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Transcript
Chapter 1-1
What is Organic Chemistry?
Organic Chemistry
What is it? Why is it a two semester course?
Why is it important to careers in health care?
Organic chemistry is essential to the understanding of the intricate details of
life The interactions and reactions of organic molecules are what define living
systems.
One needs to understand bonding, structure, properties, reactions, and
synthesis to understand natural systems.
CHAPTER 1
INTRODUCTION
Organic Chemistry: the study of carbon compounds...
organic compounds contain the elements C, H, N, O, S, Cl, Br, etc.
Organic compounds were originally thought to come only from living
organisms...thus the term organic (until about 1830)
Inorganic compounds were all those which came from non-living sources
Scientists thought a vital force was necessary to produce organic compounds
and that they could not be synthesized in the lab.
In 1828 Fredreich Woeller synthesized urea, an organic compound excreted
as waste, from ammonium cyanate, and vitalism slowly died out.
Organic compounds range from methane , CH4 (natural gas) to DNA, the
genetic coding material, and taxol, a plant derived substance which is a
potential anticancer agent.
Organic chemistry is fundamental to many scientific disciplines...biochemistry,
polymer chemistry, microbiology, botany, pharmacy, medicine...since living
systems are composed primarily of organic compounds and water.
Chem 61
Electronic Structure of Carbon
Chapter 1-2
STRUCTURAL THEORY AND BONDING
Carbon is intermediate in electronegativity, therefore it neither completely
donates or completely accepts electrons. As a result it forms covalent bonds to
itself and other atoms.
It can bond to itself to form chains and
rings...catenation. This allows the formation of a staggering number of organic
compounds. 95% of all known compounds are organic.
ATOMIC STRUCTURE
It is important to understand molecular structure to understand reactivity of
organic compounds. Molecular structure depends on atomic structure.
Atomic structure of carbon C: 1s22s22p2
that is, carbon has 2 electrons in the lowest energy level, the 1s orbital
2 electrons in the next energy level, the 2s orbital
2 electrons in the third energy level, the 2p orbital
1s, 2s, 2p orbitals
1s is spherical with the same phase throughout
1s
2s is spherical with a node
node is where ψ2 = 0
node
2s
2p... three orbitals of equal energy (degenerate)
2px
2py
2pz
Chem 61
Chapter 1-3
Rules for Electronic Configuration
Pauli Exclusion Principle: Maximum of two electrons per orbital.
Electrons have a spin of +1/2 or -1/2 which gives rise to a small magnetic
field since electrons are charged.
Repulsion between the electrons is reduced if they have opposite spins
and thus opposite magnetic moments.
If two electrons occupy the same orbital they must have opposite spins or
be paired.
Aufbau Principle: orbitals are filled from lower to higher energy
4s
3p
3s
2p
2s
1s
__
__ __ __
__
__ __ __ order of filling of atomic orbitals
__
__
THUS: Li 1s22s1
Hund's Rule: orbitals of equal energy (degenerate) receive one electron
until there is one electron in each orbital: then pairing of electrons begins.
THUS: C 1s22s22px12py1
Chem 61
Chapter 1-4
Atomic Radius and Electronegativity
ATOMIC RADIUS
Atomic Radius: distance between the nucleus and the outermost electrons
(valence electrons) or one half the bondlength of a diatomic molecule
H—H bond length is 0.74Å, atomic radius is 0.37Å
Atomic radius increases with increasing number of electron shells within an
atom and decreases with the increase in the number of protons within an
atom
Thus atomic radius decreases from left to right within the same row of the
periodic table (increasing number of protons in the nucleus) and increases
from top to bottom within a group of the periodic table (increasing number of
electronic shells)
ELECTRONEGATIVITY
Electronegativity: a measure of an atom's attraction for outer bonding
electrons
Electronegativity increases with increasing charge on the nucleus and with
decreasing distance between the nucleus and the electrons
Thus electronegativity increases from left to right within a row of the
periodic table and from bottom to top within a group in the periodic table
Chem 61
Chemical Bonding
Chapter 1- 5
CHEMICAL BONDING
G.N. Lewis put forth the first explanation of the nature of the chemical bond.
Ionic Bond..one or more electrons is completely transferred from one atom to
another creating ions: a cation (positively charged) and an anion (negatively
charged).
the ions are held together by electrostatic attractions
ionic bonds generally occur between atoms of highly different
electronegativies. e.g.
Nao – e–
Na+ + e–
Clo + e–
Cl –
Nao + Clo
Na+Cl –
Covalent Bond...one or more electrons is shared by atoms...atomic orbitals
merge into shared or molecular orbitals
covalent bonds are usually formed between atoms of similar
electronegativities
since carbon is intermediate in electronegativity it usually forms
covalent bonds
Generally...atoms not having the noble gas configuration tend to form bonds
such that each atom may obtain the stable noble gas electronic configuration
[OCTET RULE]
4H + :C:
H
H C H
H
H
C::O
H
single bond
double bond
H:C:::C:H
triple bond
Atoms of the elements of organic compounds can
form a fixed number of bonds.
..
O
..
N
..
C
H
Chem 61
Chemical Bonding
Chapter 1- 6
COVALENT BONDS
Bond lengths: the distance between two covalently bonded nuclei
Bond angle: the angle formed between two covalent bonds
Bond dissociation energy: the energy required for homolytic cleavage of
a bond;
Cl Cl
H 3C H
.
Cl + Cl
.
.
H 3C + H
.
two radicals
Heterolytic cleavage: cleavage of a bond to give a cation and an anion;
one atom which is part of the covalent bond retains both electrons from
the bond
Cl+ + Cl –
Cl Cl
H 3C H
H 3C +
cation
+ H–
anion
∆H (change in enthalpy) for homolytic cleavage of many covalent bonds
has been determined
H3C—H
H—H
Br—Br
H—OH
104 kcal/mole
104 kcal/mole
46 kcal/mole
110 kcal/mole
CH3—CH3
CH2=CH2
HC≡CH
88 kcal/mole
163 ckal/mole
230 kcal/mole
Chem 61
Chapter 1-7
Polar Covalent Bonds
POLAR COVALENT BONDS
covalent bonds between atoms of similar electronegativities are said to be
nonpolar bonds since each atom shares the electrons of the bond
approximately equally
H—H
CH3—CH3
if atoms of different electronegativities form a covalent bond, the more
electronegative atom will have a stronger attraction for the electrons and
polarize the bond giving a polar covalent bond in which one atom is partially
negatively (δ–)charged and one atom is partially positively charged (δ+)
δ+ δ−
H3C—Cl
CH3—NH2
δ+ δ−
δ+ δ−
H—Cl
δ+ δ−
H2C=O
δ+ δ−
H3C—OH
If any entire molecule has an overall dipole, then the molecule is said to be
polar.
CCl4...nonpolar, polar covalent bonds cancel each other because of tetrahedral
symmetry
HCCl3...polar
H2O, NH3...very polar
Cl
_
C
Cl
δ_
Cl
Cl
Cl
dipoles cancel
δ Cl
δ_
+
C
δ
H
δ_ Cl
δ_
δ+
dipoles add
δ_
O
δ+ H
H δ+
dipoles add
δ+
Chem 61
Chapter 1-8
Polar Covalent Bonds
ATTRACTIONS BETWEEN MOLECULES
ion-ion forces: the attraction between oppositely charged ions and
repulsion between like charges very strong...not often encountered in organic
compounds
van der Waals forces: all dipole-dipole forces are both repulsive and
attractive... the distance at which the repulsive forces are minimized and the
attractive forces are maximized is called the van der Waals radius
Induced dipole-dipole interactions or London forces: small temporary
dipoles occur and induce dipoles in another molecule due to small uneven
distribution of electron density.
Dipole-dipole interactions: permanent dipoles in molecules attract or repel
Hydrogen bonding: a specific type of dipole-dipole interaction; very strong
occur only between a hydrogen atom bonded to an electronegative atom O,
N, S and a lone pair on O, N, S
O H
:N
7 kcal /mole
N H
:N
3 kcal/mole
N H
:O
2 kcal/mole
O H
:O
5 kcal/mole
Intermolecular hydrogen bonds increase the boiling points of compounds
and increase their solubility in water.
Hydrogen bonds also can influence the shapes of biomolecules by internal
hydrogen bonding as well as hydrogen bonding between molecules.
Chem 61
Formulas and Formal Charge
Chapter 1-9
CHEMICAL FORMULAS
empirical formula: gives the types and ratios of atoms in a molecule
molecular formula: gives the type and actual number of atoms
structural formula: gives the type, number and attachment of atoms...the
actual structure
for example:
hexane: C6H14
CH3CH2CH2CH2CH2CH3
molecular formula
C3H7
empirical formula
structural formula
LEWIS STRUCTURES
1. draw the molecular skeleton
2. count the number of available valence electrons (be sure to account for
any overall charge on the species)
3. draw covalent bonds between all the atoms giving as many as possible
an octet (duet for hydrogen)
4. assign charges in the molecule
FORMAL CHARGE
formal charge on an atom = # valence electrons - # shared pairs of
3
electrons - # unshared electrons
:O:
..
.. .. 1
N:O:H
HNO3 Lewis structure:
.. ..
:O:
.. 2
N...5-4-0 = +1
O1...6-2-4 = 0
O2...6-1-6 = -1
..
O3...6-2-4 = 0
:O:
.. ..
..
H...1-1
=
O
H
SO
Lewis
Structure
-2
2
4
SO4
:O:S:O:
.. .. ..
:O:
..
O...6-1-6 = -1
S...6-4-0 = +2
overall charge (+2) + 4(-1) = -2
Chem 61
Formulas and Formal Charge
Chapter 1-10
Structural Formulas
Lewis Structures: as described above using dots for electrons
Line-bond formulas: a line is used to represent two electrons forming a bond
(a shared pair)
Condensed formulas: bonds are not always shown and atoms of the same
type bonded to another atom are grouped together
H
..
..
..
..
H..
.
.C::O
. .
H
C O
H
line-bond
H C C H
line-bond
H
H
C C
H
H
line-bond
..
..
H2C=O
Lewis
condensed
H:C:::C:H
Lewis
HC≡CH
condensed
..H
H..
.C::C
. ..
H
H
Lewis
CH2=CH2
condensed
Polygon formulas: polygon formulas are often used to represent cyclic
compounds for simplicity;
each carbon is assumed to have enough additional hydrogens to give each
carbon four bonds
cyclohexane
cyclopropane
H
H H H
C
C H
H C
H C
C H
C
HH H
H
cyclobutane
benzene
cyclopentane
decalin
Chem 61
Chapter 1-11
Acids and Bases
ACIDS AND BASES
Bronsted-Lowry:
acid: a proton donor
base: a proton acceptor
Strong acid: completely ionized or dissociated in water
e.g., HCl, H2SO4, HNO3, HBr
Weak acid: only partially dissociated in water:
carboxylic acids are weak acid
CH3CO2H
acid
+
CH3CO2-
H2O
+
conjugate base
base
H3O+
conjugate acid
amines are weak bases
CH3NH2
base
+
HO –
CH3NH3+ +
H2O
acid
conjugate acid
conjugate base
Generally: strong acids have weak conjugate bases and weak acids have
strong conjugate bases
that is, as acid strength increases, the basicity of the conjugate base
decreases
Thus the ability of the conjugate base to stabilize a negative charge
determines the strength of an acid
Conjugate Acids
H 2O
pKa
HCN
15.75
CH3CO2H
6.37
H3PO4
HCl
2.12
-7
4.75
increasing acid strength
Conjugate Bases
HO –
NC –
CH3CO2–
H2PO4 – C l –
decreasing base strength
Chem 61
Chapter 1-12
Acids and Bases
Factors affecting Acidity:
the electronegativity and the size of the atom which carries the negative
charge influence its ability to stabilize the negative charge
size of the atom
H—F
H—Cl
pKa 3.45
-7
H—Br
-9
H—I
-9.5
increasing size of halogen, increasing acid strength
electronegativity
pKa
(CH3)3C—H
50
(CH3)2N—H
35
CH3O—H
15.5
F—H
3.45
increasing electronegativity of atom, increasing acid strength
ACIDITY CONSTANTS, Ka's
for acetic acid
CH3CO2H + H2O
Ka =
CH3CO2– + H3O+
[CH3CO2–] [H+]
[CH3CO2H]
since stronger acids are more ionized, the larger Ka, the stronger the acid
pKa = -logKa, the lower the pKa, the stronger the acid
also, the higher the pKa, the weaker the acid or the stronger the base
the stronger the acid, the more stable the anion produced by ionization of the
acid.
Chem 61
Chapter 1-13
Lewis Acids and Bases
LEWIS ACIDS AND BASES
Lewis acid: electron pair acceptor: any species with an electron deficient atom
BBr3, AlCl3, H3C+
Lewis base: electron pair donor; any species with an unshared pair of electrons
..
..
H3N:, CH3CH2..OH, H2C=O:
Chem 61
Quantum Mechanics
Chapter 1-14
QUANTUM MECHANICS...Molecular Orbitals
In 1923 Louis De Broglie postulated that electrons have properties of three
dimensional waves
Later a wave equation was developed. Solutions (Ψ) to this wave equation
give the various electronic states known as atomic orbitals.
Ψ = probability of finding an electron in a certain space = electron
2
probability density
plots of Ψ give the familiar s,p,d...orbitals
2
WAVE PROPERTIES
The amplitude of a wave may be above the resting state (positive) or below
(negative)...no charge implied
A node is a point at which the amplitude is zero
Waves reinforce creating a wave of higher amplitude if they they are in
phase.
+
–
+
+
–
–
Waves interfere if they are out of phase and create a wave which is of lower
amplitude. Complete interference results in the cancelling of one wave by
another.
+
+
–
–
+
–
Chem 61
Quantum Mechanics
Chapter 1-15
1s, 2s, 2p orbitals
1s is spherical with the same phase throughout
+
2s is spherical with a node
node is where ψ2 = 0
node
+
-
2p... three orbitals of equal energy (degenerate)
node
2px
2py
4s __
3p __ __ __
3s __
2p __ __ __ order of filling of atomic orbitals
2s __
1s __
2pz
Chem 61
Chapter 1-16
Molecular Orbitals
MOLECULAR ORBITALS
Molecular orbitals = Linear combination of atomic orbitals
2 Atomic Orbitals must produce 2 Molecular Orbitals (the number of
molecular orbitals equals the number of atomic orbitals which were
combined to form them)
the hydrogen molecule
σ*
>∆E
H1s
antibonding
Ψ1 − Ψ2
H2
.
.
H1s
∆E
σ
Ψ1
bonding
Ψ2
..
Ψ1 + Ψ2
the energy of the hydrogen molecule with two electrons in the sigma orbital
is 104 kcal/mole more stable than the separate hydrogen atoms; ∆E = 52
kcal/mole
if one electron is in the sigma and one in the sigma*, the molecule is of
higher energy than the two separate atoms because the s* is slightly >∆E
higher than the s orbitals while the s is ∆E lower
Bonding orbital...high electron density between nuclei
Antibonding orbital..node between nuclei (zero electron density)
σ bond is cylindrically symmetrical
Aufbau principle...fill lowest energy orbitals first
Hund's Rule...place one electron in each degenerate orbital first, then pair
up
Pauli Exclusion Principle...two electrons in the same orbital must have
opposite spins
Chem 61
Chapter 1-17
Molecular Orbitals of Carbon
MOLECULAR ORBITALS ON CARBON
Overlap between atomic orbitals in complex molecules often results in
electron repulsions which destabilize the molecule. Hybrid orbitals allow for
better overlap and a more accurate prediction of molecular structure.
Methane CH4 has a central carbon with four equivalent bonds to hydrogen
bond angles = 109.5°
bond lengths = 1.09Å
sp3 Hybridization
carbon has electronic configuration 1s22s22p2
2p
2p
96 kcal
2s
1s
hybridize
2s
sp3
1s
1s
+
2s
+
-
+
3-3p's
109.5° apart
4-sp3's
sp3 orbitals point toward the corners of a tetrahedron, 109.5° apart
any carbon bonded to four other atoms is sp3 hybridized, e.g. CH4, H3CCH3,
C—C bond length = 1.54Å
C-H σ bonds require 104 kcal/mol to be broken
H
H
C
methane
H
H
bond angles = 109.5°; tetrahedral
Chem 61
sp2 Molecular Orbitals of Carbon
Chapter 1-18
sp2 Hybridization
sp2 hybridized carbons are trigonal planar with atoms 120° apart
2p
2p
2p
96 kcal
2s
sp2
hybridize
2s
1s
1s
1s
ethylene: trigonal planar
H
C
H
C
H
H
2p
H
H
C
C
H
H
H
C
H
120°
H
H
H
pi orbital
H sigma orbitalH
sp2
H
H
C
H
C
C
C
H
H
H
H
pi* orbital
H
C
H
sigma* orbital
bond angles = 120°; C=C bond length = 1.34Å
pi (π) bonds are formed by the side by side overlap of two p-orbitals (approx. 68
kcal/mol)
pi bonds are above and below the plane where the sigma bond is located
pi bonds make the molecule rigid between the two atoms preventing rotation
E
sigma σ*
pi π*
pi π
sigma σ
ethylene C=C double bond: one sigma, one pi
Chem 61
sp2 Molecular Orbitals of Carbon
Chapter 1-19
sp Hybridization
sp hybridized carbons are linear with atoms 180° apart
2p
2p
2p
96 kcal
2s
1s
sp
hybridize
2s
1s
1s
ethylene: trigonal planar
+
180° apart; the remaining two 2p
orbitals are 90° to the sp and
each other
+
1-2p
2s
2-sp's
Acetylene: linear; bond angles 180° C=C bond length = 1.20Å
acetylene has two perpendicular pi bonds and one sigma bond
H
H
H
2-sp's
H
C
H
acetylene σ-orbital
H
2-2p's on each carbon
combine to form pi-orbitals
E
C
C—C σ*
C—C π*
C—C π
C—C σ
H
C
C
H
acetylene π-orbitals
Chem 61
Functional Groups
Chapter 1-20
FUNCTIONAL GROUPs
CH3CH3
Oxygen: sp3
CH2 CH2
HC CH
alkanes
alkenes
alkynes
CH3 OH
CH3 OCH3
CH3 NH2
CH3 Br
alcohols
ethers
amines
alkyl halides
O
CH3 C
H
O
CH3 C
CH3
O
CH3 C
OH
O
CH3 C
OCH3
aldehydes
ketones
Chem 61
2p
2s
carboxylic acids
esters
1s
1s
O: 1s22s22p4
..
.. O H
H
Hybrid orbitals of oxygen and nitrogen
water
Nitrogen: sp3
2p
sp3
hybridize
2s
1s
1s
N: 1s22s22p3
one sp3 orbital already filled
with an unshared pair of electrons
bonding can occur to three other atoms
..
N
H
..
H
H
ammonia
N
CH3
H
methyl amine
H
sp3
hybridize
two sp3 orbitals already filled
with an unshared pair of electrons
bonding can occur to two other atoms
..
.. O
H
CH2CH3
ethanol
sp3
..
.. O CH CH
2
3
CH2CH3
diethyl ether
..
..
O
sp2
CH3
C
CH3
acetone
Resonance Structures
Chapter 1-21
RESONANCE STRUCTURES
Chapter 6; Bruice, Pages 260 - 282:
Some molecules cannot be accurately represented by one simple line-bond
formula:
they are "hybrids" of two or more structures
–
O
O C O–
O–
–
O C O–
O–
–
O C O
2/3–
O –2/3
O C O –2/3
1. Resonance structures exist only on paper...the actual structures are
hybrids of all the resonance structures
2. Resonance structures differ only in the position of electron pairs....not
atoms
3. All structures should be proper Lewis structures (exceptions)
4. All resonance structures should have the same number of unpaired
electrons
Nonequivalent Resonance Structures
1. Structures with a maximum number of octets is preferred.
2. Charges should be located on atoms with compatible electronegativity.
3. Minimize charge separation.
4. Charge separation may be enforced by the octet rule (atoms may be
charged if they have an octet.)
Chem 61
Hydrocarbons
Chapter 2-1
CHAPTER 2
Hydrocarbons: compounds containing only hydrogen and carbon: alkanes,
alkynes, alkenes
alkanes contain only C—H and C—C single bonds CnH2n+2; alkenes CnH2n
contain C— C double bonds and alkynes CnH2n-2 contain C—C triple bonds
Alkanes do not react with hydrogen, but alkenes and alkynes can react with
hydrogen under certain conditions
H2, catalyst
CH2 CH2
CH3 CH3
alkene
alkane
2H2, catalyst
CH3 C C CH3
H3C–H2C CH2–CH3
alkyne
ISOMERISM
Structural Isomers: compounds with the same molecular formula that
differ in the order in which the atoms are bonded to one another
CH3
CH3
C CH3
CH3
C5H12
2,2-dimethylpropane
H CH3
CH3 C C CH3
H H
H H H
CH3 C C C CH3
H H H
C5H12
2-methylbutane
C5H12
n-pentane
Chem 61
Nomenclature
Chapter 2-2
Chem 61
NOMENCLATURE
Alkanes are named by the following parent names
CH4
CH3CH3
CH3CH2CH3
CH3(CH2)2CH3
CH3(CH2)3CH3
CH3(CH2)4CH3
CH3(CH2)5CH3
CH3(CH2)6CH3
CH3(CH2)7CH3
CH3(CH2)8CH3
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
Cyclic alkanes are named the same but with cyclo as a prefix
cyclohexane
cyclobutane
Branched hydrocarbons are named from the parent with a substituent as a
prefix hydrocarbon
branches are named by dropping ane from the parent and adding -yl
thus
methane: methyl
ethane:
ethyl
propane: propyl , etc.
special trivial names for branches
CH3 C CH3
H
CH3 C CH3
CH3
isopropyl
tert-butyl
H CH3
C C CH3
H H
isobutyl
H
CH3 C C CH3
H H
sec-butyl
– 23 –
Nomenclature
Chapter 2-3
Chem 61
BASIC RULES OF NOMENCLATURE
1. Find the longest continuous chain (not necessarily drawn as a straight
line) and name the parent
2. Number the parent chain starting at the end nearest the branch
3. Identify the branch and its position
4. Attach the number and the name of the branch to the parent name.
H
CH3 C
3
4
H
the branch is CH3; therefore methane
CH3
C CH3
2 1
H
methyl
butane is the parent
2-methylbutane
H
H
CH3H
C
C
C
H 3C
H
CH3H
CH3
C
CH3
3,3,5-trimethylhexane
Other Functional Substitutents
—CO2H
increasing
priority
-oic acid
C H
-al
O
C
-one
O
—OH
—NR2
—C=C—
—C≡C—
R—,
C6H5—,
Cl—, Br—,
—NO2
-ol
-amine
-ene
-yne
prefix
substituents
– 24 –
Nomenclature
Chapter 2-4
Chem 61
alkenes
H
H
C
C
C
C
CH3H
H
CH3
H
H
CH3
C
C
C
HO
H
H
CH3
CH3
2,5-dimethyl-2-hexene
alcohols
CH3
H
2-pentanol
carboxylic acids
H
H
H
H
C
C
C
Br
H
H
COOH
4-bromobutanoic acid
aldehydes and ketones
CH3
H
H
CH2CH3
C
C
C
H
CH3H
CHO
2-ethyl-3-methylpentanal
CH3
H
O
CH2CH2CH3
C
C
C
H
CH3
H
4-methyl-3-heptanone
– 25 –
Alkanes
Chapter 2-5
Chem 61
ALKANES
Physical Properties
nonpolar compounds
C1 to C4 are gases; C5 to C17 are liquids; >C17 are solids
Boiling points increase about 30°C for each additional CH2 unit
branching lowers the boiling point due to disruption of van der Waals attractions
insoluble in water; soluble in organic solvents like diethyl ether, benzene
Chemical Properties
very unreactive compounds
Halogenation
CH3CH3 +
Cl2
light
CH3CH2Cl + HCl + other products
Oxidation of Alkanes
Combustion
spark
CH3CH2CH2CH2CH3
+ 8O2
5CO2 + 6H2O
oxidation: a reaction that either removes a hydrogen atom from a carbon or
adds an electronegative element to the molecule (O, N, S, halogen).
low oxidation level
CH3CH3
CH2=CH2
CH3CH2OH
CH3CH2NH2
CH3CH2Cl
high oxidation level
HC≡CH
CH3CH=O
CH3CH=NH
CH3CHCl2
CH3CO2H
CH3C≡N
CH3CCl3
CO2
CCl4
– 26 –
Oxidation and Reduction
Chapter 2-6
"Complete"/"incomplete" oxidation (combustion) of propane
5O2
CH3CH2CH3
2 CH3CH2CH3
CH3CH2CH3
7O2
2O2
3 O=C=O + 4 H2O "complete" combustion
6 CO + 8 H2O
"incomplete" combustion
3 C + 4 H2O
"incomplete" combustion
Heat of Combustion: energy released when a compound is completely
oxidized to CO2 and water; depends mostly on number of CH2 units:
approximately 157 kcal/ methylene unit
Reduction of Alkenes, Alkynes
reduction: a reaction that either adds H atoms or removes an electronegative
atom from the molecule
CH3CH3
Pd, Ni,or Pt; H2
CH2=CH2
Pd, Ni,or Pt; H2
HC≡CH
Pd, Ni,or Pt; H2
Pd, Ni,or Pt; H2
heat
no reaction (NR)
CH3CH3
CH2=CH2
Pd, Ni, Pt
H2
CH3CH3
Chem 61
Chapter 2-7
Conformations of Acyclic Hydrocarbons
Chem 61
Butane
CONFORMATIONS OF OPEN CHAIN COMPOUNDS
Molecular Mechanics
E
eclipsed methyls
Steric Energy: (isolated molecule in gas phase at 0° K): relative energy of a
conformation or stereoisomer calculated using classical mechanics (atoms
and bonds treated as balls and springs)
eclipsed
4.5 kcal
3.8 kcal
Stretch (bond length): energy associated with stretching or compressing
bonds from their optimal length
gauche
Bend (bond angle): energy associated with deforming bond angles from their
optimal angle
0
60
120
0.9 kcal
180
240
anti
300 360
Stretch-Bend: energy required to stretch two bonds involved in a severely
compressed bond angle
dipole-dipole: energy associated with interaction of bond dipoles
H
out of plane: energy required to distort a trigonal center out of planarity
H
torsional strain: destabilization from eclipsing of bonds on adjacent atoms
Ethane
Newman projection
H
C
H
H
H
H
C
H
H
H
H
H
H
H
H
H
staggered
HH
staggered
HH
eclipsed
E
∆H = 3 kcal/mole
eclipsed
Ethane
0 kcal
staggered
0
60
120
180
240
300 360
H
CH3
anti
van der waals strain: destabilization from two atoms being too close
together
dimensional
CH3
H HH
H
CH3 H
HCH3
eclipsed1
H
H
H
CH3
CH3
gauche
HH
H
H
H3CCH3
eclipsed2
H
H
CH3
H HCH3
H
CH3
gauche
H
H
HCH3
eclipsed3
Chapter 2-8
Conformations of Cyclic Hydrocarbons
CYCLIC COMPOUNDS
Strain energy
-∆H
(kcal/mole)
cyclopropane
cyclobutane
cyclopentane
cyclohexane
499.8
655.9
793.5
944.5
-∆H
per CH2
166.6
164.0
158.7
157.4
strain energy
per CH2
total strain energy
9.2
6.6
1.3
0
27.6
26.4
6.5
0
cyclopropane: bond angles 60°; tetrahedral 109.5°
sp3 orbitals are 109.5° apart
cyclopropane bond angles 60°
maximum overlap cannot be achieved
cyclobutane and cyclopentane
H
H
H
H
H
HH
H
H
H
H
H
H
puckered cyclobutane
H
H
H
H
H
envelope cyclopentane
puckering allows bond angles to be at or close to the tetrahedral angle and
minimizes torsional strain (electron—electron repulsions in eclipsed bonds)
between adjacent C—H bonds
Chem 61
Chapter 2-9
Conformations of Cyclohexane
cyclohexane
H
H
H
H
H
H
H
H
H
equatorial bonds
H
H
H
chair cyclohexane
axial bonds
half chair
boat
E
twist boat
7.1
chair
0
60
H
H
H
H
H
H H
chair
H
H
H
H
H
120
5.5
10.8 kcal
180
240
H H
H
H H
H
H
H
H
H
300
360
H
H
H
half chair
H
H
H
H
H
H
H
H H
H H
boat
chair
twist boat
Chem 61
Chapter 2-10
Conformations of Cyclohexane
Substituted Cyclohexanes
substitutents on cyclohexanes preferntially occupy equatorial positions due
to 1,3 diaxial interactions in axially substituted cyclohexanes
H
H
H
H
C
H
H
H
H
H
1,3-diaxial interactions
H
H
H
gauche
H
CH3
C
H
H
E
more stable by
1.8 kcal/mole
CH3
H
H
H
C
H
H
anti
H
H
H
C
H
CH3
C
H
H
H
axial substitution similar to gauche butane
equatorial substitution similar to anti butane
H
CH3
CH3
H CH3
C
H
H
H
H
C(CH3)3
H
H
H
H
5.6 kcal/mole
more stable
∆G = -RTlnK
∆G = -(1.98 kcal/mol°)(298)(2.94) = - (1.98 kcal/mol°)(298)(ln19)
= -1.74 kcal/mole for methylcyclohexane
Chem 61
Stereochemistry: Geometric Isomers
Chapter 3-1
Chapter 3
STEREOCHEMISTRY
Geometric isomerism in cyclic compounds
OH
OH
OH
H
OH
H
OH
OH
cis
trans-1,2-cyclohexanediol
CH3
CH3
H
H
CH3
Br
trans-1,3-dimethylcyclobutane
Br
Br
cis
cis
Geometric Isomerism in alkenes
68 kcal/mole to cleave a carbon-carbon pi bond thus no "free" rotation
CH3
H
C
CH3
C
H
CH3
H
C
C
CH3
trans-2-butene
CH3CH2
H
H
C
CH3
cis-2-butene
H
H
C
C
H
trans-2-pentene
Cl
C
Cl
cis-1,2-dichloroethene
Chem 61
Absolute Configuration
Chapter 3-2
Stereoisomers: compounds with the same structures differing only in their
arrangement of atoms in space.
not cis or trans
Cl
CH3
C
CH3CH2
F
Cl
C
C
H
C
Br
E entgegen (across)
I
Z zusammen (together)
Cahn-Ingold-Prelog Sequence rules
1. if the atoms are different, highest atomic number gets highest priority
2. if two isotopes of the same element, the one with the higher mass gets
higher priority
3. if the atoms are the same, the atomic numbers of the next atoms are used
to assign priority
4. atoms attached by double or triple bonds are given single bond
equivalencies
O
R
(O)
C R'
R
=
O
C R'
R
(O)
OH = R
C
O (C)
(C)
R
C
CR'
=
R
C (C)
R'
C (C)
(C)
C
OH
O
(C)
(C)
R
C
C—R'
R
C
H
H
H
H
C
R'
(C)
Chem 61
Stereochemistry: Chirality
Chapter 3-3
CHIRALITY
An object or molecule which cannot be superimposed on its mirror image is
said to be chiral
If an object or molecule can be superimposed on its mirror image, it is achiral.
Enantiomers: isomers which are nonsuperimposable mirror images.
I
I
H
Br
H
0000
0000
0000
0000
0000
0000
0000
0000
000
000
000
0000
000
0000
0000
0000
0000
0000
0000
0000
0000
0000
Br
Cl
000
000
000
000
000
000
000
000
000
000
000
000
Cl
000
000
000
000
0000
0000
0000
0000
0000
0000
0000
mirror plane
Cl
Cl
I
Br
H
Br
I
H
Stereogenic carbon atom: a carbon with four different groups bonded to it
(designated *).
H
Cl
H
C* CH3
CH3
C* Cl
CH2CH3
CH2CH3
enantiomers
OH
CH3
Br
C* CH2CH3
H
CH2CH2CH3
Fischer Projections
by convention: horizontal bonds come out of the paper
vertical bonds go back into the paper
CH3
H
C
OH
CH2CH3
CH3
H
OH
CH2CH3
H
H C OH
H C OH
CH3
H
H
H
OH
OH
CH3
C*
CH3
Chem 61
Stereochemistry: Chirality
Chapter 3-4
OPTICAL ROTATION
enantiomers have almost all the same physical and chemical properties
properties which differ are:
1. interaction with other chiral substances
2. interactions with polarized light
Polarimeter
ordinary
light
lamp
polarized light
polarizer
rotated light
solution of
sample
if plane polarized light is passed through a solution of a single enantiomer, the light
is rotated either to the right or the left; the opposite enantiomer will rotate the light
in the opposite direction
optically active: a compound which rotates plane polarized light
optical isomers: enantiomers
dextrorotatory: rotates plane polarized light to the right, also (+) or d
levorotatory: rotates plane polarized light to the left, also (–) or l
racemic mixture: a 1:1 (50:50) mixture of two enantiomers
does not rotate plane polarized light; therefore optically inactive
Chem 61
Stereochemistry: Absolute Configuration
Chapter 3-5
Absolute configuration: the order of arrangement of the four groups
around a stereogenic center
enantiomers have opposite configurations
(R) and (S) Cahn-Ingold-Prelog System
R: rectus or right S: sinister or left
To assign R and S to an asymmetric atom
1. Rank the four attached groups from 1 (highest) to 4 (lowest) priority
based on the Cahn-Ingold -Prelog Sequence rules
2. Project the molecule with the lowest priority group to the rear
3. Draw a semicircle from 1 to 2 to 3
4. If the direction of the semicircle is clockwise; configuration is R; if
counterclockwise; configuration is S
1
Br
3
3
CH3
CH3
C
Cl
2
2 CH3CH2
C
OH 1
H
H
4
4
R
R
If lowest priority group is out; assign the configuration as usual and then
reverse it
1
Br
2
2
Cl
Cl
C
CH3 3
H
4
lowest priority group out
3 CH3
C
Br 1
H
R
S
4
S
Chem 61
Stereochemistry: Diastereomers
Chapter 3-6
Chem 61
Molecules with Two or more Asymmetric centers
If a molecule has n asymmetric carbon atoms, it contains a maximum of 2n
isomers; may not have that many
CH2OH
CH2OH
CH2OH
H
C
OH
HO
C
H
H
C
OH
H
C
OH
HO
C
H
HO
C
H
CH3
enantiomers
CH3
CH2OH
HO
C
H
H
C
OH
CH3
CH3
enantiomers
two asymmetric carbons; 4 isomers
diastereomers: stereoisomers which are not enantiomers; may have different
physical and chemical properties
Meso compounds
meso compounds contain an internal plane of symmetry and are achiral: that
is the mirror image is identical to the original
diastereomers
meso: identical
CH2OH
CH2OH
CH2OH
H
C
OH
HO
C
H
H
C
OH
H
C
OH
HO
C
H
HO
C
H
CH2OH
CH2OH
superimposable
H
CH3
diastereomers
diastereomers
meso
CH2OH
HO
C
H
H
C
OH
CH2OH
enantiomers
CH2OH
H
CH3
mirror plane
chiral
H
CH3
H
CH3
OH
HO
H
meso
H
mirror plane
H
HO
OH
H
chiral
Stereochemistry: Asymmetric Synthesis, Resolution
Chapter 3-7
Preparation of Enantiomerically Enriched Compounds
Generating chiral compounds from achiral compounds
Enzymes
O
O
yeast
H
HO
O
OEt
OEt
>98% R
Asymmetric Reagents
H
OH
t-BuOOH
O
OH
(+)-diethyl tartrate
Ti(i-OPr)4
H
>95%
one enantiomer
Resolution of a Racemic Mixture
separated salt converted back to acid
(R) R*COOH + (S) R*NH2
+
(S) R*COOH
racemic
(R) R*COO–(S) R*NH3+
+
(R) R*COOH
(S) R*COO–(S) R*NH3+
diastereomeric salts (separable) (S) R*COOH
Chem 61
Chapter 4-1
Acids and Bases
ACIDS AND BASES
Bronsted-Lowry:
acid: a proton donor
base: a proton acceptor
Strong acid: completely ionized or dissociated in water
e.g., HCl, H2SO4, HNO3, HBr
Weak acid: only partially dissociated in water:
carboxylic acids are weak acid
CH3CO2H
acid
+
CH3CO2-
H2O
+
conjugate base
base
H3O+
conjugate acid
amines are weak bases
CH3NH2
base
+
HO –
CH3NH3+ +
H2O
acid
conjugate acid
conjugate base
Generally: strong acids have weak conjugate bases and weak acids have
strong conjugate bases
that is, as acid strength increases, the basicity of the conjugate base
decreases
Thus the ability of the conjugate base to stabilize a negative charge
determines the strength of an acid
Conjugate Acids
H 2O
pKa
HCN
15.75
CH3CO2H
6.37
H3PO4
HCl
2.12
-7
4.75
increasing acid strength
Conjugate Bases
HO –
NC –
CH3CO2–
H2PO4 – C l –
decreasing base strength
Chem 61
Chapter 4-2
Acids and Bases
Factors affecting Acidity:
the electronegativity and the size of the atom which carries the negative
charge influence its ability to stabilize the negative charge
size of the atom
H—F
H—Cl
pKa 3.45
-7
H—Br
-9
H—I
-9.5
increasing size of halogen, increasing acid strength
electronegativity
pKa
(CH3)3C—H
50
(CH3)2N—H
35
CH3O—H
15.5
F—H
3.45
increasing electronegativity of atom, increasing acid strength
ACIDITY CONSTANTS, Ka's
for acetic acid
CH3CO2H + H2O
Ka =
CH3CO2– + H3O+
[CH3CO2–] [H+]
[CH3CO2H]
since stronger acids are more ionized, the larger Ka, the stronger the acid
pKa = -logKa, the lower the pKa, the stronger the acid
also, the higher the pKa, the weaker the acid or the stronger the base
the stronger the acid, the more stable the anion produced by ionization of the
acid.
Chem 61
Chapter 4-3
Lewis Acids and Bases
LEWIS ACIDS AND BASES
Lewis acid: electron pair acceptor: any species with an electron deficient atom
BBr3, AlCl3, H3C+
Lewis base: electron pair donor; any species with an unshared pair of electrons
..
..
H3N:, CH3CH2..OH, H2C=O:
Chem 61
Chapter 5-1
Alkenes: Structure and Isomerism
ALKENES
Alkene Structure
sp2 hybridized carbons are trigonal planar with atoms 120° apart
2p
2p
2p
96 kcal
2s
sp2
hybridize
2s
1s
1s
1s
ethylene: trigonal planar
H
C
H
C
H
H
2p
H
H
C
C
H
H
H
C
H
120°
H
H
H
pi orbital
H sigma orbitalH
sp2
H
H
C
H
C
C
C
H
H
pi* orbital
H
H
H
C
H
sigma* orbital
bond angles = 120°; C=C bond length = 1.34Å
pi (π) bonds are formed by the side by side overlap of two p-orbitals (approx. 68 kcal/mol)
pi bonds are above and below the plane where the sigma bond is located
pi bonds make the molecule rigid between the two atoms preventing rotation
E
sigma σ*
pi π*
pi π
sigma σ
ethylene C=C double bond: one sigma, one pi
Chem 61
Chapter 5-2
Alkenes: Structure and Isomerism
Geometric Isomerism in alkenes
68 kcal/mole to cleave a carbon-carbon pi bond thus no "free" rotation
CH3
H
C
H
H
C
C
CH3
H
CH3
trans-2-butene
C
H
H
C
C
CH3CH2
CH3
cis-2-butene
CH3
H
C
C
Cl
H
Cl
cis-1,2-dichloroethene
trans-2-pentene
Stereoisomers: compounds with the same structures differing only in their
arrangement of atoms in space.
not cis or trans
Cl
CH3
C
CH3CH2
C
H
E entgegen (across)
F
Cl
C
Br
C
I
Z zusammen (together)
Cahn-Ingold-Prelog Sequence rules
1. if the atoms are different, highest atomic number gets highest priority
2. if two isotopes of the same element, the one with the higher mass gets
higher priority
3. if the atoms are the same, the atomic numbers of the next atoms are used
to assign priority
4. atoms attached by double or triple bonds are given single bond
equivalencies
Chem 61
Alkenes: Electrophilic Addition Reactions
Chapter 5-3
Chem 61
Alkenes: Reactivity
The reactivity of alkenes is due to their ability to donate a pair of electrons:
their Lewis basicity.
H
H
C
Consider Ethylene:
The site of reactivity is the pi-bond due to the exposed nature of the pi electrons
H
H
C
C
H
H
2p
H
H
C
C
sp
H
pi orbital
H sigma orbitalH
H
H
H
C
C
H
H
pi* orbital
C
H
sigma* orbital
The pi electrons act as a Lewis base (electron pair donor).
These electrons react with electron deficient species (Lewis acids)
Alkenes undergo electrophilic additions reactions.
Electrophile: an electron deficient ion or molecule
H+
Br+
BH3
+CH3
Nucleophile: an electron rich ion or molecule.
I–
CH3NH2
Br–
H
H
Br–
+
CH2—CH2–H
CH2=CH2
H–Br
H
C
H
H
2p
C
2
H
H
H
Br
H
H
H
H
H
H
H
C
H
120°
H
H
H
H
C
H
Br
H
H2O
Reaction Mechanisms of Electrophilic Addition Reactions
reaction mechanism: a detailed description of how a chemical reaction
occurs.
A roadmap of a reaction....curved arrows show which bonds are formed or
broken. A mechanism includes the transition states involved in making and
breaking bonds and reactive intermediates that are formed along the pathway
from reactants to products.
Br H
H2C CH2
Thermodynamics
Chapter 6-1
Thermodynamics
describes the properties of a system at equilibrium
∆G = – RT ln Keq where Keq = [reactants]
[products]
R = 1.986 X 10-3 kcal K-1 mol-1 (gas costant)
T = temperature in degrees Kelvin
∆Go = ∆Ho – T∆So
free energy = enthalpy – T x entropy
enthalpy = heat of reaction
entropy = state of disorder
transition state
Energy Diagram
∆G ‡
E
products
reactants
progress of reaction
reactants
– ∆G
products
exergonic
+ ∆G
products
reactants
endergonic
Chem 61
Kinetics
Chapter 6-2
Kinetics
describes the rate of progression of a reaction
depends on the energy of activation (the stability of the transition state):
the lower the transition state energy, the faster the reaction will be.
transition state
faster
transition state
slower
∆G ‡
E
∆G ‡
E
products
reactants
progress of reaction
reactants
products
progress of reaction
first order reaction
rate = k[A]
rate is proportional to the concentration of one reactant
second order reaction
rate = k[A] [B]
rate is proportional to the concentration of two reactants
A two step reaction
transition state
E
∆G‡
reactants
transition state
∆G‡
intermediate
products
Rate limiting step:
is the slowest step (step with the highest energy of activation)
Chem 61
Electrophilic Addition: Addition of H–X
Chapter 6-3
Electrophilic Addition Reactions
Nucleophilic addition does not occur with alkenes unless an electron-attracting
group is attached to one of the carbon atoms to cause a polarity difference.
C
No Reaction
Nu:
+
C
However, in the electrophilic addition reaction, the reagent is E+A– (E+ =
electrophile, A– = some anion)
+
E A
+
C C
–
E
+
C C
A
A
–
E
C C
Note the alkene acts as a Lewis base (or nucleophile) toward the Lewis acid
(or electrophile)
The intermediate, a carbocation, reacts with A– to yield a product in which E
and A have added to the C=C double bond.
Addition of HX (E+ = H– A– = F–, Cl–, Br–, I–)
CH3
C
CH2 + H—X
CH3
CH3
+
C
X–
CH3
CH3
X
CH3
C
CH3
CH3
Reactivity: H—I > H—Br > H—Cl > H—F
The reaction is said to be regioselective since an unsymmetrical alkene gives a
predominance of one of two possible electrophilic addition products. The term
regiospecific is used if one product is formed exclusively.
In these reactions, the halogen (A–) is found attached to the most substituted
carbon atom of the alkene (Markovnikoff's Rule):
Chem 61
Electrophilic Addition: Addition of H–X
Chapter 6-4
CH3
C
CH2 + H—X
CH3
CH3
R
C
C
X
CH3
CH3
C
CH3
CHR + H—X
R
R
R
X–
+
C
+
C
CH3
X–
X
CHR
R
R
3° carbocation
more favored
X–
+
R CH C—R
CHR + H—X
R
CH3
C
CHR
R
R
R
H
X
C
CHR
R
2° carbocation
less favored
Thus regioselectivity is explained by the lower Eact leading to the 3°
carbocation intermediate.
If a nucleophilic solvent is employed in the electrophilic addition reaction
solvent may compete with A – for the intermediate carbocation.
Cl
CH
CH2
OCOCH3
CH
HCl
CH3
CH
CH3
+
CH3COOH
(solvent)
Note the C=C's of the aromatic ring do not undergo this type of reaction.
Since a carbocation intermediate is involved, rearrangement may
sometimes occur.
+
HCl
(CH3)2CH CH CH3
(CH3)2CH CH CH2
(CH3)2CH
CH CH3
40% Cl
+
(CH3)2C
Cl
CH2 CH3
60%
Chem 61
Carbocation Stability
Chapter 6-5
Carbocation stability
Carbocations are highly reactive intermediates ; they cannot be observed
directly in the reaction mixture since they react as soon as they are formed.
Any structural feature that will disperse a positive charge will stabilize the
carbocation.
This stabilization has been quantitated by so-called hydride H:– affinity
measurements (gas phase).
The lower the value of the hydride affinity, the more stable the carbocation.
Name
Structure
methyl
primary 1°
secondary 2°
tertiary 3°
vinyl
allyl
secondary allyl
tertiary allyl
CH3+
CH3CH2+
(CH3)2CH+
(CH3)3C+
CH2=CH+
CH2=CH—CH2+
CH2=CH—CH+(CH3)
CH2=CH—C+(CH3)2
Hydride Affinity
(kcal/mol)
314
274
247
230
287
256
237
225
298
phenyl
+
233
benzyl
CH2+
secondary benzyl
+
CHCH3
226
tertiary benzyl
+
C(CH3)2
220
diphenylmethyl
+
CH
215
triphenylmethyl
+
C
210
Chem 61
Carbocation Stability
Chapter 6-6
Thus the order of carbocation stability is
triphenyl methyl > diphenylmethyl > 3° ≈ benzyl ≈ allyl > 2° > 1° >> methyl
Carbocation centers can be stabilized by overlap with adjacent pi orbitals
(resonance) or adjacent sigma orbitals (hyperconjugation).
H
+
CH3
C
CH3
+
C
H
H
p
H
σ
..
C
C
C
+
CH2
adjacent sigma bond donates some electron density
to the empty p orbital on the carbocation center
more adjacent sigma orbitals result in a more stable
carbocation thus 3° > 2° > 1° > methyl
H
H
p
sigma donation
H
H
empty p orbital ovelaps with adjacent pi orbital to
disperse the positive charge
pi
CH
CH2
CH2
CH
+
CH2
equivalent resonance structures
Since the pi orbital is closer in energy to the empty p than the sigma orbital,
the pi overlap stabilizes the cation more efficiently.
Chem 61
Electrophilic Addition: Addition of H2O
Chapter 6-7
Addition of Water (E+ = H+; A– = HOH)
use H2SO4, H2O
H+
Step 1
R 2C
+
R 2C
CHR
..
+
R 2C
Step 2
Step 3
H
O:
R 2C
H
+
:O H
:OH2
CH2 R
R 2C
..
H
:OH2
+
CH2 R
CH2 R
CH2 R
..
:O H
R 2C
CH2 R
+ H3O+
TS1
TS2
+
R 2C
Eact
CH2 R
TS3
H
+O H
R 2C
CHR
R 2C
CH2 R
O H
R 2C
CH2 R
Addition of water to alkenes is a reversible reaction and whether the alkene or
the alcohol predominates at equilibriuim depends on the reaction conditions.
Low temperatures and high concentrations of water favor the alcohol.
Higher temperatures and removal of water favor the alkene.
Chem 61
Electrophilic Addition: Oxymercuration
Chapter 6-8
If alcohol is used as the solvent in this process ethers are obtained as the
products.
Addition of Mercury acetate and Water
( E+ = +HgOCOCH3; A– = OH2)
To avoid carbocation rearrangements under the conditions used in addition of
H2O (H2SO4/H2O) a better system for the addition of water to a C=C has
been devised.
O
CH3
C
O
O
Hg O
C
O
+
CH3
Hg O
E
mercury acetate
C
CH3
CH3
CH3
C
O
CH3 +
–
O
CH2
+ +Hg O
C
or
C
CH2 Hg O C CH3
CH3
carbocation
O
+
H2O:
CH3 C
CH2 Hg O C CH3
Hg O
O
C CH3
CH2
CH3
bridged carbocation
+OH2
Since no rearrangement is observed in reactions with +HgOCOCH3 the
bridged carbocation is believed to be the true intermediate. Attack by H2O
on the more electron deficient carbon opens the bridged cation (backside
attack).
NaBH4 reduction of the C—Hg bond yields the alcohol in a separate step.
CH3
CH3
C CH2HgOCOCH3
+OH2
H 2O
CH3
CH3
C CH2 HgOCOCH3
HO
CH3
CH3
HO
NaBH4
C
CH3
O
+
CH3 C
CH3
C
+
O
CH3
Chem 61
CH3 + Hg°
CH3
C
CH3
CH2
1. Hg(OCOCH3)2, ROH
2. NaBH4
CH3
CH3
RO
C
CH3
Electrophilic Addition: Hydroboration
Chapter 6-9
Hydroboration (E+ = B, A– = H–)
Discovered by H.C. Brown in the 1950's. The reagent BH3 (borane) is used in
electrophilic addition reactions.
Borane is generated by
2 B2H6 (diborane)+ 3 NaBF4
3 NaBH4 + 4 BF3
4 BH3 (borane)
Addition of BH3 to C=C, C≡C is termed hydroboration.
R—CH=CH2 + BH3
R
H
H
B
C
C
H
H
H
R
H
H
H
B
C
C
H
H
H
H
‡
R
H
B
C
C
H
Note B is the electrophile and H acts as A-. The addition of H– and B occur
from the same face of the alkene (syn addition). Further reaction of
R—CH2–CH2—BH2 with more alkene yields first the dialkyl borane and then
the trialkyl borane as the stable product.
R—CH=CH2
+
R—CH2—CH2—BH2
( R—CH2—CH2)2BH
R—CH=CH2
( R—CH2—CH2)3B
Brown discovered that trialkylboranes were easily oxidized by alkaline
hydrogen peroxide (H2O2 in HO–). The oxidation reaction proceeds as
shown.
H
H
H
H
Chem 61
Electrophilic Addition: Hydroboration
Chapter 6-10
–
( R—CH2)3B
CH2R
OOH
RCH2
B–O
OH
RCH2
RCH2
RCH2
–
RCH2CH2
B O CH2R
+ —OH
OCH2R
RCH2
CH2R
OOH
B
O
B–O
O
–
OH
OCH2R
OOH
CH2R
CH2R
RCH2O
B
O
–
OH
CH2R
3 R—CH2—OH + B(OH)3
The overall reaction is
R—CH=CH2
1. BH3
R—CH2—CH2—OH
–
2. H2O2, OH
Note the reaction is regioselective and yields an alcohol that is isomeric
to that obtained by Hg(OCOCH3)2—NaBH4 or H2SO4—H2O.
R—CH=CH2
H3O+
or
Hg(OCOCH3)2
then NaBH4
R—CH—CH3
OH
Treatment of a trialkylborane with CH3CO2D yields a monodeuterated alkane
(CH3CH2)3B
CH3CO2D
CH3CH2D
Chem 61
Electrophilic Addition: Hydroboration
Chapter 6-11
Treatment of a trialkylborane with CH3CO2D yields a monodeuterated alkane
(CH3CH2)3B
CH3CO2D
CH3CH2D
Bromination of a trialkylborane yields the alkyl bromide.
(C6H5CH2CH2)3B
Br2
C6H5CH2CH2Br
HO–
Stereochemical Results of Hydroboration
Hydroboration is regioselective and stereospecific
CH3
H
B
BH3
H
H
CH3
cis-addition
Thus H and OH add cis to the C=C.
H2O2
HO–
OH
OH
H
H
+ H
CH3
CH3
enantiomers
H
Chem 61
Electrophilic Addition: Stereochemistry of Halogen Addition
Chapter 6-12
Stereochemistry of Halogenation
Br
+
C C
C C
Br
C C
Br
Br—
Alkenes containing alkyl substituents react with Br2 via a bridged bromonium
ion intermediate. Nucleophilic opening of the bridged ion by backside attack
of Br – at carbon gives overall anti (trans) addition of E+(Br+) and A– (Br–) to
the double bond. Chlorination reactions proceed by a similar pathway.
Some examples:
CH3
C
H
C
X
CH3 Br2
H
or Cl2
CH3
C
C
CH3
X
H
CH3 meso
C
Br2
or Cl2
C
(enantiomers)
X
H
H
+
X
H
CH3CO2H
(solvent)
CH3
C
H
C
CH3
C
X
H
X
Br2 or Cl2
X
H
H
CCl4
X
X
CH3
CH3
H
+
HH
X
C
H
CH3
Chem 61
Electrophilic Addition: Mixed Addition
Chapter 6-13
Mixed Addition
Br
C C
Br
Br
C C
Cl
Br—
C C
Br+
Br+
C C
Cl—
Br2 or Cl2 with H2O, HO– present:
C C
X+
X+
C C
HO–
X
C C
OH
Mixed Reagents
R
S+
RS—Cl +
O=N—Cl +
C
C
C
C
C
RS
C
C
Cl
NO
NO
+
C
C
C
C
C
Cl
I
I—Cl +
C
C
C
+
C
I
C
C
Cl
Chem 61
Chapter 6-14
Hydrogenation of Alkenes
Hydrogenation of Alkenes
Alkene Heat of Hydrogenation, ∆Hh (kcal/mol)
addition of H2 is cis (syn)
CH3CH2CH=CH2 -30.3
CH3CH=CHCH3, Z -28.6
CH3CH=CHCH3, E -27.6
CH3
H
H2/Pd
Thus the stability of the three alkenes is
H CH3
CH3
CH3 C C CH3
CH3CH=CHCH3, E > CH3CH=CHCH3, Z > CH3CH2CH=CH2
CH3
H2/Pd
These comparisons indicate
H
H
C C
CH3
CH3
i) increasing alkyl substitution stabilizes an alkene
ii) conjugated dienes are more stable than non-conjugated dienes,
iii) trans alkenes are more stable than cis alkenes (steric repulsions in cis)
ii) Heat of Hydrogenation Studies
The heat of hydrogenation of an alkene is the energy difference between the
starting alkene and the product alkane. It is calculated by measuring the
amount of heat released in a hydrogenation reaction.
H
H
C
C
CH3CH2
H
H
H
C
CH3CH2CH2CH3 + heat
C
CH3
CH3
CH3
H
C
CH3
C
H
CH3CH2CH=CH2
E
∆Hh
progress
Z-CH3CH=CHCH3
∆Hh
progress
Chem 61
E-CH3CH=CHCH3
∆Hh
progress
Chapter 6-15
Radical Addition Reactions
Addition of HBr in the presence of a radical initiator such as peroxides effects
so called anti Markovnikov addition of HBr
CH3
C
CH2 + H—X
X–
+
C
CH3
CH3
Reversal of regioselectivity (position of bromine and H addition)
results from the inital addition of Br radical rather than H+.
X
CH3
CH3
CH3
C
CH3
CH3
Stability of Carbon Free Radicals (Which H atom is abstracted?)
Free Radical: any atom or group of atoms that contains one or more unpaired
Stability of carbon free radicals follows the same pattern as
carbocations:
electrons.
.
A free radical is symbolized as a single dot representing the unpaired electron: Cl ,
.
Chem 61
.
Br , H3C , etc.
.
CH3
Free radicals are usually encountered as high energy short-lived, non-isolable
.
.
.
.
CH2=CH—CH2 , C6H5CH2 > (CH3)3C > (CH3)2CH > CH3CH2 >
.
allyl
benzyl
tertiary
secondary
primary
methyl
intermediates in certain reactions.
heat or light
RO
CH3
R
initiation
+ H—Br
C
CH2 + Br
RO–H + Br
.
.
CH3
CH3
C
.
propagation
Br
H–Br
CH2
CH3
CH3
H
C
Br
CH2 + Br
.
CH3
propagation
combination of any two radicals
Br
. + Br .
. Br
CH3
C
CH2
Br2
termination
+ Br
.
CH3
CH3
2 RO
.
C
R
.
Br Br
C CH2
CH3
RO–OR
.
ROOR
2 RO .
RO–OR
peroxide
CHR + H—Br
R
C
Br
CHR
R
3° carbocation
more favored
H
R
C
R
Br
CHR
Radical Reactions of Alkanes
Chapter 7-1
Free Radical Substitution Reactions
Free Radical: any atom or group of atoms that contains one or more unpaired
electrons
A free radical is symbolized as a single dot representing the unpaired electron:
.
.
.
Cl , Br , H3C , etc.
Free radicals are usually encountered as high energy short-lived, non-isolable
intermediates in certain reactions. (Recall that carbocations were seen to be
intermediates in certain reactions in Chapter 5).
CH3
a) CH3
C
CH3
Br
–
C+ Br
CH3
carbocation intermediate
CH3
CH3
double barbed arrows indicate the movement of two electrons
CH3
b) CH
3
C
CH3
Br
CH3
CH3
C
.
Br.
free radical intermediate
CH3
single barbed arrows indicate the movement of one electron
Note that the carbon in a) has lost an electron pair to the bromine atom, thus the
carbon has an sp2 configuration and a single "+" charge resulting from the empty p
orbital.
In b) the carbon atom has lost one electron; the carbon is also in the sp2
configuration with the unpaired electron in the p orbital.
CH3
CH3
C
CH3
(CH3)3C+
CH3
CH3
.
C
CH3
.
(CH3)3C
equal probability of electron
being in either lobe
Chem 61
Radical Reactions of Alkanes
Chapter 7-2
Free Radical Substitution Reactions
Chem 61
Hence CH2Cl2, CHCl3, and CCl4 are formed
Chlorination of methane in the presence of light is the classic example
c) Termination of the Chain Reaction
CH4 + Cl2
light
CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
.
.
.
terminate the reaction.
.
.
Cl
a) Initiation of the radical chain reaction
Cl
Cl Cl
light
.
.
H
Cl
i)
C
.
H—Cl + CH3
H
requires 1 kcal/mole
a methyl radical
.
ii)
.
CH3 + Cl—Cl
CH3—Cl + Cl
chlorine radical can now react with more CH4
and propogate the chain
Note: Cl has substituted for a H and therefore free radical substitution
.
iii) As the concentration of CH3—Cl increases in the reaction mixture Cl starts to
react with both CH4 and CH3Cl.
.
Cl
.
+ CH3—Cl
CH2—Cl + Cl—Cl
H—Cl +
.
CH2—Cl (chloromethyl radical)
Cl—CH2—Cl + Cl
.
CH3—Cl
+
.
.
CH2Cl
.
.
CH2Cl + CH2Cl
b) Propagation (self-perpetuating the chain reaction)
H
.
+ CH3
CH3 + CH3
2Cl requires 58 kcal/mole
two chlorine radicals
.
.
CH2Cl, CHCl2, CCl3) that disrupts the propagation of the chain will
Reaction mechanism involves three general steps:
Initiation, propagation, and termination
heat or
.
Any reaction of Cl or of the carbon free radical intermediates (CH3 ,
CH2Cl2
CH3—CH3
Cl—CH2—CH2—Cl
Reactivity of the Halogens
Chapter 7-3
Reactivity of the Halogens
F2
Bond Dissociation Energy
Cl2
Br2 I2
37 58
46
36 kcal/mole
.
.
.
Thus the energy of activation for formation of F and I is lower than Cl and Br
.
however, the order of reactivity of X. with alkanes is: F2 >> Cl2 > Br2 >> I2 (I2
does not normally react; F2 reacts explosively) Thus the rate determining step is
not:
2X.
X—X
Rather, hydrogen atom abstraction is the rate determining step.
.
X + CH4
.
H—X +
CH3
Since chlorination is faster than bromination, Eact for
.
Cl + CH4
H—Cl +
.
CH3
must be lower than the Eact for
.
.
Br + CH4
H—Br + CH3
.
[CH3 + HCl +Cl2]
.
[CH3 + HBr + Br2]
E
Eact
.
Cl + CH4
CH3Cl + Cl
Progress
.
Eact
.
Br + CH4
CH3Br + Br
.
Chem 61
Reactivity of the Halogens
Chapter 7-4
Stability of Carbon Free Radicals (Which H atom is abstracted?)
Stability of carbon free radicals follows the same pattern as carbocations:
.
.
.
.
.
CH2=CH—CH2 , C6H5CH2 > (CH3)3C > (CH3)2CH. > CH3CH2 > CH3
allyl
benzyl
tertiary
secondary
primary
methyl
Unlike carbocations, free radicals do not rearrange to a more stable free radical.
CH3
CH3
CH3
CH3
C
CH CH3
+
CH3 2°
C
+
CH3
.
C
CH CH3
CH3
CH CH3
C
CH CH3
OH CH3
CH3
3°
CH3
CH3
H 2O
CH3
Cl2
CH3
CH3
2°
C
CH CH3
CH3Cl
Since both carbocations and free radicals involve planar sp2 carbon atoms,
racemization will be observed in the products resulting from both intermediates.
H
CH3CH2 C
CH2Cl
CH3
R
Cl2
Br
CH3CH2
CH3
CH3OH
C
CH2Cl
S
Cl
CH3CH2
+ CH3
CH3CH2 C
CH2Cl
CH3
S
(50:50)
OCH3
CH3CH2 C
CH2Cl
CH3
S
C
Cl
R
CH3CH2
+ CH3
(50:50)
CH2Cl
C
CH2Cl
OCH3
R
Chem 61
Selectivity of Halogenation
Chapter 7-5
Selectivity of Hydrogen Atom Abstraction
Chem 61
Other examples of this are:
The more stable free radical should determine the nature of the product
CH3
Cl
.
CH2
Br.
CH3
CH3
.
a) 1°
.
CH3
.
c) allyl
b) 3°
.
CH3
CH2CH3 Cl2, light
CH3
d) allyl
e) 2°
Br
Br2, light
Br
and
as products.
Br
Chlorination is much less specific than bromination
Cl
CH3CH2CH3
CH3CH2CH3
Cl2
Br2
CH3
CH CH3
55%
Br
CH3
CH CH3
100%
+ CH3CH2CH2Cl
45%
.
.
The difference in selectivity of H atom abstraction by Cl and Br is explained by
the Hammond postulate. The transition state in the chlorination reaction is less
influenced by the stability of the intermediate free radical; thus both (CH3)2CH
.
.
and CH3CH2CH2 result.
In bromination the more stable free radical intermediate is highly favored; thus
.
(CH3)2CH is formed exclusively.
CH2CH2Cl
+
(56:44)
Thus intermediates c) and d) should be more stable and should lead to
CH3
CHCH3
CHCH3
100%
Halogenations
Chapter 7-6
Bromination with N-Bromosuccinimide (NBS)
O
O
CCl4
NH +
NBr +
light or
peroxide
O
O
Br
NBS acts as a Br2 source; the reaction is initiated by either light or a peroxide
(ROOR). NBS is used to introduce a Br atom at allylic or benzylic positions.
Other Sources of Free Radicals
Although light (hν) and heat (∆) are used for halogenation reactions other
reagents are often useful for initiating free radical reactions. Some of these
reagents include:
a) dibenzoyl peroxide
C6 H5
C
O
O
O
C
2 C6 H5
C 6H 5
C
O
.
O
O
b) hydrogen peroxide
H
O
O
H
2
c) alkyl hypochlorites
∆
R—O—Cl
H
.
O.
RO + Cl
.
d) azobisisobutyrylnitrile
CH3
CH3
C
N
C≡N.
CH3
CH3
N
C
CH3
2 CH3
C≡N.
Clues to Whether a Reaction Involves Free Radicals
a) Reaction requires high temperatures (>200°)
b) Reaction requires light energy (hν)
c) Reaction requires an initiator (a-d above) or oxygen.
.
C
C≡N.
+ N≡N
Chem 61
Nucleophilic Substitution Reactions
Chapter 8-1
Nucleophilic Substitution Reactions (SN1, SN2)
alkyl halides: CH3—X
methyl
RCH2—X
R2CH—X
R3C—X
primary, 1°
secondary, 2°
tertiary, 3°
X
aryl halide
vinyl halide
CH2=CH—X
do not easily undergo nucleophilic substitution reactions
Examples
CH3OH
CH3—I + Na+ – OCH3
CH3CH2CH2—Br + Na+
–
OH
CH3OH
CH3—OCH3 + Na+
+ Na
SH
SH
CH3OH
+ NaCl
Br
+
I
CH3CH2CH2—OH + Na+ – Br
Cl
+–
–
OCH3
CH3OH
+ HBr
CH3
CH3
Anatomy of a nucleophilic substitution reaction
SH
CH3
CH
I
CH3
+ Na+ – SH
CH3OH
CH3
CH
CH3
+ Na+ I –
Chem 61
Nucleophilic Substitution Reactions
Chapter 8-2
Anatomy of a nucleophilic substitution reaction
CH3
CH
CH3
+ Na+ – SH
CH3OH
I
CH3
CH
CH3
+ Na+
–
I
SH
(L) Leaving Group: any group that can be displaced from a carbon atom (in this
case I—)
(Nu) Nucleophile: the species that attacks the carbon atom bearing L and
donates the electron pair to form the nucleophile—C bond (in this case —SH)
(R—I) Substrate: the molecule, containing the leaving group that is acted on by
the nucleophile (here: 2-iodopropane)
Solvent: the medium used to dissolve the substrate and the nucleophile (in this
case methanol, CH3OH); solvent can sometimes be the nucleophile (solvolysis)
Curved arrows are used to indicate the movement of electrons during a
nucleophilic substitution reaction. By convention electron movement is written
from negative to positive.
δ+ δ −
Nu:– + R—CH2—L
electron pair for
new bond at C
R—CH2—Nu + L:–
electron pair that
accompanies L
Chem 61
Nucleophilic Substitution Reactions
Chapter 8-3
Reaction Mechanisms of Nucleophilic Substitution Reactions
Energetics of SN2
reaction mechanism: a detailed description of how a chemical reaction occurs. A
roadmap of a reaction....curved arrows show which bonds are formed or broken. A
mechanism includes the transition states involved in making and breaking bonds
and reactive intermediates that are formed along the pathway from reactants to
products.
potential
energy E
CASE 1:SN2 Substitution Nucleophilic 2nd order (bimolecular):
H
C
CH3O
CH3
δ−
CH3O
Br
H
C
H
∆H of reaction
δ−
Br
CH3O C
H
+ Br–
products
CH3
H
H
partial bonds in
transition state
Orbital Picture
H
H
H
H
C
Nu
L
H C—L sigma*
Nu nonbonding orbital
H
H
Nu
Eact
sp3
CH3
–
energy of transition state
reactants
sp2
sp3
Chem 61
+
C
Nu—C sigma
H
L
L nonbonding orbital
Nu
C
H
L
progress of reaction
Nucleophilic Substitution Reactions
Chapter 8-4
Orientation of Nu: and L in SN2:
Nucleophile approaches substrate from the backside of L
Point of highest energy along the reaction coordinate is the transition state: the
O—C bond is partially formed and the C—Br bond is partially broken
The Nu is bonded to the carbon on the opposite side of that occupied by the
leaving group L: Net inversion of the carbon atom is observed in all SN2
reactions
Reaction Rate: the time required for all substrate molecules to be converted to
product
The rate of an SN2 reaction is proportional to the concentrations of the substrate
and the nucleophile, thus the reaction is second order
Rate = k[substrate][nucleophile]
k is the proportionality constant called the rate constant
Rate and Eact
Under the same reaction conditions, the reaction with the lower Eact has a faster
rate
Reaction 2 below is faster
Eact
Eact
E
Rxn 1
Rxn 2
Increasing alkyl group substitution at the carbon atom bonded to the leaving group
hinders approach of the nucleophile. The E
act increases due to steric hindrance of
nucleophile approach and thus the SN2 rate decreases
Chem 61
Nucleophilic Substitution Reactions
Chapter 8-5
Relationship between Substrate Structure and SN2 Rate
Alkyl halide Relative rate of SN2
CH3—X 30
(CH3)2CH—X 0.025
CH3CH2—X 1
(CH3)3C—X ~0
CH3CH2CH2—X 0.4
Relative SN2 rate
CH3—Cl
+
I
–
CH3—I
+
–
Cl
93
CH3CH2—Cl + I–
CH3CH2—I + Cl–
1
(CH3)2CH—Cl + I–
(CH3)2CH—I + Cl–
0.0076
Increasing alkyl group substitution at the carbon atom bonded to the leaving group
hinders approach of the nucleophile. The E
act increases due to steric hindrance of
nucleophile approach and thus the SN2 rate decreases
SUMMARY
SN2 reactions occur by the attack of a nucleophile on substrates containing the
leaving group attached to a methyl, primary carbon, or secondary carbon.
SN2 reactions do not occur with tertiary alkyl halides
SN2 exhibit a single transition state (concerted reaction, i.e. not stepwise, no
intermediates)
Inversion of configuration at the carbon results from backside attack of the
nucleophile
The rate of SN2 reaction depends on the concentration of the nucleophile and the
substrate
The rate of SN2 follows the order:
CH3—X > CH3CH2—X > CH3CH2CH2—X > (CH
3)
2CH—X >> (CH
3)
3C—X
Chem 61
Nucleophilic Substitution Reactions
Chapter 8-6
Chem 61
CASE 2: SN1: Substitution Nucleophilic First Order (Unimolecular)
Substrates containing the leaving group attached to a tertiary carbon atom react
with weakly basic nucleophiles by an alternative nucleophilic substitution reaction
mechanism
(CH3)3C—Br + CH3OH
(CH3)3C—OCH3 + HBr
The process involves three steps:
Step 1: ionization (loss of halide ion)
(CH3)3C—Br + CH3OH
slow
(CH3)3C+
Br –
+
The transition state is pictured as (CH3)3Cδ+ ------Br
δ–
.
1
Step 2: attack by the nucleophile
..
CH3O—H
..
+
–
(CH3)3C
Br
fast
2
(CH3)3C
O+ CH3
H
H
O+ CH3
H
H
CH3
Step 3: deprotonation (loss of proton)
(CH3)3C
The transition state is pictured as (CH3)3Cδ+------δ+OCH3
(CH3)3C
O
CH3 + HBr
fast
3
Br–
Step 1 involves ionization of the alkyl halide to the halide ion and the tertiary
carbocation.
The step with the highest Eact is the slowest step (rate determining step) in the
pathway.
Ionization is aided by polar solvents (H2O, ROH) which solvate and thus stabilize
the carbocation and the leaving group anion.
Step 3 involves acid-base reaction between either Br– or more likely solvent
CH3OH in a very rapid reaction
The transition state is pictured as (CH3)3C O
+
H
O
CH3
Nucleophilic Substitution Reactions
Chapter 8-7
Stereochemistry of SN1 reaction
Relationship Between Substrate Structure and SN1 Rate
the intermediate carbocation contains an sp2 hybridized carbon atom which is planar Alkyl halide
+
CH3
C
Chem 61
CH3
CH3
Relative SN1 rate
CH3—Br
1.0*
CH3CH2—Br
1.0*
(CH3)2CH—Br
11.6
1.2 X 106
Thus attack of CH3OH on the carbocation can occur equally from the top and
bottom. If the carbon atom containing the leaving group were asymmetric,
racemization of the product would be observed
Br
CH3
H 2O
C
CH2CH2CH3
CH3CH2
R
OH
C
CH3
CH3CH2
R
+
CH2CH2CH3
CH3CH2
CH3
(50:50)
(CH3)3C—Br
*These observed reaction rates probably proceed through SN2 not SN1
SN1 rates reflect the relative Eact leading to the different carbocations
(transition state 1)
C
CH2CH2CH3
OH
S
Reaction Rate of SN1
The rate of the SN1 reaction does not depend on the concentration of the
nucleophile, but depends only on the concentration of the substrate
SN1 Rate = k[substrate]
The reaction is first order because the rate is proportional to the concentration of
only one reactant
Since the relative stability of carbocations is 3° > 2° > 1° > methyl;
tertiary alkyl halides proceed at the faster rate
Nucleophilic Substitution Reactions
Chapter 8-8
Chem 61
SUMMARY
Allyl and benzyl primary halides
Allyl and benzyl primary halides are reactive substrates in both SN1 and SN2
type reactions.
Substrate
Realtive SN1 Rate
1.0
30
CH3CH2—X
1.0
1
33
40
380
120
CH2=CH—CH2—X
C6H5—CH2—X
SN1 reactions occur with substrates that yield stabilized carbocations
(allyl, benzyl, 3°, 2°, not primary or methyl)
Relative SN2 Rate
CH3—X
SN1 reactions are first order in rate.
SN1 reactions lead to racemization.
Reactions of Carbocations
1. Reaction with a nucleophile: SN1 reactions
R+ + Nu:— or Nu:
2. Elimination of H+ on an adjacent carbon: E1 reactions
CH3
+
CH
CH2
CH3—CH=CH2 + H+
H
3. Rearrangement to a more stable carbocation: occurs whenever a more stable
carbocation can be formed
CH3
+
CH
H
C
CH3
CH3
CH3
SN1 reactions involve an intermediate carbocation which is attacked by
the nucleophile to yield the product of nucleophilic substitution.
+
CH2 C
CH3
CH3
Chapter 8-9
Elimination Reactions
Elimination Reactions (E1, E2)
Case 1: E1: Elimination First order
Formation of a carbocation intermediate in a reaction is a first order process.
Once formed the carbocation may react with a nucleophile (SN1) or
lose a proton from an adjacent carbon to form an alkene (E1)
Thus SN1 and E1 reactions are competitive
CH3
(CH3)3C—Br
CH3
–
C + Br
(CH3)3C
–
Nu +Br (SN1)
H C H
H
CH3
C
CH3
CH2
+ HBr
(E1)
E1 reactions predominate when the reaction contains only poor nucleophiles;
otherwise SN1 reactions are the more likely pathway
Since a carbocation is involved, rearangement may occur
TS1‡
TS2‡
E
Eact
R+
R–X
alkene
Chem 61
Elimination Reactions
Chapter 8-10
Chem 61
Thus
CASE 2: E2: Elimination Second Order
Bimolecular elimination results when an alkyl halide is treated with a strong
base (HO –, RO –, etc) at elevated temperatures (80°-120°).
CH3CH2CH
CH3CH2CH2 CHCH3
RO–
69%
+
80°, ROH
Br
The E2 reaction does not involve an intermediate as in the E1 reaction
E2 reactions are concerted as in the SN2 reaction
RO–
H
H
H
C
C
H
ROδ−
H
Br
ROH +
H
C
H
C
H
C
H
Brδ−
C
CH3CH2 CHCH3
H
H
Br
Brδ−
transition state for E2 Elimination
Reaction rate
E2 rate = k[base][substrate]
halides
tertiary halides > secondary halides > primary
Direction of Elimination
The more substituted (and more stable) product is normally the predominant E2
product
CH2=CH2
RCH=CH2
unsubstituted monoincreasing alkene stability
RCH=CHR
di-
If RO– is bulky (CH3)3CO– vs. CH3O–; a higher percentage of the less
substituted alkene results
CH3O –
H
H
H
ROδ−
H
H
R2C=CH2
di-
CH2
31%
+ Br–
C
H
becoming more sp2
CH3CH2CH2CH
H
C
H
CHCH3
R2CH=CHR
tri-
R2C=CR2
tetra-
CH3CH
CHCH3 +
80%
(CH3)3CO –
CH3CH CHCH3
50%
CH3CH2CH
CH2
20%
+ CH3CH2CH
50%
CH2
Elimination Reactions
Chapter 8-11
Chem 61
Stereochemistry of E2
SUMMARY
Generally, in E2 reactions the proton and the leaving group must be in an anti
orientation.
This allows for the best overlap of the developing pi orbitals.
E1 reactions result from carbocation intermediates and are competivive
with SN1 reactions.
E1 reactions may be accompanied by carbocation rearrangements.
E1 reactions generally yield the most substituted (most stable) alkene.
RO–
H
b
b
a
C
c
–
+ ROH + L
C
Relative rates: 3° > 2°.
d
a
d
E1 reactions are not stereoespecific and yield a mixture of E,Z alkenes.
c
E2 reactions result from a concerted reaction and are competitive with
SN2 reactions.
L
In order to predict the stereochemistry of the alkene resulting from an E2 reaction:
a) rotate the C atoms so that the H to be removed by RO– and the leaving group
are in the anti conformation
b) the stereochemistry of the alkene resulting from the E2 reaction is as shown
RO–
Br
H
H
C
rotate
C
Ph
CH3
Ph
C
Ph
CH3
H
C
Ph
H
Ph
Ph
C
CH3
Br
C
H
Z
In cyclohexane systems the H— and the leaving group must be trans and diaxial
Cl
RO–
D
H
H
D
H
E2 reactions generally produce the more stable alkene (except with bulky
bases or leaving groups).
E2 reactions are stereospecific and give anti elimination of H and L.
Relative rates: 3° > 2° > 1°.
Chapter 8-12
Substitution vs. Elimination Reactions
Factors Governing Substitution and Elimination Reactions
Structure of Alkyl Halide
Leaving group attached to methyl or primary carbon: SN2 reaction observed unless
Nu: is a strong base and elevated temperature employed; then observe E2.
Leaving group attached to tertiary carbon: E2 reaction observed unless Nu: is weak
base; then SN1 observed.
Solvent nucleophilicity increases with increasing electron releasing capacity of
molecule. If the solvent has low nucleophilic properties; i.e. H2SO4, H3PO4 then E1
reaction results
CH3CH2OH > H2O > CH3CO2H > CF3CH2OH > CF3CO2H
increasing solvent nucleophilicity
Leaving group attached to secondary carbon: if the nucleophile is less basic than
HO – (e.g. – CN, – N3, – SR) then the reaction will follow SN2 path
If the Nu: is HO – or more basic (e.g. RO–, R2N–, RC≡C–), then the reaction will follow
E2 path
If the Nu: is a weak nucleophile (e.g. H2O, ROH, RCO2H) the reaction will follow SN1
path with some E1
The leaving group is attached to an allylic or benzylic carbon: SN1
Solvent
Highly polar solvents (high dielectric constant) favor SN1, E1 reactions
Concentration of Nucleophile or Base
Increasing the concentration of the nucleophile has no effect on SN1, E1 but
increases SN2,E2 rates.
Chem 61
Temperature
Increase in temperature increases the rates of all reaction types
but has a larger effect on E2 reactions.
Nucleophilicity
Chapter 8-13
NUCLEOPHILICITY
General comments
Nucleophilic Constants of Various Nucleophiles
Chem 61
Nucleophile
n(CH3I)a
pKa of conjugate acid
1. Note that nucleophilicity toward CH3I does not correlate directly with
basicity. (N3–, PhO–, Br- are equivalent nucleophilicity but differ greatly in
basicity. Also N3– and CH3CO2– are nearly identical in basicity but N3– is 30
times [1.5 log units] more nucleophilic).
CH3OH
NO3–
F–
CH3CO2–
Cl–≠
(CH2)2S
NH3
N3–
PhO –
Br –
CH3O–
HO–
NH2OH
NH2NH2
(CH3CH2)3N
NC–
I–
HOO–
(CH3CH2)3P
PhS–
PhSe–
Ph3Sn–
0.0
1.5
2.7
4.3
4.4
5.3
5.5
5.8
5.8
5.8
6.3
6.5
6.6
6.6
6.7
6.7
7.4
7.8
8.7
9.9
10.7
11.5
-1.7 (CH3OH2)
-1.3
3.45
4.8
5.7
2. Among neutral nucleophiles while (CH3CH2)3N is more basic than
(CH3CH2)3P (pKa 10.7 vs. 8.69) the phosphine is 100 times (n = 8.7 vs 6.7)
more nucleophilic.
9.25
4.74
9.9
-7.7
15.7
15.7
5.8
7.9
10.7
9.3
-10.7
8.69
8.5
an(CH3I) = log(knucleophile/kCH3OH) in CH3OH 25°C
n = nucleophilic constant
3. Correlation of nucleophilicity with basicity is better if the attacking atom is
the same. Thus CH3O– > PhO– > CH3CO2–> NO3–
4. Nucleophilicity usually decreases going across a row in the periodic table.
(HO– > F– ; PhS–> Cl–). This order is determined by electronegativity.
5. Nucleophilicity usually increases going down the periodic table (I– > Br– >
Cl– > F–; and PhSe– > PhS– > PhO–) Related to weaker solvation and greater
polarizability of the heavier atoms.
Nucleophilicity
Chapter 8-14
Examples
E1: 3° > 2° > 1°
E2: 3° > 2° > 1°
SN1: 3° > 2° > 1°
SN2: 1° > 2° > 3°
Nu:–
1°
R
CH CH2
L
R
H
H
hindered
R
strong base
heat
3°
CH CH2
Nu
SN2 (inversion
CH CH2
E2 (anti elimination of HL)
NuH
C
C
H
L
C
C
C
+
C
H
–
Nu: acts
as base
C
C
–H
C
C
E2
C
C
C
H
Nu+ H
+
– H+
C
E1
C
C
H
Nu
S N1
H
S N2
2°
H
C
C
H
L
–
Nu: or base:
E2
H
H
C
+
C
C
H
Nu
C
H
poor Nu
or base
C
C
–
C
SN1 or E1
C
C
C
C
C
Chem 61
Alcohols
Chapter 9-1
Alcohols are similar to water: contain sp3 hybridized oxygen, —OH
functional group bonded to an sp3 carbon, with two lone pairs on oxygen
H
H
..
O
H
CH3CH2
..
..
water
O
..
ethanol
Alcohols can hydrogen bond thus their boiling points are higher than
similar polar compounds which cannot hydrogen bond.
Low molecular weight alcohols are soluble in water
Alcohols contain a hydrophobic alkyl group and a hydrophilic hydroxyl
group
As the alkyl group increases in length and size, the solubility in water
decreases
Acidity and Basicity
alcohols act as bases in the presence of strong acids
CH3CH2
base
..O. .
H
CH3CH2
H+
..O+
H
H
acid
and as acids in the presence of strong bases
(CH3)3C
..O. .
acid
H
+ K+H–
..
.– + + H2
(CH3)3C O
. .. K
base
CH3CH2O—H + Na0
CH3CH2O– Na+ + H2
Chem 61
Alcohols
Chapter 9-2
IUPAC Names
For alcohols drop —e from the alkane name and add —ol
CH3CH2OH: ethan + ol = ethanol
CH3CH2CHOHCH3: butan + ol; —OH on carbon 2: 2-butanol
Classification
Alcohols are classified as:
methyl, primary, secondary, tertiary, allylic, benzylic
CH3OH
CH3CH2CH2CH2CH2CH2OH
methyl
primary 1°
CH3CHOHCH3
secondary 2°
CH2=CHCH2OH allylic
CH2OH
(CH3)3COH
tertiary 3°
benzylic
REACTIONS OF ALCOHOLS
Substitution Reactions: Reaction with Hydrogen Halides (HX)
Alcohols do not undergo nucleophilic substitution by X– since HO– is a
poor leaving group.
Alcohols do undergo substitution by X – in acidic solution.
Here the alcohol is protonated and H2O, a neutral species, is the leaving
group.
R
..
OH
..
H+
R
.. +
O H
H
X–
R—X + H2O
Chem 61
Alcohols
Chapter 9-3
Chem 61
Reactivity of Hydrogen Halides
The reactivity of hydrogen halides in alcohol substitution reactions is as
follows:
HF < HCl < HBr < HI
pKa 3.45 -7 -9 -9.5
E
The reactivity is explained since the acidity of HX increases and the
nucleophilicity of X– increases going from HF to HCl to HBr to HI.
+
CH3CH2CH2
CH3CH2CH2
Reactivity of Alcohols Toward HX
OH
O H
H
progress of reaction
methyl primary secondary tertiary allylic and benzylic
increasing reactivity of ROH toward HX
All alcohols react readily with HBr and HI;
3°, allylic and benzylic react rapidly with HCl
2° and 1° alcohols require the addition of ZnCl2 for rapid reaction with HCl
Mechanism of Alcohol Substitutions
Methyl and primary alcohols follow the SN2 mechanism
Step 1 protonation of the alcohol
CH3CH2CH2
OH
H+
CH3CH2CH2
+
O H
H
Step 2 SN2 dsiplacement of Water
CH3CH2CH2
O+H
H
Br –
S N2
CH3CH2CH2
Br + H2O
CH3CH2CH2
Br
Alcohols
Chapter 9-4
Secondary and tertiary alcohols follow the SN1 mechanism
Step 1 protonation of the alcohol
CH3
CH3
C
CH3
H+
CH2CH3
CH3
C
+
OH
CH2CH3
OH2
Step 2 loss of water to for the carbocation
CH3
CH3
C
+
CH2CH3
CH3
– H 2O
CH3
OH2
C
+
CH2CH3
Step 3 attack of halide ion on the carbocation
CH3
CH3
C
+
CH3
X–
CH2CH3
CH3
S N1
C
CH2CH3
I
Rearrangement can occur
R+
E
R–O+H
R–OH
H
R–X
Chem 61
Conversion of Alcohols to Alkyl Halides
Chapter 9-5
Other Reagents for the Conversion of Alcohols to Alkyl Halides
R—OH + SOCl2
R—Cl + HCl + SO2
R—OH + PBr3
R—Br + HOPBr2
H
H
O
CH3CH2 C OH
Cl
CH3
CH3CH2 C O+ S Cl
CH3
Cl
H
S
Cl
Cl –
H
+
CH3CH2 C OSOCl
CH3
H
R3N:
O–
H
O
CH3CH2 C O S
Cl
CH3
H
Cl
C
CH3CH2
CH3
+ SO2
Esters of Alcohols
Reaction of alcohols with carboxylic acids in the presence of acid produces
esters of carboxylic acids
O
CH3CH2CH2
C OH + CH3CH2CH2OH
alcohol
carboxylic acid
H+
heat
O
CH3CH2CH2
ester
C OCH2CH2CH3
Chem 61
Reactions of Alcohols
Chapter 9-6
Inorganic esters
mineral acids such as H2SO4, HNO3 and H3PO4 form esters with alcohols
to produce important compounds
H
O
H
O
H
C
ONO2
H
C
O
P
H
C
ONO2
H
C
H
OH
H
C
ONO2
H
C
H
H
nitroglycerin, a nitrate ester
O
P
O
OH
CH3O
OH
S
OCH3
O
dimethyl sulfate,
a sulfate ester
H
a diphosphate ester
P-toluenesulfonates (tosylates) and methanesulfonates (mesylates) are
excellent leaving groups in nucleophilic substitution reactions. They are
readily prepared form an alcohol and the corresponding sulfonyl chloride.
CH3OH + ClSO2CH3
O
(CH3CH2)3N
CH3O
S
CH3 + (CH3CH2)3N+HCl –
O
a methanesulfonate (mesylate)
O
OH
+
O
CH3
S Cl
O
S
O
O
p-toluenesulfonylchloride
(tosyl chloride)
cyclohexyl tosylate
Reaction of sulfonates with nucleophiles
OTs
CH3CH2CH2O– Na+
OCH2CH2CH3
CH3
Chem 61
Dehydration of Alcohols
Chapter 9-7
Chem 61
Examples:
Elimination (Dehydration) of Alcohols
CH2 CHCH2CH3
Alcohols undergo elimination much like alkyl halides
CHCH2CH3
CH
H+
OH
Tertiary alcohols readily undergo dehydration by an E1 pathway
Secondary alcohols also follow an E1 path, but primary alcohols probably
eliminate by an E2 mechanism
OH
conc. H2SO4
(CH3)3COH
ease of
dehydration
(CH3)2CHOH
CH3CH2CH2OH
60° C
conc. H2SO4
100° C
conc. H2SO4
CH3CH=CH2 +H2O
CH3
CH3
C
CH3
CH3
C
CH3
OH
CH3
C
C+
H
CH3
C
C
CH3
CH3
H
CH3
H+
– H+
O+H2
CH3
CH3
C
C
CH3
H
CH3
CH3
C
C
CH3
H
C
C
CH3
CH3
R+
ROH2+
ROH
progress of reaction
alkene
CH3
CH3
CH3
C
C
CH3
H
CH3
+
C
1,2 shift
CH3
CH3
CH3
The most stable alkene predominates in dehydration reactions.
Rearrangements can occur.
E
OH
+
CH3
CH3
H+
CH3
– H 2O
CH3
C
CH3
CH3
CH3CH=CH2 +H2O
Mechanism:
CH3
heat
CH3
180° C
H
CH3
H2SO4
CH CH3
(CH3)2C=CH2 +H2O
C
CH3
CH3
O+H2
– H 2O
CH3
C
H
CH3
– H+
Dehydration of Alcohols
Chapter 9-8
ETHERS, EPOXIDES AND SULFIDES
Ethers are derivatives of water where both hydrogens have been replaced
by an alkyl group.
Ethers are less polar than alcohols and water and are not capable of
hydrogen bonding to themselves because of the lack of an —OH group.
The boiling points of ethers are also much lower than alcohols of
comparable molecular weight.
O
CH3CH2OCH2CH3
O
diethyl ether
CH2
tetrahydrofuran
CH2
ethylene oxide
Preparation of Ethers
Williamson Ether Synthesis
(SN2 Reaction of an Alkoxide with an Alkyl Halide)
R1O– + R2—X
S N2
R1—O—R2 + X–
Best results are obtained if the alkyl halide is methyl or primary
(2° and 3° give mostly elimination)
There are few limitations on the alkoxide
CH3CH2CH2O – + (CH3)2CHCH2—Br
O–
CH3CH2CH2O—CH2CH(CH3)2
OCH2CH3
+ CH3CH2—I
(CH3)3CO – + CH3—I
(CH3)3CO—CH3
Chem 61
Chapter 9-9
Reactions of Ethers and Epoxides
Substitution Reactions of Ethers
Ethers are relatively unreactive compounds, but they do undergo substitution
reaction when heated with hydrogen halides, particularly HI and HBr.
This is a very similar reaction to the reaction of alcohols with HX
Br –
..
..
H—Br
CH3CH2CH2—O—CH2CH2CH3
+
CH3CH2CH2—O—CH2CH2CH3
H
HBr
CH3CH2CH2—Br
CH3CH2CH2—Br + CH3CH2CH2—OH
OCH2CH(CH3)2
OH
HI
+ I—CH2CH(CH3)2
unreactive with HI
Reactions of Epoxides
Epoxides are strained much like cyclopropanes because their bond angles
(60°) are far removed from the normal tetrahedral angle (109.5°).
The orbitals have poor overlap and the bonds are weakened. The C—O
bond is also polarized and consequently, epoxides are highly reactive
compounds.
Base Catalyzed Cleavage of Epoxides
Alkoxides:
O
CH3
C
CH3
H
CH3O –
H
CH3OH
–
O
CH3 C
C
H 3C
H
C H
OCH3
HO
HOCH3
CH3
C
H 3C
CH3O –
nucleophile attacks at the least hindered carbon in an SN2 reaction
H
C H
OCH3
Chem 61
Chapter 9-10
Reactions of Epoxides
Grignard Reagents
O
H
C
H
BrMgO
CH3MgBr
H
C
H
H
H
HO
H+
C
C H
H
CH3
H
C
H
CH3–MgBr+
Acid Catalyzed Cleavage
+
CH3 O H H
C C
H CH3OH
CH3
CH3OH
CH3 C
C H
CH3O + H
– H+
H
+
O
CH3
δ+ C
CH3
H
C
H
..
CH3O—H
..
CH3OH
CH3 C
CH3O
C H
H
H
Nucleophile attacks at the most hindered carbon since the
carbon is partially positive and the C—O bond is partially
broken
H
+
O
OH
H
H
H
H
H 2O
OH trans diaxial opening
in cyclohexanes
H
C H
CH3
Chem 61
Chapter 9-11
Crown Ethers and Thiols
Crown Ethers
Crown ethers are macrocyclic ethers with repeating –OCH2CH2— units.
Depending on the ring size, they effectively chelate alkali metal ions such as
K+, Na+ or Li+
O
O
O
O
O
Na+
O
K+
O
O
O
O
O
12-crown-5
18-crown-6
Thiols and Sulfides
The sulfur analog of an alcohol is a thiol, or mercaptan. Thiols are stronger
acids than alcohols (pKa = 8, ROH = 16).
Thiols form weaker hydrogen bonds than alcohols due to the lower
electronegativity of sulfur.
Thiols and thioethers (sulfides) are prepared by substitution reaction in the
same way that alcohols and ethers are prepared.
Br
SH
HS–
CH3CH2CH2S–
SCH2CH2CH3
I
Disulfides are formed by the oxidation of thiols and are an important structural
feature in some proteins.
RSH
I2 or
K3Fe(CN)6
RS—SR
a disulfide
Chem 61
Chapter 9-12
Organometallic Compounds
Grignard Reactions
Grignard reagents are organometallic reagents derived from an alkyl halide
and magnesium
diethyl ether
R—X + Mg
Rδ−—Mgδ+X Grignard reagent
Since the carbon carries a partial negative charge, the carbon is a strong
base and a good nucleophile.
CH3CH2—Br + Mg
diethyl ether
CH3CH2δ−—Mgδ+Br
HOH
CH3CH2—H
Because carbonyl pi bonds are polarized, they can undergo a reaction
called nucleophilic addition: the addition of a nucleophile to an electron
deficient pi bond.
R1
O–
Oδ−
O–
C+
Cδ+
C
R1
R1
R1
R1
Nu
Nucleophilic
Addition
R1
Nu:
O–
Oδ−
O– MgX+
C+
Cδ+
C
R1
R1
R1
R2
R1
R1
R2
Mg-X
A Grignard reaction with
1. formaldehyde produces a primary alcohol
O
1. CH3
C
H
H
OH
Mg-X
2. H2O, H+
H
C
CH3
H
R1
Nucleophilic
Addition
Chem 61
Chapter 9-13
Organometallic Compounds
2. an aldehyde produces a secondary alcohol
O
OH
1. (CH3)CHMgBr
C
CH3CH2
H
2. H2O, H+
H
C
CH2CH3
CH(CH3)2
3. a ketone produces a tertiary alcohol
O
C
CH3CH2
1. (CH3)CHMgBr
+
CH3 2. H2O, H
OH
CH3
C
CH2CH3
CH(CH3)2
4. an ester produces a tertiary alcohol (addition of two molecules of
Grignard reagent)
O
OH
1. 2 CH3MgBr
C
OCH3
2. H2O, H+
CH3
C
CH3
5. ethylene oxide produces a primary alcohol
O
1. C6H5MgBr
2. H2O, H+
OH
Chem 61
Alkynes: Structure and Bonding
Chapter 10-1
sp Hybridization
sp hybridized carbons are linear with atoms 180° apart
2p
2p
2p
96 kcal
2s
1s
sp
hybridize
2s
1s
1s
ethylene: trigonal planar
–
+
180° apart; the remaining two 2p
orbitals are 90° to the sp and
each other
+
1-2p
2s
2-sp's
Acetylene: linear; bond angles 180° C=C bond length = 1.20Å
acetylene has two perpendicular pi bonds and one sigma bond
H
H
H
2-sp's
H
C
H
acetylene σ-orbital
H
2-2p's on each carbon
combine to form pi-orbitals
E
C
C—C σ*
C—C π*
C—C π
C—C σ
H
C
C
H
acetylene π-orbitals
Chem 61
Alkynes: Electrophilic Addition
Chapter 10-2
Reaction of Alkynes with E+A– Reagents
Reaction of alkynes with E+A– reagents proceed in the same manner as
alkenes except different intermediates are possible
R
C
C
R
E+
+
C
R
C
E
vinyl carbocation
R
E+
R
C
bridged intermediate
C
R
For H—X the vinyl carbocation is more stable than the bridged intermediate.
Thus:
R
C
C
R
E+
+
C
R
C
E
vinyl carbocation
R
E+
R
C
bridged intermediate
C
R
secondary vinyl carbocation is about the same stability as a primary carbocation
Thus the order of carbocation stability is
triphenyl methyl > diphenylmethyl > 3°≈ benzyl ≈ allyl > 2° > 1° ≈ 2° vinyl >> methyl
Chem 61
Alkynes: Electrophilic Addition
Chapter 10-3
For H–X
CH3
C
C
H
CH3
HCl
C
H
C
H
Cl
CH3
C
CH3 CH3
C
C
+
H
Cl
CH3
HCl
C CH3
C
Cl
H
C
CH3
For Hg(OCOCH3)2:
CH3CH2
Hg(OCOCH3)2
C2H5C≡CC2H5
HgOCOCH3
C
C
CH2CH3
CH3CO2
E
For Cl2, Br2:
X+
X2
R—C≡C—R
C
C6H5—C≡C—C6H5
X2
+
C
C6 H5
X
C
C
R
but,
R
R
X
R
X
X
C
C
C 6H 5
C
X
C6 H5
C
C 6H 5
+
X
C 6H 5
C
C6 H5
C
X
Thus the choice of intermediate depends on structure; alkyl groups tend to favor
the bridged ion; groups such as phenyl which stabilize the free carbocation tend
to proceed via the vinyl carbocation.
Chem 61
Alkynes: Electrophilic Addition
Chapter 10-4
Addition of Water
With alkynes this electrophilic addition reaction generates a vinyl alcohol
(also called an enol). Hg(II) ion is often used.
R—C
H2SO4
C—R
Hg(II)
H 2O
OH
O
R—C CH—R
enol
H
+OH
R—C C—R
H2O:
R—C
R—C CH2—R
keto
C—R
H2O:
Hg2+
Hg2+
H+
R—C
OH
C—R
H O
R—C C—R
Hg2+
Hg2+
H O
R—C C—R
OH
C—R
R—C
H
H
Hydroboration
H
R—C
C—H
BH3
R—C
BH2
CH
H
OH
R—C
CH
enol
H O
R—C
H
C–H
keto (aldehyde)
Chem 61
Alkynes: Acidity
Chapter 10-5
H
C
+ base
C
C
–
..
C
H
pKa = 45
Chem 61
HF
+ base—H
H
pKa
3.2
H 2O
15.7
HC≡CH
25
NH3
36
H2C=CH2
H3C–CH3
44
50
increasing acid strength
C
C
+ base
H
C
.
C.
–
+ base—H
F–
pKa = 25
The relatively high acidity of the alkyne —C≡C—H bond is associated with
the large degree of s character in the sp C—H bond (50% compared with
33% in sp2 bonds). The carbon atom is more electronegative in the sp
state; thus the C—H bond is more acidic.
The acetylide ion may be formed by such strong bases as —:NH2 (pKa
33), RMgX or RLi (pKa 45-50).
R
C
C
H
NaNH2
R
NH3
R
CH
CH2
C
C:– Na+ + NH3
acetylide ion
NaNH2
No reaction
NH3
Electronegativities
Electronegativities
acid strength
pKa
HC≡CH
H2C=CH2
sp
2
>
sp
>
N < O < F
NH3 < H2O < HF
36
15.7
3.2
H3C–CH3
sp3
HO–
HC≡C–
NH2–
H2C=CH–
increasing base strength
H3C–CH2–
Alkynes: Acetylides
Chapter 10-6
SN2 reaction with acetylide ion
NH3
–
+
R'—C≡C: Na
R'
C
+
R—CH2—C≡C—R'
R—CH2—L
O
C:– MgBr+ +
R'
C—CH2—CH2—O – MgBr+
C
H+
R'
C
C—CH2—CH2—OH
Nucleophilic addition reaction with acetylide ion.
O – MgBr+
O
R'
C
C:– MgBr+
R
C
H
R'
C
R
C
C
H
CH3CH2MgBr
R'
C
H+
OH
C—H
R'
C
C
C
H
O
R'
C
–
C: MgBr
+
+
O – MgBr+
C C R'
H+
OH
C
C R'
R
Chem 61
Spectroscopy
Chapter 12-1
Infrared and Nuclear Magnetic Resonance Spectroscopy
electromagnetic radiation: energy that is transmitted through space in
the form of waves
wavelength: (λ): the distance from the crest of one wave to the crest of
the next wave
frequency: (ν): the number of complete cycles per second
ν=
c
λ
where c = speed of light
Electromagnetic radiation is transmitted in particle-like packets called
photons or quanta. The energy is inversely proportional to the
wavelength and directly proportional to frequency.
Ε=
hc
λ
where c = speed of light; h = Planck's constant
Ε = hν
ultraviolet
visible
h = Planck's constant
infrared
radio
decreasing energy
Absorbtion of ultraviolet light results in the promotion of an electron to a
higher energy orbital.
Absorbtion of infrared results in increased amplitudes of vibration of
bonded atoms.
Intensity of radiation is proportional to the number of photons.
Chem 61
Infrared Spectroscopy
Chapter 12-2
Infrared Spectroscopy
Infrared is recorded as %T versus wavelength or frequency
When a sample absorbs at a particular wavelength or frequency,
%T is reduced and a peak or band is displayed in the spectrum.
Infrared is recorded as %T versus wavelength or frequency
When a sample absorbs at a particular wavelength or frequency, %T
is reduced and a peak or band is displayed in the spectrum.
100
%T
0
frequency
Nuclei of bonded atoms undergo vibrations similar to two balls
connected by a spring. Depending on the particular atoms bonded to
each other (and their masses) the frequency of this vibration will vary.
Infrared energy is absorbed by molecules resulting in an excited
vibrational state. Vibrations occur in quantized energy levels and thus a
particular type of bond will absorb only at certain frequencies.
Both stretching and bending vibrations can be observed by infrared.
O
CH3
CH3
stretching
O
CH3
CH3
bending
Chem 61
Infrared Spectroscopy
Chapter 12-3
Interpretation of Infrared Spectra
Chem 61
C—C and C—H Bonds
Correlation tables
Infrared spectra of thousands of compounds have been tabulated and
general trends are known. Some common functional groups are
shown below.
C=O str
OH and NH str
CH str
C—O str
C=N str
C≡N str
C=C str
NH bend
C—N str
C—C str
sp3 C—C
sp2 C=C
sp2 C—C (aryl)
sp C≡C
sp3 C—H
sp2 C—H
sp C—H
C(CH3)2
weak, not useful
1600-1700 cm–1
1450-1600 cm–1
2100-2250 cm–1
2800-3000 cm–1
3000-3300 cm–1
3300 cm–1
1360-1385 cm–1 (two peaks)
CH bend
OH bend
3500
3000
2500
2000
1500
1000 800
Alcohols and Amines
Carbonyls
One of the most useful absorbtions in infrared 1640-1820 cm-1
O—H or N—H
C—O or C—N
Ketones (saturated) C=O
Ethers
1640-1820 cm–1
C—O
Aldehydes C=O;
C—H(O)
1640-1820 cm
Carboxylic acids C=O;
C(O)—OH
1640-1820 cm–1
Esters C=O ;
C(O)—OR
1640-1820 cm–1
–1
2820-2900 and 2700-2780 cm–1 (weak but characteristic)
3330-2900 cm–1
1100-1300 cm–1
3000-3700 cm–1
900-1300 cm–1
1050-1260 strong
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-1
Nuclear Magnetic Resonance (NMR) Spectroscopy
Some atomic nuclei (1H,
have a nuclear spin.
13
C, others) behave as if they are spinning...they
Spinning of a charged particle creates a magnetic moment.
If an external magnetic field is applied, these small magnetic moments (of the
nuclei) either align with the field (α) or against the field (β), about 50% with
and 50% against the field at any one time.
β
Ho
∆E
hν
α
β
Ho
∆E
α
Ho = the external magnetic field
Resonance: the flip of the magnetic moment from parallel to antiparallel to
the external magnetic field.
Irradiation at the frequency equal to the energy difference, ∆E, causes
resonance.
∆E depends on the external magnetic field.
Protons (or other nuclei) in different magnetic environments resonate at
different field strengths.
A proton which resonates at a higher field is in a stronger magnetic
environment or shielded.
A proton which resonates at a lower magnetic field is said to be deshielded.
Different magnetic environments are created by different electron densities in
the vicinity of a proton.
Chem 61
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-2
Chem 61
Adjacent electron withdrawing groups, highly electronegative atoms, or the
hybridization of the carbon to which the proton is bonded can alter the
magnetic environment.
The pi system of benzene creates a magnetic field or ring current which
deshields the protons attached to the ring.
The local electrons create a small electric and magnetic field around a proton
and shield it.
Similarly, pi electrons in a C=O bond create a field which deshields the
proton bonded to the C=O of an aldehyde. This is also affected by the
inductive effect of the C=O.
The more electron density present around the proton, the greater the field and
the greater the shielding.
Resonances are reported in chemical shifts (δ) downfield from
tetramethylsilane (TMS) (CH3)4Si.
δ=
distance from TMS in Hz
MHz of spectrum
ppm
In methyl halides, the more electronegative the halogen, the more deshielded
the protons on the methyl. This is because F is inductively more electron
withdrawing, causing the carbon to be more positive and thus pulling more
electrons away from the hydrogen and causing it to be less shielded.
In methyl halides, the more electronegative the halogen, the more deshielded the pr
H3C—F
δ
H3C—Cl
3.0
4.3
H3C—Br
H3C—I
2.7
2.1
Pi electron effects
Magnetic fields created by pi electrons are directional and said to have an
anisotropic effect.
R
Ho
C
H
O
H
H deshielded
H deshielded
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-3
Equivalent and Nonequivalent Protons
Protons that are in the same magnetic environment are equivalent and have
the same chemical shifts.
Protons in different magnetic fields are nonequivalent and have different
chemical shifts.
Magnetic equivalence is usually the same as chemical equivalence.
Equivalence can be established by symmetry operations such as rotation,
mirror planes and centers of symmetry
Chemically equivalent protons have the same chemical shifts.
To determine if protons are chemically equivalent, replace one by a different
group, e.g. D or Br.
Then replace a different one by the same group and compare the two
compounds. If they are identical, the protons are equivalent.
H
H
H
C
C
H
H
OH
H
H
H
H
C
C
C
H
Cl H
H
equivalent, but not to CH3 protons
all six are equivalent
equivalent
Equivalent protons can be on different carbons.
Protons which are homotopic or enantiotopic resonate at the same chemical
shift in the NMR.
If protons are interconverted by rotation about a single bond, they will average
out on the NMR time scale and a single resonance will be observed.
ClH2CCH2Cl anti and gauche forms rapidly interconvert and a single
resonance is observed.
Axial and equatorial hydrogens in cyclohexane average to a single peak
because of rapid ring inversion.
Diastereotopic hydrogens are chemically nonequivalent and thus give different
chemical shifts in the NMR
Chem 61
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-4
Intergration
The spectrometer can integrate and determine the relative number of
hydrogens associated with each resonance in the NMR spectrum by
determining the area under the peaks.
Spin-Spin Coupling
for example...
3
3
CH3CH2OCH3
2
TMS
If a proton (Ha) is bonded to a carbon which is bonded to a carbon that has
one proton (Hb), Ha will appear as a doublet
Since in half the molecules, Hb will be in the α state and in half will be in the β
state, Ha will experience two different magnetic fields and two peaks (a
doublet) will appear for Ha.
Ha without an adjacent hydrogen
For one adjacent hydrogen
α or β
Ha with Hb adjacent in theβ state
Ha with Hb adjacent in the α state
Chem 61
Chapter 13-5
Nuclear Magnetic Resonance Spectroscopy
For two adjacent hydrogens: Hb, Hc
At any one time Hb or Hc could be in the α or β state (50:50) thus 4
combinations for Hb, Hc exist:
αbαc αbβc
βbαc
βbβc gives 1:2:1 triplet
When both Hb and Hc are α, a different field is observed than if both are β or
one is α and one is β.
When one is α and one is β, the field is the same. That is, βbαc and αbβc
produce the same field and a single signal for Ha is observed with twice the
intensity.
Thus three signals are observed in a 1:2:1 ratio: a so-called triplet
For three adjacent protons:
ααα ααβ αββ βββ 1:3:3:1 quartet
αβα βαβ
βαα ββα
Thus the splitting pattern of a particular proton or equivalent protons will be a
pattern with n+1 lines where n is the number of adjacent equivalent protons.
singlet 0 neighboring protons
doublet 1 neighboring protons
triplet 2 neighboring protons
quartet 3 neighboring protons
quintet 4 neighboring protons
sextet 5 neighboring protons
septet 6 neighboring protons
The separation of the peaks in a splitting pattern is called the coupling
constant, J.
Chem 61
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-6
Splitting Diagrams
Splitting patterns for protons can be constructed in diagram form by starting
with one line to represent the unsplit proton resonance.
If an adjacent proton Hb affects Ha it is split into a doublet; if another
equivalent proton to Hb is present, each line of the double will be split into a
doublet, since the coupling constant J is the same, the two center lines
overlap and a only three lines are observed with the center line twice the
height.
This can be repeated for additional adjacent protons.
Ha without an adjacent hydrogen
1
1
1
splitting diagram
Ha split by one adjacent hydrogen
1
2
Ha split by a second adjacent hydrogen
Ha split by a third adjacent hydrogen
1
3
3
1
Chemical Exchange and Hydrogen Bonding
CH3OH, methanol would be expected to give an NMR spectrum of a
doublet for the CH3 and a quartet for the OH. For a dilute sample at -40° in
CCl4 this is the case.
If the NMR spectrum is run at 25° as a more concentrated sample only two
singlets are observed. This is because the intermolecular hydrogen
bonding in methanol allows the rapid exchange of the OH proton from one
CH3OH molecule to another, effectively averaging the spin states of the OH
proton and resulting in no change in the magnetic field due to the OH.
Amines and other compounds which can undergo hydrogen bonding can
also show this effect. Thus the NMR spectra of alcohols, amines and
carboxylic acids are temperature, concentration and solvent dependent.
Chem 61
Nuclear Magnetic Resonance Spectroscopy
Chapter 13-7
CHEMICAL SHIFTS
Functional Group Shift,δ
Primary alkyl, RCH3
Secondary alkyl, RCH2R
Tertiary alkyl, R3CH
0.8-1.0
1.2-1.4
1.4-1.7
Allylic, R2C=C—CH2R
1.6-1.9
Benzylic, ArCH2R
Iodoalkane, RCH2I
Bromoalkane, RCH2Br
Chloroalkane, RCH2Cl
Ether, RCH2OR
Alcohol, RCH2OH
Ketone, RCH2C(=O)R
2.2-2.5
3.1-3.3
3.4-3.6
3.6-3.8
3.3-3.9
3.3-4.0
2.1-2.6
Aldehyde, RCH(O)
Terminal alkene, R2C=CH2
Internal alkene, R2C=CHR
Aromatic, Ar—H
Alkyne, RC≡C—H
Alcoholic hydroxy, ROH
Amine, RNH2
9.5-9.6
4.6-5.0
5.2-5.7
6.0-9.5
1.7-3.1
0.5-5.0 (variable)
0.5-5.0 (variable)
Chem 61
CHEMISTRY 61
Exam 1
Dr. M.T. Crimmins
September 15, 1998
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
I.
Nomeclature (12 points) Give the IPUAC name for the following compounds: Indicate R, S, cis, trans,
E, or Z where appropriate.
1.
CH 3
CH 2CH 2CH 2CH 3
CH 2 C H C H CH 2 CH 2 C H CH 3
CH 3
CH 3
______________________________
2.
H 3C
H
C
C
CH 3
CH 2CH 3
______________________________
3.
CH 3
H
______________________________
4.
CH(CH3)2
______________________________
II.
A. Write valid Lewis structures for the following species. Show all nonbonding (unshared) electrons
and indicate any formal charges . (6 points).
5. [H2COH]+
6. IO4 –
7. Give the hybridization of the indicated atoms in the species below (6 points)
H 3C C N:
H 3C N
.. CH 2
1
O
H 3C C
O CH 3
8. Draw three structural isomers for C3H6O. Indicate what type of functional group is represented by
each compound (e.g. carboxylic acid) (6 points).
9. Draw all the possible stereoisomers of 3-bromo-2-butanol. Indicate if they are chiral, meso or achiral and
indicate their relationship to each other. (i.e. enantiomers, diastereomers) (8 points)
10. Draw both chair conformations of trans-1,2-dimethylcyclohexane. If one is more stable than the
other, circle it. (6 points)
11. Draw an energy diagram for one 360° rotation about the C3-C4 C–C bond of hexane. Also draw a
Newman Projection of the most stable conformation. (8 points).
12. Circle the molecule(s) which have a permanent dipole. In those which have a permanent dipole, show
the direction of the overall dipole. (6 points)
O
H
C
H
Cl
Cl
B
Cl
2
H 3C O H
13. In the space to the right, indicate if each of the pairs of molecules below are identical compounds,
enantiomers, diastereomers, structural isomers, or conformational isomers. (9 points).
CH 3
CH 3
H 3C
H 3C
CH 3
a.
Br
H
H
OH
H
CH 3
C C
H 3C
H
________________________
Br
H
H
OH
b.
c.
CH 3
________________________
H 3C
CH 3
C C
H
H
________________________
14. What two effects cause cyclobutane and cyclopropane to be higher in energy than cyclohexane? (3
pts)
15. Label the species below as Lewis Acids or Lewis Bases (4 points)
Br +
H3CO–
_____________
_____________
16. Circle the following which has the highest heat of combustion per CH2 unit. (4 points)
a. cyclopentane
b. cyclopropane
c. cyclohexane
d. cyclobutane
17. Indicate the geometry of carbon in the molecules below (e.g. trigonal bipyramidal). (6 points)
a. CH3CH3
b. H2C=O
c. HC≡CH
____________
______________
______________
3
18. Circle the statement(s) which are true of enantiomers. (4 points)
a. They have a non-superimposable mirror image.
b. They have no asymmetric carbon atoms.
c. They are chiral.
d. They do not rotate the plane of polarized light.
19. Draw an energy diagram of the molecular orbitals of the C=C bond of ethylene (H2C=CH2) and label
them (e.g. σ) and indicate their relative energies. Indicate the ground state electronic configuration of
the C=C electrons. (6 points)
20. What kind of molecular orbital results (σ, σ*, π, π*) results when the pairs of orbitals show below are
combined in the indicated manner? (6 pts)
+
a.
+
+
c.
b.
4
CHEMISTRY 61
Exam 1
Dr. M.T. Crimmins
September 21, 1999
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
I.
Nomeclature (12 points) Give the IPUAC name for the following compounds: Indicate R, S, cis, trans,
E, or Z where appropriate.
1.
CH 3CH 2
CH 2CH 2CH 2CH 3
CH 2 C H C H C H CH 2 C H CH 3
CH 3
CH 2CH 3 CH 3
____________________________
2.
C(CH3)3
CH 3
_____________________________
3.
H 3C
CH 2CH 3
C
CH 2CH 2CH 3
H
_____________________________
4.
OH
_____________________________
5. A. Write valid Lewis structures for the following species. Show all nonbonding (unshared) electrons
and indicate any formal charges . (6 points).
CH3–
CH2N2
6. Give the hybridization of the indicated atoms in the species below (6 points)
H 2C C C(CH3)2
CH 3
H 3C N
.. CH 3
1
O
H 3C C
O CH 3
7.
Write structures for the each of the following having a molecular formula of C4H8O (6 points).
a. a n aldehyde
b. a cyclic alcohol
c. an ether
8. What intermolecular forces exist between molecules of each of the following. (6 points).
a. CH3CH2CH2CH2CH3
..
CH 3 S
CH 3
_________________________
O
_________________________
c. CH3CH2OH
_________________________
b.
9. Draw all the possible stereoisomers of 3,4-dibromohexane. Indicate if they are chiral, meso or achiral
and indicate their relationship to each other. (i.e. enantiomers, diastereomers) (8 points)
10.
Label the following molecules as chiral or achiral.
OH
HO
OH
H
H
_____________
H
CH3
CH2OH
______________
C
CH2CH3
CH3
______________
11. Draw both chair conformations of trans-1,2-dimethylcyclohexane. If one is more stable than the
other, circle it. (6 points)
2
12. Draw an energy diagram for one 360° rotation about the C2-C3 C–C bond of butane. Also draw a
Newman Projection of the most stable conformation. (8 points).
13. Circle the molecule(s) which have a permanent dipole. In those which have a permanent dipole, show
the direction of the overall dipole. (6 points)
14. In the space to the right, indicate if each of the pairs of molecules below are identical compounds,
enantiomers, diastereomers, structural isomers, or conformational isomers. (9 points).
CH 3
CH 3
H 3C
CH 3
H 3C
H 3C
CH 3
a.
CH 3
CH 3
H 3C
CH 3
CH 3
________________________
Br
H
H
OH
H
OH
b.
c.
Br
H
H
CH 3
C C
H 3C
H
________________________
H 3C
CH 3
C C
H
H
________________________
15. What effect(s) cause cyclobutane and cyclopentane to be non-planar? (3 pts)
3
16. Label the species below as Lewis Acids or Lewis Bases (4 points)
H3C+
H2C=CH2
_____________
_____________
17. Indicate the geometry of carbon in the molecules below (e.g. trigonal bipyramidal). (6 points)
CH 4
____________
CH 3
H 3C C
CH 3
H 2C C CH 2
______________
______________
18. Draw and label the atomic orbitals which combine to form the molecular orbitals of formaldehyde,
H2C=O. . (6 points)
19. What kind of molecular orbital results (σ, σ*, π, π*) results when the pairs of orbitals show below are
combined (mathematically) in the indicated manner? (6 pts)
–
a.
+
+
c.
b.
4
CHEMISTRY 61
Exam 2
Dr. M.T. Crimmins
October 22, 1998
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
I.
REACTIONS: Predict the major organic products of the following reactions.
INDICATE STEREOCHEMISTRY AS NEEDED. (4 points each)
1.
CH 3
HI
2.
HBr
CH 2
peroxide
3.
H
C
H
1. BH3
CH 2CH 3
2. H2O 2, NaOH
C
H
4.
CH 3
H2, Pd/C
CH 3
5.
H 3C
H 3C
C
C
CH 2
H 3C
+
CH 2
H
C
C
H
heat
CO 2 CH 3
6.
H 3C
C
C
CH 3
Cl2
1 equivalent
7.
H 2C
H
C
C
H
HBr, -80°C
CH 2
8.
CH 3
1. Hg(OAc)2, H2O
2. NaBH4
1
II.
Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (4 points each)
___9. The rate limiting step for hydration of an alkene with water and acid is
a.
b.
c.
d.
protonation of the alkene by a strong acid
addition of water to a carbocation to form the protonated alcohol
loss of a proton from the protonated alcohol to form the alcohol.
simultaneous addition of H+ and HO – to the alkene.
___10. Which of the following free radicals is the most stable?
a.
c.
c.
CH 3
.
H 3C
C
d.
CH 3
H
H 3C
.
C
CH 3
CH 3
CH 3
H
.
C
H 3C
H
.
C
___11. Which of the following indicated hydrogens is the most acidic?
a. CH3CH=CH2
c. CH3CH2CH3
b. CH3C≡C–H
d. H–CH2CH2OCH3
___12. In the addition of HBr to 1,3-butadiene the 1,2 product predominates at -80°C while
the 1,4 product predominates at 40°C. The 1,4 product is said to result from
a.
b.
c.
d.
kinetic control.
thermodynamic control.
a Diels Alder reaction.
the s-cis diene.
___13. Which of the following alkenes would have the lowest heat of hydrogenation?
a.
b.
c.
d.
H
H 3C
CH2 CH3
C
H 3C
C
H3C
CH2 CH3
H 3C
C CH
H2C
CH3
H
C
CH2
H 3C
C
CH 3
CH C
CH CH
H3C
CH3
CH3
H
H3C
CH 2
___14. What is the stereochemical relationship of the products of the following reaction?
H
H 3C
a.
b.
c.
d.
e.
C
C
H
Br2
CH 2CH 3
diastereomers
enantiomers
identical (only one stereosiomer of the product is formed).
cis-trans isomers
conformational isomers
___15. Which of the carbocations below is the most stable?
a.
b.
H
c.
H
H
C+
CH 3
C+
CH 3
2
d.
+ CH 2
C
CH3
C+
CH 3
___16. What is the hybridization of the positively charged carbon in the carbocation below?
+ CH 2
C
a. sp3
V.
b. sp2
c. sp
d. s
Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the
following reactions. Be sure to account for stereochemistry as needed. (6 points each)
17.
H3 C
H
H
C
H3 C
Br2
H
C
CH3
C
Br
CH3
C
H
Br
18.
CH3
C CH2
CH3
X+ Y–
Y X
CH3
C CH2
CH3
Syntheses: Give reagents to carry out the transformations below. (5 points each)
19.
H
H
C
H
ClCH 2–CH 2Cl
C
H
20.
O
H 3C
C
C
H
H 3C
3
C
CH 2CH 3
21. Consider the energy diagram below and answer the questions using the letters on the diagram. (10 points).
B
D
C
E
N
E
R
G
Y
F
E
M
J
H
K
N
A
L
G
PROGRESS OF REACTION
a. What point(s) in the diagram represent transition states?
_______________
b. What point(s) in the diagram represent intermediates?
_______________
c. What is the energy of activation for the reaction?
_______________
d. What is the rate-limiting step in the reaction?
_______________
e. Is the reaction endergonic or exergonic?
_______________
f. What is the free energy change of the reaction?
_______________
g. Does G or C form faster from E?
_______________
22. Draw the HOMO (highest occupied molecular obrital) and the LUMO (lowest unoccupied molecular
obrital) of 1,3-butadiene and label them. Circle the one which would interact favorably with the
ethylene orbital below in a Diels-Alder reaction. (4 points)
4
CHEMISTRY 61
Exam 2: October 21, 1999
Dr. M.T. Crimmins
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
I.REACTIONS: Predict the major organic products of the following reactions.
INDICATE STEREOCHEMISTRY AS NEEDED.
1.
CH3
1. Hg(OAc)2,
H 2O
2. NaBH4
2.
CH3
CH2CH3
1. BH3
CH3
2. H2O2, HO –
C C
H
3.
HCl (1 equiv)
CH3CH2
C C H
4.
CH3
Br2
H
5.
CH3CH2
H2
C C CH2CH3
poisoned catalyst
(Pd/BaSO4)
6.
H3 C
Br2, H2 O
C CH2
H3 C
7.
H
C
C
CH3
HBr, peroxide
H
8.
H3 C
H2O, H+
C CH2
H3 C
9.
CH3CH2
C C H
NaNH2
CH3CH2Br
1
(4 points each)
II.
Multiple Choice: Circle the best answer (only one). (3 points each)
10. Which of the following alkenes is the most stable?
CH3
CH3
b.
a.
c.
CH3
11.
CH3
d.
CH3
CH3
CH3
Which of the following is the least stable carbocation?
a. H2C=CHCH2+
b. (CH3)3C+
12.
CH2
c. C6H5(CH3)2C+
d. CH3CH2+
In the following reaction what is the relationship of the products formed?
Br2
H
CH3
a. enantiomers
b. meso compound
13.
The carbon -carbon triple bond of an alkyne is composed of_________
a.
b.
c.
d.
14.
two σ bonds and one π bond
three σ bonds
one σ bond and two π bonds
three π bonds
The free energy of reaction is
a.
b.
c.
d.
e.
15.
the difference in energy between the reactants and an intermediate in the reaction
the difference in energy between the reactants and the transition state
the difference in energy between the reactants and the products
the difference in energy between the transition state and the products
the difference in energy between the intermediate and the products
What is the hybridization of the positively charged carbon in H3C+
a. p
b. sp2
16.
e. d2sp3
f. s
c. sp
d. sp3
A secondary cation is more stable than a primary carbocation because of
a.
b.
c.
d.
17.
c. structural isomers
d. diastereomers
overlap of a filled p orbital with an adjecent σ* antibonding orbital
overlap of an empty p orbital with adjacent σ bonding orbitals
resonance
deduction
Which of the following free radicals is the most stable?
a)
H3 C
.
C
H
CH3
b)
.
c)
CH3
H3 C
.
C
CH3
CH3
2
d)
H
.
C
H
CH3
18.
Which of the following is not true of H2C=CH2?
a)
b)
c)
d)
It contains 5 σ bonds.
It has bond angles of 120°.
All the atoms are in the same plane.
It has free rotation about the C=C bond.
Syntheses: Give reagents to carry out the transformations below. (4 points each)
19.
CH 3
C CH2
OCH3
CH 3
CH 3
C CH3
CH3
20.
H3 CC C–H
CH3CH2CH2CH2CH3
21.
O
HC
22.
CH3CH2
C
CH3
Draw an energy diagram for the hypothetical exergonic reaction below where B is an unstable
intermediate.
Label the positions for A, B, and C on the diagram and indicate the energy of activation on the
diagram. (5 points).
A
23.
CCH 2CH3
fast
slow
B
C
Draw resonance structures (4 pts each) for
a) benzyl cation
b) acetate ion (CH3CO2–)
3
V.
Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the
following reactions. (4 points each)
24.
CH3
H2 O, H+
CH3
H3 C C C CH2
H
CH3
CH3
OH CH3
25.
H3 C
H
H
C
C
Br2
H3 C
H3 C
CH3
H
Br
Br
H
26.
H3 C
CH3
HBr
C
C CH CH3
CH2
H3 C
H3 C
C
Br
4
CH3
CHEMISTRY 61
Exam 3
Dr. M.T. Crimmins
November 24, 1998
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
I.
Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (4 points each)
___1. The rate limiting step for free radical halogenation is
a.
b.
c.
d.
initiation
hydrogen atom abstraction from carbon by the halogen radical
attack of carbon radical on molecular halogen
termination
___2. Which of the following are "concerted" reactions?
a. SN1
b. SN2
c. E1
d. E2
e. SN1 and E1
f. SN2 and E2
___3. Reaction of a strong base with a tertiary alkyl halide is most likely to result in:
a. no reaction
b. E2 elimination
c. SN1 substitution
d. E1 elimination
___4. Which of the following statements is correct?
a. Free radical bromination is more selective than chlorination because the transition
state is more reactant-like.
a. Free radical bromination is more selective than chlorination because the transition
state is more product-like.
c. Free radical chlorination is more selective than bromination because the transition
state is more product-like.
d. Free radical chlorination is more selective than bromination because the transition
state is more reactant-like.
___5. Which of the folowing species is the most nucleophilic?
a. NH3
b. H3P
c. H2S
d. H2O
___6. SN1 reactions lead to
a. formation of free radicals
b. retention of stereochemistry
c. racemization
d. inversion of stereochemistry
___7. Which of the following would undergo the fastest dehydration reaction in the presence of acid?
a.
b.
CH 3
c.
CH 2 O H
d.
CH 3
OH
OH
CH 3
OH
1
___8. Which of the following reactions will proceed the fastest?
a.
b.
CH 2Cl
CH 2 Br
NaOH
c.
NaOH
d.
CH 3
CH 2 Br
CH 2 Br
NaOH
NaOH
.
II.
REACTIONS: Predict the major organic products of the following reactions.
INDICATE STEREOCHEMISTRY AS NEEDED. (4 points each)
9.
CH 2CH 3
Br2, light
10.
O
C H CH 2
CH 3 CH 2 CH 2 OH, H+
CH 3
1. NaH
OH
2. CH3I
11.
12.
H
H
CH(CH3)2
(CH3)3CO–K+
Br
13.
H
O
H
H
NaOH, H2 O
C(CH3)3
2
14.
H
CH3Se– Na+
Br
H
CH 3
15.
H
H 3C
C
C
CH 3
C
CH 3
CH 3 O H
Br
CH 3
16.
1. Mg
CH 3CH 2CH 2Br
2. H2C=O
3. H2SO4
V.
Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the
following reactions. Be sure to account for stereochemistry as needed. (6 points each)
17.
CH 3
CH 3
CH 3
H3PO4, heat
OH
CH 3
18.
CH 4 + Br2
heat or light
CH 3 Br + HBr
3
19.
CH 3
CH 3CH 2
O
C
CH 3
CH 3
CH 3
HI
CH 3CH 2
excess
I
+
I
C
CH 3
CH 3
Syntheses: Give reagents to carry out the transformations below. (5 points each)
20.
I
21.
O
O
CH 3
CH 2CH 2CH 3
22.
Only one monochlorination product is obtained from an alkane with the molecular formula C5H12. What
is the structure of the alkane? (4 points)
23.
Draw the transition state for the reaction of CH3Br with HO–. (4 points)
4
CHEMISTRY 61
Exam 3
Dr. M.T. Crimmins
November 23, 1999
Name___________________________________________
Pledge: I have neither given nor received aid on this exam.
Signature________________________________________
II.
Reactions: Predict the major organic product of the following reactions. If more than one product is
formed give both and indicate the major product. Indicate stereochemistry where necessary. (4
points each)
1.
H
CO2CH3
H
CO2CH3
heat
+
2.
HBr
-80 °C
3.
CH3
H3C C CH2 CH2CH3
Br2, light
H
4.
Br
H
NaI
H
CH3
5.
H 3C
H
H 2O
low temperature
Br
CH3
6.
Br
Ph
CH3
C C Ph
H
–
(CH3)3CO K
+
H
7.
CH3
Ph
C
CH3OH
CH CH2
Br
–1–
8. Indicate if the following compounds are aromatic, non-aromatic or anti-aromatic. (2 points each)
+
+
H
H
a.
b.
c.
d.
e.
9. List three criteria for aromaticity. (6 points)
1.______________________________________
2.______________________________________
3.______________________________________
II.
Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (3 points each)
___10.
In the following solvolysis reaction what is the relationship of the products formed?
H 3C
CH2CH3
C CH2CH2CH3
Br
a. enantiomers
b. meso compound
CH3OH
c. structural isomers
d. diastereomers
___11. Which of the following reactions would proceed the fastest?
a. CH3CH2CH3 + Br2 + light → CH3CHBrCH3
b. CH3CH2CH3 + F2 + light → CH3CHFCH3
c. CH3CH2CH3 + I2 + light → CH3CHICH3
d. CH3CH2CH3 + Cl2 + light → CH3CHClCH3
___12.
Which of the following is the strongest nucleophile?
c. CH3Od. CH3Se-
a. CH3NH–
b. Cl___13.
Which of the following alkyl halides would undergo the fastest SN2 reaction?
a.
CH2I
b.
b. CH3CH2CH2CH2—I
CH2Cl
d. (CH3CH2)2CH—I
___14. Which of the following are "concerted" reactions?
a. SN1
b. Diels-Alder reaction
c. E1
d. electrophilic addition
e. SN1 and E1
–2–
___15. Reaction of a hydroxide ion (HO–) with a primary alkyl halide is most likely to result in:
a. SN2 substitution
b. E2 elimination
c. SN1 substitution
d. E1 elimination
___16. Which of the following statements is correct?
a. Free radical bromination is more selective than chlorination because the transition
state for bromination is more reactant-like.
a. Free radical bromination is more selective than chlorination because the rate limiting step for
bromination is more endothermic than for chlorination.
c. Free radical bromination is more selective than chlorination because the rate limiting step for
bromination is more exothermic than for chlorination.
d. Free radical chlorination is more selective than bromination because the transition
state for chlorination is more reactant-like.
___17. Which of the following would be the rate limiting step in a free radical halogenation?
a.
b.
c.
d.
CH3CH2. + Br2 → CH3CH2Br + Br.
Br2 → 2 Br.
2 CH3CH2. → CH3CH2CH2CH3
CH3CH3 + Br. → CH3CH2. + HBr
___18. Ionization to give a carbocation and a leaving group is the rate determining step for
a. SN1
b. SN2
c. E1 and E2
d. E1
e. SN1 and SN2
f. SN1 and E1
Syntheses. Give reagents to show how to synthesize the compounds on the right from the compounds
on the left. They may require more than one step. (4 pts each)
19.
CH3
CH2OH
20.
CH3 CH3
CH3 CH3
H 3C
C
H3 C
CH CH3
C
CH3
CH3
–3–
C
CH2
V.
Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the
following reactions. (5 points each)
21.
Br2, light
CH 3CH2CH3
CH 3CHBrCH3
22.
Br
OCH3
CH3
CH3OH
CH3
CH3
+
23.
H2C=CH CH=CH2
HBr
H3 C–CH CH=CH2
40 °C
Br
–4–
+
H3 C–CH CH–CH2 Br
24. The energy diagram for the hypothetical reaction A + B → D → G + H is shown below. (6pts).
C
E
F
E
D
G
A +B
H
reaction coordinate
a. What is the rate determining step?_____________
b. What happens to the rate if the concentration of A is doubled? _____________
c. What is the rate expression?
rate = _______________
d. If D is a charged species and A and B are neutral, what effect will a polar protic solvent have on
the rate of reaction?___________________
e. What is the kinetically favored product?__________
f. What is the thermodynamically favored product?_________
–5–