4b Flashcards
describe a baeymer villager reaction
u have a ketone and u use a peracid (think of the peroxy carboxylic acid or mcpba group,, the C=O,O-OH)
the OH lone pair attacks the C and makes the ketone bond into an -o- bond which is protonated to give oh.
then the C that is bonded to the OH,, one of its groups migrate to the per o and kick off the carboxylate group.
to give an ester
ur going from a ketone to an ester
which group migrates in a bae villager reaction
the larger R group,, aka the one that can stabilise the cationic charge better. bc the carboxylate is anionic so the rest will be cationic.
what group usually migrates in general
the one that is antiperi planar to the leaving group!!
aka the bond that is pointing in the same directions as the one that u want to get rid of.
it must be on an adjacent carbon tho!!
why is the bae villager rearangement thermo and kinetrically favourable
kinetics : good orbital overlap: the cc sigma and the o - o antibonding
thermodynamics: the o-o is a weak bond
when drawing a bae villager mechanism,, what do we neeed to think about
having constructive overlap between the o-o- antibonding (bigger lobes facing outside)
and the sigma c-c bond.
makw sure theyre parallel to eachother.
why cant u just do the bae villager thing with an oh
bc u cant kick off a H- bc its a poor lg.
also theres poor overlap between the orbitals
beckman rearrangement explanation
u have a ketone that then turns into a ketone with an N instead of an O and an oh bonded to the =N.
the Oh is then protonated to form water. this is kicked off of the N and a group bonded to the C the N is attacked to migrated to the N.
a water then attacks the carbocation and an amide is formed. the water that attacks the cc+ is the ketone o.
in the beckman rearrangement,, C=N,, what group migrates
the group trtans to the OH will migrate.
bc it needs good overlap with the N-O sigma star
curtis rearrangement start and end product
start : carboxylic accid // acyl chloride (carb acid but with a cl instead of oh)
into an amine (react with oh)
into an ester (react with alcohol)
curtis rearrangemet with OH
carb acis with SOCl2 turns into acyl chloride.
reactt with Na N3,, form an acyl nitrene ( carb acid but with an istead of an oh)
use n lone pair and migrate the r on the c to make O=C=N-R,, then attack the C with an H20 and move the double bond to the N.
u now have O=C- OH - N-R
use the lone pair on the OH to make a double bond and use the bond between the CN to break and put it onto the N,, breaking that bond.
u now have CO2 and and amine (primary amine)
curtis rearrangement with an alcohol
carb acis + SOCl2 makes a acyl chloride.
react with Na N3 to form an acyl nitrene,, use the lone pair on N and migrate the R to the N to form a O=C=N-R
attack the C with an alocol lone pair,, move the double bond to the N so u now have O=C- OR- N-R
u now have an ester // amide type thing going on
in the curtis rea,, the empty p orbital is the
the empty N p orbital
homo and lumo for the curtis rearrangement
the homo of the C-R filled sigma orbital
the lumo is the empty N p orbital
describe a hoffman rearrangement
u have an amide,, (primary amine on the end)
u then react this with NaOH to remove one of the H’s,, the double bond from the carbonyk can then be resonated to form a double bond between the C -N of the amide.
this double bond can then attack Br2,, so a Br is now attacked to the N.
the OH then attacks the other H on the amine and this forms another double bond between the C and N and an o- on the cqarbonyl o.
this is then used to kick off the Br and form an acyl nitrene
from here,, the curtis reaction can occur!!
from amide to amine using OH to deprotonate the amine and Br.
lossen rearrangement
u have a amide with an OH on the N
this reacts with TsCl to form OTs which is a rlly good lg.
u then do a bunch of stuff similar to the hoffman and get the R-N=C=O
goal for curtis,, hoffman and lossen
to form the r-n=c=o group.
