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:)

So happy you found the channel! I hope it can help you more in the future.

I am so glad you enjoyed it! More are on the way!

I appreciate the support my friend. Good luck studying!

Happy to help! Good luck on your midterm!

You’re very welcome!

Thanks, you too!

Yep that would work!

Glad it was helpful!

Great idea! That would work. Make sure you use a strong oxidizing agent for the second step.

Thanks for your question! KOtbu would defiantly work. It is a base strong enough to remove an alpha hydrogen, and it is bulky enough to direct to the major product we want so it would work perfectly. MeOH, on the other hand, would run into some issues. Because it is an alcohol, it would not be able to deprotonate the alpha carbon. It would have to go through the E1 pathway to form an alkene, which would involve the formation of a carbocation and potential rearrangements. It could work, but I would recommend sticking to E2 if presented the option for a synthesis problem on an exam because there are fewer moving parts.

Great thinking! You are correct. In cases like this, when a hydrogen could be pulled from one of two alpha carbons, the use of a bulkier base will attack the less sterically hindered hydrogen, allowing us to control for the product we want. That is why we use NaOC(CH3)3, sodium tert-butoxide, as the base for the E2 reaction in the first problem.

Yes! Great vision. That would be a successful synthesis for that problem.

Happy to help!

Good luck!

Happy to help 👍

I believe that the hardest part of this problem is identifying which conjugated carbon system is the diene and which is the dienophile. Here are a couple of videos that can help with that. th-cam.com/video/d5-05nqS05A/w-d-xo.html This one shares a very in depth explanation for why how electron donating/withdrawing groups work and why they apply to Diels-Alder reactions. th-cam.com/video/9QJ_IEFJ-rs/w-d-xo.html This one keeps it somewhat simpler and gives a few examples. Hopefully you find this helpful!

Great observation. You are correct that the reaction requires a diene in the cis configuration. The diene in the first reaction is in the trans configuration, but it is possible to flip the configuration with heat.

@@ConstantijnCole it is very possible to flip the diene to be in s-cis configuration

Or why is the eye on the left when everything implies ur looking to thru to the right. My mind rn lmao

Thanks for the question. I think there are a couple of approaches you can take to better understand this perspective. First, when we are looking at a Newman projection as it is normally drawn, we are looking down the 2-3 bond. Therefore, the eye looking from the left of the Newman projection is not looking down the 2-3 bond. It is looking across it as is seen in the skeletal structures. You can also think about it in terms of which substituents are close to or far from the eye. In this case, the OH on the front carbon and the H on the back carbon are the closes substituents, so it makes sense for them to be on a wedge in the skeletal structure. On the other hand, the NH3 on the second carbon and the H on the first carbon are far away from the eye. They are dashed in the skeletal structure because they are going away from our perspective (Note that the H's don't actually have wedges/dashes because they aren't normally indicated). Hopefully you find this helpful. Let me know if you still have questions.

@@ConstantijnCole awesome. So would this Newman technique only apply when you need to move a hydrogen off plane? I ask because I’ve had success w Newman structures using the stick figure technique (right arm and left arm)and this technique kinda reversed the latter strategy.

@@PauloJAdams If I'm understanding your question, it works either way. You would just have to adjust the stick figure based on the direction you look from. I would recommend continuing to use the method that has worked for you and keeping your perspective consistent. That way, you can limit your mistakes on exams.

Hi at 14.30 to 14.50 u talk about radical rearrangment, is it possible that we rearrange free radical becoz as far as I know rearrangment only happened in carbocation. correct me if I am wrong

Great question. You're right that you could look at this specific chair from either side because the skeletal structure could always be flipped or rotated in space anyways (this can be seen through the meso character of the structure in this case). I suggest looking from above every time in order to be consistent and ensure that all substituents are facing in the correct direction. If you were to look at the chair from bellow, you would just have to invert all of the substituents. However, it is important to add that that could only be done here because this is a meso compound (meaning that flipping all the stereocenters results in the same structure). If it were not a meso compound, looking at the chair from bellow, and in turn flipping all of the substituents could lead you to draw an enantiomer instead. So I suggest using the convention of looking at the chair conformation from above in all cases to avoid those mistakes.

Great question! You are correct in recognizing that hydride shifts are possible whenever a carbocation forms adjacent to a carbon with a free hydride, but no hydride shift happens in this reaction. Here is why. Hydride shifts happen in order to move the carbocation to a more stable position. Good job observing that the free hydride here is in a more substituted position than the carbocation. Under normal conditions, that would make it more stable and promote the hydride shift. However, the carbocation is being stabilized by resonance from the phenyl group on the left. This makes cation more stable where it is, removing the need for a hydride shift. Hope you find this helpful!

@@ConstantijnCole Thank you really appreciate it!

Exactly! That cation would be subjected to a hydride shift.

@@ConstantijnCole Got it. Thanks for the quick response.

Thank you for your question. I hope you find this explanation helpful. The trick I suggest for identifying meso compounds is as follows. As long as the carbon skeleton is symmetrical, counting carbons between chiral centers will reveal what stereochemistry is necessary to be meso. If there is an an odd number of carbons in between, the stereochemistry must the same. If there is 0 or an even number, it must be the opposite. In this case you correctly assessed that cis 2,4-difluropentane has 1 carbon between chiral centers. Therefore, it must follow the rule for odd numbers which states that it is a meso compound as long as the stereochemistry is the same. In this case, both F's are on wedges. Therefore, it is a meso compound.

@@ConstantijnCole So that means I have to see if both are wedges/dashes, and if one is dashed and the other is wedged, hence it is not a meso compound. Correct?

Great eye. However, in this case the rule does not apply because the substituents (NH3 and OH) are not identical. Had they been identical, this would have been a meso compound.

@@ConstantijnCole oh i see, thank you!!

sorry thats supposed to be difluoro.

There are a lot of ways to think about this. I can try give you three explanations. Hopefully you find one or more helpful. The first explanation is conceptual. You're definitely thinking about this the right way. I did mention that a good trick is to line up either side of the mirror so that wedges correspond to wedges and dashes correspond to dashes. However, in this case, we are looking at a meso compound. This means means that its mirror image is identical to the original structure. The second explanation is a bit of a trick. There are two ways to go about drawing an enantiomer or mirror image. The first is to establish the mirror on the page and draw reflective images on either side. The other way is to flip all of the stereocenters. That is what I did. The last way of thinking about this is to try to change your point of view. Try to think about it as if the mirror is in the plane of screen.

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