LOL I love your humor in teaching. "its as stable as it can carbocationly be" and the ITS A TRAP meme.. I will forever know not to fall for leaving groups on sp2 hybridized carbons. thank you
Do SN1 reactions show neighboring group participation? According to the ChatGPT , SN1 reactions generally do not show neighboring group participation. These reactions proceed via a two-step mechanism involving the formation of a carbocation intermediate. The rate-determining step is the dissociation of the leaving group, forming a carbocation, which then reacts with a nucleophile. Neighboring group participation is more commonly associated with SN2 reactions, where a neighboring group can assist in the displacement of the leaving group by stabilizing the transition state or intermediate. What's your opinion about it? Sir, kindly inform . Thank you!
I would suggest you leave the glorified text predictor alone and read a book instead. Me ranting aside, the "neighboring group participation" is a bit of an umbrella term. If we're talking about the anchimeric assistance, then no, SN1 reactions don't do that. However, we can easily see the carbocation intramolecularly "trapped" by heteroatoms and interfere with the mechanism and even affect the stereoselectivity and stereospecificity.
Hello Mr. Victor , for the first problem I would like to ask isn't E1 also a major product , sice CH3OH is a weak nuc. and weak base making Sn1 and E1 both equally feasible unless there is mention of temperature ?
Generally, we're only going to indicate the high temperature, which then would favor E1 reaction. If the temperature not mentioned, the assumption is that we're working in the "lower temperature" conditions, which typically favors SN1 unless you have some factors which favor E1 like a formation of a conjugated system.
Could problem number 1 undergo an E1 reaction as well since it's the Br is a 3rd degree alkyl halide and CH3OH is a weak nucleophile and weak base? Love your videos!
E1 and SN1 are always in competition. It's nearly impossible to have one and completely avoid the other. The question here is always going to be which is the major product.
It is possible. But that would make a secondary carbocation upon expansion, so there's not much of a driving force behind that even though we're making a 6-membered ring, it's still going to be less favorable compared to keeping a 3° carbocation. But it's a possibility and will probably happen to some small extent.
LOL I love your humor in teaching. "its as stable as it can carbocationly be" and the ITS A TRAP meme.. I will forever know not to fall for leaving groups on sp2 hybridized carbons. thank you
I try 😂
you are the best chemistry channel ive ever seen thank you for all this
Thank you 😊
¡Thank you! This quiestions is all I need. Some SNac or SeA would be great!, again, thank you for content!
Do SN1 reactions show neighboring group participation?
According to the ChatGPT , SN1 reactions generally do not show neighboring group participation. These reactions proceed via a two-step mechanism involving the formation of a carbocation intermediate. The rate-determining step is the dissociation of the leaving group, forming a carbocation, which then reacts with a nucleophile. Neighboring group participation is more commonly associated with SN2 reactions, where a neighboring group can assist in the displacement of the leaving group by stabilizing the transition state or intermediate.
What's your opinion about it? Sir, kindly inform . Thank you!
I would suggest you leave the glorified text predictor alone and read a book instead.
Me ranting aside, the "neighboring group participation" is a bit of an umbrella term. If we're talking about the anchimeric assistance, then no, SN1 reactions don't do that. However, we can easily see the carbocation intramolecularly "trapped" by heteroatoms and interfere with the mechanism and even affect the stereoselectivity and stereospecificity.
@@VictortheOrganicChemistryTutor THANK YOU, Sir.
In question number 4 why we haven’t done rearrengment to the secondary alkyl halide ?
Why on earth would we do that??? You're making a benzylic carbocation intermediate, this is as stable as possible of a carbocation.
@ pardon me sir 😔🙏 and thank you for your quick response ❤️
@@marcelinemoran5029 No worries and you're very welcome. I'm willing to bet many others had the same question but didn't have courage to ask.
Awesome video，very helpful，couldn't love the tricks more. Thank u！
Hello Mr. Victor , for the first problem I would like to ask isn't E1 also a major product , sice CH3OH is a weak nuc. and weak base making Sn1 and E1 both equally feasible unless there is mention of temperature ?
Generally, we're only going to indicate the high temperature, which then would favor E1 reaction. If the temperature not mentioned, the assumption is that we're working in the "lower temperature" conditions, which typically favors SN1 unless you have some factors which favor E1 like a formation of a conjugated system.
Overall, a wonder revision session. Thank you, sir.
I'm glad you found it helpful!
Could problem number 1 undergo an E1 reaction as well since it's the Br is a 3rd degree alkyl halide and CH3OH is a weak nucleophile and weak base? Love your videos!
E1 and SN1 are always in competition. It's nearly impossible to have one and completely avoid the other. The question here is always going to be which is the major product.
I couldn't understand the 1,2-methyl shift. Sir, any tips to learn it .
I have more videos on carbocations and rearrangements, you should check those out. I also have many of these tutorials written up on my website.
can we make a ring expansion in the first reaction ?
It is possible. But that would make a secondary carbocation upon expansion, so there's not much of a driving force behind that even though we're making a 6-membered ring, it's still going to be less favorable compared to keeping a 3° carbocation. But it's a possibility and will probably happen to some small extent.
In the second question, why the resonance happens?
Resonance is something you should always consider in any reaction.