That's certainly possible and will happen too. This is a video that is meant for the intro-level organic chemistry students, so I see no reason to overwhelm the audience with the details here. If we start talking about the differences in the "driving force" between the ring expansion and the formation of the 3° carbocation, we'll get to a very tight corner where it would be impossible to really predict anything without real experimental numbers. So, here I introduce the most typical style rearrangement students are likely to see in their classes. I have a dedicated video on carbocations where I talk more about the rearrangements, but even that one is a bit simplistic. At some point, I'll probably make an advanced version of the video where I'll dig into all the experimental evidence and show how all we teach to undergrads is one big lie 🤣 Carbocation rearrangements are, unfortunately, a very dark forest and it's easy to lose yourself in it.
@@VictortheOrganicChemistryTutor Totally understandable... I really appreciate your effort and I am a new subscriber... I like your channel's vibe, it's vintage and aesthetic... and your explanation is on point! Great work... keep it up!!
That was great. Victor, I remember studying acetal formation with acid catalysts and I remember the book saying it's the alcohol getting protonated first and not the carbonyl. But I guess they meant, that in a separate step alcohol gets protonated, and next carboyl will nab a proton from it and so on. That always bugged me! because they didn't explain why that happens. What you were saying around 2:30 or so, seems to be the same idea, but here it's water getting protonated first. Am I correct?
Yeah. When it comes to the mechanisms, we often simplify things and often do so unreasonably. Here, water is the reagent + solvent. What is more statistically likely to happen: direct protonation of an alkene or water? Also, from the equilibrium standpoint, what’s more favorable: formation of oxonium or a carbocation? Same with acetal formation. We often skip the step where we protonate the alcohol which then protonates the carbonyl. But in reality, the most reasonable and thermodynamically favorable step is to get your ROH2+ which *then* serves as your actual acid in the reaction.
@@VictortheOrganicChemistryTutor Thanks a lot. That makes sense and clears it up. I guess the more I learn, the more I find out I don't knw! It's like the old saying.
11:40 won't it cause ring expansion? Pentane -> Hexane since its more stable
That's certainly possible and will happen too.
This is a video that is meant for the intro-level organic chemistry students, so I see no reason to overwhelm the audience with the details here. If we start talking about the differences in the "driving force" between the ring expansion and the formation of the 3° carbocation, we'll get to a very tight corner where it would be impossible to really predict anything without real experimental numbers. So, here I introduce the most typical style rearrangement students are likely to see in their classes. I have a dedicated video on carbocations where I talk more about the rearrangements, but even that one is a bit simplistic. At some point, I'll probably make an advanced version of the video where I'll dig into all the experimental evidence and show how all we teach to undergrads is one big lie 🤣 Carbocation rearrangements are, unfortunately, a very dark forest and it's easy to lose yourself in it.
@@VictortheOrganicChemistryTutor Totally understandable... I really appreciate your effort and I am a new subscriber... I like your channel's vibe, it's vintage and aesthetic... and your explanation is on point! Great work... keep it up!!
That was great. Victor, I remember studying acetal formation with acid catalysts and I remember the book saying it's the alcohol getting protonated first and not the carbonyl. But I guess they meant, that in a separate step alcohol gets protonated, and next carboyl will nab a proton from it and so on. That always bugged me! because they didn't explain why that happens. What you were saying around 2:30 or so, seems to be the same idea, but here it's water getting protonated first. Am I correct?
Yeah.
When it comes to the mechanisms, we often simplify things and often do so unreasonably. Here, water is the reagent + solvent. What is more statistically likely to happen: direct protonation of an alkene or water? Also, from the equilibrium standpoint, what’s more favorable: formation of oxonium or a carbocation? Same with acetal formation. We often skip the step where we protonate the alcohol which then protonates the carbonyl. But in reality, the most reasonable and thermodynamically favorable step is to get your ROH2+ which *then* serves as your actual acid in the reaction.
@@VictortheOrganicChemistryTutor Thanks a lot. That makes sense and clears it up. I guess the more I learn, the more I find out I don't knw! It's like the old saying.
best explanation
Thanks!!!
Thank you
Welcome!