Retrosynthesis (Part 1): Choosing a Disconnection

[@MolecularMemory]

Better conditions for the reductive amination reaction at 9:16 would be ammonium acetate as the nitrogen source, and sodium cyanoborohydride for the reduction. These acidic conditions allow for formation of an iminium ion, which will be reduced by the hydride reagent more easily than the starting aldehyde.

[@paraspatel6783]

Nice teaching 👌 love u

[@navjotsingh2251]

This is so awesome. I hope you continue to do synthesis videos, maybe go into more advanced/interesting synthesis!!! Thanks for helping us understand and grow together, when I get a job I will certainly donate to your channel!

[@mateuszhess4549]

It's great content, very well presented. I'm looking forward to learning more from you!

[@makhosazanedlamini1249]

Wow! I wish I came across this video before writing my main exam- I would have passed my exam and now I'm just studying for my supp exam. I think I will pass :) God led me to you! You are amazing at this!

[@fahimahmad9428]

Great way to explain.. It really makes sense.

[@skiphoffenflaven8004]

Great stuff!

[@ClaireSamuelsVA]

You’re a great teacher!! This actually kind of makes sense after weeks of it baffling me.

[@sujeevankrish9979]

I also had the same...coz I have to submit a tutorial this week

[@ProfessorWaltherKotz]

Awesome video!

[@sumayyaaslamy3792]

Thank you for this video.

[@thecasualfront7432]

This channel is awesome. Keep going, channels take a while to get off the ground, this will get popular if you keep active

[@homosapiensqp3225]

Great, soothing voice and what's even better the topic of the vid. is one of my favorites. BTW. Why in the last example Hofmann rearrangement wasn't mentioned?

[@猛男-s7f]

Thank you very much, your video is very useful

[@ahmedsaadkamalmohamed9154]

Very good Thank you

[@maw9406]

wow! Thank you so much :D

[@richardramoscastellanos395]

Did you consider the possibility of a double substitution? I think it would be better to use ethylene oxide instead of dibromoethane, that will add an extra step to the synthesis but we are avoiding the double substitution.

[@MolecularMemory]

Thanks for watching! Two group disconnections are almost always superior, so I certainly like your method. I don't get to that type of two group disconnection until the second video of this series-this one is a warm up to get folks thinking about one-group disconnections. This example is from a patent for the synthesis of ICI D7114, a beta-adrenoceptor agonist, so I guess it didn't work TOO badly for them! Though, they didn't report a yield, so I'm not sure how much of the disubstituted product actually forms. Thanks again! :)

[@roblangford4593]

You could also use ethene and Br2 with a halide abstracting salt (e.g. Ag+) leaving a bromonium ion that can be attacked by the phenoxide

[@mpilow9463]

thanks, please do post more videos with some tricks since we will be writing exams so soon

[@ItzGrundy]

So to clarify the synthons don't represent the mechanism but rather a way of thinking about what "should" occur?

[@MolecularMemory]

This is exactly correct. Synthons allow us to think about what partner would be better as the nucleophile (-) and the electrophile (+). In my first example, I show a primary carbocation for the synthon-this is not a likely mechanistic intermediate! But, if it were a tertiary carbocation, this may very well be an intermediate of an Sn1 reaction. Thanks for the question, and thank you for watching!

[@aka5617]

You are good

[@MolecularMemory]

Thanks for watching!

[@tesfayeayisa1253]

thanks

[@harendra97]

Good lacture

[@yogeshkumar9976]

Mam u r good

[@fatimabatool5079]

Thanku so much Sir

[@natalijasantic8634]

Can someone explain me how she just FGI the amino group into aldehyde?

[@pffftnames9047]

The nitrogen lone pair attacks the carbonyl carbon, forming R-CH(NH3+)-OH intermediate. The oxygen leaves as H2O through proton transfer from -NH3+ and you get R-CH-(NH2-). Then the (NH2-) group donates its lone pair to form the imine. Finally it is reduced to amine by use of NaBH4.

[@sultanwardag]

This was hilarious,

[@surajborudkar632]

Thanks mam

[@llewelynmoriscorvinus6514]

Couldn't the first reaction (the Williamson Ether synthesis) have the same Problem of the dibromoethane reacting twice? Not sure at all though, that's why I'm asking. :)

[@MolecularMemory]

It technically could, but mono-alkylations with dibromoethane can occur in good yields if the reaction conditions are carefully chosen. I wanted to give you an example, so I did a quick search and found this amine alkylation that proceeds in 78% yield with dibromoethane! In this case, there's no need to add steps-that might actually lower your yield overall! :) www.beilstein-journals.org/bjoc/content/pdf/1860-5397-15-77.pdf

[@koushiknayak136]

Your speech is too easy to recognize...... mam there is a small question ..in the analysis how much reactions must be memorized?

[@MolecularMemory]

To propose a disconnection, you have to have a reaction in mind. The more reactions you're familiar with, the better. I hope this sort of answers your question. One group disconnections often involve Sn1/Sn2 reactions, so that is a good start to approaching these problems. It's also pretty easy to think about cleaving in the middle of an ester or amide, so knowing the reactions of carboxylic acid derivatives is helpful as well.

[@heenasoni1300]

Nic