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carboxylic acid derivatives
acid halide
anhydride
ester
amide
how does LG affect reactivity
acid halide > anhydride > ester > amide
as list progresses:
LG becomes less e- neg
LG becomes a less likely LG
LG becomes less acidic
resonance increases, more stable, unlikely to react
LGs and their notes
-X —> fast, even at room temp
-OCOR —> slow at RT but still reacts with nucleophile
-OR —> very slow, use catalyst or heat
-NR2 —> not observed, unles use catalyst, heat, time
resonance with acid derivatives
resonance with acid halide: (+) on X, (-) on O
minor resonance product, because (+) charge on a highly e- neg atom
resonance with amide: (+) on NR2, (-) on O
important contributor because (+) is stable enough.
C=N double bond is shorter than C=X bond in above resonance.
NMR shows 2 isomer outcomes due to barrier rotation
where on carboxylic acid is acidic and basic
alpha protons are acidic and will react with base
O is basic
acidity trend in acid derivatives
acid halide > ketone > ester > amide
acid halide most acidic bc inducitive effect of very e- neg X. X is not a good resonane donator, so little resonance stabilizer bc no N to donate e- pair.
has the least stable CB —> more acidic molecule
amide is the most basic because it has the best resonance forms as N donates e- pair. the alpha H becomes less acidic and less likely to deprotonate.
the negative charge on the O shows resonance delocalization and electron density pulled from N to O
better electron donating abilities makes the molecule more stable and less acidic —> most basic (the amide)
how do alcyl halides react
via addition elimination
are very reactive due to dipole facing O —> X
acid halide + carboxylic acid —> anhydride
acid halide + H2O —> carboxylic acid
acid halide + ROH —> ester
acid halide + H-NR2 —> amide
mech for addition-elim of alycl halides
O from ROH adds to carbbonl carbon as nucleophile, arrow from bond to O
O makes double bond, Cl LG
Cl- base deprotonates H from ROH (use pyridine or amine if doing amine)
same mech for all reactions
carboxylic acid + SOCl2 + ROH,pyridine (bulky base)
—> ester
need to use this method to make t-butyl esters
alycl halide —> ketone
R-CuLi
cant use grignard, because will react twice after product is formed whereas cuprates dont react with ketones
reduction of alcyl halides
alcyl halide —> ROH
1) LiAlH4 2) H+ workup- because very strong
alcyl halide —> aldehyde
1) LiAl(OtBu)3H 2) H+ workup
more bulky and less reactive- will react with reactive alycl halide, but not strong enough to react with less reactive aldehyde
anhydride reactions
anhydride + H2O —> acid
anhydride + ROH —> ester
anhydride + RNH —> amide
symmetrical anhydrides reactions
symmetrical anhydrides will give 2 products with reaction, because will react at both carbonyl carbons
esters properties
slow reactions, catalyst helps, OR is a poor LG
ester hydrolysis to carboxylic acid
ester + TsOH,H2O,acetone,heat —> carboxylic acid + EtOH
padped reverse reaction of ester formation
ester under basic conditions- saponification: ester + -OH,H2O,heat —>
—> carboxylate ion + EtOH
mech
-OH adds to carbonyl C, arrow to O
O to double bond, OR is LG
-OR attacks H from O-H, O gets negative charge
CH3OH byproduct
ester + 1)-OH,H2O,heat 2) H+ workup
—> carboxylic acid
needs acid workup to make this product, without it will get carboxylate ion
ester —> carboxylic acid
carboxylic acid —> ester
ester —> carboxylic acid
add H2O
carboxylic acid —> ester
add ROH
transesterification
one ester to another ester
OMe ester + EtOH,TsOH,heat —> OEt ester
converting an ester to an amide
ester + amine,heat —> amide
ester + 1) grignard 2) H+ workup
grignards add twice, so will get OH and 2 of the R groups in product
ester + 1) LiAlH4, Et2O 2) H3O+, H2O
replaces OR with OH, removes carbonyl
very strong reducing agent
ester + 1) DIBAL, -78°C 2) H3O+, H2O
makes aldehyde
replaces OR with H, keeps carbonyl
DIBAL is bulky version of LiAlH4 (Al with 2 isobutyls and an H)
is less reactive than LiAlH4 because of sterics
ester + 1) LDA, 0° 2) R-Br
adds R-Br to alpha carbon
via intermediate enolate
problem: side reaction-claisen condensation
amide reaction conditions
are the least reactive derivative so reactions will require heat
amide hydrolysis: acid vs basic
acidic: amide + H2SO4,H2O,heat —> carboxylic acid + (+)amide group with 2 extra Hs added
basic: amide + NaOH,H2O,heat —> carboxylate ion(-) + amide group with 1 extra H
amide + 1) -OH,H2O,heat 2) H3O+,H2O
carboxylic acid
even tho step 1 is base, to end with acid product, last step must be acidic
mech
PADPED like ester hydrolysis
amide reduction to ROH
amide + 1) LiAlH4, THF 2) H3O+,H2O 3) -OH,H2O
same amide molecule but carbonyl O replaced with 2 Hs
N bonds don’t break as easily
use basic workup at end to end with basic amine product
mech
from ws
amide + 1) DIBAL 2) H3O+,H2O
makes aldehyde- replaces amide group with H
because of DIBAL sterics, so less reactive
hoffmann rearrangement
amide + Br2,NaOH,H2O
C=O gets removed and replaced with the amide group
loses a carbon
CO2 byproduct
mech
from ws
amidate
amide group with one less H- so has negative charge
nitrile hydrolysis
acidic: R-CN + H2SO4,H2O,heat —> R-COOH (carboxylic acid)
basic: R-CN + 1) -OH,H2O,heat 2) H3O+, H2O —> R-COOH (cabroxylic acid)
acidic mech: PADPED
first step protonate N
basic mech: start with addition and get carboxylate ion product unless adding H+ workup
when to do PADPED vs addition-elimination
when you have a bad nucleophile like H2O, you are under acidic conditions and need to protonate- so do PADPED
when you have a good nucleophile (like -OH) you dont need to protonate so can do addition-elimination
nitrile + organometallic
R-CN + CH3Li + 1) THF, 2) H+,H2O —> remove N, becomes ketone with R group on other side
3 methods of nitrile reduction
R-CN + 1) DIBAL 2) H3O+,H2O —> aldehyde
R-CN + 1) LiAlH4,THF 2) H3O+,H2O —> replace N with NH2
R-CN + H2,cat Pt —> replace N with NH2