Buscar

14.SNA QUI02015

Esta é uma pré-visualização de arquivo. Entre para ver o arquivo original

01/10/14 
1 
1 
 
Prof. Dr. José Eduardo Damas Martins 
Reações de derivados 
de ácidos carboxílicos 
2 
 
Prof. Dr. José Eduardo Damas Martins 
SUBSTITUIÇÃO NUCLEOFÍLICA 
 
NO CARBONO ACÍLICO 
 
01/10/14 
2 
3 
 
Prof. Dr. José Eduardo Damas Martins 
É a principal reação realizada por 
derivados de ácidos carboxílicos. 
4 
 
Prof. Dr. José Eduardo Damas Martins 
Cloreto de ácido 
Anidrido 
Éster 
Amidas 
Derivados de ácido 
carboxílico 
01/10/14 
3 
5 
 
Prof. Dr. José Eduardo Damas Martins 
Adição nucleofílica 
Instável 
Relembremos a reação de formação de Hemi-acetais 
O
HO OR
OH
ROH
Hemiacetal
6 
 
Prof. Dr. José Eduardo Damas Martins 
A ins tab i l idade é dev ido à 
facilidade com que o grupo RO─ é 
expulso da molécula 
Intermediário 
 instável 
Grupo de 
saída 
01/10/14 
4 
7 
 
Prof. Dr. José Eduardo Damas Martins 
Grupos que podem ser expulsos de 
uma molécula, levando consigo a 
carga negativa, são chamados de 
“Grupos de partida ou de saída” 
Cl Br IRO RCO2RS, , , , ,
8 
 
Prof. Dr. José Eduardo Damas Martins 
Se o nucleófilo é também um grupo de 
saída, há a chance de, após a adição, 
ele ser eliminado e portanto a reação 
se torna “rreversível” 
O
HO OR
OH
ROH
Hemiacetal
01/10/14 
5 
9 
 
Prof. Dr. José Eduardo Damas Martins 
O mesmo pode acontecer caso o 
material de partida contenha um 
potencial grupo de saída 
Instável 
Cetona 
continua 
reagindo 
Substituição nucleofílica acílica 
10 
 
Prof. Dr. José Eduardo Damas Martins 
Substituição nucleofílica acílica 
R X
O
+ Nu
R Nu
O
+ X
Derivado 
de ác. 
carboxílico 
01/10/14 
6 
11 
 
Prof. Dr. José Eduardo Damas Martins 
Mecanismo 
R X
O
Nu
R Nu
O
+ X
X
O Nu
Grupo de
 saída
Intermediário 
tetraédrico 
instável 
12 
 
Prof. Dr. José Eduardo Damas Martins 
Hidrólise 
R OR
O
R O
O
+ RO
OR
O OH
Grupo de
 saída
OH H2O H
R O
O
Na
NaOH
+ ROH
Sal de ácido 
carboxílico 
01/10/14 
7 
13 
 
Prof. Dr. José Eduardo Damas Martins 
Intermediário 
tetraédrico 
instável 
R Cl
O
R Cl
O OCH3
HOCH3
Py
H Py
R Cl
O OCH3
Observe a reação 
14 
 
Prof. Dr. José Eduardo Damas Martins 
Possibilidades: 
R Cl
O OCH3
R OCH3
O
+ Cl
1)
R Cl
O OCH3
R Cl
O
2) + OCH3
01/10/14 
8 
15 
 
Prof. Dr. José Eduardo Damas Martins 
Cl─ ─OCH3 
Cloreto Metóxi 
16 
 
Prof. Dr. José Eduardo Damas Martins 
O ânion mais estável tem preferência 
a sair 
01/10/14 
9 
17 
 
Prof. Dr. José Eduardo Damas Martins 
“Geralmente”o melhor grupo de 
saída é aquele que for a base mais 
estável, ou seja, a base mais fraca 
R Cl
O OCH3
R OCH3
O
+ Cl
18 
 
Prof. Dr. José Eduardo Damas Martins 
Como determinar quem é a base 
mais fraca? 
01/10/14 
10 
19 
 
Prof. Dr. José Eduardo Damas Martins 
Para isso, pode-se utilizar os 
valores de pKaH 
20 
 
Prof. Dr. José Eduardo Damas Martins 
Ácido forte = Base conjugada fraca 
Ácido fraco = Base conjugada forte 
HA + H2O H3O + A
ácido base base conjugada
ácido
 conjugado
01/10/14 
11 
21 
 
Prof. Dr. José Eduardo Damas Martins 
pka = -log Ka 
Quanto menor o pka, maior é a 
acidez 
[H3O ] [A]Ka =
[HA]
22 
 
Prof. Dr. José Eduardo Damas Martins 
pKaH é o pKa da espécie protonada (ácido conjugado) 
pKaH de X = pKa HX
pKaH de Cl = pKa HCl = -7
Exemplo: 
01/10/14 
12 
23 
 
Prof. Dr. José Eduardo Damas Martins 
R Cl
O OCH3
R OCH3
O
+ Cl
Cl HCl
CH3O CH3OH
Ac. conjugado
Ac. conjugado
pKaH = pKa = 7−
pKaH = pKa = 15CH3OHCH3O-
Cl HCl-
24 
 
Prof. Dr. José Eduardo Damas Martins 
O HCl é um ácido mais forte que 
o metanol, portanto sua base 
conjugada, Cl− , é mais fraca. 
01/10/14 
13 
25 
 
Prof. Dr. José Eduardo Damas Martins 
portanto o Cl− , é melhor grupo 
de saída que o CH3O− . 
26 
 
Prof. Dr. José Eduardo Damas Martins 
Geralmente, quanto menor o 
pkaH, melhor o grupo de saída 
01/10/14 
14 
27 
 
Prof. Dr. José Eduardo Damas Martins 
Reatividade de derivados 
de ácido carboxílico 
28 
 
Prof. Dr. José Eduardo Damas Martins 
A reatividade depende da estrutura 
do derivado de ácido carboxílico 
R X
O
01/10/14 
15 
29 
 
Prof. Dr. José Eduardo Damas Martins 
Mais reativo 
Menos reativo 
30 
 
Prof. Dr. José Eduardo Damas Martins 
Por que as amidas são os 
derivados de ácidos carboxílicos 
menos reativos ? 
R NH2
O
01/10/14 
16 
31 
 
Prof. Dr. José Eduardo Damas Martins 
A estabilização por ressonância 
explica a baixa reatividade 
Carbonila de amidas é menos eletrofílica 
32 
 
Prof. Dr. José Eduardo Damas Martins 
the hierarchy. We’ve seen that this hierarchy is partly due to how good the leaving group is (the ones
at the top are best), and partly due to how good the nucleophile needed to make the derivative is (the
ones at the bottom are best).
most reactive
acid chlorides
(acyl chlorides)
acid anhydrides
esters
amides least reactive
Delocalization and the electrophilicity of carbonyl compounds
All of these derivatives will react with water to form carboxylic acids, but at very different rates.
Hydrolysing an amide requires boiling in 10% NaOH or heating overnight in a sealed tube with
concentrated HCl. Amides are the least reactive towards nucleophiles because they exhibit the great-
est degree of delocalization. You met this concept in Chapter 7 and we shall return to it many times
more. In an amide, the lone pair on the nitrogen atom can be stabilized by overlap with the π* orbital
of the carbonyl group—this overlap is best when the lone pair occupies a p orbital (in an amine, it
would occupy an sp3 orbital).
The molecular orbital diagram shows how this interaction both lowers the energy of the bonding
orbital (the delocalized nitrogen lone pair), making it neither basic nor nucleophilic, and raises the
energy of the π* orbital, making it less ready to react with nucleophiles. Esters are similar but,
because the oxygen lone pairs are lower in energy, the effect is less pronounced.
The greater the degree of delocalization, the weaker the C=O bond becomes. This is most clearly
R NH2
O
R OR1
O
R O
O
R1
O
R Cl
O
Not all carboxylic acid derivatives are equally reactrive 287
R
O
Cl R
O
OH
R
O
O R
O
OH
O
H2O
fast at 20 ˚C
H2O
slow at 20 ˚C
R
O
OEt
R
O
NH2
R
O
OH
R
O
OH
H2O only on heating with 
acid or base catalyst
H2O only on prolonged heating 
with strong acid 
or base catalyst
H N C O
R
H
H
N O
H
R
H
N O
H
R molecular orbital diagram shows how energy of 
orbitals changes as lone pair and C=O π* interact
isolated
lone pair
 on N
isolated C=O 
π* orbital
new, stabilized lower-energy lone pair
orbitals overlap
empty π* 
orbital
lone pair in
 p orbital
new higher-energy π* orbital
allow orbitals to 
interact
Velocidades de hidrólise de derivados de ácido carboxílico 
Rápida a 20 0C 
the hierarchy. We’ve seen that this hierarchy is partly due to how good the leaving group is (the ones
at the top are best), and partly due to how good the nucleophile needed to make the derivative is (the
ones at the bottom are best).
most reactive
acid chlorides
(acyl chlorides)
acid anhydrides
esters
amides least reactive
Delocalization
and the electrophilicity of carbonyl compounds
All of these derivatives will react with water to form carboxylic acids, but at very different rates.
Hydrolysing an amide requires boiling in 10% NaOH or heating overnight in a sealed tube with
concentrated HCl. Amides are the least reactive towards nucleophiles because they exhibit the great-
est degree of delocalization. You met this concept in Chapter 7 and we shall return to it many times
more. In an amide, the lone pair on the nitrogen atom can be stabilized by overlap with the π* orbital
of the carbonyl group—this overlap is best when the lone pair occupies a p orbital (in an amine, it
would occupy an sp3 orbital).
The molecular orbital diagram shows how this interaction both lowers the energy of the bonding
orbital (the delocalized nitrogen lone pair), making it neither basic nor nucleophilic, and raises the
energy of the π* orbital, making it less ready to react with nucleophiles. Esters are similar but,
because the oxygen lone pairs are lower in energy, the effect is less pronounced.
The greater the degree of delocalization, the weaker the C=O bond becomes. This is most clearly
R NH2
O
R OR1
O
R O
O
R1
O
R Cl
O
Not all carboxylic acid derivatives are equally reactrive 287
R
O
Cl R
O
OH
R
O
O R
O
OH
O
H2O
fast at 20 ˚C
H2O
slow at 20 ˚C
R
O
OEt
R
O
NH2
R
O
OH
R
O
OH
H2O only on heating with 
acid or base catalyst
H2O only on prolonged heating 
with strong acid 
or base catalyst
H N C O
R
H
H
N O
H
R
H
N O
H
R molecular orbital diagram shows how energy of 
orbitals changes as lone pair and C=O π* interact
isolated
lone pair
 on N
isolated C=O 
π* orbital
new, stabilized lower-energy lone pair
orbitals overlap
empty π* 
orbital
lone pair in
 p orbital
new higher-energy π* orbital
allow orbitals to 
interact
Lenta a 20 0C 
 
Cloreto de 
ácido 
Anidrido 
01/10/14 
17 
33 
 
Prof. Dr. José Eduardo Damas Martins 
the hierarchy. We’ve seen that this hierarchy is partly due to how good the leaving group is (the ones
at the top are best), and partly due to how good the nucleophile needed to make the derivative is (the
ones at the bottom are best).
most reactive
acid chlorides
(acyl chlorides)
acid anhydrides
esters
amides least reactive
Delocalization and the electrophilicity of carbonyl compounds
All of these derivatives will react with water to form carboxylic acids, but at very different rates.
Hydrolysing an amide requires boiling in 10% NaOH or heating overnight in a sealed tube with
concentrated HCl. Amides are the least reactive towards nucleophiles because they exhibit the great-
est degree of delocalization. You met this concept in Chapter 7 and we shall return to it many times
more. In an amide, the lone pair on the nitrogen atom can be stabilized by overlap with the π* orbital
of the carbonyl group—this overlap is best when the lone pair occupies a p orbital (in an amine, it
would occupy an sp3 orbital).
The molecular orbital diagram shows how this interaction both lowers the energy of the bonding
orbital (the delocalized nitrogen lone pair), making it neither basic nor nucleophilic, and raises the
energy of the π* orbital, making it less ready to react with nucleophiles. Esters are similar but,
because the oxygen lone pairs are lower in energy, the effect is less pronounced.
The greater the degree of delocalization, the weaker the C=O bond becomes. This is most clearly
R NH2
O
R OR1
O
R O
O
R1
O
R Cl
O
Not all carboxylic acid derivatives are equally reactrive 287
R
O
Cl R
O
OH
R
O
O R
O
OH
O
H2O
fast at 20 ˚C
H2O
slow at 20 ˚C
R
O
OEt
R
O
NH2
R
O
OH
R
O
OH
H2O only on heating with 
acid or base catalyst
H2O only on prolonged heating 
with strong acid 
or base catalyst
H N C O
R
H
H
N O
H
R
H
N O
H
R molecular orbital diagram shows how energy of 
orbitals changes as lone pair and C=O π* interact
isolated
lone pair
 on N
isolated C=O 
π* orbital
new, stabilized lower-energy lone pair
orbitals overlap
empty π* 
orbital
lone pair in
 p orbital
new higher-energy π* orbital
allow orbitals to 
interact
Só ocorre com 
aquecimento e catálise 
 ácida ou básica 
the hierarchy. We’ve seen that this hierarchy is partly due to how good the leaving group is (the ones
at the top are best), and partly due to how good the nucleophile needed to make the derivative is (the
ones at the bottom are best).
most reactive
acid chlorides
(acyl chlorides)
acid anhydrides
esters
amides least reactive
Delocalization and the electrophilicity of carbonyl compounds
All of these derivatives will react with water to form carboxylic acids, but at very different rates.
Hydrolysing an amide requires boiling in 10% NaOH or heating overnight in a sealed tube with
concentrated HCl. Amides are the least reactive towards nucleophiles because they exhibit the great-
est degree of delocalization. You met this concept in Chapter 7 and we shall return to it many times
more. In an amide, the lone pair on the nitrogen atom can be stabilized by overlap with the π* orbital
of the carbonyl group—this overlap is best when the lone pair occupies a p orbital (in an amine, it
would occupy an sp3 orbital).
The molecular orbital diagram shows how this interaction both lowers the energy of the bonding
orbital (the delocalized nitrogen lone pair), making it neither basic nor nucleophilic, and raises the
energy of the π* orbital, making it less ready to react with nucleophiles. Esters are similar but,
because the oxygen lone pairs are lower in energy, the effect is less pronounced.
The greater the degree of delocalization, the weaker the C=O bond becomes. This is most clearly
R NH2
O
R OR1
O
R O
O
R1
O
R Cl
O
Not all carboxylic acid derivatives are equally reactrive 287
R
O
Cl R
O
OH
R
O
O R
O
OH
O
H2O
fast at 20 ˚C
H2O
slow at 20 ˚C
R
O
OEt
R
O
NH2
R
O
OH
R
O
OH
H2O only on heating with 
acid or base catalyst
H2O only on prolonged heating 
with strong acid 
or base catalyst
H N C O
R
H
H
N O
H
R
H
N O
H
R molecular orbital diagram shows how energy of 
orbitals changes as lone pair and C=O π* interact
isolated
lone pair
 on N
isolated C=O 
π* orbital
new, stabilized lower-energy lone pair
orbitals overlap
empty π* 
orbital
lone pair in
 p orbital
new higher-energy π* orbital
allow orbitals to 
interact
Só ocorre com aquecimento 
prolongado e catálise com 
 ácidos ou bases fortes 
Éster 
Amida 
34 
 
Prof. Dr. José Eduardo Damas Martins 
1 – X deve ser melhor grupo de saída que Nu 
 
2 – Nu deve ser um nucleófilo forte o suficiente para 
realizar a adição à carbonila 
 
3 - deve ser um eletrófilo suficientemente 
forte para reagir com Nu 
Condições para que a reação ocorra 
R X
O
Nu
R Nu
O
+ X
R X
O
01/10/14 
18 
35 
 
Prof. Dr. José Eduardo Damas Martins 
Esterificação 
de Fisher 
36 
 
Prof. Dr. José Eduardo Damas Martins 
Esterificação de Fischer 
01/10/14 
19 
37 
 
Prof. Dr. José Eduardo Damas Martins 
Mecanismo 
38 
 
Prof. Dr. José Eduardo Damas Martins 
Mecanismo ? 
01/10/14 
20 
39 
 
Prof. Dr. José Eduardo Damas Martins 
Mecanismo ? 
40 
 
Prof. Dr. José Eduardo Damas Martins 
Devido à formação de água 
durante a reação, a mesma 
d e v e s e r r e m o v i d a 
utilizando-se um Dean-stark 
ou qualquer outro agente 
secante para que o equilíbrio
seja deslocado no sentido 
de formação do éster. 
Dean-Stark 
01/10/14 
21 
41 
 
Prof. Dr. José Eduardo Damas Martins

Teste o Premium para desbloquear

Aproveite todos os benefícios por 3 dias sem pagar! 😉
Já tem cadastro?

Outros materiais