Mechanism of the blue bottle experiment

Question

I have researched multiple mechanisms for this experiment yet all of them suggest different versions or are too vague. None of them are consistent since the final overall equation seems to be missing a few elements. Can someone help me out?

Step 1: $\ce{O2(g) -> O2(aq)}$ (Fast)

Step 2: $\ce{O2(aq) + MBred (aq) -> MBox (aq)}$ (Fast)

Step 3: $\ce{MBox (aq) +C-(aq) -> MBred (aq) + X-(aq)}$ (Slow)

The general mechanism that showed up frequently is very vague. The $\ce{X-}$ is always labeled as the oxidation products of the carbohydrate. I can't seem to specifically find the product for any of the carbohydrate (glucose, fructose etc.). On top of that, from my understanding of the reaction mechanisms is that each step must be balanced and in the end, intermediate species are canceled out. However, the same issue always comes out with the leftover elementary species $\ce{O2}$ where the overall equation is then not balanced. For example:

Prior to the addition of MB: $$\ce{C6H12O6(aq) + OH-(aq) <=> C6H11O6-(aq) + H2O(l)}$$

Step 1: $\ce{O2(g) -> O2(aq)}$ (Fast)

Step 2: $\ce{O2(aq) + MBred (aq) -> MBox (aq)}$ (Fast)

Step 3: $\ce{MBox(aq) + C6H11O6-(aq) -> MBred (aq) + C6H10O6(aq)}$ (Slow)

is the best possible mechanism I could put together and step 3 makes sense to me since the 2 $\ce{H+}$ ions will be used to reduce the methylene blue. but then the overall equation will look like:

$$\ce{O2 + C6H11O6-(aq) -> C6H10O6(aq)}$$

where it is now missing the balance in hydrogen and oxygen.

Answer 1

According to Shakhashiri and coworker's 2012 J. Chem. Ed. paper, the linear form of glucose forms an endiolate which gets oxidized to a dicarbonyl (on carbon 1 and 2) and then further reacts with the dioxygen product, $\ce{HO2-}$ (deprotonated hydrogen peroxide), in an oxidative decarboxylation reaction to form arabinoic acid:

The blue color comes from the oxidized form of methylene blue. The "bleaching" of the colored dye through reaction with glucose is slow, while (in the presence of dissolved oxygen), the reaction of dioxygen to regenerate the dye is fast. This leads to the observed behavior of the blue bottle reaction mixture. The last step of the mechanistic hypothesis does not directly influence color changes, so visual observations don't tell us about how fast this step happens.

On the other hand, according to Wikipedia, substituting ascorbic acid or benzoin for glucose, while showing similar kinetics, results in the diketones without further reaction.

Each dioxygen needs to gain 4 electrons to yield water. Reducing the dye is a two-electron transfer yielding peroxide, which can then react with the dicarbonyl. (I guess only when it is at the end of the chain, ready to decarboxylate, so ascorbic acid does not undergo decarboxylation).

Some sources say that the product of glucose oxidation is the sugar acid gluconate, but that hypothesis seems not to be supported by experimental evidence.

Answer 2

Bassam Shakhashiri's book entitled "Chemical Demonstrations, A Handbook for Teachers of Chemistry" gives a detailed study of the mechanism of the blue bottle, in his Volume $2$, page $145$ ($5$ volumes are available).

First glucose $\ce{C6H12O6}$ reacts with $\ce{NaOH}$ in a basic solution according to $$\ce{C6H12O6 + OH^- -> C6H11O6^- + H2O}$$ Now methylene blue $\ce{C14H12N3S^+}$ is added into the glucose solution. The glucose molecular ion is oxidized to gluconic acid $\ce{C6H12O7}$ (or its basic form the gluconate ion $\ce{C6H11O7^-}$) according to : $$\ce{C6H11O6^- + C14H12N3S^+ + H2O -> C6H11O7^- + C14H14N3S^+}$$ Simultaneously methylene blue is reduced into methylene white $\ce{C14H14N3S^+}$. Later on, the dissolved oxygen molecules reoxide the methylene white into methylene blue, according to $$\ce{C14H14N3S^+ + 1/2 O2 -> C14H12N3S^+ + H2O}$$ The oxidation does not stop on the gluconic acid. This acid (and its deprotonated ion) is further oxidized into glucaric acid and other more oxidized structures that will not be studied here.

The detailed structures of the interacting molecules are more interesting than their formula.

Let's first consider glucose and its oxidized forme. In Fischer's projection, glucose is a linear molecule with an upper aldehyde group $\ce{-CHO}$ on top. The gluconic acid has the same structure except the upper atomic group which is transformed into an acidic group $\ce{- COOH}$.

Let's now describe the methylene blue structure. The methylene blue is made of two $N$-dimethylaniline structures connected by a =$N-$ bridge in their $para$ position and a $\ce{-S-}$ bridge in $meta$. One of the benzene ring has a quinonic structure, with a $+$ charge on the outer $\ce{N}$ atom. Let's now consider the reduced form called methylene white. Its rings are both benzenic. The first added $\ce{H}$ atom is attached on the internal $\ce{N}$ bridge; and the second is fixed on the outer and positively charged $\ce{N}$ atom, which destroys the quinonic structure.

Ref.: Bassam Z. Shakhashiri, Chemical Demonstrations, A Handbook for Chemistry Teachers, The University of Wisconsin Press, 1985, p.142, Vol. 2, ISBN 0-299-10130-4