Whenever we add $\ce{H+}$/ Ethanol to glucose in its hemiacetal form, why doesn't pinacone/pinacolone rearrangement take place in place of nucleophillic substitution. The product would lead to formation of a carbonyl bond and so, should be the favoured product.

The actual pathway is:enter image description here

But, why can't the reaction occur like this with ethanol as a polar solvent?:enter image description here

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    $\begingroup$ Not on topic, but I feel like it would be a disservice to you to not point this out: your curly arrows for the first step (protonation and loss of water) aren't correct. The protonation and loss of water should also be two separate steps. I would also prefer to depict carbocations adjacent to oxygen with the alternative resonance form on the right: $$\ce{>C^{+}-OH <-> >C=O^+H}$$ $\endgroup$
    – orthocresol
    Aug 8 '17 at 8:53

First of all, there's no pinacole in the first place, so pinacole/pinacolone rearrangement isn't possible. Your rearrangement reaction/hydride shift wouldn't be called pinacole/pinacolone rearrangement.

Second, we have to think about why this would happen. Such rearrangements happen because we produce a more stable carbocation. In case of pinacole/pinacolone, if we do the rearrangement we end up with:

![enter image description here

Which is well stabilized and after deprotonation a stable compound.

In case of sugars it's a little bit different, since it's not an alcohol in 1-position, it's a hemiacetale. This is extremly important since it has quite different reactivities.

enter image description here

We can cleave of water under much weaker acidic and mild conditions than in case of alcohols. The produced carbocation is quite stable since we got an oxygen stabilizing the cation, therefore no rearrangement is happening.

enter image description here


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