L
TOPIC 5. ORGANIC REACTIONS:
ACIDS AND BASES (Chapter 3)
OBJECTIVES
1. Classify types of reaction:
-Addition, Substitution, Elimination, Rearrangement
2. Define the concept of “Mechanism”
3. Discuss the thermodynamics (equilibrium) and kinetics (rate) of organic
reactions
4. Describe acid-base reactions
5. Develop relationships between structure and acidity/basicity
6. Take a first look at acid-promoted reactions
D.M. Collard 2007
S:3.1
CLASSIFYING REACTIONS
Reactions are conveniently classified as substitutions, additions,
eliminations and rearrangements. These terms describe the overall
process, simply comparing the structure of starting materials and products.
They do not indicate anything about the pathway (“mechanism”) by which
the reaction proceeds.
Substitutions
Additions
Eliminations
Rearrangements
(often in combination
with another type of reaction)
Substitution
NaI
CH3OH
D.M. Collard 2007
CH3CH2 Br
+
acetone
(solvent)
O
+
H 3C
C
Cl
pyridine
(solvent)
Addition
H3C
C
CH2
H
H2O
H2SO4
H2O
O
H2SO4
Elimination
H
OH
C
H3C
D.M. Collard 2007
CH3
H2SO4
Rearrangement
H
H3C
H
C
C
H3C
H3PO4
C
CH3
H
A Preview of Reactivity
su
t
bs
Substn
H
C C
Br
Substn
Substn
El
Addn
H
Substn
Ad
dn
im
n
C C
ns
tio
itu
H
H
Elimn
ad
C C
OH
bs
su
ns
tit
ut
tio
di
io
ns
C C
D.M. Collard 2007
Practical Aspects of Running a Reaction
BEGINNING
END
Starting material
(“substrate”)
Reagents
Catalysis
Time
Temperature
(lowers activation energy)
Solvent
(heating/cooling)
(allows for mixing)
Environment
(avoid O2, H2O?)
Stirring
Mode of addition
% yield =
moles of product
moles of limiting reactant
Product
Byproducts
Unreacted reactants
Catalyst
Solvent
SEPARATION
PURIFICATION
IDENTIFICATION
x 100
Representations of reactions
Reactant(s)
Product(s)
+ H2
Substrate
(organic starting
material)
reagent
...more commonly written as….
Substrate
reagent(s)
catalyst(s)
(organic starting
material)
solvent(s)
Product ( + byproduct(s))
H2, Pt
EtOH
H2O
H2SO4
D.M. Collard 2007
OH
WHAT IS A MECHANISM?
A mechanism is a proposal for a step-by-step pathway by which a reaction
proceeds. Each step involves bond making and/or breaking. The mechanism
takes into account all currently available evidence (kinetics, formation of
byproducts, effect of structure on reactivity). Any new data collected must be
consistent with the proposed mechanism, or the mechanism itself must be
modified to account for the new finding.
An understanding of common mechanistic steps can be applied to new
combinations of reagents to predict the outcome of a new reaction. As such,
development of an understanding of mechanisms will save you from
memorizing a huge amount of material.
While you must develop a familiarity with reactions, do not try to pass this
course by just memorizing the outcome of reactions!
Electrophiles and Nucleophiles
Nucleophile
Electron-rich, “nucleus-loving” species
Electrophile
Electron-deficient, “electron-loving” species
D.M. Collard 2007
S:3.4
CURVED ARROWS
Prob
3.20,25,
31,36,37,40
examples of 2e- processes
A B
Nu
E+
A B
H+
Polarity and Heterolytic Bond Cleavage
(CH3)3C Br
H3 C Li
D.M. Collard 2007
S:3.3
(CH3)3C
CH3
Br
Li
Problem 3.20(c). Provide curved arrows to account for the changes
in bonding in the following reaction step.
H
H
C
H
C
H
F
H
H
H
C
C
H
H
F
H
Problem 3.31(d). Show the curved arrows to account for the following
reaction step.
H
O
H3C
I
H
O
CH3
Problem: Show the curved arrows to account for the following
reaction step.
O
H3C
C
H
CH3
C N
D.M. Collard 2007
O
H3C
H
C CH3
C
N
I
–
Problem: What is wrong with each of the following mechanistic steps,
suggested by students in previous classes? [consider what the curved
arrow is meant to depict, or draw the products of the suggested flow of
electrons and comment on why that product is not stable]
CH 3
Br
H3C N CH3
CH 3
H
Cl
Cl
H
H
CH3
C
H3C
CH3
C Br
Br
Problem: What is wrong with each of the following mechanistic steps
suggested by students in previous classes?
O
H3C
C
CH3
H 3C
Cl
D.M. Collard 2007
O
H Cl
H
C
H
Cl
CH3
Cl
H
L
S:3.2; 3.6
3.15
ACID-BASE REACTIONS
Prob
3.15,17,19,
22-24
Definitions
Brønsted Acidity: ability to donate a proton
Brønsted Acid: proton donor
Brønsted Base: proton acceptor
H—A + :B
Strong acids have weak conjugate bases
Conjugate base strength: Br – < HO–
e.g., Acid strength: HBr > H2O
Lewis Acidity
Lewis Acid: electron pair acceptor
Lewis Base: electron pair donor
Ph3P: + BF3
pKa and Acid Strength
A–H
acid
K eq =
Ka =
[A - ][H3O+ ]
[HA][H2O]
[A - ][H3O+ ]
[HA]
+
H2O
Keq
A
base
conjugate
base
H2O–H
conjugate
acid
since H2O is solvent, [H2O] = 55 M
Ka ↑ , acid strength ↑
pKa = –logKa
Bottom line: pKa ↑ , acid strength ↓
D.M. Collard 2007
+
S:3.5
THE STRENGTH OF ACIDS AND BASES
pKa values
NH4+
C6H5OH
HCO3–
CH3NH3+
► H2O
Ö CH3CH2OH
(CH3)3COH
CH3COCH3
„ HC≡CH
H2
► NH3
„ CH2=CH2
►„CH3CH3
¾ HI
-10
H2SO4
-9
¾ HBr
-9
¾ HCl
-7
(CH3)2C=OH+
-2.9
+
CH3OH2
-2.5
H3O+
-1.74
HNO3
-1.4
CF3CO2H
0.18
¾►HF
3.2
H2CO3
3.7
Ö CH3CO2H
4.75
CH3COCH2COCH3 9.0
Prob
3.16,18,21
27-29
9.2
9.9
10.2
10.6
15.7
16
18
19.2
25
35
38
44
55
Species that can act as either an acid or as a base
CH3NH2
+
pKa = 38
CH3NH3+
H2O
pKa = 15.7
+
HO¯
+
H3O+
pKa = 10.6
versus
CH3NH2
+
pKa = 38
CH3COOH
H2O
CH3NH¯
pKa = 15.7
+
pKa = 5.5
pKa = -1.74
H2O
CH3COO¯
H3O+
+
pKa = 15.7
pKa = -1.74
versus
CH3COOH
pKa = 5.5
D.M. Collard 2007
+
H2O
pKa = 15.7
CH3COOH2+
+
HO¯
Amine + Acid
CH3COOH
+
pKa = 5.5
CH3NH2
CH3COO¯
+
CH3NH3+
pKa = 38
Amino acids
pKa = 10.6
O
H2N
O
H3N
OH
O
Amino acids in aqueous solution
O
H 3N
O
H 3N
OH
O
H2N
O
O
Predicting Equilibrium Constants
A–H
acid
+
B:–
base
Keq
A:–
+
conjugate
base
B–H
conjugate
acid
Equilibrium lies of side of the weaker acid and weaker base (the weaker
acid and weaker base will always be on the same side)
log Keq = pKa (conjugate acid) – pKa (acid)
Keq =
HCl
+
NaOH
NaCl
+
pKa = -7
pKa = 15.7
Keq =
CH3OH
D.M. Collard 2007
H2O
+
H2SO4
+
CH3OH2
pKa = -9
pKa = -4
+
HSO4–
STRUCTURE-ACIDITY RELATIONSHIPS
S:3.7;
3.8A;3.12
In order to assess the relative strengths of acids, consider the ability of the acid
to donate a proton (ability to break the H-A bond) and for the conjugate base to
accommodate negative charge. Stronger acids have weaker conjugate bases...
A:– + H–B
A–H + B:–
We have ways to assess the ability of ions (anions and cations) to
accommodate negative charge based on:
– Inductive effects (substituents donate or withdraw electron
density via sigma bonds)
– Resonance effects (electron donation or withdrawal by pibonds)
– Hybridization
Acidity: Across a Row of the Periodic Table
pKa
H–CH3
48
CH3 –
H–NH2
38
H–OH
15.7
H–F
3.2
conjugate bases
NH2 –
OH –
F–
►
► Bonds strengths are similar:
Acidity primarily depends on electronegativity
Acidity: Down a Column of the Periodic Table
H–F
H–Cl
H–Br
H–I
D.M. Collard 2007
pKa
3.2
-7
-9
conjugate base
F–
Cl–
Br–
-10
I–
¾ Bonds strength decreases dramatically from H-F to H-I:
Acidity primarily depends on bond strength.
¾
Ö
The Effect of Resonance
CH3CH2OH
CH3CO2H
pKa
16
4.75
conjugate base
CH3CH2O─
CH3CO2─
CH3CH2 O
O
H3C
Ö
O
O
H3C
O
Resonance stabilization of conjugate base and
increases acidity
pKa values
H
He
H-H
55
Li
acid strength ↑
Be
B
C
N
H-CH3 H-NH2
35
38
K
D.M. Collard 2007
Mg
Ca
Al
Ga
Si
Ge
P
As
F
H-OH
15.7
H-F
3.2
S
Cl
H-SH
7.04
H-Cl
-7
Se
Br
H-SeH
3.9
H-Br
-9
Te
I
H-TeH
2.6
H-I
-10
Ne
acid strength ↑
Na
O
Ar
Kr
„
The Effect of Hybridization
Acidity
R C C H
>
R CH CH H
pK a=25
>
R CH2 CH2 H
pK a=44
pK a=50
Conjugate bases
R C C
R CH CH
„
R CH2 CH2
Lone pair in lower energy hybrid orbital stabilizes
conjugate base and increases acidity
ORGANIC BASES
Organic bases have either lone pairs of electrons or pi-bonding electrons
H
R
N
H
Cl
H
Cl
H
H
H
C
H
C
H
Strong bases commonly used in organic chemistry:
potassium tert-butoxide: K+-OC(CH3)3
sodium amide:
Na+NH2sodium hydride:
Na+H-
D.M. Collard 2007
The Effect of Structure on Basicity
Effect of charge
RO
>
ROH
Effect of electronegativity
RNH2
>
RNH
ROH
>
RO
Effect of hybridization
>
R CH2 NH2
R CH NH
Effect of resonance
O
CH3CH2 O
>
H3C
O
O
CH3CH2 NH2
>
H3C
NH2
Effect of alkyl substituents
R3N > R2NH > RNH2 > NH3
(gas phase basicity)
D.M. Collard 2007
>
R C N
ACID-BASE REACTIONS IN
AQUEOUS AND NON-AQUEOUS MEDIA
S:3.11;
3.14
Prob
3.38,39
H3O+ is the strongest acid, and HO– is strongest base, that can be present in
water.
HCl
+
pKa = -7
NaH
+
H2O
H3O+
pKa = 15.7
pKa = 1.74
H2O
HO–
pKa = 15.7
+
Cl–
+
H2
pKa = 35
In any solvent, the strongest acid [base] which can be present is the conjugate
acid [base] of the solvent. The choice of solvent for acid-base reactions is
important…
conjugate
conjugate
acid
base
R C C H
pK a=25
NaNH2
R C C H
pK a=25
NaNH2
hexane
NH3
pK a=55
pK a=38
H2O
NH3
pK a=38
pK a=16
e.g., CH3OH2+ is the strongest acid possible in methanol (i.e., H2SO4, a
stronger acid, is completely dissociated in methanol). CH3O- is the strongest
base present in methanol (i.e., NH2-, a stronger base, is completely protonated
by methanol).
D.M. Collard 2007
L
PRACTICAL APPLICATIONS OF ACID-BASE
CHEMISTRY
Separation of Neutral, Acidic and Basic Compounds
Challenge: Separate a mixture of hexanoic acid and nonane.
NaOH, H2O
Et2O
O
O
O
OH
Na
Neutral organic
component in Et2O
1. Separate the layers, distill off the ether to
isolate nonane
2. Acidify the separated aqueous solution,
extract neutral hexanoic acid into another
portion of ether. Separate these two
layers, remove ether.
sodium carboxylate
in aqueous base
Problem: Design a procedure to separate hexylamine (C6H13NH2) from nonane.
Provide Greater Water-solubility to Drugs
Naproxen sodium” (Aleve)
OH
O
H3CO
Na
OH
O
H3CO
Go directly to jail…..
H3C
N
Cl
H3C
O
N
H O
OCH3
H Cl
OCH3
O
Na2CO3
O
Ph
O
D.M. Collard 2007
Na
O
Ph
O
+
H2O
Catalysis of Reactions
Proton transfer is usually fast. Protonation (or deprotonation) of an organic
starting material with an acid (or base) often catalyzes reactions which do not
take place in the absence of catalyst.
Bases deprotonate molecules and make them better (i.e., more reactive)
nucleophiles
Acids protonate molecules and make them better electrophiles
ENERGY CHANGES AND EQUILIBRIA:
THERMODYNAMICS
Equilibria
Reactants (R)
Keq
Enthalpy
ΔH° = H°products – H°reactants
based on changes in bonding
D.M. Collard 2007
Products (P)
Keq =
[products]
[reactants]
S:3.8-3.9
Prob 3.35
Calculating Heats of Reaction, ΔH
H
H
C C
H H
152
H
kcal mol-1
H H
H C 88 C H
H H
104
H
bonds broken
(energy IN)
organic stuff
C,H,O (+N,S,P,...)
bonds made
(energy OUT)
O2
O=C=O + H2O
Representative Bond Lengths and Strengths
H-H
H-F
H-Cl
H-Br
H-I
H-O
H-C
C-C
C=C
C≡C
C-O
C=O
C-F
C-Cl
C-Br
C-I
bond length
Ǻ
0.74
0.92
1.27
1.41
1.61
0.97
1.10
1.55
1.33
1.20
1.43
1.21
1.38
1.77
1.95
2.14
1 Ǻ = 10-10 m = 100 pm
D.M. Collard 2007
bond strength
kcal/mol
kJ/mol
104
435
136
571
103
432
87
366
71
289
110
460
99
414
88
368
152
636
200
837
80
355
191
799
110
461
79
330
67
280
57
240
1 kcal = 4.18 kJ
net change
Problem. Calculate the ΔH for the following reaction.
CH3
CH3
H3C C OH
H3C C
H Br
Br
H OH
CH3
CH3
Relationship Between Keq and ΔG°
Equilibrium constant depends on changes in enthalpy and entropy (change
in disorder)
Change in Gibbs free energy for a reaction:
ΔG° = ΔH° – T ΔS°
G
ΔG° = G°products – G°reactants
P
ΔG° = –RT lnKeq = –2.303 RT logKeq
R = 8.314 J/K.mol = 1.987 cal/K.mol
T = temperature (in K)
ΔG° > 0
(positive)
R
endothermic
Keq = [products]/[reactants]
When ΔG° = 0; Keqm
When ΔG° >0; Keqm
When ΔG° < 0; Keqm
eqm
G
eqm
R
ΔG° < 0
(negative)
If ΔG° < -13 kJ
D.M. Collard 2007
mol-1,
K >100
exothermic
P
For the equilibrium:
A
ΔG o(kcal/mol)
ΔG o = –RT lnKeq = –2.303 RT logKeq
B
Keq
100
%A
at equilibrium
0.0002
0.001
0.006
0.03
0.2
1
5
29
159
862
4670
80
99.98
99.88
99.38
96.71
84.42
50
16
3.3
0.63
0.12
0.02
%A
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
-5
60
40
20
0
-5 -4 -3 -2 -1 0
1
2 3
4
5
ΔG (kcal/mol)
» A small change in the ΔH of a reaction has a large influence on Keq
If ΔG° < -13 kJ mol-1, K >100
Thermodynamics tells us the extent to which a reaction CAN occur, but nothing
about how FAST it will be
KINETICS
P
Reaction Coordinate
Energy
R
Energy
Energy
Transition State Theory: Energy-Reaction Coordinate Diagram
for a One-Step Reaction
Reaction Coordinate
Rate constant: k = Ae-Ea/RT
Catalysis -
D.M. Collard 2007
Reaction Coordinate
k = Ae-Ea/RT
Effect of Ea on Rate (at 25 °C)
14
logk
Ea
5
10
15
20
25
logkrel
kcal/mol
14.7
11.0
7.3
3.7
0 (i.e., krel=1)
12
10
8
6
4
2
0
5
10
15
20
Ea (kcal/mol)
25
Effect of Temperature on Rate
For reaction:
A → B (for Ea = 25 kcal/mol)
krel
0.05
0.17
0.56
1
1.75
5.05
13.7
34.9
84.2
193
424
893
10
8
krel
T/°C
0
10
20
25
30
40
50
60
70
80
90
100
6
4
2
0
0
10
20
30
40
50
T/°C
» A small change in the Ea of a reaction has a very large influence on k
» A small increase in temperature can have a large influence on k
Energy-Reaction Coordinate Diagram for Two-Step
Reaction
Intermediate
Energy
Reactants
Reaction Coordinate
D.M. Collard 2007
Product
A MULTISTEP MECHANISM IN WHICH THE
FIRST STEP IS AN ACID-BASE REACTION
Substitution of tert-Butyl Alcohol
Overall Reaction
OH
H Br
Br
How do changes in bonding take place?
Mechanism
– Protonation
OH
H Br
– Heterolytic Cleavage
H
O
H
–Nucleophilic Addition
Br
D.M. Collard 2007
H OH
S:3.13
TOPIC 5 FOR EXAM 3
Types of Questions
- Classify organic reactions
- Compare acid (or base) strength
- Show movement of electrons
The problems in the book are good examples of the types of problems on
the exam!
Preparing for Exam 3:
- Get up to date now!
- Work as many problems as possible
- Work in groups
- Do the “Learning Group Problem” at the end of the chapter
D.M. Collard 2007