How can any of these compounds give same SN1 & SN2 product?

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The Attempt at a Solution


Is there some hindrance caused by the π bond in (c)?
How do I approach the problem anyway?
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  • #2
TeethWhitener
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What does the geometry of the carbon center do in the SN2 mechanism that is different from the SN1 mechanism?
 
  • #3
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What does the geometry of the carbon center do in the SN2 mechanism that is different from the SN1 mechanism?
It allows only the backside attack of the nucleophile, thus preventing the front side attack.
But front side attack isn't possible in (c) due to hindrance from the π bond?
Hence it'll give the same SN1 & SN2 product?
 
  • #4
TeethWhitener
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It allows only the backside attack of the nucleophile, thus preventing the front side attack.
It isn’t clear what you mean by this. Forget for a second things like steric hindrance, etc. that determine which mechanism dominates kinetically. Just focus on the geometries of the intermediates. What are the intermediates in SN1 and SN2 and what are their geometries?
 
  • #5
epenguin
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Think stereochemistry.
I would even think products. I would think for a simple substitution for which of the reactions can there be different products?
If there can't that would save me some thinking - overthinking in fact. And if there can be I could probably think what they are.
 
  • #6
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What are the intermediates in SN1 and SN2 and what are their geometries?
It's a carbocation (trigonal planar) in SN1.
In SN2, it's a transition state in which the base, the leaving group and the carbon atom bearing it are coplanar. The carbon is partially bonded to 5 atoms.
 
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  • #7
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The real problem is that I used to think that same SN1 & SN2 products are possible only if a molecule doesn't have a stereogenic center. What am I missing?
 
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  • #8
TeethWhitener
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The real problem is that I used to think that same SN1 & SN2 products are possible only if a molecule doesn't have a stereogenic center. What am I missing?
More precisely, if the substitution center isn't stereogenic. So which compound does that apply to?
 
  • #10
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So which compound does that apply to?
So do you mean to say it only applies to aliphatic hydrocarbon derivatives?? (Coz that's what I had dealt with uptil now)
 
  • #11
TeethWhitener
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So do you mean to say it only applies to aliphatic hydrocarbon derivatives?? (Coz that's what I had dealt with uptil now)
You're getting off-track. Look at the four compounds you were given. Which of them has a non-stereogenic substitution center?
 
  • #12
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Which of them has a non-stereogenic substitution center?
That would be (b). But that's not the answer they've given.
 
  • #13
TeethWhitener
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That would be (b). But that's not the answer they've given.
Well, it's the correct answer. I'd ask them why they think that b isn't correct.
 
  • #14
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Well, it's the correct answer. I'd ask them why they think that b isn't correct.
Well, they've given (c)
 
  • #15
TeethWhitener
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Well, they've given (c)
The only way c is correct is if the starting material is racemic in the first place. But that's also true of a and d. As you've already noted, b is the only compound where the substitution center is not stereogenic, meaning that the racemizing aspect of the SN1 mechanism and the stereoinverting aspect of the SN2 mechanism are irrelevant and give the same product.
 
  • #16
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Well, they've given (c)
Are you sure that the π bond won't hinder the back side attack (with respect to the leaving group) in (c)??
 
  • #17
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The only way c is correct is if the starting material is racemic in the first place. But that's also true of a and d. As you've already noted, b is the only compound where the substitution center is not stereogenic, meaning that the racemizing aspect of the SN1 mechanism and the stereoinverting aspect of the SN2 mechanism are irrelevant and give the same product.
Yes, that does seem convincing but I still think that the π bond in (c) would somehow prevent the leaving group and the nucleophile from being coplanar (as in the SN2 transition state).
 
  • #18
TeethWhitener
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Yes, that does seem convincing but I still think that the π bond in (c) would somehow prevent the leaving group and the nucleophile from being coplanar (as in the SN2 transition state).
1) The question merely asks about the products from SN1 vs. SN2, not the feasibility of the mechanisms.
2) The pi bond won't hinder anything. Take a look at cyclohexene:
https://en.wikipedia.org/wiki/Cyclohexene
Nothing is hindered at the allyl center.
 
  • #19
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Well, assuming (b) to be correct, I now realise how easy the question was. I somehow just didn't realise that the substitution center in (b) wasn't stereogenic ( Probably due to anxiety during the exam). I should've got this one.
Thank you
 
  • #20
TeethWhitener
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No problem. And if it turns out that the professor has a good reason why c is correct, I'm curious to know what it is. :smile:
 
  • #21
Ygggdrasil
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I don't think the problem has to do with inversion of stereocenters in SN2 vs SN1 reactions (plus none of the reactions would result in optically active products). Rather, the correct answer has to do with thinking about the various rearrangements that the carbocation intermedia would undergo in an SN1 reaction vs an SN2 reaction.
 
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  • #22
TeethWhitener
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(plus none of the reactions would result in optically active products)
How do you figure? a, b, and c all have one stereocenter, and d has 2.
 
  • #23
Ygggdrasil
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How do you figure? a, b, and c all have one stereocenter, and d has 2.
Would these be meso compounds because the plane of the page is a plane of symmetry?
 
  • #24
TeethWhitener
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@Ygggdrasil is definitely right here. As long as you assume that c is a racemic mixture to start with (not an unreasonable assumption), consideration of carbocation rearrangement leads to the answer c. (If you don’t assume a racemic mixture, all of them change.)
Would these be meso compounds because the plane of the page is a plane of symmetry?
I don’t think so. The symmetry is broken by the out of plane chloro groups in a and c and by the chloromethyl group in b. D is only meso if it’s R-chloro-S-methyl (or vice versa). (Edit: taking rearrangement into account,) The SN1 products all end up being achiral except for b (and there might be a mixture of alkyl and hydride shift in b).
 
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  • #25
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(Edit: taking rearrangement into account,) The SN1 products all end up being achiral except for b (and there might be a mixture of alkyl and hydride shift in b).
How do they end up being achiral? Do you mean by racemisation?
 
  • #26
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@Ygggdrasil is definitely right here. As long as you assume that c is a racemic mixture to start with (not an unreasonable assumption), consideration of carbocation rearrangement leads to the answer c.
But what will the carbocation rearrange to?And why?
There can be a 3° carbocation in preference to a 2° only in (d) & (b).
 
  • #27
TeethWhitener
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How do they end up being achiral? Do you mean by racemisation?
No, by rearrangement. Racemization is the deciding factor for compound c. It also happens in a but it's not the deciding factor; also, the rearrangement in d gives an achiral product.

But what will the carbocation rearrange to?And why?
There can be a 3° carbocation in preference to a 2° only in (d) & (b)
Remember the order of carbocation stability: allyl/benzyl > 3° > 2° > 1° > methane

Hey, tell me something. Why'd everybody go cold?
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I also occasionally sleep.
 
  • #28
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No, by rearrangement. Racemization is the deciding factor for compound c. It also happens in a but it's not the deciding factor; also, the rearrangement in d gives an achiral product.


Remember the order of carbocation stability: allyl/benzyl > 3° > 2° > 1° > methane
But even (a) can give an allyl carbocation (hydride shift). So how do you decide whether reacemisation is the deciding factor or rearrangement?
 
  • #29
TeethWhitener
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But even (a) can give an allyl carbocation (hydride shift). So how do you decide whether reacemisation is the deciding factor or rearrangement?
You kind of answered your own question here. In a, the dominant SN1 intermediate will be the (rearranged) allyl cation, which will give a different product from the SN2 mechanism, without even having to consider stereochemistry. In c, no rearrangement has to occur to give the allyl cation as the SN1 intermediate, so the only difference between SN1 and SN2 would be due to racemization/stereoinversion. But that is only relevant if you assume that c wasn't a racemic mixture to begin with.
 
  • #30
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You kind of answered your own question here. In a, the dominant SN1 intermediate will be the (rearranged) allyl cation, which will give a different product from the SN2 mechanism, without even having to consider stereochemistry. In c, no rearrangement has to occur to give the allyl cation as the SN1 intermediate, so the only difference between SN1 and SN2 would be due to racemization/stereoinversion. But that is only relevant if you assume that c wasn't a racemic mixture to begin with.
Oh yes, I get it fully now. :smile:
But is there a general rule so as to decide which factor dominates under what conditions?
 

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