Chem 232 Fall 2015 UIUC Notes
Chem 232 Fall 2015 UIUC Notes
Chem 232 Fall 2015 UIUC Notes
Elementary step?
Elementary steps describe the
process of change between
adjoining minima on the MEP of
an overall reaction. Many
reactions have several minima
(i.e.,
intermediates)
located
between starting and ending
points. Thus, the overall reaction
mechanism requires a sequence of
elementary steps to describe the
entire process.
Each elementary step has:
A reaction arrow () connecting
the starting and ending structures
of that step
Balanced charge
Balanced stoichiometry
Nu
a nucleophile
an electrophile
reaction arrow
Nu
The sa
of "ac
used t
Nu
Nu
a nucleophile
Nu
an electrophile
Nu
E
Nu
phenol
water
hydroxide
phenolate
O
OH
hydroxide
phenol
O
water
phenolate
sink
OH
source
+
phenol
The hydroxide anion is the source. A lone pair of electrons on the
hydroxide anion forms a new bond with the phenolic hydrogen atom
making water. Simultaneously the O-H bond in phenol breaks and
this electron pair moves onto oxygen to make a phenolate anion.
phenolate
Filled n
1st arrow
OH
H
2nd arrow
New
bond
[pt]
An n *
-type interaction
Curved Arrows Imply the Frontier Orbitals
Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals)
involved in an elementary step. The tail of the first arrow implies
the filled orbital (HOMO) is a non-bonded electron pair, n. The
head of the first arrow points to the empty orbital (LUMO, *) that
accepts the electron pair. The empty orbital may not be obvious
until you realize that phenols O-H sigma bond breaks, as implied
by the tail of the second arrow. Sigma bond breaking would result
upon filling * with the electron pair from n. The frontier orbitals
align coaxially (a -type interaction).
empty
filled
"a
"
" a
"
" na
n "
" a
"
H2O
HO
n
n
(b)
(a) reactants
(b) transition state
(c) products
(d) G Gibbs standard free energy change
(e) G free energy of activation
(e)
(a)
free energy
(d)
(c)
The essential components of a reaction coordinate diagram for a one-step reaction are shown above. Most important is
the MEP on this diagram. The particular MEP on the above diagram might, for example, apply to the proton transfer
from phenol to hydroxide. The reactants (a) are higher in energy than the products (c), so there is a favorable driving
force (d) to promote the change. This driving force is the Gibbs standard free energy and its directly related to the
equilibrium constant (Keq) by G = -RT ln Keq where T is the temperature in Kelvin and R is the gas constant.
The transition state (b) is a transient structure caught in the act of undergoing electron reconfiguration. The
symbol TS represents the transition state structure. Notice that the energy associated with TS is higher than the
reactants. This maximum on the energy profile represents a barrier that resists the transformation of reactants to
products. The higher the barrier is, the slower the rate of the reaction. The coefficient, k, is proportional to the
reaction rate (k and G are related by k = 2.084x1010Te-(G/1.986T)).
Activated-Rate Theory
Getting from reactants to products requires energy, e.g.,
energy to surmount a transition state barrier or energy to
become an unstable intermediate on the MEP. Where
does this energy come from?
In order to surmount barriers on the MEP, reactants
must somehow acquire energy from their surroundings.
This energy might come from heat, light, electrical
potential, or mechanical force. The barrier is thus only
a temporary one, because as soon as the reactants
acquire sufficient energy from their surroundings,
electron reconfiguration can take place. In the case of
proton transfer, the barrier is not very high and is easily
surmounted by the thermal energy (i.e., random,
Brownian motion) available to the reactants at room
temperature.
The energy transduction network (diagram at the right)
reminds us that potential energy of one form (e.g.,
electrical potential) can be converted into another form
(e.g., chemical potential). When molecules acquire
energy from their surroundings, they can surmount the
barriers on an MEP and reactions will proceed. This is
the basic idea behind the activated-rate theory which
describes how fast reactants transform into products.
The theory says that reactants must become activated to
their transition state structures by acquiring energy
from their surroundings.
developing
charge
developing
bond
OH
OH
H
O
diminishing
bond
diminishing charge
signifies transition
state structure
Case 1: the beginning state is higher than the ending state In this case
Hammonds postulate tells us that the TS will more closely resemble
the beginning state. The energy of the TS must be closer to the
beginning state. Theres just no other way to draw the diagram. We call
this TS early since the structure of the transition state has not
evolved far from its starting point.
Case 2: the ending state is higher than the beginning state In this case
Hammonds postulate tells us that the TS will more closely resemble
the ending state. The energy of the TS must be closer to the energy of
the ending state. We call this TS late since the structure of the
transition state has nearly evolved to its ending point.
OH
Keq
OH
free
energy
Gf
H
G
Gb
O
H
H
progress of the reaction
Summary
1.
2.
3.
4.
Curved arrow notation shows the changes in electron configuration for each
step of a reaction mechanism. !
Curved arrows indicate the partial bonds formed or broken in the transition
state (e.g., an arrow tail drawn from a bond represents a bond that is
breaking; an arrow head drawn between two atoms represents a bond being
made).!
The sum of the formal charges on each side of a reaction arrow must
balance. !
The atoms in transition state structures may have developing or diminishing
charge. Developing or diminishing charge in transition states is represented
by the partial charge symbol (e.g., + or ). If the formal charge on a
particular atom changes from one side of the reaction arrow to the other,
that atom is assigned a partial charge in the transition state.
Structure-Reactivity Relationships
A change in structure corresponds to a
chemical reaction. Reactivity is the term we
use to describe the potential of a structure to
undergo chemical change. Some structures
are more prone to react than others. Structures
that are likely to undergo chemical change are
said to be reactive.
Structures that are
resistant to chemical change are said to be
unreactive. We sometimes choose to speak
about the complementary property known as
stability. Structures that are highly reactive
are generally unstable. Structures that are
unreactive are generally stable.
Potential energy (specifically, chemical potential) is the link between structure and reactivity.
The greater a structures chemical potential, the greater is its reactivity. Understanding the link
between structure and reactivity will help us decide which, among several possible changes, are
most reasonable (e.g., pathways having intermediates with exceedingly high energies are not
likely to be reasonable and should therefore be avoided, especially if a pathway with lower
energy intermediates is available).
structure
chemical potential
reactivity
free energy
AB
GA
G = GB - GA
GB
(4) Coulombic
_______________
Like charges repel; opposite charges attract.
Si
Cl
Br
104
99
83a
93
73b
39
111
86c
53e
47
135
116d
65
45
37
76
72
83
65
103
81
46
52
87
68
71
52
48
56
91
61
58
74
52
56
108
135
53
60
In CF4
46
c
36
H
C
N
O
F
Si
S
Cl
Br
I
C C
H
H H
H
H Br
H C C Br
H H
the HBr and C=C double bond are broken. The energy of the C=C double bond is
not entirely lost, however, as the double bond becomes a CC single bond. Thus,
we will take this change into account by assuming that a C=C bond is broken and a
CC bond is made. Other new bonds include CH and CBr. The energies of all
bonds broken and all bonds made are then summarized as shown in the table below.
The reaction is favorable by 17 kcal/mol as a result of the increased net bond
strength. Note that this calculation is for bond enthalpies (H), not G.
Bonds changes (reactants)
bond
Energy (kcal/mol)
C=C
-146
H-Br
-87
energy
*
*
a
n
Typical nucleophiles
HO
Cl
R
C C
H2O
CH3NH2
Nu 1
Nu 2
Nu 3
Nu 4
energy
*
*
a
n
Typical electrophiles
R
O CH3
R C
H
H
H B
H
+
R R
H3C O
Cl Cl
E+ 1
E+ 2
E+ 3
E+ 4
Theres a good analogy between electrophiles and acids. Just as some acids are stronger than others,
some electrophiles are stronger than others. Whereas we speak of acidity to describe the strength of
acids, we speak of electrophilicity to describe the strength of electrophiles.
Given the HOMO energy
level of Nu as indicated:
Which frontier orbital
interaction will be the
strongest?
Which empty orbital is
most easily accessed?
Which electrophile has the
highest electrophilicity?
Type I
Type II
HX
Ka
acid
HX
acid
proton
Ka
proton
base
X
base
Acids:
Must have a hydrogen atom to donate
The most acidic hydrogen is often bound to a heteroatom (N, O, S, X)
In rare (but important!) cases, the acidic hydrogen is bound to carbon
Will be neutral or positively charged
Bases:
Must have a lone pair to accept a hydrogen
Will be neutral or negatively charged
Or
[A-]
[HA]
pKa
pH
[HA]
[A-]
[HA]
+ log
[A-]
Henderson-Hasselbalch eqn.
pH
Keq
Keq can be calculated by analyzing this reaction as a sum of two acid / base equilibria
Relevant
equilibria
_______________
Ka(1)
Ka(2)
1
Keq = Ka(2)
Ka(1)
pKa
____
-2.5
-1.7
pKa = - log Ka
10-(-1.7)
Keq = = 0.16
10-(-2.5)
6.
7.
8.
Each side of the equation will have one proton donor and one proton acceptor (i.e., one acid
and one base will be on each side of the equation).
Charge must be balanced.
Stoichiometry must be balanced.
Draw all electron lone pairs on all structures to see what bonding takes place in the proton
transfer process. In cases where the proton donor is a C-H group, the implied hydrogen
atom must be explicitly drawn.
Some molecules have more than one acidic (or basic) site. Use the rules of charge stability
to determine the location to add (or remove) a proton. If necessary, check your guess with
the pKa calculator.
The position of the equilibrium is predicted (qualitatively) by consideration of charge
stability. If necessary, the pKa calculator can be used to check your guess.
The equilibrium arrows ( ) indicate the reaction is reversible. A longer arrow is drawn
toward the species favored at equilibrium.!
Curved arrows are added to show the flow of electrons.!
circle one
CH3
H
H3C
N
H
H
O
H
or
STEP 7 - Once the arrows are drawn, finish drawing the structures. Take
advantage of ACE shortcuts. For example, double click on the structure
and copy it to the product box. Adjust the structure as required, but
dont draw it from scratch.
oxygen bases
hydrochloric acid
Et
Et
Et
pyridine
(Py)
triethylamine
(TEA)
NaOH
sodium hydroxide
NaOMe
sodium methoxide
NaOEt
sodium ethoxide
KOtBu
potassium tert-butoxide
nitric acid
organic acids
strong bases
O
H3C
OH
sodium hydride
NaNH2
p-toluenesulfonic acid
(PTSA, TsOH)
sodium amide
CH3CH2CH2CH2Li
CH3COOH
acetic acid
O
F3C
S
O
OH
trifluoromethanesulfonic acid
(triflic acid, TfOH)
NaH
CH3
H3C
butyllithium
(nBuLi)
CH3
N
Li
CH3
lithium diisopropylamide
(LDA)
O
OCH3
+ LiI
O
DMF ,
O
OCH3
LiI
DMF ,
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
CH3Nu + Br
CH3Nu + Br
Nu
C L
[SN2]
Nu C
An n *
-type interaction
Here bond making and breaking take place at the same time - a concerted process. Any one-step process is its
own elementary step. So well call the elementary step that results from concerted combination of [AN] + [DN] the
[SN2] step, which stands for nucleophilic, bimolecular substitution. Its bimolecular because two species must
come together. We need to re-consider the frontier orbitals involved in this new elementary step. The tail of the
first arrow implies the filled orbital (HOMO) is a non-bonded electron pair, n. The head of the first arrow points to
the empty orbital (LUMO, *) that accepts the electron pair. The empty orbital may not be obvious until you
realize that the C-L sigma bond breaks, as implied by the tail of the second arrow. Sigma bond breaking would
result upon filling * with the electron pair from n. The frontier orbitals align coaxially (a -type interaction).
C L
[DN]
[AN]
Nu
Nu
C
Filled
-L is the
leaving group
C L
[DN]
Empty a
[DN] involves a a
no-bond interaction
Curved Arrows Imply the Frontier Orbitals
Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals)
involved in an elementary step. The tail of the arrow implies the
filled orbital (HOMO) is electron pair. The head of the arrow
points to an atom on the leaving group (L) suggesting that an
atom-centered empty orbital (LUMO, a) accepts the electron pair.
The process produces no new bonding interactions. A good
leaving group will have a high energy and low lying a.
[DN] involves
sigma bond
breaking
+ L
empty
filled
#a
# a
# na
n #
# a
is the nucleophile
This symbol means an
association of a nucleophile
Filled n
Nu
[AN]
[AN] involves
sigma bond
making
Nu C
Empty a
[AN] involves an n a
-type interaction
Curved Arrows Imply the Frontier Orbitals
Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals)
involved in an elementary step. The tail of the arrow implies the
filled orbital (HOMO) is non-bonded electron pair, n. The head of
the arrow points to an empty orbital on the carbocation,
suggesting that an atom-centered empty orbital (LUMO, a)
accepts the electron pair. A good nucleophile will have a high
energy n.
empty
filled
#a
# a
# na
n #
# a
Br
HO
Br
HO
http://csi.chemie.tu-darmstadt.de/ak/immel/misc/oc-scripts/animations.html?structure=sn2
CH3 C+Br
slow
step
CH3
CH3 C
CH3
CH3
H
O
fast
step
CH3
CH3
CH3 C OH
Br
fast
step
CH3
H
H
H3O+
CH3 H
C O
CH3 H
H
[SN2] ---[SN1]
http://www.chemtube3d.com/
Governed by
electronic factors
Carbocation stability
favors
SN2
-7
-9
-10
This general trend holds for [SN2], [SN1] and other reactions that well study
3.2
Increasing basicity X
pKa
___
Increasing acidity HX
Nu
TS for SN2
+C
TS for SN1
>
>
>>
>
Increasing reactivity
increasing nucleophilicity
Nucleophile Reactivity
good
moderate
poor
<
<
conjugate
acid
pKa
4-5
15.7
17
<
<
<
basicity
row 3
<
<
DMSO
Reason: Cations are effectively solvated
but anions are poorly solvated and thus
the nucleophile is not shielded by solvent.
<
<
<
basicity
Polar protic
<
<
<
Summary
1.
2.
3.
4.
Comparing nucleophilic atoms that are in the same row, the stronger
the base, the stronger the nucleophile.!
Comparing nucleophilic atoms that are in the same column, the larger
the atom, the stronger the nucleophile.!
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
H C H
H C CH2CH3
KOH
Br
CH3
CH3 C Br
CH3
CH3
H2O,
C
C
C
C
KOH
KBr
CH2CH3
CH3
+
H
OH
HOH
HOH
Br
Alkenes are hydrocarbons that contain the double bond functional group. Alkenes are also called olefins and they
are said to be unsaturated compounds meaning that their carbon atoms do not have the maximum number of
allowed substituents (4 for carbon). As explained on the following page, heats of hydrogenation can be used to
determine stability for alkenes that differ by their degree of alkene substitution. The general conclusion is that
the more alkyl substituents that are bonded to sp2 carbons of an alkene, the more stable is the double bond.
H2 +
1.0 kcal/mol
trans
isomer
CH3
cis
isomer
-28.6 kcal/mol
-27.6 kcal/mol
H H
butane
CH3 C C CH3
H H
The heat produced by a reaction is accurately measured by a calorimeter. Lets compare the hydrogenation of cis and
trans 2-butene. Both reactions yield the same product, butane. The enthalpy change for the trans isomer is 27.6 kcal/
mol whereas for the cis isomer the enthalpy change is 28.6 kcal / mol. Since both reactions yield the same product,
we know that the energy difference comes from the difference of the starting alkenes. The diagram illustrates this
result, and shows that the trans isomer is more stable than the cis isomer by about 1.0 kcal/mol. The origin of this
stability difference is a steric interaction between the methyl groups. You can use MarvinSketch to convince yourself
of the close proximity of the methyl groups in the cis isomer.
Filled E
C C
[DE]
C C
Empty a
[DE] is a a
-type interaction
Curved Arrows Imply the Frontier Orbitals
Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals)
involved in the elementary step. The tail of the arrow implies the
filled orbital (HOMO) is an electron pair in a -bond. The head of
the arrow points between two carbon atoms and suggests a new
-bond. An empty orbital on the carbocation is available to
accommodate this electron pair implying that the LUMO is an
atom-centered, empty orbital (a). The HOMO (-orbital) is beside
(and nearly parallel to) the LUMO (a) allowing for a -type
interaction.
empty
[DE] involves
-bond breaking,
-bond making
filled
$a
$ a
$ na
n $
$ a
Empty * L
Filled n
C C
[E]
C C
[E] involves
-bond making
-bond breaking
[E] is an n *
-type interaction
empty
filled
$a
$ a
$ na
n $
$ a
$
L
BH
[pt]
B:
[DE]
[E1]
[DN] then [DE]
(two-step)
[E]
BH + L
[E1cb]
[pt] then [E]
(two-step)
[E2]
[pt] + [DN]
(concerted)
The three elimination pathways differ in the timing of the leaving groups departure. In the unimolecular elimination,
[E1], the leaving groups departs in the first step via a [DN] process. In the [E1cb] (conjugate base variant of the
unimolecular elimination), the leaving group departs in the second step via a [E]. In the [E1cb] mechanism, a
strong base must be present, the leaving group will generally be poor, and the beta hydrogen must be acidic. In the
bimolecular elimination pathway, [E2], the leaving group leaves at the same time as the beta hydrogen is loss. This is
a one-step, concerted process with [pt] + [DN] happening in unison.
http://www.chemtube3d.com/index.html
'
'
'
major
minor
"
"
http://www.chemtube3d.com/Elimination%20-%20E1.html
major
major
major
major
http://www.chemtube3d.com/Elimination%20-%20E2.html
Summary
1.
2.
3.
E2 Promoting Factors
Br
check the boxes corresponding to the conditions that favor the E2 mechanism
strong base (CH3O)
CH3
CH3CH2
CH3
H
C CH3
CH3
C CH3
X
heat ( )
no heat
http://wps.prenhall.com/wps/media/objects/725/742549/0044f.html
Starting with a
single enantiomer
planar
carbocation
(achiral)
http://www.chemtube3d.com/
http://www.chemtube3d.com/EliminationE2regioselectiveA.html
nar
a
l
p
eri xists
p
i
t
e
n
No a onship
ti
rela
ring-flip
http://www.chemtube3d.com/EliminationE2regioselectiveB.html
Which -H is this?
cis
http://www.chemtube3d.com/
Summary
Orbital alignment requirements in the [SN2] and [E2] pathways have
stereochemical consequences.
The
* sigma-type interaction of the [SN2] pathway results in
inversion of configuration at carbon.
In the [E2] pathway, the C H bond and the C X bonds break
simultaneously; these bonds must be parallel to one another to allow
the orbitals involved to be properly aligned.
Acyclic compounds can orient the two bonds in two ways:
synperiplanar (eclipsing at a 0 dihedral angle) or antiperiplanar
(staggering at a 180 angle). The antiperiplanar conformation is
generally lower in energy and the reaction usually proceeds
through this conformation.
In cyclic compounds, there are much greater restrictions on
conformational flexibility. In six-membered rings, the
antiperiplanar requirement for [E2] elimination is satisfied
when both the leaving group and the adjacent H atom are axial.
In principle, any lone pair can act as either a base or a nucleophile
toward C(sp3)X. Thus, it is sometimes difficult to decide if a
reaction will follow the [E2] or [SN2] pathway. The competition
between substitution vs. elimination depends on two factors: (1) the
nucleophilicity vs. basicity of the lone-pair and (2) the identity of the
alkyl halide substrate (1, 2, 3). The table summarizes the behavior.
Carbocation Intermediates
Carbocations are common intermediates in organic chemistry. We have
encountered carbocations in [SN1] substitution and [E1] elimination.
Although carbocations are never present in very high concentrations
(because they are unstable and highly reactive), they do open up a
reaction pathway through which important chemistry ensues.
Carbocations have a carboncentered empty p-orbital
C
carbocation
6-electron species
trigonal planar geometry
sp2 hybridized
empty p orbital on carbon
H
H
methyl cation
least stable carbocation
bonds directly attached to
the carbocation do not
share their electron density
H
H2C
eth
stabilized b
bonds dire
carbocatio
share their
Increasing stability
H
H
H
H2C
H
H
methyl cation
least stable carbocation
ethyl cation
stabilized by hyperconjugation
http://www.chemtube3d.com/index.html
H3C
CH3H
CH3
a long-distance
migration is the result of
successive 1,2-shifts
a secondary
carbocation
CH3H
a tertiary
carbocation
H
1,2-shift
H3C
H3C
H
C
CH3H
a secondary
carbocation
CH3
1,2-shift
CH3
Filled
Y
C C
[1,2R]
[1,2R] involves
-bond breaking,
-bond making
C C
Empty a
The 1,2-shift involves migration of a sigma bond (its electrons and
the attached atoms to that bond) from a carbon that is adjacent to
the carbocation onto the carbocation. It is illustrated for the
generic group labeled Y. The 1,2-numbering reflects the fact
that the migration takes place from atom labeled 1 to the
adjacent atom labeled 2.
[1,2R] involves a a
-type interaction
empty
filled
$a
$ a
$ na
n $
$ a
[1,2R]
CH3 H
a strained
4-membered ring
CH3
C C
CH3 H
C CH3
C
H
CH3
less strained
5-membered ring
Step 1
either way
is fine
Step 4
Step 5
2 carbocation
3 benzylic
carbocation
H3C CH3
CH3
CH3OH,
Cl
H3C
CH3
CH3
Br
CH3
CH3
CH3OH,
CH3
Breaking CO bonds:
OH is a poor leaving group and does NOT directly
participate in [SN1], [SN2], [E1] or [E2] reactions.
Why OH is a Poor LG
-10.0
-9.0
15.7
Protonation of an Alcohol
Improves the Leaving Group Ability
___
LG
_____________
Conjugate Acid
pKa of
____________
Conjugate
Acid
H 2O
H3O+
-1.7
[SN1]
Nucleophilic attack
on phosphorus by
oxygen
PBr3 is an electrophile
OH,
Nucleophilic
attack on sulfur
by oxygen
sulfonyl
chloride
sulfonate
ester
-1.0
(tosyl
abbreviated Ts)
(mesyl
abbreviated Ms)
(triflyl
abbreviated Tf)
pKa = ?
-3.6
Summary:
Converting
Alcohols,
Ethers and
Amines
into Leaving
Groups
http://wps.prenhall.com/wps/media/objects/725/742646/0048f.html!
Leaving Groups
Br
CH3CH2O H
P Br
CH3CH2O P
Br
a bromophosphite
Br
O
CH3O H
Cl
Cl S Cl
CH3O S
a chlorosulfite
Cl
O S O
CH3CH2CH2O H
C N
S
a tosylate
O
O
+
H
OH
H O S OH
H2SO4
Cl P Cl
Cl
O P Cl
Cl
CH3
CH3 O C CH3
H CH3
CH3 O C CH3
CH3
CH3CH2CH2NH2
CH3
CH3I
K2CO3
H H CH3
CH3 C C N CH3
[pt]
+
+
[SN2]
Regiochemistry Rationalization
carbocation character in the
Because there is some ____________
TS, attack occurs at the carbon better able to bear a
partial positive charge.
protonated
intermediate
2nd best
resonance contributor
Option 1
[AN] at O+
H
O H
CH3
Nu
H
O Nu
CH3
oxygen
exceeds octet
Option 2
[SN2] at O+
H
O H
CH3
Nu
H
O Nu
charges are
CH3 not reasonable
Summary of
Epoxide Reactivity
Complete the diagrams (use the link for help)
http://wps.prenhall.com/wps/media/objects/725/742646/AADFQQH0.html
acidic conditions
O
OCH3
H2SO4 (cat.)
CH3OH
OCH3
OH
HOSO3H
HOCH3
O
1
OCH3
basic conditions
OH
O
CH3ONa
CH3OH
OH
OCH3
O
O
product
HOCH3
Carbon-Based Nucleophiles
A special class of carbon-centered nucleophiles
organometallic compounds
are _____________
M is less electronegative than C
"
+"
<
<
Increasing stability
<
Grignard (Organomagnesium)
New CC bond
Summary
1) Heres an important general rule to be followed in writing all reaction mechanisms. If a reaction is executed
under acidic conditions, no strong bases can be present. Any negatively charged species must be a weak
base (e.g., Cl). If a reaction is executed under basic conditions, no strong acids can be present. Any
positively charged species must be a weak acid.
2) Under basic conditions, a hydroxyl group cannot be substituted directly because HO is too poor of a leaving
group. The hydroxyl group must first be transformed into a derivative that is a suitable leaving group.
3) Under acidic conditions, the R3CX+ bond can ionize spontaneously to give X: and R3C+. The likelihood of
ionization depends on the stability of the carbocation product, the leaving group and the solvent.
4) Under acidic conditions, a lone pair nucleophile bearing a proton (e.g., H2O, ROH, RCO2H, RNH2) is always
deprotonated after it adds to the electrophile (e.g., carbocation).
5) Under acidic conditions, epoxides open by attack of the nucleophile onto the protonated ring. The
nucleophile attacks the most highly substituted carbon.
6) Under basic conditions, epoxides open by nucleophilic attack onto the least substituted carbon (the reaction
resembles an SN2 reaction; the leaving group is a poor one, but the process is driven by the release of ring
strain).
7) If carbocations are intermediates, rearrangement is possible.
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
CH3
+ HBr
1
Br
H2C CHCH3
+ H2O
2
H H
H
H2C C CH3
H OH
a non-nucleophilic
solvent (i.e., inert)
Br Br
CH2Cl2
Br
cyclopentene
Br
a nucleophilic solvent
participates in the reaction
+
cyclopentene
Cl Cl
H2O
H
Cl
a chlorohydrin
H
OH
+ HCl
Empty a
[AE]
Filled
[AE] involves
-bond breaking and
-bond making
[AE] involves a a
-type interaction
empty
filled
#a
# a
# na
n #
# a
[AE]
E
C C
[AN]
E
C C
Nu
Nu
When E+ has no
lone pair (e.g., H+)
[AE]
E
C C
[SN2]
Nu
E
C C
Nu
Depending on whether or not the electrophile has an electron lone pair will determine the nature of the
intermediate resulting from the [AE] step. If the electrophile has a lone pair, then a 3-membered intermediate
will form in order to satisfy the octet rule. If no lone pair exists on E+, then [AE] results in a carbocation.
[AE] Drawn
in ACE
Addition of H-Br
When the Electrophile Has No Lone Pair,
[AE] Gives a Carbocation Intermediate
H
Draw the
1st arrow
After the
2nd arrow
C C
H
[AE]
Check the
1st arrow
H H
H
H Br
H C C Br
H H
H
H
H
C C
[AN] Drawn
in ACE
Check the
arrow
[AN]
H Br
intermediate
Draw the
arrow
http://aceorganic.pearsoncmg.com/epoch-plugin/public/mechmarvin.jsp
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Clean in 2D
free energy
H
H
+ H
C
H
H
H
H
H2C CH2
C C
H
H
H
+ H
C C H
Br
H
H Br
intermediate
HBr
CH3 CH2Br
Unsymmetrical Alkenes
CH3
CH3
HBr
this carbon bears
no hydrogen atoms
an
unsymmetrical
alkene
CH3
+
Br
H H
major
H Br
minor
The first step of the mechanism involves the formation of a carbocation intermediate. Because
the alkene is unsymmetrical, there are two possible carbocations that can form. The relative
stability of these carbocation intermediates (technically, the transition state leading to these
intermediates) determines which product is formed the fastest. Hammonds postulate allows us
to predict the relative energy of the TSs leading to the two possible intermediates. Draw a
representation of the transition state structures for both modes of reactivity. Your representations
should show that one begins to develop character of a secondary carbocation, while the other
develops character of a tertiary carbocation. The TS leading to the secondary carbocation is a
higher energy species comparied to the TS leading to the tertiary carbocation intermediate.
Thus, the secondary carbocation pathway represents a higher barrier to reaction. In contrast, the
TS leading to the tertiary carbocation is a lower energy species, and thus represents a lower
barrier to the reaction pathway. The reaction coordinate diagram illustrates this logic and helps
explain why the major observed product results from attack of bromide on a tertiary carbocation.
Start from the reactants. Notice the pair of diverging pathways, one leading to the major product
and the other leading to the minor product.
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H2O
Balance charge
at every step
Step 1 combines
neutral reactants
and H+. The
charge at the end
of step 1 must be
+1.
CH3
H
C C
H
H
H
step 1
OH
H
CH3
[AE]
H2O
[AN]
step 2
step 3
CH3
[pt]
Step 2 combines
the positively
charged
intermediate
from step 1 and
neutral H2O.
The 2nd
intermediate
must also have a
+1 charge.
Notice first, that the stoichiometry is balanced. Water is probably the solvent of this reaction, and in this case, it
participates in the reaction, too. Such behavior where the solvent gets incorporated into the product is fairly common in
organic chemistry. In contrast to water, the H+ above the equilibrium arrow does not enter into the stoichiometry.
However, H+ does play an important role. If you were to combine water and an alkene without any acid, no change will
take place for a long time, at least not at lower temperatures. These observations point to the notion that the reaction is
catalyzed by H+, meaning that acid enhances the rate of the reaction, but does not show up in the stoichiometry. So, when
you see H+ or HO listed above the arrow, but not enter the stoichiometry, think about acid or base catalyzed reactions.
Complete the mechanism, reasoning by analogy to the [AdE2] mechanism for H-Br addition to an alkene.
Br Br
Br
[AE]
CH2Cl2
Br
Br
[SN2]
Br
H
Br
[AN]
H
H
Br
Br
This carbocation does not
satisfy the octet rule, yet
there is a way to do so.
Br
This bromonium ion satisfies the octet rule.
It's true that it is highly strained and has a
positive charge on an electronegative atom;
nonetheless, this is a better structure than
an electron deficient carbocation.
Br Br
[AE]
H
Br
Br
Draw the
1st arrow
After the
2nd arrow
Draw the
3rd arrow
After the
3rd arrow
Clean in 2D
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H
Br
Br
Br
Br
Br
Br
Br
Br
second best contributor
Br
best contributor
Br
second best contributor
CH3CH2C CH
[AdE2]
HBr
H
CH3CH2C C
Br
[AE]
H
CH3CH2C C
H
a vinylic cation
intermediate
http://wps.prenhall.com/wps/media/objects/724/742058/0019f.html
Br
[AN]
CH3CH2C CH2
CH3CH2CH CH
more stable
less stable
The vinylic cation possesses a positively charged carbon atom having only two electron pair
domains (EPDs); the carbon bearing the positive charge is thus sp-hybridized. The addition of the
electrophilic H+ takes place so as to place the positive charge on the most highly substituted carbon,
just as for alkenes. As before, the more highly substituted the vinylic cation, the greater its
stability. This preference determines the regioselectivity of the addition. As a consequence, the
bromide is located on the more highly substituted carbon (i.e, the internal carbon). It turns out that
the vinyl cation is much less stable (i.e., much more reactive) than the corresponding alkyl
cation. One reason is because sp hybrid orbitals, consisting of 50% s character, are less well-suited
to stabilize positive charge compared to sp2 hybrid orbitals that have only 33% s character (the s
orbital resides closer to the positively charged nucleus).
primary alkyl cation
R C H
H
RCH CH
3-EPDs
(sp hybridized)
2
2-EPDs
(sp hybridized)
Successive Addition
Since the addition product to an alkyne is a vinyl bromide a type of alkene it
should not be surprising that subsequent reaction involving further addition is
possible. In fact, successive addition tends to be the normal mode of reactivity for
alkynes, as addition to the alkene is generally faster than the initial alkyne addition.
H
CH3CH2C C
Br
Br
HBr
CH3CH2C C H
Br H
a geminal dibromide
a vinyl bromide
(an alkene)
One reason why the subsequent reaction is faster is because of the special stability of the carbocation
intermediate that results when an electrophile adds to a vinyl bromide. The lone pairs on the bromo
group can donate to the electron deficient carbocation. Although this places a positive charge on the
electronegative bromo group, all atoms have an octet of electrons making this a very important
resonance contributor.
H
CH3CH2C C H
Br H
n a -type
resonance interaction
H
CH3CH2C C H
Br H
the most important resonance
contributor (each atom has an
octet of electrons)
Br2
CH3CH2C CCH3
CH3CH2
Br
Br
CH3CH2
Br
CH3
a second addition
usually occurs
CH3
Br Br
Br
a bromonium ion
intermediate
CH3CH2C CCH3
a tetrabromide
Br Br
H3C
CH3
Bonds Made
Bonds Broken
C1=O
C2-H
C2-H
C1C2
H-O
H-O
+ HOH
H3C
H2SO4
O
1 2 CH3
H
The addition of water to an alkyne initially results in an enol. The enol, whose name is derived from alkene + alcohol, is
generally not the form that is most stable. Enols tend to undergo a process known as tautomerization, an equilibrium
reaction which shifts the position of a double bond together with a C-H, N-H or O-H bond (in this case, an O-H). A bond
energy calculation will show why the equilibrium favors the keto form. What we are after right now is not the knowledge
that this reaction is a tautomerization, but to understand the mechanism of how tautomerization occurs.
H O
O
1
CHCH3
H3C
an enol
H
H3C 1
tautomerization
CH3
H H
a ketone
The acid-catalyzed tautomerization mechanism is included with the discussion problem. You can fully
understand this mechanism using elementary steps that you previously learned.
To begin, it should not be surprising that the enol is a nucleophilic -bond capable of
undergoing [AE]. This -bond nucleophile finds an electrophilic partner in H+ (from H2SO4).
The new C-H bond must be made at carbon atom C2, consistent with the need for a new C2-H
bond. The resulting carbocation is stabilized by resonance with the lone pair of oxygen (Box 4).
This resonance contributor has an octet of electrons on every atom. Moreover, this resonance
contributor has the necessary C=O bonding of the final product. All that remains to realize the
final product is dissociation of a proton, breaking the final O-H bond in our bonds made / bonds
broken list. Notice how H2SO4 is regenerated each time it is used, consistent with its use as a
catalyst.
O
oleic acid (an unsaturated fatty acid)
Pd-C
H
H H
H H
hydrogen
OH
The addition of H2 occurs only in the presence of
a metal catalyst (Pd, Pt or Ni) that is adsorbed
onto a finely divided inert solid, such as charcoal.
The catalyst 10% Pd on carbon is common. The
reaction takes place at the metals surface but we
wont study the details.
O
steric acid (an saturated fatty acid)
OH
H2
Pt / C
H
CH3CH2CH C
H
H2
Pt / C
H
CH3CH2CH2 C H
H
Hydrogen adds to alkynes in an analogous way as with alkenes. The initially formed
alkene generally continues on to the saturated alkane under the reaction conditions.
CH3CH2C CCH3
H2
Lindlar's
catalyst
H
CH3CH2
H
CH3
a cis alkene
The addition of H2 to triple bonds is somewhat faster than to alkenes; thus, it is possible to stop
the reaction at the alkene using a type of poisoned or deactivated catalyst known as Lindlars
catalyst. Lindlars catalyst is stereoselective for the formation of the cis alkene.
http://wps.prenhall.com/wps/media/objects/724/742058/0024f.html
CH3 C C H
an internal alkyne
a terminal alkyne
compared to other CH
bonds this one is acidic
CH3 C C H
CH3 C C
an acetylide anion
pKa = 25
H
CH2 C
H
H
H
CH2
a vinylic anion
pKa = 44
The fishhook
arrow tool.
How it looks
in ACE
http://aceorganic.pearsoncmg.com/epoch-plugin/public/mechmarvin.jsp
HBr
tert-Butyl peroxide
heat
Br
n-butyl bromide
The thermally induced bond homolysis of peroxides is a common way to initiate radical reactions.
The oxygen-oxygen bond of peroxides is a weak covalent bond of 47 kcal/mol and hence is readily
susceptible to thermal dissociation. The result is a pair of oxygen-centered radicals.
Initiation
RO OR
RO
a peroxide
Thermal homolysis of peroxides results in a pair of oxygen-centered radicals. The next step in
initiation involves HBr. A newly formed oxygen-centered radical abstracts a hydrogen atom
from H-Br resulting in a bromine-centered radical and a hydroxyl group. The bromine-centered
radical then enters the propagation cycle.
RO
H Br
RO H
Br
Propagation
The first step of the propagation cycle involves the addition of the bromine-centered radical to an alkene. This
step creates a carbon-centered radical, and it determines the isomer that is formed (i.e., n-butyl vs. sec-butyl
bromide). The formation of the most stable carbon-centered radical is what governs the outcome of this crucial
step. By having the bromine-centered radical add to the terminal position of the alkene, a secondary carboncentered radical results. This pathway is lower energy than the addition that leads to a primary radical.
The second half of propagation involves the carbon-centered radical abstracting a hydrogen atom from another
molecule of HBr. This regenerates a bromine-centered radical along with the alkene addition product. The brominecentered radical is ready to reenter a new chain propagation cycle.
H
CH3
Br
H Br
H
CH3
H
Br
Br
Termination
Termination events are the ways that radicals annihilate themselves. All propagation steps involve
a radical combining with a non-radical species. Through combination of a pair of radicals, the
radicals will disappear. Because the concentration of radicals tends to be very low, these events
are relatively rare. Some examples of termination reactions are shown below. All of them involve
the combination of two radicals, and they produce non-radical products.
Br
Br
Br Br
H
Br
CH3
Br
H
CH3
Br
CH3
H
Br
H
Br
CH3
Br
Br
Br
CH3
CH3
H2
Pd / Al2O3
CH3CO2H
CH3
CH3
+
CH3
CH3
anti
syn
73%
27%
The specific set of conditions written above and below the reaction arrow describe to a chemist
important details about how the reaction was performed. In this particular case, palladium metal
was deposited on solid aluminum oxide and the reaction was conducted in a solvent of acetic acid
(CH3CO2H) in the presence of gaseous hydrogen. Please note that it is NOT important for you to
memorize these specialized conditions or even understand the mechanistic details of how these
conditions produce anti products. What is important is that you have the ability to examine a
reactant / product pair and determine the mode of addition (syn vs. anti).
Br
Br
Br2, CH2Cl2
trans 2-pentene
Br
Br
enantiomeric pair
Br
Br
Br2, CH2Cl2
cis 2-pentene
Br
enantiomeric pair
http://www.chemtube3d.com/
Br
[SN2]
[SN2]
bromonium ion
from cis-2-pentene!
A Non-Stereospecific Example
Reactions that proceed by non-stereospecific mechanisms will result in a much greater array of
stereoisomers. Consider the reaction of HBr adding to an alkene (recall that the addition of HBr to an
alkene proceeds via either a cationic or radical intermediate). As a specific case, the addition of HBr to
cis-3,4-dimethylhex-3-ene results in all four possible stereoisomers as shown below. Those isomers
related as enantiomers must be formed in equal amounts. Those isomers related as diastereomers may be
present in unequal amounts. Think through the mechanism and decide if trans-3,4-dimethylhex-3-ene
will produce the same stereoisomer mixture as cis (will the proportions of the various isomers be the
same for cis vs. trans?).
CH2CH3
H3C
H3C
Br
CH2CH3
enantiomers
CH3
Br
CH3
CH2CH3
CH3CH2
CH3CH2
CH3
HBr
diastereomers
CH2CH3
diastereomers
diastereomers
CH3
CH2CH3
cis-3,4-dimethylhex-3-ene
H
H3C
CH3
Br
CH2CH3
CH2CH3
enantiomers
H3C
Br
H
CH3
CH2CH3
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
Aromatic Substitution:
Four Examples
I. Electrophilic Aromatic Substitution
O
OH
HNO3
H2SO4
Cl
NO2
O
NO2
NaOH
H2O, 100 C
NO2
NO2
H OH
Aromatic Substitution:
Four Examples
III. Substitution via arenediazonium ion
CH3
Na
NH2
CH3
NH2
+
NH2
NaCl
E
[AE]
H
[DE]
Nu
Nu
EWG
EWG
Nu
[AdN]
Nu LG
[E]
Nu LG
LG
LG
MO picture of
the anionic
intermediate
Nu LG
EWG
EWG
EWG
resonance stabilized anionic intermediate
The HOMO shows positions
of surplus electron density
Nu
+
H
[AN]
[DN]
N2
H
H
Nu
N N
H
H
H
side
H
an aryl cation
The LUMO shows
positions of
electron deficiency
Cl
NH2
NH2
[E2]
[pt]
NH3 + Cl
H NH2
H
H
benzyne
side
NH2
[AdN]
Two views of the
benzynes LUMO
top
NH2
H
H
NH2
H
NaCl
LUMO
LUMO
electron
withdrawing
F
F
F
C
reference
HOMO
electron
donating
H
H
H
C
HOMO
HOMO
top
side
LUMO
top
side
HOMO
top
side
LUMO
top
side
EWG
EWG
RDS
RDS
E
E
EDG
EDG
E
E
A B C
TS resembles
the intermediate
free energy
A B C
starting
products
substrates
progress of reaction
Comparing Intermediates
For Rings with an EWG vs. EDG
EWG
O
E H
EDG
CH3
deactivating
CH3
activating
CH3
E H
CH3
CH3
CH3
E H
X
H
+ E
or
or
E
X is EWG
or EDG
ortho
meta
E
para
Ortho / para
directing
From P. Y. Bruice
Organic Chemistry (5th Ed.)
Summary
The bonds in aromatic compounds behave as nucleophiles and are thus reactive toward
electrophiles. In the first step of electrophilic aromatic substitution, the aromatic ring attacks an
electrophile in an [AE] fashion to give a delocalized carbocation intermediate."
In the second step of electrophilic aromatic substitution, the carbocation undergoes [DE] with loss
of H+ from the same carbon to which the electrophile added, re-forming the benzene ring (the net
result is substitution at this C)."
The electronic nature of substituents on an aromatic ring has profound effects on the rate of
reactivity and the regiochemistry of electrophilic aromatic substitution reactions."
Draw your curved arrows correctly. Arrows should not show the H+ flying off into space! If H+ loss
via [DE] bothers you, show a weak base making a new bond to hydrogen at the same time the CH
bond is cleaved (technically now a [pt] step). Often the base is assumed to be present but need not be
shown."
B
[DE]
CORRECT
[pt]
CORRECT
H WRONG
Which best represents the transition state for the first step in the electrophilic aromatic substitution pathway?"
A"
+
+
+
+
+
E
+
E
E
H
E
B"
C"
H
H
+
D!
+
+
Bromination
Br
Br Br
Lewis Base
Br2
FeBr3
Br
Fe
Br
Br
Lewis Acid
[AN]
HBr
Br
Br Br Fe Br
Br
Other Halogenations
Chlorine is similar to bromine
Cl
FeCl3
Cl2
HCl
HNO3
I2
HI
OH
Here the active electrophilic
N
O
I
species is presumed to be ________
J.
H
O
H2SO4
OH
HNO3
O
H OH
N
O
H O S O H
[pt]
O
H
N
O
+
H
[E]
H2O
O N O
http://www.chemtube3d.com/
O S O H
O
Sulfonation
OH
O S O
H2SO4
H2O
O
The active electrophilic species is H O S
O
O
H O S O H+
O
H O S O H
[pt]
[DN]
H O S
O
http://www.chemtube3d.com/
H O S O
O
O
H2O
O S O H
O
Friedel-Crafts Acylation
O
O
R
Cl
acyl chloride
+
H
R
O
R
acid anhydride
1) AlCl3
2) H2O
1) AlCl3
2) H2O
http://www.chemtube3d.com/
R C O
AlCl4
Cl
O
R C Cl
Cl
Al
[AN] (or
[AdN])
[DN] (or
[E])
[AN]
ion pair
dissociation /
formation
Cl
R C Cl Al
Cl
Cl
Cl
Friedel-Crafts Alkylation
H
R Cl
AlCl3
HCl
Example:
H
Cl
isobutyl
chloride
+
tert-butylbenzene
HCl
Cl
CH3 CH CH2Cl
Cl
Al
CH3
Cl
[AN]
Cl
CH3 C
CH2Cl Al
Cl
Cl
[1,2R] + [DN]
H
CH3
[AE]
CH3
CH3
http://www.chemtube3d.com/
AlCl4
HNO3
H2SO4
NO2
1:1
HNO3
H2SO4
NO2
NO2
1 : 4.5
NO2
CH3
CH3
+
NO2
p-nitrotoluene
HNO3
H2SO4
NO2
NO2
2,4-dinitrotoluene
Summary
Electrophiles preferentially attack ortho or para to electron-donating substituents such as
(RO, R2N, RCONH) and halo substituents, whereas electrophiles attack meta to
electron-withdrawing groups such as carbonyl, CN, NO2, and SO3H.!
The directing ability of a group is closely related to its ability to stabilize or destabilize
the intermediate carbocation.!
Give the major product of these reactions.
CH2CH3
+ Br2
CH2CH3
CH2CH3
CH2CH3
Br
Br
CH2CH3
FeBr3
Br
Br
Br
NO2
+ HNO3
Br
B
Br
C
Br
Br
Br
NO2
H2SO4
NO2
OCH3
CH2CH2Br
OCH3
NO2
OCH3
OCH3
NH2
Sn, HCl
NO2
NH2
Pd/C, H2
Mechanism of Diazotization
O N O
sodium
nitrite
[pt] H
H O N O
nitrous
acid
[pt] H
[pt]
NH2
N N O
[AN] or [AdN]
H
H
[pt] H
N N O
H
H
N N O
[E] or
H2O [DN]
O N O
N O
N O
nitrosonium ion OR
nitrosyl cation
[pt]
H
H
N N O
[pt]
H
H
N N
H2O
[E]
H
H
N N O
H
Mechanistic Considerations:
Substitution of Arenediazonium Ions
In these reactions the leaving group leaves before the Nu adds
(compare to electrophilic aromatic substitution mechanism)
H
H
H
H
N N
[DN]
H
H
H
H
aryl cation
Nu
[AN]
Nu
H
H
Benzene as an Electrophile:
Nucleophilic Aromatic Substitution
Cl
NO2
O
NO2
NaOH
H2O, 100 C
NO2
NO2
addition / elimination
This reaction goes by an ___________________
pathway,
analogous to nucleophilic acyl substitution (recall that substitution
reactions at sp2 carbon do not follow the [SN2] pathway).
Mechanism of Nucleophilic
Aromatic Substitution
Cl
NO2
Cl OH O
N
OH
[AdN]
addition step
(slow step)
NO2
Cl OH O
N
Cl
O
NO2
Cl OH O
N
O
O
[E]
elimination step
(fast step)
NO2
Cl OH O
N
ortho or _____
The EWG must be located ______
para positions.
A variety of nucleophiles can be used. The incoming
group must be a stronger base than the outgoing
leaving group.
OCH3
Na
Br
NH2
OCH3
Br
Na
NH2
Na
NaBr
NH2
OCH3
OCH3
NH2
OCH3
NaBr
NH2
+
NH2
NaBr
H
aldehyde
ketone
thioketone
N
C
imine
O
OH
carboxylic
acid
O
NH2
amide
nitrile
O
O
carboxylic
ester
O
S
thioester
N
H2N
NH2
urea (or guanidine)
O
carboxylic
anhydride
O
Cl
acid chloride
, unsaturated
carbonyls
Leaving group
(substitution
reactions)
Basic site:
protons add here
Electrophilic site:
nucleophiles
attack here
Acidic hydrogen:
deprotonation creates
a nucleophile
O H
Nu
Nu
O
R
O
R
Nu
H H
H E
The X group is
replaced by the
Nu group. This
is substitution at
an sp2 carbon.
Overall result is
substitution of -H at
the -carbon by -E.
+
2nd best
resonance
contributor
O
C
formaldehyde
Reason:
Steric factors
Better electrophile
Reason:
Electronic factors;
LUMO energy level
better electrophile
O
C
add
H+
O
C
Circle the
stronger base
A pKa = -7.2
(stronger acid)
B pKa = -6.5
(weaker acid)
stronger
acid
free energy
O
C C
H
+B
C C C
H
+B
reaction progress
http://www.chemtube3d.com/
These are good rules to follow when you write any organic reaction,
but they are especially common in carbonyl chemistry (We will see
specific examples in the next several lessons).
2. Proton-transfer/beta-elimination
3. A series of redox reactions converts the chromium from the 4+ oxidation state to the 3+
oxidation state
Summary
Tautomers are isomers in which a double bond and a hydrogen atom change their locations. The
most important kinds of tautomers are carbonyl-enol tautomers. Tautomerization is a chemical
equilibrium that occurs very rapidly in acidic or basic media; it should not be confused with
resonance (resonance is not an equilibrium process).
O
OH
CH2
CH3
<1%
C
H2
CH3
> 99 %
An organic acid, HA, is much more acidic when the lone pair of the conjugate base can be
stabilized by resonance (e.g., PhOH is more acidic than EtOH). A C-H bond is acidic when the lone
pair of the conjugate base (a carbanion) can be delocalized into a carbonyl group, and even more so
when it can be delocalized into two carbonyl groups.
Carbonyl compounds are some of the most important organic compounds. To help understand their
behavior, remember these pKa values:
O
H
CH3
N
H
pKa = 5
O
CH3
pKa = 16
H
H
C
H
CH3
pKa = 20
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
Filled n
Nu
C O
[AdN]
Empty *
C O
Nu
[AdN] involves
-bond making and
-bond breaking
[AdN] is a n *
-type interaction
empty
filled
#a
# a
# na
n #
# a
pathway
H
Nu
[pt]
C O
[AdN2]
Nu
C O
H
alkoxide
adduct
O
C
[pt]
[AdN]
C O
C O
H
Nu
Nu
[AdN2]
[pt] then [AdN]
(two-step, acidic
conditions)
The two addition pathways differ in the timing of the proton transfer and nucleophile addition. Both are two-step
mechanisms and both are called [AdN2] (bimolecular nucleophilic addition to a polarized -bond). The
bimolecular [AdN] step is rate-determining for both. In the top case, the two steps are often actually performed
sequentially (i.e., in the laboratory, a proton source is subsequently introduced to quench the initially formed
adduct). For the top case nucleophilic addition is under basic conditions and an acid then neutralizes the initially
formed alkoxide adduct. In the bottom case, reversible proton transfer takes place followed by [AdN]. Unlike the top
case, this two-step process occurs without changing conditions (no intervention on the part of the chemist is needed).
Reacts like a
carbanion Nu:
http://www.chemtube3d.com/
CH3 C C H
CH3 C C
an acetylide anion
pKa = 25
H
CH2 C
H
H
sodium acetylide
H
CH2
a vinylic anion
pKa = 44
http://www.chemtube3d.com/
a cyanohydrin
http://www.chemtube3d.com/
Nucleophilic Addition to
, Unsaturated Carbonyl Compounds
Filled n
Nu
EWG
C C
Empty *
[AdN] is a n *
-type interaction
EWG
[AdN]
C C
Nu
[AdN] involves
-bond making and
-bond breaking
H
H
Nu
[pt]
C C
Nu
C C
C O
C O
enolate anion
C C
C O
Nu
[pt]
[AdN]
C C
[taut]
C C
C O
C O
Nu
Nu
enol
H
C C
C O
Nucleophilic addition to a polarized carbon-carbon -bond takes place by analogous pathways. The top case is just
like carbonyl addition except that [AdN] occurs at the C site. The initially formed enolate anion is resonance
stabilized (via electron donation into the withdrawing group (e.g., C=O)). For the bottom case, the proton first adds to
the carbonyl oxygen. Then, [AdN] occurs at the C site giving rise to an enol. The enol is usually unstable and
undergoes spontaneous tautomerization to produce the addition product.
C N
CH3CO2H
H H
HO
Li
http://www.chemtube3d.com/
C
H
OH
C
H
Nu
fast
O
C
H
C
H
slow
O
H
C
CH2 C
Nu
C
Nu
fast
C
H
C
H
C
H
kinetic product
slow
O
H
C
CH2 C
Nu
thermodynamic product
Summary
Alkenes and alkynes that are substituted with electron-withdrawing groups such as carbonyl,
nitro, or sulfonyl groups are electrophilic. Many nucleophiles react with these alkenes to give
conjugate addition (i.e., 1,4-addition) products.!
The enol resulting from 1,4-addition undergoes spontaneous tautomerization to give alkene
addition as the net result.!
Nucleophiles that are weak bases such as alcohols, thiols, amines, and halides, add reversibly
to ,-unsaturated carbonyls and tend to give the thermodynamic product (i.e., typically the
product of conjugate addition). Nucleophiles that are strong bases add irreversibly and give
direct addition (i.e., 1,2-addition) products.!
1,2- vs. 1,4-Additions to ,-Unsaturated Carbonyl Compounds
O
nucleophile
CH3CH2 MgBr
H3O+
C N
H3O+
H3O+
http://wps.prenhall.com/wps/media/objects/725/743227/0089f.html
acid
acid
+ X
base
form bond
-XY
4) Substitution - replace Cs
substituent (-X) with another (-Y),
neither being -H
Y
replaces
X
Y
X
C
Y
C
+ X
elimination
Z
addition
break bond
+XY
Z = C, N, O
7) Rearrangement - isomerization
process (no atoms lost or gained);
results in new bonding connectivity
(one of many examples shown as
there is no generic representation).
CH3
R
R
R stands for a generic "residue"
CH3 O
CH3 C Cl
CH3 C OCH3
+ Cl
H
CH3 NH2
CH3 NH
H
CH3
CH3 NH2
O
O
CH3
NHCH3
O
CH3
+
O
CH3 OH
CH3 C OH
esterification
H
O
CH3 C OCH3
H OH
CH3
Nu
Nu
Nu
C O
[AdN]
C O
[E]
H
base
Nu
O
C
[AdN]
[pt]
B
H Nu
C O
L
H B
Nu
Nu
C O
[pt]
H
[pt]
H L
[AdN]
C O
L
C O
[E]
Nu H
H Nu
[pt]
C O
L
H
Nu
C O
H L
[E]
Nu
C O
H
All three pathways include [AdN] and subsequently [E] steps (i.e., addition followed by elimination,
proceeding through tetrahedral intermediate [TI]). The three paths differ by the role of [pt]. Which pathway is
best will depend on the pH, the nucleophile strength, and the leaving group quality. A rough generalization is
that the top pathway is for basic, the middle for neutral, and the bottom for acidic conditions. Slight variations
on these pathways will be encountered.
Empty * L
-L is a LG;
Z: is O or N
Filled n
C C
L
C
Z
Y
[E]
[E]
C C
Z
C Y
[E] involves
-bond making
-bond breaking
[E] is an n *
-type interaction
We first encountered [E] when we studied the [E1cb] elimination pathway (top equation). In the second
equation, [E] is now shown for the generalized case where Z: may be an oxygen or nitrogen atom. Many
leaving groups (-L) participate in [E], even poor ones (e.g., -OH). This step is the reverse of [AdN].
empty
filled
%a
% a
% na
n %
% a
O
R
CH3O
Cl
addition
[AdN]
Cl
OCH3
elimination
O
R
OCH3
[E]
Cl
tetrahedral
intermediate
[TI]
Heres a related mechanism on the
formation of amides via acid chlorides.
http://www.chemtube3d.com/
free energy
free energy
relative basicities:
Nu > L
[TI]
O
R C Nu
R C L
+ L
+ Nu
reaction progress
leaving
group
(conjugate
acid pKa)
R C Nu
+ L
R C L
+ Nu
reaction progress
O
Cl
O C R
(-2.2)
O
derivative
[TI]
R C Cl
acyl chloride
(4.8)
O
OH
(15.7)
R C O C R
R C OH
acid anhydride
acid
O R
(16)
O
NH2
(35)
O
R C O R R C NH2
ester
amide
O
CH3 C OCH3
addition
OH
CH3 C OCH3
carboxylic ester
OH
[E]
elimination
O
CH3 C O
carboxylate
CH3OH
CH3 C OH
OCH3
carboxylic acid
Note that under basic conditions, all intermediates are either neutral or negatively
charged. All steps are reversible, but the last step lies very far to the side of the
carboxylate.
This nearly irreversible step makes formation the reverse process
(formation of esters from acids under basic conditions) nearly impossible.
ester
[pt]
pyridine
[E]
[AdN]
carboxylic
ester
Note that under acidic conditions,
all intermediates are either neutral
or positively charged. All steps are
reversible so formation and
hydrolysis are related as the
forward and backward directions of
these pathways.
For ways to
manipulate the equilibrium see the
link below.
[E]
http://www.chemtube3d.com/
O
R C NH2
H OH
R C OH
HCl
+ NH4Cl
H
H
R C NH2
R C NH2
R C OH
R C NH2
H
+ NH3
elimination
addition
H
R C NH2
O
O H
R C N H
H
O H
Decreasing reactivity!
http://wps.prenhall.com/wps/media/objects/725/743131/0087f.html
thionyl
chloride
chlorosulfite
Summary
Many carbonyl compounds including esters, acyl chlorides, and acid anhydrides
have leaving groups attached to the carbonyl C, and many reactions proceed with
substitution of this leaving group by a nucleophile.!
Substitutions at the carbonyl C usually occur by an addition-elimination
mechanism. The nucleophile (Nu) adds to the electrophilic C of the carbonyl
group to make a tetrahedral intermediate. The leaving group (L) then leaves in
the elimination step to give a new carbonyl compound. Note that either L or Nu
may be expelled from the tetrahedral intermediate. Which one that is expelled
depends on their relative leaving group abilities and the reaction conditions.
Expulsion of Nu gives back the starting material (which is not very productive!).!
(1) What products are observed from the following reaction (the
oxygen of methanol is labeled with the 18O isotope)?!
A!
B!
C!
D!
E!
F!
Nu
O
C
addition
Nu
Nu
H
Nu
C O
L
L
C O
Nu
The carbonyl product resulting from acyl substitution may be susceptible to further attack by
the nucleophile. These sequential processes follow pathways that have previously been
discussed. Whenever substitution is slower than addition, stopping the reaction at the end of
the substitution reaction can be nearly impossible. The result will be two equivalents of
nucleophile attaching to the carbonyl carbon (the first by substitution, the second by
addition). In general, the substitution-addition sequence takes place for very strong
nucleophiles such as hydrides and Grignard reagents. A source of H+ must intentionally be
added by the chemist to quench the reaction after formation of the addition adduct (i.e., in
chemical equations this is often written in H+ workup).
+ CH3CH2
CH3CH2 C CH3
C O
following
H+ workup
CH3
HO
OH
O
LiHAlH4
CH3CH2 C H
CH3CH2 C O
CH3
following
H+ workup
HO
CH3
O
NH2
1) LiAlH4
2) H+ workup
H H
NH2
Substitution-Addition Mechanism
for Grignard Nucleophiles
OH
O
CH3 MgBr
CH3CH2 C O
CH3CH2 C CH3
CH3
[AdN]
OH
O
[MgBr]
CH3CH2 C O
[pt]
CH3
[E]
O
CH3CH2 C CH3
[AdN]
CH3 MgBr
add
H
O [MgBr]
CH3CH2 C CH3
CH3
Substitution-Addition Mechanism
for Hydride Nucleophiles & Esters
OH
H
Li
H Al H
H
CH3CH2 C O
CH3CH2 C H
[AdN]
H
OH
O Li
CH3CH2 C O
H
add
H
[pt]
[E]
[AdN]
CH3CH2 C H
H
Li
H Al H
Li
O Li
CH3CH2 C H
H
Substitution-Addition Mechanism:
LiAlH4 with Primary or Secondary Amides
[AdN]
[pt]
[AN]
[E]
[AdN]
Substitution-Addition Mechanism:
LiAlH4 with Tertiary Amides
H
tertiary
amide
H Al H
H
N CH3
[AdN]
R C H
Al
H3O
low temp.
hydrolysis
R C N CH3
[E]
H CH3
This TI is
relatively
stable to [E]
CH3
H
O
Al
R C N
H
H
CH3
an iminium ion
[pt]
[AdN]
CH3
R C N
H Al H
H
H
an aldehyde
[AN]
CH3
CH3
add
H+
H
H Al H
O H
C N H
C
addition
C
C
H
addition
O H
N R
C
imine
2 amine
R N H
O
elimination and
[pt] from N
elimination and
[pt] from C
C C N R
O H
C C N R
R
O H
H
enamine
When primary amines are combined with ketone and aldehyde groups, the carbonyl C=O is
replaced by the C=N (imine) group. The mechanistic sequence is addition followed by
elimination of water (the second NH proton is necessary for imine formation). When secondary
amines are combined with aldehydes and ketones, there is no second proton on nitrogen to permit
imine formation. In this case, elimination of water is achieved via proton transfer from a HC
bond (assuming that the aldehyde or ketone substrate has at least one -hydrogen). The functional
group that results is called an enamine.
NH2
H
(cat.)
N
primary
amine
imine
H H
H
(cat.)
O H
N
H
secondary
amine
H
enamine
O H
H
Derivative Name
H2N OH
hydroxylamine
oxime
H2N NHR
hydrazine
hydrazone
Structure
N
semicarbazide
NHR
H
H2N N C NH2
OH
semicarbazone
NH
C NH2
NH2
H2O
N
H
O
[pt]
[pt]
[AdN]
HN
H
H2O
H H
HO N
H H
H O N
[E]
http://www.chemtube3d.com/
[pt]
N
H
[pt] (or [DE])
[pt]
H
O
H2O
H
H
H H
N
H
H N
H2O
[E]
[AdN]
H
O
[pt]
H
H
H
H
http://www.chemtube3d.com/
H2O
+
H
N
C
H
O
CH3
NH2
CH3
NH3
NaBH(OAc)3
pH 6
H+ + H
N H
C H
hydride
H NH2
CH3
CH3
excess C=O
NaCNBH3
pH 6
CH3
CH3
Imines and iminium ions are reduced by hydride sources to make amines. The process is known as reductive
amination. The hydride reducing reagents in the above equations are carefully chosen so as to remain stable
under the mildly acidic (pH=6) conditions and to selectively react with protonated imines rather than ketones.
R O
R O
O H
O
C
addition
C O
elimination
then addition
H R
aldehyde
O R
C O
H R
hemiacetal
O H
acetal
H
H
R O
R O
O H
O
C
ketone
addition
C O
elimination
then addition
R
hemiketal
O R
C O
R
O H
ketal
H
http://www.chemtube3d.com/
O
C
a ketone
H3C
an aldehyde
H3C
2 CH3OH
+
H
H3CO
+
CH3
a diol
+ H2O
an acetal
OH
HO
OCH3
H3C
O
CH3
a cyclic ketal
http://www.chemtube3d.com/
+ H2O
+
CH3
CH3O OCH3
2 CH3OH
CH3
CH3
CH3
CH3
addition
CH3 C CH3
CH3 C CH3
[AdN]
O
O
CH3
HOCH
3
CH3
H2O
CH3
HOCH3
addition
[AdN]
H2O
H
[pt]
H
[pt]
O
CH3
elimination
[E]
H
[pt]
H
[pt]
CH3 C CH3
CH3 C CH3
CH3 C CH3
CH3
CH3
CH3
Ph
Ph
Ph P CH2
P CH2
Ph
a phosphonium ylide
Ph
(carbon nucleophile)
This elimination
involves breakdown of
the 4-membered ring
Ph3P O
[AN]
a betaine
CH2
Ph3P CH2
[AdN]
Ph =
O
Ph3P
an oxaphosphetane
Ph3P=O
Formation of Ylides
Ph3P +
CH3
triphenylphosphine
[SN2]
H
Ph3P
C H
H
methyltriphenylphosphonium
iodide
[pt]
Ph3P CH2
LiI
CH3CH2CH2CH3
butane
CH3CH2CH2CH2 Li
butyllithium
(strong base)
http://www.chemtube3d.com/
Summary
Primary amines (RNH2) add reversibly to ketones or aldehydes to give imines (Schiff
bases) and related compounds via the intermediate hemiaminals. The position of the
equilibrium depends on the structure of the amine and the carbonyl compound. With
alkylamines, the equilibrium favors the carbonyl compound, but it can be driven to the
imine by removal of water.
With hydrazines (R2NNH2) and hydroxyl and
alkoxylamines (RONH2), the equilibrium greatly favors the hydrazone, oxime, or
oxime ether.!
Secondary amines (R2NH) can form hemiaminals and iminium ions, but they cannot
form imines. Removal of the -hydrogen from iminium ions gives enamines. The
equilibrium can be driven to favor the enamine by removal of water from the reaction
mixture.!
Water and alcohols add reversibly to aldehydes and ketones under either acidic or
basic conditions to give hydrates, hemiacetals, or hemiketals. Only under acidic
conditions can further reaction take place to give acetals and ketals. !
Acetals and ketals are usually thermodynamically unfavorable, so these reactions are
usually driven to completion by removing H2O from the reaction mixture.!
CH2
CH3
Recall, the larger the pKa, the weaker the acid. The
data below shows that of all the CH bonds, the
carbonyl CH is the strongest acid. The unusually
high acidity of this C-H bond gives it some very
important reactivity (e.g., deprotonation of C-H
makes the alpha position a carbon nucleophile).
bond
carbons
CH3CH2O H
pKa
16
OH
CH2
CH3
C
H2
CH3
<1%
> 99 %
enol tautomer
keto tautomer
CH2 H
20
CH3 C C H
25
H2C C H
H
44
CH3CH2 H
51
CH3 C
O
RCH C R
enolate anion
O
O
enol
tautomer
RCH C R
HO
HO
+
H
RCH C R
H
RCH C R
H
H
H3O
enol tautomer
pKa = 35
H
diisopropyl amine
CH3CH2CH2CH2 Li
n-BuLi
N
Li
LDA
CH3CH2CH2CH3
pKa = 50
Keq = 5 x 10-5
O
CH3 C CH3
pKa = 20
NaOH
O Na
CH2 CCH3
H2O
pKa = 15.7
Keq = 1 x 1015
O
CH3 C CH3
pKa = 20
O Li
LDA
CH2 CCH3
N
H
pKa = 35
O
CH3
LDA
slight
excess
0 C
O
H
CH3
B
1%
A
99%
O
H
O
CH3
LDA
0 C
A
10%
CH3
+
slight
excess
CH3
CH3
B
90%
O OR"
HO OR"
R'
R
[pt]
then base
R"OH
[AdN]
base
R'
[pt]
then
+H
H2O [DN]
+
OR"
[AN]
+ R"OH then
H+
[pt]
[AN] + R"OH
then
H+
[pt]
O
R carbonyl
R'
[pt]
+H+
R'
OH
R"O OR"
R'
R'
acetal / ketal
+H+
[AE]
base
OH
O
R
enolate
-H+
[DE]
R'
base
R'
R
enol
COLOR KEY
intermediates generated under
basic conditions are blue
neutral compounds are black
intermediates generated under
acidic conditions are red
pay careful attention to which
side of the equilibrium arrow (i.e.,
forward or backward direction)
the base or acid lies
enol
RCH C R
tautomer
RCH C R
E
H
H
O
RCH C R
E -substituted
product
enolate anion
Under basic conditions, the nucleophilic form will be the ____________
enol
Under acidic conditions, the nucleophilic form will be the ____
LDA
Li
(CH3)3Si Cl
"O"-alkylation
Si(CH3)3
LiCl
These are
[SN2] reactions
O
CH3 I
"C"-alkylation
CH3
LiI
enolate HOMO
As the HOMO indicates, enolates are ambident nucleophiles, being reactive at
both C and oxygen. Consistent with the largest lobe of the HOMO being on C,
most electrophiles react at carbon; some very powerful electrophiles (and oxophilic
electrophiles) react mainly at oxygen (e.g., silicon is strongly oxophilic).
[SN2]
Br
iminium
bromide
Br
an enamine from
cyclohexanone
HCl / H2O
hydrolysis
O
CH3 C Cl
Cl
O
CH3
an enamine from
cyclohexanone
Enamine nucleophiles are mild enough
to give high yields of acylation products
without interference from over acylation
or addition products. Hydrolysis of the
iminium ion involves the [AdN] addition of
water, [pt], and [E] elimination.
an
iminium
ion
HCl / H2O
N
H2
O
1
O
2
3 CH
3
a 1,3-dicarbonyl
O
"
CH2
+ Br2
CH3CO2H
H
H
enol
formation
OH
CH2
Br
Br Br
H
CH2Br
[AE]
OH
C C
+X
C C
X
R C CH3
3 Br2
3 OH
R C CBr3
+ 3 Br +
3 H2O
OH
R C O
CHBr3
+$ +
O
H
Nu H
H
C
H C
Nu C
O
1
O
2
Nu
Draw a (----) line across the new CC bond in the product
Identify the nucleophilic carbon
Assign partial charges to atoms of the electrophile to indicate their state of polarization
in the reactant
Number the carbonyl carbons in the product and use this to recognize Michael adducts
O
R
O
H
O
OH
CH3
CH3
CH3
CH3
H
H H
H
OH
[pt]
OH
[pt]
H2O
H
O
CH3
[AdN]
CH3
CH3
H
CH3
H
For the acid-catalyzed version of this reaction see:
http://www.chemtube3d.com/
,unsaturated compound
O
Reactants for
the Michael
Reaction
nucleophile
O
aldehyde
CH3
CH3
-diketone
H H
O
O
CH3
ketone
CH3
OCH3
-diketoester
H H
O
OCH3
-diketonitrile
H H
NH2
amide
O
N
CH3
ester
O
-diester
EtO
OEt
nitrile
H H
nitro
O
CH3
enamine
Summary
The conjugate base of a carbonyl compound, an enolate, is generated under
basic conditions. Enolates are nucleophilic at the -carbon (and at oxygen).
Enolates are better nucleophiles than enols. They react with alkyl halides
(alkylation) by the [SN2] mechanism. They react with ,-unsaturated carbonyl
compounds to give conjugate addition products (another version of the Michael
reaction).!
Enolates can be prepared quantitatively by deprotonation with a strong base
such as lithium diisopropylamide (LDA).!
Enols are good nucleophiles that can be generated under acidic conditions.
Under acidic conditions, carbonyl compounds are thus electrophilic at the
carbonyl C and nucleophilic at the -carbon (and also at oxygen).!
Enols react with electrophiles (e.g., bromine) leading to net substitution of one
-hydrogen by the electrophilic atom(s).!
Enols are particularly reactive toward electrophilic alkenes such as ,unsaturated carbonyl compounds to give conjugate addition products (the
Michael reaction). This is one of the mildest, most versatile, and most efficient
methods for forming C-C bonds.!
Na OH
O
H C
H C CH3
[pt]
O
O
H C CH2
H C CH2
enolate anion
H OH
C CH3
OH
OH
CH2
H
a -hydroxyaldehyde
OH
H C CH3
[AdN]
[pt]
O
H C
O
CH2
C CH3
H
H OH
H C CH2
enol
[pt]
H OH
H
H C CH3
C CH3
H O
H O
[taut]
CH2
[pt]
OH
H C
H C CH3
OH
[AdN]
H C
OH
CH2
C CH3
H
OH
CH3 CH
CH
CH3 CH
H
[taut]
OH
CH C H
H
O
OH
CH3 CH CH C H
enol tautomer
H
[pt]
H
H
[pt]
The acid-catalyzed
dehydration is shown.
The base-catalyzed
dehydration follows
the [E1cb] pathway.
+
O
warm
acid or base
C H
H2O
CH3 CH
H2O
H2O
[E]
H
CH3 CH
O
CH
C H
O
CH
C H
O
CH2 +
H
EtO
CH3
CH3
H
O
O
CH3
a -ketoester
O
O
CH2
CH3
EtO
O
H
EtO
[pt]
EtOH
O
CH2
CH2
an ester enolate
CH3
H H pK = 10.7
a
[E]
EtO
CH3
[AdN]
CH3
O
O
O
O
O
C OEt
CH3
a tetrahedral intermediate
excess
warm
CO2
O
3-oxobutanoic acid
keto-enol
tautomerization
O
a cyclic 6-membered TS
C
O
O
O
1) CH3CH2O (deprotonation)
O
O
malonic ester
2) RBr (alkylation)
3) H / H2O / heat (hydrolysis
& decarboxylation)
RCH2 COH
heat
CO2
O
1)
HO
O
O
OH
R
2)
O
O
R
hydrolysis
Alkylation of -Ketoesters
(Acetoacetic Ester Synthesis)
Here is a general synthetic route to methyl ketones.
1) EtO Na
OEt 2)
acetoacetic ester
O
OEt
Br
3) NaOH / H2O
4) HCl / H2O
The product of
acetoacetic ester
synthesis is a
substituted acetone.!
CO2
heat
O
OH
Aldol
Aldol + dehydration
Claisen
Acetoacetic acid
synthesis
Malonic acid
synthesis