5c Flashcards
when u have an e- poor aromatic and a nucleophile,, where can the nucleophile attack
it can attack the C with a halogen bonded to it.
bc that would make the C slightly positive and electrophilic.
the double bond would then be resonated onto an EWG or a C. but an EWG is preferred as it can stabilise the anionic charge.
how can we remove a ketone,, remmeber that turning it into an OH wont work bc OH is a bad lg
add POCl3
the N lone pair will conjugate and use the ketone double bond to attack the P of POCl3 and form a O-. the O- is then used to kick of a Cl off the POCl3
the Cl is used to attack the C the ketone was on and it moves the double bond and neutralises the N.
the N: then kicks off the POCl2 and u have an aromatic with a Cl bonded to the C where the ketone was.
this occurs via the Vilsmeier Haack reaction.
under what conditions do we get a H- leaving group
when we have a strong, nucleophile, an e- poor aromatic and a high temperature.
what strong nucs can be used to kick off a H- leaving group when u have an e- poor aromatic and heat
u can use NaNH2
or PhLi.
when u have a e- poor aromatic, a strong nuc like PhLi or NaNH2 and ur removing a H- as a LG. describe the reaction that occurs
the nuc attacks the molecule and is added to it.
the N // EWG now has an anionic charge.
this charge is used to kick off the H- that is bonded to the C the nuc is bonded to.
the H is substituted for a nucleophile.
for PhLi, Ph is substituted and Li isnt
for NaNH2,, NH2 is sbstituted and Na is not.
what is a C metallated heterocycle
where a C is bonded to a metal.
metal will be cationic and C will be anionic.
if the C metallated heterocycle has an anionic charge on a C. what can we use it as
it can be used as a nucleophile.
it can attack partial positively charged things with its negative charge.
what happens when a C metallated heterocycle attacks an amide
the anionic C attacks the partially positive C of the amide ad it forms an O-.
this O- is used to kick of the amine part of the amide.
leaving behind an aldehyde on the heterocycle.
what happens when the C metallated heterocycle attacks an RSSR
the anionic charge attacks an S,, the SS bond is then moved onto the S of the SR.
this gives a heterocycle with SR on it, and SR with an anionic charge floating around.
how do we prepare a C metallated heterocycle
we need to perform a metal halide exchange or use a gringard reaget (Mg)
describe the preparation of a C metallated heterocycle using a metal halide or gringard reagent
its fast and convenient
describe a metal halide
smt like an alkyl lithium reagent or an alkyl MgBr
and then mix it wih an aromatic halide bc its more nucleophilic.
okay so lets make alkyl lithium using pyridine with a Br attached in the 3 position
u add BuLi and Et2O
and a temp of -78
this gived pyridine with an Li where the Br was
u can then add a CN Ph which forms an immine on the position where the Br and Li were.
hydrolysis will turn the imine into a carbonyl with the same substituent bonded to it (aka the subs of the imine will stay on it and form a carbonyl with the same substituent)
okay so now we have a pyridine with a Cl and I on it,, which halogen will then gringard reagent attack and substitute
the one with the weaker bond.
I will be substituted and not Cl,, bc Cl has a stronger bond bc greater orbital overlap bc the orbitals are less diffuse.
so u have MgBr where the halogen was.
the C- MgBr bond can then be used to attack an electrophile // double bond and this will be added to the C where the gringard was,, removing the gringard.
okay so what is directed ortho metalation
where a metal is added in the ortho position
how does directed ortho metalation work
u have a directive metalatting group on one of the carbons.
this makes the metals be added to the ortho position of itself.
what is substituted off when we have a directing orther metallation reaction + how does this occur
the hydride is substituted for a metal.
the DMG bonds to the metal.
the orther C bonds to the metal whilst its bonded to the DMG
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what counts as a DMG
anything with a lone pair tbhhhh
what normally happens to pyridine: is it normally alkylated or deprotonated
pyridine is normally alkylated.
aka a R group is added to it.
this is due to the N or pyrifine being bonded to like a Cl or smt,, which is also bonded to the R group. then the C ortho to the N of the pyridine attacks the other part of the group, ana the R group the Cl is attached to.
and an R group is added to the ortho C,, alkylating the moelcule instead of deprotonating it.
pyrrole and BuLi,, what happens
then what happens when u add a cO2 to this
then what happens when u add BuLi to this again
then add an electrophile to this
then the addition of water.
u have pyrrole with an Li bonded to the N.
u form a carboxylate on the N
u have the carboxylate on the N.. and an Li ortho to the N.
the Li is substituted by the electrophile so now theres an electrophile ortho to the N and not an Li.
adding H20 will then make N be bonded to H,, and not the carboxylate
miscellaneous reactions
1. pyrrole + ROCl + NaH gives what and what is the mechanism
pyrrole
- take away the H on the N (most acidic H)
- (-) on the N then attacks the C of the RCOCl carbonyl
- this gets moved up onto the ocygen
- the oxygen anion then uses its negative charge to kick of the Cl.
- leaves u with the N bonded to a carbonyl with an R group attached to it.
whats special about pyrrole
its the most e- rich heterocycle.
which carbons on pyrrole are e- rich,, and what structures can be made bc of this
the carbons alpha to the N are e- rich
meaning H’s can bond here
and other pyrrols can also bond here to form polymers
when smt is in a solution made up of MeOH,, can H’s be removed from it
no
H’s cannot be released when smt is in a solution of NeOH.
pyrrole but with an O rather than an N + Br + solvents ,, whats special about the solvents and what product is formed
the solvents must be non nucleophilic.
Br will be added to every Carbon
pyrrol but with an O instead of an N. + Br2 + MeOH
OMe added to the e- rich carbons,, aka the ones alpha to the ‘O’ that is normally an ‘N’ on pyrrole.
Sn1 type reaction
OMe attacks the double bonds and these neutralise the O+
pyrrole with an ‘O’ where the “N” normally is and with R groups bonded to the e- rich carbons + H+
it breaks the pyrrole and u have a ketone on two sides of the chain type thing.
