Products formed on cleavage of epoxides using HI

There seems to be multiple forms of $$\ce{HI}$$ used to cleave epoxides:

conc. $$\ce{HI}$$, anhydrous $$\ce{HI}$$, cold $$\ce{HI}$$, hot $$\ce{HI}$$, cold conc. $$\ce{HI}$$, hot conc. $$\ce{HI}$$, etc.

So, does concentration of $$\ce{HI}$$ and temperature have any significance in determining the product?

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded.

If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms.

If the medium is basic, it will proceed via the SN2 pathway.

In other words, the type of cleavage doesn’t depend on the concentration of $$\ce{HI}$$, the only thing that matters is the acidity or basicity of the medium.

• This seems to be a repeat of the previous answer.. what about temperature? See Martin's comment below.. – Safdar Sep 19 '20 at 12:25
• Temperature shouldnt matter. This is a competition between sn1 and sn2, and temperature should be considered only when elimination is involved, like in the case of beta unsaturated epoxides, where elimination will lead to a conjugated product. – Waterloo Biscuit Sep 19 '20 at 13:30

for unsymmetrical epoxides or oxiranes reaction takes place in this way:

in acidic medium- ring on reaction with HI ring opens from more hindered side i.e., the more hindered C-O bond. So the carbon in which more number of groups are attached which is bonded to the oxygen in ring gets -I.

in basic medium- the ring opens from less hindered side.

temperature is not a much important factor in this reaction.

• Welcome to chemistry.se! While there is certainly some truth to what you have written, I think it would be better to beck it up with some references; especially the last sentence. I also think this does not address the question completely. – Martin - マーチン Mar 25 '19 at 15:20