thank you for your videos Chad!

@ChadsPrep

3 жыл бұрын

Glad you like them, genesis - you are welcome!

for the last example, what happens to the other cl- ion (the one that doesn't get attached). does it attach to another molecule, or...?

Would the product of halohydrin formation have enantiomers? How many products would it have? (It's about the last part of the halohydrin formation mechanism explanation)

Thank you so much for your time put into this video. I just have one question. Why is the CH3 on a wedge for the very last example? can I keep it in a regular line instead of making it a wedge? Thank you!

@veselinavolik3002

Ай бұрын

i think its because of the back side attack. since its anti addition, you have to show that the addition is happening on opp sides of the molecule. don't rely 100% on me tho, I'm also learning

you are the best

@ChadsPrep

8 ай бұрын

Thank you

beautiful

@ChadsPrep

2 ай бұрын

Thanks

Hey Chad I've been watching a lot of your videos and have been wondering an idea which you cover in your videos. With one of retrosynthetic analysis videos I believe it's 11.1 Introduction to Organic Synthesis | Retrosynthesis there's an example you do with an epoxide where you state that the less substituted carbon will be attacked because it has less steric hinderance. However in this example with a similar process going on you tell us that the more substituted carbon will be attacked because it is more positively charged. Can you help with this inconsistencies or at least explain when which is more important than the other?

@ChadsPrep

2 жыл бұрын

Hi Eric! Great question. It gets covered in lesson 13.6 on Ring Opening of Epoxides. kzread.info/dash/bejne/iIqfr9uzisfZpMY.html If the 3rd member of the 3-membered ring is positively charged the result is the two carbons bonded to it are a little 'carbocation' like and the more substituted one will share more of the positive charge which increases its electrophilicity. We often oversimplify this and simply say the more substituted one will be attacked, but the truth is it depends (see video). If the 3rd member of the 3-membered ring is not positively charged then it is all about the less hindered back side attack which is why attack will occur in this instance at the less substituted carbon. I go into greater detail in the video. Hope this helps!

That awkward moment at 6:54 when you and Chad just stare at each other in silence

@ChadsPrep

Жыл бұрын

Indeed! 🙃🙃🙃

When the water molecule performs a backside attack, is there preference to which carbon it will bond?

@ChadsPrep

Жыл бұрын

Yes, the more substituted carbon. Both carbons bonded to the halogen have a partial positive charge, but the more substituted one bears more of a partial positive charge (just as a more substituted carbocation is more stable, we can explain this along the same line of reasoning) and is therefore more electrophilic and more reactive with water here (acting as a nucleophile). Hope this helps!

Why isn't alkene halogenation syn, producing a transition state with a partial X-X, partial C-C, and 2 partial C-X bonds? Is it just unlikely that the X2 will collide in a way that both X atoms are close enough to the pi bond? Even if so, how can this reaction proceed at a reasonable rate if we're generating 2 ions in nonpolar solvent, from 2 neutral molecules?

@ChadsPrep

3 жыл бұрын

Hello Dennis! You may not like my answer, but because. We actually study the kinetics and energetics of this reaction as well as the stereoselectivity and use what we learn to infer a mechanism that is consistent with all of this information. We know it is an anti-addition based upon the stereochemistry of the products it creates. No syn-addition occurs which allows us to rule out the mechanism you proposed. Now why that doesn't happen is a completely different question. The data simply tells us that it doesn't. We also know that it never undergoes rearrangements which rules out any sort of carbocation intermediate. But we can also infer that it doesn't occur in a concerted mechanism but occurs in multiple steps. The two-step mechanism involving the halonium ion intermediate is consistent with all of this and is also consistent with the mechanisms of other reactions that are similar like the ring-opening of epoxides (chapter 13). Now you also question forming ions in nonpolar solvent. Now I would much rather form ions in a polar aprotic solvent and even more so in a polar protic solvent, but don't rule it out forming certain ions in a nonpolar solvent completely. When we discuss SN1 and E1 reactions we explain that in order to form a carbocation the reaction has to take place in a polar protic solvent to stabilize the carbocation with ion-dipole forces otherwise the carbocation wouldn't be stable enough to form. Now in most SN1/E1 reactions we also form a halide ion when the carbocation is formed, but the focal point solvent-wise is on the carbocation since the halide ion is much, much, much more stable than a carbocation. You're right that a lot of ions won't form in nonpolar solvents and the key is the the less stable the ion the less likely it is going to form in any solvent and especially in a nonpolar solvent. But Cl-, Br-, and I- are very stable anions and the halonium ions are quite a bit more stable than carbocations and apparently stable enough to form even in nonpolar solvents. Hope this helps!

@dennisgrassl2219

3 жыл бұрын

@@ChadsPrep Yes, that helped a lot! Thank you! I'm still getting used to thinking like a chemist and your videos have been a great resource.

👏🏻👏🏻👏🏻👏🏻👏🏻

@ChadsPrep

8 ай бұрын

Many thanks

Roses are red Violets are blue Chaddy daddy is my hero, it's true.

@ChadsPrep

Жыл бұрын

Glad the channel is helping you, Laura.

you look like bill burr if he was smart

@ChadsPrep

2 жыл бұрын

😃😃😆