1. Organic Chemistry
Chapter 22

2. 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

3. 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

4. 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

5. 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

6. Alkanes
General formula – CnH2n+2
Examples
Methane
Butane

7. Naming Prefixes 1. Name the longest chain first 1 C - meth 2 C - eth
3 C - prop
4 C – but
** If there are two longest chains of the same length give priority to the one with the most branching
5 C – pent
6 C - hex
7 C - hept
8 C -oct

8. Alkanes Carbon compound + oxygen  carbon dioxide + water
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

9. Naming Single bonds – -ane Double bonds – -ene Triple bonds – -yne
2. Identify the type of bonding that is present
Single bonds – -ane
Double bonds – -ene
Triple bonds – -yne

10. Naming Alkane – -e Alcohol – -ol
3. Identify any functional groups
Alkane – -e
Alcohol – -ol
Halogenalkane – chloro-, bromo- or iodo-
Aldehyde – -al
Ketone – -one

11. Naming Carboxylic acid – -oic acid Ester – -oate Amide – -amide
3. Identify any functional groups
Carboxylic acid – -oic acid
Ester – -oate
Amide – -amide
Amine – -amine

12. Naming Alkanes with ring-like structures
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

13. Naming Called arenas Based off of benzene rings
**One more side note – aromatic compounds
Called arenas
Based off of benzene rings
Use the smallest numbers possible

14. Nomenclature Practice
Name this compound
CH3
H3C CH3 Cl
H3C

15. 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.

16. Alkenes and Alkynes Alkenes – have at least one double bond
Alkynes – have at least one triple bond
Both are said to be unsaturated

17. Aromatic Hydrocarbons
Cyclic unsaturated hydrocarbons with delocalized electrons
The simplest aromatic hydrocarbon is benzene (C6H6)
OR…

18. 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

19. Alcohols Change ending of the main chain to -ol General formula – R-OH
Naming
Change ending of the main chain to -ol

20. Amines General formula – RNH2 Naming
Add amino- to the beginning or –amine to the ending of the name

21. Aldehydes General formula O R – C - H Naming
Change the ending of the main chain to -al

22. Ketones General formula O R – C - R Naming
Change the ending of the main chain to -one

23. Carboxylic acids General formula O Naming R – C - OH
Change the ending of the main chain to –oic acid

24. Esters General formula O R – C – O – R’ Naming Name the R’ chain first
Name the R chain and change the ending to -oate

25. Amides
General formula O
R – C – NH2
Naming
Change ending to -amide

26. Ethers General formula R – O – R’ Naming
Name both R groups as side chains and add ether to the end

27. Isomers Example – pentane Carboxylic acid and ester
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

28. Electrochemistry
Chapter 17

29. 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

30. Electrochemistry Terminology
Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e-
Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2Cl-
LEO says GIR
Loss of electrons = oxidation
Gain of electrons = reduction

31. Electrochemistry Terminology
Oxidizing agent
The substance that is reduced is the oxidizing agent
Reducing agent
The substance that is oxidized is the reducing agent

32. Redox reactions 0 0 +1 -1 2Na + Cl2  2NaCl
2Na + Cl2  2NaCl
Each sodium atom loses one electron:
Na  Na+ + e-
sodium is oxidized
Each chlorine atom gains one electron:
Cl + e-  Cl-
Chlorine is reduced

33. 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

34. Balancing a Redox reaction in Acidic Solutions
Write separate equations for the oxidation and reduction half-reactions.
For each half reaction
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.

35. 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) + 12e-
12e- + 28H+(aq) + 2Cr2O72-(aq)  4Cr3+(aq) + 14H2O(l)
16H+(aq) + C2H5OH(aq) + 2Cr2O72-(aq)  2CO2(g) + 4Cr3+(aq) + 11H2O(l)

36. Balancing a Redox Reaction in Basic Solution
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

37. Example – Basic Solution
Ag(s) + CN-(aq) + O2(g)  Ag(CN)2-(aq)
4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 4e-
4e- + 4H+ + O2(g)  2H2O(l)
4H2O(l) + O2(g) + 4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 2H2O(l) + 4OH-
2H2O(l) + O2(g) + 4Ag(s) + 8CN-(aq)  4Ag(CN)2-(aq) + 4OH-
Cu + Cr2O7-2  Cu+2 + Cr+3
3Cu  3Cu+2 + 6e-
6e- + 14H+ + Cr2O7-2  2Cr+3 + 7H2O
14H+ + 3Cu + Cr2O7-2  2Cr+3 + 7H2O + 3Cu+2
14H2O + 3Cu + Cr2O7-2  2Cr+3 + 7H2O + 3Cu OH -

38. Electrode Potentials and Half Cells
When a metal comes into contact with a solution containing its own ions, an equilibrium is set up
Mx+(aq) + xe-  M(s)
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
Size and sign of the potential depends on the ability to lose or gain electrons
The system is a half cell

39. 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.

40. 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

41. Line Notation Zn(s) | Zn2+(aq)|| Cu2+(aq) | Cu(s)
An abbreviated representation of an electrochemical cell
Zn(s) | Zn2+(aq)|| Cu2+(aq) | Cu(s)
Anode Anode Cathode Cathode
material solution solution material

42. 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

43. Zn-Cu Galvanic Cell
The less positive, or more negative reduction potential becomes the oxidation…
Zn  Zn2+ + 2e E = +0.76V
Cu2+ + 2e-  Cu E = +0.34V
Zn + Cu2+  Zn2+ + Cu E0 = V

44. 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)

45. Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
1 V = 1 J/coulomb
Zn + Cu2+  Zn2+ + Cu E0 = V
G0 = -(2mol e-)(96485 coulombs)( J )
mol e coulomb
= J = -212 kJ

46. Gibb’s Free Energy and EcellK Eo
DGo
Conclusion
> 1
Positive
Negative
Spontaneous cell reaction
= 1
At equilibrium
< 1
Non-spontaneous cell reaction; the reaction is spontaneous in the reverse direction

47. The Nernst Equation E = Eo – RT ln(Q) nF
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 e-
Eo = the voltage at standard conditions
Q = the reaction quotient

48. The Nernst Equation Example
At 25˚C, the middle term can be simplified to
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 – log (5.0 M) = V (0.3 M)

49. Calculating an Equilibrium Constant from a Cell Potential
Zn + Cu2+  Zn2+ + Cu E0 = V
0 volts = 1.10 – log(K) = (1.10)(2) = log(K)
= 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 – log(K) n

50. 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

51. 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)

52. Concentration Cell
Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).
Zn2+ (1.0M)  Zn2+ (0.10M)
E = Eo – log(Q) n
Eo = 0.0 volts n = Q = (0.10)/(1.0)
E = 0.0 – log(0.10) = volts

53. Electrolysis Electrolytic processes are NOT spontaneous. They have:
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)

54. Electroplating of Silver
Anode reaction:
Ag  Ag+ + e-
Cathode 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

55. 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 C
= 2.2 x 102s

56. The Nucleus: A Chemist’s View
Chapter 18

57. U Nuclear Symbols Mass number, (A) (p+ + no) Element symbol
Atomic number, (Z)
(number of p+) U
235 92

58. Radioactivity
Spontaneous decay of certain atoms with the evolution of a, b and g particles
Comes from the nucleus

59. Radioactivity Alpha (α) Beta (β) Gamma (γ) Nature Charge +2 -1 Mass
Helium Nucleus
Negative particle with no mass
High energy & frequency radiation
Charge +2 -1
Mass
4 amu
1/1840 amu
Movement in electronic field
To negative
To positive
None
Penetrating Power
Weak
Intermediate
Strong

60. Balancing Nuclear Equations
SAreactants = SAproducts
= (1)
= (0)
SZreactants = SZproducts
n 
U +
Ba +
Kr + 3n

61. Balancing Nuclear Equations #2 2.
Ra 
a + ?
U + n 
1 + 2n + ?

62. 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

63. Beta Decay Converts a neutron into a proton.
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.

64. 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

65. 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

66. 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

67. 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
N k k
N = number of nuclides remaining at time t
N0 = number of nuclides present initially
k = rate constant
t = elapsed time

68. Half Life: 1/2n Examples: 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 Another sample of carbon gave a count rate of Calculate the age of the second sample.

69. 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

70. 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: DE = Dmc2
Dm = mass defect
DE = change in energy
c = speed of light
Because c2 is so large, even small amounts of mass are converted to enormous amount of energy.

71. Nuclear Fission and Fusion
Fusion: Combining two light nuclei to form a heavier, more stable nucleus.
Fission: Splitting a heavy nucleus into two nuclei with smaller mass numbers.
He + H  He + e
n U  Ba + Kr + 3 n

72. Fission Processes
A self-sustaining fission process is called a chain reaction.
Neutrons
causing
Event fission Result
subcritical < reaction stops
Critical = sustained reaction
Supercritical > violent explosion

73. A Fission Reaction

74. 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