I think I understand, but need some reconfirmation. Since the base abstracts H from the beta carbon, in that case a methyl halide does not undergo elimination reactions since there's only the alpha Carbon. And primary, secondary, and tertiary halides undergo E1/E2 depending on weak base/strong base, steric hindrance is not a factor
the first part is correct, a methyl halide can't possibly undergo elimination. otherwise elimination can occur and the mechanism will depend essentially just on the strength of the base, and steric hindrance isn't a factor for determining the mechanism, but it is a factor in determining which proton is abstracted (zaitsev/hoffman) products.
Cool Prof. Dave ... nice and short. Hydroxide ion steals a Hydrogen off beta carbon on Bromopropane molecule , also releasing the Bromine, turning it into Propene. With Water molecule. Thanks. Well done !!
I'm a little confused. You said that the electrons will have nowhere to go when a proton is removed from the alpha position, but isn't there an electronegative bromine atom it's bonded to that would be perfectly happy to accept those electrons? Why can't this happen?
@@ProfessorDaveExplains Oh, I get it now. The carbon-bromine bond supplies bromine with electrons, so it won't be accepting any more. That took me embarrassingly long to figure out. Thanks!
What are other conditions for sn1 or e1 to occur other than temperature and ... The steric hinderence of reagent and substrate both responsible for sn1 and e1 .....
this might be a dumb question but why does the leaving group (Br) leave the molecule? Like why does it's bond weaken? Isn't the (Br) stronger than the proton?
well it's quite stable in solution on its own, and the collision between the base and the proton was of sufficient energy for the reaction to occur, and for the pi bond to form, the leaving group must leave.
Prof. Dave Ji ( In Malayalam Dave = Dev ( The God) . You mechanically acted like God to us guys though. Am from Kerala S. India. Am a Tutor and love to view your videos for finalized notes. Teaching upto plus two. Expecting your valuable videos still. Corona is not a joke. Really creepy when the educational arena is considered. You do rock Sir. Take Care please. Saareeyakal Sir !!! 🙏🙏🙏
merci! je veux savoir si t'as une video qui explique quand utiliser SN1, E1, SN2, E2 rxns? :-D je sais que SN2 et E2 sont rxns competitif... egalement pour SN1 et E1.
i don't really speak french but i think you're asking for a clip that explains when to choose which mechanism? i did that already! find it in my organic chemistry playlist.
Je vois que ce comment etais la depuis 7 moins, mais si ca marche pour vous ou quel-qu'un d'autre il y a une video la ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-IOViLLuDMTs.html
hi! I don't quite understand why removing a proton from the alpha carbon doesn't work (1:25) - couldn't the electrons just form a lone pair on the carbon?
Why is it specific that bulky bases like tertiary butyl alcohol and if Flourine give only Hoffman products? Is it that they give only Hoffman? Or do they also give other minor products??
the bulkier the base the more the hoffman product will be preferred for reasons of kinetics. the activation barrier associated with generating the hoffman product is lower than that associated with zaitsev. but there is typically a mixture, depending on precisely how bulky.
With the transition state you have drawn, does this imply that the antiperiplanar nature of the hydrogen and bromine atoms is the best conformation for an E2 reaction?
Ashwangg in fact, it is the only possible conformation from which E2 can occur, because it is the only way for the developing p orbitals to generate the overlap that will result in the pi bond.
If you are referring to the orientation of the beta proton, that is covered in my tutorial on E2 on cyclic systems. If you actually meant regiochemistry, that is covered in my tutorial on Zaitsev vs. Hofmann elimination products.
+pk sagara depends on various factors like solvent system, temperature. this is just one mechanism that could happen. in fact, unless otherwise specified, you would certainly get a mixture of SN2 and E2 products.
@@ProfessorDaveExplains in the video the alpha is attached to the bromine and then the beta is attached to a hydrogen and methyl group so I guess I just don't understand how to identify what is what. Is there a general rule to it?
They're just relative terms, in this case they reference the halogen. Alpha means one away, so the carbon attached to the halogen is alpha to the halogen, and any carbon one further away than the alpha carbon is a beta carbon.
+AstridMPerez it certainly could! i'm just outlining the E2 mechanism. to decide which would occur you'd need to know more about solvent, temperature, etc, and if just given these compounds in a general sense, both mechanisms would occur and you'd get a mixture of products.
it's just because the transition state is showing all the geometry, before we were simplifying and making it look flat. yep the proton being abstracted by the base is the same as the one that is drawn in before.
check out the VSEPR tutorial in general chemistry, and maybe the first IUPAC tutorial in the organic chemistry playlist as well to make sure you understand line notation
sure, polar protic solvents make mechanisms with carbocation intermediates more favorable, they also reverse the nucleophilicity trend of the halides, stuff like that.
Thank you very much sir Can I ask you a question please? If I have 2-bromobutane in an elimination reaction then what would the product be But-1-en or But-2-en or both? Thanks in advance 🌺🌺
Hi Professor Dave, I'm having trouble doing an E2 reaction when given a molecule in a Fischer projection. My current method is to convert the Fischer projection to a sawhorse, then to a Newman, then finally the final product. However, this process is very tedious and I always end up making mistakes. Is there a better way of doing this starting with a Fischer projection?
Fischers are annoying, huh? I've never done eliminations that way, you definitely don't need Newmans, but the best way to be sure is to convert into line notation and do it that way. Otherwise you can probably do them right on the Fischer, you just have to make sure the stereochemistry works.
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