SN1/SN2 Racemic Mixture

[Pisiform]

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Can someone please explain me about racemization in SN1 and SN2, the main idea we need to know for the mcat. And also this question please:

The reaction R—Br + Br*–→ R—Br* + Br– is always accompanied by inversion. If this reaction is carried out on an optically pure sample of a chiral compound, which of the following statements will be true? [Note: Br* represents a radioactive isotope of bromine.]

A - The rate of Br* incorporation is half the rate of racemization.
B - The rate of Br* incorporation is equal to the rate of racemization.
C - The rate of Br* incorporation is twice the rate of racemization.
D - The relation between the rate of Br* incorporation and the rate of racemization cannot be determined.

Thanks

 

[Pisiform]

forgot to post the answer, answer is A

 

[MedPR]

Can someone please explain me about racemization in SN1 and SN2, the main idea we need to know for the mcat. And also this question please:

The reaction R—Br + Br*–→ R—Br* + Br– is always accompanied by inversion. If this reaction is carried out on an optically pure sample of a chiral compound, which of the following statements will be true? [Note: Br* represents a radioactive isotope of bromine.]

A - The rate of Br* incorporation is half the rate of racemization.
B - The rate of Br* incorporation is equal to the rate of racemization.
C - The rate of Br* incorporation is twice the rate of racemization.
D - The relation between the rate of Br* incorporation and the rate of racemization cannot be determined.

Thanks

Is the answer D?

I think it would depend on what kind of "inversion" they are talking about.

If it's SN2, the inversion may or may not lead to an inversion of absolute configuration, and thus may not change the optical activity and therefore wouldn't produce racemization.

If "inversion" means "inversion of absolute configuration" then I think the answer would be A.

In SN1 racemization the reaction goes from 100% (R) or (S) to 50-50 (R) and (S). So I would assume the answer is that the rate of racemization is twice the rate of Br* incorporation. If you have 100 molecules, you only have to incorporate 50 Br* to have 100% racemization.. I don't know if that's right but that is how I would approach the problem.

Edit: Btw I've been hoping you would post a question so I could help you out because you've helped me a ton so far and I'm only 20ish days in. I was also hoping your eventual question would be one that I had a definitive answer for 🙁

 

[Pisiform]

Is the answer D?

I think it would depend on what kind of "inversion" they are talking about.

If it's SN2, the inversion may or may not lead to an inversion of absolute configuration, and thus may not change the optical activity and therefore wouldn't produce racemization.

If "inversion" means "inversion of absolute configuration" then I think the answer would be A.

In SN1 racemization the reaction goes from 100% (R) or (S) to 50-50 (R) and (S). So I would assume the answer is that the rate of racemization is twice the rate of Br* incorporation. If you have 100 molecules, you only have to incorporate 50 Br* to have 100% racemization.. I don't know if that's right but that is how I would approach the problem.

Edit: Btw I've been hoping you would post a question so I could help you out because you've helped me a ton so far and I'm only 20ish days in. I was also hoping your eventual question would be one that I had a definitive answer for 🙁

see, I chose D, because I thought SN2 has nothing to do with racemic mixture ... i dunno ... here is the explanation that didnt make sense to me 😕

The question states that the reaction is always accompanied by inversion. An inversion of stereochemistry can only result from an SN2 reaction. Each time an alkyl halide reacts with a radioactive bromine atom, it will be inverted from an R to S configuration or vice versa. Each molecule inverted therefore “cancels out” one of the original molecules. Also, each reaction leads to the racemization of two molecules of alkyl halide. When half of the radioactive bromine is incorporated, the entire solution is racemic; the rate of radioactive bromine incorporation is equal to half of the rate of racemization.

 

[chiddler]

Can someone please explain me about racemization in SN1 and SN2, the main idea we need to know for the mcat. And also this question please:

The reaction R—Br + Br*–→ R—Br* + Br– is always accompanied by inversion. If this reaction is carried out on an optically pure sample of a chiral compound, which of the following statements will be true? [Note: Br* represents a radioactive isotope of bromine.]

A - The rate of Br* incorporation is half the rate of racemization.
B - The rate of Br* incorporation is equal to the rate of racemization.
C - The rate of Br* incorporation is twice the rate of racemization.
D - The relation between the rate of Br* incorporation and the rate of racemization cannot be determined.

Thanks

Because SN2 is the only way for inversion to occur, the half of the SN2's that cause inversion is equal to the rate of incorporation of Br*. Therefore, Br* incorporation = half SN2 rate so the answer must be A.

I'm never certain about my answers except in bio so...this is what makes sense to me. :x This took a lot of thinking. I really hope mcat is not as brutal.

 

[Pisiform]

so does this mean if we add 100 molecules of substrate and 50 molecules of nulceophile in SN2 ... we would get racemic mixture?

so does both SN1 and SN2 causes racemization?

 

[MedPR]

If there's no inversion in SN1, then how does SN1 produce racemization?

If you have (R) reactant and SN1 nuc attacks the carbocation from the opposite side of the original leaving group, then you have (S), no? I guess it isn't considered inversion because of the carbocation intermediate?

I've also never heard of SN2 being responsible for racemization, though I guess it makes sense if your nucleophile is the limiting reagent and you 2 moles of reactant per 1 mole of nucleophile, thus inverting 50% and leaving the other 50% and creating a racemic mixture, which is what Pisiform said above.

 

[SaintJude]

Yes, SN2 CAN lead to racemization. See my note:

MCAT Essentials
Yes, SN1 leads to racemization.
Sn1 involves the formation of a carbocation, so it generally leads to a racemic product. Why? B/c of its reaction mechanism. Recall that first the leaving group leaves to form a sp^2 carbocation, which is trigonal planar. Therefore the nucleophile can attack that carbocation from either side, and the reaction would produce equal amounts of all possible configurations.

Sn2 always leads to an inversion of relative configuration change and so the product will always be optically active if the reactant was too.

Note on Racemization via SN2 rx:
Since MCAT loves explaining interesting circumstances, I'll add this too:
In case the leaving group and the nucleophile are the same molecule in an SN2 rx, the reaction will become reversible, an equilibrium will be set up leading to racemization. (Ex: 2-iodoctane + sodium iodide)

On to the question : If rx is always accompanied by an inversion, I assumed it was an SN2 reaction. And what's happening is actually the phenomenon that I just explained in my "special note"--this is a racemization via an SN2 equilibrium. Step-by-step logic:

1. Every time the an SN2 rx happens, an inversion occurs.

2. Racemization in a solution occurs when we have an equal amount of all configuration. So 50-50 in this case.

3. So racemization will therefore be complete when HALF of the material gets inverted. B/c then you'll have 50% inverted and 50% of "starting" configuration.
4.So rate of racemization is 2x of incorporation of Br*.

 

[MedPR]

Ok so we're all basically saying the same thing at this point. Good to know.

 

[Pisiform]

Thanks MedPR and StJude .... this def makes alot more sense. I was born with orgo curse lol

 

[chiddler]

Ok so we're all basically saying the same thing at this point. Good to know.

Except for me! I was wrong.

Question: How do we know that the said reaction is SN2?

NEVERMIND SN1 generally only occurs with tertiaries.

I understand.

 

[MedPR]

Except for me! I was wrong.

Question: How do we know that the said reaction is SN2?

NEVERMIND SN1 generally only occurs with tertiaries.

I understand.

It doesn't apply here, since SN2 is implied by "always involves inversion," but in future problems R-Br could be tertiary.