[davidenarb]

Hi all,

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

[Dan]

You must show you have attempted the question, this is a Forum Rule.

What do you think the mechanism is? Based on that, which product do you predict?

[TheOrganic]

Have you read the mechanism by which epoxide rings open? What are the two usual classifications? Are any catalysts present here?

You should be in a position to answer the above questions after you read a bit. Based on the answers, do you think the strength of the nucleophile would affect where it would attack, in this case?

[davidenarb]

First, sorry about not showing that I didn't attempted the problem.

For me, the product B will be formed because a Sn2 back side attack will occur, so we would expect an inversion of configuration.

Can anyone confirm to me?
Thank you

[mjc123]

A is the product with inversion of configuration. It is the one I would expect.

[davidenarb]

The configuration of the chiral carbon on the right of the starting material is R
The configuration of A is R, and the configuration of B is S. Therefore, we would expect the B product.

Where is my mistake please

[davidenarb]

please help, I cannot go on orgo if I don't understand such a basic configuration problem.

[Dan]

Where is my mistake please

The starting material is R,R

The product from invertive backside attack is B A.

B A is also R,R.

This may seem counterintuitive - it is because the the substituent priorities have changed. The CN substituent is second priority at that centre in the product, and it replaced a first priority substituent in the starting material - this means that you can't just look at the R/S designation to identify the product easily.

It is an issue arising from cutting corners. I actually used to give a similar example to my students to point out the dangers of shortcuts. Inversion does not always mean R S swap. Draw the mechanism with the implicit H at the reactive centre drawn out. Draw the S_N2 transition state. Don't cut any corners and things should become clear.

[davidenarb]

is B R,R  ?

[davidenarb]

B is also R,R.

I understand the fact that inversion of configuration doesn't mean R and S swap, but I would like please to confirm that B is R,S right?

[Dan]

is B R,R  ?

B is also R,R.

I understand the fact that inversion of configuration doesn't mean R and S swap, but I would like please to confirm that B is R,S right?

Sorry, I made a mistake and have corrected my post above. The product is the R,R product, which is A (not B).

[davidenarb]

I would like just to understand something.

Inversion does not always mean R S swap.

Yes, I agree with you Dan, but is the reverse statement true ? That is,

if we have  R S swap, then it means that we had an inversion of configuration

[Dan]

I would like just to understand something.

Inversion does not always mean R S swap.

Yes, I agree with you Dan, but is the reverse statement true ? That is,

if we have  R S swap, then it means that we had an inversion of configuration

No, that's the opposite statement. Both cannot be true. An R S swap tends to accompany S_N2, but not always - it is not a requirement of inversion.

If the the leaving group has CIP priority X, and the incoming nucleophile becomes a substituent with CIP priority Y, then S_N2 always results in an R S swap if X = Y. Normally, X = Y = 1.

If X ≠ Y, then R S swap does not necessarily occur. X = 1, Y = 2 (as in this case) results in no R S switch.