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Oxidation of hydroxyl groups of glucose
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Mojo Jojo
Oxidation of hydroxyl groups of glucose
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:
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):
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 $\ce{C=O}$:
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.
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:
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):
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 $\ce{C=O}$:
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.
I still dont 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 isnt too strong an oxidising agent?More
@ user55119 : True. But I am concentrating OPs concern about not oxidizing hydroxyl groups (the title and the first sentence of OPs text). So, I think my answer is appropriate as well.More
Nice answer to some other question but it doesnt address the OPs question about nitric acid.More
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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.
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.
Thats 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...More
Thank you so much for the answer! But why wouldnt 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?More
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:
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):
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 $\ce{C=O}$:
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.
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:
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):
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 $\ce{C=O}$:
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.
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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.
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.
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