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For the first synthesis, is there a need for the elimination of the alkyl halide? Is it not possible for the alkyl halide to directly undergo nucleophilic substitution in NaOH to form the alcohol?
Hello! Thank you so much for the video. Just a quick question. For the first problem, when you brominate the starting material in light, isn't it true that the bromination would occur likely towards the tertiary carbon, as radical bromination is considerably more selective than radical chlorination? Regardless, it should still work if the elimination step doesn't utilize a bulky base such as t-BuOK. Just wanted to clarify. Thank you!!
You bring up an excellent point here! You’re also spot on that both pathways allow for the formation of the same alkene depending on base. Here’s how I understand this type of reaction: Radical Chlorination, almost certainly would favor the primary methyl group. Radical bromination typically favors tertiary>secondary>primary. Therefore, at least in my own reasoning, since there are 2 equivalent secondary carbons, the statistical probability would likely favor the secondary carbon, since there are 2 of them and only 1 tertiary carbon. I would wager that a real-life experiment would produce a mixture of products but your line of thinking is perfectly reasonable as well!
Good video, few points: the alkyl halide at 4:28 can be hydrolyzed directly to the alcohol. Alkynylation is new to me; how to prevent the carbonyl from deprotonating instead of the alkyne at 9:45? And at 15:45 is the para preference because of sterics at ortho? Thanks :)
Thanks Daniel! Great questions and points. For alkynylation, I recommend checking out the wikipedia page: en.wikipedia.org/wiki/Alkynylation I haven't made a video about it yet but it's a great reaction to extend carbon chains. On the wikipedia link I shared, they showcase a similar reaction with ketone. For the para EAS reactino to add bromine, you're exactly right. Statistically, since there are 2 ortho positions, you often see ortho favored over para substitution. However, a group like isopropyl has enough steric encumbrance to block those sites!
As others have commented, the initial bromination is not the best proposal: you are likely to get a mixture of products, and the most reasonable-looking position for bromination would be the more nucleophilic tertiary C-H, since the Br radical is pretty electrophilic. Also, in the video, you keep referring to this step in terms of Markovnikov’s rule, which describes alkene additions - this chemistry is totally different, and is guided by polarity matching and/or sterics in the HAT step, depending on your H-atom acceptor. It might look like this shouldn’t matter for the route, since you want to eliminate, but the 5-ring creates an issue with exo- vs endo-alkene formation: for 5-rings, exo-alkenes are preferred due to strain. It might be better to start with the bromide or the alkene, for simplicity. The other comment, suggesting that hydroxide would get to the alcohol more quickly, might be correct, but you also risk a lot of side reactions (especially elimination). If you have the bromide, a nicer idea might be to use a Kornblum oxidation to go directly to the ketone. From the alkene, a Tsuji-Wacker would be ideal.
Won't radical bromine react selectively on the tertiary carbon rather than this secondary carbon you've suggested in this first step? Chlorine may generate a mixture of isomers but i dont think you'll be able to get that isomer from bromine in this fashion.
This is an excellent question that actually has come up by several commenters on this video. The way I understand radical Halogenation is basically that chlorination would lead to a mixture of isomers, as you mentioned. Bromination of Secondary alkyl groups is significantly more likely over primary carbons because the activation energy? At least in propane, is 3 kcal lower, leading to about a 99:1 Secondary bromination. There have been far less studies looking at tertiary radical bromination. In truth, I imagine an actual experiment would yield a variety of products in a mixture. Importantly, the next step is achievable by either position depending on which base you use for the elimination reaction. Either way, your head is in the right place!
If you enjoyed this video, don't forget to give it a thumbs up, subscribe for more educational content, and hit the notification bell to stay updated! Have questions or specific topics you'd like me to cover? Drop them below, and let's keep the chemistry conversation going! #ScienceCommunity #LearningTogether #OrganicChemistry
This is so helpful! Thanks, Dr. Rojas!
Glad I could help out and thanks for watching!
For the first synthesis, is there a need for the elimination of the alkyl halide? Is it not possible for the alkyl halide to directly undergo nucleophilic substitution in NaOH to form the alcohol?
That’s actually a phenomenal route! Even shorter than mine. Nice work!
Hello! Thank you so much for the video. Just a quick question. For the first problem, when you brominate the starting material in light, isn't it true that the bromination would occur likely towards the tertiary carbon, as radical bromination is considerably more selective than radical chlorination? Regardless, it should still work if the elimination step doesn't utilize a bulky base such as t-BuOK. Just wanted to clarify. Thank you!!
You bring up an excellent point here! You’re also spot on that both pathways allow for the formation of the same alkene depending on base. Here’s how I understand this type of reaction: Radical Chlorination, almost certainly would favor the primary methyl group. Radical bromination typically favors tertiary>secondary>primary. Therefore, at least in my own reasoning, since there are 2 equivalent secondary carbons, the statistical probability would likely favor the secondary carbon, since there are 2 of them and only 1 tertiary carbon. I would wager that a real-life experiment would produce a mixture of products but your line of thinking is perfectly reasonable as well!
1. Amazing vid
2. mCBPA does not contain at most 6 carbons, but 7
Daaaaang!!!! That’s a great call! So how are we making the epoxide?
@@rojaslab I only participate in the chemistry Olympiad (at a high level), but epoxide are never taught so idk
@@keikazazic3296 Oh that's really cool. Good luck!!
Thanks
@@rojaslab dmdo possibly, although not commercially available, but very effective. lots of epoxidation agents
Thanks for the explanation
It's the best and it's helpful to me 👏👏👏
That’s so kind! Thank you for watching!
mCPBA. Great video btw :)
Haha! You know I tell my students that for whatever reason my brain cannot compute the correct order of letters in that reagent!
Good video, few points: the alkyl halide at 4:28 can be hydrolyzed directly to the alcohol. Alkynylation is new to me; how to prevent the carbonyl from deprotonating instead of the alkyne at 9:45? And at 15:45 is the para preference because of sterics at ortho? Thanks :)
Thanks Daniel! Great questions and points. For alkynylation, I recommend checking out the wikipedia page: en.wikipedia.org/wiki/Alkynylation I haven't made a video about it yet but it's a great reaction to extend carbon chains. On the wikipedia link I shared, they showcase a similar reaction with ketone. For the para EAS reactino to add bromine, you're exactly right. Statistically, since there are 2 ortho positions, you often see ortho favored over para substitution. However, a group like isopropyl has enough steric encumbrance to block those sites!
As others have commented, the initial bromination is not the best proposal: you are likely to get a mixture of products, and the most reasonable-looking position for bromination would be the more nucleophilic tertiary C-H, since the Br radical is pretty electrophilic. Also, in the video, you keep referring to this step in terms of Markovnikov’s rule, which describes alkene additions - this chemistry is totally different, and is guided by polarity matching and/or sterics in the HAT step, depending on your H-atom acceptor. It might look like this shouldn’t matter for the route, since you want to eliminate, but the 5-ring creates an issue with exo- vs endo-alkene formation: for 5-rings, exo-alkenes are preferred due to strain. It might be better to start with the bromide or the alkene, for simplicity.
The other comment, suggesting that hydroxide would get to the alcohol more quickly, might be correct, but you also risk a lot of side reactions (especially elimination). If you have the bromide, a nicer idea might be to use a Kornblum oxidation to go directly to the ketone. From the alkene, a Tsuji-Wacker would be ideal.
Thanks for the comments! Sounds like you have a lot of experience with organic chemistry!!
Won't radical bromine react selectively on the tertiary carbon rather than this secondary carbon you've suggested in this first step? Chlorine may generate a mixture of isomers but i dont think you'll be able to get that isomer from bromine in this fashion.
This is an excellent question that actually has come up by several commenters on this video. The way I understand radical Halogenation is basically that chlorination would lead to a mixture of isomers, as you mentioned. Bromination of Secondary alkyl groups is significantly more likely over primary carbons because the activation energy? At least in propane, is 3 kcal lower, leading to about a 99:1 Secondary bromination. There have been far less studies looking at tertiary radical bromination. In truth, I imagine an actual experiment would yield a variety of products in a mixture.
Importantly, the next step is achievable by either position depending on which base you use for the elimination reaction. Either way, your head is in the right place!
It is mCPBA
Extra points if you know what all the letters stand for!
@@rojaslab hehe i work with it 😏
@@bigseleba That's cool! I've never done an epoxidation. I have started with epichlorohydrin though with the pre-installed epoxide.
@@rojaslab not exactly pleasant reagent. You never know the exact purity of it without a titration, also not the easiest work-up