That might have been the best chemistry lecture I ever had. You are genuine, funny, interesting and straight to the point. Thank you for your excellent teaching sir, it helped me tremendously
the OCH3 that is now dashed, since it is inverted, does that mean it is going away or coming toward you? i cannot tell where the point of the dash is headed. Thank you.
I have a question, so idk if this is just a professor preference thing but for acid catalyzed ring opening my professor has the CH3OH on the top and then HCl (cat) on the bottom. So I was wondering of the placement of this means anything at all or if it doesn’t matter. Because your way makes a lot more sense.
What would happen with the methyl group during the ring opening if it was neither dashed nor wedged from the start? Would it remain unchanged at the end of the reaction or would it become dashed or wedged?
Why would the positively charged pronated epoxide not first lose the bond to the carbon to become neutral and causing a full carbocation on the tertiary carbon. Then the weak nucleophiel could attack either side like a typical SN1 reaction. Why does this not happen?
Hey there, I mention a few times in the video that it can occur in an SN1 type reaction, where the O-C bond on the more substituted side will be longer and weaker than the other side (as if the ring were starting to open in that way before the nucleophile attacks) leaving a more positive tertiary carbon.
H2SO4 was just used to protonate the alcohol (CH3-OH), I assume it can also activate the epoxide without having to protonate the alcohol first but both scenario works.
That might have been the best chemistry lecture I ever had. You are genuine, funny, interesting and straight to the point. Thank you for your excellent teaching sir, it helped me tremendously
You are very welcome, MangeUlver - thank you for saying so!
Thank you so much! This was the best lecture on ring opening of epoxides that I've ever seen.
You're welcome, Rajika Robinso - glad it helped!
i keep saying this man needs an award.........and i'mma keep saying it!!!!
Big up from Ghana
Thank you, from U.S.A. - Happy Studying!
Thank you so much! I've been scratching my head for an hour with two different textbooks but this sorted it all out clearly. Ur amazing!
Glad you found the channel and the video so helpful, Amal Gregory - Happy Studying!
This is the best breakdown of epoxide reactions online
Glad to hear it, Emilio!
the OCH3 that is now dashed, since it is inverted, does that mean it is going away or coming toward you? i cannot tell where the point of the dash is headed. Thank you.
You explained this concept so well! Thank you.
You're welcome!
this was so helpful! great video
Glad you think so, Cararose - thanks for commenting!
great stuff as always
Thank you
Could you give an example of where Acids are more reactive to primary Carbons in an epoxide than secondary Carbons? Thanks!
Thank you-this helped me so much!
You're welcome, Kanchan. Thank you for the feedback.
I have a question, so idk if this is just a professor preference thing but for acid catalyzed ring opening my professor has the CH3OH on the top and then HCl (cat) on the bottom. So I was wondering of the placement of this means anything at all or if it doesn’t matter. Because your way makes a lot more sense.
Do you mean above/below the arrow? There is no set rule, but the reagent tends to be written above the arrow and anything else, solvent etc below.
Idk what it is about these videos but they make so much sense
Glad they do!
Amazing lecture sir, thank you
You're welcome and Thank You.
What would happen with the methyl group during the ring opening if it was neither dashed nor wedged from the start? Would it remain unchanged at the end of the reaction or would it become dashed or wedged?
This is the video that made me subscribe
Welcome to the channel :)
hi! what happen to the H2SO4 number 2 example?
It was just used to protonate the alcohol (CH3-OH)
Absolutely worth it 👍
Thank you
Why would the positively charged pronated epoxide not first lose the bond to the carbon to become neutral and causing a full carbocation on the tertiary carbon. Then the weak nucleophiel could attack either side like a typical SN1 reaction. Why does this not happen?
Hey there, I mention a few times in the video that it can occur in an SN1 type reaction, where the O-C bond on the more substituted side will be longer and weaker than the other side (as if the ring were starting to open in that way before the nucleophile attacks) leaving a more positive tertiary carbon.
I LOVE THIS GUY MAN
Thanks!
thank you very much for great lecture
You're welcome.
Thanks 😃💖 you're best 😎👍
Thanks and you're welcome!
I’ve a doubt, what happend with H2SO4?
H2SO4 was just used to protonate the alcohol (CH3-OH), I assume it can also activate the epoxide without having to protonate the alcohol first but both scenario works.
Love you!
Glad the channel is helping you
The best, thank you sir...
You're welcome!
Great video keep up
Thanks, Stephen!
Thanks alot sir!
Very welcome.
This guy never fails to reply to comments 🙃
Trying....not....to....comment....CAN'T!! 😀 😀 😀
YOU ARE AMAZING WOW
THANKS!
Hey i'm back 🧤
Welcome back and Happy Studying!
My exam is coming up, can I rub your head for good luck?
If you mean by rubbing my head that you watch every video 3 times, then GO FOR IT! 🙂