Chemistry 51, Organic 1
Prerequisite: Chemistry 2
Grading Scheme
Recitation Grade:
100 pts average 75
Laboratory Grade
100 pts average 75
Lecture Exams
100 pts
Final
100 pts
Total
400 pts
Old exams and quizzes:
academic.brooklyn.cuny.edu/chem/howell/jhowell.htm
Safety: goggles, pregnancy
Cheating: F
jhowell@brooklyn.cuny.edu
Goals of the course:
1. Structure of organic molecules, relation of
structure to reactivity
2. Organic reaction patterns
3. Mechanism of reactions
4. Synthesis of organic compounds
5. Techniques of the trade
Scaling of Raw Grades for Recitation and Lab
1. Omit the student, B,
that dropped the course.
2. Get the average to
75
Multiply these sums by
4.9342
or
Note that student D
was absent from two
labs and will be low for
that reason. We will
deal with D later.
Multiply these sums by
0.49342
In both cases get the
same result…….
Now we have to spread the grades out so that there is a reasonable distribution
about the desired average of 75. This is done by expanding or contracting the set
of grades around the average so that the maximum turns out to be 95.
Now for the student D who is outside of the desirable limits (50 – 95). We
move his raw grade up to 70 so that D does not skew the scaling process. D
receives a scaled grade of about 51.
Atomic Structure
Electrons occupy orbitals which are grouped into subshells
and then shells.
Carbon atom: 1s22s22p2
Nitrogen: 1s22s22p3
Valence Shell Electrons: Used in chemical bond
formation
C
N
The pairing of electron spins
Electron Configuration of Atoms
• Pauli Exclusion Principle:
– No more than two electrons may be present in an
orbital. If two electrons are present, their spins must
be paired.
• Hund’s Rule:
– When orbitals of equal energy are available but there
are not enough electrons to fill all of them, one
electron is added to each orbital before a second
electron is added to any one of them; the spins of the
electrons in degenerate orbitals should be aligned.
• Aufbau (“Build-Up”) Principle:
– Orbitals fill in order of increasing energy from lowest
energy to highest energy.
Atomic Structure
•1 shell consists of
•1s subshell holding the
•1s orbital
•2 shell consists of
•2s subshell holding the
•2s orbital and the
•2p subshell holding the
•2px,
•2py, and
•2pz orbitals.
s orbital
p orbital
Each orbital can
hold two electrons.
An orbital can be
•an atomic orbital (on an atom) or
•a molecular orbital (on a molecule)
Octet Rule, Lewis Structures
Electrons are stabilized by bond formation.
H atom can stabilize up to two electrons in
the valence shell.
CF can stabilize up to 8 electrons in the
valence shell.
Complete octet: two electrons around H;
eight electrons for C  F in the valence
shell. Now we are providing maximum
stabilization.
Electron Configuration of Atoms
Completing the Octet
Ionic Bonding: Electrons can be transferred
to an atom to produce an anion and
“complete the octet”.
Covalent Bonding: Electrons can be shared
between atoms providing additional
stabilization. The shared electrons
typically “complete the octet” for each
atom.
Number of Bonds
Define: Unused stabilization of atom = (number of electrons that
can be held in valence shell) – (number of electrons already
present)
Neutral atoms
Some positive ions
Some negative ions
H: 1 more electron
H+ 2 more
H- 0 more
C: 4 more
C2+ 6 more
C- 3 more
N: 3 more
N+ 4 more
N- 2 more
O: 2 more
O+ 3 more
O- 1 more
F: 1 more
F+ 2 more
F- 0 more
Bonds make use of the unused stabilizing capability of the atoms.
# Bonds in molecule= (Sum of unused stabilizing capability)/2
Formal Charge: the charge an atom
in a Lewis bonding diagram.
Recipe to calculate the Formal Charges.
Assign each valence shell electron to an atom:
– Non-bonding electrons are assigned to the atom on
which they reside.
– Bonding electrons are divided equally between the
atoms of the bond.
Formal charge = (# valence shell electrons in neutral atom)
- (# nonbonding electrons)
- ½ (# bonded electrons)
Formal Charge and Bonding
Patterns. If the octet rule obeyed…
Formal
charge
C
N
O
+1
N
C
0
N
C
-1
N
C
O
O
O
What is “wrong” with the yellow box? Give examples of common molecules.
Lewis Diagrams
Typical Problem: Given a compound of molecular formula CH3CHCH2 draw a
Lewis bonding structure.
H
C
(3 * 4 + 6 * 1) / 2 = 9 bonds
How many bonds in the molucule?
Draw a bonding structure making use of single bonds to hold the molecule together.
H
H
C
C
H
H
C
H
H
9 – 8 = 1 bond left
How many bonds left to draw?
Put remaining bond(s) any place where the octet rule is not violated.
H
H
C
C
H
H
C
H
H
Another Example
Draw Lewis structure for CH2N2 with formal
charges.
Unused stabilizing
capacities.
# bonds = ½ (4 + 2 *1 + 2*3) = ½ x 12
Let’s characterize the valence electrons
# electrons present = 4 + 2*1 + 2*5 = 16
# bonding electrons = 12
# non-bonding electrons = 16 – 12 = 4 =
2 lone pairs
CH2N2 (continued)
H
6 bonds, 2 lone pairs
Skeleton uses 4 bonds:
N
C
N
H
H
H
OR
C
H
N
C
N
N
N
H
Which of these two structures is more important? Which provides the better description?
More important due to negative charge being on electronegative N
Curved Arrows to Reposition Electrons
Curved arrows are the standard way of showing
electron reorganizations, repositioning of lone pairs
and bonds.
Transforms the lone pair into a bonding pair. Still
counts towards terminal N but also towards middle
N. Too many electrons around middle N.
H
H
C
H
N
C
N
N
N
H
Bonding pair must retreat from the middle N and become
a lone pair on the C.
Some principals of using curved arrows.
•Reactions: Restructure the molecular system, creating bonds between
unbonded atoms.
•Resonance: Moving pi bonds and lone pairs but leaving single bond
framework intact.
An overly simplified presentation involves simply the motion shown below and
its reverse.
A
B
A
B
A lone pair on B is converted to bonding electrons.
What happens at B: The octet count at B is not changed but it does become more
positive by 1. The actual charge depends on the initial charge.
What happens at A: Two more electrons are added to the octet count of A. We
cannot exceed the octet count of 8.
•If A originally had an octet count of only 6 then we have simply completed the octet.
•If the octet of A was originally complete at 8 then we must withdraw some other pi
electrons from A to make room for the newly formed bond. A becomes more negative
by 1.
Now for its reverse
The reverse motion, conversion of bond into a lone pair, is shown below.
A
B
A
B
Here we convert a pi bond to a lone pair.
What happens at B: The pi electrons move away from B and no
longer count towards either the octet of B or its formal charge. B
becomes more positive by 1 and no longer satisfies the octet rule
(electron deficient).
What happens at A: The octet count is not changed at A but A does
become more negative by 1.
Example: resonance in amides
Formamide, NH2CHO, could be written with two electron diagrams
as shown below.
O
O
H
H
N
H
H
N
H
H
Each structure obeys the octet rule (provides maximum stabilization
by bonding). The structure on the right is less stable because of the
charges that are present.
The problem is how to smoothly interconvert one to the other.
We have to make room for the
incoming electrons by retreating
the pi electrons of the CO
double bond onto the O.
Proceeding with our elementary steps
O
O
H
H
N
H
N
H
H
But this structure violates the
octet rule on the high side. 10
ELECTRONS AT THE
CARBON!! FORBIDDEN!!
H
O
O
H
H
N
H
H
N
H
H
These two steps would be combined into one (always) and avoid that
repugnant structure which violates the octet rule on the high side.
O
O
H
H
N
H
H
N
H
H
As the lone pair on the N moves in towards the central C the
pi electrons in the C=O double bond retreat onto the O.
HO-CH-NH2+
Maximum number of bonds permitted using valence shell orbitals:
4H
4 more
1O
2 more
1C
4 more
1 N+
4 more
14/2 = 7 bonds
N is positive to allow for overall
positive charge. Could have
made any atom positive.
Set-up single bond framework.
H
6 bonds used in single bond framework.
O
H
N
C
H
1 left to assign. It will be a pi bond, part
of a double bond. The additional bond is
used to provide maximum stabilization;
completing the octet.
H
Where does it go? Only two possibilities….
How are they interconverted?
H
O
H
H
N
C
H
O
H
H
N
C
H
H
So what is the third structure?
It won’t have the double bond. Will not obey the octet rule.
Can collapse the pi bond into a lone pair.
H
H
O
N
H
C
O
H
N
H
C
H
H
H
Could we have collapsed in other direction?
H
No lone pair on this
N. High energy
structure; not very
important.
H
O
H
2
N
C
O
H
H
N
C
H
H
H
Our three structures…..
H
O
H
H
O
N
C
H
H
Intermediate stable with
positive charge on more
electronegative oxygen.
H
H
N
C
O
H H
H
Most stable with
positive charge on
the electropositive
atom, N.
N
C
H
H
Does not obey octet
rule. Least stable,
least important.
Examine one of the interconversions shown earlier.
Looking at it in reverse.
H
H
O
H
O
N
C
H
Higher energy structure
H
H
N
C
H
H
Lower energy, one more bond.
This is a common and significant resonance motion.
A supply of electrons moving towards a place where needed and providing a
new bond in the process.
Exceptions to Octet Rule
Example: sulfate ion.
Working as usual (use S2- and four O): # bonds = 1/2 (0 + 4*2) = 4
O
Result:
-1
+2
O
S
O
-1
-1
O
-1
Sulfur (2+) is very electronegative pulling the high energy lone pairs
from oxygen into itself. A better representation would be
O
-1
S
O
But the S violates the octet rule by having more than 8
electrons around the S. Valence shell is 3s and 3p. S
now would have to use additional atomic orbitals
O
-1
O
This means S would have to use the 3d shell orbitals.
Luckily, they are low in energy and can be utilized.
Octet is expanded due to use of 3d orbitals.
Donation from O to S.
o
-1
+2
o
S
-1
o
-1
o
-1
Sulfur (2+) is very electronegative pulling the high energy lone pairs
from oxygen into itself. A better representation would be
S
Empty 3d
O

Filled 2p
Observe that this is another
example of donation of
available electrons (lone pair)
into a neighboring acceptor
orbital.
Survey of Classes of Molecules
1. Hydrocarbons, C & H
A Alkanes: only single bonds
1. Acyclic (no rings) Alkanes: CnH2n+2
suffix: -ane
2-methylpropane
butane
2. Cyclic alkanes: example cyclohexane
CnH2n+2-2(# rings) Presence of a ring eliminates two
hydrogens
H2
C
CH3
CH3
-2H
CH2
2. Alkenes: double bonds, pi bonds
a.
Suffix: -ene
i acyclic
2-methylpent-1-ene
4-methylpent-1-ene
ii cycloalkenes
cyclohexene
A pi bond eliminates two hydrogens
C4H10 (butane)  C4H8 (butene) + 2 H
general formula: CnH2n+2-2(# rings) -2(#pi bonds)
Alkynes: C to C triple bond = 2 pi bonds + sigma
CH3CH2CCH
but-1-yne
hepta-1,3-dien-6-yne
Oxygen Containing Molecules
Incorporation of an oxygen into a molecule
does not change the relationship
between the number of carbons and
number of hydrogens.
CnH2n+2-2(#pi bonds) – 2(# rings)Oa
H
H
C
H
Watch how this
makes sense.
O
H
H
Hydrogen Deficiency
CnH2n+2 - 2(#pi bonds) – 2(# rings)
hydrogen deficiency.
2n + 2: Number of hydrogens for an acyclic compound
with n carbons and no pi bonds.
If a compound has a different number of hydrogens it must
be due to the presence of pi bonds or rings.
# pi bonds
+
# rings
= 1/2 ( (2n + 2) - (actual number of hydrogens present) )
Singly bonded Oxygen
Alcohols, ROH, -ol
OH
CH3CH2OH
ethanol
Two OH groups: a diol
OH
OH
2
5
1
hex-5-ene-1,2-diol
Useful Classiification of Alcohols
(and other things): Primary,
Secondary, Tertiary
Alcohols and many other groups may be
classified as
Primary: one carbon directly bonded to the
C bearing the –OH. CH3CH2CH2OH
Secondary: two carbons directly bonded to
the C bearing the –OH. (CH3CH2)2CHOH
Tertiary: three carbons directly bonded to
the C bearing the –OH. (CH3CH2)3COH
1o, 2o, 3o
Ethers: ROR, isomeric with alcohols
O
H3C
CH3
dimethyl ether
methoxymethane
O
H3C
C2H5
methyl ethyl ether
methoxyethane
Methoxy group (alkoxy)
Doubly Bonded Oxygen
Carbonyl Group
O
This is a pi bond, resulting in the elimination of two
hydrogens.
Aldehydes, RCHO, -al
O
H3C
H
ethanal
acetaldehyde
Still more complex, now
A more complex example
we put a double bond in
the parent chain.
4
3
O
Double
Bonds part of
Parent Chain
2
1
2-methylbutanal
2-methylbut-3-enal
Start numbering
the chain at
CHO
Ketones, R(CO)R, -one
O
Simple example
pentan-2-one
More complex example:
Double bond
included as part of
parent chain
Both carbonyls must be
included in parent chain
–dione.
Butyl Side chain.
1
O
O
3-butyl-5-methylhex-5-ene-2,4-dione
Carboxylic Acids, RCO2H, -oic acid
Ka = 10-5
CO2-
CO2H
pentanoic acid
pentanoate carboxylate anion
Another more complex problem
Methyl group
Both carboxylic
groups must be in
parent chain
Start numbering here
+ H+
CO2H
CO2H
Vinyl side chain
2-ethyl-3-methyl-4-vinylpentanedioic acid
Ethyl Group
Attached to or part of the Parent
Chain are:
• Functional Groups: -OH, carbonyl
(aldehydes, ketones), carboxylic acids
• Substituents: alkyl groups, halogens,
alkoxy groups
Some Simple Substituents
• Alkyl, -R: An alkane minus one hydrogen
(providing the point of attachment to the parent
chain).
-CH2CH3: ethyl (-Et);
-CH2CH2CH3: propyl;
-CH(CH3)2: isopropyl
• Alkoxy, -OR. –OCH2CH3: ethoxy (-OEt)
• Halo, -X. –Cl: chloro
More complex substituents
Systematically named.
Note that the atom directly
attached to the parent chain is
atom 1 in the substituent.
Substituents on the
butyl substituent!!
CH2Cl
1
6
2
3
4
OCH3
Substituent
Parent chain
6-(2-(chloromethyl)-4-methoxy-1-methylbutyl)
Indicates position of attachment on parent chain
Functional Groups
Non-Functional Groups (alkyl, alkoxy, halide) will
always appear as substituent and as a prefix.
Functional Groups (pi bonds, alcohol, aldehyde,
ketone, carboxylic acid) will determine the suffix of
parent chain. If more than one functional group then
we must prioritize. Highest priority determines
suffix. The rest ( alcohol, aldehyde, ketone,
carboxylic acid) appear a prefix. But note
unsaturation (pi bonding) in the parent chain is
always specified as a suffix.
Functional Group Priorities
Highest priority at bottom
Group
Suffix if highest Prefix if not highest
Triple bond
-yne
NA. Always as suffix
Double Bond
-ene
NA. Always as suffix
Alcohol
-ol
hydroxy
Ketone
-one
oxo
Aldehyde
-al
oxo
Carboxylic Acid
-oic Acid
OH
1
O
O
alcohol
hydroxy prefix
ketone
oxo prefix
aldehyde
highest -al suffix
4-hydroxy-3-oxopentanal
One more
example.
1
HO2C
2
HO
3
4
CHO
CO2H
5
Highest Priority Functional Group: two carboxylic acids. Pentanedioic acid
Double bond in parent chain: Pent-2-enedioic acid
Ethyl side chain bearing a hydroxy at position 1:
3-(1-hydroxyethyl)
Ethyl side chain bearing an oxo at position 2: 4-(2-oxoethyl)
Butyl side chain with double bond at position 2: 2-(but-2-enyl)
2-(but-2-enyl)-3-(1-hydroxyethyl)-4-(2-oxoethyl)pent-2-enedioic acid
Polar Bonds and Electronegativity
H
Li
Be
B
C
N
O
F
Na
Mg Al
Si
P
S
Cl
Electronegativity increases towards the
upper right.
Electronegativity differences results in polar
bonds.


Example:
H
Cl
Polar Bonds can lead to Polar Molecules
Combine the bond dipoles to get the overall dipole by means
of vector addition.
Bond dipoles are put tail to head to provide an overall result.
Resultant Vector
Combine these three vectors
Some Examples
O
C
+
O
=
resulting molecular dipole = 0
O
+
H
H
=
zero
Predicting Molecular dipoles
Clearly, molecular geometry is very
important in determining molecular
dipoles.
Problem: Rank the following for size of
molecular dipole. Bond angles are 120o.
F
F
F
H
H
H
A
B
A > C > B(0)
H
F
F
F
H
C
H
Molecular geometry (VSEPR)
The number of groups of electrons around
an atom determines the shape.
Each lone pair, single, double, or triple bond
counts as a group.
# groups
2
3
4
5
hybridization
sp
sp2
sp3
dsp3
Geometry of
the groups
Linear
Trigonal
planar
Tetrahedral Trigonal
bipyramid
Example 1
• CH4
• Four bonds =
H
H
C
Four groups
• Tetrahedral
H
H
Example 2
Two
lone pairs
O
Four groups of electrons.
H
H
Tetrahedral geometry for
groups.
Bent Molecule
Two
bonds
H
104o
O
H
Nature of Chemical Bonds
So far, we have the octet rule which tells us
how many bonds we can make. But how
do we understand the nature of the
bonds?
Three models for bonding: ionic, valence
bond, molecular orbitals.
Ionic Bonding
Requires very different electronegativities to
make the complete transfer of electrons worthwhile.
Not discussed further although it occurs in salts
of organic acids. For example sodium acetate.
Quantum or Wave Mechanics
• Albert Einstein: E = hn (energy is quantized)
– light has particle properties.
• Erwin Schrödinger: wave equation
– wave function, : A solution to a set of equations that
depicts the energy of an electron in an atom.
– each wave function is associated with a unique set of
quantum numbers.
– each wave function represents a region of threedimensional space and is called an orbital.
–  2 is the probability of finding an electron at a given
point in space.
Quantum or Wave Mechanics
• Characteristics of a wave associated with
a moving particle. Wavelength is
designated by the symbol l .
Quantum or Wave Mechanics
• When we describe orbital interactions, we are referring
to interactions of waves. Waves interact
– constructively or
– destructively.
• When two waves overlap, if they are of the same sign
then they combine constructively, build-up. Opposite
sign overlap combines destructively, meaning they
cancel.
Shapes of Atomic s and p Orbitals
– All s orbitals have the
shape of a sphere with
the center of the
sphere at the nucleus.
– Figure 1.8 (a)
Calculated and (b)
cartoon
representations
showing an arbitrary
boundary surface
containing about 95%
of the electron density.
Shapes of Atomic s and p Orbitals
– Three-dimensional representations of the 2px, 2py,
and 2pz atomic orbitals. Nodal planes are shaded.
Shapes of Atomic s and p
Orbitals
2px, 2py, and 2pz atomic orbitals.
Molecular Orbital Theory
• MO theory begins with the hypothesis that
– electrons in atoms exist in atomic orbitals
and
– electrons in molecules exist in molecular
orbitals.
Molecular Orbital Theory
• Rules:
– Combination of n atomic orbitals (mathematically
adding and subtracting wave functions) gives n MOs
(new wave functions).
– MOs are arranged in order of increasing energy.
– MO filling is governed by the same rules as for atomic
orbitals:
• Aufbau principle: fill beginning with lowest
energy orbital
• Pauli exclusion principle: no more than 2e- in a
MO
• Hund’s rule: when two or more MOs of
equivalent energy (degenerate) are available,
add 1e- to each before filling any one of them
with 2e-.
Molecular Orbital Theory
• MOs derived from combination by (a) addition and (b)
subtraction of two 1s atomic orbitals.
Covalent Bonding
• Bonding molecular orbital: A MO in which
electrons have a lower energy than they
would have in isolated atomic orbitals.
• Sigma (s) bonding molecular orbital: A MO
in which electron density is concentrated
between two nuclei along the axis joining
them and is cylindrically symmetrical.
Covalent Bonding
• A MO energy diagram for H2. (a) Ground
state and (b) lowest excited state.
Covalent Bonding
• Antibonding MO: A MO in which electrons
have a higher energy than they would in
isolated atomic orbitals.
VB: sp3 Hybridization of Atomic Orbitals
Energy
– The number of hybrid orbitals formed is equal to the
number of atomic orbitals combined.
– Elements of the 2nd period form three types of hybrid
orbitals, designated sp3, sp2, and sp.
– The mathematical combination of one 2s atomic
orbital and three 2p atomic orbitals forms four
equivalent sp3 hybrid orbitals.
2p
sp3
2s
sp 3 Hybridization, with electron population for
carbon to form four single bonds
VB: sp3 Hybridization of Atomic Orbitals
• sp3 Hybrid orbitals. (a) Computed and (b) cartoon threedimensional representations. (c) Four balloons of similar
size and shape tied together, will assume a tetrahedral
geometry.
VB: sp2 Hybridization of Atomic Orbitals
Energy
• The mathematical combination of one 2s atomic orbital
wave function and two 2p atomic orbital wave functions
forms three equivalent sp2 hybrid orbitals.
2p
2p
sp2
2s
sp 2 Hybridization, with electron
population for carbon to form double
bonds
VB: sp2 Hybridization of Atomic Orbitals
• Hybrid orbitals and a single 2p orbital on an sp2
hybridized atom.
VB: sp Hybridization of Atomic Orbitals
Energy
• The mathematical combination of the 2s atomic orbital
and one 2p atomic orbital gives two equivalent sp hybrid
orbitals.
2p
2p
sp
2s
sp Hybridization, with electron
population for carbon to form
triple bonds
VB: sp Hybridization of Atomic Orbitals
• sp Hybrid orbitals and two 2p orbitals on an sp
hybridized atom.
Combining VB & MO Theories
• VB theory views bonding as arising from electron pairs
localized between adjacent atoms. These pairs create
bonds.
• Further, organic chemists commonly use atomic orbitals
involved in three hybridization states of atoms (sp3, sp2,
and 2p) to create orbitals to match the experimentally
observed geometries.
• How do we make orbitals that contain electrons that
reside between adjacent atoms? For this, we turn back
to MO theory.
Combining VB & MO Theories
• To create orbitals that are localized between adjacent
atoms, we add and subtract the atomic and hybrid
orbitals on the adjacent atoms, which are aligned to
overlap with each other.
• Consider methane, CH4. The sp3 hybrid orbitals of
carbon each point to a 1s orbital of hydrogen and,
therefore, we add and subtract these atomic orbitals
to create molecular orbitals.
• As with H2, one resulting MO is lower in energy than the
separated atomic orbitals, and is called a bonding s
orbital. The other is higher in energy and is antibonding.
Combining VB & MO Theories
• Molecular orbital mixing diagram for creation of a C-C s
bond.
Combining VB & MO Theories
• A double bond uses sp2 hybridization.
• Consider ethylene, C2H4. Carbon (and other secondperiod elements) use a combination of sp2 hybrid orbitals
and the unhybridized 2p orbital to form double bonds.
• Now the atomic and hybrid orbitals before mixing into
MOs.
Combining VB & MO Theories
• MO mixing diagram for the creation of C-C p bond.
Present in double and triple bonds.
Combining VB & MO Theories
• A carbon-carbon triple
bond consists of one s
bond formed by overlap of
sp hybrid orbitals and two
p bonds formed by the
overlap of parallel 2p
atomic orbitals.
Kinds of Hybridization
spn hybridization obtained by mixing the 2s
atomic orbital with n different 2p atomic
orbitals to yield (n+1) hybrids.
sp
geometry
VSEPR
groups
linear
2
Orbitals
Where
found
C
#
Pi bonds
2
C
sp2
trigonal
planar
3
sp3
tetrahedral
4
1
C
C
0
Hybrids are in black, colored orbitals are p orbitals not used in hybridization.
Example of how hybridization
determines geometry
Assign hybridization
CH2=C=CH2
sp2
Match up p
orbitals for pi
bonds
sp2
sp
H
H
C
C
C
H
H