9.AN a aldeídos e cetonas QUI02015 5

Prévia do material em texto

09/09/14 
1 
211 
 
Prof. Dr. José Eduardo Damas Martins 
Redução de 
aldeídos e cetonas 
212 
 
Prof. Dr. José Eduardo Damas Martins 
R
H2
C OH R C H
O
R C
H
R
OH
R C R
O
Oxidação
Oxidação
Álcool primário
Álcool secundário
Aldeído
Cetona
NOX C = −1 NOX C = +1
NOX C = 0 NOX C = +2
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Prof. Dr. José Eduardo Damas Martins 
R
H2
C OHR C H
O
R C
H
R
OH
R C R
O
Redução
Redução
Álcool primário
Álcool secundário
Aldeído
Cetona
NOX C = −1NOX C = +1
NOX C = 0NOX C = +2
214 
 
Prof. Dr. José Eduardo Damas Martins 
A redução de aldeídos e cetonas é uma das 
formas mais simplês de síntese de álcoois 
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215 
 
Prof. Dr. José Eduardo Damas Martins 
AGENTES REDUTORES 
216 
 
Prof. Dr. José Eduardo Damas Martins 
H2 gasoso 
Hidrogenação 
catalítica 
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217 
 
Prof. Dr. José Eduardo Damas Martins 
R R
O
R R
OH
H2, 2 atm
Catalisador
Aldeído ou Cetona Álcool
218 
 
Prof. Dr. José Eduardo Damas Martins 
O catalisador 
Suporte 
metálico 
H H H H
Metais utilizados: Pt, Rh, Pd, Ni 
R
R
O
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Prof. Dr. José Eduardo Damas Martins 
Geralmente, aquecimento e pressão são 
necessários 
O Catalisador de Ni de raney é um dos mais 
utilizados 
220 
 
Prof. Dr. José Eduardo Damas Martins 
O OH
H2/Ni
40 0C, 2 atm
H2/Ni = Níquel de Raney 
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221 
 
Prof. Dr. José Eduardo Damas Martins 
Desvantagem 
Também reduz duplas 
ligações C=C 
222 
 
Prof. Dr. José Eduardo Damas Martins 
H2/Pd
40 0C, 2 atm
O OH
Ausência da dupla 
ligação C=C 
09/09/14 
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Prof. Dr. José Eduardo Damas Martins 
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ligação C=C 
224 
 
Prof. Dr. José Eduardo Damas Martins 
Redução com hidretos metálicos 
É o método mais comum de 
se reduzir aldeídos e cetonas 
09/09/14 
8 
225 
 
Prof. Dr. José Eduardo Damas Martins 
Um hidreto consiste em um ânion do átomo 
de hidrogênio 
H
226 
 
Prof. Dr. José Eduardo Damas Martins 
No entanto, o hidreto por si só não é um bom 
nucleófilo, pois é muito pequeno, possuindo 
somente um orbital 1s. 
09/09/14 
9 
227 
 
Prof. Dr. José Eduardo Damas Martins 
to the carbonyl carbon atom. As the carbonyl carbon atom changes from sp2 to sp3, its bond angles
change from about 120° to about 109°—in other words, the substituents it carries move closer
together. This reduction in bond angle is not a problem for aldehydes, because one of the sub-
stituents is just a (very small) hydrogen atom, but for ketones, especially ones that carry larger alkyl
groups, this effect can disfavour the addition reaction. Effects that result from the size of substituents
and the repulsion between them are called steric effects, and we call the repulsive force experienced
by large substituents steric hindrance.
The angle of nucleophilic attack on aldehydes and ketones
Having introduced you to the sequence of events that makes up a nucleophilic attack at C=O (inter-
action of HOMO with LUMO, formation of new σ bond, breakage of π bond), we should now tell
you a little more about the direction from which the nucleophile approaches the carbonyl group. Not
only do nucleophiles always attack carbonyl groups at carbon, but they also always approach from a
particular angle. You may at first be surprised by this angle, since nucleophiles attack not from a
direction perpendicular to the plane of the carbonyl group but at about 107° to the C=O bond. This
approach route is known as the Bürgi–Dunitz trajectory after the authors of the elegant crystallo-
graphic methods that revealed it. You can think of the angle of attack as the result of a compromise
between maximum orbital overlap of the HOMO with π* and minimum repulsion of the HOMO by
the electron density in the carbonyl π bond.
Any other portions of the molecule that get in the way of (or, in other words, that cause steric hin-
drance to) the Bürgi–Dunitz trajectory will greatly reduce the rate of addition and this is another rea-
son why aldehydes are more reactive than ketones. The importance of the Bürgi–Dunitz trajectory
will become more evident later—particularly in Chapter 34.
Nucleophilic attack by ‘hydride’on aldehydes and ketones
Nucleophilic attack by the hydride ion, H–, is not a known reaction. This species, which is present in
the salt sodium hydride, NaH, is so small and has such a high charge density that it only ever reacts
as a base. The reason is that its filled 1s orbital is of an ideal size to interact with the hydrogen
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 139
!
Steric hindrance (not hinderance)
is a consequence of repulsion
between the electrons in all the
filled orbitals of the alkyl
substituents.
!The Bürgi–Dunitz angle
Bürgi and Dunitz deduced this
trajectory by examining crystal
structures of compounds containing
both a nucleophilic nitrogen atom and
an electrophilic carbonyl group. They
found that, when the two got close
enough to interact, but were not free to
undergo reaction, the nitrogen atom
always lay on or near the 107°
trajectory decribed here. Theoretical
calculations later gave the same 107°
value for the optimum angle of attack.
C O
Nu
C O
Nu
C O
Numaximum overlap with π* 
perpendicular to C=O bond
repulsion from filledπ orbital 
forces nucleophile to attack at 
obtuse angle
π*
π
combined effect: 107°
nucleophile attacks 
C=O at 107° angle
!
Although we now know precisely
from which direction the
nucleophile attacks the C=O
group, this is not always easy to
represent when we draw curly
arrows. As long as you bear the
Bürgi–Dunitz trajectory in mind,
you are quite at liberty to write any
of the variants shown here,
among others.
HR
O
HR
O
H
R
O
Nu
Nu
Nu
never happens
O
Me Me
H
Me Me
H O
H X
H2 XHX
H– always reacts as a basenucleophilic attack by H–
+
Não ocorre 
desta forma 
228 
 
Prof. Dr. José Eduardo Damas Martins 
O Hidreto atua melhor como base forte 
to the carbonyl carbon atom. As the carbonyl carbon atom changes from sp2 to sp3, its bond angles
change from about 120° to about 109°—in other words, the substituents it carries move closer
together. This reduction in bond angle is not a problem for aldehydes, because one of the sub-
stituents is just a (very small) hydrogen atom, but for ketones, especially ones that carry larger alkyl
groups, this effect can disfavour the addition reaction. Effects that result from the size of substituents
and the repulsion between them are called steric effects, and we call the repulsive force experienced
by large substituents steric hindrance.
The angle of nucleophilic attack on aldehydes and ketones
Having introduced you to the sequence of events that makes up a nucleophilic attack at C=O (inter-
action of HOMO with LUMO, formation of new σ bond, breakage of π bond), we should now tell
you a little more about the direction from which the nucleophile approaches the carbonyl group. Not
only do nucleophiles always attack carbonyl groups at carbon, but they also always approach from a
particular angle. You may at first be surprised by this angle, since nucleophiles attack not from a
direction perpendicular to the plane of the carbonyl group but at about 107° to the C=O bond. This
approach route is known as the Bürgi–Dunitz trajectory after the authors of the elegant crystallo-
graphic methods that revealed it. You can think of the angle of attack as the result of a compromise
between maximum orbital overlap of the HOMO with π* and minimum repulsion of the HOMO by
the electron density in the carbonyl π bond.
Any other portions of the molecule that get in the way of (or, in other words, that cause steric hin-
drance to) the Bürgi–Dunitz trajectory will greatly reduce the rate of addition and this is another rea-
son why aldehydes are more reactive than ketones. The importance of the Bürgi–Dunitz trajectory
will become more evident later—particularly in Chapter 34.
Nucleophilic attack by ‘hydride’on aldehydes and ketones
Nucleophilic attack by the hydride ion, H–, is not a known reaction. This species, which is present in
the salt sodium hydride, NaH, is so small and has such a high charge density that it only ever reacts
as a base. The reason is that its filled 1s orbital is of an ideal size to interact with the hydrogen
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 139
!
Steric hindrance (not hinderance)
is a consequence of repulsion
between the electrons in all the
filled orbitals of the alkyl
substituents.
!The Bürgi–Dunitz angle
Bürgi and Dunitz deduced this
trajectory by examining crystal
structures of compounds containing
both a nucleophilic nitrogen atom and
an electrophilic carbonyl group. They
found that, when the two got close
enough to interact, but were not free to
undergo reaction, the nitrogen atom
always lay on or near the 107°
trajectory decribed here. Theoretical
calculations later gave the same 107°
value for the optimum angle of attack.
C O
Nu
C O
Nu
C O
Numaximum overlap with π* 
perpendicular to C=O bond
repulsion from filled π orbital 
forces nucleophile to attack at 
obtuse angle
π*
π
combined effect: 107°
nucleophile attacks 
C=O at 107° angle
!
Although we now know precisely
from which direction the
nucleophile attacks the C=O
group, this is not always easy to
represent when we draw curly
arrows. As long as you bear the
Bürgi–Dunitz trajectory in mind,
you are quite at liberty to write any
of the variants shown here,
among others.
HR
O
HR
O
H
R
O
Nu
Nu
Nu
never happens
O
Me Me
H
Me Me
H O
H X
H2 XHX
H– always reacts as a basenucleophilic attack by H–
+
09/09/14 
10 
229 
 
Prof. Dr. José Eduardo Damas Martins 
Como fazer o hidreto adquirir 
caráter nucleofílico ? 
230 
 
Prof. Dr. José Eduardo Damas Martins 
Através do uso de hidretos metálicos, 
como o NaBH4 e o LiAlH4, é possível 
reduzir aldeídos e cetonas a alcoois 
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09/09/14 
11 
231 
 
Prof. Dr. José Eduardo Damas Martins 
O Boro hidreto de sódio NaBH4 
Redutor seletivo 
232 
 
Prof. Dr. José Eduardo Damas Martins 
Reduz aldeídos e cetonas com ótimos 
rendimentos 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehydeis reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
09/09/14 
12 
233 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
234 
 
Prof. Dr. José Eduardo Damas Martins 
O NaBH4 pode ser utilizado, inclusive, em 
solução aquosa 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
09/09/14 
13 
235 
 
Prof. Dr. José Eduardo Damas Martins 
É um dos mais simplês reagentes redutores 
utilizados para aldeídos e cetonas 
236 
 
Prof. Dr. José Eduardo Damas Martins 
atom’s contribution to the σ* orbital of an H–X bond (X can be any atom), but much too small to
interact easily with carbon’s more diffuse 2p orbital contribution to the LUMO (π*) of the C=O
group.
Nevertheless, adding H– to the carbon atom of a C=O group would be a very useful reaction, as
the result would be the formation of an alcohol. This process would involve going down from the
aldehyde or ketone oxidation level to the alcohol oxidation level (Chapter 2, pp. 25–36) and would
therefore be a reduction. It cannot be done with NaH, but it can be done with some other com-
pounds containing nucleophilic hydrogen atoms.
The most important of these compounds is sodium borohydride, NaBH4. This is a water-soluble
salt containing the tetrahedral BH4
– anion, which is isoelectronic with methane but has a negative
charge since boron has one less proton in the nucleus than does carbon.
But beware! The boron’s negative charge doesn’t mean that there is a lone pair on boron—there
isn’t. You cannot draw an arrow coming out of this charge to form another bond. If you did, you
would get a pentacovalent B(V) compound, which would have 10 electrons in its outer shell. Such a
thing is impossible with a first row element as there are only four available orbitals (1 × 2s and 3 ×
2p). Instead, since all of the electrons (including that represented by the negative charge) are in B–H
σ orbitals, it is from a B–H bond that we must start any arrow to indicate reaction of BH4
– as a nucle-
ophile. By transferring this pair of electrons we make the boron atom neutral—it is now trivalent
withjust six electrons.
What happens when we carry out this reaction using a carbonyl compound as the electrophile?
The hydrogen atom, together with the pair of electrons from the B–H bond, will be transferred to the
carbon atom of the C=O group.
Though no hydride ion, H–, is actually involved in the reaction, the transfer of a hydrogen atom
with an attached pair of electrons can be regarded as a ‘hydride transfer’. You will often see it
described this way in books. But be careful not to confuse BH4
– with the hydride ion itself. To make it
quite clear that it is the hydrogen atom that is forming the new bond to C, this reaction may also be
helpfully represented with a curly arrow passing through the hydrogen atom.
The oxyanion produced in the first step can help stabilize the electron-deficient BH3 molecule by
adding to its empty p orbital. Now we have a tetravalent boron anion again, which could transfer a
second hydrogen atom (with its pair of electrons) to another molecule of aldehyde.
140 6 . Nucleophilic addition to the carbonyl group
Me Me
O
Me Me
H O
Me Me
H OH
H
reduction of a ketone to an alcohol
?
!
In Chapter 4, we looked at isoelectronic
BH3 and CH3+. Here, we have effectively
just added H– to both of them.
H
B
H
H H
H
C
H
H H
borohydride anion methane
!
Just as we have used Nu– to indicate
any (undefined) nucleophile, here E+
means any (undefined) electrophile.
E
H
B
H
H H
H
B
H
H
H
EX
impossible structure: ten 
electrons in B–H bonds
eight electrons 
in B–H bonds
arrow cannot start on negative 
charge: no lone pair on B
E B
H
H HH
B
H
H H
EH
electrons must be 
transferred from a bond
eight electrons in 
B–H bonds
six electrons in B–H 
bonds and one empty p 
orbital
H
B
H
H H H R
O
H R
B
H
H H H O
H
B
H
H H H R
O
H R
B
H
H H H O
!
The reason that H– never acts as
a nucleophile is that its 1s orbital
is too small. The orbitals involved
in borohydride reductions are the
π* of the C=O group as the LUMO
and a B–H σ orbital as the HOMO,
so there is a much better orbital
match.
C O
H
B
H
H
H
HOMO = B–H σ
LUMO = π*
Mecanismo 
TRANSFERÊNCIA DE 
HIDRETO 
09/09/14 
14 
237 
 
Prof. Dr. José Eduardo Damas Martins 
atom’s contribution to the σ* orbital of an H–X bond (X can be any atom), but much too small to
interact easily with carbon’s more diffuse 2p orbital contribution to the LUMO (π*) of the C=O
group.
Nevertheless, adding H– to the carbon atom of a C=O group would be a very useful reaction, as
the result would be the formation of an alcohol. This process would involve going down from the
aldehyde or ketone oxidation level to the alcohol oxidation level (Chapter 2, pp. 25–36) and would
therefore be a reduction. It cannot be done with NaH, but it can be done with some other com-
pounds containing nucleophilic hydrogen atoms.
The most important of these compounds is sodium borohydride, NaBH4. This is a water-soluble
salt containing the tetrahedral BH4
– anion, which is isoelectronic with methane but has a negative
charge since boron has one less proton in the nucleus than does carbon.
But beware! The boron’s negative charge doesn’t mean that there is a lone pair on boron—there
isn’t. You cannot draw an arrow coming out of this charge to form another bond. If you did, you
would get a pentacovalent B(V) compound, which would have 10 electrons in its outer shell. Such a
thing is impossible with a first row element as there are only four available orbitals (1 × 2s and 3 ×
2p). Instead, since all of the electrons (including that represented by the negative charge) are in B–H
σ orbitals, it is from a B–H bond that we must start any arrow to indicate reaction of BH4
– as a nucle-
ophile. By transferring this pair of electrons we make the boron atom neutral—it is now trivalent
with just six electrons.
What happens when we carry out this reaction using a carbonyl compound as the electrophile?
The hydrogen atom, together with the pair of electrons from the B–H bond, will be transferred to the
carbon atom of the C=O group.
Though no hydride ion, H–, is actually involved in the reaction, the transfer of a hydrogen atom
with an attached pair of electrons can be regarded as a ‘hydride transfer’. You will often see it
described this way in books. But be careful not to confuse BH4
– with the hydride ion itself. To make it
quite clear that it is the hydrogen atom that is forming the new bond to C, this reaction may also be
helpfully represented with a curly arrow passing through the hydrogen atom.
The oxyanion produced in the first step can help stabilize the electron-deficient BH3 molecule by
adding to its empty p orbital. Now we have a tetravalent boron anion again, which could transfer a
second hydrogen atom (with its pair of electrons) to another molecule of aldehyde.
140 6 . Nucleophilic addition to the carbonyl group
Me Me
O
Me Me
H O
Me Me
H OH
H
reduction of a ketone to an alcohol
?
!
In Chapter 4, we looked at isoelectronic
BH3 and CH3+. Here, we have effectively
just added H– to both of them.
H
B
H
H H
H
C
H
H H
borohydride anion methane
!
Just as we have used Nu– to indicate
any (undefined) nucleophile, here E+
means any (undefined) electrophile.
E
H
B
H
H H
H
B
H
H
H
EX
impossible structure: ten 
electrons in B–H bonds
eight electrons 
in B–H bonds
arrow cannot start on negative 
charge: no lone pair on B
E B
H
H HH
B
H
H H
EH
electrons must be 
transferred from a bond
eight electrons in 
B–H bonds
six electrons in B–H 
bonds and one empty p 
orbital
H
B
H
H H H R
O
H R
B
H
H H H O
H
B
H
H H H R
O
H R
B
H
H H H O
!
The reason that H– never acts as
a nucleophile is that its 1s orbital
is too small. The orbitals involved
in borohydride reductions are the
π* of the C=O group as the LUMO
and a B–H σ orbital as the HOMO,
so there is a much better orbital
match.
C O
H
B
H
H
H
HOMO = B–H σ
LUMO = π*
Pode-se expressar também da seguinte forma 
Flexa corta o hidrogênio 
indicando que o par de 
elétrons foi transferido 
Mecanismo 
238 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
Hexamples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
Mecanismo 
O Processo continua até 
que todos os hidretos 
sejam transferidos 
09/09/14 
15 
239 
 
Prof. Dr. José Eduardo Damas Martins 
Na prática, uma molécula de NaBH4 pode 
reduzir quatro moléculas de um aldeído ou 
cetona 
B
O
O
OO
R
R
R R
R
R
R R
Na H2O
R R
OH
4 Na B(OH)4+
240 
 
Prof. Dr. José Eduardo Damas Martins 
A reação, geralmente, não produz tanta 
eficiência de tal forma que costuma-se 
utilizar um excesso do reagente. 
09/09/14 
16 
241 
 
Prof. Dr. José Eduardo Damas Martins 
Forma simplif icada de representar o 
mecanismo 
H3B H
R R
O
R R
H O
R R
H OHHNa Na
242 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
Mecanismo ? 
09/09/14 
17 
243 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
Mecanismo ? 
244 
 
Prof. Dr. José Eduardo Damas Martins 
Seletividade… 
O NaBH4 não reduz ésteres, ácidos 
carboxílicos e amidas 
09/09/14 
18 
245 
 
Prof. Dr. José Eduardo Damas Martins 
R OR
O
R OH
O
R NH2
O
Éster
Ác. carboxílico
Amida
NaBH4 Não ocorre reação
246 
 
Prof. Dr. José Eduardo Damas Martins 
Isto se deve à baixa reatividade do NaBH4, 
aliada ao baixo caráter eletrofílico dos 
ésteres, ácidos carboxílicos e amidas. 
09/09/14 
19 
247 
 
Prof. Dr. José Eduardo Damas Martins 
R OR
O
R O
O
R
R OH
O
R O
O
H
R NH2
O
R N
O
H
H
Éster
Ác. carboxílico
Amida
Carbonila 
menos 
eletrofílica 
devido à 
ressonância 
248 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
NaBH4 Não reduz éster 
09/09/14 
20 
249 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
NaBH4 Não reduz grupo 
Nitro (NO2) 
250 
 
Prof. Dr. José Eduardo Damas Martins 
We shall use this synthesis as a basis for discussion on chemoselectivity in reductions. In the first
step, sodium borohydride leaves the black carbonyl group of the ester untouched while it reduces the
ketone (in yellow); in the last step, lithium aluminium hydride reduces the ester (in black). These
chemoselectivities are typical of these two most commonly used reducing agents: borohydride can
usually be relied upon to reduce an aldehyde or a ketone in the presence of an ester, while lithium
aluminium hydride will reduce almost any carbonyl group.
Each reduction gives an alcohol, apart from the reduction of an amide with LiAlH4, which gives
an amine, which we shall explain next. We shall return to the salmefamol synthesis later to explain
the reductions with hydrogen gas catalysed by palladium.
Reduction of carbonyl groups
We should now look in detail at reductions of carbonyl compounds, and in doing so we shall intro-
duce a few more specialized reducing agents. Then we will come back to the other type of reduction
in the salmefamol synthesis—catalytic hydrogenation.
How to reduce aldehydes and ketones to alcohols
We don’t need to spend much time on this—sodium borohydride does it very well, and is a lot easier
to handle than lithium aluminium hydride. It is also more selective: it will reduce this nitroketone,
for example, where LiAlH4 would reduce the nitro group as well.
You met borohydride in Chapter 6, where we discussed the mechanism of its reactions. Sodium
borohydride will reduce only in protic solvents (usually ethanol, methanol, or water) or in the pres-
ence of electrophilic metal cations such as Li+ or Mg2+ (LiBH4 can be used in THF, for example).
The precise mechanism, surprisingly, is still unclear, but follows a course something like this with the
dotted lines representing some association, perhaps coordination or bond formation.
Reduction of carbonyl groups 617
HO
MeO
O
N
Ph
O
Ph
HO
MeO
OH
N
Ph
O
Ph
HO
MeO
OH
H
N
O
OMe
OMe
O
HO
HO
OH
H
N
OMe
LiAlH4
NaBH4
salmefamol
H2, Pd/C, H
!
In general, it’s best to use the
mildest conditions possible for
any particular reaction—the
potential for unwanted side-
reactions is lessened. What is
more, NaBH4 is a lot easier to
handle than LiAlH4—for example,
it simply dissolves in water while
LiAlH4 catches fire if it gets wet.
NaBH4 is usually used to reduce
aldehydes and ketones, even
though LiAlH4 also works.
R H
O
R R
O
R OR
O
R NR2
O
R OH
O
aldehyde ketone ester amide carboxylic acid
> > > >
reduced by NaBH4 not reduced by NaBH4
reduced by LiAlH4
R1 R2
O
R1 R2
OH
R1 H
O
R1 OH
NO2
O
NO2
OHNaBH4
EtOH
NaBH4 Não reduz grupo 
Nitro (NO2) 
09/09/14 
21 
251 
 
Prof. Dr. José Eduardo Damas Martins 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
HH R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminium is more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
NaBH4 não promove a eliminação 
de Bromo α−carbonila 
252 
 
Prof. Dr. José Eduardo Damas Martins 
Redução de 
iminas 
09/09/14 
22 
253 
 
Prof. Dr. José Eduardo Damas Martins 
R R
N
R'
R R
HN
R'
NaBH4
CH3OH
Imina Amina
254 
 
Prof. Dr. José Eduardo Damas Martins 
Imina Amina
N
F
 NaBH4
CH3OH
HN
F
Mecanismo ? 
09/09/14 
23 
255 
 
Prof. Dr. José Eduardo Damas Martins 
Aminação redutiva 
É uma forma de sintetizar aminas à partir de 
aldeídos e cetonas em uma única etapa. 
256 
 
Prof. Dr. José Eduardo Damas Martins 
carbonyl compounds. For example, iminium ions are reduced rapidly by the mild reducing agent
sodium cyanoborohydride (NaCNBH3), while carbonyl compounds are not.
An alternative to Na(CN)BH3 is NaBH(OAc)3 (sodium triacetoxy-borohydride)—somewhat
safer because strong acid can release HCN from Na(CN)BH3.
Amines from imines: reductive amination
A useful way of making amines is by reduction of imines (or iminium ions). This overall process,
from carbonyl compound to amine, is called reductive amination. This is, in fact, one of the few suc-
cessful ways, and the best way, of making secondary amines. This should be your first choice in
amine synthesis.
This can be done in two steps, provided the intermediate is stable, but, because the instability of
many imines makes them hard to isolate, the most convenient way of doing it is to form and reduce
the imine in a single reaction. The selective reduction of iminium ions (but not carbonyl com-
pounds) by sodium cyanoborohydride makes this possible. When NaCNBH3 is added to a typical
imine-formation reaction it reacts with the products but not with the starting carbonyl compound.
Here is an example of an amine synthesis using reductive amination.
In the first step, the ketone and ammonia are in equilibrium with their imine, which, at pH 6, is
partly protonated as an iminium ion. The iminium ion is rapidly reduced by the cyanoborohydride
to give the amine. Reactions like this, using ammonia in a reductive amination, are often carried out
with ammonium chloride or acetate as convenient sources of ammonia. At pH 6, ammonia will be
mostly protonated anyway.
In the second step of the synthesis, amine plus formaldehyde gives an imine, present as its proto-
nated iminium form, which gets reduced. Formaldehyde is so reactive that it reacts again with the
secondary amine to give an iminium ion; again, this is reduced to the amine.
354 14 . Nucleophilic substitution at C=O with loss of carbonyl oxygen
!
Sodium cyanoborohydride
contains the cyanoborohydride
anion, whose structure is
It is a ‘toned down’ version of
sodium borohydride—the
electron-withdrawing cyano group
decreases the ease with which
hydride is transferred.
N
Me Me
N
Me MeO
H
pH 6 pH 6
NaCNBH3
no reaction 90%
yield
NaCNBH3
CN
B
H
H
H
R2
O
H
R2
N
H
R1
R2
HN
R1
H
H
secondary amine
H
R1NH2
2 x [H]
imine
Ph Me
O
Ph Me
NH2
Ph Me
N
Me Me
pH 6
86% yield 81% yield
NH3
NaCNBH3
pH 6
CH2=O
NaCNBH3
Ph Me
O
Ph Me
NH2
H
B
H
H
CN
Ph Me
NH2
H
NH3
pH 6
!
You will again meet the highly
electrophilic iminium ions produced by
reaction of formaldehyde with amines
in Chapter 27, where we introduce you
to the Mannich reaction.
Redução 
Aminação redutiva 
09/09/14 
24 
257 
 
Prof. Dr. José Eduardo Damas Martins 
O
+ H2N R
1- H, remoção de água
2- NaBH4 ou NaBH3CN
HN
R
Amina
Aminação redutiva 
258 
 
Prof. Dr. José Eduardo Damas Martins 
carbonyl compounds. For example, iminium ions are reduced rapidly by the mild reducing agent
sodium cyanoborohydride (NaCNBH3), while carbonyl compounds are not.
An alternative to Na(CN)BH3 is NaBH(OAc)3 (sodium triacetoxy-borohydride)—somewhat
safer because strong acid can release HCN from Na(CN)BH3.
Amines from imines: reductive amination
A useful way of making amines is by reduction of imines (or iminium ions). This overall process,
from carbonyl compound to amine, is called reductive amination. This is, in fact, one of the few suc-
cessful ways, and the best way, of making secondary amines. This should be your first choice in
amine synthesis.
This can be done in two steps, provided the intermediate is stable, but, because the instability of
many imines makes them hard to isolate, the most convenient way of doing it is to form and reduce
the imine in a single reaction. The selective reduction of iminium ions (but not carbonyl com-
pounds) by sodium cyanoborohydride makes this possible. When NaCNBH3 is added to a typical
imine-formation reaction it reacts with the products but not with the starting carbonyl compound.
Here is an example of an amine synthesis using reductive amination.
In the first step, the ketone and ammonia are in equilibrium with their imine, which, at pH 6, is
partly protonated as an iminium ion. The iminium ion is rapidly reduced by the cyanoborohydride
to give the amine. Reactions like this, using ammonia in a reductive amination, are often carried out
with ammonium chloride or acetate as convenient sources of ammonia. At pH 6, ammonia will be
mostly protonated anyway.
In the second step of the synthesis, amine plus formaldehyde gives an imine, present as its proto-
nated iminium form, which gets reduced. Formaldehyde is so reactive that it reacts again with the
secondary amine to give an iminium ion; again, this is reduced to the amine.
354 14 . Nucleophilic substitution at C=O with loss of carbonyl oxygen
!
Sodium cyanoborohydride
contains the cyanoborohydride
anion, whose structure is
It is a ‘toned down’ version of
sodium borohydride—the
electron-withdrawing cyano group
decreases the ease with which
hydride is transferred.
N
Me Me
N
Me MeO
H
pH 6 pH 6
NaCNBH3
no reaction 90%
yield
NaCNBH3
CN
B
H
H
H
R2
O
H
R2
N
H
R1
R2
HN
R1
H
H
secondary amine
H
R1NH2
2 x [H]
imine
Ph Me
O
Ph Me
NH2
Ph Me
N
Me Me
pH 6
86% yield 81% yield
NH3
NaCNBH3
pH 6
CH2=O
NaCNBH3
Ph Me
O
Ph Me
NH2
H
B
H
H
CN
Ph Me
NH2
H
NH3
pH 6
!
You will again meet the highly
electrophilic iminium ions produced by
reaction of formaldehyde with amines
in Chapter 27, where we introduce you
to the Mannich reaction.
Aminação redutiva 
Mecanismo ? 
09/09/14 
25 
259 
 
Prof. Dr. José Eduardo Damas Martins 
O hidreto de lítio e alumínio LiAlH4 
Redutor não-seletivo 
260 
 
Prof. Dr. José Eduardo Damas Martins 
Redutor 
Altamente 
reativo 
09/09/14 
26 
261 
 
Prof. Dr. José Eduardo Damas Martins 
Reduz aldeídos, cetonas, ésteres, amidas, 
ác. Carboxílicos, nitro compostos, iminas, 
oximas, nitrilas, entre outros grupos 
262 
 
Prof. Dr. José Eduardo Damas Martins 
The essence of the reaction is the transfer of a hydrogen atom with two electrons(called hydride
transfer though no hydride ion is involved). In addition, the developing negative charge on oxygen
gets help from the alcohol or the sodium ion or both and a molecule of alcohol adds to the boron
during or immediately after the reduction. The by-product, an alkoxyborohydride anion, is itself a
reducing agent, and can go on to reduce three more molecules of carbonyl compound, transferring
step-by-step all of its hydrogen atoms.
How to reduce esters to alcohols
LiAlH4 is often the best reagent, and gives
alcohols by the mechanism we discussed
in Chapter 12. As a milder alternative (LiAlH4
has caused countless fires through careless handling), lithium borohydride in alcoholic solution will
reduce esters—in fact, it has useful selectivity for esters over acids or amides that LiAlH4 does not
have. Sodium borohydride reduces most esters only rather slowly.
How to reduce amides to amines
Again, LiAlH4 is a good reagent for this transforma-
tion. The mechanism follows very much the same course
as the reduction of esters, but there is a key difference at
the steps boxed in yellow and in green.
How to reduce carboxylic acids to alcohols
The best reagent for this is borane, BH3. Borane is, in
fact, a gas with the structure B2H6, but it can be ‘tamed’
as a liquid by complexing it with ether (Et2O), THF, or
dimethyl sulfide (DMS, Me2S).
618 24 . Chemoselectivity: selective reactions and protection
R
R
O
H3B HR R
O
R R
OH
H OEt
H OEt
Na
NaBH4
+ EtOBH3
still a reducing agent
EtOH
R1 OR2
O
R1 OH
R2OH
H H
+
!
Why not try writing the mechanism out
now to make sure you understand it,
before checking back to p. 000? In a
moment, we will show you a slightly
more sophisticated version, in which
we account for the fate of the Li and Al
species.
!
The ester mechanism has rather more
detail than the simplified one we
presented to you in Chapter 12.
MeO2C CO2H CO2HHO
H HLiBH4
EtOH
2R
1 NR2
O
R1 NR2
H H
2
R OR
O
Li
R OR
O
Li
R H
O
Li
H3Al H
H
R H
O
Li
H
AlH3
H3Al H
R H
O
H
AlH3
R H
OH
H
R NR2
O
Li
H3Al H
R H
NR2
HR NR2
O
Li
H
AlH3
R NR2
O
H
AlH3
R H
NR2
H3Al H
R NR2
O
Li
H
DOES NOT HAPPEN
NR2 is not eliminated in this step
LiAlH4 reduction of esters
LiAlH4 reduction of amides
H
 quench
tetrahedral intermediate 
collapses to give an aldehyde
tetrahedral intermediate collapses 
to give an iminium ion
R OH
O
R OH
H H!
These complexes are Lewis salts: BH3
is a Lewis acid that accepts a lone pair
of electrons from the basic ether or
sulfide.
09/09/14 
27 
263 
 
Prof. Dr. José Eduardo Damas Martins 
H
EtO
O
O
H
OH
OH
H
H
O
N
O
O
LiAlH4
THF seco
LiAlH4
THF seco
H
OH
H
H2N
264 
 
Prof. Dr. José Eduardo Damas Martins 
O
Br LiAlH4
Et2O seco
OH
H
H H
H
O
LiAlH4
THF seco
OH
09/09/14 
28 
265 
 
Prof. Dr. José Eduardo Damas Martins 
Razões da alta reatividade ? 
266 
 
Prof. Dr. José Eduardo Damas Martins 
O Alumínio é mais eletropositivo que o boro, 
portanto está mais apto a transferir um 
hidreto. 
The essence of the reaction is the transfer of a hydrogen atom with two electrons (called hydride
transfer though no hydride ion is involved). In addition, the developing negative charge on oxygen
gets help from the alcohol or the sodium ion or both and a molecule of alcohol adds to the boron
during or immediately after the reduction. The by-product, an alkoxyborohydride anion, is itself a
reducing agent, and can go on to reduce three more molecules of carbonyl compound, transferring
step-by-step all of its hydrogen atoms.
How to reduce esters to alcohols
LiAlH4 is often the best reagent, and gives
alcohols by the mechanism we discussed
in Chapter 12. As a milder alternative (LiAlH4
has caused countless fires through careless handling), lithium borohydride in alcoholic solution will
reduce esters—in fact, it has useful selectivity for esters over acids or amides that LiAlH4 does not
have. Sodium borohydride reduces most esters only rather slowly.
How to reduce amides to amines
Again, LiAlH4 is a good reagent for this transforma-
tion. The mechanism follows very much the same course
as the reduction of esters, but there is a key difference at
the steps boxed in yellow and in green.
How to reduce carboxylic acids to alcohols
The best reagent for this is borane, BH3. Borane is, in
fact, a gas with the structure B2H6, but it can be ‘tamed’
as a liquid by complexing it with ether (Et2O), THF, or
dimethyl sulfide (DMS, Me2S).
618 24 . Chemoselectivity: selective reactions and protection
R
R
O
H3B HR R
O
R R
OH
H OEt
H OEt
Na
NaBH4
+ EtOBH3
still a reducing agent
EtOH
R1 OR2
O
R1 OH
R2OH
H H
+
!
Why not try writing the mechanism out
now to make sure you understand it,
before checking back to p. 000? In a
moment, we will show you a slightly
more sophisticated version, in which
we account for the fate of the Li and Al
species.
!
The ester mechanism has rather more
detail than the simplified one we
presented to you in Chapter 12.
MeO2C CO2H CO2HHO
H HLiBH4
EtOH
2R
1 NR2
O
R1 NR2
H H
2
R OR
O
Li
R OR
O
Li
R H
O
Li
H3Al H
H
R H
O
Li
H
AlH3
H3Al H
R H
O
H
AlH3
R H
OH
H
R NR2
O
Li
H3Al H
R H
NR2
HR NR2
O
Li
H
AlH3
R NR2
O
H
AlH3
R H
NR2
H3Al H
R NR2
O
Li
H
DOES NOT HAPPEN
NR2 is not eliminated in this step
LiAlH4 reduction of esters
LiAlH4 reduction of amides
H
 quench
tetrahedral intermediate 
collapses to give an aldehyde
tetrahedral intermediate collapses 
to give an iminium ion
R OH
O
R OH
H H!
These complexes are Lewis salts: BH3
is a Lewis acid that accepts a lone pair
of electrons from the basic ether or
sulfide.
09/09/14 
29 
267 
 
Prof. Dr. José Eduardo Damas Martins 
Devido à alta reatividade, o LiAlH4 reage 
VIOLENTAMENTE, com a água. 
This process can continue so that, in principle, all four hydrogen atoms could be transferred to
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol
from the alkoxide.
Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride,
LiAlH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones,
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times
faster than acetophenone in isopropanol.
Sodium borohydride does not react at all with less reactive carbonyl compounds such as
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be
reduced.
The next two examples illustrate the reduction of aldehydes and ketones in the presence of other
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide
in the second.
Nucleophilic attack by ‘hydride’ on aldehydes and ketones 141
H R
B
H
H H
H O
H R
H O
B
H
H H
H R
O
H R
H O
B
H
H
H R
H O
NaBH4
H
O
MeO HMeO
OH
H
examples of reductions with sodium borohydride
H2O
NaBH4
O HO H
MeOH
O OH
NaBH4
i -PrOH
Ph H
O
Ph Me
O
benzaldehyde acetophenone
!
Aluminiumis more electropositive
(more metallic) than boron and is
therefore more ready to give up a
hydrogen atom (and the
associated negative charge),
whether to a carbonyl group or to
water. Lithium aluminium hydride
reacts violently and dangerously
with water in an exothermic
reaction that produces highly
flammable hydrogen.
H
Al
H
H H
Li
H
OH
H
Al
H
H H2 LiOH
violent 
reaction!
H
O
Et
O
O
H
O
Et
OH
O
H
NaBH4
EtOH
NaBH4H
O
O2N
OH
O2N
H H
NaOH, H2O, MeOH
O H OH
Br Br
NaBH4
MeOH, 25 °C
268 
 
Prof. Dr. José Eduardo Damas Martins 
Portanto, as reações com LiAlH4 devem ser 
conduzidas em solvente seco (sem água) e 
sob atmosfera de gás inerte para evitar a 
umidade do ar. 
09/09/14 
30 
269 
 
Prof. Dr. José Eduardo Damas Martins 
Hidreto de Alumínio e Lítio 
Redução de ésteres 
270 
 
Prof. Dr. José Eduardo Damas Martins 
Hidreto de Alumínio e Lítio 
Redução de amidas 
The essence of the reaction is the transfer of a hydrogen atom with two electrons (called hydride
transfer though no hydride ion is involved). In addition, the developing negative charge on oxygen
gets help from the alcohol or the sodium ion or both and a molecule of alcohol adds to the boron
during or immediately after the reduction. The by-product, an alkoxyborohydride anion, is itself a
reducing agent, and can go on to reduce three more molecules of carbonyl compound, transferring
step-by-step all of its hydrogen atoms.
How to reduce esters to alcohols
LiAlH4 is often the best reagent, and gives
alcohols by the mechanism we discussed
in Chapter 12. As a milder alternative (LiAlH4
has caused countless fires through careless handling), lithium borohydride in alcoholic solution will
reduce esters—in fact, it has useful selectivity for esters over acids or amides that LiAlH4 does not
have. Sodium borohydride reduces most esters only rather slowly.
How to reduce amides to amines
Again, LiAlH4 is a good reagent for this transforma-
tion. The mechanism follows very much the same course
as the reduction of esters, but there is a key difference at
the steps boxed in yellow and in green.
How to reduce carboxylic acids to alcohols
The best reagent for this is borane, BH3. Borane is, in
fact, a gas with the structure B2H6, but it can be ‘tamed’
as a liquid by complexing it with ether (Et2O), THF, or
dimethyl sulfide (DMS, Me2S).
618 24 . Chemoselectivity: selective reactions and protection
R
R
O
H3B HR R
O
R R
OH
H OEt
H OEt
Na
NaBH4
+ EtOBH3
still a reducing agent
EtOH
R1 OR2
O
R1 OH
R2OH
H H
+
!
Why not try writing the mechanism out
now to make sure you understand it,
before checking back to p. 000? In a
moment, we will show you a slightly
more sophisticated version, in which
we account for the fate of the Li and Al
species.
!
The ester mechanism has rather more
detail than the simplified one we
presented to you in Chapter 12.
MeO2C CO2H CO2HHO
H HLiBH4
EtOH
2R
1 NR2
O
R1 NR2
H H
2
R OR
O
Li
R OR
O
Li
R H
O
Li
H3Al H
H
R H
O
Li
H
AlH3
H3Al H
R H
O
H
AlH3
R H
OH
H
R NR2
O
Li
H3Al H
R H
NR2
HR NR2
O
Li
H
AlH3
R NR2
O
H
AlH3
R H
NR2
H3Al H
R NR2
O
Li
H
DOES NOT HAPPEN
NR2 is not eliminated in this step
LiAlH4 reduction of esters
LiAlH4 reduction of amides
H
 quench
tetrahedral intermediate 
collapses to give an aldehyde
tetrahedral intermediate collapses 
to give an iminium ion
R OH
O
R OH
H H!
These complexes are Lewis salts: BH3
is a Lewis acid that accepts a lone pair
of electrons from the basic ether or
sulfide.
09/09/14 
31 
271 
 
Prof. Dr. José Eduardo Damas Martins 
Estereoseletividade nas reduções 
272 
 
Prof. Dr. José Eduardo Damas Martins 
Majoritário 
O OH
H
H
OH+NaBH4
MeOH
Endo Exo
Norbornanona
09/09/14 
32 
273 
 
Prof. Dr. José Eduardo Damas Martins 
O OH
H
H
OH+NaBH4
MeOH
Endo ExoCânfora
Majoritário 
274 
 
Prof. Dr. José Eduardo Damas Martins 
Majoritário 
NaBH4
MeOH
O OH OH
+
cis trans
09/09/14 
33 
275 
 
Prof. Dr. José Eduardo Damas Martins 
Face menos 
impedida 
H
H
H
H
CH3
O BH3H
276 
 
Prof. Dr. José Eduardo Damas Martins 
H
H
H
H
CH3
O BH3H
CH3
OH
H
H
OH
09/09/14 
34 
277 
 
Prof. Dr. José Eduardo Damas Martins 
Outros agentes redutores 
278 
 
Prof. Dr. José Eduardo Damas Martins 
L-Selectride 
Tri-isobutil-lítio borohidreto 
09/09/14 
35 
279 
 
Prof. Dr. José Eduardo Damas Martins 
280 
 
Prof. Dr. José Eduardo Damas Martins 
NaBH3CN 
Sódio ciano borohidreto 
09/09/14 
36 
281 
 
Prof. Dr. José Eduardo Damas Martins 
Estereoquímica da adição 
nucleofílica em aldeídos e cetonas 
282 
 
Prof. Dr. José Eduardo Damas Martins 
C O
R2
R1
R1 ≠ R2
Nu
Ataque por cima
Ataque por baixo
Cetona 
 pró-quiral 
09/09/14 
37 
283 
 
Prof. Dr. José Eduardo Damas Martins 
OH
Nu R1
R2 Nu
OH
R2
R1
+
Produto do 
ataque 
por cima 
Produto do 
Ataque 
 por baixo 
Enantiômeros
284 
 
Prof. Dr. José Eduardo Damas Martins 
O OH OH
+NaBH4
MeOH
50 % 50 %
Enantiômeros
Mistura racêmica
09/09/14 
38 
285 
 
Prof. Dr. José Eduardo Damas Martins 
Cetona 
 pró-quiral 
Faces 
enantiotópicas 
C O
Ataque por cima
BH H
H
H
Na
286 
 
Prof. Dr. José Eduardo Damas Martins 
Como diferenciar as faces ? 
09/09/14 
39 
287 
 
Prof. Dr. José Eduardo Damas Martins 
O
1
2 3
Sentido anti-horário 
 
Face “ si” 
(de frente para a folha) 
Sentido horário 
 
Face “ re” 
(atras da folha) 
288 
 
Prof. Dr. José Eduardo Damas Martins 
O
HO
OH
NaBH4
MeOH
Ataque na face "si"
Ataque na face "re"
H
H
(R)
(S)
09/09/14 
40 
289 
 
Prof. Dr. José Eduardo Damas Martins 
O Uso de reagentes quirais pode “favorecer" 
o ataque a uma das faces 
290 
 
Prof. Dr. José Eduardo Damas Martins 
O HO H
(S)
Enzima
ee > 99 %
Face “ re” 
Foi atacada 
09/09/14 
41 
291 
 
Prof. Dr. José Eduardo Damas Martins 
H
O
CH3
OH
OH
N(CH3)2
Zn(CH3)2
ee > 95 %
(S)
Face “ si” 
Foi atacada