My book (NCERT India) states that n-hexane is formed when glucose is heated in the presence of hydrogen iodide. That is, the following reaction takes place: $$\ce{C6H12O6 + HI ->[{\Delta}] C6H14}$$ This seems a bit strange to me because I only know of three possibilities that can occur when a compound containing hydroxyl groups is treated with a strong acid:
A substitution reaction, where all the $\ce{-OH}$ groups are replaced by $\ce{-I}$
A dehydration reaction, where $\ce{H2O}$ molecules would be eliminated and it would end up with few double bonds.
Formation of an ether.
There are also other sources which mention this reaction (thanks to Karan Singh, Tyberius, and user55119 for pointing these out). A Michigan State University webpage states that:
Hot hydriodic acid (HI) was often used to reductively remove oxygen functional groups from a molecule, and in the case of glucose this treatment gave hexane (in low yield).
Also, Dietary Sugars (edited by Preedy, V. R.) states (p 80) that
Glucose is reduced with concentrated hydriodic acid in the presence of red phosphorus to form a trace amount of n-hexane. This indicates an unbranched chain of six carbon atoms [...]
It goes on to cite the 5th edition of Principles of Biochemistry by Nelson et al. (published 2008), but I could not find this statement in any recent edition of this book.
Lastly, a similar reaction was carried out by Kiliani (Ber. Dtsch. Chem. Ges. 1885, 18 (2), 3066–3072) where the cyanohydrin of fructose was reduced to to 2-methylhexanoic acid using hot HI and phosphorus.
I would like to know:
- Is phosphorus required for this reaction, or not?
- What is the mechanism by which this reaction proceeds, with or without phosphorus?
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