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

In summary, certain compounds can give the same product in both SN1 and SN2 reactions due to their similar reaction mechanisms. These compounds, known as ambident nucleophiles, have multiple reactive sites that can be attacked by nucleophiles in both SN1 and SN2 reactions. Additionally, solvent effects and steric hindrance can also play a role in determining the final product of these reactions. This highlights the importance of understanding the underlying chemistry and factors that influence these reactions in order to predict and control the outcome.
  • #1
baldbrain
236
21

The Attempt at a Solution


Is there some hindrance caused by the π bond in (c)?
How do I approach the problem anyway?
20180606_230918.jpg
 

Attachments

  • 20180606_230918.jpg
    20180606_230918.jpg
    20.9 KB · Views: 1,402
Physics news on Phys.org
  • #2
What does the geometry of the carbon center do in the SN2 mechanism that is different from the SN1 mechanism?
 
  • #3
TeethWhitener said:
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
HPPAS said:
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
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.
 
  • Like
Likes baldbrain
  • #6
TeethWhitener said:
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.
 
Last edited:
  • Like
Likes TeethWhitener
  • #7
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?
 
  • Like
Likes TeethWhitener
  • #8
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?
More precisely, if the substitution center isn't stereogenic. So which compound does that apply to?
 
  • #9
TeethWhitener said:
More precisely, if the substitution center isn't stereogenic.
Yes, that would be more precise.
 
  • #10
TeethWhitener said:
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
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)
You're getting off-track. Look at the four compounds you were given. Which of them has a non-stereogenic substitution center?
 
  • #12
TeethWhitener said:
Which of them has a non-stereogenic substitution center?
That would be (b). But that's not the answer they've given.
 
  • #13
HPPAS said:
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.
 
  • Like
Likes baldbrain
  • #14
TeethWhitener said:
Well, it's the correct answer. I'd ask them why they think that b isn't correct.
Well, they've given (c)
 
  • #15
HPPAS said:
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
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)??
 
  • #17
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.
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
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).
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
Well, assuming (b) to be correct, I now realize how easy the question was. I somehow just didn't realize 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
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
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.
 
  • Like
Likes TeethWhitener
  • #22
Ygggdrasil said:
(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
TeethWhitener said:
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
@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.)
Ygggdrasil said:
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).
 
  • Like
Likes baldbrain and Ygggdrasil
  • #25
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).
How do they end up being achiral? Do you mean by racemisation?
 
  • #26
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.
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
HPPAS said:
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.

HPPAS 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)
Remember the order of carbocation stability: allyl/benzyl > 3° > 2° > 1° > methane

HPPAS said:
Hey, tell me something. Why'd everybody go cold?
1. Job
2. Family
I also occasionally sleep.
 
  • Like
Likes baldbrain
  • #28
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
But even (a) can give an allyl carbocation (hydride shift). So how do you decide whether reacemisation is the deciding factor or rearrangement?
 
  • #29
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?
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.
 
  • Like
Likes baldbrain
  • #30
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.
Oh yes, I get it fully now. :smile:
But is there a general rule so as to decide which factor dominates under what conditions?
 
  • #31
HPPAS said:
But is there a general rule so as to decide which factor dominates under what conditions?
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).
 
  • #32
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).
Oh. Yes, that last bit was very helpful thanks to @Ygggdrasil .
All my concepts are quite clear, but I still **** up during exams (probably due to haste). I also solve a lot of problems, I can get them quicker when there's no time limit.
Sometimes, the opposite happens. I get the good problems, and **** up the easy ones. Can you give me some advice regarding this?
 

1. How do SN1 and SN2 reactions produce the same product?

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.

2. Can you explain the differences between SN1 and SN2 reactions?

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.

3. How do the structures of the compounds affect the SN1 and SN2 products?

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.

4. Is there a way to predict whether a compound will undergo SN1 or SN2 reaction?

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.

5. Can the same compound undergo both SN1 and 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.

Similar threads

  • Biology and Chemistry Homework Help
Replies
2
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
  • Biology and Chemistry Homework Help
Replies
1
Views
630
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
  • Biology and Chemistry Homework Help
Replies
6
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
10
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
7K
Back
Top