4
$\begingroup$

In all oxidation reactions of glucose, it seems that the aldehyde group alone gets oxidised and none of the hydroxyl groups. In one reaction with nitric acid, the aldehyde group and the terminal hydroxyl group get oxidised but none of the remaining hydroxyl groups. I was wondering why this happens. Are the hydroxyl groups less susceptible to oxidation for any reason? If so why would that be? And also is there a difference is ease of oxidation of the terminal hydroxyl and remaining hydroxyl groups?

$\endgroup$
1
  • 1
    $\begingroup$ When there are various pathways then what is oxidised depends more on the oxidising reagent - as Matthew mentioned, you can oxidise or reduced in various ways. Do you want to know about mechanism of oxidation with nitric acid? $\endgroup$ – Mithoron Aug 22 '20 at 15:25
4
$\begingroup$

Aldehydes are easier to oxidize via radical mechanisms because of the ease with which one can break the C-H bond. The resulting radical is stabilized via delocalization of an oxygen lone pair.

Such stabilization is absent in an alcohol. The alcohol shows a similar moiety, but I think the stabilization in the aldehyde is greater because of the shorter C-O bond, which improves orbital overlap.

In a laboratory setting, alcohols tend to be more stable to air, whereas aldehydes can be prone to air oxidation.

$\endgroup$
3
  • 1
    $\begingroup$ Thank you so much for the answer! But why wouldn't the hydroxyl group be able to stabilise a carbon radical attached to it because it too has a lone pair on it, would the attached hydrogen to the oxygen in hydroxyl affect it ability to stabilise the radical? Of so why? Also, what if the oxidation reaction is based on an ionic mechanism, how would difference in ease of the two groups be explained in this case? $\endgroup$ – Sid Aug 22 '20 at 1:08
  • $\begingroup$ That's a really good question. I think added an edit. I would think that the two-electron mechanisms more readily oxidize an alcohol (many reagents stop at the aldehyde, e.g., DMP). However, I would think that nitric acid operates via a two-electron mechanisms though I could be wrong here. If it were a two-electron mechanism, my guess is that the steric factors involved with being terminal versus non-terminal were at play... $\endgroup$ – Zhe Aug 23 '20 at 13:03
  • $\begingroup$ The steric factors and orbital overlap concepts make so much sense! Thanks a lot for the answer! $\endgroup$ – Sid Aug 24 '20 at 6:24
3
$\begingroup$

OP's statement of "In all oxidation reactions of glucose, it seems that the aldehyde group alone gets oxidised and none of the hydroxyl groups" is not entirely true. For example, the step by oxidative cleavage with periodic acid ($\ce{HIO4}$) has been successfully applied in the constitutional analysis of sugars. The presence of several pairs of vicinal diols such as those in glucose during oxidation with $\ce{HIO4}$ can lead to the formation of complex product mixtures:

Oxidation of diol

Accordingly, the oxidation reaction of D-glucose with 5 equivalents of $\ce{HIO4}$ gives five equivalents of formic acid and one equivalent of formaldehyde (Figure A). An analogous degradation of D-fructose yields three equivalents of formic acid, two equivalents of formaldehyde, and one equivalent of carbon dioxide (Figure B):

Periodate oxidation of glucose & Fructose

This oxidation is believed to be step-wise. For example, glucose (a hexose) contains a terminal aldehyde group ($\ce{C}$1) adjacent to a hydroxylated carbon. This setup tends to be oxidized to formic acid, $\ce{H-C(OH)2-CH(OH)R -> H-C(=O)OH + (O=)CHR }$, and an another aldose, which contains one less carbon atom than original sugar (here, it is the corresponding pentose). This formation can be explaned by a $\ce{C-C}$ cleavage followed by each $\ce{C-OH}$ oxidation to $\ce{C=O}$:

Stepwise periodate oxidation of glucose & Fructose

Now, this corresponding pentose also contains a terminal aldehyde group adjacent to a hydroxylated carbon. Thus process continues until $\ce{C}$6. The $\mathrm{R}$-group of $\ce{C}$6 is $\ce{H}$. Thus, final product is $\ce{R-CHO \ # \ H-CHO}$, which is formaldehyde. Additional description of this oxidation can be found in here.

$\endgroup$
3
  • 2
    $\begingroup$ Nice answer to some other question but it doesn't address the OP's question about nitric acid. $\endgroup$ – user55119 Aug 21 '20 at 19:38
  • $\begingroup$ @ user55119 : True. But I am concentrating OP's concern about not oxidizing hydroxyl groups (the title and the first sentence of OP's text). So, I think my answer is appropriate as well. $\endgroup$ – Mathew Mahindaratne Aug 21 '20 at 19:44
  • 1
    $\begingroup$ I still don't understand why the aldehydes are easier to oxidise than the hydroxyl groups with weaker oxidising agents like say Benedicts reagent, and also why terminal hydroxyl group alone is oxidised when aldehyde is protected and not the other hydroxyls? And what about nitric acid, why the selectivity of oxidation of only aldehyde and terminal hydroxyl groups? Is it because it isn't too strong an oxidising agent? $\endgroup$ – Sid Aug 22 '20 at 1:21

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.