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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:

  1. A substitution reaction, where all the $\ce{-OH}$ groups are replaced by $\ce{-I}$

  2. A dehydration reaction, where $\ce{H2O}$ molecules would be eliminated and it would end up with few double bonds.

  3. 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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  • $\begingroup$ It looks wrong, if HI could reduce alcohols to hydrocarbons chemistry would be different ;) $\endgroup$ – Mithoron Feb 19 '16 at 19:12
  • $\begingroup$ I don't think it is wrong, I think it is right. But I'm curious about the book anyway. $\endgroup$ – Curt F. Feb 19 '16 at 19:15
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    $\begingroup$ HI is a strong reducing agent. It can reduce iodoalkanes to alkanes through the formation of I2. $\endgroup$ – Curt F. Feb 19 '16 at 19:21
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    $\begingroup$ beilstein-journals.org/bjoc/articles/8/36 See scheme 1 and text. $\endgroup$ – Zhe Sep 7 '18 at 12:56
  • $\begingroup$ @Zhe can you provide the DOI? I am unable to view the article from your link. $\endgroup$ – A.K. Sep 9 '18 at 18:06
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  1. Hydroiodic acid is a reducing agent. As Wikipedia says:

    Although harsh by modern standards, HI was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by HI reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane (see the article on cyclohexane). As first reported by Kiliani,[10] hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility.[11] In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by HI can provide synthetically useful yields of the corresponding hydrocarbon product (ROH + 2HI → RH + H 2O + I2).[8] This process can be made catalytic in HI using red phosphorus to reduce the formed I2.[12]

  2. When $\ce{HI}$ acts as a reducing agent, the oxidized product is $\ce{I2}$. To regenerate $\ce{HI}$ from the $\ce{I2}$, phosphorus can be used. Thus, phosphorus is not strictly required for the reaction, as long as you are OK with using $\ce{HI}$ in superstoichiometric amounts and/or getting very low yields.

  3. The overall scheme for $\ce{HI}$ reduction of an alcohol to an alkane is first displacement of the (protonated) alcohol by iodide, leading to formation of alkyl iodide. Then, $\ce{HI}$ catalyzes the reduction of the alkyl iodide to the alkane, leading to $\ce{I2}$ as a byproduct. The reduction step apparently follows a radical mechanism, at least under some conditions.

  4. I'm not 100% sure how $n$-hexane would form from glucose. It is unclear to me how $\ce{HI}$ would catalyze the reduction of the glucose aldehydic (or anomeric) carbon to an alcohol. But, that is only one carbon out of six. The other carbons have bona fide hydroxyl groups, and as such can be deoxygenated and reduced by $\ce{HI}$ to form alkyl carbons. For example, Lv et al. report that hydrocarbons can be obtained from the five-carbon sugar alcohol xylitol in yields above 60%. Unlike glucose, xylitol has no aldehydic/anomeric carbon, and only has alcohol groups. This process did use phosphorous acid as an in situ iodine recycling agent. Also, Yang et al. were able to obtain 5-methylfurfural from fructose by using $\ce{HI}$, without using an in situ agent to recycle $\ce{}$ to $\ce{HI}$.

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Yes, phosphorus is required for this reaction. As stated in your literature and in the comment, red phosphorous is the stoichiometric reducing agent. I assume that $\ce{PI3}$ and $\ce{P2I4}$ are formed. Both are known to be deoxygenating agents (#1, #2).

I propose the mechanism to be like this, where the oxophilicity of phosphorous is the driving force of the reaction. The mechanism starts with an electrophilic attack of phosphorous to an oxygen, which is then substituted by iodide. $\ce{C-O}$ bond cleavage leaves an alkyl iodide. These must then be reduced somehow.

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  • $\begingroup$ Most of the members of this community are award of $\ce{HI, P}$ reduction and this would be obvious to most. The question is about using just $\ce{HI}$ or literature explaining why phosphorous got dropped. $\endgroup$ – A.K. Sep 11 '18 at 19:21
  • $\begingroup$ @A.K. oh i'm sorry. Here's the DOI you requested in the comments above: doi:10.3762/bjoc.8.36 As far as i can say, the Beilstein JOC is a reputable journal. $\endgroup$ – Chris Sep 11 '18 at 19:34
  • $\begingroup$ I think you missed my point. There are many sources that present this reaction so they have either propagated a myth or glucose is special. That is what we want answered. $\endgroup$ – A.K. Sep 12 '18 at 12:34
  • $\begingroup$ I just provided the DOI you asked Zhe for. I have not enough reputation to directly comment in comments of the question. $\endgroup$ – Chris Sep 12 '18 at 12:58

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