Thanks. Sometimes I myself disagree with some of my mechanisms but I show what is the common mechanistic convention as it is taught in the northern American schools as it’s my main audience.

Thank you! Comments like this encourage me to start making these videos again.

Haha, thanks. I'm slowly working on it.

Glad it helped!

Will do 👍

Glad it helped!

Sure, you could cherry pick the actual rate information when you’re watching a video, but you won’t have the internet on the exam. This question was a *generalization* of the approach how this topic is taught and what type of answer we would expect on the test. Unless the instructor specifically points out how really bulky the t-Bu group is, students would only focus on the positioning on the leaving group. But yes, if we compare the actual experimental data, A is slower than C. But as I’ve mentioned, most classes and textbooks won’t go too deep into this topic. As far as I remember, Karty, and maybe one other textbook even mention that things a a bit more complicated than a simple 1° vs 2° vs 3° comparison lol 😆 Perhaps, I should’ve chosen a slightly different set for that example.

@@VictortheOrganicChemistryTutor You’re fine, they’re both really close, I just wanted to confirm! Thanks for the reply

The majority of your molecules will be in the more stable conformation, so when we're talking about the kinetics of reactions, you should always be comparing the more stable conformations. Here, the more stable conformation is the one with the isopropyl in the equatorial position for both molecules. If you need a refresher on the chair conformations and how to draw those, I have a tutorial dedicated to that too.

Allylic is the position next to a double bond. Equatorial and Axial refer to specific positions of atoms on (typically) a chair conformation of a cyclohexane, although they can also be used to describe 3D orientations of groups in other cyclic molecules as well.

The inversion happens regardless if the atom is chiral or not. In this case, while the atom is not chiral, you still have a very specific orientation of the leaving group, and since we have to do the back-side attack, you'll end up with the incoming nucleophile looking in the opposite direction.

@@VictortheOrganicChemistryTutor I actually was about to ask the same question. I think there's another sample question in one of your review videos where one carbon bond doesn't rotate, but the leaving group bond is re-arranged to an inversion angle. This is quite tricky....

@@rockubabe Yeah, I purposefully include questions like this in my reviews as I know they trip up students 😆 The key here to remember that the inversion will always happen in SN2, whether it's going to be relevant for the molecule or not, is a different question. To be on the safe side, I recommend you just flip the stereochemistry of the atom you're attacking regardless, and then check if the resulting molecule(s) has a meaningful stereochemistry. Also, keep in mind that you don't need to have chiral atoms to have enantiomers or diastereomers! This is another tricky point a lot of students forget about 😉

Within a scope of a regular sophomore organic chemistry course, you can pretty much ignore the spectator ions completely unless your instructor asks for balanced equations.

It doesn't. Oxygen is the second most electronegative element. Br's larger atomic radius and bigger orbitals allow it to stabilize the negative charge better than the oxygen.

@@VictortheOrganicChemistryTutor ohh thanks for clearing doubts.

🧠 here 😂