At 6:37 you show both the LG and the attacking Nucleophile in the same stereochemistry. Although I see that one is a R and the other a S configuration, I thought the Nu would have to be wedged? am I wrong?
At 6:00 the other substituents were "pushed" to the other side, so the nucleophile was shown on a wedge like of the leaving group. at 7:50 the placement of the other substituents was retained and the nucleophile was shown at the reverse side of the place of the leaving group. if you flip the molecule at 6:00, it will be the same molecule at 7:50. the two are just different ways to show the inversion that happened.
They are unrelated to each other. In each situation the key is to recognize what is happening. If the nucleophile comes in from a backside attack, make sure to draw the opposite configuration compared to the starting molecule. If the leaving group is on dashes, SCH3 will be on a wedge, if the leaving group is on a wedge, SCH3 will come in on dashes
Hey, im a bit late on this video but I have one question. I dont quite understand why polar aprotic solvents cant stabilize any charges in solution? I always thought the "polar" part was responsible for that. A polar aprotic solvent still has a partial negative and positive charge... Why cant it stabilize charges with these partial charges? Or am I missing something? Water stabilizes the other charges with its partial charges on the O and H does it not?
This is a really good question one that I address in my polar protic/aprotic solvents video. In short, take a look at the aprotic solvents and you'll notice that while they do have a partial positive, it tends to be hidden on the inside such as a carbonyl carbon. They are also less polar compared to the protic solvents. So while that can somewhat stabilize charged intermediates, it tends to be the partial negatives that are less hindered which will not stabilize your nucleophiles or leaving groups
Hey Leah, Just wanted to say great video series! But I am confused on the stereo portion of the video-- from my understanding of the video (and previous knowledge) nucleophilic attack occurs trans (i.e. the nucleophile attacks on the side opposite to the leaving group). You showed that, but in the final product, showed that the nucleophile ended up on wedge-- the same side the leaving group was on. How can this be possible if the nucleophile attacked oppisite (i.e. at the dashed position)? I thought that the hydrogen would be pushed to wedge during the attack as the nucleophile went wedge... Thanks for any clarification! Again, awesome series!
You are superb! I guess the best orgo teacher in the internet.... but as per my knowledge arent R and S ENANTIOMERS.... but in substitution reactions the product and reactants arent even ISOMERS...... so shouldn't you be using +/- or d/l to expain the stereochemistry.... only the configuration changes but they don't becoms R/S ENANTIOMERS . RIGHT?
R and S is just configuration. You can use them to check whether 2 molecules are enantiomers or not . However they are defined for every chiral molecule and not just isomers.
Reactants and products aren't even isomers, they have completely different atoms in most cases. Instead we're looking just at the chirality of the starting molecule to understand what happens when the SN2 reaction occurs via backside attack
At 6:37 you show both the LG and the attacking Nucleophile in the same stereochemistry. Although I see that one is a R and the other a S configuration, I thought the Nu would have to be wedged? am I wrong?
how come at 6:00 you show that SCh3 is on a wedge but at 7:50 SCh3 is on dashes im kinda confused about that
At 6:00 the other substituents were "pushed" to the other side, so the nucleophile was shown on a wedge like of the leaving group. at 7:50 the placement of the other substituents was retained and the nucleophile was shown at the reverse side of the place of the leaving group. if you flip the molecule at 6:00, it will be the same molecule at 7:50. the two are just different ways to show the inversion that happened.
They are unrelated to each other. In each situation the key is to recognize what is happening. If the nucleophile comes in from a backside attack, make sure to draw the opposite configuration compared to the starting molecule. If the leaving group is on dashes, SCH3 will be on a wedge, if the leaving group is on a wedge, SCH3 will come in on dashes
Hey, im a bit late on this video but I have one question. I dont quite understand why polar aprotic solvents cant stabilize any charges in solution? I always thought the "polar" part was responsible for that. A polar aprotic solvent still has a partial negative and positive charge... Why cant it stabilize charges with these partial charges? Or am I missing something? Water stabilizes the other charges with its partial charges on the O and H does it not?
This is a really good question one that I address in my polar protic/aprotic solvents video.
In short, take a look at the aprotic solvents and you'll notice that while they do have a partial positive, it tends to be hidden on the inside such as a carbonyl carbon. They are also less polar compared to the protic solvents. So while that can somewhat stabilize charged intermediates, it tends to be the partial negatives that are less hindered which will not stabilize your nucleophiles or leaving groups
very helpful video, you're honestly a hero.
So glad it helped! Thanks for watching.
Hi! Why Acetone and not other solvent link DMSO or DMF? Thanks!!
Random example and not a limitation
Hey Leah,
Just wanted to say great video series! But I am confused on the stereo portion of the video-- from my understanding of the video (and previous knowledge) nucleophilic attack occurs trans (i.e. the nucleophile attacks on the side opposite to the leaving group). You showed that, but in the final product, showed that the nucleophile ended up on wedge-- the same side the leaving group was on. How can this be possible if the nucleophile attacked oppisite (i.e. at the dashed position)? I thought that the hydrogen would be pushed to wedge during the attack as the nucleophile went wedge... Thanks for any clarification! Again, awesome series!
+Juan Romero I apologize-- i asked my question before finishing the video. I'm silly ! THank you, nevermind!
Awesome, glad the video helped!
Thank u so much for your great work
You're so welcome!
thank you ma'am, thank you very much!
You're very welcome!
Excellent!! 😊
Thank you!
Thank you Leah!!!!!You are jus the best!!!!!!!!!!!!!!!!
Awww you're so welcome!
Hi Leah, Is rearrangement only possible in Sn1 reactions?
Yes because in Sn2 the reaction is fast and single step where carbonation doesn't form.
And E1. Rearrangement requires a carbocation intermediate and you'll never see a carbocation form in a fast '2 type' reaction like SN2 and E2
You are superb! I guess the best orgo teacher in the internet.... but as per my knowledge arent R and S ENANTIOMERS.... but in substitution reactions the product and reactants arent even ISOMERS...... so shouldn't you be using +/- or d/l to expain the stereochemistry.... only the configuration changes but they don't becoms R/S ENANTIOMERS . RIGHT?
R and S is just configuration.
You can use them to check whether 2 molecules are enantiomers or not . However they are defined for every chiral molecule and not just isomers.
Reactants and products aren't even isomers, they have completely different atoms in most cases. Instead we're looking just at the chirality of the starting molecule to understand what happens when the SN2 reaction occurs via backside attack
You are the best!
Thanks!!!
Really want to meet u once in a life...many many love and thanks
So nice of you! who knows one day! You are very much welcome! :)
You are from?
New York
thank you ma'am, thank you very much!
You're very welcome!