In the first example with the 'wrong' minor products before rearrangement, even though we have an X through them I see you labeled them enantiomers. Are those not diastereomers? There are two chiral centers. The wedge methyl group is R and remains R while the substituted carbocation, without rearrangement is R and S. Please correct me if I'm wrong. Of course, it isn't that important, I just want to make sure I'm seeing this correctly. Also 6:03 "so groups can hop from carbon to another, cool beans!" I love it! So engaging. 🙂Excellent video again.
Hi! At around 9:20, why did you choose to do the hydride shift rather than the alkyl shift (since there's a methyl on the tertiary carbon, can't you do the alkyl shift too?) Thanks!
The general migratory aptitude goes: H>Ph>alkyls, so you would normally prioritize the hydride shift over the alkyl shift. The only time when the alkyl shift is more favorable is in the case of small rings that can undergo ring expansion and reduce the ring strain, or when the molecule is conformationally locked making a hydride shift impossible due to the orbital symmetry limitations, but we don’t typically talk about latter in a sophomore course. So, for our purposes always prioritize the hydride shift when possible.
Hey Doctor Vector, in the last example the product we got it is said that it has 2 chiral carbons, but the first carbon has two phenyl groups, while a chiral carbon shall have 4 different groups, so how come it is chiral? I will be glad for an explanation, thanks.
@@SueCao-i3i If I've made any mistakes, I've added notes on those in the description. You're not going to learn anything wrong here, those mistakes are mostly typos.
It depends on the molecule 🤷♂️ If you have a carbocation intermediate that makes a chiral carbon upon reaction with a nucleophile, then you would show the corresponding stereochemistry using the wedge-dash notation, if you end up with a non-chiral carbon, showing dashes and wedges is pointless for that specific atom.
Dear Victor, why does this reaction not proceed via a phenonium intermediate with intramolecular contribution from the phenyl ring? I have seen similar reactions with -OTs instead of -OH in AcOH where this occurs as a major pathway (retention of stereochemistry).
It does. As I've mention in the video, it's an oversimplification of the actual mechanism. I do talk about phenonium ions in my video on tricky rearrangements. And to answer your question, it really depends on the substrate and conditions of the reaction. Phenonium ion can either be a pathway of rearrangement or it can be doing, essentially, an anchimeric assistance function, hence the retention of stereochemistry. Within the scope of an intro organic chemistry, we don't go into those aspects at all, so I didn't include those here either. Would've made the video too advanced and difficult to comprehend with all those details included.
for the 2nd example, why wasn't the polar protic solvent used. also for the first example how did u know to use 2 of the ethanol's and not just stop at one.
What is more nucleophilic, the acetic acid or the acetate anion? Solvent is always in excess in SN1 reactions, so statistically speaking, it's the solvent that's going to act as a base and not bromide. Coincidentally, if we check the corresponding pKa values for the conjugate acids, we'll see that alcohols are actually more basic than the bromide anion!
You can certainly show Br- doing the proton transfer. Some instructors will be perfectly fine with that but some will require you to write a more statistically and thermodynamically likely process, which is the deprotonation with methanol here. I prefer to write the reactions with the more statistically likely steps, but that's me. In cases like this, I always suggest double checking with your instructor to see how they want you to write those since at the end of the day, they are the one giving you your grade.
Very organized and helpful!!!
In the first example with the 'wrong' minor products before rearrangement, even though we have an X through them I see you labeled them enantiomers. Are those not diastereomers? There are two chiral centers. The wedge methyl group is R and remains R while the substituted carbocation, without rearrangement is R and S. Please correct me if I'm wrong. Of course, it isn't that important, I just want to make sure I'm seeing this correctly.
Also 6:03 "so groups can hop from carbon to another, cool beans!" I love it! So engaging. 🙂Excellent video again.
Yes, they are diastereomers. Sort of crept in when I was making changes from the drafts 😆 I should prolly need to add the note in the description.
Thank you for the video. It helps me a lot.
You are welcome! 👍
Hi! At around 9:20, why did you choose to do the hydride shift rather than the alkyl shift (since there's a methyl on the tertiary carbon, can't you do the alkyl shift too?) Thanks!
The general migratory aptitude goes: H>Ph>alkyls, so you would normally prioritize the hydride shift over the alkyl shift. The only time when the alkyl shift is more favorable is in the case of small rings that can undergo ring expansion and reduce the ring strain, or when the molecule is conformationally locked making a hydride shift impossible due to the orbital symmetry limitations, but we don’t typically talk about latter in a sophomore course. So, for our purposes always prioritize the hydride shift when possible.
Okay, thank you so much!!@@VictortheOrganicChemistryTutor
Hey Doctor Vector, in the last example the product we got it is said that it has 2 chiral carbons, but the first carbon has two phenyl groups, while a chiral carbon shall have 4 different groups, so how come it is chiral?
I will be glad for an explanation, thanks.
That’s an error. I’ve made a few mistakes in the videos in this series, so I’m redoing the entire series this semester.
@@VictortheOrganicChemistryTutor are the rest of the videos in this playlist now updated? Just checking before I study something wrong... :)
@@SueCao-i3i If I've made any mistakes, I've added notes on those in the description. You're not going to learn anything wrong here, those mistakes are mostly typos.
@@VictortheOrganicChemistryTutor Thank you so much your videos are truly the best!
How do you know when to draw the wedge and dashes for the products? Is it because the sp2 hybridatzation?
It depends on the molecule 🤷♂️ If you have a carbocation intermediate that makes a chiral carbon upon reaction with a nucleophile, then you would show the corresponding stereochemistry using the wedge-dash notation, if you end up with a non-chiral carbon, showing dashes and wedges is pointless for that specific atom.
Dear Victor, why does this reaction not proceed via a phenonium intermediate with intramolecular contribution from the phenyl ring? I have seen similar reactions with -OTs instead of -OH in AcOH where this occurs as a major pathway (retention of stereochemistry).
It does. As I've mention in the video, it's an oversimplification of the actual mechanism. I do talk about phenonium ions in my video on tricky rearrangements. And to answer your question, it really depends on the substrate and conditions of the reaction. Phenonium ion can either be a pathway of rearrangement or it can be doing, essentially, an anchimeric assistance function, hence the retention of stereochemistry. Within the scope of an intro organic chemistry, we don't go into those aspects at all, so I didn't include those here either. Would've made the video too advanced and difficult to comprehend with all those details included.
@@VictortheOrganicChemistryTutor Great thank you, love your videos.
for the 2nd example, why wasn't the polar protic solvent used. also for the first example how did u know to use 2 of the ethanol's and not just stop at one.
What is more nucleophilic, the acetic acid or the acetate anion?
Solvent is always in excess in SN1 reactions, so statistically speaking, it's the solvent that's going to act as a base and not bromide. Coincidentally, if we check the corresponding pKa values for the conjugate acids, we'll see that alcohols are actually more basic than the bromide anion!
in the last example, why was a shift necessary at all? Aren't we going from a 3 carbon to a 3 carbon?
If you can make a more stable carbocation, then there’s a driving force for the rearrangement.
does it matter of the proton transfer takes place with Br instead at example 1
You can certainly show Br- doing the proton transfer. Some instructors will be perfectly fine with that but some will require you to write a more statistically and thermodynamically likely process, which is the deprotonation with methanol here. I prefer to write the reactions with the more statistically likely steps, but that's me. In cases like this, I always suggest double checking with your instructor to see how they want you to write those since at the end of the day, they are the one giving you your grade.
Hey sir give me plz some named group reaction for carbocation
Thankyou so much
Most welcome 😊