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- เผยแพร่เมื่อ 13 ต.ค. 2024
- joechem.io/vid... for video on jOeCHEM and attached worksheet + solution (below video on jOeCHEM aka the link)
Worksheet, Solutions, & Walkthroughs:
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worksheets.joec...
worksheets.joec...
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Study Guide: (Not available yet, but Joe's working on it!)
In this video, we go through Enamine Formation from a carbonyl, both the forward and reverse reaction directions. I know it's a pain, but know BOTH directions like the back of your hand (and if you don't know the back of your hand, get well acquainted, and then learn this reaction that well).
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thanks! you're saving lives
Excellent explanation! You have a gift at making organic chemistry relatable. Thank you for the video, you have a new subscriber.
Mai Rav Mai, thank you for putting a big ole’ smile on my face. If you need anything more Ochem related (videos or worksheets), please check out my site joechem.io. And a million thanks you for the sub 😁
@@jOeCHEM Thank you so much, and anytime!
Excellent explanations! you made the emine hydrolysis reaction mechanism seem so easy!
I'm so glad! Thanks for watching :) for guided practice, make sure to check out the link in the description (goes to my website), where I have a guided worksheet + answers, for free.
when the teachers got the looks, so you just come in the class to look at them 💀
This is amazing. You make these things look so easy. Subbed :) -- Thank you!
Aditya! Thank you for the kind words 😁Make sure to check out my actual website joechem.io where all my videos as well as worksheets + solutions are available for free. And if you'd subscribe to my channel (if you haven't already), I'd massively appreciate it!
@@jOeCHEM Definitely have subscribed, man! Keep it up
JOECHEM LETS GOOOOOOOO THANKS SO MUCH FOR THIS
How do you know which side the double bond will go to (left or right) of where the original ketone was? Or will you get a mixture of products? Will one be a major?
Assuming there is NOT a symmetrical structure, attempt to form the double bond on the molecule where you can lock the structure in a conformation that is sterically favorable. In this example, the double bond is formed between carbons 2, 3 because a E double bond can be formed and the methyl group can be locked into place, creating a minimized steric situation.
Go to the worksheet link on this video (on my website):
joechem.io/videos/37
Here is the worksheet: s3.us-east-2.amazonaws.com/prod-joechem/worksheets/Reactions%20of%20Carbonyls
Here is the solution to the worksheet: s3.us-east-2.amazonaws.com/prod-joechem/worksheets/ANSWERS%20-%20Reactions%20of%20Carbonyls
just amazing, thanks bro !!☺️
at 2 min 48 sec, is the hydronium a h30+? Can we protonate the OH with an H30+ or do we need to use that specific hydronium? -Gabbie
Hi Gabbie! In an acidic environment, you can either use a specific acid if you’re provided with one (e.g. H2SO4) or you can generically use H3O+. At 2:48, I used H3O+ generically, but I drew it such that I could expose a proton to pick off.
Does that answer your question?
@@jOeCHEM yes it does thank you!
Bro is getting me through ochem
for the double bond placement for enamine, why is it on the left? The carbon on the right is more substituted so more stable, so shouldnt we put the double bond there?
Hi Directioner Here!
If you're talking about the ring problem at 11:57, the double bond is on the *less* substituted side in this case because if you were to "lock" in the right hand side (with the tertiary carbon with the methyl group), you'll likely incur more steric strain since you're working in the context of a ring as opposed to a 4:38 where we had a straight chain situation where you can lock the double bond into an E conformation, minimizing steric strain.