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Organic 331
Basic information for all students taking Organic Chemistry
These notes are keyed to the textbook. This will let you know exactly where you should be in
the textbook.
These notes contain many examples to give you additional practice. Some of these are for
additional practice and not covered in detail in lecture.
Some sections say “Review.” These sections will significantly help you to understand the
material. You should plan on spending more time studying these sections.
Some sections say “Read.” These sections supply important background material to help you
understand classroom/test material. You will not need to study these sections extensively.
You should pay attention to any “Thought questions,” “Exam questions,” and “asides.” You will
not see the exact same question on an exam, but an understanding of these will help you to
prepare. Some of the Thought questions an some of the Exam questions have answers. Try to
answer the question before looking at the answer.
Each exam concentrates on the material covered since the last exam. However, every exam
throughout Chemistry 331 and Chemistry 332, in reality, is comprehensive. Adjust your study
habits accordingly.
Much of Chapter 1 is a review of Chemistry 133/134. You may wish to review sections in the
Chemistry 133/134 sections of the website.
Chapter 1 THE BASICS: Bonding and Molecular Structure
1.1 Organic Chemistry and Life
Organic Chemistry
Vital Force Theory (Vitalism)
F. Wöhler (1828)
1.2 The Structural Theory of Organic Chemistry
Read – be very careful
1.3 Isomers: The Importance of Structural Formulas
Read – Constitutional isomers are a common source of error
1.4 Chemical Bonds: The Octet Rule
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1.4A Chemical Bonding
Electrovalent or ionic
Ionic compounds have
1.4B
Covalent – two nonmetals
1. Pure covalent
2. Coordinate covalent
3. Polar covalent
Electronegativity –
Polar molecules
1.5 Writing Lewis Structures
Note – in organic compounds, atoms other than hydrogen strongly favor the octet rule.
It is important to keep track of the valence electrons.
Valence electrons = those electrons in the outer shell
Example
H2S
There are methods/ formulas to allow a prediction of the number of bonds.
These only work if all the atoms obey the octet rule.
They are useful/trivial for organic molecules.
Example
CO2
Example NH4+
Example NO3Thought questions: Answer the following questions concerning H2SO4
What will be the central atom?
Short answer = S
Where will the oxygen atoms be?
Short answer = around the S
Where will the hydrogen atoms be? Short answer = attached to O’s
Answer
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There are 32 total electrons.
Initially:
This accounts for the 6 bonds and for 12 of the 32 electrons (leaving 20).
(The hydrogen atoms can be on any 2 oxygen atoms.)
The S has an octet.
The oxygen atoms with hydrogen atoms attached have 4 electrons and need 4 more.
(8 total)
The oxygen atoms without hydrogen atoms attached have 2 electrons and need 6 more.
(12 total)
8 + 12 = 20
Example CaSO4
1.6 Exceptions to the Octet Rule
Other than H, these are not common in basic organic chemistry.
Boron is a common example, and we will see boron again.
1.7 Formal Charge
Example
Cl
Example
Br+ Bromonium ion
Example
CH3+
Example
CCl3-
Example
CH3
Example
CCl2 Dichlorocarbene
Example
HNO3 nitric acid
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Example
HCN
1.8 Resonance Theory
This is very important to both Chemistry 331 and 332
Resonance Structures
SO3
sulfur trioxide (not sulfite ion)
Rules for resonance
1. Resonance structures are a way of understanding stability; they do not really
exist.
2. When writing the resonance structures, it is only permissible to move π
electrons (think of the π electrons as a swinging gate)
3. All of the structures must have reasonable Lewis structures. This includes the
same overall charge
4. There should be no change in the number of unpaired electrons
5. All atoms that are part of the delocalized π-system must coplanar
6. The presence of resonance leads to resonance stabilization, which means the
species is more stable than any of the contributing structures are.
7. Equivalent resonance structures are equally important to the overall structure
8. The more stable the resonance structure, the more important (more it
contributes) to the hybrid.
8a. More bonds = more stable – the “best” structures have the most bonds
8b. Structures with all atoms having complete valence shells are very
important. (In most cases, this is an octet.)
8c. Structures that place unlike charges close together or like charges are
apart are more stable. Structures doing the opposite are less important,
they are not necessarily impossible.
8d. A negative charge on an electronegative atom is better than a negative
charge on a less electronegative atom.
1.9 Quantum Mechanics and Atomic Structure
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1.10 Atomic Orbitals and Electron Configuration
Atomic Structure
Quantum Mechanical Terms
Orbital =
Shell =
Spectroscopic notation = electron configuration
The s orbitals are spherical and there is one per shell
The p orbitals have a “dumbbell” shape and there are three per shell.
Aufbau Principle
1s
2s
3s
4s
5s
6s
7s
C = 1s22s22p2
3d
4d 4f
5d 5f
6d 6f
(2s22p2 are the valence electrons)
Pauli Exclusion Principle =
Hund’s Rule =
1.11 Molecular Orbitals
Molecular Orbitals and Covalent Bonding
H• + •H Æ H-H + 435 kJ/mole
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2p
3p
4p
5p
6p
7p
6
Homolytically =
Atomic orbitals
Molecular orbital
Molecular Orbital Energy Diagram
Bonding molecular orbital = σ = ψ1s + ψ1s
Antibonding molecular orbital = σ* = ψ1s – ψ1s
1.12 The Structure of Methane and Ethane: sp3 Hybridization
Orbital Hybridization
However, carbon is tetravalent (CH4) with 4 equivalent bonds.
ψ2s + ψ2p + ψ2p + ψ2p = 4
All four hybrid orbitals are equal and at 109°28’
The 109° 28’ angle is the tetrahedral angle.
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Indicating the three dimensional structure
Or
In plane
Front
Back
Thought question: Why is the dipole moment of CCl4 = 0 (μ = 0)?
Answer
The molecule does not have a positive end and a negative end. It has a positive center and a
negative exterior.
Thought question: Use a picture to explain why the dipole moment of CH2Cl2 ≠ 0 (μ ≠ 0)?
Answer
There is a net bond moment (a vector quantity).
Other examples with a “tetrahedral” geometry
NH3
H2O
Thought question: What is the geometry of CH3-?
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Thought question: What is the geometry of CH3OH?
Answer
H
O
C
H
H
H
The geometry around the oxygen atom is similar to H2O.
The geometry around the carbon atom is similar to CH4.
1.13 The Structure of Ethene (Ethylene): sp2 Hybridization
The carbon atoms are sp2 hybridized
Molecular orbitals
Note the difference between the σ and σ* (seen earlier) and the π and the π* seen here.
Note: It is possible to move an electron from the highest occupied molecular orbital (HOMO) to
the lowest unoccupied molecular orbital (LUMO) by adding sufficient energy. This energy is
normally in the ultraviolet portion of the spectrum.
Bond strength
σ
kJ/mole
π__________ kJ/mole
kJ/mole
Rotation is restricted – very important
BF3 boron trifluoride
BeCl2
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1.14 The Structure of Ethyne (Acetylene): sp Hybridization
H-C≡C-H sp hybridized and linear
End on view:
Why is the hydrogen acidic?
1.15 A Summary of Important Concepts that Come from Quantum Mechanics
Review
Know
spx hybridization
sp
sp2
sp3
s character
50 %
33 %
25 %
p character
50 %
67 %
75 %
1.16 Molecular Geometry: The Valence Shell Electron Pair Repulsion Model
Review (Some is above)
Know
# of attached atoms
+
# of lone pairs
2
3
4
Hybridization
sp
sp2
sp3
Bond angles
(ideal)
180°
120°
109.5°
1.17 Representation of Structural Formulas
Read – Will cover later
1.18 Applications of Basic Principles
Review – underlying principle of many things we will see later
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Molecular
geometry
Linear
Trigonal planar
Tetrahedral
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Chapter 2 REPRESENTATIVE CARBON COMPOUNDS: Functional Groups,
Intermolecular Forces, and Infrared (IR) Spectroscopy
There will some sections from Chapter 3 included here.
2.1 Carbon-Carbon Covalent Bonds
Representative Carbon compounds
There are a large number of carbon compounds for two reasons.
2.2 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds
2.2A
Alkanes
Draw the structures
Condensed
Line
Structural (extended)
Alkanes, the paraffin hydrocarbons
All alkanes are saturated hydrocarbons (saturated in H)
1.
2.
3.
Methane = CH4
2.2B
Ethane = CH4 + CH2 Propane = CH4 + CH2 + CH2
Alkenes
These are unsaturated (have less than the maximum number of hydrogen atoms).
Ethene
When a carbon atom bonds to three other atoms, the hybridization is sp2.
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The C=C is the focus of chemical reactivity. It is a functional group.
2.2C
Alkynes
These are also unsaturated.
Ethyne (acetylene) is the simplest, and then comes propyne.
Alkanes, Alkenes, and Alkynes are aliphatic hydrocarbons.
Alicyclic (aliphatic + circle)
Cyclobutane
2.2D
Aromatic hydrocarbons
The most common is benzene, C6H6.
Note: Do not make the mistake of treating aromatic hydrocarbons like alkenes
A benzene ring attached to something is a phenyl group.
Phenyl group = C6H5- = Ph- = ϕGeneral
CH4 = methane = alkane
CH3- = methyl group = alkyl group
All alkyl groups = R
So an alkane = R-H
CH3CH2CH3CH2CH2-
2.3 Polar Covalent Bonds
Covered earlier – will see repeatedly
2.4 Polar and Nonpolar Molecules
Will be covered repeatedly – review now and expect to see it over and over
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2.5 Functional Groups
Most organic reactions will involve functional groups containing either multiple bonds or
heteroatoms.
This is VERY important.
In general, R =
In general, Ar =
Be careful of the benzyl group (Ar-CH2-) = Bn
2.6 Alkyl Halides or Haloalkanes
Alkyl halides =
CH3Br
CH3CH2Cl
2.7 Alcohols
Alcohols
R-OH
(-OH = hydroxyl)
CH3OH = wood alcohol
CH3CH2OH = grain alcohol
1° alcohol
= rubbing alcohol
2° alcohol
3° alcohol
Propyl alcohol and isopropyl alcohol are structural and positional isomers.
Isomers (iso = same + meros = parts) = same composition but different structures.
CH3CH2CH2OH = C3H8O as is
Thought question: Is HOCH2CH2CH3 another isomer?
2.8 Ethers
Ethers are structural and functional isomers of alcohols.
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Ethers have C-O-C
2.9 Amines
Amines contain C, H, and N
Amines are related to ammonia like alcohols are related to water.
Beginning with ammonia (NH3)
Primary (1°) amine Æ
Secondary (2°) amine Æ
Tertiary (3°) amine Æ
This is not the same system as for carbon atoms.
2.10 Aldehydes and Ketones
Aldehydes and Ketones have a carbonyl group.
Aldehydes have at least one hydrogen atom attached to the carbonyl group.
Ketones have two carbon atoms attached to the carbonyl group (no hydrogen atoms).
Thought question: Is the following compound and aldehyde or a ketone?
Thought question: Is the following compound and aldehyde or a ketone?
Notice this structure is the condensed form of
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Notice that every carbon atom has four bonds.
This requires two hydrogen atoms on the carbon atoms of the right-hand ring.
2.11 Carboxylic Acids, Esters, and Amides
Carboxylic acids have a carboxyl functional group (carbonyl + hydroxyl)
Alternatives: RCO2H and RCOOH
Derivatives (replace the –OH)
Esters use –OR and amides use –NH2
Recall Acids/Bases
Thought question: What is the conjugate acid of HSO4-?
Thought question: What is the conjugate base of HSO4-?
Other examples:
The strongest acid that can exist in water is H3O+.
The strongest base that can exist in water is OH-.
Acids react with bases:
NaOH(aq) + HBr(aq) Æ NaBr(aq) + H2O(l)
Thought question: What is the actual reaction?
Answer
The Na+ and the Br- are spectator ions.
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Spectator ions do not participate in a chemical reaction; they are present to maintain electrical
neutrality.
Acetic acid is a weak acid
Keq =
Ka =
= 1.8 x 10-5
Sometimes it is easier to use pKa (-log Ka)
Amines are bases.
See Table 3.1 p101, in the textbook as a reference. Do not memorize it.
If the basicity order is:
CH3- > NH2- > OH- > Cl-
Then the acidity order is:
CH4 < NH3 < H2O < HCl
Trends:
3.2B
Lewis acids and bases
A Lewis acid is
A Lewis base is
Example
B (BF3) is electron loving (= electrophile)
N (NH3) is electron rich and is attracted to electron deficient materials
(= nucleophile)
3.2C
Review and keep this in mind
2.12 Nitriles
Not too common = -CN
2.13 Summary of Important Families of Organic Compounds
KNOW Table 2.3, pages 68-69
2.14 Physical Properties and Molecular Structure
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Characteristics of the different states:
Solid
Liquid
Gas
Intermolecular forces (between molecules)
2.14A
1. Ion-Ion (not really molecules)
2.14B
2. Dipole-Dipole
2.14C
3. Hydrogen bonding
2.14D
4. van der Waals forces (= London forces = dispersion forces)
Examples:
H2O
CH3CH2OH
C5H12 (pentane)
CH3-O-CH3
MW
18
46
72
46
BP (Do not memorize numbers.)
100°C
78°C
35°C
-25°C
H2O
CH3CH2OH = EtOH
C5H12 (pentane)
CH3-O-CH3
2.14E Solubility
Like dissolves like
2.14F Guidelines for solubility in water
1.
2.
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(Fundamental rule of solubility)
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Thought question: The organic acid CH3(CH2)6CO2H is insoluble in water, but soluble in
aqueous base. Why?
Answer:
The acid reacts with the base to become ionic and soluble.
Water and alcohols are polar protic solvents. (They have ionizable hydrogen atoms.)
Polar aprotic solvents are also important.
DMSO, DMF, and HMPT (HMPA)
2.15 Summary of Attractive Electric Forces
Review – you will see this multiple times
2.16 Infrared Spectroscopy: An Instrumental Method for Detecting Functional Groups
Read for now.
This will be covered in more detail later.
Begin learning Table 2.7 on page 79 in the textbook. For Exam 1, know those that are strong.
The exact numbers are too important. For example, all C=O are between 1630 and 1780 cm-1.
(Look at spectra. Do not let IR spectra overwhelm you. Try to assign some of the peaks. You
will not be able to assign all the bands; so do not try)
2.17 Applications of Basic Principles
Read
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Chapter 4 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND
CYCLOALKANES
4.1 Introduction to Alkanes and Cycloalkanes
Read
4.2 Shapes of Alkanes
Review
4.3 IUPAC Nomenclature of Alkanes, Alkyl Halides, and Alcohols
Know the names of the first 10.
Structural isomerism begins with butane. There are two possibilities butane (or n-butane)
and isobutene.
(n-butane, the “n” means normal (not branched), “straight” chained)
There are three isomers of pentane (n-pentane, isopentane, and neopentane).
Extended structural formulas
Show all atoms and bonds
Condensed formula
Carbon skeleton
Never use on an
Exam
Useful in finding isomers
IUPAC system (International Union of Pure and Applied Chemistry)
This is a systematic method.
Example What are the isomers of C6H14?
Note: Be careful, this is still hexane:
Example
CH3
CH3 C
CH2 CH2 CH3
CH3
1
2
3
First –
Second –
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Thought question: Draw and name the nine isomers of heptane.
Example
Name the following compound:
“Reference to system”
8 primary (1°) C’s Æ 24 primary H’s (-CH3)
4 secondary (2°) C’s Æ 8 secondary H’s (-CH2)
4 tertiary (3°) C’s Æ 4 tertiary H’s (-CH)
1 quaternary (4°) C’s Æ 0 quarternary H’s (-C)
Group Names
Alkyl groups CnH2n+1
NEVER part of the parent chain
NEVER alone
Note
4 Butyls (Know)
(Only 2 C4H10 isomers – skeletons)
sec-butyl because is on a secondary carbon.
Do not use sec- for anything else.
4.3E
RX Æ Alkyl halides Æ Common names
Haloalkane Æ IUPAC
Know common names for 4 butyls (see above) and 4 pentyls
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The attachment for the iso- is on the end furthest from the branch.
Note: “amyl-“ may replace “pentyl-“
4.3F
Review, we will cover alcohols in more detail later
4.4 Nomenclature of Cycloalkanes
4.4A
Cycloalkanes CnH2n
CH3
CH3
4.4B
Fused and bridged ring systems
4.5 Nomenclature of Alkenes and Cycloalkenes
Modify the rules for alkanes
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IUPAC
ethene
propene
1-butene
2-butene
methylpropene
CH2=CH2
CH3-CH=CH2
CH3-CH2-CH=CH2
CH3-CH=CH-CH3
(CH3)2C=CH2
1
2
CH3 CH
3
C
1
α and β give the location
of the C=C
CH2 CH2 CH3
4
CH3 C
CH3 C
Common
ethylene
propylene
α-butylene
β-butylene
isobutylene
5
H
CH3
6
CH3
Longest chain containing C=C
Numbering to give lowest number to C=C
In cyclic structures the double bond is between C #1 and C #2
The substituents determine the direction of numbering.
5 3
2
1
1
1
4
4
Cl
5
1
CH3
3
Cl
2
Correct numbers
Incorrect numbers
Geometric isomers (different structures around the double bonds)
CH3-CH=CH-CH3
On an exam if a structure is drawn to indicate the geometry – then it must be named accordingly.
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The presence of two identical groups (-CH3) on one carbon means there is no cis or trans.
For cis and trans isomers to exist; the two groups on each side of the double bond must be
different.
A ≠ B and D ≠ E
trans
4.6 Nomenclature of Alkynes
Nomenclature – 2 systems
IUPAC
H-C≡C-H
ethyne or acetylene
H-C≡C-CH3
propyne
CH3-C≡C-CH3
2-butyne
H-C≡C-CH2CH3
1-butyne
This is no isobutyne or cis/trans
Common
acetylene
methyl acetylene
dimethyl acetylene
ethyl acetylene
The C≡C takes priority in numbering
The triple bond is on the end = terminal acetylene
The terminal acetylene has an acetylenic hydrogen atom (acidic).
4.7 Physical Properties of Alkanes and Cycloalkanes
1. Boiling points and melting points increase with molecular weights.
Note Impurities will increase the boiling point and decrease the melting point.
Note Lowing the pressure will lower the boiling point, but have minimal effect
on the melting point.
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2. The more symmetrical (compact) the structure; the lower the surface area and the
lower the boiling point.
3. The density (specific gravity) goes up to ≈0.8 (floats on water.
4. Insoluble in water
5. Viscosity (resistance to flow) increases with chain length
4.8 Sigma Bonds and Bond Rotation
Ethane has free rotation about the C-C (σ bond).
An easier method is to look down the C-C bond as a Newman Projection.
Energy
E
Rotation angle
ASIDE – Other energy concerns
Energy is the ability or capacity to do work.
Kinetic energy is energy of motion. It is possible to interconvert these
Potential energy is stored energy.
Bond energies are potential energies.
H• + •H Æ H-H
ΔH° = -435kJ/mol (Enthalpy change)
– = exothermic
The products are more stable than the reactants.
+ = endothermic
The products are less stable than the reactants.
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Bond energies are homolytic – the bond breaks down the middle with each atom getting one-half
of the bonding electrons.
END ASIDE
4.9 Conformational Analysis of Butane
Example Ethane has two conformations (staggered and eclipsed).
Example C4H10
E
Rotation Angle
Thought question: Why is there a peak between anti and gauche?
See Figure 4.8, page 152 in the textbook
Torsional strain = barrier to rotation
Van der Waals Repulsion = electron cloud repulsion from interpenetration
This occurs when the distance between the atoms is less than the sum of the van
der Waals radii. This creates a barrier to rotation.
4.10 The Relative Stabilities of Cycloalkanes: Ring Strain
4.11 The Origin of Ring Strain in Cyclopropane and Cyclobutane: Angle Strain and
Torsional Strain
Bayer strain theory (1890) –
Example
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Observations:
CH4 + 2 O2 Æ CO2 + 2 H2O
ΔH = -803 kJ/mol
(Heat of combustion)
Empirical data
See Table 4.5, page 154 in the textbook
Approximate heats of combustion in kJ/CH2
697
686
664
659
662
4.12 Conformations of Cyclohexane
4.13 Substituted Cyclohexanes: Axial and Equatorial Hydrogen Atoms
Further
Too close
H
H
H
C
H
H
H
H
C
H
1,3-diaxial interaction
[unfavorable]
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More stable
H
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4.14 Disubstituted Cycloalkanes: Cis-Trans Isomerism
Note: It is easier to see if drawn as if it were flat.
(Recall – cis = same side and trans = opposite side)
It is harder to see if drawn in the puckered form, but it is more realistic.
4.15 Bicyclic and Polycyclic Alkanes
Example Decalin
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4.16 Chemical Reactions of Alkanes
In general, these are not very reactive.
This means reaction conditions will be relatively “extreme.”
Like virtually all organic compounds, they will combust.
Note: This is a commonly forgotten reaction
4.17 Synthesis of Alkanes and Cycloalkanes
Synthesis of alkanes and cycloalkanes
Hydrogenation of alkenes
CH2=CH2 + H2
C
CH3CH3
Catalyst = Ni or Pt or Pd
Exam question:
What is the product of the following reaction?
Exam question:
What reactant might you use in the following reaction?
Hydrogenation of Alkenes
1. Hydrogenation (addition of hydrogen) or reduction
a. Nonselective
b. Selective (C≡C to C=C)
1.
2.
3.
Note: All the above are cis
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Note: A common mistake on exams:
4.
2. Zinc reduction of RX
On an exam, this might appear as:
Answer:
3. Corey-House Synthesis
Exam question:
atoms or less?
Answer
How do you prepare 2-methylbutane from alkyl halides of three carbon
Steps to get to the answer:
1. Draw the product
2. The pieces must be three carbon atoms or less. So divide the molecule into fragments.
For example:
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The isopropyl group is 2°, so it cannot be R’; therefore it must be R.
The ethyl group is 1°, so it can be either R or R’. (It must be R’ in this case.)
3. The reaction (names are unnecessary)
Any halide will work.
4.18 Structural Information from Molecular Formulas and the Index of Hydrogen
Deficiency
Index of Hydrogen Deficiency (IHD)
Alternate (important names) = “degree of unsaturation” = “number of doublebond equivalencies”
This works for compounds containing C, H, X, N, O, or S
IHD = #C - #H/2 - #X/2 + #N/2 + 1
Alternative equation
Degree of unsaturation =
n = number of carbon atoms
x = number of hydrogen atoms or other univalent atom (F, Cl, Br, or I)
O and S have no effect
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If N is present, replace each N by one C and one H
Exam question:
Draw the isomers of C4H8.
Answer:
IHD = 4 C – 8 H/2 + 1 = 1 = 1 ring or 1 C=C
CH3
CH3CH=CHCH3
CH2 C
CH3
CH3
etc.
C6H10 Æ IHD = 2 = 6 C – 10 H/2 + 1
IHD = 2 this gives the following choices:
1. 2 rings or bicyclic
2. 2 C=C
3. 1 C=C plus 1 ring
4. 1 C≡C
To decide which is the correct choice requires more information
For example:
The compound reacts with H2 in the presence of a catalyst to produce C6H12.
This reaction will eliminate all double and triple bonds.
The product, C6H12, has an IHD =1 Æ a ring
Therefore, the original compound contained a ring and a double bond.
C8H18O Æ IHD = 8 C – 18 H/2 + 1 = 0
There are no rings or double bonds present. The compound must be either an
ether or an alcohol.
4.19 13C NMR Spectroscopy-A Practical Introduction
Covered later – see Chapter 9
4.20 Applications of Basic Principles
Review
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Chapter 10 RADICAL REACTIONS
10.1 Introduction
10.2 Homolytic Bond Dissociation Energies
Bond dissociation energy (DE) –
For a reaction:
ΔH =
Example
It is possible to do this stepwise in a mechanism
10.3 The Reactions of Alkanes with Halogens
There are not very many reactions.
Halogenation
∆
∆
CH4 + Cl2
CH3Cl + Cl2
CH2Cl2 + Cl2
Each step produces an HCl
∆
CHCl3 + Cl2
The general reaction is:
R-H + X2
∆
RX + HX
This is a substitution reaction.
Reactivity:
Thought question:
CH4 + Br2(xs)
∆
?
Answer:
Excess (xs) Br2 gives a continued reaction to CBr4
Exam question:
What is a reaction where the major product is CH3Br?
∆
Answer
CH4(xs) + Br2
CH3Br
The Br2 is “limited” (no special symbol) so CH4(xs)
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∆
CCl4
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10.4 Chlorination of Methane: Mechanism of Reaction
Reaction mechanism ≡ reaction pathway
For CH4 + Cl2 there are three observations:
1. Heat or light is required
2. hν gives several CH3Cl molecules per photon
3. O2 slows the reaction
A mechanism must account for these
Initiation
Propagation
Termination – there are two free radicals combining
Note all the termination steps involve a combination of two free radicals.
This is a substitution process – the full name is: A free radical substitution process or A
free radical chain reaction.
O2 slows the reaction because it causes another termination
The inhibition of a process by O2 is indicative of a free radical process.
10.5 Chlorination of Methane: Energy Changes
For chlorination, ΔH ≈ -106 kJ/mol (more exothermic than bromination)
Energy of activation ≡ Ea ≡
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General rules (there are some exceptions):
If ΔH is low and either low positive or negative, then Ea is small
Ea > ΔH
The measurement of Ea is empirical
Thought question: What is the Ea for F2?
10.6 Halogenation of Higher Alkanes
For the higher alkanes, it is necessary to expand on the CH4 mechanism.
Initiation
Propagation
Termination
Note: ΔH for the slow step is exothermic for chlorine and fluorine (chlorine plus methane
is an exception), but for bromine, it is endothermic.
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Example
Bromine is less reactive – more selective
The product distribution depends on two factors:
1.
2.
1. Ease of formation [stability] of alkyl radical
RH + X• Æ R• + HX
Note: this is the order of stability of free radicals
Stability continuum Æ
Alkyl radicals 3°
2°
1°
Me
Most stable Æ Least stable
Least reactive Æ Most reactive
Lowest energy Æ Highest energy
2. Probability factor (statistics)
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The reactivity factor is relatively constant
For Cl2
5.0:3.8:1 for 3°:2°:1°
The relative amount of an isomer is equal to
(Probability Factor) x (Reactivity Factor)
Example
Relative amounts:
3°:
1°:
Total =
Distribution:
3°:
1°:
Example
Determine the relative amounts of 1-chloro, 2-chloro, and 3-chloro products resulting
from the following reaction:
CH3-CH2-CH2-CH2-CH3
Answer:
,C
?
1-chloro ≈
2-chloro ≈
3-chloro ≈
Bromination has a different set of reactivity factors
1600:82:1 for 3°:2°:1°
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Bromine is less reactive, so it is more selective.
See Figure 6.10, p243 in the textbook
Hammond-Leffler Postulate:
This means:
Br-H••••R
Cl••••H-R
R-H
+ Cl•
*
E
*
E
R-H
+ Br•
R• + HBr
Reaction Coordinate
Thought question:
R• + HCl
Reaction Coordinate
What reactant would be the best to use for the following reaction?
Thought question: Why does the monochlorination of either ethane or cyclopentane only give
one product?
10.7 The Geometry of Alkyl Radicals
Read
10.8 Reactions that Generate Tetrahedral Chirality Centers
Review
10.9 Radical Addition to Alkenes: The Anti-Markovnikov Addition of Hydrogen Bromide
Free radical addition of HBr (NOT HCl or HI)
An alternative to polar molecule addition
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Initiation:
Propagation:
Termination:
Any two free radicals
Key aspects to HBr/Peroxide addition to alkenes
Intermediates
All intermediates are radicals resulting from hemolytic bond
cleavage
Rearrangement
This is impossible, because there are no carbocations, and radicals
do not rearrange
Stereochemistry
The planar carbon radical can bond to a hydrogen attacking from
either side of the unhybridized p orbital
Note
Only HBr behaves this way. HCl and HI do not show the peroxide
effect.
10.10 Radical Polymerization of Alkenes: Chain-Growth Polymers
SPECIAL TOPIC A: CHAIN-GROWTH POLYMERS
Review both this section and Special Topic A.
Note: Draw at least 3 reacting molecules.
Example
Addition of alkenes:
Ethylene
Polyethylene
Note: This is free radical polymerization.
A conjugated diene will undergo free radical polymerization to form a polymer
containing double bonds.
Commonly used to produce synthetic rubber
1,3-butadiene
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isoprene
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Alkenes containing electron withdrawing groups, such as nitrile, react via anionic
polymerization.
Note: Anionic polymerization occurs when a strong base reacts with an alkene
containing an electron withdrawing group (such as –CN).
Review this section and work Review Problems 10.21 and 10.22.
10.11 Other Important Radical Reactions
In general, the cycloalkanes are like other alkanes.
The only cycloalkane that is different is cyclopropane.
Exam question:
What is the product of this reaction?
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Chapter 3 AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR
MECHANISMS: Acids and Bases
3.1 Reactions and Their Mechanisms
Read over to get some idea of the goal we are seeking to achieve. This is a good overview.
Do not worry about specific reactions; just how to identify.
Keep this in mind for every reaction in Chem 331 and Chem 332.
Homolytic bond breaking
Heterolytic bond breaking
3.2 Acid-Base Reactions
Review – more detail below
3.3 Heterolysis of Bonds to Carbon: Carbocations and Carbanions
Read Covered elsewhere
Carbocation = C+ (= carbonium ion) (Electron deficient)
A carbon atom with three bonds
Carbanion = C-
May be a carbon atom with three bonds
3.4 The Use of Curved Arrows in Illustrating Reactions
Read Covered elsewhere
The arrow ALWAYS points towards the most positive center.
There MUST be one or two electrons at the tail of the arrow.
The electrons may be a single electron, a lone pair, or a covalent bond.
3.5 The Strength of Acids and Bases: Ka and pKa
All weak acids have an equilibrium like:
pKa = - log Ka
Larger Ka (or smaller pKa) = stronger acid.
Bases are the opposite of acids.
A weak acid has a strong conjugate base.
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Note: These are the general relationships for acid strength:
HI
HBr
HCl
HNO3
H2SO4
Strong
acids
O
R
O
S
OH
R
OH
carboxylic
acids
O
sulf onic
acids
sp hybridized
C-H bonds
alcohols
and
water
C
sp2 hybridized
C-H bonds
sp3 hybridized
C-H bonds
3.6 Predicting the Outcome of Acid-Base Reactions
A stronger acid will donate an H+ to a stronger base to form a weaker acid and a weaker base.
Consider the following reaction/equilibrium:
HCl + C2H3O2- ' HC2H3O2 + ClIf the conjugate acid or conjugate base is an ion, the species will be more soluble in water than
the unionized form.
This is a useful way of increasing the solubility of organic compounds.
3.7 The Relationship between Structure and Acidity
There is no simple way of predicting the strength of a binary acid.
For ternary acids, predictions will be limited to oxyacids.
H–O–Cl
H–O–Cl–O
Note: Remember that the base trend is the reverse of the acid trend.
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The greater the “s” character of a hybrid orbital, the greater the electronegativity of the atom.
The more electronegative, the more acidic
3.8 Energy Changes
Review
3.9 The Relationship between the Equilibrium Constant and the Standard Free-Energy
Change, ΔG°
Review
3.10 The Acidity of Carboxylic Acids
The most common organic acids are the carboxylic acids.
The presence of resonance (in the conjugate base) favors its formation.
Any inductive effect that removes electron density from the hydrogen atom will increase
the strength of the acid.
3.11 The Effect of the Solvent on Acidity
Read
3.12 Organic Compounds as Bases
The presence of a lone pair makes it possible for a substance to behave as a base.
3.13 A Mechanism for an Organic Reaction
You will see this elsewhere – multiple times
Note: This is very important – the concepts are here.
Do not worry about the specifics only the general concepts.
3.14 Acids and Bases in Nonaqueous Solutions
In aqueous solutions, the strongest acid is H+(aq) and the strongest base is OH-.
Stronger acids and bases may exist in other solvents.
3.15 Acid-Base Reactions and the Synthesis of Deuterium- and Tritium-Labeled
Compounds
The examples in this section give a good idea of what to expect
3.16 Applications of Basic Principles
Review
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Chapter 6 IONIC REACTIONS: Nucleophilic Substitution and Elimination Reactions
of Alkyl Halides
6.1 Organic Halides
Review structure and nomenclature of alkyl halides etc
6.2 Nucleophilic Substitution Reactions
Introduction to nucleophilic substitution and elimination reactions of alkyl halides
X may leave as X-.
6.3 Nucleophiles
A nucleophile (-) attacks an electrophile (+)
Definitions here – clarification coming
Pay attention to what makes a good nucleophile
6.4 Leaving Groups
Definitions here – clarification coming
Note: It is important to pay attention to how good a leaving group is
6.5 Kinetics of a Nucleophilic Substitution Reaction: An SN2 Reaction
6.6 A Mechanism for the SN2 Reaction
Reaction Types (SN2, SN1, E2, and E1)
1. SN2 – S (substitution) – N (nucleophilic) – 2 (bimolecular)
Example
Note – there is a conservation of charge
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In general,
Note: The inversion of the arrangement about the C.
Since the nucleophile is attacking on the side opposite the leaving group, this is a backside
attack.
There is a pentavalent carbon atom in the transition state. The R1, R2, and H are trigonal planar,
sp2. The Nu and L are sharing an unhybridized p-orbital.
The result is that the molecule turns “inside out,” this is a Walden inversion (of configuration).
Example
Using orbitals:
Reaction Kinetics
Halogens are good leaving groups – so a large number of nucleophiles (Lewis
bases) will work.
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Note: The bromide may still be present as the bromide ion. See NH3 reaction.
Thought questions: – Would NPh3 react better than NH3?
How would PPh3 compare in reactivity to NPh3? Note: P+ is more stable than N+.
6.7 Transition State Theory: Free-Energy Diagrams
Kinetics
1. Collision frequency –
2. Orientation factor –
3. Energy factor –
Fraction of
collisions
with a given
energy
Molecules with
suf f icient energy
Ea
E
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Example
Br2 + CH4
Cl2 + CH4
Increase T:
°C
°C
Rate = 1 = RBr
Ea ≈ 4 EaCl
Rate = 430000 RBr = RCl
Ea = 1 = EaCl
Note: the large change in rate gives relative change in the Ea
Fraction of
collisions
with a given
energy
T1
T2
Ea
E
T2 > T1
See figure 6.3, page 230 in the textbook
Note: Remember that the barrier does not move
In general, a 10° increase in temperature doubles the rate.
6.8 The Stereochemistry of SN2 Reactions
Key concepts of an SN2 reaction:
Reactivity of ROH 1° > 2° >> 3° (due to steric hindrance)
Rearrangement
This never happens because there is no carbocation present
Solvent affects
The presence of polar, aprotic solvents favor SN2
Stereochemistry
There is complete stereochemical inversion about the attacked
carbon
Kinetics
Rate = k [nucleophile] [electrophile]
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6.9 The Reaction of tert-Butyl Chloride with Hydroxide Ion: An SN1 Reaction
2. SN1 – S (substitution) – N (nucleophilic) – 1 (unimolecular)
6.10 A Mechanism for the SN1 Reaction
Example
Solvolysis – broken apart by the solvent
Hydrolysis – broken apart by water
Note The faster the rate of solvolysis the fast the reaction rate.
The mechanism:
The carbocation has a carbon atom with 3 σ bonds (sp2); it is a trigonal planar (flat) intermediate
Note: A combination of a positive ion and a neutral molecule, so the result must be positive
(determine from formal charges).
Energy Reaction Diagram
[ ]* – rate determining step (highest transition state)
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[Rδ+---Xδ-]*
6.11 Carbocations
A carbocation is a sp2 hybridized atom.
It is trigonal planar with an unhybridized p-orbital perpendicular to the plane.
Stability continuum Æ
Carbocations 3°
2°
1°
Me
Alkyl radicals 3°
2°
1°
Me
Carbanions Me
1°
2°
3°
Most stable Æ Least stable
Least reactive Æ Most reactive
Lowest energy Æ Highest energy
Note these do not depend on the “strength” of the other reactant.
6.12 The Stereochemistry of SN1 Reactions
Note: If it occurs at a chiral center, racemization (see later) will result.
6.13 Factors Affecting the Rates of SN1 and SN2 Reactions
1. Steric condition of the starting material (substrate)
See, Table 6.4, p 241 in the textbook
2. Nature of the nucleophile
a.
OH- vs H2O
EtO- vs EtOH
Example
EtBr + EtO-/EtOH
EtBr + EtOH
b.
°
f
EtO- > OH- >> RCO- >> EtOH > H2O
Strongest
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Weakest
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Base
c.
Base
I- > Br- > Cl- >> FWhy?
Example
Example
Example
Nu- + ROH Æ RNu + OH- (No Reaction)
d. Solvation effects
See Table 6.5, p 247 in the textbook
Solvation is a stabilization effect.
Relate this to:
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*
Hsolv
E
RX
+ Nu
Hsolv
RNu + X-
Reaction Coordinate
(progress of reaction)
An ionic nucleophile in a polar solvent:
Polar aprotic solvents are good for solvating cations not anions
Hint on an exam, a polar aprotic solvent is a clue that the reaction is SN2.
Most transition states are nonpolar so they are solvated by nonpolar solvents:
There is no positive/negative end Æ so it is nonpolar
As a general trend, nonpolar solvents will speed up an SN2 reaction (not always).
Exception
Et
Et
N
Et
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CH3
+ CH Br
2
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The transition state is polar, so it is best solvated by a polar solvent.
Relative rates:
Solvent
Benzene
CH3OH
Nitrobenzene
Relative Rate
1
Least polar
9
13
Most polar
Factors influencing SN1 reactions
1. The primary factor – the relative stability of the carbocation (carbonium ion)
The order of stability is:
Compare
3° > 2° > 1° >> CH3
The higher stability of the 3° carbocation is due to an inductive effect (a qualitative to
semiquantitative measure of the pushing or pulling of electron density through σ bonds).
2. Solvation of the intermediate (and the transition state leading to it)
SN1 versus SN2
This is only a problem when considering 2° carbon atoms.
For 3° carbon atoms, SN2 is strongly favored because of the stability of the carbocation.
For 1° and methyl, SN2 is strongly favored because of the lack of steric hindrance.
For 2° carbon atoms:
1. A high concentration of the nucleophile favors SN2. This increases the chance
of the nucleophile-substrate encounter (needed for the rate determining step).
2. A more reactive nucleophile favors SN2 – again this is because it is easier to go
to the transition state (lower activation energy).
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Test for SN2 – Finkelstein test (NaI/acetone)
RX (X = Br or Cl)
RX
N I/
RI + NaX(s)
NaCl and NaBr are insoluble in acetone.
R:
1° Æ smooth SN2 (test for 1° RX also)
2° Æ slow
3° Æ no reaction
Also for classifying RX, use AgNO3/EtOH test
RX
1° or CH3 Æ requires heat and time
2° Æ needs a few minutes
3° Æ immediate precipitation of AgX
3. A strongly ionizing solvent favors SN1
4. Polar aprotic solvents favor SN2
(Polar aprotic solvent = a non-ionizing solvent that only solvates the cations.)
If both cations and anions are solvated Æ SN1
Key concepts of an SN1 reaction:
Reactivity of ROH 3° > 2° >> 1° (due to stabilization of carbocation)
Rearrangement
Possible, if it occurs, it will be to form a more stable carbocation.
Solvent affects
The presence of protic solvents (hydrogen bonding) favor SN1,
because they stabilize the carbocation.
Stereochemistry
Normally racemization occurs because attack can be from either
side. However, residual shielding by the leaving group leads to a
small amount of net inversion, increasing solvent polarity enhances
this.
Kinetics
Rate = k [electrophile]
Note: Rearrangements are rare in most other mechanisms.
6.14 Organic Synthesis: Functional Group Transformations Using SN2 Reactions
Note: There is no substitution (SN1 or SN2) for vinyl halides or aryl halides
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Neither of these will form a carbocation; thus, no SN1
No inversion is possible; thus, no SN2
In these cases, attack of a nucleophile must be through an electron rich region,
which repels the electrons donated by the nucleophile.
6.15 Elimination Reactions of Alkyl Halides
Example
A strong base is necessary
a. Na+-OEt, sodium ethoxide (sodium salt of ethyl alcohol) is a very strong base
(the conjugate base of a very weak acid).
Preparation:
Na° + ROH Æ RO- Na+ + ½ H2
Potassium t-butoxide = K+ -O-t-Bu
b. KOH in alcohol may also be used.
Note: On the next exam, there will be problems under the heading retrosynthetic analysis.
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Exam question: (Retrosynthetic analysis)
Answer:
This is the only method
to get an alkene, so far.
Therefore, it must be here.
This is the only
(so far)
6.16 The E2 Reaction
Mechanisms for elimination processes:
3. E2
Bimolecular elimination Æ two species are in the rate determining step
Key concepts of an E2 reaction:
Reactivity of R-X
3° > 2° > 1° (due to product stability)
Solvent affects
Polar; aprotic solvents (non-hydrogen bonding) favor E2 reactions
Products
The anti arrangement of the β-proton and the leaving group
determines the stereochemistry of the alkene.
Kinetics
Rate = k [base] [R-X]
Conditions
Strong bases and high temperatures favor elimination over
substitution
6.17 The E1 Reaction
4. E1
Example
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Mechanism
CH3
CH3 C
Br
slow
-Br-
CH3
Note DO NOT USE
Key concepts of an E1 reaction:
Reactivity of R-X
3° > 2° >> 1° (due to carbocation stability)
Solvent affects
Protic solvents (hydrogen bonding) favor E1 reactions (as these
solvents stabilize the carbocation)
Products
More substituted alkenes are favored over less substituted
Kinetics
Rate = k [R-X]
Conditions
Weak bases and high temperatures favor elimination over
substitution
6.18 Substitution versus Elimination
1. SN1 versus E1 –
2. SN2 versus E2
a. With a nucleophile that is also a good base, then steric hindrance on the
substrate is important.
1° RX
2° RX – it is less accessible to nucleophilic attack
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3° RX – not accessible to nucleophilic attack
b. Increasing the bulk of the nucleophile (base) favors E
RO- is nucleophilic (seeks C+) and basic (seeks H+)
t-butoxy is large, so it tends to pick off H+ at a distance instead of
coming close to the α–C.
methoxy is small, so it can get close to the α–C.
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NH2- is an exceedingly strong base, so E predominates with RX
c. Weakly basic and/or highly polarizable ions favor substitutions
Example
Acetate, cyanide, CH3S-, I-, and PPh3
3° halides
SN1 versus E2
Solvolysis conditions – EtOH/H2O
Without a good base Æ SN1
With a good base Æ E2
General trends:
6.19 Overall Summary
Summary
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Chapter 7 ALKENES AND ALKYNES I: Properties and Synthesis.
Elimination
Reactions of Alkyl Halides
Do not make the mistake of treating benzene, or other aromatic compounds like an alkene.
7.1 Introduction
7.2 The (E)-(Z) System for Designating Alkene Diastereomers
Already covered so review
7.3 Relative Stabilities of Alkenes
1. cis-trans
2. The more heavily substituted a C=C is, the more stable the compound is.
Compare:
H
CH3
C
CH3
C
3 substituents
H is the smallest = most stable
CH3
CH3
C
CH2 CH3
2 substituents
CH2
CH3
CH
CH
CH2
1 substituent
CH3
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H is the largest = least stable
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Exam question:
Which is more stable (has a higher ΔH of hydrogenation)?
Answer
#1 has 1 olefinic H Æ 3 carbon atoms Æ total 4
The 3 carbon atoms mean that the compound is trisubstituted (more stable, lower ΔH).
#2 has 2 olefinic H Æ 2 carbon atoms Æ total 4
The 2 carbon atoms mean that the compound is disubstituted (less stable, higher ΔH).
The total of four (check) refers to:
An alternative would be to use the heats of combustion. This will give the same answers for the
same reasons.
7.4 Cycloalkenes
Review and relate to non-cyclic systems
7.5 Synthesis of Alkenes via Elimination Reactions
Review
Note: In general, these will be previously seen reactions
7.6 Dehydrohalogenation of Alkyl Halides
Preparation of Alkenes
Dehydrohalogenation (halogen + β-H)
The major product predicted by Zaitsev’s Rule.
Zaitsev’s Rule (variations in spelling) –
This is an empirical rule = A Crutch (so it is never an explanation on an exam)
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The mechanism involves this transition state:
Notice that the loss of the H and the Br are from opposite sides. This is antiperiplanar.
If a larger bulky base (i.e., K t-BuO-) is used the major product is:
7.7 Acid-Catalyzed Dehydration of Alcohols
Dehydration of alcohols (dehydration = loss of H2O)
Mechanism
H
CH3 C
CH3
OH
H+
Always use arrows to indicate electron flow.
Note: Remember the rate-determine step.
The symbolism “-H+” is acceptable. Actually, the H+ goes to the H2O.
Note: This is NOT acceptable:
H
H
CH3 C
C
H
H
Note: The biggest problem with mechanisms is when one tries to do too much at one time.
Rate of dehydration Æ 3° > 2° > 1°
CH3CH2CH2OH requires con H2SO4 and 170°
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H3PO4 will also dehydrate.
7.8 Carbocation Stability and the Occurrence of Molecular Rearrangements
What is the mechanism for the preceding reaction?
Notice: The products imply an intermediate of the form:
Note: Acidic conditions nearly always use H+ in the first step.
CH3
CH3 C
CH
CH3
CH3 OH
H+
Thought question: – How would 3-methyl-2-butanol react via this mechanism?
Put in arrows for the mechanism for the dehydration of 1-butanol to 2-butene (major product).
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Note: Any dehydration gives the thermodynamically most stable product.
Note: A rearrangement indicates a carbocation intermediate.
The move is a hydride shift or migration. A :H- has moved.
Symbolize as ~H-
The major product is due to a rearrangement.
Thought question: Write a complete step-wise mechanism for this reaction
Answer
Stereochemistry
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Exam question: on retrosynthetic analysis: Outline the synthesis of
7.9 Synthesis of Alkynes by Elimination Reactions
Preparation of acetylene
Debromination of vic-dibromides
Adjacent carbon atoms
vic Æ vicinal (vicinity)
Same carbon atom
gem Æ geminal (twin)
The reaction occurs on the surface of the zinc metal.
Summary of preparations of alkynes
1. From acetylide salts with alkyl halides
2. From alkenes
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CH3 CH
CH
CH3
Br2/CCl4
2 NaNH2
KOH/alcohol
NaNH2
Two dehydrohalogenations
The C≡C is always between the carbon atoms that originally held the X’s.
Thought question:
How do you prepare CH3C≡C-CH3 from CH3CH=CH2?
Answer
7.10 The Acidity of Terminal Alkynes
Order of acidity:
R-C≡C-H > R-CH=CH-H > R-CH2CH2-H
Obviously, the reverse order for the base strength
R-C≡C- < R-CH=CH- < R-CH2CH2Acetylide ion
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Remember – Na+ is a spectator ion – do not show it reacting
7.11 Replacement of the Acetylenic Hydrogen Atom of Terminal Alkynes
Formation of acetylide salts
Recall – SN2 works for 1° and Me alkyl halides; 2° alkyl halides will undergo E2.
Test for terminal C≡C
Oxidative cleavage
The same types of reactions as for alkenes with some minor differences
CH3 CH2 C
C
CH3
1) KMn
O
2) H +
4 /OH -,
1) O3
2) H2O
hot, con
c
Note – acids form
Note – the second step was H2O not Zn/H2O as with alkenes. Zn could be used, but it is not
necessary here.
Thought question:
What is the structure of C7H10 if it reacts as follows?
Answer
IHD (degree of unsaturation) = 7 – 10/2 + 1 = 3 Æ
1-hepten-5-yne
These are the smallest numbers. There are no priority considerations of C≡C to
C=C unless both are equal distance from the end – then C=C has higher priority.
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7.12 Alkylation of Alkynide Anions: Some General Principles of Structure and Reactivity
Illustrated
Review
7.13 Hydrogenation of Alkenes
7.14 Hydrogenation: The Function of the Catalyst
Mechanism:
The catalyst is necessary to lower the activation energy by providing an alternate pathway.
No Catalyst
Catalyst
C
C
+ H2
H (exothermic)
Alkane
Adam’s catalyst
Raney nickel
7.15 Hydrogenation of Alkynes
Syn-addition
Anti-addition
Note the different products for these two types of addition.
7.16 An Introduction to Organic Synthesis
Major consideration – review well
There are numerous exceedingly useful hints here
Begin to examine how the information given in class addresses these issues.
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Chapter 5 STEREOCHEMISTRY: Chiral Molecules
Models really help
5.1 The Biological Significance of Chirality
Read
5.2 Isomerism: Constitutional Isomers and Stereoisomers
Stereochemistry (3-dimentional structure)
Isomerism
1. Constitutional (= Structural) – different connectivity
2. Stereoisomers (Configurational and geometric) – same connectivity
5.3 Enantiomers and Chiral Molecules
Optical activity and chiral centers
Chiral molecules have nonsuperimposable mirror images.
A carbon atom with four different groups attached to it is a chiral center (chiral Æ handlike, i.e., left and right hands). Older texts refer to this as an asymmetric carbon.
Otherwise, the carbon atom is achiral.
5.4 More about the Biological Importance of Chirality
5.5 Historical Origin of Stereochemistry
Read these sections
5.6 Tests for Chirality: Planes of Symmetry
Note: Watch out when more than one chiral center is present.
5.7 Nomenclature of Enantiomers: The R,S-System
MODELS KITS – help initially (Make models of labeled compounds in the textbook.)
The Cahn-Ingold-Prelog system (CIP system) or E-Z system
3-methyl-trans-2-pentene
It is not easy to decide if this is cis or trans.
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CIP Æ
C #2
C #3
cis or trans?
Know: Group names:
CH2=CHCH2=CH-CH2-
vinyl
allyl
Example
CH2=CHCl
CH2=CH-CH2-Br
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vinyl chloride
allyl bromide
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trans-1,2-divinylcyclopropane
CH2= = methylene
methylenecyclohexane
Reference to the system
CH3
H
H
two vinyl or olef inic hydrogen atoms
three allylic hydrogen atoms
H
H
H
Vinyl or olefinic hydrogen atoms are hydrogen atoms attached directly to the
carbon atoms involved in the C=C (these are olefinic carbon atoms).
Allylic hydrogen atoms are the hydrogen atoms attached directly to the carbon
atoms adjacent to the C=C (these are the allylic carbon atoms).
ASIDE
N-bromosuccinimide (NBS) will substitute a bromine atom for an allylic hydrogen atom.
N-bromosuccinimide (NBS)
Thought question: How would you prepare 3-bromocyclohexene from bromocyclohexane?
Answer: Convert the bromocyclohexane to cyclohexene via a dehydrohalogenation. Treatment
of cyclohexene with NBS yields 3-bromocyclohexene.
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END ASIDE
Nomenclature to distinguish the forms
Name the arrangement about the chiral center.
Rule 1
This gives:
Rule 2
Rule 3
Configuration = S
Example
C
Multiple bond, assign as C
H
C
Br
1
Br
2
CH3
3 CH3
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C
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Example
5.8 Properties of Enantiomers: Optical Activity
Enantiomers – have the same physical properties
Plane polarized light – light normally vibrates in all directions perpendicular to
the path of propagation – these are a combination of two components (↑ and →),
the separation of one gives plane polarized light.
Polarimeter:
Source Æ Polarizer Æ sample Æ Polarizer Æ observer
A measure of the rotating power = specific rotation (a physical constant)
Specific rotation = α
α TD =
T
D
Note a +90° is indistinguishable from a -90°, so dilute the sample and re-measure
Remember
R ≠ + (d)
S ≠ - (l)
The CIP system is NOT an alternative to the experimental determination of d and l.
If you have a compound, and learn that it rotates +; then you know that its
enantiomer will rotate -.
Specific rotation = [α] =
5.9 The Origin of Optical Activity
See examples below
5.10 The Synthesis of Chiral Molecules
Synthesis of enantiomers
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Lactic acid dehydrogenase = LAD
Prochiral –
Once the molecule fits into the active site, only one side is open to attack – so
only one isomer forms. (With H2/Ni – either side could be attacked Æ 50:50 mixture)
Chiral centers in cyclic species
Example
The product has 2 chiral centers.
OH
CH2
S
OH
C
C
H
H
CH2
R
H
CH2
H
C
C
OH
OH S
R
CH2
Note: Fischer projection conventions do not apply to cyclic molecules.
Example
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Example
Stereospecific reactions =
Example
Rotating this molecule about an axis in the plane gives the other molecule:
Rotation about the C-C bond changes it to:
Rotation places like groups together
Example
Syn-hydroxylation
Here are two views of one of the molecules:
CH3
OH
CH3
C
H
OH
H
C
CH3
H
C
OH
HO
C
H
CH3
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Reactions with inversion of configuration
Note: Even though it works in this case, these are not always/necessarily true:
R-I
I
R-I* + IObservations:
1. Radioactive iodine (I*) is incorporated.
2. Optical activity is lost.
Thought question: — Why does number 2 occur?
SN1 reactions at the chiral center lead to racemization.
However, SN1 may not give complete racemization.
H
MeOH
C
solvolysis
Br
Me
Retention of configuration
Example
Thionyl chloride = SOCl2
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Unimportant mechanism step:
For a cyclic system
Allene
Cl Cl
H
C
Cl
C
C
H
C
H
C
C
H
Cl
5.11 Chiral Drugs
Read
5.12 Molecules with More than One Chirality Center
Molecules with more than one chiral center
2n Rule – the maximum number of stereoisomers is given by 2n, where n = the
number of nonidentical chiral centers.
Example
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Diastereomers – completely different molecules = different physical properties
Diastereomers –
1. Stereoisomers that are not enantiomers (too broad)
2. Differ in configuration about at least one of the chiral centers and are
not enantiomers
What is the form of compound “E”?
Rules for rearranging structures (Fischer projections)
DO not do anything else
(Horizontal = towards the observer, and vertical = away from the observer, this
must be maintained)
1. Move the molecule in the plane
2. Rotate the molecule by 180° about an axis perpendicular to the plane of the
diagram.
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Example
What if the chiral centers have the same type of groups?
Also:
R, R
Enantiomers
S, S
Diastereomers
Enantiomers
R, S
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5.13 Fischer Projection Formulas
This is review – use it
5.14 Stereoisomerism of Cyclic Compounds
Read
5.15 Relating Configurations through Reactions in Which No Bonds to the Chirality Center
Are Broken
Reactions where bonds to the chiral center are not broken proceed with retention of
configuration – the stereochemical nomenclature may change, however.
Example
Example
5.15A
Rational
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O
OH
HNO2
C
H
C
OH
HNO2/HBr
CH2NH2
(+)-isoserine
Since both D-(+)-glyceraldehyde and (+)-isoserine both give (-)-glyceric acid, they must
have the same stereochemistry.
Note: For exams etc. – you should be able to follow the above reactions – but it is not necessary
to memorize them or the compounds other than glyceraldehyde.
Thought question: – What is the structure of (-)-lactic acid?
Answer
5.16 Separation of Enantiomers: Resolution
Separation of optically active mixtures = resolution
5.16A
5.16B Use a resolving agent (Pasteur also)
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5.17 Compounds with Chirality Centers Other than Carbon
See Figure 5.22, page 216 in the textbook
Allene – pay attention to this (seen earlier)
5.18 Chiral Molecules that Do Not Possess a Chirality Center
Summary
no
2 molecules
Same
molecular
f ormula? yes
not
isomers
ISOMERS
Same
bond
connectivity?
no
Constitutional
Isomers
yes
Stereoisomers
Mirror
no images? yes
Diastereomers
no
Identical?
Enantiomers (R/S)
Dif f er in absolute conf iguration
at a single chiral center
Meso Compounds
Dif f er in orientation of substituents
around a ring on a double bond
Epimers Geometric
Isomers
Due to
(cis/trans, E/Z)
ring
closure?
yes
Anomers
( )
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Chapter 8 THE CHEMISTRY OF ALKENES AND ALKYNES II: Addition Reactions
Do not make the mistake of treating benzene, or other aromatic compounds like an alkene.
8.1 Introduction: Addition to Alkenes
8.2 Electrophilic Addition of Hydrogen Halides to Alkenes: Mechanism and
Markovnikov’s Rule
8.3 Stereochemistry of the Ionic Addition to an Alkene
1. Hydrogenation – syn addition of H2 – review under alkanes
2. Addition of polar molecules
CH3
CH3 C
HBr
CH2
Which gives the more stable intermediate?
CH3
CH3 C
CH2
+ H+ (electrophile)
Markovnikov Addition – when adding a polar molecule to a carbon-carbon double bond, the
positive end goes to the carbon within the π bond that has more hydrogen atoms.
Note: If an exam question asks you, why does this product form? The answer “because of
Markovnikov’s Rule is NOT acceptable. You must discuss why it works (i.e., the more stable
carbocation intermediate).
Thought question: – What is the energy diagram for the preceding mechanism?
Answer
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Thought question: – The following reaction will not work. CH3CH2CH3 + I2 Æ CH3CHICH3
What is an alternate procedure to prepare this product using only three steps? One of the
intermediates must be an alkene.
Answer
Example
Key aspects of H-X electrophilic addition to an alkene
Nucleophile
π bond
Electrophile
HX
Intermediate
Protonation of the π bond leads to the most stable carbocation
(Markovnikov addition).
Rearrangement
Possible, leading to the formation of a more stable carbocation
Stereochemistry
The halide can attack either side of the unhybridized p orbital. If
the reaction site is chiral, a racemic mixture will result.
Note as before rearrangements are rare, they appeared in SN1 mechanisms.
8.4 Addition of Sulfuric Acid to Alkenes
Example
Example
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Note: In mechanisms under acidic conditions, all structures are neutral or positive; under basic
conditions, all are neutral or negative.
Note: ' refers to the Principle of microscopic reversibility. The mechanism of the forward
reaction is the same as for the reverse reaction (more or less).
3. Dimerization Æ self addition
(CH3)2C=CH2
% H SO
C8H12 or (C4H8)2
8.5 Addition of Water to Alkenes: Acid-Catalyzed Hydration
Hydration (= addition of H2O; not hydrolysis, which is the breaking apart by H2O)
Note: This is tautomerism (“internal acid-base reaction”)
Thought question:
What is the product of the following reaction?
8.6 Alcohols from Alkenes through Oxymercuration-Demercuration: Markovnikov
Addition
Oxymercuration-Demercration
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Advantages:
The mechanism involves the mercurium ion:
Example
Example
8.7 Alcohols from Alkenes through Hydroboration-Oxidation: Anti-Markovnikov Syn
Hydration
Hydroboration oxidation
No matter what hydroboration agent is used, it is usually simplified to BH3 in the mechanism.
8.8 Hydroboration: Synthesis of Alkylboranes
8.9 Oxidation and Hydrolysis of Alkyboranes
BH3 is a good Lewis acid.
There are two reasons why the attack is on the terminal carbon.
1.
2.
Recall:
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Normal hydration gives the Markovnikov alcohol, while oxidation of the
alkyl borane gives the anti-Markovnikov alcohol.
The 1) and 2) refer to the first and second steps of a reaction sequence.
Skip the intermediate products. This is a sequence of reactions.
Note: THE NUMBERS ARE NECESSARY
Example What will be the product of the following reaction sequence?
There are alternate ways of writing the structures:
No dots Æ no attempt to indicate the stereochemistry
Hydroboration/oxidation
Diglyme is a useful organic solvent (an ether).
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Thought question: – What are the byproducts of this reaction?
8.10 Summary of Alkene Hydration Methods
Review
8.11 Protonolysis of Alkyboranes
Review
8.12 Electrophilic Addition of Bromine and Chlorine to Alkenes
Halogenation of alkenes
Recall: RH + Br2
RBr + HBr(g)
Substitution
The release of HBr causes a fog above the solution that will turn blue
litmus paper red.
For alkenes:
Key aspects of halogen addition to alkenes
Nucleophile
π bond
Electrophile
X2
Intermediate
A bridged halonium ion. The positive charge is on the bridging
halogen atom.
Rearrangement
Impossible since there is no carbocation intermediate
Stereochemistry
The addition of X2 to the π bond is anti.
8.13 Stereochemistry of the Addition of Halogens to Alkenes
Halogenation primarily refers to bromination.
Cl2 requires special conditions, as it is too reactive.
Stereochemistry:
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General mechanism:
8.14 Halohydrin Formation
(Halohydrin = halogen and OH on adjacent carbon atoms)
Initial attack is by the Br2:
CH3 C
CH2
CH3
Br2
Note that the OH is on the tertiary carbon. No primary alcohol forms:
No
OH
CH3
CH3
C
CH2
Br
Exam question:
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Answer:
This looks like a simple replacement of an H with an OH – there is no such thing.
The product is a halohydrin. This requires the formation of the halohydrin from an alkene.
8.15 Divalent Carbon Compounds: Carbenes
This is where carbon (temporarily) has two bonds.
Carbene chemistry – do not go much beyond the lecture detail
Generation of carbenes
a.
b. Dihalocarbenes are much more stable, so they are better for synthesis.
Example
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c. Simmons-Smith Reaction
CH2I2 + Zn(Cu) Æ carbenoid
Zn(Cu) = copper coated zinc
ASIDE
Alkenes will also add to conjugated dienes.
This reaction is the Diels-Alder Reaction (Chapter 13).
The general process is:
The double bond ends up between the locations of the original pair of C=C’s.
The presence of an electron withdrawing group (pulls electron density
away from the double bond) speeds up the reaction (a lower temperature is
necessary).
The Diels-Alder Reaction is a concerted syn addition with the stereochemistry of the
dienophile preserved in the stereochemistry of the product.
If the dienophile is trans, then the product will be trans also. (the same for cis)
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Exam question:
What is the Diels-Alder product of the following reaction?
A dienophile may be either an alkene or an alkyne
Answer:
If the product has two C=C, then the double bond next to the electron
withdrawing group was originally part of the dienophile.
The most complicated Diels-Alder reactions:
Two possible cis products:
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This is an Alder-endo addition – Alder said the endo product would form:
Note: This is the Alder endo rule.
For addition to 3
Look at the largest remaining ring (1)
The cis addition may be close to it (endo) or away from it (exo)
The endo product will form.
END ASIDE
8.16 Oxidations of Alkenes: Syn 1,2-Dihydroxylation
Epoxide formation (Epoxidation)
Note: MCPBA is a widely used peroxy acid.
Common names are as an oxide of an alkene:
O
C
O
OH
/CH2Cl2
The oxidizing agent is peroxybenzoic acid, and the product is cyclohexene oxide.
Mechanism (not on the exam)
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Chemistry of epoxides – the three membered ring is unstable, so epoxides are reactive.
The ring juncture is cis.
Antihydroxylation:
Oxidation of alkenes
a. Syn hydroxylation – the Baeyer test for unsaturation
CH3 CH
CH2
KMnO4/OHcold, dilute
MnO4-
A better reaction occurs with OsO4 (but toxic and expensive)
Note: There will be more retrosynthetic reactions on the next exam.
Exam question:
Answer:
This is NOT an H/OH replacement
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b. Ozonolysis – breaking apart with ozone (O3)
Note: It is possible to substitute dimethyl sulfide (Me2S) for the zinc/acid.
Exam question: What is the starting material if the products of ozonolysis are the following?
Answer:
“Block” the hydrocarbon portions:
Remove the carbonyl oxygen atoms and connect the pieces.
Exam question:
What is X if ozonolysis only produces acetaldehyde?
Exam question:
Roadmap
Determine the structure of C6H10 if it reacts as follows.
Answer:
The original compound has IHD (degree of unsaturation) = 2 (= 6–10/2+1)
IHD = 2 means
2 C=C O3 would give more than one product
1 C≡C O3 would give more than one product
2 rings O3 would not react, but it could give one product
1 ring and 1 C=C O3 could open the ring and give one product
1 C=C (with a CH3 attached) + 1 ring:
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ASIDE
Simple Chemical Tests – these may help with roadmaps also
Reagent
Test
red-brown Æ
1. Br2/CCl4
Colorless (light yellow)
2. H2SO4, cold, con solution (all protonated Æ ionic)
3. KMnO4, cold, dil purple (MnO4-) Æ brown (MnO2)
4. Ag(NH3)2+OH- or precipitate
Cu(NH3)2+OH5. AgNO3(alcoholic) precipitate
6. H2/cat (not simple) Pressure drop (qual or quant)
7. Sodium fusion
Gives a positive test
C=C or C≡C
C=C or C≡C or ROH
C=C or C≡C
C≡C-H
3° RX
C=C or C≡C
RX
END ASIDE
c. Oxidative cleavage – hot KMnO4(aq)/OH- (or other reagents)
CH3
1) KMnO4/OH-/
2) H+
Example
What is the structure of X?
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8.17 Oxidative Cleavage of Alkenes
Review (already covered)
8.18 Electrophilic Addition of Bromine and Chlorine to Alkynes
Bromination
8.19 Addition of Hydrogen Halides to Alkynes
Hydrohalogenation
Note: The presence of lone pairs on the atom adjacent to the atom with the positive charge will
help stabilize the charge (see the other contributing structure (left)).
8.20 Oxidative Cleavage of Alkynes
Review and compare to alkenes
8.21 Synthetic Strategies Revisited
Very Important Review
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Chapter 9 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY:
Tools for Structure Determination
2.16 Infrared Spectroscopy: An Instrumental Method for Detecting Functional Groups
Review Table 2.7 on page 79 in the textbook.
Infrared spectroscopy – IR – less energy and longer wavelength than visible
Light energy = E = hc/λ α 1/λ = ΰ (wave number – not frequency = ν)
Stretching and bending vibrations occur at certain quantized frequencies. When one shines light
on the molecule, which is of the same frequency as one of the stretching or bending vibrations,
light is absorbed and the amplitude of that vibration is increased. This gives rise to
FUNDAMENTAL bands. In order for a particular vibration to result in the absorption of
infrared energy, the vibration must cause a CHANGE IN DIPOLE MOMENT OF THE
MOLECULE.
Typical Spectrum:
Around 3000 cm-1 is important Æ C-H stretch
A compound with a C-H will give a band in his region.
For a vibration to absorb energy there must be a change in dipole moment.
1400-900 cm-1 = fingerprint region
Memorize
Note: Finish learning Table 2.7, page 79 in the textbook
Stretching:
C-H
≈3000 sp > sp2 > sp3
sp3 is just below 3000, the others are above
sp and sp2 mean the H is attached to a C that is involved in
multiple bonding or aromatics
C=C 1680-1640
Aromatic ring 1600 and 1500
Bending:
-CH2- 1470-1430
-CH3 1470-1430
And 1395
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(scissoring of 2 H’s)
(scissoring of 3 H’s)
96
Benzene ring Different substitution patterns give different patterns in the
850-650 cm-1 region. These are out-of-plane H bends. (Do
not memorize the specific patterns.)
The most useful aspect Æ functional group analysis
O large
strong band
C +
For example, the C=O band in butanone is stronger than the 8 C-H bands
≈1700 cm-1
-O-H stretch in alcohols AND water = 3600-3200 cm-1
-C-O stretch 1150 (3°) – 1050 (1°) (2° ≈1100 cm-1)
3500-3300 cm-1 (Usually sharper than alcohol)
R-NH2
2 H’s = 2 Bands (RNH2)
1 H = 1 band (R2NH)
9.1 Introduction
Read
9.2 Nuclear Magnetic Resonance (NMR) Spectroscopy
9.3 Interpreting Proton NMR Spectra
9.4 Nuclear Spin: The Origin of the Signal
Schematic of a simple NMR instrument
A typical spectrum
Practical Applications
Possible observations
a. Number of signals (kinds of hydrogen)
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b. Position of signal –
c. Intensity
d. Splitting of the signal
Number of signals –
9.5 Detecting the Signal: Fourier Transform NMR Spectrometers
Read
9.6 Shielding and Deshielding of Protons
Read and see below
9.7 The Chemical Shift
Chemical shift Æ the position of the signal with respect to TMS
TMS = tetramethyl silane – originally chosen because it did not interfere (no longer true)
Hydrocarbons
Electron density around the hydrogen gives an induced field around the
atom that opposes the external field – the signal is shielded
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Electron withdrawing Æ deshielded
9.8 Chemical Shift Equivalent and Nonequivalent Protons
CH3-O-CH2CN
a
b Å more shielded = downfield
The relative increase Î relative number of hydrogen atoms
10:15
2:3
CH2:CH3
May also be done by lines/H
If 5 H’s = 25 lines then the conversion is 5 lines/H
9.9 Signal Splitting: Spin-Spin Coupling
Spin-spin coupling of a signal
Coupling:
The three adjacent hydrogen atoms split the signal into a quartet
Points:
1. Aromatic hydrogen atoms in hydrocarbons are usually singlets
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2. The signals will “point” to what is causing the splitting
3. Equivalent hydrogen atoms do not split each other
4. n-hydrogen atoms cause n + 1 peaks (splitting)
9.10 Proton NMR Spectra and Rate Processes
Read
9.11 Carbon-13 NMR Spectroscopy
13
C-NMR (cover lightly)
13
C is about 1 % of carbon
Factors:
1. A wide range of chemical shifts
≈200 ppm versus ≈20 for 1H
2. 13C-13C not a factor since the isotope abundance is so low
3. Proton decoupled spectra Æ everything occurs as singlets
There are three types of carbon atoms.
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The carbonyl is furthest downfield because of the electron
withdrawing O.
The ratio is 1:1:3
4. Proton Off-resonance Decoupling
The hydrogen atoms directly bonded to each carbon atom split the signal
9.12 Two-Dimensional (20) NMR Techniques
Skip
9.13 An Introduction to Mass Spectrometry
9.14 Formation of Ions: Electron Impact Ionization
Result Æ “typical” spectrum
The base peak is adjusted so its intensity is 100 %. It represents the most stable carbocation.
1. M+ is the exact molecular weight of the molecule
2. Some of the structural units may be deduced.
3. It is possible to derive the molecular formulas from (M + 1)+ and (M + 2)+
H
1
H and 2H
relative abundances
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100:0.016
(or 100000:16)
101
C
N
12
14
C and 13C
N and 15N
100:1
10000:15
M+ (molecular ion) is composed of the most abundant isotopes of each element present
(M + 1)+ has a mass 1 greater than M+ – mostly due to 13C
(M + 2)+ has a mass 2 greater than M+ – mostly due to Cl, Br, or S
Element
H
C
N
Cl
Br
S
O
Relative Abundance of Principle Isotopes
Mass Number
Relative Abundance
1
100
2
0.016
12
100
13
1.08
14
100
15
0.38
35
100
37
32.5
79
100
81
98.0
32
100
34
4.40
16
100
17
0.04
18
0.20
The study of fragmentation patterns is exceedingly interesting.
Example
CH4
[12CH4]+• cation radical
Example
CH3CH3
Nitrogen Rule: A molecule with an even numbered molecular weight must contain either no
nitrogen or an even number of nitrogen atoms; an odd molecular weight requires an odd number
of nitrogen atoms.
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Example:
Determine the structure of the following compound. It contains carbon,
hydrogen, and a halogen.
m/e
14
15
26
27
28
29
30
127
128
Relative
Intensity
0.11
0.14
2.19
18.7
2.53
35.6
0.79
15.0
5.71
m/e
140
141
142
155
156
157
Relative
Intensity
0.49
1.18
0.01
0.56
100.0
2.25
Example:
Determine the structure of the following compound. It contains carbon,
hydrogen, and a halogen.
m/e
25
26
27
37
38
39
50
51
52
73
74
75
76
77
78
79
Relative
Intensity
0.20
0.61
1.21
2.72
10.1
1.89
17.80
7.65
0.33
2.06
7.31
5.41
4.32
100.
12.2
5.49
m/e
104
106
117
119
130
132
155
156
157
158
159
160
Relative
Intensity
0.29
0.28
0.32
0.30
0.34
0.33
0.40
91.6
6.45
89.9
5.95
0.17
Thought question:
A compound has M+ = 202 with a relative abundance of (M + 1)+ = 13.99. How many carbon
atoms are there?
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9.15 Depicting the Molecular Ion
9.16 Fragmentation
9.17 Determination of Molecular Formulas and Molecular Weights
9.18 Mass Spectrometer Instrument Designs
9.19 GC/MS Analysis
9.20 Mass Spectrometry of Biomolecules
Read these sections
ASIDE
UV-Vis (Ultraviolet-Visible) Spectroscopy (More detail in Chapter 13)
Also known as electronic spectroscopy
If sufficient energy is supplied, it is possible to more an electron from the π to the π*.
This energy is normally in the ultraviolet of visible region of the spectrum.
The greater the number of conjugated double bonds, the lower the energy requirement.
A lower energy requirement = a longer absorption.
The term “Conjugated double bonds” (Chapter 13) refers to alternating single and double bonds.
A typical alkene absorbs below 200 nm (far ultraviolet). Conjugated dienes absorb at longer
wavelengths. It there are sufficient double bonds conjugated; the absorption may move into the
visible portion of the spectrum.
END ASIDE
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Chapter 11 ALCOHOLS AND ETHERS
11.1 Structure and Nomenclature
1. As alcohols (common)
Phenol is not an alcohol
2. As carbinols (methyl alcohol = carbinol)
3. IUPAC
a. Parent chain contains the OH
b. OH gets the smaller number
c. Drop –e and add –ol
4. Halohydrins
5. Salts of alcohols (“alkoxide salts”)
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6. Glycols (diols) – vic diols (vic = vicinal)
Aluminum isopropoxide
Nomenclature of phenols (ArOH)
OH
OH
NO2
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Nomenclature of ethers
Ethers have the general formula ROR or ArOAr or ROAr
As “ethers”
Et-O-Et
CH3-O-CH(CH3)2
O
As an “alkoxy” compound
Br
O
CH2 CH3
CH3
O
C
CH3
A few “common” names
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11.2 Physical Properties of Alcohols and Ethers
It is best to make comparisons to compounds with similar molecular weights
1. Boiling Point
Note Impurities will increase the boiling point and decrease the melting point.
2. Solubility in H2O
3. Solubility in con H2SO4 – indicates the presence of N or O
4. Alcohols are denser than hydrocarbons with comparable carbon content.
5. Ethers are good for extracting organics (the density is less than water)
11.3 Important Alcohols and Ethers
Important examples
Methanol – Wood alcohol
Ethanol – Grain alcohol
95% – a solvent
A constant boiling azeotrope
100% EtOH b.p. = 78.3 °C
95% EtOH
b.p. = 78.15 °C
100% EtOH b.p. = 100 °C
Simple distillation Æ 190 proof (proof is from frontier days)
100% – Absolute
a. Use a ternary azeotrope to dry
b. Mg + H2O Æ Mg(OH)2(s) + H2(g)
Ethylene glycol
Diethyl ether – “Ether”
Autoxidation (Chapter 10)
11.4 Synthesis of Alcohols from Alkenes
Some parts covered earlier
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Preparation of alcohols
1. Hydration of alkenes
2. Fermentation of sugars
EtOH
Sugar (grain)
3. Catalytically from CO and H2
Non-Industrial
Hydration of alkenes
Hydroboration-oxidation
Review:
Exam question: What is the product of the following reaction?
Answer:
OH
H
H
CH3
trans-2-methylcyclopropane
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Anti-Markovnikov addition
The stereochemistry is from the syn addition of
Reduction of carbonyls
See later
Hydrolysis of methyl or 1° RX
CH3CH2CH2Br
KOH⁄H O
CH3CH2CH2OH
Polyhydroxy alcohols
Diols
a. Vicinal (vic) diols
Alternative:
1) OsO4
2) Na2SO3/EtOH
Very efficient, but toxic
Reaction with periodate
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OH
O
C
C
HIO4
b. Geminal (gem) diols
O
CH3 C
CH3
or
H218O
18O
CH3 C
CH3
Triols
Glycerol = Glycerin = 1,2,3-propantriol
11.5 Reactions of Alcohols
Dehydration – an E1 process
Why is it never an E2 process?
Recall – the alcohols are not strong bases
Example
Example
CH3
CH
CH3
OH
H+
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H2SO4
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11.6 Alcohols as Acids
11.10 Tosylates, Mesylates, and Triflates: Leaving Group Derivatives of Alcohols
1. Alcohols as acids
Versus OH-/H2
Na+ -OCH3 + H2O Æ
Versus NH2-/NH3
CH3OH + Na+NH2- Æ
Versus R-/RH
CH3CH2δ- Liδ+ + CH3CH2OH Æ
Active metal:
Also:
NaH + EtOH Æ
Zerewittinoff (or Zerevitinov) determination of active hydrogen
ROH + CH3MgBr Æ CH4 + ROMgBr
Positive test = formation of methane
2. Esterification
It is useful to have a good leaving group
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Alternatives – Mesylate, Mesyl, and Triflate
Example
CH3
H
C
OH + TsCl
CH2
CH3
Basicity of alcohols (Cleaving the C-O bond)
Because of the free pairs of electrons on the oxygen atom, alcohols will react quite
readily with Brønsted-Lowry of Lewis acids.
11.7 Conversion of Alcohols into Alkyl Halides
11.8 Alkyl Halides from the Reaction of Alcohols with Hydrogen Halides
Conversion to alkyl halide
ROH Æ RX
Can use HX, NaBr/H2SO4, HCl/ZnCl2 PBr3, SOCl2
CH3CH2CH2OH + HBr (anhydrous) Æ CH3CH2CH2Br
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CH3
CH
CH3
CH2 CH2 OH
NaBr
1° so SN2
NaBr/H2SO4
The reaction takes place because H2O is a better leaving group than OH-
Expect rearranged products (often by a ~H- mechanism).
Overall mechanism:
SN1 except for CH3OH and most 1° ROH’s, which go SN2
allyl, benzyl > 3° > 2° > 1° < Me
The Lucas test
Alcohols with six carbon atoms or less (soluble in reagent mixture) react with
HCl/ZnCl2 to form an alkyl chloride, which is insoluble in the reaction mixture
(the solution turns cloudy).
3° alcohols turn immediately
2° alcohols require five minutes
1° alcohols require 24 hours and heating
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11.9 Alkyl Halides from the Reaction of Alcohols with PBr3 or SOCl2
PBr3 (or PI3) still SN2 needs a mild base
1. Inversion of configuration
2. NO rearrangement (No carbonium ion)
Thionyl chloride (SOCl2)
11.11 Synthesis of Ethers
This only works for symmetrical ethers.
ASIDE
How it is possible to modify the course of organic reactions by changing reaction conditions?
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END ASIDE
1. Dehydration of alcohols
This reaction competes with alkene formation
It will give the symmetrical ether (R2O)
H2SO4/140°
2 CH3CH2OH Æ CH3CH2-O-CH2CH3 + H2O
However:
H2SO4/180°
CH3CH2OH Æ CH2=CH2 + H2O
For 2° and 3° alcohols
Alkene production >> Ether production
2. Williamson Synthesis
An SN2 process
L is a good leaving group such as X, sulfonate, sulfate ester
Examining R’-X
1. It is an alkyl halide.
2. It is not vinyl or aromatic (no backside attack)
3. It cannot be R°.
This would give elimination.
Examining RO-Na+ and ArO-Na+
1. Formed by the reaction of ROH + Na° or phenol plus base
2. The anion is both a base and a nucleophile.
Example
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3. Epoxidation
4. t-butyl ethers – used as protecting groups
5. Alkoxymercuration – Demercuration
This is a variation of a reaction seen earlier.
6. The use of methyl sulfate for making methyl ethers
11.12 Reactions of Ethers
Peroxides and ethers
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Ether Reactions
1. Cleavage by HX (HX can be HI, HBr, or HCl and HF to a lesser extent)
2° and 3° are SN1 others are SN2
RO- is a poor leaving group. RO+HR is better.
Mechanism
2. Cold concentrated sulfuric acid
3. Solution of metal complexes
11.13 Epoxides
1. Catalytic oxidation of alkenes
2. Peroxidation of an alkene
O
C
CH3
CH
CH3
OOH
C
CH3
MCPBA
Cl
ether or CHCl3
MCPBA = metachloroperoxybenzoic acid
CF3COO2H will also work
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3. From halohydrin
11.14 Reactions of Epoxides
11.15 Anti 1,2-Dihydroxylation of Alkenes via Epoxides
1. Acidic cleavage
Trans
Mechanism:
2. Nucleophilic cleavage
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CH3NH2
-
1)
O
+
MgBr
CH2 CH2
2) H2O
1)
O- Na+
2) H+
Note the first step is attack by a strong nucleophile, followed by acid cleavage.
Key aspects to the formation and reactions of epoxides
Nucleophile
π bond
Electrophile
Peroxy oxygen furthest from carbonyl group
Rearrangement
There is no carbocation intermediate so this in not possible.
Stereochemistry
Both acidic and basic hydrolysis produces the trans-diol
Note “Green Chemistry” refers to processes done under environmentally friendly
conditions.
Thought question: What other nucleophiles might work?
11.16 Crown Ethers: Nucleophilic Substitution Reactions in Relatively Nonpolar Aprotic
Solvents by Phase-Transfer Catalysis
Read
11.17 Summary of Reactions of Alkenes, Alcohols, and Ethers
Review
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Spectra of alcohols and ethers
Recall:
IR
C-H stretch
O-H stretch
C-O stretch
~3000
~3600-3200 cm-1
~1150 (3°) – 1050 (1°) cm-1
(Strong for ethers)
Ar-C-O stretch
C-O stretch ~1275-1200 cm-1
NMR
O is deshielding
-CHO
δ
3.3-4
ArOH
δ
6-8
ROH
δ
1-5
ArOH and ROH involve rapid exchange of protons and are solvent dependant
UV-VIS
All absorptions are characteristic of other parts of the molecule.
Example
C4H10O
Singlet, δ4.35
Doublet, δ3.42
Multiplet, δ1.75
Doublet, δ0.89
1H
2H
1H
6H
Example
C8H10O
Singlet, δ7.25
Quartet, δ4.70
Singlet, δ3.37
Doublet, δ1.31
5H
1H
1H
3H
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Chapter 12 ALCOHOLS FROM CARBONYL COMPOUNDS: Oxidation-Reduction
and Organometallic Compounds
12.1 Introduction
Review – it will prepare you
12.2 Oxidation-Reduction Reactions in Organic Chemistry
Organic Redox
Oxidation = adding oxygen/removing hydrogen
Reduction = removing oxygen/adding hydrogen
Normally, some inorganic material is also oxidized or reduced.
General Oxidation
General Reduction
O
H
Add an oxidizing agent
KMnO4, K2Cr2O7, CrO3, or OsO4
Add a reducing agent
H2/catalyst, LiAlH4, or NaBH4
It is not necessary to know the procedures for balancing redox equations.
12.3 Alcohols by Reduction of Carbonyl Compounds
Reduction of carbonyls
1. Carboxylic acids produce primary alcohols
O
1) LiAlH4/Et2O
C
OH
2) H2O
Be careful: LiAlH4 Æ LAH Æ reacts violently with water
The actual reduction is by H:2. Esters produce primary alcohols
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Or use
Hydrogenolysis
3. Aldehydes give primary alcohols, and ketones give secondary alcohols
a. H2/Ni
b. LAH
c. NaBH4 (this is not strong enough for acids or esters)
Reactions of alcohols
12.4 Oxidation of Alcohols
1. 1° alcohols
O
O
R-CH2OH
RCHO
RCOOH
a. ROH Æ RCHO
1. Cu/Δ (actually a dehydrogenation)
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2. Dipyridine-chromium(VI) oxide
[CrO3•2py] = Sarett’s reagent of Sarett-Collins reagent
CrO3 + 2 py Æ [CrO3•2py]
3. Cr2O72-/H+/Δ
It is difficult to isolate the aldehyde before further oxidation to the carboxylic acid.
b. ROH Æ RCOOH
KMnO4(aq)/OH- (Purple (MnO4-) to brown (MnO2))
CH3
1) KMnO4/OH-/
CH3 CH
CH2 OH
2) H+
2. 2° alcohols
ROH Æ ketone
a. Cu/Δ
b. Cr2O72-/H+/Δ
c. KMnO4(aq)/OHAlso – Jones Oxidation
3. 3° alcohols
This would require the cleavage of C-C bonds
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These are extreme conditions, so this is not synthetically useful
Alcohol reactions on the O-H bond
Oxidation using I2/NaOH (NaOI) – the Iodoform test
The iodoform test is a test for the presence of a CH3 adjacent to a carbonyl or a -(CHOH)Example
Positive:
Negative:
3° so it cannot
be oxidized
Iodoform, CHI3, is a yellow precipitate with a sweet smell.
12.5 Organometallic Compounds
12.6 Preparation of Organolithium and Organomagnesium Compounds
12.7 Reactions of Organolithium and Organomagnesium Compounds
12.8 Alcohols from Grignard Reagents
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Review all these sections
1. Grignard or organolithium
Details of Grignard Synthesis
a. Preparation of the reagent
RX + Mg/dry Et2O Æ Rδ-Mgδ+X
b. Reaction of the reagent with a carbonyl group:
c. Hydrolysis of the salt
Some specific examples:
Rδ-Mgδ+X +
Any other aldehyde
Any ketone
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Exam question: What are three methods of preparing the following compound via a Grignard
reagent?
OH
C
CH3
CH2CH3
Consider:
Extensions:
Rδ-Mgδ+X +
Limitations:
General reaction:
Z may be –R, -Ar, -OR, or –Cl
Z may not be –CO2H, -OH, -NH2, -SO3H, -CO2R, -CN, or –NO2
In general, acidic groups are a problem.
12.9 Protecting Groups
Read and see earlier
FIRST REVIEW PROBLEM SET
Review here and again just before the first Chemistry 332 exam
Note: Exam 4 will contain a significant amount of review for the Final.
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