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How is a racemic mixture possible at 3:08 even though there were no chiral carbon in that compound formed after hydride shift? Isn't a chiral carbon mandatory in Sn1 reaction for a racemic mixture to be formed?
Would it be also possible for e.g. a chloride shift? Chloride is Electronegative (partial negative) and thus pulled towards the Kation. Shouldn't it be even easier for chloride because the Cl-Ion state is accessed more easily, so lower activation energy?
Hey ! I was studying the hydration of alkynes, and i wanted to ask if a carbocation rearrangement could happen in a vinyl carbocation (for example with a hydride shift). Thanks a lot !
since there is a tertiary and secondary carbon wouldn't the hydride ion stay for longer with the secondary carbon than the tertiary carbon giving it a lower charge than the tertiary carbon?
Hi Dave ! Your YT channel is amazing, really nice job ! Just a quick question : in the methanide shift, right after the rearrangement, is the molecule configuration correct or am I missing something? I mean, after the shift, the new carbocation has 3 bonds (CH3, CH3, R, nothing) and the former carbocation has 4 bonds (CH3, CH3, R, implied H). Thus shouldn't one of the 2 methyl groups on the former carbocation be going towards (or away from) the viewer? Not very important but I want to make sure I don't miss something here. Thanks for enlightening me :)
Nope you're absolutely right! Sometimes we write groups as though they are flat to indicate that the stereochemistry is not important, or not known, or even just out of laziness! It is indeed tetrahedral, as opposed to the other carbon, which actually is newly trigonal planar, with a formal charge.
trans 2 phenyl 1 bromocyclopentane+alcoholic koh,can you please explain why the carbocation does not rearrange and move on the tertiary carbon attaching the phenyl group to Cyclo Pentane?
Professor Dave Explains ,this was asked in the year 2006 in Aieee(all india engineering entrance examination),so according to the key 3 phenyl cyclopentene is being formed whereas if rearrangement would have taken place,the product should have been 1phenyl cyclopentene,could it be that it may have undergone E2 mechanism,if yes,could you please explain why?
oh sorry, i didn't fully understand the question, yes i think with hydroxide present there would be no chance for E1 or SN1 to take place, so there would be no carbocation intermediate. E2 is most likely, depending on temperature. however, one might still expect the tertiary proton to be extracted over any secondary ones, so the final product is still a little puzzling.
i suppose it's possible that a little bit more electron density is getting to the tertiary carbon, not necessarily by resonance but by induction, and this makes its proton slightly less acidic? if the base were bigger i would say sterics were a factor, but its hydroxide.
The shift is impossible if the Hydroxy is connected to a primary carbon right? Somewhere (on the internet) is saw a guy doing a concerted shift, because the primary carbokation isnt stable, but shouldnt that shift be impossible and therefore solely a Sn2/E2 with no shift(because its a primary substrate)? PS: Substrate was was the same, except one carbon shorter on the right
Sure it can, and I've shown that possibility here. Temperature would be the main element of control to steer things towards substitution or elimination products.
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I don't know how to think you professor
This is really helpful with details and great explications
I'm writing every single information hhhhh I don't care about being late to study the whole course because I can never forget those chemical reactions
Thank you very much 😍😍😊
thank*
explanations*
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How is a racemic mixture possible at 3:08 even though there were no chiral carbon in that compound formed after hydride shift? Isn't a chiral carbon mandatory in Sn1 reaction for a racemic mixture to be formed?
Yes that particular example is achiral, I was just speaking generally.
@@ProfessorDaveExplains Oh,ok. Sorry about that, I didn't realize it initially, anyways a great tutorial as always!
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So if a secondary carbocation is not adjacent to a tertiary or quaternary carbon then it most likely does not rearrange?
yep!
You are the best man.
Would it be also possible for e.g. a chloride shift? Chloride is Electronegative (partial negative) and thus pulled towards the Kation. Shouldn't it be even easier for chloride because the Cl-Ion state is accessed more easily, so lower activation energy?
thank you dave good explanation
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Does the hydride shift occur more spontaneously than the methanide shift?
hmm, i suppose so, since given the option of either it would prefer hydride?
Professor Dave Explains thank you sir
Thanks sir helped a lot for me
Hey !
I was studying the hydration of alkynes, and i wanted to ask if a carbocation rearrangement could happen in a vinyl carbocation (for example with a hydride shift). Thanks a lot !
no not for vinyl, because of the resonance!
Nice explanation sir, Thanks a lot.
Just like you said rearrangements occur among adjacent carbons, so can a primary carbocation transform itself into a tertiary carbocation? Plz help
since there is a tertiary and secondary carbon wouldn't the hydride ion stay for longer with the secondary carbon than the tertiary carbon giving it a lower charge than the tertiary carbon?
Hi Dave ! Your YT channel is amazing, really nice job !
Just a quick question : in the methanide shift, right after the rearrangement, is the molecule configuration correct or am I missing something?
I mean, after the shift, the new carbocation has 3 bonds (CH3, CH3, R, nothing) and the former carbocation has 4 bonds (CH3, CH3, R, implied H). Thus shouldn't one of the 2 methyl groups on the former carbocation be going towards (or away from) the viewer? Not very important but I want to make sure I don't miss something here.
Thanks for enlightening me :)
Nope you're absolutely right! Sometimes we write groups as though they are flat to indicate that the stereochemistry is not important, or not known, or even just out of laziness! It is indeed tetrahedral, as opposed to the other carbon, which actually is newly trigonal planar, with a formal charge.
isn't Sn1 reaction supposed to produce racemic mixture only when there is chiral center?
yep
But in the example the center is achiral because it's connected to 2 identical methyl groups
ah yes, so in this case it wouldn't matter, but modify a methyl and the stereochemistry becomes relevant
trans 2 phenyl 1 bromocyclopentane+alcoholic koh,can you please explain why the carbocation does not rearrange and move on the tertiary carbon attaching the phenyl group to Cyclo Pentane?
i would assume that it would rearrange in that way. where are you seeing that it does not?
Professor Dave Explains ,this was asked in the year 2006 in Aieee(all india engineering entrance examination),so according to the key 3 phenyl cyclopentene is being formed whereas if rearrangement would have taken place,the product should have been 1phenyl cyclopentene,could it be that it may have undergone E2 mechanism,if yes,could you please explain why?
oh sorry, i didn't fully understand the question, yes i think with hydroxide present there would be no chance for E1 or SN1 to take place, so there would be no carbocation intermediate. E2 is most likely, depending on temperature. however, one might still expect the tertiary proton to be extracted over any secondary ones, so the final product is still a little puzzling.
Yeah,that is true too,I just realised that the tertiary hydrogen could also be extracted, is the phenyl's resonance is affecting this in anyway?
i suppose it's possible that a little bit more electron density is getting to the tertiary carbon, not necessarily by resonance but by induction, and this makes its proton slightly less acidic? if the base were bigger i would say sterics were a factor, but its hydroxide.
The shift is impossible if the Hydroxy is connected to a primary carbon right? Somewhere (on the internet) is saw a guy doing a concerted shift, because the primary carbokation isnt stable, but shouldnt that shift be impossible and therefore solely a Sn2/E2 with no shift(because its a primary substrate)?
PS: Substrate was was the same, except one carbon shorter on the right
hmm yeah if the alcohol is primary i'm thinking water won't leave, at least not typically
ok, thanks for the reply
Wouldnt water as a nucleophile react via Sn1 since it a small Nucleophile?
Sure it can, and I've shown that possibility here. Temperature would be the main element of control to steer things towards substitution or elimination products.
Professor Dave Explains Thx man
Thx king
Zaitsev will dominate over the E1 reaction but will it also dominate over the SN1 reaction?
Zaitsev/Hofmann refer to alkene products, so the terminology does not apply to substitution reactions
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what drives a hydride over a methanide shift?
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