# Configuration change at chiral carbon in SN2 reaction

Consider $\ce{A}=\ce{CH(CH3)(C2H5)(C3H7)}$ in R configuration. Suppose somehow in a $\mathrm{S_N2}$ reaction, $\ce{Br-}$ attacks and replaces $\ce{-H}$ (this compound we call $\ce{C}$). Now we expect change of configuration (Walden inversion), i.e. it to become S. If the structure was inverted without replacing $\ce{-H}$, we would have got the S-stereoisomer of $\ce{A}$, call this $\ce{B}$. Replace the $\ce{-H}$ in $\ce{B}$ with $\ce{Br}$, we obviously get $\ce{C}$. Now since $\ce{H}<\ce{C}<\ce{Br}$ in the Cahn-Ingold-Prelog rules, $\ce{C}$ is R. But $\ce{C}$ should have been S.

Where is my logic wrong?

EDIT: As noted by A.K. below, there is no problem in this case, and $\ce{C}$ is in-fact S. But everything depends on the atomic number of $\ce{Br}$. Is there a general proof of correctness for the scheme?

• I think you need to draw out the molecule if you look at C you will see that the methyl group now has the low priority and that C is in fact S-configured.
– A.K.
Dec 24 '16 at 8:15
• I worked it out, and C is in-fact S-configured. But it all depends on the atomic number of the atom we use for substituting the $-H$. Can we prove this in general? Dec 24 '16 at 8:25
• This is not $\mathrm{S}_{N}2$. There's no leaving group.
– Zhe
Dec 24 '16 at 15:34
• @Zhe the $S_N2$ is hypothetical and irrelevant.
– A.K.
Dec 25 '16 at 5:09
• C is (R)-configured unless I am seriously mistaken.
– Jan
Dec 25 '16 at 10:34

I am going to use ‘Pseudo-Fischer’ structures in this post, i.e. anything vertical should be considered behind the PC screen plane and anything horizontal in front of it.

Your original molecule and the hypothetic $\mathrm{S_N2}$-type inversion is here:

   Pr            Pr
|             |
H-C--Me ---> Me-C--Br
|             |
Et            Et

(A)           (C)


If you determine both compounds’ absolute configuration, naturally we will have (R)-A but look closely at C and realise that it is also (R)-C. This is because of one of two fallacies students often have with CIP priorities (I’m presenting the other one here just for completeness):

• A chemical inversion of a specific stereocentre does not have to mean an inversion of the stereodescriptor.

• Just because a stereocentre does not participate in a reaction does not mean its stereodescriptor remains identical.

In our case, we can easily arrive at that conclusion by the two-step process you outlined. First, invert the original structure A to give B, then replace (under retention) the hydrogen in B to give C.

   Pr            Pr            Pr
|             |             |
H-C--Me ---> Me-C--H  ---> Me-C--Br
|             |             |
Et            Et            Et

(A)           (B)           (C)


We already established that A is (R)-A. Going to be, this is a strict inversion and only in this case of a strict inversion will the fallacy of bullet point one actually hold true; thus, it is (S)-B. Finally, we replace the atom of lowest priority with one that will automatically become the highest priority. To explain what happens requires further elaboration.

First, assume without loss of generalisation that whichever group of B had second priority (in our case: ethyl) is pointing upwards in a Pseudo-Fischer structure while whichever group originally had third priority in B (here: methyl) is pointing downwards. This gives us depiction 1.

   Et
|
Pr-C--H
|
Me

(1)


As we have the highest priority on the left above and the lowest priority on the right above, it is easy to see that this compound has an (S) configuration. If the hydrogen and propyl groups had been switched, we would have had (R). Now, perform the replacement in a manner such that the former fourth priority becomes first. The original third priority is now fourth, second became third and first became second. This is shown in depiction 2.

   Et
|
Pr-C--Br
|
Me

(2)


Now, since our lowest priority is pointing downwards, to assign a stereodescriptor in 2 we must look from above. The first and second priorities are above the PC screen plane and on the left and right-hand sides. The third priority is behind the plane and centred. It is rather easy to see that in the case of 2 as drawn, with the right-hand group having a higher priority than the left-hand group, the stereodescriptor is (R). If they were reversed, it would be (S). We see that replacing the fourth priority with something of highest priority will invert the stereochemistry.

What happens if you replace the lowest priority with something of intermediate priority (i.e. something that will come in second or third)? The third case is easy; you are just switching two groups. Switching two groups will always give you a formal inversion. What if our replacement group takes second priority? I refer back to depiction 1 but this time replace hydrogen with $\ce{BEt} = \ce{(pin)B-C2H4}$ giving depiction 3.

   Et
|
Pr-C--BEt
|
Me

(3)


Again, we must look from above. The highest priority is on the left-hand sinde in 3, the second is on the right-hand side, the third is in the centre and behind. Our journey is counterclockwise or (S). Had the original configuration been inverted it would be identical. Thus, replacing a fourth priority with something of second priority retains the stereochemistry.

This is a one step analysis on a per-case basis. A.K. showed the usage of the switching method (two groups switch at a time) to its fullest; thereby, these independently established rules can also be derived as was shown.

• I guess the two fallacies you outlined just solved my problem. A Walden inversion does not always imply a change from R to S or vv , if I got it right? Dec 25 '16 at 11:03
• @MrigankaBasuRoyChowdhury Absolutely.
– Jan
Dec 25 '16 at 11:06

Yes you can generalize the flip in chirality but with math rather than chemistry.

Given: For a four-coordinated molecule with one chiral center, if any two of the groups switch position, then the chirality of the molecule flips.

1.)From this, we can infer that twice switching any two groups produces the original molecule. Generalizing further we get that an odd number of two-group switches results in the enantiomer, while an even number of switches yields the starting molecule.

2.) Molecule A can have the positions ordered from low priority to high priority you get a set ike: [A, B, C, D] with order corresponding to position (shown below, ignore the hydrogens) and letter corresponding to priority (A-low, D-High). Molecule A (R-configuration)

3.) Your new molecule (C) can be represented as [D', A', B', C']. the question here is how many two-group switches does it take to get the molecule in the form of [A', B', C', D'] without any rotations of the molecule. Molecule C as substituted

4.) Crunching through we get:
[D', A', B', C'] (0)
[C', A', B', D'] (1)
[A', C', B', D'] (2)
[A', B', C', D'] (3)
It took 3 two-group switches to return the molecule to an R-configuration, therefore the molecule was in the S-configuration.

• I can see what you are trying to say, but can you please explain how you get [D',A',B',C'] as the new structure? I'm a new student, so I'm having difficulties. Thanks! Dec 25 '16 at 6:41
• Because the bromine substituted the hydrogen the lowest became the highest and the rest shifted down on priority.
– A.K.
Dec 25 '16 at 14:28