E1/SN1 question

[Ashley3323]

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hello everyone,
I have these following questions and I cant understand why they are E2 and SN1 . please excuse me if they are very simple , this is my weak area.

 

[hockey553]

For the E2 one there is a strong small base, the substrate is secondary and hindered by the R group on the cyclo ring. Therefore, it is E2.

The second one is SN1 because of the secondary substrate combined with a weak nucleophile. It would also be E1, Sn1 and E1 always compete. There is no heat, however, E1 products would still form.

 

[Ashley3323]

For the E2 one there is a strong small base, the substrate is secondary and hindered by the R group on the cyclo ring. Therefore, it is E2.

The second one is SN1 because of the secondary substrate combined with a weak nucleophile. It would also be E1, Sn1 and E1 always compete. There is no heat, however, E1 products would still form.

Thank you for responding hockey . My main issue is that ive seen questions where ch3oo-na+ is considered a good nucleophile ? am I wrong? and the first question , do we ignore the polar protic solution?

 

[hockey553]

Thank you for responding hockey . My main issue is that ive seen questions where ch3oo-na+ is considered a good nucleophile ? am I wrong? and the first question , do we ignore the polar protic solution?

http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

Read this site it explains SN1/SN2/E1/E2 well

And CH3OO- is a weak/moderate nucleophile due to resonance. The negative charge is spread out and stabilized. E2 would be favored anyway if the substrate is secondary and there is a strong small base. If the substrate is primary and you use a small strong base like CH3O- then SN2 is almost always favored.

• Quick N’ Dirty Rule #5: Polar protic solvents tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2)

 

[Ashley3323]

oh wow ! that's perfect!!! thank you !!!

 

[DAT Destroyer]

View attachment 192982 hello everyone,
I have these following questions and I cant understand why they are E2 and SN1 . please excuse me if they are very simple , this is my weak area.

The first reaction is a secondary halide being treated with a strong base; this is the classical E2 reaction conditions. Make sure to put it in the chair conformation with the halogen going in the axial position.

The 2nd reaction is written poorly since no solvent has been specified. When acetate reacts with the secondary halide the SN2 mechanism is usually preferred. The texts by Klein and Carey offer excellent reviews of this topic. Marc Loudon also does a nice job.


Hope this helps.


Dr. Jim Romano

 

[510586]

The first reaction is a secondary halide being treated with a strong base; this is the classical E2 reaction conditions. Make sure to put it in the chair conformation with the halogen going in the axial position.

The 2nd reaction is written poorly since no solvent has been specified. When acetate reacts with the secondary halide the SN2 mechanism is usually preferred. The texts by Klein and Carey offer excellent reviews of this topic. Marc Loudon also does a nice job.


Hope this helps.


Dr. Jim Romano

Chad says that NaOCH3 can be a strong nuc as well so it could be both SN2 and E2 apparently. Question of the day for June 8th had a question like this and the explanation was that if it is a tie between E2 and SN2, E2 is favored in protic, SN2 in aprotic. That's what Ari told me when I emailed him about it as well. @Ari Rezaei "Think of protic solvents as a spider web. They keep interacting with the nuc and get tangled up in the protic sea, so they can't get enough energy to push through with a SN2 reaction. In an aprotic solvent, this isn't a problem and there is no spider web to get tangled up in, so it gets more energy and can SN2 more easily"

 

[DAT Destroyer]

Chad says that NaOCH3 can be a strong nuc as well so it could be both SN2 and E2 apparently. Question of the day for June 8th had a question like this and the explanation was that if it is a tie between E2 and SN2, E2 is favored in protic, SN2 in aprotic. That's what Ari told me when I emailed him about it as well. @Ari Rezaei "Think of protic solvents as a spider web. They keep interacting with the nuc and get tangled up in the protic sea, so they can't get enough energy to push through with a SN2 reaction. In an aprotic solvent, this isn't a problem and there is no spider web to get tangled up in, so it gets more energy and can SN2 more easily"

Perhaps you need to consult a book written by a Ph.D in Chemistry.

The David Klein books explains this in nice detail. NaOCH3 can indeed do an SN2 reaction if on a primary halide. If you have a secondary halide,,,,, Methoxide and Ethoxide favor Sn2 reactions on secondary halides, and tertiary halides. I did these reactions many times while in Grad school , and they give great yields of Zaitsev alkenes. Indeed , some Sn2 competition can result.

In the book by written Maitland Jones, PhD , Professor Emeritus of Princeton who ran the reaction found approximately 70% of the Zaitsev product.

In the Paula Bruice textbook, she ran a secondary halide under these conditions and found 75% alkene....all these by the E2 pathway.

There is no debate here,,,,,,this is a common " nuts and bolts " type of question. Don't focus on the solvents methanol or ethanol when looking at this.....these strong nucleophiles do E2 mechanism. In the new Klein book, Dr. Klein gives a nice summary table on page 389 of the second edition. He also states that a secondary halide gives E2 as the major product when treated with sodium methoxide/methanol.....or sodium ethoxide/ethanol.

In real life.....we can turn up the heat and get even more E2 product.

See the attachment from the text written by PhD Organic Chemist, Dr. Seyhan Ege of the University of Michigan.

I hope this helps.

 

[510586]

Perhaps you need to consult a book written by a Ph.D in Chemistry.

The David Klein books explains this in nice detail. NaOCH3 can indeed do an SN2 reaction if on a primary halide. If you have a secondary halide,,,,, Methoxide and Ethoxide favor Sn2 reactions on secondary halides, and tertiary halides. I did these reactions many times while in Grad school , and they give great yields of Zaitsev alkenes. Indeed , some Sn2 competition can result.

In the book by written Maitland Jones, PhD , Professor Emeritus of Princeton who ran the reaction found approximately 70% of the Zaitsev product.

In the Paula Bruice textbook, she ran a secondary halide under these conditions and found 75% alkene....all these by the E2 pathway.

There is no debate here,,,,,,this is a common " nuts and bolts " type of question. Don't focus on the solvents methanol or ethanol when looking at this.....these strong nucleophiles do E2 mechanism. In the new Klein book, Dr. Klein gives a nice summary table on page 389 of the second edition. He also states that a secondary halide gives E2 as the major product when treated with sodium methoxide/methanol.....or sodium ethoxide/ethanol.

In real life.....we can turn up the heat and get even more E2 product.

See the attachment from the test written by PhD Organic Chemist, Dr. Seyhan Ege of the University of Michigan.

I hope this helps.
View attachment 193053

Yea it definitely does. Hopefully the people who write the DAT go this in depth then.. It's hard to know what to go by! Various sources that were used by different people who have scored exceptionally high sometimes say different things. Thank you for this in depth response.

 

[DAT Destroyer]

Yea it definitely does. Hopefully the people who write the DAT go this in depth then.. It's hard to know what to go by! Various sources that were used by different people who have scored exceptionally high sometimes say different things. Thank you for this in depth response.

Keep up the good work, Organic Chemistry is very hard and I have been studying if for over 30 years, I continue to study to this very day so don't feel so bad on topics that are not so straight forward. There are text books written on solvents and nucleophiles that go over a 1000 pages.

Dr. Romano

 

[510586]

Keep up the good work, Organic Chemistry is very hard and I have been studying if for over 30 years, I continue to study to this very day, don't feel so bad.

Dr. Romano

Thanks! By the way, I apologize if some of the stuff I say on these forums seems to contradict destroyer answers or questions them. I am not trying to point fingers or accuse things of being right or wrong. I am often just trying to get to the bottom of certain topics and make sure what I am studying goes along with what the DAT thinks is the correct answer. I really appreciate the destroyer and your help on these forums!

 

[DAT Destroyer]

Thanks! By the way, I apologize if some of the stuff I say on these forums seems to contradict destroyer answers or questions them. I am not trying to point fingers or accuse things of being right or wrong. I am often just trying to get to the bottom of certain topics and make sure what I am studying goes along with what the DAT thinks is the correct answer. I really appreciate the destroyer and your help on these forums!

Albert Einstein said it best

" The important thing is not to stop questioning"

Dr. Romano

 

[orgotutor2015]

I completely agree with Dr. Romano on this one. SN2 and E2 reactions will compete here, but E2 will definitely prove to have the majority of the product. Even though it is not given, temperature can really push which way a reaction will go (lower temp - SN2, higher temp - E2). The solvent as mentioned before can also influence what type of product is formed (aprotic - sn2, protic - E2). Happy studying!

 

[Ashley3323]

The first reaction is a secondary halide being treated with a strong base; this is the classical E2 reaction conditions. Make sure to put it in the chair conformation with the halogen going in the axial position.

The 2nd reaction is written poorly since no solvent has been specified. When acetate reacts with the secondary halide the SN2 mechanism is usually preferred. The texts by Klein and Carey offer excellent reviews of this topic. Marc Loudon also does a nice job.


Hope this helps.


Dr. Jim Romano

Thank you very much for the explanation!!

 

[510586]

Perhaps you need to consult a book written by a Ph.D in Chemistry.

The David Klein books explains this in nice detail. NaOCH3 can indeed do an SN2 reaction if on a primary halide. If you have a secondary halide,,,,, Methoxide and Ethoxide favor Sn2 reactions on secondary halides, and tertiary halides. I did these reactions many times while in Grad school , and they give great yields of Zaitsev alkenes. Indeed , some Sn2 competition can result.

In the book by written Maitland Jones, PhD , Professor Emeritus of Princeton who ran the reaction found approximately 70% of the Zaitsev product.

In the Paula Bruice textbook, she ran a secondary halide under these conditions and found 75% alkene....all these by the E2 pathway.

There is no debate here,,,,,,this is a common " nuts and bolts " type of question. Don't focus on the solvents methanol or ethanol when looking at this.....these strong nucleophiles do E2 mechanism. In the new Klein book, Dr. Klein gives a nice summary table on page 389 of the second edition. He also states that a secondary halide gives E2 as the major product when treated with sodium methoxide/methanol.....or sodium ethoxide/ethanol.

In real life.....we can turn up the heat and get even more E2 product.

See the attachment from the text written by PhD Organic Chemist, Dr. Seyhan Ege of the University of Michigan.

I hope this helps.

View attachment 193053

Sorry for bumping this old post, but when you said "The David Klein books explains this in nice detail. NaOCH3 can indeed do an SN2 reaction if on a primary halide. If you have a secondary halide,,,,, Methoxide and Ethoxide favor Sn2 reactions on secondary halides, and tertiary halides." did you mean it favors E2 on secondary and tertiary? I came back to this post because I was working a problem that had to do with it.