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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?
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.
It's a carbocation (trigonal planar) in SN1.TeethWhitener said:What are the intermediates in SN1 and SN2 and what are their geometries?
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?
Yes, that would be more precise.TeethWhitener said:More precisely, if the substitution center isn't stereogenic.
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?
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)
That would be (b). But that's not the answer they've given.TeethWhitener said:Which of them has a non-stereogenic substitution center?
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.
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.
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)
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)
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.
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).
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)
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.
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?
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).
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.
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?
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
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?
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.
It's not a matter of one factor dominating over another (at least not for this question). Rearrangements will happen, and stereochemical effects will happen. I unfortunately didn't think of rearrangements when I initially answered your question, but Ygggdrasil did, and his answer was therefore on the money. If you consider only rearrangements, then c will be the correct answer. If you consider rearrangements and stereochemistry, c is still correct, assuming that you start with a racemic mixture of c. If you assume that you start with one enantiomer of c, then there is no correct answer (SN1 and SN2 give different compunds in every case).HPPAS said:But is there a general rule so as to decide which factor dominates under what conditions?
Oh. Yes, that last bit was very helpful thanks to @Ygggdrasil .TeethWhitener said:It's not a matter of one factor dominating over another (at least not for this question). Rearrangements will happen, and stereochemical effects will happen. I unfortunately didn't think of rearrangements when I initially answered your question, but Ygggdrasil did, and his answer was therefore on the money. If you consider only rearrangements, then c will be the correct answer. If you consider rearrangements and stereochemistry, c is still correct, assuming that you start with a racemic mixture of c. If you assume that you start with one enantiomer of c, then there is no correct answer (SN1 and SN2 give different compunds in every case).
SN1 and SN2 reactions both involve the substitution of a leaving group with a nucleophile, resulting in the same product. However, the mechanisms of these reactions are different.
SN1 reactions involve a two-step mechanism, where the leaving group first dissociates to form a carbocation intermediate, which is then attacked by the nucleophile. On the other hand, SN2 reactions involve a one-step mechanism, where the nucleophile attacks the substrate while the leaving group is still attached.
The structure of the substrate can greatly affect the outcome of the SN1 and SN2 reactions. For SN1 reactions, a more stable carbocation intermediate is favored, while for SN2 reactions, a less hindered substrate is preferred to allow for easier nucleophilic attack.
Yes, there are several factors that can help predict the mechanism of a reaction, such as the strength of the nucleophile, the steric hindrance of the substrate, and the solvent used. Generally, more polar solvents favor SN1 reactions, while less polar solvents favor SN2 reactions.
Yes, some compounds can undergo both SN1 and SN2 reactions, depending on the reaction conditions. For example, a primary substrate may undergo SN2 reaction in the presence of a strong nucleophile, but can also undergo SN1 reaction in the presence of a polar solvent and weak nucleophile.