I know sugars have $\ce{-OH}$ groups, and that sulfuric acid is a strong acid. What I'm failing to see is how exactly sulfuric acid dehydrates sugars. Does it protonate the hydroxyl groups, making them good leaving groups? What about the case of sucrose?

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For example, how are the oxygens "taken care of" by the sulfuric acid? Are they doubly protonated? Is there some established mechanism for the dehydration of sugars by sulfuric acid, because I can't find any - I just find general equations and I've looked on Google and through textbooks.

After some further research I learned that the gas, sulfur dioxide, is evolved. This suggests to me that this reaction is also redox in nature. So is this reaction both an acid-base reaction and a redox reaction? Sulfur in sulfuric acid has an oxidation state of +6; sulfur in sulfur dioxide has an oxidation state of +4. Plus the elemental carbon formed has an oxidation state of 0 while in sucrose, carbon has various oxidation states.

  • 5
    $\begingroup$ I doubt sugar dehydrates into pure carbon. Afaik both glukose and fructose can be relatively easy dehydrated into furan derivatives, that are known for their ability to polymerize under sever conditions and condense with pretty much any carbonyl compound, including furfural derivatives, else produced in this conditions. So, in sulfuric acid the product is a complex resin with significant ( 10+%) of non-carbon atoms. Most I wrote is pretty easily googlable, so feel free to educate yourself. $\endgroup$
    – permeakra
    Aug 19 '14 at 19:00
  • $\begingroup$ When I googled I found several reaction equations implying that "C" was formed, or pure carbon. $\endgroup$
    – Dissenter
    Aug 19 '14 at 19:08
  • 2
    $\begingroup$ Pure carbon is actually quite hard to obtain. Deep carbonization usually occurs around 700-1500 Celsium and higher . Charcoal, usually assumed being pure carbon actually may contains up to ~10% non-carbon atoms, and given that sugar-sulfuric acid reaction occurs at temperatures up to 200-300 C I can safely conclude the product contains a good measure of non-carbon atoms. $\endgroup$
    – permeakra
    Aug 19 '14 at 21:30
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    $\begingroup$ @Dissenter, the acid catalyzed dehydration of alcohols is usually and E1 reaction. This paper is one of the top Google results for "dehydration of glucose" and shows 5-hydroxymethylfurfural as one of the major products, among others: sciencedirect.com/science/article/pii/S0896844606002117 $\endgroup$
    – jerepierre
    Aug 19 '14 at 21:32
  • 1
    $\begingroup$ Some sites claim that some of the produced carbon is entirely oxidized to CO2. I'm doubtful; I suspect the aldehyde moiety of reducing sugars or of the resultant furfurals is oxidized to an alcohol, which if further dehydrated would result in C-H bonds in the elemental carbon. In accordance with this, the C:H ratio in sugar carbonized with H2SO4 was found to be ~14:1 mol:mol in this paper. $\endgroup$
    – Curt F.
    Feb 5 '15 at 22:06

$\ce{H2SO4}$ protonates the alcohol to make $\ce{R-OH2+}$ and $\ce{HSO4-}$, the $\ce{HSO4-}$ then takes a hydrogen from the carbon atom adjacent to the alcohol, yielding an alkene and water. Alcohol dehydration reactions using acid are just elimination reactions.

Additionally, at temperatures above 340 degrees Celsius, sulfur trioxide exists in equilibrium with sulfuric acid in significant concentration. The presence of sulfur trioxide explains the presence of sulfur dioxide (via dissociation), or some redox chemistry occurring with the sugar.

  • $\begingroup$ They are E1 reactions. Not E2. There is no base. $\endgroup$
    – Zhe
    Jun 30 '17 at 14:02
  • $\begingroup$ HSO4- is the base. $\endgroup$
    – AS_1000
    Jun 30 '17 at 14:13
  • $\begingroup$ Not nearly strong enough for E2. $\endgroup$
    – Zhe
    Jun 30 '17 at 14:22
  • $\begingroup$ Depends on substrate, works for primary. $\endgroup$
    – AS_1000
    Jun 30 '17 at 14:26

In strong mineral acid conditions, sucrose, splitted to glucose/fructose, undergoes dehydration (elimination of water by OH groups protonation), mainly to 5-(hydroxymethyl)furan-2-carbaldehyde (or HMF, 5-hydroxymethylfurfural).[1]

Fig.1 - HMF formation

(Stereochemistry omitted. Glucose is more stable than fructose, is reported to follow the path by isomerization to fructose (via acyclic endiol) in this conditions, or a via another mechanism proposed:[2] C-2 (in glucose numbering, i.e. C-3 in pyrane numbering) with protonated OH group is attacked by pyrane oxygen, forming intermediate oxiranium ring…)

(Pentose carbohydrates, like xylose, form furfural itself.)

HMF then polymerizes (possibly together with other byproducts like levulinic acid) to form rather insoluble humin, whose fragment may look like:[3]

Fig.2 - humin

I don't know if it's possible for such structure to be further dehydrated to carbon, under non-pyrolytic conditions.


  1. Feather, M. S.; Harris, J. F. Dehydration Reactions of Carbohydrates. Advances in Carbohydrate Chemistry and Biochemistry 1973, 28, 161–224.
  2. Dee, S. J.; Bell, A. T. A Study of the Acid-Catalyzed Hydrolysis of Cellulose Dissolved in Ionic Liquids and the Factors Influencing the Dehydration of Glucose and the Formation of Humins. ChemSusChem 2011, 4 (8), 1166–1173.
  3. van Zandvoort, I.; Wang, Y.; Rasrendra, C. B.; van Eck, E. R. H.; Bruijnincx, P. C. A.; Heeres, H. J.; Weckhuysen, B. M. Formation, Molecular Structure, and Morphology of Humins in Biomass Conversion: Influence of Feedstock and Processing Conditions. ChemSusChem 2013, 6 (9), 1745–1758.

I think sulfuric acid is hygroscopic. When you put sugar in it, it will take out water of crystallization from it and if not, most probably it will dehydrate it by taking $\ce{-OH}$ groups from it.


The sulfuric acid brakes up the bonds of the sugar and absorbs the water from the crystallization

  • $\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, you can ‎visit the help center or take a ‎‎tour of the website.‎|| I see no problems in your answer, though the community might think otherwise. You might wanna expand it a little bit, just in case. $\endgroup$
    – M.A.R.
    Feb 5 '15 at 21:24

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