In alkyl halides, the carbon has a partial positive charge
Chapter 6
This carbon will be susceptible to attack by nucleophiles
Ionic Reactions
Nucleophilic Substitution and
Elimination Reactions of Alkyl Halides
Carbon-halogen bonds get less polar, longer and weaker in going
from fluorine to iodine
Role of Nucleophile
Nucleophilic Substitution Reactions
A characteristic reaction of alkyl halides is reaction at the electron
deficient carbon with the unshared electron pair of a nucleophile
The nucleophile reacts at the electron deficient carbon
Net result: a leaving group is substituted by a nucleophile
A nucleophile may be any molecule with an unshared electron pair
t Examples of nucleophilic substitution
Leaving Groups
Leaving Groups
A leaving group is a substituent that can leave as a relatively
stable entity
It can leave as an anion….
Good anionic leaving groups are generally the
conjugate bases of strong acids
H-A
strong acid
+ H2O
+
H3O
+
A:anion
very stable
…or as a neutral molecule
A - = Cl , Br , I
1
Neutral molecules as leaving groups
Poor leaving groups can
be turned into good
leaving groups by
proton ation.
Hydroxide ion is a
poor leaving group
because it is the anion
of a weak acid, H2O.
CH3-OH
In the presence of a strong acid,
: :
CH3 OH
+
CH3 OH +
H
H2SO 4
+
+
CH3OH
+
CH3 OH
H
+
CH3OCH3
H
There are two important considerations in any chemical reaction:
(1) The extent to which it takes place---how completely the reactants
are converted into products after equilibrium is reached.
This analysis is based on therm odynamic param eters.
H SO 4
(2) The rate of the reaction---how fast (moles per liter per second)
the reactants are converted to products. The kinetics of the
reaction means the rate of the reaction.
a nucleophilic substitution reaction occurs:
nucleophile
Thermodynamics and Kinetics of Chemical Reactions
+
H2O
leaving group
good leaving group
Thermodynamics: tells us if the reaction is possible
l Exergonic reaction: negative ∆ Go (products favored)
Products are lower energy than starting materials
l Endergonic reaction: positive ∆ Go (products not favored)
l The reaction of chloromethane with hydroxide is highly
exergonic
Kinetics = the rate of a chemical reaction
Even if a rea ction has a very favorable ∆Go (nega tive), the rate of the
reaction ma y still be very slow.
Th e reaction
2H2(g) + O2(g)
2H2O(l)
ha s a very fa vorable free energy chang e: ∆Go = -237 kJ/mol
l The equilibrium constant is very large
Yet when hydrogen and oxygen ga ses are mixed, no reactio n occurs.
If there is a high energ y source (f lam e or spark), an ex plosive reaction
occurs .
The conversion of diamonds to graphite is exergonic at room
temp and pressure, but…..
Reactions require a pathway, a mechanism to overcome energy barr iers
Kinetics = the rate of a chemical reaction
Diamonds are
“unstable”
Kinetics of Nucleophilic Substitution: An SN2 Reaction
Let us observe the initial rate of the following reaction:
Higher energy
than graphite at
normal T and P
Or could be
combustible
Doubling the concentration of either reagent doubles the rate
The rate is directly proportional to the initial concentrations of
both methyl chloride and hydroxide
2
Kinetics of Nucleophilic Substitution: An S N2 Reaction
Graphical Determination o f Rea ctio n Ra tes
The pro gre ss o f the che mical re act ion is fo llowe d gra phica lly.
Fundamental Law of kinetics: if the reaction rate is dependent
on the concentration of any particular reagent, that reagent
must be involved in or prior to the rate-determining step
Increa se in product
concentration.
Decrease in rea ct ant
concentration.
concentration, M
At time = T, the concentration
of a reactant is measured..
The ra te o f the reaction ca n be
expressed as t he amo unt of
reactant (moles/L) consumed
per seco nd:
Called a second order reaction: dependent on two concentrations
time
[Rea cta nt] = Rate = - ∆
∆T
SN2 = Substitution, nucleophilic, bimolecular
where the notation
[Reactant]T=T - [Reactant]T=0
Tim e in seconds
∆ me ans the differ enc e.
Method of In iti al Rates
Example: The reaction of methyl bromide and sodium hyd roxide
The ma thema tical form of the rate expressio n ca n be ex perimentally
determined by t he method o f initia l ra tes. The rate o f t he rea ction is
measured just a fter the rea ction begins .
CH 3Br + NaOH
This initial rate is ca lculat ed directly f rom the slo pe .
initial conc.
T= O
Initia l Rate = - ∆ [Reactant]
∆T
( from slo pe)
con c. at
T= T
concentration, M
CH 3OH + NaBr
The reaction of methyl bromide and sodium hydroxide in a solvent of
80% ethanol and 20% water at 550 C has a k = 0.0214 M -1 s -1 . What is
the rate of formation of methyl alcohol if [CH 3Br] = 0.5 M and [HO -] =
0.5 M?
Rate = k[CH 3Br][HO -] = (0.0214 M-1 s-1 ) (0.5 M) (0.5 M)
Ra te = 0 .00 53 5 mo l per lite r pe r second
No te tha t k is defined f or a so lvent a nd t empera ture.
time
A Mechanism for the S N2 Reaction
The Mechanism for the SN2 Reaction
Proposed by Hughes and Ingold in 1937
δ−
H
HO-
C
H
HO
Cl
H
H
C
H
δ−
Cl
HO
C
+ Cl
H
HH
H
Starting materials
Transition state
Products
t ransition state
t The transition state is the energetic high point of the reaction
l An unstable entity with a very brief existence (10-12 s)
l About the time of one molecular vibration
Nucleophile approaches on the backside of C-X bond
Reaction proceeds over an energy barrier in a single step directly to product
t In the transition state of this reaction bonds are partially formed
Nu
-
:
C :
L
Nu
:
dev eloping
bo nd
C
:
L
and broken
l Chloromethane and hydroxide are both involved in the transition state
weakening
bond
l Explains why the reaction is second order
3
In an endergonic reaction the energy barrier is usually
even higher (?G‡ is very large)
Energy
diagram of a
typical SN2
reaction
An energy barrier is evident because energy must be supplied to
make the reaction proceed
The difference in energy between starting material and the
transition state is the free energy of activation (∆ G‡ )
The difference in energy between starting molecules and
products is the free energy change of the reaction, ∆ Go
Is this reaction even possible?
Why does rate increase so rapidly with temperature?
The higher the temperature, the faster the rate
Boltzman distribution (kinetic molecular theory)
As T increases, distribution of molecular energy broadens
fraction of molecules
There is a direct relationship between ∆G‡ and the
temperature of a reaction
l Near room temperature, a 10oC increase in temperature
causes a doubling of rate
l Higher temperatures cause more molecules to collide
with enough energy to reach the transition state and
react
T2
T1
T 2 > T1
?G‡
T2
# molecules
with enough
energy to reach
Transition State
kinetic energy (1/2 mv2 )
Enthalpy and Entropy of Activation
Energy diagram for the reaction of CH3Cl with hydroxide :
The overall rate depends on ?G ‡ , but additional information is obtained by
separating its two components, the enthalpy and entropy of activation.
∆G = ∆H
-
T ∆S
?H ‡ measures the enthalpy change from reactants to transition state
A reaction with ∆ G‡ above 84 kJ mol-1 will require heating to proceed at a
reasonable rate
t This reaction has ∆ G‡ = 103 kJ mol-1 so it will require heating
t
1.
Bonding changes (mostly)
2.
Solvation effects
?S‡ measures the change in order as reactants move to transition state
•
Usually a negative factor because T.S. is more ordered
4
Example: the Sn2 reaction of pyridine with methyl iodide
25 oC
N: + CH 3I
+
N C H3 + I -
ac etonitri le
pyrid ine
Transition State = the pass thru the energy mountains
When reactants collide in so lution, there are numerous possible orientations
wit h pa ths leading to products. The operating tra nsit io n sta te is the most
energetically favo rable pa th throug h an "energy m ountain ra nge. "
N-m ethylpyri dini um iodi de
∆G
o
= -50.2 kJ/mol Exergonic
The mounta ins on either side
of the pass represent
energetically less fa vorable
reaction pathways, described
in terms o f bonding cha nges,
so lv ation eff ect s, etc.
By examining rate as a function of temperature, we can split the free
energy of activation into its components of enthalpy and entropy
∆ G = + 92.1 k J/mol ∆ H = + 53.6 k J/mol
∆ S = -129.8 J K -1 mol-1
Both enthalpy and entropy make significant contributions
∆G = ∆H
-
T∆S
A very wide pass me ans many possible orientations of react ants will
lead to products...a less restrictive t ransit ion state structure wit h a less
negative entropy of act ivation t erm.
at 25 o C, 92.1 kJ/mol = 53.6 kJ/mol + 38.5 kJ/mol
Understanding Reaction Energy Diagrams
The Stereochemistry of SN2 Reactions
Sn2 mechanism is based on two main pieces of evidence:
1.
Kinetic = second order
2.
Inversion of configuration at reaction center
The Stereochemistry of SN2 Reactions
Later proof of inversion
Inversion also demonstrated in cyclic compounds
H3 C
H3C
TsCl
H
OH
C6H13
[a] = - 9.9o
H
C6 H13
Ch3
OT s
OH HO
H
C6 H13
[a] = + 9.9o
5
Reaction of tert-Butyl Chloride and Hydroxide Ion
tert-Butyl chloride undergoes substitution with hydroxide to produce
an alcohol, the same kind of product as with methyl chloride
tert-Butyl Chloride and Hydroxide Ion: the S N1 Reaction
Therefore a different mechanism than the earlier Sn2 reaction
What can we say about it?
The rate is independent of hydroxide concentration and depends only
on concentration of tert-butyl chloride
Doubling or even increasing the concentration of hydroxide ion t enfold has no effect on the rate of reaction!
1.
2.
3.
4.
Reaction must be multi-step. Why?
Because only the alkyl halide is involved at or prior to the rat edetermining step
Hydroxide ion is part of reaction, must be a subsequent step
Therefore there must be an intermediate species
SN 1 reaction: Substitution, nucleophilic, 1st order (unimolecular)
Multistep Reactions and the Rate-Determining Step
Energy Diagram for a two step reaction with slow first step
A-B + C - à
B -C + A -
t In multistep reactions, the rate of the slowest step will be the
rate of the entire reaction
t This is called the rate determining step
In a multi-step reaction, if
k1<<k2 or k3, the first step
is rate determining
First step is endothermic, but overall reaction is exothermic
First step has the higher T.S. and is therefore the rate -determining step
Mechanism for t-butyl chloride reaction with water
1. Formation of a carbocation
intermediate along with a chloride ion
Ions will be solvated
2. Water as a Lewis base and nucleophile bonds to the Lewis acid cation
3. Another water molecule acts as Lewis base
to remove a proton from the oxonium ion
6
Carbocations
Carbocation stability: 3o > 2o > 1o > methyl
l Positively charged reactive intermediates
The more alkyl groups, the more stable the carbocation
l Highly reactive
l Formation helped by ion-dipole solvation effects
l Only three groups attached to C
l Only 6 electrons around the C
l sp 2 hybridized, with an empty p orbital
Stabilized by
1. inductive effects
t
2.
Hyperconjugation
Sigma bonds on the carbons adjoining the positive carbon can
donate electrons by partial overlap with the vacant p orbital
alkyl group is electron-donating relative to a hydrogen
hyperconjugation
Superacids : evidence for carbocations
t Carbocations have been observed under “superacid
conditions”
t Super acid = SbF5 + FSO3H
t pK a = -25 ( 1019 more acidic than conc. H 2SO4)
More alkyl groups on C+
provide more
hyperconjugation
t Under such extreme conditions, there is no nucleophile
and carbocations can be observed
t George Olah , Nobel Prize, 1994
The Stereochemistry of S N1 Reactions
When the leaving group leaves from a stereogenic center of an
optically active compound, the products are not optically active
Mechanism of Sn1 explains the racemization
Step 1 produces a planar
achiral carbocation
Racemization = transformation of an optically active
compound to a racemic mixture
Step 2: both sides of carbocation attacked equally
7
Stereoselective Reactions
I
CH3
CH3
If there is more than one stereogenic center, the products are diastereomers rather
than enantiomers, and one product may be favored over another
I
(CH 3)3C
(CH3) 3C
H
H2O
aceton e
trans
acetone
S N1
Hydrolysis of either cis or trans-4-tert-butyl-1-methylc yclohexyl iodide
yields the same unequal mixture of cis and trans- 4-tert-butyl-1methylcyclohexanol. This reaction is stereoselective.
H
H2O
cis
SN1
Path A
+ CH3
H3C
I
H3 C
OH
(CH3) 3C
Path A
(H3C)3C
H
cis or trans
(H3 C)3C
H
sa me mixture of diastereo meric
alcohols from either starting iodide
Path B
H
Path B
A common inte rmediate r eacts
with H2O via c ompeting paths A
and B. The two paths are
diastere ome ric ally re lated.
CH3
OH
CH3
(CH3) 3C
same product mixture
from either source
OH
(CH3)3C
H
H
tra ns
cis
Factors Affecting the Rate of SN1 and SN2 Reactions
Solvolysis in substitution reactions
t Often a molecule of the solvent is the nucleophile in a
1. The Effects of the Structure of the Substrate
l In SN 2 reactions alkyl halides show the following general order
of reactivity
substitution reaction
l If the solvent is water the reaction is a hydrolysis
Steric hinderance: the spatial arrangement of the atoms or groups at or near
a reacting site hinders or retards a reaction
l
Factors Affecting the Rate of SN1 and SN2 Reactions
In tertiary and neopentyl halides, the carbon is too sterically hindered to react
The Hammond -Leffler Postulate
1. The Effects of the Structure of the Substrate
t The transition state for an exergonic reaction looks very much
In SN 1 reactions, the order follows the order of carbocation stability,
namely 3o > 2o > 1o
t The transition state for an endergonic reaction looks very much
like starting material
like product
t In other words, the transition state looks most like
the species it is closest to in energy
Except for solvolysis, only the tertiary systems
normally react by SN1
To understand why the stability of the cation affects the rate so
much, let us look more at the nature of transition states
8
Hammond-Leffler postulate: another look
TS
reactant state
free energy
Application to the S N1 reaction
TS resembles product state
TS
free energy
resembles
reactant sta te
prod uct s tate
highly
exergonic step
high ly en dergon ic step
prod uct s tate
high ly en dergon ic step
reactant s tate
product state
reactant s tate
reaction coordinate
reaction coordinate
According to the Hammond-L effler Postulate, since SN1 pro cesses are
highly enderg onic, the tra nsition state comes late along the reactio n
coordina te and resem bles the carbo cation state. Factors that sta bilize
carbocations will stabilize the TS in an SN 1 reaction lea ding to a faster rate.
Since the transition state looks very much like the carbocation:
Properties of Transition State (TS) will resemble those of the intermediate
Summary of competition between S N2 and S N1
l The transition state is stabilized by the factors that stabilize
carbocations
l The transition state leading to tertiary carbocations is more stable and
lower in energy than transition states leading to other carbocations
SN2 dominates
S N2
SN1 dominates
Log of Rate
competitive
mechanisms
C H3
C H3 CH2
(CH 3) 2CH
overall reactivity in
nucleophilic
substitution
SN 1
(CH3 )3 C
RX Type
Effect of the nucleophile
Example of competitive SN1 and SN2 mechanism
A change in the concentrati on of the nucleophile will affect the
rate of an SN2 reaction but wi ll have no effect on an SN1 reaction.
H
H
SN 1
R-X
slow step
R+ + X-
R+ + Nu:-
fast step
R-Nu
H3 C
Br
CH3
H3C
OH
CH3
Rate = k [R-X]
SN 2
R -X + Nu: -
single step
R-Nu +
X-
Rate = k [R-X] [N u:- ]
The conversion of isopropyl bromide t o isopropyl alcohol may occur by
either an SN 1 or SN 2 mechanism. This secondary alkyl bromide is a
"borderline" system.
In water the SN 1 m echa nism dom ina tes. But with a dded HO- , the SN 2
mecha nism becom es increa singly co mpetitive.
Since the concentration of nucl eophile appears only in the rate
expression for the S N2 reaction, the concentration of nucleophile
only affects that reaction rate.
9
Strength of the Nucleophile
Rate comparison of competing reactions
SN 1 Reaction rate does not depend on identity of nucleophile, but it
may affect which product is formed
In the SN 2 reaction, the rate is directly affected by the strength of
the nucleophile (Stronger nucleophiles react faster)
In 80% ethanol-20% water, at 55o C, isopropyl br omide r eacts
ac cording to the comprehensive rate expr ession:
Rat e =
k 2 [ RX] [H O- ]
+
SN 2 m echan ism
Rate =
k 1 [ RX]
1.
S N1 mecha nism
(4.7 x 10-5) [RX] [HO-] + (0.24 x 10 -5 ) [RX]
2.
A negatively charged nucleophile is always more reactive than its
neutral conjugate acid
When comparing nucleophiles with the same nucleophilic atom,
nucleophilicities parallel basicities
At [HO -] = 0.001 M, the per centage of the nucleophilic substitution
going by the SN 2 reaction is
% SN 2 =
(4.7 x 10 -5) [RX] [0.001 M]
x 100
l
(4.7 x 10-5 ) [RX] [0.001 M] + (0.24 x 10 -5) [RX]
% SN2 =
For example, methoxide is a much better nucleophile than
methanol
1.95%
But at [HO-] = 0.10 M, % SN2 = 66.1%
Solvent Effects on SN2 Reactions: Polar Protic and Aprotic Solvents
Quiz Chapter 6 Section 1 3
t Polar Protic Solvents
l Polar solvents have a hydrogen atom attached to strongly
electronegative atoms
The nucleophilic substitution reaction
R-X
+ Nu:-
R-Nu + X:Protic solvents solvate
nucleophiles
and make them less reactive
may proceed by either the SN 1 or S N2 mechanism.
With which mechanism is each of the following observations associated?
Addition of alkyl groups around the reacting
center slows down the rate of the reaction.
SN 2
When the reaction takes place at a stereocenter, racemization is
observed with optically active substrates.
SN 1
Increas ing the concentration of the nucleophile results in
an increase in the rate of the reaction.
SN2
Larger nucleophilic atoms are less solvated via hydrogen bonding and more
polarizable, to donate electron density to reactive carbon
Relative nucleophilicity in protic solvents
Polar AproticSolvents
Polar aprotic solvents arepolar solvents that do NOT have
a hydrogen atom attached to an electronegative element
Effects of Polar AproticSolvents
t They solvate cations well but leave anions unsolvated because
positive centers in the solvent are sterically hindered
t
Polar protic solvents lead to generation of “naked” and very reactive
nucleophiles
Trends for nucleophilicity are the same as for basicity
t They are excellent solvents for SN 2 reactions
t
10
The Nature of the Leaving Group
Solvent Effects on S N1 Reactions
The best leaving groups are weak bases which are relatively stab le
t Polar protic solvents are excellent solvents for SN 1 reactions
l Can be an anion or a neutral molecule
because they stabilize both anions and cations
Leaving group ability of halides:
t Polar protic solvents stabilize the carbocation-like transition
state leading to the carbocation thus lowering ∆ G‡
t Water-ethanol and water-methanol mixtures are most common
This trend is opposite to basicity:
Other very weak bases which are good leaving groups:
The poor leaving group hydroxide can be changed into the good leaving
group water by protonation
Summary S N1 vs. S N2
Functional Group Transformations Using S N2 Reactions
In both types of reaction alkyl iodides react the fastest
because of superior leaving group ability
Organic Synthesis:
Clarifications on reactive alkyl halides
Functional Group Transformations Using SN2 Reactions
t Stereochemistry can be controlled in SN 2 reactions
Benzyl and allyl halides are very reactive by both SN 2 and SN 1 mechanisms:
Primary and unhindered facilitates SN 2
Resonance stabilization of carbocation facilitates SN 1
+
+
CH 2
X
etc.
SN2 more widely useful than SN1 for synthetic purposes
Benzyl halide
X
+
+
Allyl halide
11
Elimination Reactions
Clarifications on reactive alkyl halides
Another major class of organic reactions is elimination reactions
Phenyl and vinyl halides are unreactive by both SN 2 and SN 1 mechanisms:
In an elimination reaction, groups X and Y are lost from a larger molecule
Sp2 bond holds leaving group more strongly
Carbocation would be much less stable
C C
X Y
Backside attack for SN 2 impossible in phenyl and high energy in vinyl
C C
When X and Y are on adjacent C’s, this is called 1,2-elimination or beta -elimination
X
X
Phenyl halide
(-XY)
t
For example, loss of HX from an alkyl halide produces an alkene
Loss of HX is called
dehydrohalogenation
Vinyl halide
Elimination Reactions of Alkyl Halides
The alkoxide bases are made from the corresponding alcohols
By adding metallic Na or K to carefully dried alcohol
t Useful for the synthesis of alkenes
t Strong bases such as alkoxides favor elimination
Or by using a stronger base like sodium hydride
p Ka = 16
Mechanism of dehydrohalogenation: the E2 Reaction
The reaction of isopropyl bromide with sodium ethoxide shows second order kinetics
p Ka = 35
E2 requires anti -coplanar geometry in Transition State
δ−
CH3CH2O
CH 3CH2 OH
H
H C
H
Stereochemistry will also demonstrate concerted removal of the proton, formation of the
double bond, and departure of the leaving group
C
H
CH3
Br δ−
H CH CH H
H
CH3
Br -
The two carbons are rehybridizing - from sp3 toward sp2 - in T.S.
Energy requirements are reduced by the partial formation of the double
bond as incipient p orbitals overlap to form new p bond
12
Energy diagram of the E2 reaction
Another look at E2 geometry
CH 3CH 2O -
CH3CH2O-
Newman
H CH 3 Projection
C
H
Br
H
H
H
H
C
CH3
H
H
Br
C 2 H5
O 1
/2
-
The alkyl bromide reacts
best from its staggered, anti
coplanar geometry
H
H
C H3
H
H
Br 1
/2
Concerted means a single step, with no intermediates – like SN2
-
tElimination can also be unimolecular: the E1 Reaction
Mechanism for E1
The E1 reaction competes with the SN 1 reaction and likewise
goes through a carbocation intermediate
Slow step is ionization
of C-Cl bond, aided
by solvent
Water can act as
nucleophile (Sn1)
or as base (E1)
Free Energy Diagram for competing pathways
(C H3)3CCl
Free Energy Diagram of the Product Forming Steps
The com petition between th e E1 and SN1 path s, and the resulting
product mixture, is determined b y the relative magnitu des of th e
activation energies, ∆Gelim and ∆Gsub , for the fast steps.
(CH3)3C+
partitioning factor
83:17
∆G
E1
CH 2=C(CH 3)2
SN1
(CH 3 )3COR
Free Energy
Free Energy
RDS
TS
∆Gsu b (CH3) 3C+∆Gelim
(CH3)3COR
SN 1
CH 2=C(CH3) 2
E1
The RDS leads to the carbocation intermediate
that sit s in a "saddle po int" where a partitio ning
occurs between the two product forming paths.
13
Substitution versus Elimination – effect of substrate
Substitution versus Elimination
Bec ause strong nucleophiles are usua lly a lso stro ng ba ses, t he S N 2 and E 2
react ions are usua lly com peting pa thway s as are the SN 1 and E1 .
SN2 versus E2
SN2
E2
H C C X
(a)
(b)
(b)
Nu: -
(a)
C=C
H C C Nu
Rate = kelim
[RX] [Nu-]
Primary substrate (nature of the alkyl group)
Rate = ksub[RX][Nu-]
t
Rate equations are identical, so competition between SN 2 and E2 is not
affected by changing concentration of either reagent
If the base is small, SN 2 competes strongly because approach at
carbon is unhindered
However, this competition does ver y
much de pe nd on the structure of RX
and choice of base .
Effect of changing the base/nucleophile
Substitution versus Elimination – effect of substrate
Secondary substrate
Size: Large sterically hindered bases favor elimination
Approach to carbon is sterically hindered and E2 elimination is favored
t
t
Access to a ß-hydrogen is less crowded than the a-carbon
Potassium tert-butoxide is an extremely bulky base and is routinely used
to favor E2 reaction
CH 3 CH 2C H 2B r
Tertiary substrate
+ CH 3 CH 2 ON a
e tha nol
CH3CH=CH 2 + CH3CH2CH 2OCH2CH3
(9 %)
(91%)
Approach to carbon is extremely hindered and elimination
predominates especially at high temperatures
CH3
t-bu tyl alc ohol
CH3 CH 2CH 2B r + CH3 CO K
25o C
CH3
potas si um te r t-butoxi de
(a hi nd er e d b ase )
CH3 CH=CH 2 + CH 3CH2CH 2OC (C H3)3
(8 5%)
(15%)
55o C
Weaker bases favor substitution
CH3CHCH3 + CH 3CH2ONa
Br
e thanol
55o C
S N2
(21% )
But ,
=
O
C H3
CH3C HCH 2Br + CH 3CH 2ON a
is obutyl bromide
ethanol
55o C
CH3
CH3
CH3CHCH2OCH2CH3 + CH2=C
CH3
(40%)
branching slows down SN2
CH3
CH3 CHCH2 Br
SN1
CH3 CHCH3 (~100%)
OCCH3
=
CH 3C ONa
sodium acetate
(a weaker base)
(79%)
+ I-
acetone
CH 3
CH 3CHC H2I
+ Br -
(1 00%)
(60%)
S N2
CH3X
yes,
very f ast
RCH2X
mostly
O
+
CH 3CHCH 3
Br
SN 2
Overall Summary of Substitution/Elimination
Reactions of Alkyl Halides
CH3CHCH 3 + CH 2=CHCH3
E2
OCH 2CH3
R'
RCHX
R'
RCX
R''
very
little
very
favorable
mostly SN2
with weak
bases
none
E1
E2
but hindered
bases give
mostly alke nes
very
little
always
competes
with SN 1
strong bases
pr omote E2
st rong bases
promot e E2 path
Io dide ion is a very wea k bas e and als o is m uch more
po larizable than alkox ide ions . Substitution is fa vored
by nucleophiles with high polarizability.
14
Quiz Chapter 6 Section 18
For the foll owing reacti ons, which mechanistic pathway will
domi nate among S N1, SN2, E1 and E2?
CH3
CH3CHCH2Br + KOBu-t
Br
CH3CCH3
CH3
Br
CH3CH2CHCH3 +
NaI
t -BuOH
E2
CH3OH
SN 1
acetone
SN2
15