mechs Flashcards
(28 cards)
baeyer villiger reaction,, what is it,, what do u need and what do u have to do: u need carbonyl - o - o - r btwwww
ur going from ketone to an esterrrr!!!
u then have a C with an OH and O O in a row
u migrate one of the R groups from C onto O and u kick off the last O
baeyer villiger: aka it takes a village,, ur going from the C onto the O.
OH . COOR (va) bc in poland u live in a village, the R goes from the C to the O and u kick off the other O. u go from ketone to ester!!
OH. COOR(va)
bayer villiger,, ur going from a ketone to an ester.
and u have to switch the R group on the C to the O,, then kick off the other O bc u have 3 in total. and obvs u need to form a ketone so u need to keep one of the O’s
OH.COOR(VA) U NEED TO SWAP THE R GROUP FROM IT BEING ON THE C TO THE O. aka u going from ur house to zosias house
ur gonna start from a ketone first tho!! so dont get too streesed,, so tbh it should be O.COOR(va)
describe the beckman rearrangement!! youve got this girl u know this
ur gonna have an immine 💗 with an OH on top.
and u want to remove the OH in the form of H2O.
as u do this tho the R group trans to the OH2 is gonna attach to the N
so as ur breaking a bond with N ur also forming one.
then u get like a R-C=N-R bond ,, and the C is +,, so then u use the OH2 to attack it. and then u form an OH. this OH is then used to form a ketone and u protonate the C=N double bond
u get an amide!!!! 💗
when we see an immine and we’re asked to rearrange // when we see a C=N- OH
- protonate the OH
- R group trans to the OH2 gets moved onto N
- OH2 gets kicked off
- u get a CC+
- the OH2 attacks the CC+,, the OH2 is used to form a carbonyl and the C=N is protonate to form an amide bond.
pinacol rearrangement2 leaving groups on adjacent carbons
u protonate and remove one
u form a CC+
and R group atatched to the C of the other C moves to that CC+.
the OH turns into a carbonyl to stabilise the newly formed CC+.
u move the group that will give u the most stable CC+
carboxulic acid + SOCl
turns it into Carbonyl with Cl as a subs
carbonyl Cl subs with NaN3
ur gonna attach the N=N=N onto the carbonyl C and then kick them off the give a nitrene : aka C-N:
once u get a C-N u have to use the N of the lone pair to attack the C and then the R group moves onto the N
so u have O=C=N-R
then OH2 attacks the C and makes C=N into C-N and the C obvs has an OH2 attached to it too
u basically get a carboxylic ester // amide on the other sideeee
making a Nitrene aka carbonyl and N when u have a carbonyl and NH2
u need to deprot the H off NH2 and form an enolate,, then u attack a Br,, then u remove the other H and attack another Br. then u kick off the Brs to give u a nutrene
what if u have a carbonyl with a N and an OH in a row and a TsCl (SO2Cl)
u form a carbonyl with an N and OTs attached to it.
the OH attacks the S and u kick off Cl,, the Cl then attacks the H to deprot OH. then u form an enolate and u kick off the SO2R group
a cyclohexane with a Br attached and a PPH3,
the PPH3 kicks off the Br and u get C- and PPH3+
the carbon then attacks a carbonyl and the O of the carbonyl attacks the P of PPH3 to form a yilide.
u get a little box, then u fold the box to give u two diff products, the O goes to the P to give P=O and the C-P bond breaks to form a CC bond
a stable yilide : one with an ewg near it,, gives u a E alkene!!
unstable yilide: one without an ewg near it,, forms a z alkene
what if u have a hydrazine with a carbonyl
a hydrazine is where u have
Ar - N - NH2,, the NH2 attacks the carbonyl,, the OH is protonated and water is kicked off,, u have C=N but want C=C,, so u move the double bond. then u form C=N again,, break the N-N bond and use the NH to attack the C and kick off the unwanted N.
NO2 + H2 and Pd
form NH2
name a reuing agent
MnO2
when u have basic conditions how do u protonate stuff
u have H-OH
going from an amide to a carbonyl R using POCl3
the N attacks the C and the C=O attacks the P,, Cl is kicked off
the Cl then attacks the molecyle to kick off the POCl
u then add ur R group to neutralise N.
Cl is kicked off and then u add H20 and this is used for form a carbonyl!!!
describe how we go from a alkene to a syn dihydroxylation,, aka OH othat are both wedged
OsO4!! Os with double bonds onto each O!!
double bond attacks O ,, e- from double bond go onto the Os and then the e- from the other Os=O go onto the other side of the alkene.
so now u have both C of the alkene attached to an O which are both attached to the Os
OH2 then comes to attack the Os and kicks one of the alkene O’s off to give u O-,, this is protonated to give OH on the syn side.
another OH2 comes to attack the Os and then kicks the other alkene O off and this is also protonated to give u OH on both sides syn on the alkene.
okay if u have an alkene with OH on both of the alkene C’s how do we break that bond to give us carbonyls on each alkene C
u use IO4
the OH attcks the I which forms O- on the reagent,, the OH of the alkene are then deprot to give an ether type thing.
once both O’s are bonded to the I,, u break the bond between the C’s to give u a carbonyl,, and the I-O bond is also broken to give u a carbonyl,, this gives u a broken alkene with carbonyls on them. 💗
alkene with O3
u have O-O=O
the alkene attacks the end O which conjugates to neutralise the middle O
the O- attacks the other side of the alkene ,, giving u a little ring
this then breaks to give u carbonylsssss!!
alkene + O3 + acid
carboxylic acids
alkene + O3 + base
alcohols on both sides of the alkeneee
homo and lumo on hoffman prearrangement
homo = RC sigma bond
lumo = empty P orbital on N
wait does hoffman end on the carbonyl compound thing
nopeu have to remove CO2 to get an amine by itself
diff between pinacol and semipinacol
semipinacol = sn2 aka 1 step
so from filled sigma to an antibonding orbital
pinacol = sn1 aka 2 steps
so loss of lg to then rearangement
so filled sigma to empty p orbital
homo and lumo for curtius rearrangement
homo = filled Cr
lume = empty p on N
