How to determine the most reactive hydroxyl group of α-ᴅ-glucopyranose?

In the following reaction, only one of the hydroxyl groups in D reacts under the given conditions:

I was wondering which OH group reacts and why? Also I have been trying to determine the structure of E based on which hydroxyl group reacts.

My thoughts are that OH in the side group can react via SN2 with the reactants, but the OH right adjacent to the ether O can react leaving a carbocation that can be stabilised by lone pair from O. This would mean SN1 and a completely different product.

Can someone please explain the correct mechanism?

• There is one hydroxyl group that is significantly different from all others, and another one, that is mildly different from the remaining. Does that hint help you? By significantly I mean rea~lly significantly – Jan Jun 17 '15 at 19:41
• @Jan Do you mean the hydroxyl on the side group that would undergo SN2? It is a Primary Alcohol..the others are secondary... – justbehappy Jun 17 '15 at 19:46
• That's the mildly different hydroxyl group. Now look for the significantly different one. – Jan Jun 17 '15 at 19:47
• You're on the right track with the hydroxyl on the same carbon as the ether. You've described the mechanism (at least the beginning) in words already, with the correct rationale for why that hydroxyl reacts. Write it up and you could answer your own question. – jerepierre Jun 17 '15 at 20:07
• Just so we're clear. The $S_{\mathrm{N}}2$ reaction with an alcohol is crap. Your yields will not be good. When you see these kinds of conditions, you should rule out $S_{\mathrm{N}}2$. – Zhe Dec 2 '16 at 22:10

$\ce{EtOH}$ is a rather bad nucleophile and we have the assistance of a lone pair of the neighboring oxygen atom (the positive charge can hence be delocalized over the carbon and oxygen) which greatly speeds up $\ce{S_{N}1}$ in comparison to $\ce{S_{N}2}$. So what will probably happen is protonation by $\ce{HCl}$ of the alcohol next to the ether oxygen followed by nucleophilic ($\ce{S_{N}1}$) attack of $\ce{EtOH}$ on the formed (delocalized) secondary carbocation. In comparison to the possible reactions at the other secondary centers (having the formulation of the question in mind) this primary reaction will be fast enough to justify the ignoring of them.
• Might be worth mentioning that the big difference of the $\alpha$ carbon is its oxidation state: Formally it's an aldehyde, not an alcohol. And in the OP's figure, one would call it hemi-acetal. – Jan Jun 18 '15 at 17:02