Ring opening reaction of epoxide with strong nucleophile

[davidenarb]

Hi chemists,

 

I would like to know whether we have compound A or B in this reaction and why:

 

[Elite Engineer]

You have compound B. This is due to the fact that the oxygen atom gravitates to the more substituted side of the molecule for stability during transition states.

 

The left side has a tertiary carbon atom which provides higher stability during transition, while the right side (where the cyano group attaches) has only a secondary carbon, offering less stability.

 

Not to mention, the cyano group being a very strong nucelophile and the nature of the reaction being a substitution reaction kicks out the oxygen atom as well.

 

~EE

[hypervalent_iodine]

You have compound B. This is due to the fact that the oxygen atom gravitates to the more substituted side of the molecule for stability during transition states.

 

The left side has a tertiary carbon atom which provides higher stability during transition, while the right side (where the cyano group attaches) has only a secondary carbon, offering less stability.

 

Not to mention, the cyano group being a very strong nucelophile and the nature of the reaction being a substitution reaction kicks out the oxygen atom as well.

 

~EE

Hm, I think you've missed the difference in the products. The carbon that is substituted is the same in both cases, the difference is the direction that the incoming nucleophile attacks and thus the stereochemistry of that centre, which is determined by steric influences.

 

To the OP, given my brief explanation above, which do you think would be the case? It may help you to build a model and have a look at where the best place would be for a nucleophile to attack.