wanted to put my ideas down somewhere so whatever.
the 10% and 5% not being equivalent may be for a few reasons.
A change in rate of reaction does matter in some cases, and the faster reaction may mean more heat or gas is generated. with more heat being
generated, it can break activation energy thresholds, creating unwanted side products. so using a lower concentration of catalyst can be better if
this is the case.
along with the first issue, the added heat might be enough to break the palladium on carbon down, ruining the reaction.
I couldn't find anything about what filler is used for Pd/C, but the heat might also break that down as well, causing issues.
this is all assuming that heat is an issue in this scenario, and I would suggest just using 200g 5% for the reaction rather than the 100g 10% if you
have it on hand, just in case.
I have literally zero experience with Pd/C, and this entire reply is probably completely incorrect, so please correct me where I am inevitably wrong.
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Pumukli
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The 10% Pd means not only "twice the surface covered" but different size-distribution of the metal "islands" (grains) on the carrier C. It can lead to
different reaction speeds of concurent reactions hence differing quantities of different products. E.g. 5% PdC may not removing the halogen X from a
molecule in appreciable quantities while 10% PdC does it well at the same temperature and H2 pressure...
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BromicAcid
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Just a note, patents are an interesting field. Many chemistry patents will take a scattershot approach and name wide ranges of temperatures /
solvents / pressures to obfuscate what the actual target conditions are. Examples in the experimental section are likely to omit critical details.
The patent net is cast quite wide but to enforce a patent is tricky so some of the details are usual left as trade secrets so that it's specifically
not a recipe to copy it one to one. Not saying it is the case with this patent but it has been the case for many patents where I have tried to
replicate the work.
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Laurelium
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I appreciate all of your replies. Much more research on how Pd/Pt catalysts work is clearly awaiting me. If I ever were to do experiments, it is now
obvious that more serious small scale testing of the various % of metal load, solvents & conditions would be mandatory.
I do agree that quite a handful of patents (including this one) often lack the "truer" details. I haven't yet and honestly do not expect to get such
precise details from the patent without experimenting as the best conditions displayed seem pretty general in contrast to reductions of similar
molecules.
Although not considerably hard to find, this one was the first one out of at least 3-4 to confirm that pressures below 1000PSI could lead to a
relatively acceptable yield when using a Pd/C catalyst.
[Edited on 12-5-2023 by Laurelium]
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