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SN1 and E1 Practice Questions

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Organic Chemistry with Victor

Organic Chemistry with Victor

Күн бұрын

Пікірлер: 19
@berilalbayrak7524
@berilalbayrak7524 Жыл бұрын
you are the best chemistry channel ive ever seen thank you for all this
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor Жыл бұрын
Thank you 😊
@OrcjaEnglish
@OrcjaEnglish 3 ай бұрын
Overall, a wonder revision session. Thank you, sir.
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor 2 ай бұрын
I'm glad you found it helpful!
@joshuahering2041
@joshuahering2041 Жыл бұрын
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
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor Жыл бұрын
I try 😂
@carolinelecter9663
@carolinelecter9663 Жыл бұрын
Awesome video,very helpful,couldn't love the tricks more. Thank u!
@cieloguadalupeperezolivare1905
@cieloguadalupeperezolivare1905 2 жыл бұрын
¡Thank you! This quiestions is all I need. Some SNac or SeA would be great!, again, thank you for content!
@steveng724
@steveng724 3 ай бұрын
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 ?
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor 3 ай бұрын
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.
@OrcjaEnglish
@OrcjaEnglish 3 ай бұрын
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!
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor 3 ай бұрын
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.
@OrcjaEnglish
@OrcjaEnglish 3 ай бұрын
@@VictortheOrganicChemistryTutor THANK YOU, Sir.
@OrcjaEnglish
@OrcjaEnglish 3 ай бұрын
I couldn't understand the 1,2-methyl shift. Sir, any tips to learn it .
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor 3 ай бұрын
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.
@marounxd3871
@marounxd3871 5 ай бұрын
can we make a ring expansion in the first reaction ?
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor 5 ай бұрын
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.
@carolinelecter9663
@carolinelecter9663 Жыл бұрын
In the second question, why the resonance happens?
@VictortheOrganicChemistryTutor
@VictortheOrganicChemistryTutor Жыл бұрын
Resonance is something you should always consider in any reaction.
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