SN2 Practice Questions
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- Опубликовано: 6 авг 2024
- In this video I'm going to go over some of the most typical SN2-related questions you might see on the exam. I'm also including some tricky ones so you know what to be prepared for.
00:00 Intro and review of the SN2 reactions
01:48 Question 1 - predicting the products
06:02 Question 2 - tricky questions
11:07 Question 3 - rankings
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brilliant in question 2 (iii) was asked in my exam and i did it correct thank you sir
Make more of these exercises because are so helpful. I enjoyed listening to your explanation is so incredible and full of information. Thanks
Thank you! Comments like this encourage me to start making these videos again.
Very well explained! 😊
Your video's are top notch. Very rarely I disagree with any of the mechanism you propose. I will highly recommend them to my students. All the best and thanks.
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.
Seriously, Very Lucid way of teaching.... Really want to see videos on all mechanism types with all the anomalies.
Haha, thanks. I'm slowly working on it.
Amazing wish you do more of those
Will do 👍
Thank you so much the ranking problems really helped!
Glad it helped!
omggg thank u very muchhh , very easy to understand for me
Glad it helped!
hello for example #1 we don’t have to write the specter ion in our ending product after the cn (cn + I-)?
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.
For Question 3 of the rankings, how could you deduct that the Isopropyl group is in an Equitorial position? Thank you!
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.
For question 1 at 5:52, why did we change the configuration from wedge to dash, if the carbon with the leaving group isn't a chiral center?
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 😉
what is carbon position 'allele, equator, axial'?
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.
At this point I just want his brain for my exam tomorrow :')
🧠 here 😂
Ohh only bromine is a good leaving group at clip 5:53 because it is more electronegative than oxygen according to linus pauling scale on the periodic table.
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.
At 12:53, looking at A and C, I think C should be second place and A should be third place. When looking online at the relative rates depending on the alpha and beta carbons, a secondary alpha carbon (C) has a relative rate of 0.008, but a tertiary beta carbon (A) has a relative rate of 0.00001, much smaller and nearly neglible.
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
i love you