chrisski
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Atom Economy vs % Yield
I have a basic idea of how to distinguish the two. However, I have a specific question that I am not positive on. Can you have a low % yield and a
high atom economy? Instinctively I would say no, that a low percent yield would never be able to accommodate a high economy. I tried google and the
only thing I found was some guy saying he knew of one instance in which you could have this combo. However, I am not convinced. Thanks in advance for
the insight.
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Arrhenius
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Atom economy is not typically a calculated figure, but merely a concept concerned with generation of chemical waste. For instance, the Mitsunobu reaction or Wittig olefination would be considered poorly atom economical, because the reactions necessarily produce 1equivalent of triphenylphospine oxide,
which has a fairly high molecular weight. Yield is, of course, a calculated value based on the theoretical outcome of a given stoichiometric
reaction.
Now, to answer your question. Let's say you have cyclohexene and you dissolve it in ethanol, add some palladium on charcoal and stir it up with some
hydrogen. But for some reason you only isolate cyclohexane in 40% of theory. You haven't really suffered poor atom economy, because the reaction is
catalytic, and because you haven't really generated much waste. Perhaps you only added 0.4eq of H2, in which case you would have perfect atom economy
for the given outcome, but poor yield.
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DJF90
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Not quite right Arrhenius... In the Wittig and Mitsunobu reactions, yes 1eq. Ph3P=O is produced in each instance, and this material is chemical waste,
but this does not affect the atom economy of the process as Phosphorus is not incorporated into the product.
Now take for instance the formation of Sodium Iodate (V) via the reaction of Iodine with hot Sodium hydroxide:
6NaOH + 3I2 => 5NaI + NaIO3 + 3H2O
Considering NaI is a waste product (seeing as we are making Sodium iodate), and 5/6 of the iodine is used to make it, this means only 1/6 of the
iodine ends up in the desired product. This is bad atom economy. If we wanted to make Sodium iodide, and not sodium iodate, then this would be much
better economy, as 5/6 of the iodine would be converted into the desired product.
Yeild is something slightly different. Using the balanced equation, you work out how much can be theoretically made (the theoretical yield) based on
the reactant in limited quantity (in the above example that would generally be the iodine). Then you weigh how much you made, and divide that by the
theoretical yield to get the *actual* yield. Using the above example, you may do the reaction and get a 95% yield of NaIO3 (compared to theoretical,
but the atom economy will still only be 1/6 (i.e. 16.67%)
[Edited on 16-4-2010 by DJF90]
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Arrhenius
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Well, atom economy as I've been taught involves all species, not just your product. It considers the atom count of all products relative to the
desired product, thus placing catalytic reactions as the most atom economical.
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HydroCarbon
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To answer your question "Can you have a low % yield and a high atom economy?" the answer is yes. The two values are completely unrelated. Atom
economy is defined for a reaction, but %yield can be any value.
Atom economy represents the percentage of your starting matter that ends up in your final target product if your reaction reaches 100% yield. This
can be calculated before a reaction is even done.
Atom economy is calculated as:
(atomic mass of reaction products) / (combined atomic mass of all reagents minus solvent and/or catalysts)
Atom economy gives insight into how much waste is generated by a particular reaction.
% yield is the percentage of products collected after a reaction compared to how much would have been collected if your reaction went to 100%. This
can only be calculated after you collect the products from a successful reaction.
% yield is calculated as:
(Moles of product collected) / (moles of limiting reagent used x stoichiometric factor)
The stoichiometric factor is how many moles of product you get per mole of limiting reagent in the reaction. for example in the reaction:
4 H2 + O2 --> 2 H2O
If dioxygen is your limiting reagent then you will have a stoichiometric factor of 2. If dihydrogen is your limiting reagent then your stoichiometric
factor is 0.5
% yield tells you how well your reaction occurs, and how efficient you are as a chemist.
The two can also be combined to tell you the actual percentage of the mass of your starting materials in a final product.
Calculated as:
(atomic mass of reaction products x %yield) / (combined atomic mass of all reagents minus solvent and/or catalysts)
This is basically the atom economy calculation for an actualized reaction. It factors in unrealized product as waste. If you follow this calculation
then %yield and atom economy are related values.
Hope that helps
[Edited on 17-4-2010 by HydroCarbon]
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