Racemic Mixture OChem question

[weanprednisone]

Question:

Answer: A
My answer was different from what the Kaplan answer was, any thoughts +reasoning?

My reasoning:
My answer is D.
With the polar aprotic solvent, this leads us to a SN2 reaction. Meaning N3- (nucleophile) will attack from the backside or front side. Iodo- is the best leaving group; I- is a stable ion in the medium.
Since SN2 is able to attack from both sides, then wouldn't R and S be possible, therefore making a racemic mixture? A racemic mixture is also possible because the end product is chiral.

What are your thoughts

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[Necr0sis713]

SN2 cant attack from both sides. It always does backside attack, because it has to kick off the leaving group at the same time its attacking. The leaving group can't leave from the same face where the nucleophile is attacking; makes no sense.

 

[mourinho]

As above, SN2 can't attack from both sides, so the answer should be C.

 

[Necr0sis713]

To make it clearer: the only reason why you get a racemic mixture from sn1 is because the leaving group leaves BEFORE the attack, so the nucleophile can attack from anywhere. There is essentially a naked carbocation open for attack.

 

[Nevadatim89]

I think of it like this. In an Sn2 rxn stereochemistry is usually inverted. The only way this does not happen is when there happens to be a substituent on the electrophile that has a higher priority than the newly added nucleophile. In this problem the electrophile started in an R configuration. The F has the highest priorty (4), then I > R group > H.... the incoming Nitro nucleophile replaced the I with an N group. This new molecule still has the high priorty F then N, R group and H as least prioritized. Since the I and N are both prioratized under F, the stereochemistry isnt changed.

 

[mourinho]

If they don't say if the attack is front or back, how are we supposed to know if the molecule retains its configuration or not?

 

[theonlytycrane]

If they don't say if the attack is front or back, how are we supposed to know if the molecule retains its configuration or not?

The polar aprotic solvent indicates it's SN2

 

[mourinho]

I don't have a problem with it being SN2 but it remaining R configuration and not S.

 

[Nevadatim89]

I don't have a problem with it being SN2 but it remaining R configuration and not S.

Ignore my previous post. It remains in R configuration because the new nucleophile has a higher priority than the F substituent, so just draw it out and label the inverted molecule. You will see that even when its inverted on paper it will still be R configuration.

 

[PassEdge]

There's two parts to this. Identifying that it's SN2, and figuring out the orientation of the final product. Messing up either will give you the wrong result.

Choosing the racemic mixture means that you thought it would be SN1 which is incorrect. Usually the group can't just leave by itself unless it's a tertiary carbon.

 

[weanprednisone]

So in terms of solvents. Can I conclude:
Aprotic solvents: Sn2 Never will have a racemic mixture.
Protic solvents: Sn1 (backside and frontside attack) Can have racemic mixture only if the molecule is chiral.
But the solvent can never conclude the stereochemistry.

 

[Nevadatim89]

So in terms of solvents. Can I conclude:
Aprotic solvents: Sn2 Never will have a racemic mixture.
Protic solvents: Sn1 (backside and frontside attack) Can have racemic mixture only if the molecule is chiral.
But the solvent can never conclude the stereochemistry.

Protic solvents produce sn1 rxns which could have racemic mixtures if the carbocation intermediate is chiral. The ansers A though

 

[aalamruad]

Ignore my previous post. It remains in R configuration because the new nucleophile has a higher priority than the F substituent, so just draw it out and label the inverted molecule. You will see that even when its inverted on paper it will still be R configuration.

No, the new nucleophile does not have a higher priority than the F substituent (9>7 : F>N). The reason it remains R is because it has a LOWER priority + inversion. If it had a higher priority like you said, then there would be conversion from R to S with inversion (which is what occurs in most cases of SN2).

 

[plukfelder2017]

polar aprotic solvent on a 1 prime substrate. You do not form primary carbocations because they are energetically unfavorable. You have an excellent leaving group and an excellent nucleophile. How can this be racemic? This could only proceed via backside attack. Once you deal with your Cahn Ingold priority assignment, you realize that N is a lower priority than F because you assign priority based off of atomic number. This means that you end up with a retention of configuration only due to your assignment of Cahn Ingold priority and you only obtain this after you invert configuration when you allow N3 to perform a backside attack on the substrate. Kapiche?

 

[weanprednisone]

Thank you for all the replies. In case anyone was wondering, the answer is A. All the correct reasoning are present.