SN1/SN2/E1/E2 - working through problems! 

Aww HC, thank you very much for this comment. I made these for my students but glad you also found them helpful. I wish I had time now to make them better.

Aww Mohammad! I miss you. You always asked great questions in class, and I enjoyed our hiking trip that summer!

Yes- when both SN2 and E2 are possible, secondary and tertiary alkyl halides favor E2 because of less sterics but also the stability of alkene being formed, over primary halide being favored for SN2

@nicholas - thank you so much! You are very kind!

HI Zoe! Thanks for your comment! Yes, you just change the configuration at the chiral center IF your nucleophile is attacking it at that precise carbon. So you don't have to label whether it's R or S, but just change it to the opposite configuration and make sure your drawing indicates that inversion of stereochem. For example, if you started with R-2-propyliodide, make sure it's S-2-propylwhatever at the end (or if it's wedged, make sure it ends dashed, or vice versa). However, please note that the chiral center may not be the carbon that is attacked - in which case, you would not change the chiral center! An example is R-1-bromo-2-methylbutane... after reaction with iodide, you should get R-1-iodo-2-methylbutane, because the chiral center is on carbon 2 and not carbon 1, where the nucleophilic attack is occurring! I hope this makes sense and that I answered your question but if not, please feel free to write back!

​@@orgotime6196 Okay thank you very much for your explanations!!

Hi Tushar! Yes - nucleophilicity and basicity are not exactly the same. Nucleophilic strength is based on kinetics - when nucleophiles are branched/bulky, the electron repulsion of these electron clouds create steric hindrance and slow down the reaction rate... thus tBuO- is a weaker nucleophile than OH- even though it's a stronger base than OH-. I hope this helps!

Great question! The effect of “hiding” the anion is much greater when the anion is acting as a nucleophile, rather than as a base (think of the solvent cage as making the Nu more sterically hindered). Therefore Sn2 will be more negatively impacted by protic solvent than E2. This might help: www.chemistrysteps.com/sn1-sn2-e1-e2-effect-of-solvent/

Strong base weak nucleophile goes for E2 for tertiary. She said weak nucleophile but she also said strong base. However the strong base is too bulky to attack a carbocation, so that means it will attack a hydrogen which means elimination.

I go into it a little at around 10:15min mark of this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ivkquiO9Hgs.html I also recommend this: www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/

one year later reply lol it's because DMSO is a polar aprotic solvent sothere will be no solvatation

Hi Dr Kora, sorry I am only getting your message now! If it’s not too late you can email me: gershie@gmail.com

I mean the notes pdf

Hi Steven! That's great - I'd love to check it out!

Hi Soumya - right, if only there wasn't a strong base in the solution. Since tBuO- is a strong base, it will not wait for carbocation - it is very reactive and will start an E2 mechanism instead. So it's important to think of every reagent in terms of: strong Nu (favor SN2), weak Nu (favor SN1) AND ALSO strong base (favor E2), weak base (favor E1). Here we have tBuO- which is a weak Nu but strong base - E2 (faster) is favored over SN1/E1.

@@orgotime6196 Thank you so much maam...

The key is understanding the reagent on top of the arrow. It’s a good Nu so it favors SN2 over SN1, and it’s a strong base which favors E2 over E1. Because it’s a secondary alkyl halide, both are possible. The chart on this page from Klein might help you with knowing whether the reagent is a strong or weak Nu and strong or weak base: www.bartleby.com/solution-answer/chapter-1010-problem-1027p-organic-chemistry-as-a-second-language-first-semester-topics-4th-edition/9781119110668/cd1360b4-5079-11e9-8385-02ee952b546e

Actually E2 is favored in protic solvent- that is a big difference between E2 and SN2. Further reference about this: www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

Hi Rick - we can get quantitative if you want, but most of the organic chem at this level is presented qualitatively to keep it accessible (I think). We try to recognize patterns from the data but I agree that omitting these data lead to feelings of hand-wavyness... Yes - I can understand your point. In general, what qualifies as enough steric hindrance to predict "no reaction" is going to be where the nucleophilic carbon is tertiary (and in some cases, secondary, or next to a tertiary carbon). I usually see 99.9% examples where something considered "branched" (isopropyl, isobutyl, or in a cyclic ring like DBN) to be hindered enough to predict no reaction or at best, minor products. We could get into the weeds on this and look up research articles siting repulsion energies if you'd like, but most treatments of this material remain qualitative. Advanced organic chem texts will go more into more detail.

​@@orgotime6196 i appreciate your reply, but this is the problem with all of western STEM education, we dont teach from first principles. after youve had OC and memorized the hand-waving, you have to unlearn it all and start from scratch having wasted another year of your life. i'm not blaming you or your channel, i'm saying the whole subject is a useless waste of time that purely serves as a filter class. same with linear algebra. either teach by historical development, or teach by first principles (whatever those are in present era.) /rant

@@rickdeckard1075A lot of the time that isn’t practical for career considerations and different needs of students

@@coolexplorer1015 yes thats why they cant get and keep a job that doesnt require being told what chem to put in what testtube