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Reatividade de Compostos Orgânicos – Aula 2 
1 
UNIVERSIDADE FEDERAL DA GRANDE DOURADOS 
 
FACULDADE DE CIÊNCIAS EXATAS E TECNOLOGIA 
 
Curso de Licenciatura e Bacharelado em QUÍMICA 
 
Aula 2 – Substituição 
 
 Nucleofílica a C saturado 
 
 
Profa. ROZANNA MARQUES MUZZI 
Reatividade de Compostos Orgânicos – Aula 2 
Ex.: 
(CH3)3CBr +CH3OH à (CH3)3COCH3 +HBr 
EQUAÇÃO GERAL: 
 
R-GS + Nu R-Nu + GS 
t-butyl bromide Nu 
methyl t-butyl 
ether 
2 
Reatividade de Compostos Orgânicos – Aula 2 
1. MECANISMO SN1: 
 SN1: substitution, nucleophilic, unimolecular 
velocidade = k[substrato] - REAÇÃO DE 1ª. ORDEM 
The substrate forms a carbocation intermediate. 
The carbocation is planar with an empty 2p orbital 
available for bonding. 
3 
Reatividade de Compostos Orgânicos – Aula 2 
POSSÍVEIS MECANISMOS: 
A) EM DUAS ETAPAS - SN1 
 
 
 
 
 
 
 
 
B) EM UMA ETAPA - SN2 
4 
MECANISMO SN1 
Reatividade de Compostos Orgânicos – Aula 2 
rate = k[(CH3)3CBr] 
5 
Reatividade de Compostos Orgânicos – Aula 2 
rate = k [alkyl halide] 
 
First-order kinetics implies a unimolecular 
rate-determining step. 
6 
Reatividade de Compostos Orgânicos – Aula 2 
The carbocation intermediate quickly reacts with the 
nucleophile. 
 
 
 
Attack can be from either side. 
+ 
7 
Reatividade de Compostos Orgânicos – Aula 2 
8 
Reatividade de Compostos Orgânicos – Aula 2 
CH3Br 
CH3CH2Br 
(CH3)2CHBr 
(CH3)3CBr 
Reatividade decrescente para SN1 
Reatividade de Compostos Orgânicos – Aula 2 
10 
Table 10.1 Relative Rates of SN2 Reactions for Several Alkyl Halides
SN2R Br Cl−+ R Cl Br−+
CCH3
CH3
CH3
CH3
Br
CH3 Br
CH3CH2 Br
CH3CH2CH2 Br
CH3CH Br
 Alkyl halide Class of alkyl halide Relative rate
methyl 1200
primary 40
primary 16
secondary
tertiary too slow to measure
 1
Section 10.2 The Mechanism of an Reaction 363SN2
Movie:
Bimolecular reaction
Edward Davies Hughes
(1906–1963) was born in North
Wales. He earned two doctoral
degrees: a Ph.D. from the University
of Wales and a D.Sc. from the
University of London, working with
Sir Christopher Ingold. He was a
professor of chemistry at University
College, London.
The rate law tells us which molecules are involved in the transition state of the rate-
determining step of the reaction. From the rate law for the reaction of methyl bromide
with hydroxide ion, we know that both methyl bromide and hydroxide ion are involved
in the rate-determining transition state.
The reaction of methyl bromide with hydroxide ion is an example of an reac-
tion, where “S” stands for substitution, “N” for nucleophilic, and “2” for bimolecu-
lar. Bimolecular means that two molecules are involved in the rate-determining step.
In 1937, Edward Hughes and Christopher Ingold proposed a mechanism for an 
reaction. Remember that a mechanism describes the step-by-step process by which
reactants are converted into products. It is a theory that fits the experimental evi-
dence that has been accumulated concerning the reaction. Hughes and Ingold based
their mechanism for an reaction on the following three pieces of experimental
evidence:
1. The rate of the reaction depends on the concentration of the alkyl halide and on
the concentration of the nucleophile. This means that both reactants are involved
in the rate-determining step.
2. When the hydrogens of methyl bromide are successively replaced with methyl
groups, the rate of the reaction with a given nucleophile becomes progressively
slower (Table 10.1).
3. The reaction of an alkyl halide in which the halogen is bonded to an asymmetric
carbon leads to the formation of only one stereoisomer, and the configuration of
the asymmetric carbon is inverted relative to its configuration in the reacting
alkyl halide.
Hughes and Ingold proposed that an reaction is a concerted reaction—it takes
place in a single step, so no intermediates are formed. The nucleophile attacks the car-
bon bearing the leaving group and displaces the leaving group.
Because a productive collision is a collision in which the nucleophile hits the car-
bon on the side opposite the side bonded to the leaving group, the carbon is said to un-
dergo back-side attack. Why does the nucleophile attack from the back side? The
simplest explanation is that the leaving group blocks the approach of the nucleophile
to the front side of the molecule.
+HO CH3 OHCH3Br + Br
− −
mechanism of the SN2 reaction
SN2
SN2
SN2
SN2
Sir Christopher Ingold
(1893–1970) was born in Ilford,
England. In addition to determining
the mechanism of the reaction,
he was a member of the group that
developed nomenclature for
enantiomers. (See p. 188.) He also
participated in developing the theory
of resonance.
SN2
BRUI10-359_399r2 18-02-2003 2:50 PM Page 363
Reatividade de Compostos Orgânicos – Aula 2 
2. ESTEREOQUÍMICA NO MECANISMO SN1 
11 
RACEMIZACÃO 
Reatividade de Compostos Orgânicos – Aula 2 
12 
HEt
Me
Lv
Lv
-
Me
Et
H
+ sp2 hybridized
Sn1 reactive chemical intermediate
ESTEREOQUÍMICA NO MECANISMO SN1 
Reatividade de Compostos Orgânicos – Aula 2 
13 
ESTEREOQUÍMICA NO MECANISMO SN1 
Me
Et
HHEt
Me
Lv
+
nucleophile adds
at the top
Nu
HEt
Me
Nu
nucleophile adds
at the bottomNu
Nu
HMe
Et
Mirror images 
Reatividade de Compostos Orgânicos – Aula 2 
Stereochemistry of the SN1 Reaction 
•  Because the carbocation is planar, the nucleophilic 
attack can come from either side. 
-H+ 
-H+ 
The products have more 
inversion, because the LG 
partially blocks that side. 
Reatividade de Compostos Orgânicos – Aula 2 
• SN1 reactions are exothermic. 
 
• SN1 reactions occur in at least two steps. 
 
• The first step is slow. It is the formation of the 
carbocation intermediate, which is a very strong 
electrophile. 
 
• The second step is very fast. It is the reaction of 
the carbocation with the nucleophile. 
 
• Reaction is not stereospecific: racemization in 
reactions of optically active alkyl halides 
 
15 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions 
²  Structure of the substrate 
² Can a stable carbocation be formed? 
²  Strength of the nucleophile 
²  Nature of the leaving group 
²  The solvent in which the reaction is 
run. 
² Must be able to stabilize the carbocation 
and the LG (which is usually an ion). 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions – 
Structure of the Substrate 
²  The carbocation is stabilized by any alkyl 
groups that are bonded to the 
electrophilic C atom. 
–  inductive effect from σ bonds 
–  hyperconjugation with p orbital 
²  Resonance will also stabilize the 
carbocation. 
•  Relative rates for SN1: 3°>2°>1°
(resonance)>”1°” 
Reatividade de Compostos Orgânicos – Aula 2 
ESTABILIDADE DO CARBOCÁTION 
This is the most important factor influencing the rate 
of SN1 reactions. 
Stability: 3° > 2° >1° > methyl 
> > > 
3° 2° 1° methyl 
18 
Reatividade de Compostos Orgânicos – Aula 2 
19 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions 
Strength of the Nucleophile 
•  The rate is not much affected by the 
nucleophile. 
•  It may be weak, moderate, or strong. 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions – 
the Leaving Group 
•  The LG should be excellent. 
•  Review: a good LG must be 
²  electron withdrawing, to polarize the bond and make 
the C atom electrophilic, 
²  polarizable, to stabilize the transition state, and 
²  stable in the solvent(so it cannot be a strong base). 
Best LGs are neutral species or anions with a 
stabilized charge. 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions – 
the Leaving Group 
•  The leaving group (LG) serves two 
purposes in an SN1 reaction. These are 
the same as for the SN2 reaction. 
–  It polarizes the bond that makes the C atom 
electrophilic. 
–  It carries away a pair of electrons from the 
electrophilic C atom. 
Reatividade de Compostos Orgânicos – Aula 2 
Factors Affecting SN1 Reactions – 
Solvent Effects 
•  The solvent must be capable of 
dissolving both the carbocation and the 
leaving group. 
•  SN1 reactions require highly polar 
solvents that strongly solvate ions. 
•  Typical solvents: water, an alcohol, 
acetone (to help the alkyl halide to 
dissolve), or a mix. 
Reatividade de Compostos Orgânicos – Aula 2 
SN1 Reactions - Summary 
² The structure of the carbocation is the most 
important factor: 
² Relative rates for SN1: 3°>2°>”1°”. 
² The nucleophile is typically weak or moderate in 
strength. 
² The LG should be excellent. 
² The solvent be polar and protic to stabilize the 
carbocation and LG. 
² Products will exhibit racemization and possibly 
rearrangements. 
Reatividade de Compostos Orgânicos – Aula 2 
• 1
7 
2. MECANISMO SN2: 
 
SN2: substitution, nucleophilic, bimolecular 
velocidade = k[substrato].[Nu] 
REAÇÃO DE 2ª. ORDEM 
Reatividade de Compostos Orgânicos – Aula 2 
26 
Reaction coordinate diagrams for (a) the SN2 reaction of 
methyl bromide and (b) an SN2 reaction of a sterically 
hindered alkyl bromide 
Reatividade de Compostos Orgânicos – Aula 2 
27 
CH3Br 
CH3CH2Br 
(CH3)2CHBr 
(CH3)3CBr 
Decreasing SN2 Reactivity 
Reatividade de Compostos Orgânicos – Aula 2 
28 
 
Stereochemistry of SN2 Reactions 
Inversion of Configuration 
nucleophile attacks carbon 
from side opposite bond 
to the leaving group 
three-dimensional 
arrangement of bonds in 
product is opposite to 
that of reactant 
Reatividade de Compostos Orgânicos – Aula 2 
29 
Inversion of configuration (Walden inversion) in an 
SN2 reaction is due to back side attack 
Reatividade de Compostos Orgânicos – Aula 2 
30 
A stereospecific reaction is one in which 
stereoisomeric starting materials give 
stereoisomeric products. 
The reaction of 2-bromoctane with NaOH 
(in ethanol-water) is stereospecific. 
 (+)-2-Bromooctane (–)-2-Octanol 
 (–)-2-Bromooctane (+)-2-Octanol 
Stereospecific Reaction 
Reatividade de Compostos Orgânicos – Aula 2 
31 Q
U
ÍM
IC
A
 O
R
G
Â
N
IC
A
 2
 
 -
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ro
fa
. 
R
o
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a
n
n
a
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a
rq
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s
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i
SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO
CC
HH
CHCH33
BrBr
CHCH33(CH(CH22))55
CC
HH
CHCH33
HOHO
NaOHNaOH
((SS)-(+)-2-Bromooctane)-(+)-2-Bromooctane
Stereospecific Reaction
Stereospecific Reaction 
Reatividade de Compostos Orgânicos – Aula 2 
32 
A bulky substituent in the alkyl halide reduces the 
reactivity of the alkyl halide: steric hindrance 
Q
U
ÍM
IC
A
 O
R
G
Â
N
IC
A
 2
 
 -
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ro
fa
. 
R
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a
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a
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a
rq
u
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s
 M
u
z
z
i
SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO
A bulky substituent in the alkyl halide reduces the
reactivity of the alkyl halide: steric hindrance
Q
U
ÍM
IC
A
 O
R
G
Â
N
IC
A
 2
 
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ro
fa
. 
R
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i
SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO
A bulky substituent in the alkyl halide reduces the
reactivity of the alkyl halide: steric hindrance
A velocidade da SN2 é governada por fatores estéricos 
Reatividade de Compostos Orgânicos – Aula 2 
33 
RBr + LiI RI + LiBr 
Effect of chain branching on rate of SN2 substitution 
Alkyl Structure Relative 
bromide rate 
Ethyl CH3CH2Br 1.0 
Propyl CH3CH2CH2Br 0.8 
Isobutyl (CH3)2CHCH2Br 0.036 
Neopentyl (CH3)3CCH2Br 0.00002 
Reatividade de Compostos Orgânicos – Aula 2 
34 
Factors Affecting SN2 Reactions 
The Leaving Group 
Section 10.3 Factors Affecting Reactions 367SN2
The weaker the base,
the better it is as a leaving group.
10.3 Factors Affecting Reactions
The Leaving Group
If an alkyl iodide, an alkyl bromide, an alkyl chloride, and an alkyl fluoride (all with
the same alkyl group) were allowed to react with the same nucleophile under the same
conditions, we would find that the alkyl iodide is the most reactive and the alkyl fluo-
ride is the least reactive.
The only difference among these four reactions is the nature of the leaving group.
From the relative reaction rates, we can see that the iodide ion is the best leaving
group and the fluoride ion is the worst. This brings us to an important rule in organic
chemistry—one that you will see frequently: The weaker the basicity of a group, the
better is its leaving ability. The reason leaving ability depends on basicity is because
weak bases are stable bases—they readily bear the electrons they formerly shared
with a proton (Section 1.18). Because weak bases don’t share their electrons well, a
weak base is not bonded as strongly to the carbon as a strong base would be and
a weaker bond is more easily broken (Section 1.13).
We have seen that the halide ions have the following relative basicities (or relative
stabilities) because larger atoms are better able to stabilize their negative charge
(Section 1.18):
Because stable (weak) bases are better leaving groups, the halides have the following
relative leaving abilities:
Therefore, alkyl iodides are the most reactive of the alkyl halides, and alkyl fluorides are
the least reactive. In fact, the fluoride ion is such a strong base that alkyl fluorides
essentially do not undergo reactions.
In Section 10.1, we saw that it is the polar carbon–halogen bond that causes alkyl
halides to undergo substitution reactions. Carbon and iodine, however, have the same
electronegativity. (See Table 1.3 on p. 10.) Why, then, does an alkyl halide undergo a
substitution reaction? We know that larger atoms are more polarizable than smaller
atoms. (Recall from Section 2.9 that polarizability is a measure of how easily an
atom’s electron cloud can be distorted.) The high polarizability of the large iodine
atom causes it to react as if it were polar even though, on the basis of the electronega-
tivity of the atoms, the bond is nonpolar.
relative reactivities of alkyl halides in an SN2 reaction
RI > RBr > RCl > RF least reactivemost reactive
SN2
relative leaving abilities of the halide ions
I− > Br− > Cl− > F− worst leaving
group
best leaving
group
relative basicities of the halide ions
I− < Br− < Cl− < F−
strongest base,
least stable base
weakest base,
most stable base
RCH2I RCH2OH+ +HO− I−
relative rates of reaction
30,000
RCH2Br RCH2OH+ +HO− Br− 10,000
RCH2Cl RCH2OH+ +HO− Cl− 200
RCH2F RCH2OH+ +HO− F− 1
SN2
Stable bases are weak bases.
BRUI10-359_399r2 18-02-2003 2:50 PM Page 367
Reatividade de Compostos Orgânicos – Aula 2 
35 
Section 10.3 Factors Affecting Reactions 367SN2
The weaker the base,
the better it is as a leaving group.
10.3 Factors Affecting Reactions
The Leaving Group
If an alkyl iodide, an alkyl bromide, an alkyl chloride, and an alkyl fluoride (all with
the same alkyl group) were allowed to react with the same nucleophile under the same
conditions, we would find that the alkyl iodide is the most reactive and the alkyl fluo-
ride is the least reactive.
The onlydifference among these four reactions is the nature of the leaving group.
From the relative reaction rates, we can see that the iodide ion is the best leaving
group and the fluoride ion is the worst. This brings us to an important rule in organic
chemistry—one that you will see frequently: The weaker the basicity of a group, the
better is its leaving ability. The reason leaving ability depends on basicity is because
weak bases are stable bases—they readily bear the electrons they formerly shared
with a proton (Section 1.18). Because weak bases don’t share their electrons well, a
weak base is not bonded as strongly to the carbon as a strong base would be and
a weaker bond is more easily broken (Section 1.13).
We have seen that the halide ions have the following relative basicities (or relative
stabilities) because larger atoms are better able to stabilize their negative charge
(Section 1.18):
Because stable (weak) bases are better leaving groups, the halides have the following
relative leaving abilities:
Therefore, alkyl iodides are the most reactive of the alkyl halides, and alkyl fluorides are
the least reactive. In fact, the fluoride ion is such a strong base that alkyl fluorides
essentially do not undergo reactions.
In Section 10.1, we saw that it is the polar carbon–halogen bond that causes alkyl
halides to undergo substitution reactions. Carbon and iodine, however, have the same
electronegativity. (See Table 1.3 on p. 10.) Why, then, does an alkyl halide undergo a
substitution reaction? We know that larger atoms are more polarizable than smaller
atoms. (Recall from Section 2.9 that polarizability is a measure of how easily an
atom’s electron cloud can be distorted.) The high polarizability of the large iodine
atom causes it to react as if it were polar even though, on the basis of the electronega-
tivity of the atoms, the bond is nonpolar.
relative reactivities of alkyl halides in an SN2 reaction
RI > RBr > RCl > RF least reactivemost reactive
SN2
relative leaving abilities of the halide ions
I− > Br− > Cl− > F− worst leaving
group
best leaving
group
relative basicities of the halide ions
I− < Br− < Cl− < F−
strongest base,
least stable base
weakest base,
most stable base
RCH2I RCH2OH+ +HO− I−
relative rates of reaction
30,000
RCH2Br RCH2OH+ +HO− Br− 10,000
RCH2Cl RCH2OH+ +HO− Cl− 200
RCH2F RCH2OH+ +HO− F− 1
SN2
Stable bases are weak bases.
BRUI10-359_399r2 18-02-2003 2:50 PM Page 367
Reatividade de Compostos Orgânicos – Aula 2 
36 
368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
The Nucleophile
When we talk about atoms or molecules that have lone-pair electrons, sometimes we
call them bases and sometimes we call them nucleophiles. What is the difference be-
tween a base and a nucleophile?
Basicity is a measure of how well a compound (a base) shares its lone pair with a
proton. The stronger the base, the better it shares its electrons. Basicity is measured by
an equilibrium constant (the acid dissociation constant, ) that indicates the tenden-
cy of the conjugate acid of the base to lose a proton (Section 1.17).
Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to
attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k).
In the case of an reaction, nucleophilicity is a measure of how readily the nucle-
ophile attacks an hybridized carbon bonded to a leaving group.
When comparing molecules with the same attacking atom, there is a direct relation-
ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For
example, a species with a negative charge is a stronger base and a better nucleophile
than a species with the same attacking atom that is neutral. Thus is a stronger
base and a better nucleophile than 
stronger base, weaker base,
better nucleophile poorer nucleophile
When comparing molecules with attacking atoms of approximately the same size,
the stronger bases are again the better nucleophiles. The atoms across the second row of
the periodic table have approximately the same size. If hydrogens are attached to the
second-row elements, the resulting compounds have the following relative acidities
(Section 1.18):
Consequently, the conjugate bases have the following relative base strengths and
nucleophilicities:
Note that the amide anion is the strongest base, as well as the best nucleophile.
When comparing molecules with attacking atoms that are very different in size, an-
other factor comes into play—the polarizability of the atom. Because the electrons are
farther away in the larger atom, they are not held as tightly and can, therefore, move
more freely toward a positive charge. As a result, the electrons are able to overlap from
farther away with the orbital of carbon, as shown in Figure 10.5. This results in a
greater degree of bonding in the transition state, making it more stable. Whether the
greater polarizability of the larger atoms makes up for their decreased basicity depends
on the conditions under which the reaction is carried out.
relative base strengths and relative nucleophilicities
−NH2 > HO− > F−strongest
base
best nucleophile
relative acid strengths
NH3 < H2O < HFweakest
acid
HO-
CH3O-
-NH2
CH3CH2NH-
 
7
7
7
7
 
H2O
CH3OH
NH3
CH3CH2NH2
H2O.
HO-
sp3
SN2
Ka
BRUI10-359_399r2 18-02-2003 2:50 PM Page 368
368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
The Nucleophile
When we talk about atoms or molecules that have lone-pair electrons, sometimes we
call them bases and sometimes we call them nucleophiles. What is the difference be-
tween a base and a nucleophile?
Basicity is a measure of how well a compound (a base) shares its lone pair with a
proton. The stronger the base, the better it shares its electrons. Basicity is measured by
an equilibrium constant (the acid dissociation constant, ) that indicates the tenden-
cy of the conjugate acid of the base to lose a proton (Section 1.17).
Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to
attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k).
In the case of an reaction, nucleophilicity is a measure of how readily the nucle-
ophile attacks an hybridized carbon bonded to a leaving group.
When comparing molecules with the same attacking atom, there is a direct relation-
ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For
example, a species with a negative charge is a stronger base and a better nucleophile
than a species with the same attacking atom that is neutral. Thus is a stronger
base and a better nucleophile than 
stronger base, weaker base,
better nucleophile poorer nucleophile
When comparing molecules with attacking atoms of approximately the same size,
the stronger bases are again the better nucleophiles. The atoms across the second row of
the periodic table have approximately the same size. If hydrogens are attached to the
second-row elements, the resulting compounds have the following relative acidities
(Section 1.18):
Consequently, the conjugate bases have the following relative base strengths and
nucleophilicities:
Note that the amide anion is the strongest base, as well as the best nucleophile.
When comparing molecules with attacking atoms that are very different in size, an-
other factor comes into play—the polarizability of the atom. Because the electrons are
farther away in the larger atom, they are not held as tightly and can, therefore, move
more freely toward a positive charge. As a result, the electrons are able to overlap from
farther away with the orbital of carbon, as shown in Figure 10.5. This results in a
greater degree of bonding in the transition state, making it morestable. Whether the
greater polarizability of the larger atoms makes up for their decreased basicity depends
on the conditions under which the reaction is carried out.
relative base strengths and relative nucleophilicities
−NH2 > HO− > F−strongest
base
best nucleophile
relative acid strengths
NH3 < H2O < HFweakest
acid
HO-
CH3O-
-NH2
CH3CH2NH-
 
7
7
7
7
 
H2O
CH3OH
NH3
CH3CH2NH2
H2O.
HO-
sp3
SN2
Ka
BRUI10-359_399r2 18-02-2003 2:50 PM Page 368
368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
The Nucleophile
When we talk about atoms or molecules that have lone-pair electrons, sometimes we
call them bases and sometimes we call them nucleophiles. What is the difference be-
tween a base and a nucleophile?
Basicity is a measure of how well a compound (a base) shares its lone pair with a
proton. The stronger the base, the better it shares its electrons. Basicity is measured by
an equilibrium constant (the acid dissociation constant, ) that indicates the tenden-
cy of the conjugate acid of the base to lose a proton (Section 1.17).
Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to
attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k).
In the case of an reaction, nucleophilicity is a measure of how readily the nucle-
ophile attacks an hybridized carbon bonded to a leaving group.
When comparing molecules with the same attacking atom, there is a direct relation-
ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For
example, a species with a negative charge is a stronger base and a better nucleophile
than a species with the same attacking atom that is neutral. Thus is a stronger
base and a better nucleophile than 
stronger base, weaker base,
better nucleophile poorer nucleophile
When comparing molecules with attacking atoms of approximately the same size,
the stronger bases are again the better nucleophiles. The atoms across the second row of
the periodic table have approximately the same size. If hydrogens are attached to the
second-row elements, the resulting compounds have the following relative acidities
(Section 1.18):
Consequently, the conjugate bases have the following relative base strengths and
nucleophilicities:
Note that the amide anion is the strongest base, as well as the best nucleophile.
When comparing molecules with attacking atoms that are very different in size, an-
other factor comes into play—the polarizability of the atom. Because the electrons are
farther away in the larger atom, they are not held as tightly and can, therefore, move
more freely toward a positive charge. As a result, the electrons are able to overlap from
farther away with the orbital of carbon, as shown in Figure 10.5. This results in a
greater degree of bonding in the transition state, making it more stable. Whether the
greater polarizability of the larger atoms makes up for their decreased basicity depends
on the conditions under which the reaction is carried out.
relative base strengths and relative nucleophilicities
−NH2 > HO− > F−strongest
base
best nucleophile
relative acid strengths
NH3 < H2O < HFweakest
acid
HO-
CH3O-
-NH2
CH3CH2NH-
 
7
7
7
7
 
H2O
CH3OH
NH3
CH3CH2NH2
H2O.
HO-
sp3
SN2
Ka
BRUI10-359_399r2 18-02-2003 2:50 PM Page 368
Reatividade de Compostos Orgânicos – Aula 2 
37 
Section 10.3 Factors Affecting Reactions 369SN2
H
XC
H H
C
H
XC
H H
I−
transition state
H
XC
H H
C
H
XC
H H
F−
transition state
more bonding
little bonding
++
++
δ−
δ−
> Figure 10.5
An iodide ion is larger and more
polarizable than a fluoride ion.
Therefore, the relatively loosely
held electrons of the iodide ion can
overlap from farther away with the
orbital of carbon undergoing
nucleophilic attack. The tightly
bound electrons of the fluoride ion
cannot start to overlap until the
atoms are closer together.
If the reaction is carried out in the gas phase, the direct relationship between basici-
ty and nucleophilicity is maintained—the stronger bases are still the best nucleophiles.
If, however, the reaction is carried out in a protic solvent—meaning the solvent mole-
cules have a hydrogen bonded to an oxygen or to a nitrogen—the relationship between
basicity and nucleophilicity becomes inverted. The largest atom is the best nucleophile
even though it is the weakest base. Therefore, iodide ion is the best nucleophile of the
halide ions in a protic solvent.
PROBLEM 4!
a. Which is a stronger base, or 
b. Which is a better nucleophile in an aqueous solution?
The Effect of the Solvent on Nucleophilicity
Why, in a protic solvent, is the smallest atom the poorest nucleophile even though it is
the strongest base? How does a protic solvent make strong bases less nucleophilic?
When a negatively charged species is placed in a protic solvent, the ion becomes sol-
vated (Section 2.9). Protic solvents are hydrogen bond donors. The solvent molecules
arrange themselves so that their partially positively charged hydrogens point toward
the negatively charged species. The interaction between the ion and the dipole of the
protic solvent is called an ion–dipole interaction. Because the solvent shields the nu-
cleophile, at least one of the ion–dipole interactions must be broken before the nucle-
ophile can participate in an reaction. Weak bases interact weakly with protic
solvents, whereas strong bases interact more strongly because they are better at shar-
ing their electrons. It is easier, therefore, to break the ion–dipole interactions between
SN2
RS-?RO-
increasing
basicity
increasing
nucleophilicity
in the gas phase
increasing
size
increasing
nucleophilicity
in a protic
solvent
F−
Cl−
Br−
I−
A protic solvent contains a hydrogen
bonded to an oxygen or a nitrogen; 
it is a hydrogen bond donor.
BRUI10-359_399r2 18-02-2003 2:50 PM Page 369
Reatividade de Compostos Orgânicos – Aula 2 
38 
Section 10.3 Factors Affecting Reactions 369SN2
H
XC
H H
C
H
XC
H H
I−
transition state
H
XC
H H
C
H
XC
H H
F−
transition state
more bonding
little bonding
++
++
δ−
δ−
> Figure 10.5
An iodide ion is larger and more
polarizable than a fluoride ion.
Therefore, the relatively loosely
held electrons of the iodide ion can
overlap from farther away with the
orbital of carbon undergoing
nucleophilic attack. The tightly
bound electrons of the fluoride ion
cannot start to overlap until the
atoms are closer together.
If the reaction is carried out in the gas phase, the direct relationship between basici-
ty and nucleophilicity is maintained—the stronger bases are still the best nucleophiles.
If, however, the reaction is carried out in a protic solvent—meaning the solvent mole-
cules have a hydrogen bonded to an oxygen or to a nitrogen—the relationship between
basicity and nucleophilicity becomes inverted. The largest atom is the best nucleophile
even though it is the weakest base. Therefore, iodide ion is the best nucleophile of the
halide ions in a protic solvent.
PROBLEM 4!
a. Which is a stronger base, or 
b. Which is a better nucleophile in an aqueous solution?
The Effect of the Solvent on Nucleophilicity
Why, in a protic solvent, is the smallest atom the poorest nucleophile even though it is
the strongest base? How does a protic solvent make strong bases less nucleophilic?
When a negatively charged species is placed in a protic solvent, the ion becomes sol-
vated (Section 2.9). Protic solvents are hydrogen bond donors. The solvent molecules
arrange themselves so that their partially positively charged hydrogens point toward
the negatively charged species. The interaction betweenthe ion and the dipole of the
protic solvent is called an ion–dipole interaction. Because the solvent shields the nu-
cleophile, at least one of the ion–dipole interactions must be broken before the nucle-
ophile can participate in an reaction. Weak bases interact weakly with protic
solvents, whereas strong bases interact more strongly because they are better at shar-
ing their electrons. It is easier, therefore, to break the ion–dipole interactions between
SN2
RS-?RO-
increasing
basicity
increasing
nucleophilicity
in the gas phase
increasing
size
increasing
nucleophilicity
in a protic
solvent
F−
Cl−
Br−
I−
A protic solvent contains a hydrogen
bonded to an oxygen or a nitrogen; 
it is a hydrogen bond donor.
BRUI10-359_399r2 18-02-2003 2:50 PM Page 369
370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
An aprotic solvent does not contain a
hydrogen bonded to either an oxygen
or a nitrogen; it is not a hydrogen bond
donor.
an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger
base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in
a protic solvent (Table 10.2).
Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol-
vent because there would be no ion–dipole interactions between the ion and the nonpo-
lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they
can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl-
sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it
does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi-
tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic
polar solvent have a partial negative charge on their surface that can solvate cations, but
the partial positive charge is on the inside of the molecule, which makes it less accessi-
ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol-
vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water.
PROBLEM 5!
Indicate whether each of the following solvents is protic or aprotic:
a. chloroform c. acetic acid 
b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3)
(CH3COOH)(CHCl3)
δ−δ−
δ−
δ−
δ+
δ+
δ+
δ+
δ−
δ+
δ+
δ+
δ+ H
H OY−
H
H
O
H
H
O
H
HO
ion–dipole interactions 
between a nucleophile 
and water
3-D Molecules:
N,N-Dimethylformamide;
Dimethyl sulfoxide
N,N-dimethylformamide
DMF
C
H CH3
Oδ−
CH3
N
dimethyl sulfoxide
DMSO
H3C CH3
the δ− is on the surface
of the molecule
 S
Oδ−
δ+
δ+
the δ+ is not
very accessible
δ+
δ+
δ+
δ−
δ−δ+
δ−
CH3
CH3 CH3
CH3 CH3
CH3 CH3
CH3
S
S
O
δ−
O OK+
O
S S
DMSO can solvate a cation better 
than it can solvate an anion 
increasing nucleophilicity
Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol
RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH
BRUI10-359_399r2 18-02-2003 2:50 PM Page 370370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
An aprotic solvent does not contain a
hydrogen bonded to either an oxygen
or a nitrogen; it is not a hydrogen bond
donor.
an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger
base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in
a protic solvent (Table 10.2).
Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol-
vent because there would be no ion–dipole interactions between the ion and the nonpo-
lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they
can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl-
sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it
does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi-
tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic
polar solvent have a partial negative charge on their surface that can solvate cations, but
the partial positive charge is on the inside of the molecule, which makes it less accessi-
ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol-
vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water.
PROBLEM 5!
Indicate whether each of the following solvents is protic or aprotic:
a. chloroform c. acetic acid 
b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3)
(CH3COOH)(CHCl3)
δ−δ−
δ−
δ−
δ+
δ+
δ+
δ+
δ−
δ+
δ+
δ+
δ+ H
H OY−
H
H
O
H
H
O
H
HO
ion–dipole interactions 
between a nucleophile 
and water
3-D Molecules:
N,N-Dimethylformamide;
Dimethyl sulfoxide
N,N-dimethylformamide
DMF
C
H CH3
Oδ−
CH3
N
dimethyl sulfoxide
DMSO
H3C CH3
the δ− is on the surface
of the molecule
 S
Oδ−
δ+
δ+
the δ+ is not
very accessible
δ+
δ+
δ+
δ−
δ−δ+
δ−
CH3
CH3 CH3
CH3 CH3
CH3 CH3
CH3
S
S
O
δ−
O OK+
O
S S
DMSO can solvate a cation better 
than it can solvate an anion 
increasing nucleophilicity
Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol
RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH
BRUI10-359_399r2 18-02-2003 2:50 PM Page 370
Reatividade de Compostos Orgânicos – Aula 2 
39 
390 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
SOLVATION EFFECTS
The tremendous amount of energy that is provid-
ed by solvation can be appreciated by considering
the energy required to break the crystal lattice of sodium chlo-
ride (table salt) (Figure 1.1). In the absence of a solvent, sodium
chloride must be heated to more than 800 °C to overcome the
forces that hold the oppositely charged ions together. However,
sodium chloride readily dissolves in water at room temperature
because solvation of the and ions by water provides
the energy necessary to separate the ions.
Cl-Na+
Stabilization of charges by solvent interaction plays an important role in organic re-
actions. For example, when an alkyl halide undergoes an reaction, the first step is
dissociation of the carbon–halogen bond to form a carbocation and a halide ion. Ener-
gy is required to break the bond, but with no bonds being formed, where does the en-
ergy come from? If the reaction is carried out in a polar solvent, the ions that are
produced are solvated. The energy associated with a single ion–dipole interaction is
small, but the additive effect of all the ion–dipole interactions involved in stabilizing a
charged species by the solvent represents a great deal of energy. These ion–dipole in-
teractions provide much of the energy necessary for dissociation of the carbon–halo-
gen bond. So in an reaction, the alkyl halide does not fall apart spontaneously, but
rather, polar solvent molecules pull it apart. An reaction, therefore, cannot take
place in a nonpolar solvent. It also cannot take place in the gas phase, where there are
no solvent molecules and, consequently, no solvation effects.
SN1
SN1
SN1
Table 10.7 The Dielectric Constants of Some Common Solvents
Solvent Structure Dielectric constantAbbreviation
(ε , at 25 °C)
Boiling point
(°C)
Water H2O 79
Formic acid HCOOH 59
Methanol CH3OH 33
Ethanol CH3CH2OH 25
tert-Butyl alcohol (CH3)3COH 11
Acetic acid CH3COOH 6
Dimethyl sulfoxide (CH3)2SO 47
Acetonitrile CH3CN 38
Dimethylformamide (CH3)2NCHO 37
Hexamethylphosphoric acid triamide [(CH3)2N]3PO 30
Acetone (CH3)2CO 21
Eth
Dichloromethane
Tetrahydrofuran
yl acetateCH
CH2Cl2
3COOCH2CH3 6
9.1
Diethyl ether CH3CH2OCH2CH3 4.3
Benzene 2.3
Hexane CH3(CH2)4CH3 1.9
100
100.6
64.7
78.3
82.3
117.9
189
81.6
153
233
56.3
77.1
40
7.6 66
34.6
80.1
68.7
—
—
MeOH
EtOH
tert-BuOH
HOAc
DMSO
MeCN
DMF
HMPA
Me2CO
THF
EtOAc
—
Et2O
O
—
—
Protic solvents
Aprotic solvents
BRUI10-359_399r2 18-02-2003 2:50 PM Page 390
Reatividade de Compostos Orgânicos – Aula 2 
40 
370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
An aprotic solvent does not contain a
hydrogen bonded to either an oxygen
or a nitrogen; it is not a hydrogen bond
donor.
an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger
base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in
a protic solvent (Table 10.2).
Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol-
vent because there would be no ion–dipole interactions between the ion and the nonpo-
lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they
can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl-
sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it
does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi-
tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic
polar solvent have a partial negative charge on their surface that can solvate cations, but
the partial positive charge is on the inside of the molecule, which makes it less accessi-
ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol-
vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water.
PROBLEM 5!
Indicate whether each of the following solvents is protic or aprotic:
a. chloroform c. acetic acid 
b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3)
(CH3COOH)(CHCl3)
δ−δ−
δ−
δ−
δ+
δ+
δ+
δ+
δ−
δ+
δ+
δ+
δ+ H
H OY−
H
H
O
H
H
O
H
HO
ion–dipole interactions 
between a nucleophile 
and water
3-D Molecules:
N,N-Dimethylformamide;
Dimethyl sulfoxide
N,N-dimethylformamide
DMF
C
H CH3
Oδ−
CH3
N
dimethyl sulfoxide
DMSO
H3C CH3
the δ− is on the surface
of the molecule
 S
Oδ−
δ+
δ+
the δ+ is not
very accessible
δ+
δ+
δ+
δ−
δ−δ+
δ−
CH3
CH3 CH3
CH3 CH3
CH3 CH3
CH3
S
S
O
δ−
O OK+
O
S S
DMSO can solvate a cation better 
than it can solvate an anion 
increasing nucleophilicity
Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol
RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH
BRUI10-359_399r2 18-02-2003 2:50 PM Page 370
370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
An aprotic solvent does not contain a
hydrogen bonded to either an oxygen
or a nitrogen; it is not a hydrogen bond
donor.
an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger
base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in
a protic solvent (Table 10.2).
Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol-
vent because there would be no ion–dipole interactions between the ion and the nonpo-
lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they
can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl-
sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it
does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi-
tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic
polar solvent have a partial negative charge on their surface that can solvate cations, but
the partial positive charge is on the inside of the molecule, which makes it less accessi-
ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol-
vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water.
PROBLEM 5!
Indicate whether each of the following solvents is protic or aprotic:
a. chloroform c. acetic acid 
b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3)
(CH3COOH)(CHCl3)
δ−δ−
δ−
δ−
δ+
δ+
δ+
δ+
δ−
δ+
δ+
δ+
δ+ H
H OY−
H
H
O
H
H
O
H
HO
ion–dipole interactions 
between a nucleophile 
and water
3-D Molecules:
N,N-Dimethylformamide;
Dimethyl sulfoxide
N,N-dimethylformamide
DMF
C
H CH3
Oδ−
CH3
N
dimethyl sulfoxide
DMSO
H3C CH3
the δ− is on the surface
of the molecule
 S
Oδ−
δ+
δ+
the δ+ is not
very accessible
δ+
δ+
δ+
δ−
δ−δ+
δ−
CH3
CH3 CH3
CH3 CH3
CH3 CH3
CH3
S
S
O
δ−
O OK+
O
S S
DMSO can solvate a cation better 
than it can solvate an anion 
increasing nucleophilicity
Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol
RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH
BRUI10-359_399r2 18-02-2003 2:50 PM Page 370
Reatividade de Compostos Orgânicos – Aula 2 
41 
Section 10.3 Factors Affecting Reactions 371SN2
ethoxide ion
tert-butoxide ion
Nucleophilicity Is Affected by Steric Effects
Base strength is relatively unaffected by steric effects because a base removes a
relatively unhindered proton. The strength of a base depends only on how well the
base shares its electrons with a proton. Thus, tert-butoxide ion is a stronger base
than ethoxide ion since tert-butanol is a weaker acid than ethanol
Steric effects, on the other hand, do affect nucleophilicity. A bulky nucleophile
cannot approach the back side of a carbon as easily as a less sterically hindered nucle-
ophile can. Thus, the bulky tert-butoxide ion, with its three methyl groups, is a poorer
nucleophile than ethoxide ion even though tert-butoxide ion is a stronger base.
PROBLEM 6 SOLVED
List the following species in order of decreasing nucleophilicity in an aqueous solution:
SOLUTION Let’s first divide the nucleophiles into groups. We have one nucleophile
with a negatively charged sulfur, three with negatively charged oxygens, and one with a
neutral oxygen. We know that in the polar aqueous solvent, the one with the negatively
charged sulfur is the most nucleophilic because sulfur is larger than oxygen. We also know
that the poorest nucleophile is the one with the neutral oxygen atom. So now, to complete
the problem, we need to rank the three nucleophiles with negatively charged oxygens in
order of the ’s of their conjugate acids. A carboxylic acid is a stronger acid than
phenol, which is a stronger acid than water (Section 7.10). Because water is the weakest
acid, its conjugate base is the strongest base and the best nucleophile. Thus, the relative
nucleophilicities are
PROBLEM 7!
For each of the following pairs of reactions, indicate which reaction occurs faster:
a.
b.
c.
d. CH3CH2Cl + I− CH3CH2Br + I−or
(in ethanol)
CH3CH2Cl + CH3O− CH3CH2Cl + CH3S−or
CH3CHCH2Br CH3CH2CHBr+ HO−
CH3
+ HO−or
CH3
CH3CH2Br CH3CH2Br+ H2O + HO−or
SN2
CH3CO−>CH3S− > > >HO− CH3OHO−
O
pKa
CH3CO−CH3OH HO−O− CH3S−
O
ethoxide ion
better nucleophile
tert-butoxide ion
stronger base
CH3CO−CH3CH2O−
CH3
CH3
1pKa = 15.92.
1pKa = 182
BRUI10-359_399r2 18-02-2003 2:50 PM Page 371 Section 10.3 Factors Affecting Reactions 371SN2
ethoxide ion
tert-butoxide ion
Nucleophilicity Is Affected by Steric Effects
Base strength is relatively unaffected by steric effects because a base removes a
relatively unhindered proton. The strength of a base dependsonly on how well the
base shares its electrons with a proton. Thus, tert-butoxide ion is a stronger base
than ethoxide ion since tert-butanol is a weaker acid than ethanol
Steric effects, on the other hand, do affect nucleophilicity. A bulky nucleophile
cannot approach the back side of a carbon as easily as a less sterically hindered nucle-
ophile can. Thus, the bulky tert-butoxide ion, with its three methyl groups, is a poorer
nucleophile than ethoxide ion even though tert-butoxide ion is a stronger base.
PROBLEM 6 SOLVED
List the following species in order of decreasing nucleophilicity in an aqueous solution:
SOLUTION Let’s first divide the nucleophiles into groups. We have one nucleophile
with a negatively charged sulfur, three with negatively charged oxygens, and one with a
neutral oxygen. We know that in the polar aqueous solvent, the one with the negatively
charged sulfur is the most nucleophilic because sulfur is larger than oxygen. We also know
that the poorest nucleophile is the one with the neutral oxygen atom. So now, to complete
the problem, we need to rank the three nucleophiles with negatively charged oxygens in
order of the ’s of their conjugate acids. A carboxylic acid is a stronger acid than
phenol, which is a stronger acid than water (Section 7.10). Because water is the weakest
acid, its conjugate base is the strongest base and the best nucleophile. Thus, the relative
nucleophilicities are
PROBLEM 7!
For each of the following pairs of reactions, indicate which reaction occurs faster:
a.
b.
c.
d. CH3CH2Cl + I− CH3CH2Br + I−or
(in ethanol)
CH3CH2Cl + CH3O− CH3CH2Cl + CH3S−or
CH3CHCH2Br CH3CH2CHBr+ HO−
CH3
+ HO−or
CH3
CH3CH2Br CH3CH2Br+ H2O + HO−or
SN2
CH3CO−>CH3S− > > >HO− CH3OHO−
O
pKa
CH3CO−CH3OH HO−O− CH3S−
O
ethoxide ion
better nucleophile
tert-butoxide ion
stronger base
CH3CO−CH3CH2O−
CH3
CH3
1pKa = 15.92.
1pKa = 182
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42 
372 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
10.4 The Reversibility of an Reaction
Many different kinds of nucleophiles can react with alkyl halides. Therefore, a wide
variety of organic compounds can be synthesized by means of reactions.
It may seem that the reverse of each of these reactions can also occur via nucleophilic
substitution. In the first reaction, for example, ethyl chloride reacts with hydroxide ion
to form ethyl alcohol and a chloride ion. The reverse reaction would appear to satisfy the
requirements for a nucleophilic substitution reaction, given that chloride ion is a nucle-
ophile and ethyl alcohol has an leaving group. But ethyl alcohol and chloride ion
do not react.
Why does a nucleophilic substitution reaction take place in one direction, but not in
the other? We can answer this question by comparing the leaving tendency of in
the forward direction and the leaving tendency of in the reverse direction. Com-
paring leaving tendencies involves comparing basicities. Most people find it easier to
compare the acid strengths of the conjugate acids (Table 10.3), so we will compare the
acid strengths of HCl and HCl is a much stronger acid than which means
that is a much weaker base than (Remember, the stronger the acid, the
weaker is its conjugate base.) Because it is a weaker base, is a better leaving
group. Consequently, can displace in the forward reaction, but cannot
displace in the reverse reaction. The reaction proceeds in the direction that al-
lows the stronger base to displace the weaker base (the best leaving group).
If the difference between the basicities of the nucleophile and the leaving group is
not very large, the reaction will be reversible. For example, in the reaction of ethyl bro-
mide with iodide ion, is the leaving group in one direction and is the leaving
group in the other direction. The reaction is reversible because the values of the
conjugate acids of the two leaving groups are similar ( of of
see Table 10.3).
You can drive a reversible reaction toward the desired products by removing one of
the products as it is formed. Recall that the concentrations of the reactants and
products at equilibrium are governed by the equilibrium constant of the reaction
+ +CH3CH2Br Br−CH3CH2II−
an SN2 reaction is
reversible when the
basicities of the leaving
groups are similar
HI = -10;
HBr = -9; pKapKa
pKa
I-Br-
HO-
Cl-Cl-HO-
Cl-
HO-.Cl-
H2O,H2O.
HO-
Cl-
HO-
NN
CR
an alcohol
a thiol
CH3CH2Cl CH3CH2OH Cl−+ +HO−
CH3CH2Br CH3CH2SH Br−+ +HS−
an ether
a thioether
CH3CH2I CH3CH2OR I−+ +RO−
CH3CH2Br CH3CH2SR Br−+ +RS−
a primary amine
CH3CH2Cl CH3CH2NH2 Cl−+ +−NH2
an alkyne
a nitrile
CH3CH2Br CH3CH2C Br−CR+ +−C
CH3CH2I CH3CH2C I−+ +−C
SN2
SN2
An reaction proceeds in the
direction that allows the stronger 
base to displace the weaker base.
SN2
BRUI10-359_399r2 18-02-2003 2:50 PM Page 372
Reatividade de Compostos Orgânicos – Aula 2 
43 
PROBLEM 22!
Which alkyl halide would you expect to be more reactive in an reaction with a given
nucleophile? In each case, you can assume that both alkyl halides have the same stability.
a. or
b. or
c.
d.
e.
f.
g.
PROBLEM 23!
For each of the pairs in Problem 22, which compound would be more reactive in an 
reaction?
PROBLEM 24!
Show the configuration of the products obtained from the following reaction:
a. under conditions that favor an reaction
b. under conditions that favor an reaction
10.9 Competition Between and Reactions
The characteristics of and reactions are summarized in Table 10.5. Remem-
ber that the “2” in and the “1” in refer to the molecularity—how many
molecules are involved in the rate-determining step. Thus, the rate-determining step of
an reaction is bimolecular, whereas the rate-determining step of an reaction isSN1SN2
“SN1”“SN2”
SN1SN2
SN1SN2
CH3OHCH3O−+CHCH2BrCH3CH
SN1
SN2
SN1
CH3CH CCH3 or
Br
CH3CH CHCHCH3
Br
CH2Br or Br
CH2CH2Br or CH2CHCH3
Br
CH3CH2CH2CHBr
CH3
CH3CH2CHCH2Br
CH3
or
CH3CH2CHBr or
CH3
CH3CH2CHBr
CH2CH3
CH3OCH2ClCH3CH2CH2Cl
CH3CH2CH2ICH3CH2CH2Br
SN2
Section 10.9 Competition Between and Reactions 385SN1SN2
Table 10.5 Comparison of and ReactionsSN1SN2
A one-step mechanism A stepwise mechanism that forms a carbocation
intermediate
A bimolecular rate-determining A unimolecular rate-determining
step step
No carbocation rearrangements Carbocation rearrangements
Product has inverted configuration Products have both retained and inverted
relative to the reactant configurations relative to the reactant
Reactivity order: Reactivity order:
methyl 3° 7 2° 7 1° 7 methyl7 1° 7 2° 7 3°
SN1SN2
BRUI10-359_399r2 18-02-2003 2:50 PM Page 385
PROBLEM 22!
Which alkyl halide would you expect to be more reactive in an reaction with a given
nucleophile? In each case, you can assume that both alkyl halides have the same stability.
a. or
b. or
c.
d.
e.
f.
g.
PROBLEM 23!
For each of the pairs in Problem 22, which compound would be more reactive in an 
reaction?
PROBLEM 24!
Show the configuration of the products obtained from the following reaction:
a. under conditions that favor an reaction
b. under conditions that favor an reaction
10.9 Competition Between and Reactions
The characteristics of and reactions are summarized in Table 10.5. Remem-
ber that the “2” in and the “1” in refer to the molecularity—how many
molecules are involved in the rate-determining step. Thus, the rate-determining step of
an reaction is bimolecular, whereas the rate-determining step of an reaction isSN1SN2
“SN1”“SN2”
SN1SN2
SN1SN2
CH3OHCH3O−+CHCH2BrCH3CH
SN1
SN2
SN1
CH3CH CCH3 or
Br
CH3CH CHCHCH3
Br
CH2Br or Br
CH2CH2Br or CH2CHCH3Br
CH3CH2CH2CHBr
CH3
CH3CH2CHCH2Br
CH3
or
CH3CH2CHBr or
CH3
CH3CH2CHBr
CH2CH3
CH3OCH2ClCH3CH2CH2Cl
CH3CH2CH2ICH3CH2CH2Br
SN2
Section 10.9 Competition Between and Reactions 385SN1SN2
Table 10.5 Comparison of and ReactionsSN1SN2
A one-step mechanism A stepwise mechanism that forms a carbocation
intermediate
A bimolecular rate-determining A unimolecular rate-determining
step step
No carbocation rearrangements Carbocation rearrangements
Product has inverted configuration Products have both retained and inverted
relative to the reactant configurations relative to the reactant
Reactivity order: Reactivity order:
methyl 3° 7 2° 7 1° 7 methyl7 1° 7 2° 7 3°
SN1SN2
BRUI10-359_399r2 18-02-2003 2:50 PM Page 385
386 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides
Table 10.6 Summary of the Reactivity of Alkyl Halides in Nucleophilic 
Substitution Reactions
Methyl and 1° alkyl halides only
Vinylic and aryl halides Neither nor 
2° alkyl halides and 
1° and 2° benzylic and 1° and 2° allylic halides and 
3° alkyl halides only
3° benzylic and 3° allylic halides onlySN1
SN1
SN2SN1
SN2SN1
SN2SN1
SN2
When an alkyl halide can undergo both an reaction and an reaction, both
reactions take place simultaneously. The conditions under which the reaction is carried
out determine which of the reactions predominates. Therefore, we have some experi-
mental control over which reaction takes place.
When an alkyl halide can undergo substitution by both mechanisms, what condi-
tions favor an reaction? What conditions favor an reaction? These are
important questions to synthetic chemists because an reaction forms a single sub-
stitution product, whereas an reaction can form two substitution products if the
leaving group is bonded to an asymmetric carbon. An reaction is further compli-
cated by carbocation rearrangements. In other words, an reaction is a synthetic
chemist’s friend, but an reaction can be a synthetic chemist’s nightmare.
When the structure of the alkyl halide allows it to undergo both and reactions,
three conditions determine which reaction will predominate: (1) the concentration of the
nucleophile, (2) the reactivity of the nucleophile, and (3) the solvent in which the reaction
is carried out. To understand how the concentration and the reactivity of the nucleophile
affect whether an or an reaction predominates, we must examine the rate laws
for the two reactions. The rate constants have been given subscripts that indicate the 
reaction order.
The rate law for the reaction of an alkyl halide that can undergo both and re-
actions simultaneously is the sum of the individual rate laws.
From the rate law, you can see that increasing the concentration of the nucleophile
increases the rate of an reaction but has no effect on the rate of an reaction.SN1SN2
contribution to the rate
by an SN2 reaction
rate = +k2[alkyl halide][nucleophile] k1[alkyl halide]
contribution to the rate
by an SN1 reaction
SN1SN2
Rate law for an SN2 reaction = k2 [alkyl halide] [nucleophile]
Rate law for SN1 reaction = k1 [alkyl halide]
SN1SN2
SN1SN2
SN1
SN2
SN1
SN1
SN2
SN2SN1
SN2SN1
unimolecular. These numbers do not refer to the number of steps in the mechanism. In
fact, just the opposite is true: An reaction proceeds by a one-step concerted mech-
anism, and an reaction proceeds by a two-step mechanism with a carbocation
intermediate.
We have seen that methyl halides and primary alkyl halides undergo only reac-
tions because methyl cations and primary carbocations, which would be formed in an
reaction, are too unstable to be formed in an reaction. Tertiary alkyl halides
undergo only reactions because steric hindrance makes them unreactive in an 
reaction. Secondary alkyl halides as well as benzylic and allylic halides (unless they
are tertiary) can undergo both and reactions because they form relatively
stable carbocations and the steric hindrance associated with these alkyl halides is
generally not very great. Vinylic and aryl halides do not undergo either or 
reactions. These results are summarized in Table 10.6.
SN2SN1
SN2SN1
SN2SN1
SN2SN1
SN2
SN1
SN2
BRUI10-359_399r2 18-02-2003 2:50 PM Page 386
Reatividade de Compostos Orgânicos – Aula 2 
44 
Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383
10.8 Benzylic Halides, Allylic Halides,
Vinylic Halides, and Aryl Halides
Our discussion of substitution reactions, to this point, has been limited to methyl
halides and primary, secondary, and tertiary alkyl halides. But what about benzylic,
allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben-
zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary
benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in 
reactions because of steric hindrance.
Benzylic and allylic halides readily undergo reactions because they form
relatively stable carbocations. While primary alkyl halides (such as and
) cannot undergo reactions because their carbocations are too
unstable, primary benzylic and primary allylic halides readily undergo reactions
because their carbocations are stabilized by electron delocalization (Section 7.7).
If the resonance contributors of the allylic carbocation intermediate have different
groups bonded to their carbons, two substitution products will be obtained.
Vinylic halides and aryl halides do not undergo or reactions. They do not
undergo reactions because, as the nucleophile approaches the back side of the 
carbon, it is repelled by the electron cloud of the double bond or the aromatic ring.
a vinylic halide an aryl halide
nucleophile
C C 
Cl
HH
R
Br
nucleophile
a nucleophile is repelled
by the pi electron cloud
p
sp2SN2
SN1SN2
sp2
SN1
SN1CH3CH2CH2Br
CH3CH2Br
SN1
benzyl chloride
1-bromo-2-butene
an allylic halide
2-buten-1-ol
benzyl methyl ether
CH2Cl CH3O−+
CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+
CH2OCH3 Cl−+
SN2 conditions
SN2 conditions
SN2
SN2
CH3OHSN1CH2Cl CH2
Cl−+
CH2OCH3 H++
SN1 CH2CH2CH
Br−+
H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2
+
+ +
+
+
CH3CH CHCH2Br CH3CH CHCH2
+
CH3CHCH CH2
CH3CHCH CH2CH3CH CHCH2OH
H+
+ Br−
H2O
+ H+OH
H2O
SN1
Benzylic and allylic halides undergo 
and reactions.SN2SN1
Vinylic and aryl halides undergo 
neither nor reactions.SN2SN1
3-D Molecule:
Benzyl cation
3-D Molecule:
Allyl cation
BRUI10-359_399r2 18-02-2003 2:50 PM Page 383
Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383
10.8 Benzylic Halides, Allylic Halides,
Vinylic Halides, and Aryl Halides
Our discussion of substitution reactions, to this point, has been limited to methyl
halides and primary, secondary, and tertiary alkyl halides. But what about benzylic,
allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben-
zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary
benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in 
reactions because of steric hindrance.
Benzylic and allylic halides readily undergo reactions because they form
relatively stable carbocations. While primary alkyl halides (such as and
) cannot undergo reactions because their carbocations are too
unstable, primary benzylic and primary allylic halides readily undergo reactions
because their carbocations are stabilized by electron delocalization (Section 7.7).
If the resonance contributors of the allylic carbocation intermediate have different
groups bonded to their carbons, two substitution products will be obtained.
Vinylic halides and aryl halides do not undergo or reactions. They do not
undergo reactions because,as the nucleophile approaches the back side of the 
carbon, it is repelled by the electron cloud of the double bond or the aromatic ring.
a vinylic halide an aryl halide
nucleophile
C C 
Cl
HH
R
Br
nucleophile
a nucleophile is repelled
by the pi electron cloud
p
sp2SN2
SN1SN2
sp2
SN1
SN1CH3CH2CH2Br
CH3CH2Br
SN1
benzyl chloride
1-bromo-2-butene
an allylic halide
2-buten-1-ol
benzyl methyl ether
CH2Cl CH3O−+
CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+
CH2OCH3 Cl−+
SN2 conditions
SN2 conditions
SN2
SN2
CH3OHSN1CH2Cl CH2
Cl−+
CH2OCH3 H++
SN1 CH2CH2CH
Br−+
H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2
+
+ +
+
+
CH3CH CHCH2Br CH3CH CHCH2
+
CH3CHCH CH2
CH3CHCH CH2CH3CH CHCH2OH
H+
+ Br−
H2O
+ H+OH
H2O
SN1
Benzylic and allylic halides undergo 
and reactions.SN2SN1
Vinylic and aryl halides undergo 
neither nor reactions.SN2SN1
3-D Molecule:
Benzyl cation
3-D Molecule:
Allyl cation
BRUI10-359_399r2 18-02-2003 2:50 PM Page 383
Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383
10.8 Benzylic Halides, Allylic Halides,
Vinylic Halides, and Aryl Halides
Our discussion of substitution reactions, to this point, has been limited to methyl
halides and primary, secondary, and tertiary alkyl halides. But what about benzylic,
allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben-
zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary
benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in 
reactions because of steric hindrance.
Benzylic and allylic halides readily undergo reactions because they form
relatively stable carbocations. While primary alkyl halides (such as and
) cannot undergo reactions because their carbocations are too
unstable, primary benzylic and primary allylic halides readily undergo reactions
because their carbocations are stabilized by electron delocalization (Section 7.7).
If the resonance contributors of the allylic carbocation intermediate have different
groups bonded to their carbons, two substitution products will be obtained.
Vinylic halides and aryl halides do not undergo or reactions. They do not
undergo reactions because, as the nucleophile approaches the back side of the 
carbon, it is repelled by the electron cloud of the double bond or the aromatic ring.
a vinylic halide an aryl halide
nucleophile
C C 
Cl
HH
R
Br
nucleophile
a nucleophile is repelled
by the pi electron cloud
p
sp2SN2
SN1SN2
sp2
SN1
SN1CH3CH2CH2Br
CH3CH2Br
SN1
benzyl chloride
1-bromo-2-butene
an allylic halide
2-buten-1-ol
benzyl methyl ether
CH2Cl CH3O−+
CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+
CH2OCH3 Cl−+
SN2 conditions
SN2 conditions
SN2
SN2
CH3OHSN1CH2Cl CH2
Cl−+
CH2OCH3 H++
SN1 CH2CH2CH
Br−+
H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2
+
+ +
+
+
CH3CH CHCH2Br CH3CH CHCH2
+
CH3CHCH CH2
CH3CHCH CH2CH3CH CHCH2OH
H+
+ Br−
H2O
+ H+OH
H2O
SN1
Benzylic and allylic halides undergo 
and reactions.SN2SN1
Vinylic and aryl halides undergo 
neither nor reactions.SN2SN1
3-D Molecule:
Benzyl cation
3-D Molecule:
Allyl cation
BRUI10-359_399r2 18-02-2003 2:50 PM Page 383
Reatividade de Compostos Orgânicos – Aula 2 
45 
Recall that protic solvents contain a hydrogen bonded to an oxygen or a 
nitrogen, so protic solvents are hydrogen bond donors. Aprotic solvents, 
on the other hand, do not have a hydrogen bonded to an oxygen or a 
nitrogen, so they are not hydrogen bond donors. 
Section 10.10 The Role of the Solvent in and Reactions 389SN1SN2
What percentage of the reaction takes place by the mechanism when these conditions
are met?
a. b.
SOLUTION TO 26a
10.10 The Role of the Solvent in 
and Reactions
The solvent in which a nucleophilic substitution reaction is carried out also influences
whether an or an reaction will predominate. Before we can understand how a
particular solvent favors one reaction over another, however, we must understand how
solvents stabilize organic molecules.
The dielectric constant of a solvent is a measure of how well the solvent can insu-
late opposite charges from one another. Solvent molecules insulate charges by cluster-
ing around a charge, so that the positive poles of the solvent molecules surround
negative charges while the negative poles of the solvent molecules surround positive
charges. Recall that the interaction between a solvent and an ion or a molecule dis-
solved in that solvent is called solvation (Section 2.9). When an ion interacts with a
polar solvent, the charge is no longer localized solely on the ion, but is spread out to the
surrounding solvent molecules. Spreading out the charge stabilizes the charged species.
Polar solvents have high dielectric constants and thus are very good at insulating
(solvating) charges. Nonpolar solvents have low dielectric constants and are poor
insulators. The dielectric constants of some common solvents are listed in Table 10.7.
In this table, solvents are divided into two groups: protic solvents and aprotic
solvents. Recall that protic solvents contain a hydrogen bonded to an oxygen or a
nitrogen, so protic solvents are hydrogen bond donors. Aprotic solvents, on the other
hand, do not have a hydrogen bonded to an oxygen or a nitrogen, so they are not
hydrogen bond donors.
δ−δ−
δ−
δ−
δ+
δ+
δ+
δ+
δ−
δ+
δ+
δ+
δ+ H
H OY− Y+
H
H
O
δ−
δ−
δ−
δ+
δ+ δ+
δ+
H
H
HH
δ+ δ+HH
O
O
O
H
H
O
δ−
δ+
δ+
H
H
O
ion–dipole interactions
between a negatively
charged species and water
ion–dipole interactions
between a positively
charged species and water
H
HO
SN1SN2
SN1
SN2
 = 96%
 = 3.20 * 10
-5
3.20 * 10-5 + 0.15 * 10-5
* 100 = 3.20 * 10
-5
3.35 * 10-5
* 100
 =
3.20 * 10-5[2-bromobutane]11 # 002 * 100
3.20 * 10-5[2-bromobutane]11 # 002 + 1.5 * 10-6[2-bromobutane]
percentage by SN2 =
SN2
SN2 + SN1
* 100
[HO-] = 0.001 M[HO-] = 1.00 M
SN2
AU: OK as
changed?
BRUI10-359_399r2 18-02-2003 2:50 PM Page 389

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