Another question for ya: Theres the sodium ethoxide on top of the arrow and the ethanol on the bottom - with reactions like these, how do we know which one is going to be acting as the base or nucleophile? is it whatever is on top? and what does that make the one that is not the base/nucleophile? a solvent? Thanks!
Hey Jaclyn! In reality there are just a load of molecules all mixed together in solution, in this case we have sodium ethoxide: NaOCH2CH3 which is a strong electrolyte so will dissociate to form Na+ and -OCH2CH3 ions. Then we have ethanol in the mix as well: CH2CH3OH. Our 2 options for a base are -OCH2CH3 and CH2CH3OH. A strong base will have a dense concentration of negative charge so which would you choose? (the other is the solvent, which also has an important role, because polar aprotic favors substitution whereas polar protic solvents favor elimination)
Thanks for another great video. I have a quick question though, at 10:00 you mentioned that sn2 is the major product. Why isn’t E2 a major as proton transfer happens relatively quick?
Hello Lan! Proton transfer is relatively fast but E2 isn't simply proton transfer, but is a concerted mechanism involving simultaneous proton transfer, pi bond formation, and loss of leaving group. So referring to it as 'proton transfer' could be a little misleading. Second, it turns out that SN2 is relatively fast as well with most methyl and primary halides but much slower for secondary halides (and so slow for tertiary halides that we say they don't happen). There's not a big why here but simply we know empirically when we perform the reactions like the one you refer to here that SN2 leads to the major product for methyl and most primary halides whereas E2 tends to lead to the major product for most secondary and all tertiary halides. If you notice my language I've left room for exceptions for primary and secondary halides. Primary halides with lots of beta branching may not favor SN2 over E2, and secondary halides that are also allylic or benzylic may favor SN2 over E2. We like providing students with 'clean' sets of rules. The exceptions get discussed to varying degrees. Hope this helps!
Oopsies...one more question... at 10:50 ish, sodium hydroxide has the positive Na and the negative O, so wouldnt that "cancel each other out" in terms of charge? If so, why is it under OH- in the strong nuc/strong base category and not ROH in the weak nuc/weak base category?
Hey Jaclyn :) If we isolated NaOH and nothing else, then yes the overall charge would be zero. But the ionic bond between Na+ and OH- is simply just the attraction of their opposite charges rather than being a covalent bond. If you put NaOH in any reaction solution they are pulled apart by the solvent and separate immediately. Then we just have a load of Na+ and OH- ions swimming around rather than NaOH sticking together and its those OH- ions which act as the nucleophile/base
Did you mean inversion rather than rearrangement? Rearrangements are possible for SN1 reactions which have a carbocation intermediate but not for SN2 reactions.
Its the night before my O-Chem exam is this has been the unit keeping me up for days. I really appreciate all your help, especially the charts!
Glad you are finding them useful!
Best video till now I have ever visited. Thank you so much sir.
You're welcome, SP - thanks for saying so!
Best explanation I've heard!
Glad it was helpful, Sydney!
This is an amazing video to clear your concepts and revise at the last moment literally saved me today ! Thank you 😊😊😊
You're welcome and Thank You!
THANK YOU, YOU ATE!
You're welcome!
I m from India & I found ur video very much helpful
Glad you found the video helpful, Anupam - thanks for commenting!
thank you soooo much chad, you totally are a homie
You're welcome Rachel! And thanks (I think?)! Best in your studies!
Excellent
Thanks
Another question for ya:
Theres the sodium ethoxide on top of the arrow and the ethanol on the bottom - with reactions like these, how do we know which one is going to be acting as the base or nucleophile? is it whatever is on top? and what does that make the one that is not the base/nucleophile? a solvent?
Thanks!
Hey Jaclyn! In reality there are just a load of molecules all mixed together in solution, in this case we have sodium ethoxide: NaOCH2CH3 which is a strong electrolyte so will dissociate to form Na+ and -OCH2CH3 ions. Then we have ethanol in the mix as well: CH2CH3OH. Our 2 options for a base are -OCH2CH3 and CH2CH3OH. A strong base will have a dense concentration of negative charge so which would you choose? (the other is the solvent, which also has an important role, because polar aprotic favors substitution whereas polar protic solvents favor elimination)
This is Great!
Thank you !!!
You're very welcome!
Thanks for another great video. I have a quick question though, at 10:00 you mentioned that sn2 is the major product. Why isn’t E2 a major as proton transfer happens relatively quick?
Hello Lan! Proton transfer is relatively fast but E2 isn't simply proton transfer, but is a concerted mechanism involving simultaneous proton transfer, pi bond formation, and loss of leaving group. So referring to it as 'proton transfer' could be a little misleading. Second, it turns out that SN2 is relatively fast as well with most methyl and primary halides but much slower for secondary halides (and so slow for tertiary halides that we say they don't happen). There's not a big why here but simply we know empirically when we perform the reactions like the one you refer to here that SN2 leads to the major product for methyl and most primary halides whereas E2 tends to lead to the major product for most secondary and all tertiary halides. If you notice my language I've left room for exceptions for primary and secondary halides. Primary halides with lots of beta branching may not favor SN2 over E2, and secondary halides that are also allylic or benzylic may favor SN2 over E2. We like providing students with 'clean' sets of rules. The exceptions get discussed to varying degrees. Hope this helps!
Oopsies...one more question...
at 10:50 ish, sodium hydroxide has the positive Na and the negative O, so wouldnt that "cancel each other out" in terms of charge? If so, why is it under OH- in the strong nuc/strong base category and not ROH in the weak nuc/weak base category?
Hey Jaclyn :) If we isolated NaOH and nothing else, then yes the overall charge would be zero. But the ionic bond between Na+ and OH- is simply just the attraction of their opposite charges rather than being a covalent bond. If you put NaOH in any reaction solution they are pulled apart by the solvent and separate immediately. Then we just have a load of Na+ and OH- ions swimming around rather than NaOH sticking together and its those OH- ions which act as the nucleophile/base
Min 5:02 - I thought when there’s heat only elimination can be done?
@10:27 How come rearrangement didn't occur for the Sn2 product?
Did you mean inversion rather than rearrangement? Rearrangements are possible for SN1 reactions which have a carbocation intermediate but not for SN2 reactions.
I love you
Thanks Hon! Glad you found this lesson helpful!
Chad... has anyone told you that you're a homie?
Nope, but thanks I think!