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I wonder if the azide could compete with hydroxide in the last step, forming an acyl azide as a side product. I guess it will depend on the amount of NaOH used. As for the next monday's reaction, is an haloform reaction: the alcaline media will favor the enolate, which attacks the iodine and formas an alpha-iodocarbonyl and NaI; this will repeat 3 times, forming three C-I bonds. Then, the hydroxide will attack the carbonyl position and, in an addition-elimination step, the CI3- group will be kicked out as an anion (stabilized by 3 halogens), forming the carboxilic acid (which will be deprotonated by more OH) and iodoform after workup
You got both! You’re right about the concentration, in this case the solvent is aqueous NaOH, which would mean it’s in incredible excess. Crushed the next one too!
As with most things, there’s a subtle mix of considerations regarding steric and electronic effects. You hit precisely on the electronic side of things, but the steric encumbrance on the carbon with the chloro groups makes this transformation more likely!
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Most underrated channel on chemistry
Haha. I appreciate that! I’m trying!
these videos are so much fun to watch
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This is a nice concept of Monday mechanism, it's now been a part of my every Monday😊
Haha. Thank you so much for the kind words! I’ll see you next Monday!
I wonder if the azide could compete with hydroxide in the last step, forming an acyl azide as a side product. I guess it will depend on the amount of NaOH used.
As for the next monday's reaction, is an haloform reaction: the alcaline media will favor the enolate, which attacks the iodine and formas an alpha-iodocarbonyl and NaI; this will repeat 3 times, forming three C-I bonds. Then, the hydroxide will attack the carbonyl position and, in an addition-elimination step, the CI3- group will be kicked out as an anion (stabilized by 3 halogens), forming the carboxilic acid (which will be deprotonated by more OH) and iodoform after workup
You got both! You’re right about the concentration, in this case the solvent is aqueous NaOH, which would mean it’s in incredible excess.
Crushed the next one too!
❤❤
Why the azide attacks the secondary carbon? The other one is more δ+
As with most things, there’s a subtle mix of considerations regarding steric and electronic effects. You hit precisely on the electronic side of things, but the steric encumbrance on the carbon with the chloro groups makes this transformation more likely!
@@rojaslab if we use a non-nucleophilic base like NaH, can we get an acyl azide with an azide group on the alpha Carbon?
Which is product itself.@@francisnaveen9041
@@francisnaveen9041 I think you may be on to something actually!
haloform
Now that’s a fancy word! I wonder how many have heard of it!
the way you pronounce azide is painful dude but you worked with buchwald so you get a pass.
🤣 funny because when people say “Ay-Zyde” it makes me cringe.