1
Solvation
• When NaCl is added to water, the electropositive hydrogen atom is attracted to the negative
chloride ion and the electronegative oxygen atom is attracted to the positive sodium.
• As the polarized hydrogen atom in water hydrogen bonds to the chloride, it “pulls” on the chloride
as the polarized oxygen atom of water associates with and thereby “pulls” on the sodium,
represented by 16.
• Therefore, in the presence of water the charges begin to separate.
• Eventually, water molecules encroach between the two atoms and the sodium ion is completely
surrounded by water (17), as is the chloride ion (see 18).
• This phenomenon is called solvation and it occurs because the hydrogen atoms (δ
(δ+) of water are
attracted to the chlorine ion and the oxygen atoms (δ
(δ–) of water are attracted to the sodium ion.
••
••
• A polar, protic solvent like water has two "poles" so it is possible to solvate both cations and
anions (17
(17 and 18).
18).
O
H
H
H
H H
O
H
H
O H
O
H
H
O
H
••
O
H
O
H
Cl
••
H
H
H
H
••
Cl
•• H
O Na
H
O
O Na
H H
••
Na Cl
O
O
O
H
••
H
H
H
H
H
O
H
O
H
O
H
H
O H
H
HH
O H
16
17
18
••
••
••
••
Water solvates and separates ions
H
H
O
H
Na
••
Cl
••
••
H
••
••
••
••
Na Cl
••
H
O
H
O
O
H
O
H
O
H
H
H
H
O
H
H
O
H
O
O
H
H
O
13
H
H
H
O
H
O
H H
H
12
H
••
Cl
••
O
H
H
••
H
Na
H
H
••
O
H
H
H
O
H
O
H
O
14
All protic solvents are capable of solvation
only water is efficient. In alcohol solvents
solvation is quite slow
H
Aprotic Solvents
Aprotic solvents can solvate (+) ions, but not
(–) ions - therefore, little or no solvation
No charge separation
Et
O
Et
Na
••
Cl
••
O
Et
Et
••
Et
••
Et
••
••
••
Na Cl
••
Et
O
Et
O
O
Et
Et
O
Et
little or no
solvation
of chloride
Et
Et
O
Et
15
An aprotic solvent such as diethyl ether will solvate cations
but not anions (see 15) so
there can be no separation of charges (ions).
Solvents and the SN2 Reaction
What about the SN2 reaction?
Ion + neutral
CH3
!"
CH3Br
I–
Ion + neutral
I
!+
C
!"
Br
CH3I
Br–
H H
Opposite charges
• protic solvents will separate charge, slow down approach of iodide
• protic solvents will inhibit the transition state, also slowing down the reaction
• aprotic solvents do not solvate anions, therefore, a little easier for
the iodide to approach C
• aprotic solvents do not destabilize the transition state
Therefore, aprotic solvents are preferred for SN2 reaction
Protic solvents slow down SN2 of a charged + a neutral
5
Solvent in SN2 Reactions
• If the solvent is changed to diethyl ether, the negatively charged nucleophile is
not well solvated, and it is easier to collide with the carbon atom.
• This suggests that an aprotic solvent is preferred in SN2 reactions because
the initial reaction with the nucleophile and the carbon atom occurs more
readily.
• The most common polar aprotic solvents used in SN2 reactions are diethyl
ether, tetrahydrofuran (THF), dimethyl sulfoxide (DMSO; Me2S=O), and
dimethylformamide (DMF).
• An example that uses DMF as a solvent is the reaction of 1chlorophenylmethane (benzyl chloride, 19) with potassium iodide, to give
benzyl iodide, 20.
KI , 50°C
Cl
19
Me
Me
O
N
I
H=
DMF
20
Solvents for SN2
O
O
diethyl ether (ether)
tetrahydrofuran (THF)
O
H
C
N
O
CH3
CH3
N,N-dimethylformamide
(DMF)
H3C
S
CH3
dimethyl sulfoxide
(DMSO)
7
Solvation
• A polar and protic solvent such as water can be contrasted with an aprotic solvent such
as ether.
• The oxygen atom is δ–, and it can certainly coordinate to cations.
• The δ+ atom is a tetrahedral carbon, however, and it is difficult due to steric repulsion
between the atoms for a negative anion or a negatively polarized atom to approach the
carbon.
• Therefore, an aprotic solvent such as diethyl ether may solvate cations but not anions
and there can be no separation of charges (ions).
• A simple experiment to examine the difference between these two solvents is to
observe that sodium chloride will dissolve in water, but not in diethyl ether.
••
Et
••
••
••
•• Et
Na Cl
••
O
little or no charge separation
O
Et
Et
Et
Et
O
Et
••
little or no
Cl
O Na
Et
••
O solvation
of chloride ion
Et
O
Et
Et
Et
Et
O Et
8
Amines
• An interesting SN2 reaction occurs when 1-bromopropane (21) reacts with dimethylamine (22).
• The product of the reaction with the amine is 23 and if it proceeds by a SN2 mechanism, the
transition state is 24.
• The transition state for the reaction between iodide ion and 1-bromopropane is 25, which
analogous to transition state 11.
• A comparison of 24 and 25 shows a different charge distribution. In 25, a polar protic solvent such
as water will surround the incoming negatively charged nucleophile and inhibit attack at carbon, as
just described, and slow the reaction.
• In the reaction with dimethylamine, however, the incoming nucleophile is neutral.
• In transition state 24 the nitrogen takes on a positive charge and the bromide takes on a negative
charge.
• A solvent that separates charge will therefore increase the rate of reaction since it leads to
products.
• If the reaction of 21 and 22 to give ammonium salt 23 is done in water, the rate of the reaction is
faster than when it is done in an aprotic solvents such as THF.
Me
Br
21
H
N
N
22
Me
H
H
Br
23
Me
Me
Me
!+
Me
N
Et
C
H H
24
Et
!"
Br
!"
I
!+
C
H H
25
!"
Br
Halide Nucleophiles
9
• In general, the order of nucleophilic strength for halides in SN2
reactions is: I- > Br- > Cl- > F-.
• In fact, iodide ion will react with most primary and secondary
alkyl bromides or alkyl chlorides to give an alkyl iodide.
• When the reagent is NaI in the solvent acetone (2-propanone;
see chapter 16, section 16.2), this transformation is known as the
Finkelstein reaction, named after Hans Finkelstein (Germany,
1885-1938).
• Since iodide is a better nucleophile than the bromide ion or the
chloride ion, it is unlikely that bromide or chloride will displace
iodide ion from an alkyl iodide to give the alkyl bromide or the alkyl
chloride.
• In other words, iodide can displace bromide or chloride, but
bromide or chloride will not displace iodide.
• In general fluoride is a poor leaving group in the SN2 reaction
and it will not be used.
10
Halide Nucleophiles
• Iodide (a good nucleophile) can displace another iodide, which is a good leaving group.
• An early experiment by Edward Hughes and co-workers reacted radiolabeled NaI with
2S-iodooctane (33), and the initial product was 2R-iodooctane, 34, where the iodine
atom has the radiolabel.
• The ability to distinguish the two iodine atoms allowed them to follow how racemization
occurred, because the final product was racemic 2-iodooctane (a 1:1 mixture of 33 and
34).
• This result is explained by the fact that the product accompanying 2-iodooctane is the
iodide ion.
• While iodine is a leaving group, the iodide ion is also a nucleophile, so once 34 is
formed, it reacts with the nucleophilic I– to give 33.
• Since 33 reacts to give 34, and subsequently 34 reacts to give 33, the 2-iodooctane
that is isolated from the reaction is racemic.
• Each individual SN2 reaction proceeds with inversion, but because both forward and
reverse reactions are equally possible, the net result is a racemic mixture of 33 and 34.
I
I
II-
33
34
Halide nucleophiles, a cautionary note
I-
I
I
I-
27
RACEMIC
28
Iodide is a good nucleophile and a good leaving group
and the reaction is reversible (generates racemate)
Nucleophilic strength: I– > Br– > Cl–
Therefore, iodide will displace bromide but bromide will not
displace iodide.
i.e. The reaction is reversible, and energetics favor the iodide. This fact
Is interpreted by saying that iodide is a stronger nucleophile than bromide.
Br
I–
I
Br–
12
Alkoxide Nucleophiles
• When an alcohol (ROH) reacts with a base, the product is an alkoxide (RO–).
• The alkoxide is formally the conjugate base of the alcohol, and is of course a base. In
the presence of electrophilic carbon compounds, alkoxides are also nucleophiles.
• Alkoxides react with primary and secondary halides to form ethers in what is known as
the Williamson ether synthesis, named after Alexander W. Williamson (1824-1904;
England).
• Another example places the focus on the alcohol, 1-hexanol (35), which is treated with
sodium metal to give the conjugate base, alkoxide 36, which is then treated with
iodoethane to give 1-ethoxyhexane (37) in 46% yield.
• Alkoxide 36 is the nucleophile in this reaction and the SN2 reaction with the
electrophilic carbon of iodoethane leads to the substitution product, 37.
CH3CH2I
Na
35
OEt
O Na
OH
36
37
13
Amine Nucleophiles
• Amines react as nucleophiles with alkyl halides in SN2 reactions, as seen for the
conversion of 21 to 23.
• The isolated product is tertiary amine 38 rather than ammonium salt 23.
• Displacement of bromide ion by the nucleophilic nitrogen in dimethylamine (22) leads
to formation of N,N-dimethylammonium-1-aminopropane, 23 as described above.
• This salt is a weak acid, and it is formed in the presence of amine 22, which is a base
as well as a nucleophile.
• A simple acid-base reaction occurs between 23 and 22 to generate the neutral amine
(N,N-dimethylaminopropane, 38) along with dimethylammonium bromide.
Br
21
Me2NH
22
- Br-
H
N
Me
23 Me Br
Me2NH
N
38
Me
Me +
Me2NH2 Br
14
Exhaustive Methylation
• There are problems when primary amines react with structurally simple alkyl halides.
• If a primary amine such as butanamine (39) reacts with iodomethane, the initial product
is N-methylbutanamine, 40 via the corresponding ammonium salt.
• This secondary amine is more reactive than primary amine 39 (it is a stronger
nucleophile and a stronger base), so it can competitively react with iodomethane to form
a tertiary amine (N,N-dimethylbutanamine, 41), also via an ammonium salt.
• It is also possible for 41 to react with iodomethane to give N,N,Ntrimethylbutanammonium iodide, 42.
• When this reaction occurs with an amine and iodomethane (or any Me-X substrate) it is
called exhaustive methylation.
NH2
CH3I
NMe2
NHMe CH3I
39
41
40
CH3I
NMe3
42
I
15
Amine Surrogates: Phthalimide
• Nitrogen-containing nucleophiles that are not amines are known as amine
surrogates.
• One such modification uses the molecule phthalimide (43), which is the imide (see
chapter 20, section 20.6.E) derived from phthalic acid (47).
• If 43 is treated with a strong base such as sodium amide, the phthalimide anion (44) is
formed as the conjugate base (ammonia is the conjugate acid of this reaction), and it is
a good nucleophile in reactions with alkyl halides.
• If 44 reacts with benzyl bromide, the product is N-benzylphthalimide (45) via a
straightforward SN2 displacement.
• To generate the amine, the imide can be hydrolyzed by acid-base reactions (1.
aqueous base 2. aqueous acid) to give the amine (46) and phthalic acid (47).
O
N
43
O
H
NaNH2
O
N
O
44
Na
O
1. aq. OH2. aq. H+
N
PhCH2-Br
COOH
H2N
Ph
Ph
45
O
46
+
COOH
47
16
Amine Surrogates: Phthalimide
• A better procedure has been developed that treats 45 with
hydrazine (NH2NH2) to generate amine 46 and a molecule known
as a hydrazide, 48.
• Hydrazide 48 is easily separated from amine 46.
O
N
43
O
H
NaNH2
N
O
PhCH2-Br
O
44
O
Na
1. aq. OH2. aq. H+
N
COOH
H2N
Ph
Ph
45
46
O
+
COOH
47
O
O
NH2NH2
N
H2N
Ph
Ph
45 O
46
N
+
N
48
O
H
H
17
Amine Surrogates: Azide
• Azides are formed by the reaction of an alkyl halide with the nucleophile, the
azide ion, 51.
• Azide ion reacts with alkyl halides via a SN2 reaction to first give an alkyl
azide, which can subsequently be converted to an amine.
• The azide anion (57) is commercially available as sodium azide (NaN3).
• 1-Bromopropane (21) reacts with sodium azide in THF, and the product is
alkyl azide 52 and sodium bromide (NaBr).
• For the azide group to function as an amine surrogate, the N3 group must be
reduced with sodium borohydride, NaBH4 (see chapter 19 for reduction
reactions) to give amine 53.
Na+ N3 = N=N=N
51
Br
21
1. NaN3 , H2O
C6H13
Br
54
2. NaBH4
3. 10% aq. HCl
52
NaBH4
or
N3 H2 , Pd
+ NaBr
C6H13
53
NH2 88%
55
NH2
18
Amine Surrogates: Cyanide
Cyanide, a bidentate nucleophile
C is nucleophilic
:C
N:
N is nucleophilic
• The cyanide ion, N≡C–, is the conjugate base of hydrocyanic acid (HCN,
pKa, 9.31).
• Cyanide is a relatively weak base, but it is a good nucleophile and it gives
SN2 reactions with alkyl halides.
• Indeed, reaction of 1-bromohexane (56) with NaCN in aqueous methanol
gives a 73% yield of heptanenitrile, 57.
• There is an electron pair on both carbon and nitrogen, so the CｺN unit is a
bidentate nucleophile.
• This means that both the carbon and the nitrogen atoms may react as
nucleophiles. In general, the nitrogen is more electronegative and is less able
to donate electrons when compared to carbon.
NaCN , heat
Br
56
MeOH/H2O
C!N
57
19
Amine Surrogates: Cyanide
• The stereochemistry of the product is predictable because it is
an SN2 reaction.
• The reaction of the p-toluenesulfonate (tosyl) ester of 4-phenyl2S-butanol (58) with KCN, for example, leads to nitrile 59 (the
potassium salt of the tosylate anion is the other product).
• Replacing the C-O bond in 58 with the C-C bond in 65 changes
the priorities of the groups attached to the stereogenic carbon,
but inversion does occur.
Me
O
O
NC
S
(R)
O
Ph
(S)
58
KCN , THF
Ph
59
+
K
O
O
S
O
Me