Hey, I'm new to this channel. Great Video. In the problem with 1-methyl-1-iodocyclohexane, the steps you've shown lead to an aldehyde on one side and a ketone on the other. My solution is first add naoch3 to get methylcyclohexene and to shift the double bond first add HBr with peroxides and then use bulky base like LDA to get 3-methylcyclohexene and then use reductive ozonolysis.
Thanks for the exercices. Always nice to have that type of content to practice. I think there's a mistake on the O3 exercise. Ozonolysis as described would lead to the methyl group being not ramified from the main 6-C chain. The double bond in the precursor has to be 1 C away from the methyl to obtain the desired product
For the first question, why is sodium methoxide used for the E2 reaction? Wouldn't a bulky base be preferred as we are trying to get the main product to have the double-bond on the outer most C-C bond, therefore we want a base that is more sterically hindered? Thank you! I appreciated this video
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.
Thann you Contantijn for a wonderful video. Quick question. For the first problem, why wouldn't something like MeOH or KOtbu work to turn the br into an Alkene?
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.
About the problem with deuterium - I thinks is it possible to make a ketone first, and then add NaD. Afterwards we simply get rid of the OH group by reduction, leaving us with the product. Am I correct, or I miss something?
Hello! Thank you for the video, but I have to ask... for the last problem, is it possible for us to add h2so4 in order to make an alkene and then do the anti-markovnikov hydroboration with O-CH2-CH3 to get the product?
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For the first problem would e1 not work because it forms a carbocation?
Exactly! That cation would be subjected to a hydride shift.
@@ConstantijnCole Got it. Thanks for the quick response.
Hey, I'm new to this channel. Great Video. In the problem with 1-methyl-1-iodocyclohexane, the steps you've shown lead to an aldehyde on one side and a ketone on the other.
My solution is first add naoch3 to get methylcyclohexene and to shift the double bond first add HBr with peroxides and then use bulky base like LDA to get 3-methylcyclohexene and then use reductive ozonolysis.
Thanks for the exercices. Always nice to have that type of content to practice. I think there's a mistake on the O3 exercise. Ozonolysis as described would lead to the methyl group being not ramified from the main 6-C chain. The double bond in the precursor has to be 1 C away from the methyl to obtain the desired product
Great content Constantijn! Keep up the good work!
Finally I understand!!! Thank you!
For the first question, why is sodium methoxide used for the E2 reaction? Wouldn't a bulky base be preferred as we are trying to get the main product to have the double-bond on the outer most C-C bond, therefore we want a base that is more sterically hindered? Thank you! I appreciated this video
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.
Great video!
Thann you Contantijn for a wonderful video. Quick question. For the first problem, why wouldn't something like MeOH or KOtbu work to turn the br into an Alkene?
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.
About the problem with deuterium - I thinks is it possible to make a ketone first, and then add NaD. Afterwards we simply get rid of the OH group by reduction, leaving us with the product.
Am I correct, or I miss something?
Yep that would work!
Hello! Thank you for the video, but I have to ask... for the last problem, is it possible for us to add h2so4 in order to make an alkene and then do the anti-markovnikov hydroboration with O-CH2-CH3 to get the product?
Yes! Great vision. That would be a successful synthesis for that problem.
the problem before the last one(the one with ozonlaysis) i made the double bond one carbon next to the methyl group is this correct?
In 2nd problem can't we simply do oxidation of secondary alcohol to ketones and then ketone oxidation to carboxylic acid
Great idea! That would work. Make sure you use a strong oxidizing agent for the second step.
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