3
$\begingroup$

I am aware that $\ce{HCl}$ can be used to help speed up the hydrolysis of sucrose but I haven't really been able to find out why. I've read somewhere that the glycosidic link in sucrose is broken off and that the $\ce{H+}$ ions from the $\ce{HCl}$ is responsible for this.

I've also come across the ether link during my research and would like to know how this plays a part in the hydrolysis of sucrose and if the $\ce{HCl}$ interacts with it.

$\endgroup$
4
$\begingroup$

The mechanism really isn't much more complicated than what you describe.

Sucrose hydrolysis mechanism

The $\ce{H+}$ binds to the acetal linkage. Water then binds to the fructose ring, breaking its bond with the acetal linking $\ce{O}$. Finally, the water molecule on the fructose loses a proton, restoring the catalyst. As Jan mentioned in the comments, the specific acid used is irrelevant so long as it is strong enough to sufficiently acidify the solution. (Credit to Khan Academy for the image)

$\endgroup$
0
$\begingroup$

The answers that have been provided so far are correct in one sense, but they skip over the key to your question. The key to why a reaction is catalyzed by a substance has to do with the mechanism, in particular the rate determining step. The answers above do not show that step or discuss it, but rather they show a step which precedes it, namely the protonation of the bridge oxygen. It is true that this step is important, and that it involves the hydrogen ion. But it is the following C-O bond cleavage step that is more likely the rate determining step.

For a detailed answer to your question see this post which references a publication that solved the mechanism details: Hydrolysis of Sucrose over Sn1Ac or Sn2Ac

Based on that reference, I do not believe @Tyberius is correct in saying, “Water then binds to the fructose ring, breaking its bond with the acetal linking O”. Rather, water only comes in after the oxocarbonium ion has been formed in the C-O bond cleavage step. The detailed answer to your question is that the presence of an acid makes that C-O bond cleavage step happen quicker.

$\endgroup$
  • $\begingroup$ I don't think you can say they "solved the detailed mechanism". There are multiple feasible pathways, and computational studies show the energetic differences are quite subtle, see chemistry.stackexchange.com/a/80353 $\endgroup$ – Karsten Theis Sep 21 '19 at 13:50
  • $\begingroup$ @KarstenTheis You make a good point. It depends how one interprets the word "solved", and I did not mean to imply that no other solutions were possible. I fully agree that other solutions are possible, but for the purpose of my answer I was not focusing on those issues but rather the key question related to the original post; namely, the C-O bond cleavage step. By the way, I am an author on that paper and you will see that in it we specifically mention other possible solutions. $\endgroup$ – Tony M Sep 21 '19 at 14:13
  • $\begingroup$ Congrats on the paper! I enjoyed exploring how you managed to distinguish between mechanistic hypotheses. No matter whether SN1 or SN2, and no matter which glycosidic bond ends up being cleaved, the reaction starts with a protonation step, and that is why the rate depends on the pH. Even though it is not the rate determining step, the hydronium concentration is expected to appear in the rate law because it is a fast equilibrium generating a species that is a reactant in the rate determining step. $\endgroup$ – Karsten Theis Sep 21 '19 at 17:09
  • $\begingroup$ @KarstenTheis Yes, I agree concerning H+. To observe the initial products revealing the mechanism, finding the exact right [H+] was the key. It turned out to be the same pH used all the time in the commercial hydrolysis of sucrose. Although noted in ref 34, what's not mentioned is that it took over a year to figure that out ;) $\endgroup$ – Tony M Sep 21 '19 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.