daragh8008
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Sugar + H2SO4
I know this is a very common demonstration experiment so apologies for asking about it but I have been looking at a different version. C12H22O11 +
H2SO4 -> 12C + 12H20 The H2SO4 is acting as a catalyst and dehydrates the sugar to leave elemental carbon + H2O. How does the H2SO4 do this. in
this paper doi:10.1016/j.surfcoat.2004.10.039 they use a sugar water ethanol solution with a dash of h2so4 to form a thin carbon film. Would the
h2so4 not be dissociated of does this matter. The solution is left to evaporate, is this to drive off the water and ethanol so that only the h2so4
remains or is the sugar being dehydrated in the solution? Any help appreciated.
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Paddywhacker
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The dehydrating agent will remove an OH from one carbon and a H from an adjacent carbon to give two carbons joined by a double bond.
Repeat all over the place to give lots of double bonds.
The double bonds polymerize to make tars that continue to dehydrate until all that is left is essentially carbon.
Furfural can be prepared by dehydrating pentose carbohydrates, and dehydrating hexoses like glucose and fructose can give hydroxymethyl furfural.
Does anybody have a decent method for preparing hydroxymethyl furfural?
Furfural itself is fun. Smells nice, and is an aromatic aldehyde like benzaldehyde but is easier to make.
Edit: Furfural, like benzaldehyde, undergoes the Cannizaro reaction, forms a bisulfite complex, and undergoes aldol condensations, for example with
acetone, see http://www.rsc.org/publishing/journals/AN/article.asp?doi=an...
And, I would expect it to undergo an analogue of the benzoin condensation to give the analogue of benzoin, and from there the analogues of benzil and
benzilic acid. These will all be more water-soluble than the benzaldehyde-derived compounds, so workup may be a bit different. Endless scope for
experimentation.
[Edited on 1-8-2009 by Paddywhacker]
[Edited on 1-8-2009 by Paddywhacker]
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Ozone
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C12H22O11, that is CnH2n-1On-1 or C12 + 11H2O. Concentrated sulfuric acid a very strong dehydration catalyst (that is, H+ is 2N and dissociation is
~100%). This is driven by the dissociation constant and the 98+ % concentration. Ergo, much heat of dissociation goes forward to drive the rxn even
faster, e.g. 2x/10°C. You get, stoichiometric amounts of steam (which can be condensed weighed) and carbon.
The odor of caramelization is the result of furfurals and maltols forming as intermediate species. Without the H2SO4, this will still occur, albeit
very slowly (hours of induction at 90°C). But, once it gets going, the sucrose hydrolizes (inverts) to yield fructose and glucose which are
dehydrated to yield (via fructose) 5-hydroxymethyl-2-furaldehyde. This, in turn, can re-hydrolyze, in-situ, to yield levulinic and formic acids. These
acids are catalytic making the reaction sequence self-catalytic. That is, it will accelerate under isothermal conditions as the amount of catalyst
increases.
This differs from the sucrose:H2SO4 case where there is insufficient water for inversion to occur and so the rings dehydrate directly (we suppose).
Caramelization does not occur, per se, until the mixture is sufficiently dilute.
Hexoses yield HMF and maltols, pentoses yield 2-F. Neither is a substitute for benzaldehdyde. 2-F smells like moldy corncobs (not my favorite) and HMF
doesn't smell like much (to me). The maltols, resulting from dehydration of the hexose-form, particularly ethyl maltol, smell quite a bit like caramel
(cotton candy). These compounds, in minute quantities, are generated during the sugar refining process. They are responsible for the odor and flavor
that make granulated refined sugar superior (organoleptically) to analytical grade sucrose (which while sweet, is flavorless).
Cheers,
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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daragh8008
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Thanks Folks,
so it is the dissociation constant of sulphuric acid that breaks down the sugar. The paper that I was reading says that the h2so4 is mixed with
ethanol and water + sugar first with a large excess of water in comparison to the sulphuric acid. I’m assuming that the h2so4 is completely
dissociated in this instance. So when the solution is allowed to dry alot of the water is lost. But surly it will retain water easier than pulling it
out of a sugar molecule? I suppose what I'm asking is if the h2so4 is really necessary. On the plus side the paper was totally correct in that you can
use this route to make groovy porous carbon films!
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Ozone
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Not exactly.
The dissociation constant of H2SO4 is large in that it is thermodynamically downhill (exothermic). This means that it is strongly dehydrating. As it
is hydrated (to yield 2H+ and SO42-), large amounts of heat are evolved. Heat makes it (further reactions) go faster. Additionally, as H2SO4 is
hydrated, the amount of dissociated H+ (the catalytic specis) increases. This also makes it go faster.
The water cannot be lost unless an unusual amount of energy is put into the system. In other words, the water is not "free" it is bound in solvation.
Until the water exceeds equilibrium (which is 2:1, H2O to H2SO4, not accounting for solvent caging), driven dehydration will continue. After this, the
reactions will be consistent with any aqueous acid catalyzed system (with a HUGE amount of catalyst).
Apologies for the lack of discussion, I am at work.
Cheers
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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daragh8008
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Thanks Ozone for clearing that up. It explains nicely why the system was heated the first time round.
Thanks again
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