How can any of these compounds give same SN1 & SN2 product?
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Discussion Overview
The discussion revolves around the mechanisms of SN1 and SN2 reactions and how certain compounds can yield the same products through both pathways. Participants explore the influence of molecular geometry, stereochemistry, and the presence of π bonds on the reaction mechanisms and outcomes.
Discussion Character
- Debate/contested
- Technical explanation
- Conceptual clarification
Main Points Raised
- Some participants question the role of hindrance caused by π bonds in affecting the SN2 mechanism compared to SN1.
- There is a discussion on the geometries of intermediates in SN1 (carbocation, trigonal planar) and SN2 (transition state, coplanar arrangement) reactions.
- Some participants suggest that the same products from SN1 and SN2 are possible if the substitution center is not stereogenic.
- Others argue that the presence of a racemic mixture in the starting material is crucial for the correctness of the proposed answers.
- There is a debate on whether certain compounds can lead to achiral products due to rearrangements in the carbocation intermediates.
- Some participants express uncertainty about the conditions under which racemization or rearrangement is the deciding factor in product formation.
Areas of Agreement / Disagreement
Participants express differing views on the influence of stereogenic centers and the role of π bonds in determining the products of SN1 and SN2 reactions. The discussion remains unresolved, with multiple competing perspectives on the mechanisms and outcomes.
Contextual Notes
There are limitations in the assumptions made regarding the stereogenic nature of substitution centers and the implications of carbocation stability on the reaction pathways. The discussion also highlights the complexity of determining the dominant mechanism based on molecular structure.
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It allows only the backside attack of the nucleophile, thus preventing the front side attack.TeethWhitener said:What does the geometry of the carbon center do in the SN2 mechanism that is different from the SN1 mechanism?
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?
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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?HPPAS said:It allows only the backside attack of the nucleophile, thus preventing the front side attack.
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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.
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It's a carbocation (trigonal planar) in SN1.TeethWhitener said:What are the intermediates in SN1 and SN2 and what are their geometries?
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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More precisely, if the substitution center isn't stereogenic. So which compound does that apply to?HPPAS said: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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Yes, that would be more precise.TeethWhitener said:More precisely, if the substitution center isn't stereogenic.
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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)TeethWhitener said:So which compound does that apply to?
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You're getting off-track. Look at the four compounds you were given. Which of them has a non-stereogenic substitution center?HPPAS said:So do you mean to say it only applies to aliphatic hydrocarbon derivatives?? (Coz that's what I had dealt with uptil now)
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That would be (b). But that's not the answer they've given.TeethWhitener said:Which of them has a non-stereogenic substitution center?
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Well, it's the correct answer. I'd ask them why they think that b isn't correct.HPPAS said:That would be (b). But that's not the answer they've given.
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Well, they've given (c)TeethWhitener said:Well, it's the correct answer. I'd ask them why they think that b isn't correct.
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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.HPPAS said:Well, they've given (c)
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Are you sure that the π bond won't hinder the back side attack (with respect to the leaving group) in (c)??HPPAS said:Well, they've given (c)
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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).TeethWhitener said: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.
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1) The question merely asks about the products from SN1 vs. SN2, not the feasibility of the mechanisms.HPPAS said: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).
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.
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Thank you
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How do you figure? a, b, and c all have one stereocenter, and d has 2.Ygggdrasil said:(plus none of the reactions would result in optically active products)
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Would these be meso compounds because the plane of the page is a plane of symmetry?TeethWhitener said:How do you figure? a, b, and c all have one stereocenter, and d has 2.
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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).Ygggdrasil said:Would these be meso compounds because the plane of the page is a plane of symmetry?
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How do they end up being achiral? Do you mean by racemisation?TeethWhitener said:(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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But what will the carbocation rearrange to?And why?TeethWhitener said:@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.
There can be a 3° carbocation in preference to a 2° only in (d) & (b).
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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.HPPAS said:How do they end up being achiral? Do you mean by racemisation?
Remember the order of carbocation stability: allyl/benzyl > 3° > 2° > 1° > methaneHPPAS said:But what will the carbocation rearrange to?And why?
There can be a 3° carbocation in preference to a 2° only in (d) & (b)
1. JobHPPAS said:Hey, tell me something. Why'd everybody go cold?
2. Family
I also occasionally sleep.
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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?TeethWhitener said: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
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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.HPPAS said:But even (a) can give an allyl carbocation (hydride shift). So how do you decide whether reacemisation is the deciding factor or rearrangement?
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Oh yes, I get it fully now.TeethWhitener said: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.
But is there a general rule so as to decide which factor dominates under what conditions?
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