2
$\begingroup$

I recently came across a question in which we have to find out the number of moles of $\ce{HIO4}$(Periodic acid) which are required for cleavage of $\ce{C-C}$ bonds in sucrose.

I went to https://en.m.wikibooks.org/wiki/Structural_Biochemistry/Carbohydrates/Carbohydrate_Structure_Elucidation_through_Periodic_Acid_Cleavage . But this article is rather incomplete and doesnot answer the question. In general what are the things to be kept in mind in order to find out the number of moles of periodic acid required for the structural determination of carbohydrates. Any ideas? Thanks.

$\endgroup$
4
  • 1
    $\begingroup$ You need to be thinking about titrating your periodic acid before and after reaction. $\endgroup$ Commented Apr 1, 2017 at 8:04
  • 1
    $\begingroup$ @user1945827 The amount of periodic acid used before and after reaction is to be neglected in the question. $\endgroup$
    – Pink
    Commented Apr 1, 2017 at 13:10
  • 1
    $\begingroup$ you're contradicting your original question, "we have to find out the number of moles of..." vs " The amount of periodic acid used before and after reaction is to be neglected". The number of moles is the amount. ... But I've just realized that this is a homework question. Work out the stoichiometry for each C(OH)-C(OH) group in glucose ... Hope that helps. $\endgroup$ Commented Apr 1, 2017 at 22:09
  • 2
    $\begingroup$ The correct term of "number of moles" is amount of substance. $\endgroup$ Commented Apr 17, 2017 at 9:36

1 Answer 1

3
+50
$\begingroup$

First of all, lets be familiar with what per-iodic acid does to organic compounds.

It selects the carbon-carbon bonds that have either an -OH group, a =O(carbonyl) group or if it is the functional group carbon of a carboxylic acid. Both the carbons of the C-C bond must satisfy this condition. Here are some examples of the selection points of per-iodic acid: enter image description here

Now that per-iodic acid has chosen the center, it cleaves the C-C bond and adds an -OH at each carbon. For every C-C bond, one per-iodic acid molecule is consumed. I am avoiding going throughout the actual mechanism to keep things simple. If any of the carbons has two -OH groups, they will spontaneously dehydrated to give a =O bond instead. For the examples shown above, the final products of the cleavage-oxidation would be: enter image description here

Now, coming to your main question, the structure of sucrose would be: enter image description here

By what we have seen, lets mark the C-C bonds that per-iodic acid can cleave: enter image description here

Since we have 3 centers, so thats why 3 molecules of per-iodic acid are consumed per molecule of sucrose. Now you can even proceed and find out the oxidation products.

I hope this answer helps. Happy carbohydrate breaking!

$\endgroup$
4
  • 1
    $\begingroup$ That is a great explanation, Balagopal I had just a small doubt... Acetals [including sucrose] can be hydrolysed in acidic medium. Is HIO4 sufficiently acidic to bring about hydrolysis of sucrose... I doubt, if this happens, we need to reconsider the number of moles of HIO4 consumed completely. $\endgroup$
    – Che Mistry
    Commented Apr 21, 2017 at 17:57
  • $\begingroup$ @Che mistry, acetals are hydrolysed? As per Wikipedia, it wasn't mentioned that per-iodic acid cleaves acetals, and neither does my textbook. $\endgroup$ Commented Apr 22, 2017 at 2:01
  • $\begingroup$ @Che Mistry, Acetal hydrolysis doesn't involve a $\ce{HIO4}$ molecule. Simply an acidic medium that is. The $\ce{H^+}$ will be returned after hydrolysis is over. $\endgroup$ Commented Apr 22, 2017 at 2:04
  • $\begingroup$ @Pritt , but i have seen that only syn diols reacts with HIO4, as when we see mechanism there is formation of ring with IO4- that is possible only with syn configuration of OH OH, how can anti react ? $\endgroup$ Commented Jan 19, 2021 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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