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Organic Chemistry
Chapter 22
What is organic chemistry?



The study of carbon compounds
Carbon can form stable bonds with itself so it
can make long chains and ring structures
Carbon atoms must make four bonds,
oxygen must make two bonds, nitrogen
must make three bonds and hydrogen must
make one bond in organic compounds
Characteristics of Organic
Compounds

Molecular formula – showing the actual number of
atoms in each element in that compound


Example – C2H4O2 for ethanoic acid
Structural formula – showing how the atoms are
arranged

Example CH3CO2H for ethanoic acid meaning 3 H atoms
are attached to 1 C atom, which is attached to another C
which has 2 O atoms attached to it, one of which has an
H attached to it
Characteristics of Organic
Compounds

Graphical formula – showing how these
atoms are arranged in space and the bonds
between them. Lines represent covalent
bonds (shared electrons) between the atoms

Example – on board for ethanoic acid
Characteristics of Organic
Compounds

Skeletal formula – showing an abbreviated
form of the carbon chain, with each line
segment understood to have a carbon atom at
each end. May or may not show terminal
carbons


Example – on board for ethanoic acid
Name – based on IUPAC nomenclature

Example - CH3CO2H is called ethanoic acid
Alkanes


General formula – CnH2n+2
Examples
 Methane
 Butane
Naming
1. Name the longest chain first
 Prefixes
 1 C - meth
5 C – pent
6 C - hex
 2 C - eth
7 C - hept
 3 C - prop
8 C -oct
 4 C – but
** If there are two longest chains of
the same length give priority to the
one with the most branching
Alkanes


Unreactive because the C-C bonds and
C-H bonds are relatively strong
Most common reaction is combustion
(exothermic)

Carbon compound + oxygen 
carbon dioxide + water
Naming
2. Identify the type of bonding that is
present
 Single bonds – -ane
 Double bonds – -ene
 Triple bonds – -yne
Naming
3. Identify any functional groups
 Alkane – -e
 Alcohol – -ol
 Halogenalkane – chloro-, bromo- or
iodo Aldehyde – -al
 Ketone – -one
Naming
3. Identify any functional groups
 Carboxylic acid – -oic acid
 Ester – -oate
 Amide – -amide
 Amine – -amine
Naming
4. Use numbers to indicate where groups
or bonds are in a chain
** A side note – cycloalkanes
 Alkanes with ring-like structures
 Side chains are named by replacing
-ane with –yl
Naming
**One more side note – aromatic
compounds
 Called arenas
 Based off of benzene rings
 Use the smallest numbers possible
Nomenclature Practice

Name this compound
CH3
H3C
CH3
Cl
H3C
Nomenclature Practice




Step #1: For a branched hydrocarbon, the longest
continuous chain of carbon atoms gives the root
name for the hydrocarbon
Step #2: When alkane groups appear as
substituents, they are named by dropping the -ane
and adding -yl.
Step #3: The positions of substituent groups are
specified by numbering the longest chain of carbon
atoms sequentially, starting at the end closest to
the branching.
Step #4: The location and name of each substituent
are followed by the root alkane name. The
substituents are listed in alphabetical order
(irrespective of any prefix), and the prefixes di-, tri-,
etc. are used to indicate multiple identical
substituents.
Alkenes and Alkynes



Alkenes – have at least one double bond
Alkynes – have at least one triple bond
Both are said to be unsaturated
Aromatic Hydrocarbons


Cyclic unsaturated hydrocarbons with
delocalized electrons
The simplest aromatic hydrocarbon is
benzene (C6H6)
OR…
Geometric Isomerism in Aromatics



ortho (o-) = two adjacent substituents
o-dichlorobenzene
meta (m-) = one carbon between
substituents
m-dichlorobenzene
para (p-) = two carbons between
substituents
p-dichlorobenzene
Alcohols


General formula – R-OH
Naming
 Change ending of the main chain to
-ol
Amines


General formula – RNH2
Naming

Add amino- to the beginning or –amine to
the ending of the name
Aldehydes

General formula

Naming

O
R–C-H
Change the ending of the main chain to
-al
Ketones

General formula

Naming

O
R–C-R
Change the ending of the main chain to
-one
Carboxylic acids

General formula
O
R – C - OH

Naming

Change the ending of the main chain to
–oic acid
Esters

General formula

Naming
O
R – C – O – R’
Name the R’ chain first
 Name the R chain and change the ending
to -oate

Amides

General formula

Naming

O
R – C – NH2
Change ending to -amide
Ethers


General formula R – O – R’
Naming

Name both R groups as side chains and add
ether to the end
Isomers


Structural isomers – have the same
molecular formula but a different structural
formula; similar chemical properties
 Example – pentane
Functional group isomers – same molecular
formula but different functional groups;
different chemical and physical properties

Examples:
Carboxylic acid and ester
 Aldehyde and ketone

Electrochemistry
Chapter 17
Oxidation and Reduction (Redox)



Electrochemistry deals with nonspontaneous
Redox reactions
Electrons are transferred
Spontaneous redox rxns can transfer energy



Electrons (electricity)
Heat
Non-spontaneous redox rxns can be made to
happen with electricity
Electrochemistry Terminology



Oxidation – A process in which an element
attains a more positive oxidation state
Na(s)  Na+ + eReduction – A process in which an element
attains a more negative oxidation state
Cl2 + 2e-  2ClLEO says GIR


Loss of electrons = oxidation
Gain of electrons = reduction
Electrochemistry Terminology


Oxidizing agent
The substance that is reduced is the
oxidizing agent
Reducing agent
The substance that is oxidized is the
reducing agent
Redox reactions


0
0
+1 -1
2Na + Cl2  2NaCl
Each sodium atom loses one electron:
0 +1
Na  Na+ + esodium is oxidized
Each chlorine atom gains one electron:
0
-1
Cl + e-  ClChlorine is reduced
Trends in Oxidation and Reduction

Active metals:




Lose electrons easily
Are easily oxidized
Are strong reducing agents
Active nonmetals:



Gain electrons easily
Are easily reduced
Are strong oxidizing agents
Balancing a Redox reaction in Acidic
Solutions
Write separate equations for the oxidation and
reduction half-reactions.
For each half reaction
1.
2.
a.
b.
c.
d.
3.
4.
Balance all the elements except H and O
Balance O by using H2O
Balance H by using H+
Balance the charge using e-
If necessary, multiply one or both balanced half
reactions to equalize the e- transferred.
Add the half reactions canceling identical species.
Examples – Acid Solution

MnO4-(aq) + Fe2+(aq)  Fe3+(aq) + Mn2+(aq)

5e- + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4H2O(l)
5Fe2+(aq)  5Fe3+(aq) + 5e-

5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)


H+(aq) + Cr2O72-(aq) + C2H5OH(aq)  Cr3+(aq) + CO2(g) + H2O(l)



4H2O(l) + C2H5OH(aq)  2CO2(g) + H2O(l) + 12H+(aq) + 12e12e- + 28H+(aq) + 2Cr2O72-(aq)  4Cr3+(aq) + 14H2O(l)
16H+(aq) + C2H5OH(aq) + 2Cr2O72-(aq)  2CO2(g) + 4Cr3+(aq) + 11H2O(l)
Balancing a Redox Reaction in
Basic Solution
1.
2.
3.
Use the half reaction method like we did for
acidic solutions
To both sides of the equation obtained in #1,
add a number of OH- ions that is equal to the
number of H+ ions
Form H2O on the side containing both H+
and OH- ions, and eliminate the number of
H2O molecules that appear on both sides of
the equation
Example – Basic Solution

Ag(s) + CN-(aq) + O2(g)  Ag(CN)2-(aq)





4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 4e4e- + 4H+ + O2(g)  2H2O(l)
4H2O(l) + O2(g) + 4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 2H2O(l) + 4OH2H2O(l) + O2(g) + 4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 4OH-
Cu + Cr2O7-2  Cu+2 + Cr+3




3Cu  3Cu+2 + 6e6e- + 14H+ + Cr2O7-2  2Cr+3 + 7H2O
14H+ + 3Cu + Cr2O7-2  2Cr+3 + 7H2O + 3Cu+2
14H2O + 3Cu + Cr2O7-2  2Cr+3 + 7H2O + 3Cu+2 + 14OH -
Electrode Potentials and Half Cells

When a metal comes into contact with a solution containing its
own ions, an equilibrium is set up





Some reactive metals will lose electrons easily – equilibrium
will lie to the left
A less reactive metal will show less tendency to ionize –
equilibrium will lie to the right
Whenever an element is placed with a solution containing its
own ions, an electrical charge will develop on the metal, or in
the case of non-metals on the inert conductor placed in
solution
The charge is an electrode potential


Mx+(aq) + xe-  M(s)
Size and sign of the potential depends on the ability to lose or gain
electrons
The system is a half cell
Electrochemical Activity Series



Elements at the top of the series gain electrons
most readily, have more positive Eo values and
are the best oxidizing agents
Elements at the bottom lose electrons more
readily, have more negative Eo values and are the
best reducing agents
The more positive the Eo1/2, the more it tends to
occur.
Electrochemical Cells



An apparatus for generating electrical energy from a
spontaneous REDOX reaction
Connect two half cells with different electrode potentials
A salt bridge connects the two halves

Allows the transfer of ions between the two solutions and completes the
circuit

Electrons flow through the wire toward the more positive half
cell

Anode: The electrode where oxidation occurs
Cathode: The electrode where reduction occurs


Red Cat: Reduction at the Cathode
Line Notation

An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq)|| Cu2+(aq) | Cu(s)
Anode Anode
Cathode
Cathode
material solution
solution
material
Standard Electrode Potential




Eo is the variable
Measured relative to a standard hydrogen
electrode (Eo = 0.00V)
Under standard conditions: 25oC, 1atm and 1M
solutions
For the Example:



From a table of reduction potentials:
Zn2+ + 2e-  Zn
E = -0.76V
Cu2+ + 2e-  Cu
E = +0.34V
Zn-Cu Galvanic Cell

The less positive,
or more negative
reduction
potential
becomes the
oxidation…
Zn  Zn2+ + 2eCu2+ + 2e-  Cu
Zn + Cu2+  Zn2+ + Cu
E = +0.76V
E = +0.34V
E0 = + 1.10 V
Galvanic (Electrochemical) Cells


Spontaneous redox processes have:
 A positive cell potential, E0
 A negative free energy change, (-G)
A nonspontaneous redox process have:

A negative cell potential

A positive free energy change, (+G)
Calculating G0 for a Cell





G0 = -nFE0
n = moles of electrons in balanced redox
equation
F = Faraday constant = 96,485 coulombs/mol
e1 V = 1 J/coulomb
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
G0 = -(2mol e-)(96485 coulombs)(1.10 J
mol e-
= -212267 J = -212 kJ
coulomb
)
Gibb’s Free Energy and Ecell
K
Eo
Go
Conclusion
>1
Positive
Negative
Spontaneous cell
reaction
=1
0
0
At equilibrium
<1
Negative
Positive
Non-spontaneous cell
reaction; the reaction
is spontaneous in the
reverse direction
The Nernst Equation

Standard potentials assume a concentration of 1 M.
The Nernst equation allows us to calculate potential
when the two cells are not at standard conditions.
E = Eo – RT ln(Q)
nF






R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol eEo = the voltage at standard conditions
Q = the reaction quotient
The Nernst Equation

Example



At 25˚C, the middle term can be simplified to 0.0591
If the reaction Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq) is
carried out with 5.00M Zn2+ and 0.300M Cu2+ at
298K, what is the actual cell voltage?
E = 1.10 V – 0.0591 log (5.0 M) = 1.064 V
2
(0.3 M)
Calculating an Equilibrium Constant
from a Cell Potential

Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
0 volts = 1.10 – 0.0591 log(K) = (1.10)(2) = log(K)
2
0.0591
1037.2 = K = 1.58 x 1037

At equilibrium, forward and reverse reactions
occur at equal rates, therefore:


The battery is “dead” ; The cell potential, E, is zero volts
Modifying the Nernst Equation (at 25 C):
0 volts = E0 – 0.0591 log(K)
n
Concentration Cell


Both sides have the
same components but at
different concentrations.
Step 1: Determine
which side undergoes
oxidation, and which
side undergoes
reduction.

We’re going to use Zn
instead of Ag
Concentration Cell


Both sides have the same components but at
different concentrations.
The 1.0 M Zn2+ must decrease in
concentration, and the 0.10 M Zn2+ must
increase in concentration
Zn2+ (1.0M) + 2e-  Zn (reduction)
Zn  Zn2+ (0.10M) + 2e- (oxidation)
Zn2+ (1.0M)  Zn2+ (0.10M)
Concentration Cell



Step 2: Calculate cell potential using the
Nernst Equation (assuming 25 C).
Zn2+ (1.0M)  Zn2+ (0.10M)
E = Eo – 0.0591 log(Q)
n
Eo = 0.0 volts n = 2
Q = (0.10)/(1.0)
E = 0.0 – 0.0591 log(0.10) = 0.030 volts
2
1.0
Electrolysis




Process in which electrical energy is used to
cause non-spontaneous REDOX reactions
Opposite of an electrochemical cell
Uses and electrolytic cell
Electrolytic processes are NOT
spontaneous. They have:
 A negative cell potential, (-E0)
 A positive free energy change, (+G)
Electroplating of Silver









Anode reaction:
Ag  Ag+ + eCathode reaction:
Ag+ + e-  Ag
Electroplating requirements:
1. Solution of the plating metal
3. Cathode with the object to be plated
2. Anode made of the plating metal
4. Source of current
Solving an Electroplating Problem



How many seconds will it take to plate out 5.0
grams of silver from a solution of AgNO3 using
a 20.0 Ampere current?
1 Ampere = 1 coulomb/second
Ag+ + e-  Ag
5.0g Ag x 1mol Ag x 1mol e- x 96485C x 1s
107.87g 1mol Ag 1mol e- 20.0C
= 2.2 x 102s
The Nucleus: A Chemist’s View
Chapter 18
Nuclear Symbols
Mass number, (A)
+
(p
+
o
n)
235
92
Atomic number, (Z)
(number of p+)
U
Element
symbol
Radioactivity
Spontaneous decay of certain
atoms with the evolution of a, b
and g particles
 Comes from the nucleus

Radioactivity
Alpha (α)
Beta (β)
Gamma (γ)
Nature
Helium
Nucleus
Negative
particle with
no mass
High energy &
frequency
radiation
Charge
+2
-1
0
Mass
4 amu
1/1840 amu
0
Movement in To negative
electronic field
To positive
None
Penetrating
Power
Intermediate Strong
Weak
Balancing Nuclear
Equations
SAreactants = SAproducts
235 + 1
235
1
=
U+ n 
92
0
92 + 0
=
142
142
+
91 +
91
3(1)
1
Ba + Kr + 3n
56
56
36
+ 36 +
SZreactants = SZproducts
0
3(0)
Balancing Nuclear
Equations #2
1. 226
Ra
88
2.

U+n
4
a
2
+?
1 + 2n + ?
Alpha Decay
Alpha production (a): an alpha
particle is a helium nucleus
 Alpha decay is limited to heavy,
radioactive nuclei
 Limited to VERY large nuclei.
 Can be shown by α or He symbols
as products

Beta Decay

Beta production (b): A beta particle is an
electron ejected from the nucleus

Beta emission converts a neutron to a
proton

Converts a neutron into a proton.
Gamma Decay



Gamma ray production (g):
Gamma rays are high energy
photons produced in association with
other forms of decay.
Gamma rays are mass-less and do
not, by themselves, change the
nucleus
Other Types of Emission

Positron Emission
Positrons are the anti-particle of the electron
 Positron emission converts a proton to a
neutron


Electron Capture
Inner-orbital electron is captured by the
nucleus
 Electron capture converts a proton to a
neutron

Half Life


The time taken for half of the atoms of a
sample to decay
Three methods to determine half life
 Graphically
 Use of the expression: ln(N/No) = - kt
 Use of the expression: fraction of
remaining activity = 1/2n
Half Life: ln(N/No) = - kt





Decay occurs by first order kinetics (the rate
of decay is proportional to the number of
nuclides present)
ln N = -kt
t1/2 = ln(2) = 0.693
N0
k
k
N = number of nuclides remaining at time t
N0 = number of nuclides present initially
k = rate constant
t = elapsed time
Half Life:
n
1/2

Fraction of the remaining activity = 1/2n
n = the number of half lives

Examples:



After 21 days a radioactive isotope has only one
eighth of its original activity. What is the half life
of the radioactive isotope?
The half life of 14C is 5585 years. A sample of
carbon gave a count rate of 2.02. Another sample
of carbon gave a count rate of 1.71. Calculate the
age of the second sample.
Nuclear Stability & Decay





Decay will occur in such a way as to return a
nucleus to the band (line) of stability.
The most stable nuclide is Iron-56
If Z > 83, the nuclide is radioactive
Stable (non-radioactive) nuclei tend to have a
neutron: proton ratio of 1:1
A radioactive nucleus reaches a stable state by a
series of steps
Energy and Mass


Nuclear changes occur with small but measurable
losses of mass. The lost mass is called the mass
defect, and is converted to energy according to
Einstein’s equation: E = mc2
m = mass defect
E = change in energy
c = speed of light
Because c2 is so large, even small amounts of mass
are converted to enormous amount of energy.
Nuclear Fission and Fusion


Fusion: Combining two light nuclei to form a
heavier, more stable nucleus.
3
1
4
0
2 He + 1 H 
2 He + 1e
Fission: Splitting a heavy nucleus into two nuclei
with smaller mass numbers.
1
235
142
91
1
0 n + 92 U  56 Ba + 36 Kr + 3 0 n
Fission Processes
A self-sustaining fission process is called a
chain reaction.
Neutrons
causing
Event
fission
Result
subcritical
<1
reaction stops
Critical
=1
sustained reaction
Supercritical
>1
violent explosion

A Fission Reaction
Use of Radiation





Medicine
Isotopic dating
Thickness control in engineering
 Radioactive source is placed on one side and
the detector on the other
Leak detection
Nuclear fission – power and atomic bomb