I will start by thanking you so much for the videos. They were so helpful. I have a question tho... Concerning the first exercise, why is there only one product? From my understanding, there have to be two products formed since a carbocation is sp2 hybridized. Thanks in advance for your reply.
In the second problem, why does the oxygen on the left attack the carbocation? There are two oxygens with the same amount of lone pairs so why can't the oxygen with double bond attack the carbocation?
@@SherryKao16 is it also partly because I -ion will take H+ from OH which causes O to be O negative and that's why the main product is ester and minor product is HI?
Thank you very much this are onlyone vudeis on youtube ehich explains nucleophilic reactions to be understood and on the end I know it solve by myself. So thank you very much.
I have a question, I realized on eqn 1 and 4, the final products had lone pairs but the final products for 2 and 3 didn't have. Is there a specific reason or you just chose not to put them there?
There are ways of answering "where did the hydrogen go" ? 1. Missing acid-base reactions: In the first reaction, H2O attacks the carbocation and becomes -OH2(+). One of the hydrogens is then removed in an acid-base reaction by another molecule of H2O, leading to H3O(+)Cl(-) [drawn in red] and the neutral alcohol product. In the interest of space, the acid-base reactions that give the neutral products are omitted in the 2nd, 3rd, and 4th reactions. 2. "Hidden" or "implied" hydrogens not drawn in 4th reaction. Note that HOCH2CH3 attacks the carbocation, and a hydrogen is drawn in, but not drawn in the next diagram. However it is still there - this is an example of a "hidden" or "implied" hydrogen from a line diagram. In the figure where the new C-O bond is a "wedge", there is a "hidden" hydrogen on a dash. In the figure where the new C-O bond is a "dash", there is a "hidden" hydrogen on a wedge. They were not drawn in to save space, although in retrospect I should have put them in. Thanks for your comment!
Thank you! If this video was helpful, you may want to check out our month-to-month MOC membership for access to over 1500 quizzes, Flashcards, the Reaction Guide, and more. Check it out here: bit.ly/2YctxPb
These videos are really awesome. Great job!
i wish you were my teacher. Thank you so much.
keep making this type of they are just awesome
It was helpful. Thank you, Sir.
In the last one why did you not do a hydride shift. Since it is adjacent to an allylic carbon which is much more stable then a secondary
I will start by thanking you so much for the videos. They were so helpful. I have a question tho... Concerning the first exercise, why is there only one product? From my understanding, there have to be two products formed since a carbocation is sp2 hybridized. Thanks in advance for your reply.
Because there is no stereocenter! If there were, then it would be 2! Hope this helps:)
It’s achiral, so it isn’t possible to form a racemic mixture of an achiral molecule as they are optically inactive
@@toritavernier419 thank you
@@helenamarek7926 thanks alot
You're welcome! If you found this video helpful, Our MOC membership has a lot to offer. Check it out here: bit.ly/2YctxPb
In the second problem, why does the oxygen on the left attack the carbocation? There are two oxygens with the same amount of lone pairs so why can't the oxygen with double bond attack the carbocation?
Because the other oxygen has a double bond and that's harder to break
@@SherryKao16 is it also partly because I -ion will take H+ from OH which causes O to be O negative and that's why the main product is ester and minor product is HI?
Thank you very much this are onlyone vudeis on youtube ehich explains nucleophilic reactions to be understood and on the end I know it solve by myself. So thank you very much.
how do you know when your carbocation is flat or not?
please help URGENT 🙏🙏🙏🙏
@@lisanation2203 hybridization of a carbocation is sp2 and sp2 is palnar
in the second exercise, are there no stereoisromers bc there is no asymetrc center?
Thanks a lot✌🏼
Why is the alcohol from problem 1 not deprotonated but it is in problems 3 and 4?
thank you so much🌸
why didn't we draw the strereoisomeres?
thank you!
hey i have Q i could use the proton transfor for cl to make HCl
Cool
Awesome
Thank you so much!!!! I actually understand all of this now!!!! Its nice to finally not feel clueless in orgo lol
same here man
I have a question, I realized on eqn 1 and 4, the final products had lone pairs but the final products for 2 and 3 didn't have. Is there a specific reason or you just chose not to put them there?
The lone pairs are implied in 2 and 3.
Yeah, I'm pretty sure he just felt lazy didnt wanna put them lol .
where did the hydrogen go in 1st,3rd,4th reactions?
There are ways of answering "where did the hydrogen go" ?
1. Missing acid-base reactions: In the first reaction, H2O attacks the carbocation and becomes -OH2(+). One of the hydrogens is then removed in an acid-base reaction by another molecule of H2O, leading to H3O(+)Cl(-) [drawn in red] and the neutral alcohol product. In the interest of space, the acid-base reactions that give the neutral products are omitted in the 2nd, 3rd, and 4th reactions.
2. "Hidden" or "implied" hydrogens not drawn in 4th reaction. Note that HOCH2CH3 attacks the carbocation, and a hydrogen is drawn in, but not drawn in the next diagram. However it is still there - this is an example of a "hidden" or "implied" hydrogen from a line diagram. In the figure where the new C-O bond is a "wedge", there is a "hidden" hydrogen on a dash. In the figure where the new C-O bond is a "dash", there is a "hidden" hydrogen on a wedge. They were not drawn in to save space, although in retrospect I should have put them in. Thanks for your comment!
I heard from my teacher that sn 1 or 2 reaction only occurs in alkyl halide
Yes
I love you
I think im gona get another A....!!!!!!!
❓❔❔❔❓
Awesome
Thank you! If this video was helpful, you may want to check out our month-to-month MOC membership for access to over 1500 quizzes, Flashcards, the Reaction Guide, and more. Check it out here: bit.ly/2YctxPb