Organic Chemistry
Tutorial
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IUPAC
IUPAC, The International Union of Pure and Applied Chemistry, is a
worldwide organization which was established in1918. It’s main
purpose is to aid in chemical sciences as well as their applications
with humankind. Organic nomenclature, being a large focus of
IUPAC, was established in 1892 when chemists created a list of rules
called the Geneva rules. This group of chemists ultimately became
IUPAC and is who we should thank for organic nomenclature.
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IUPAC Nomenclature Rules
IUPAC has 4 fundamental steps when naming organic compounds.
The following steps should be considered:
(1) Identify and name the parent chain.
The Parent Chain is usually the single largest continuous chain of
carbon atoms within an organic molecule. Its name is dependent on
the number of carbon atoms. Its name should be determined by the
following prefixes.
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Parent Chain Prefixes
Carbon
Atoms
Parent
Prefix
Carbon
Atoms
Parent
Prefix
1
Meth
11
Undec
2
Eth
12
Dodec
3
Prop
13
Tridec
4
But
14
Tetradec
5
Pent
15
Pentadec
6
Hex
20
Eicos
7
Hept
30
Triacont
8
Oct
40
Tetracont
9
Non
50
Pentacont
10
Dec
100
Hect
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IUPAC Nomenclature Rules
(2) Identify and name the substituent’s.
The Substituent’s are the chain like
projections from the parent chain.
The majority of these chains follow
the parent chain prefixes. They contain the
Suffix “yl” added to the prefix.
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You can see that the longest possible
chain is 9 without the incorporation of the
2 carbon chain. This is an example of a
substituent.
IUPAC Nomenclature Rules
(3) Number the parent chain and assign a locant to each substituent.
A Locant is the carbon number of the substituent attached to the
parent chain. It is important that you number a parent chain such
that the sum of all locant’s is the least possible. This standard is used
to prevent the same molecule being named twice simply by false
numbering.
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IUPAC Nomenclature Rules
(4) Assemble the substituent's in alphabetical order.
You should only consider the alphabetical order of the parent chain
prefix, not additional sub-prefixes (di, tri, etc.) which are attached to
the beginning of a prefix for a substituent if there exist more than
one on a parent chain of the same type.
Know that you have the 4 fundamental IUPAC steps…
Let’s start naming!
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Alkanes
Alkanes are organic compounds that contain fully saturated parent
chains. That is, they are the hydrocarbons that contain the most
hydrogen atoms possible (No Double or Triple Bonds). These
molecules contain Sigma bonds between all C-C Bonds. When
naming alkanes, the suffix “ane” is attached to the parent chain
prefix.
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Cycloalkanes
Cycloalkanes are the alkanes which exhibit a ring structure. These
molecules will contain the “cyclo” suffix added to the parent chain
prefix if the parent chain being considered is the ring structure. The
“cyclo” suffix can also be conjoint with the parent chain suffix if
considered an Alykl Group (substituent's that are of alkane essence).
For the second case, the parent chain will be named as normal
alkanes.
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Question Section
(1) QUESTION:
Name the following Organic Molecule:
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ANSWER
HELP
Question Section
(1) ANSWER:
STEP 1: Identify and name the parent.
Since there are 6 carbons there, the parent name should be hexane.
And since there are no substituent's, there is not need to follow the
next 3 steps.
ANSWER: Hexane
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ANSWER
HELP
Question Section
(2) QUESTION:
Name the following Organic Molecule in two ways:
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ANSWER
HELP
Question Section
(2) ANSWER:
Consider the Cycloalkane as the Parent
STEP 1: Identify and name the parent.
The parent is the 6 carbon atom cycloalkane. Which should be named cyclohexane.
STEP 2: Identify and name the substituent's.
The only substituent's is an 5 carbon atom alkyl group. Which should be name pentyl.
Since the other two steps aren’t required for this molecule we achieve a name of
ANSWER: Pentyl Cyclohexane
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ANSWER
HINT
Question Section
(2) ANSWER:
Consider the Alkane as the Parent
STEP 1: Identify and name the parent.
The parent is the 5 carbon atom alkane. Which should be named pentane.
STEP 2: Identify and name the substituent's.
The only substituent's is the 6 carbon cycloalkane. Which should be name cyclohexyl.
Since the other two steps aren’t required for this molecule we achieve a name of
ANSWER: 1 - Cyclohexyl Pentane
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ANSWER
HELP
Question Section
(3) QUESTION:
Name the following Organic Molecule:
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ANSWER
HINT
Question Section
(3) ANSWER:
STEP 1: Identify and name the parent.
The parent chain is the 7 carbon chain
Which should be named HEPTANE.
STEP 2: Identify and name the substituent's.
The two substituent's that exist are ETHYL and the CYCLOHEXYL
STEP 3: Number the parent chain and assign a locant to each. It is clear by the diagrams below that the
best option is 1 for cyclohexyl and 4 for
Ethyl since the sum is less than 4 and 7.
STEP 4: Alphabetize
1
2
3
4
5
6
7
7
6
ANSWER: 4-ETHYL-1-CYCLOHEXYLHEPTANE
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ANSWER
HELP
5
4
3
2
1
Question Section
(4) QUESTION:
Alkanes are typically straight chain molecules due to the alignment
of sigma bonds.
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ANSWER
HINT
Question Section
(4) ANSWER:
Alkanes aren’t typically straight chain molecules. Straight chain
molecules are due to the overlapping of pi bonds and are usually
found in alkynes.
ANSWER: FALSE
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ANSWER
HELP
Complex Substituents
The following structures have special names associated with their
type of bonding:
Isopropyl
3-Carbon
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Sec-Butyl
Tert-Butyl
4-Carbon
Isobutyl
Phenyl
Benzene
Alkyl Halides
These substituents are those in the halogen column of the periodic
table:
Fluoro
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Chloro
Bromo
Iodo
Alkenes
Alkenes are organic compounds that contain at least one
unsaturated carbon. That is, they are the hydrocarbons that contain
at least one double bond. These molecules contain Sigma bonds
between all C-C Bonds and a Pi bond between all C=C. When naming
alkenes, the suffix “ene” is attached to the parent chain prefix as
well as the appropriate location of the double bond.
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Alkynes
Alkynes are organic compounds that contain at least one triple
bond. These molecules contain Sigma bonds between all C-C Bonds
and a 2 Pi bond between all C≡C. When naming alkynes, the suffix
“yne” is attached to the parent chain prefix as well as the
appropriate location of the triple bond.
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Question Section
(5) QUESTION:
Name the following Organic Molecule:
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ANSWER
HINT
Question Section
(5) ANSWER:
STEP 1: Parent chain is 10 Carbons long which must include the triple
bond. We will include the double bond as a side chain for ease of
naming and since the triple bond has that higher authority.
STEP 2: 4 substituents (3 alkyl halides: 2 chlorines and 1 fluorine, 1
double bond ethyl side chain)
STEP 3: Chloro at 7 and 9 (must count from the carbon
closest to the triple bond) , Fluoro at 6, allyl (the name for the ethyl
double bond) at 8.
STEP 4: Alphabetize (allyl, dichloro, fluoro)
ANSWER: 8-allyl-7,9-dichloro-6-fluoro-2-decyne
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ANSWER
HELP
Question Section
(6) QUESTION:
Count the number of carbon’s,
Hydrogens.
Name this molecule.
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ANSWER
HINT
Question Section
(6) ANSWER:
STEP 1: Parent chain is 9 Carbons long which must
include the double bond.
STEP 2: 4 substituents (1 alkyl halide, 1 Tert Butyl,
1 alkyl, 1 cycloalkyl)
STEP 3: iodo at 7, tert Butyl at 6, methyl at 3,
Cyclohexyl at 2.
STEP 4: Alphabetize (tert Butyl, cyclohexyl, Iodo,
methyl)
ANSWER: 6-tert Butyl-2 cyclohexyl-7-iodo-3 methyl-1,3-nondiene (20 CARBONS, 33 HYDROGENS)
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ANSWER
HELP
Question Section
(7) QUESTION:
Name this molecule:
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ANSWER
HINT
Question Section
(7) ANSWER:
STEP 1: Parent chain is 7 Carbons long.
STEP 2: 6 substituents (4 phenol, 2 alkyl halides)
STEP 3: Phenyl at 1,2,6,7 and chloro at 3,5.
STEP 4: Alphabetize (chloro, phenyl)
ANSWER: 3,5-dichloro-1,2,6,7-tetraphenyl-heptane
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ANSWER
HELP
Alcohols and Phenols
Alcohols are organic compounds that contain an OH, hydroxyl
group, and are named according to an ending in “ol”.
Phenols are alcohols comprised of a hydroxyl group attached
directly to a phenyl ring.
When numbering the parent chain, alcohol should receive lowest
numbering despite presence of alkyl substituent's or pi bonds.
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Ethers
Ethers are compounds that contain an oxygen atom bonded
between two R groups, where each R group can be an alkyl, aryl
(aromatic compounds like benzene), or vinyl (double bonded side
chains).
Two methods of naming:
1) Name the 2 substituents alphabetically followed by “Ether”
2) Place “oxy” between a side chain and a parent chain
respectively.
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Epoxides
Epoxides are cyclic ethers that contain an oxygen within a
cycloalkane. A special type is Oxirane, which is the triangular
epoxide that is more reactive than other ethers due to a significant
ring strain.
When naming as parents use the following Parent Chains:
Oxirane
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Oxetane
Oxolane
Oxane
When naming as
side chain, merge
epoxy with the
position of the
“overlapping
carbons”.
Question Section
(8) QUESTION:
Name this molecule in two ways:
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ANSWER
HINT
Question Section
(8) ANSWER:
OPTION 1
Sides chains are butyl and pentyl
Thus: Butyl Pentyl Ether
OPTION 2
Take Pentane as the Parent chain and Butyl as the side chains
Thus: Butoxypentane
ANSWER: Butyl Pentyl Ether
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Butoxypentane
ANSWER
HELP
Question Section
(9) QUESTION:
Name all the following molecules. Using your knowledge from
previous chemistry courses, order the following molecules in order
of increasing boiling point.
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ANSWER
HINT
Question Section
(9) ANSWER:
1 hydrogen bond interaction per
molecule
Compound 1: 1-Butanol
2 hydrogen bond interactions per
molecule
Compound 2: 1,4-Butandiol
No hydrogen bonds no major
electronegative change
Compound 3: butane
Slight electronegative change
between the oxygen atom and carbon
atoms, creates dipole on the oxygen.
Compound 4: methoxyethane
ANSWER: Butane < Methoxyethane < 1-Butanol < 1,4-Butandiol
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ANSWER
HELP
Question Section
(10) QUESTION:
Name all the following molecule:
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ANSWER
HINT
Question Section
(10) ANSWER:
STEP 1: Parent chain is 12 carbons long with
A triple bond, thus dodecyne.
STEP 2: There are 5 substituents (2 isopropyl,
1 epoxy, 1 phenyl, 1 methyl)
STEP 3: isopropyl is at the 6 and 9 position,
The epoxy at 7 and 8, phenyl at 10 and methyl at 2.
STEP 4: Alphabetize (epoxy, isopropyl, methyl, phenyl)
ANSWER: Butane < Methoxyethane < 1-Butanol < 1,4-Butandiol
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ANSWER
HELP
Thiols and Sulfides
Since Sulphur is under oxygen in the periodic table, it is often the
case the oxygen contain organic molecules will have a sulphur
derivative. Alcohols, oxygen bound to hydrogen and carbon, is
referred to as a THIOL when replaced with sulphur. Add the term
“thiol” to the parent chain for SH containing molecules. If other
functional groups are present, use “mercapto” as the substituent
name. Ethers, oxygen bound to two carbons on either side, is
referred to as a SULFIDE when replaced with sulphur. Replace
“sulfide” with ether or replace “ylthio” with “oxy” for parent chains
and substituents respectively.
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Thiols and Sulfides
Here are some examples of these molecules:
1,4-Pentanedithiol
Diethyl Sulfone
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Ethylthioethane
Diethyl Sulfide
Diethyl Sulfoxide
4,6-dimercapto-3-octyne
Sulfides that undergo
oxidation reactions
are called Sulfoxide
(for one oxygen) and
Sulfone (for two
oxygen)
Aldehydes and Ketones
A CARBONYL group is an oxygen group double bonded to a Carbon.
Within an organic molecule this can happen at two locations, First,
the very end or very beginning of the molecule. This type of
molecule is referred to as an ALDEHYDE. Attach the suffix “al” to the
parent chain when naming. Use “carbaldehyde” when an aldehyde
group is attached on the first carbon after a cyclic compound.
Secondly, anywhere except the very end and the very beginning.
This type of molecule is referred to as a KETONE. Attach the suffix
“one” to the parent chain when naming. You may also use the
method similar to ether, where you will place the parent name of
each chain and followed by “ketone”.
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Aldehydes and Ketones
Here are some examples of these molecules:
Methyl propyl ketone
2-pentanone
Methanal
Formaldehyde
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Pentanal
Cyclopentanecarbaldehyde
Formaldeyhde is a popular preservative as it has the
ability to prevent the growth of common bacteria and
fungi. It is the simplest form of aldehyde and is a
significant consideration to human health as it may be
carcinogenic.
Question Section
(11) QUESTION:
Name the molecule:
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ANSWER
HINT
Question Section
(11) ANSWER:
STEP 1: Parent chain is 7 Carbons long.
STEP 2: substituents ( 1 phenyl, 2 carbonyl groups, 1 thiol)
STEP 3: Phenyl at 3, Carbonyl groups at 2, 6, Thiol at 4.
STEP 4: Alphabetize (mercapto, phenyl)
ANSWER: 4-mercapto-3-phenyl-2,6-heptandial
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ANSWER
HELP
Question Section
(12) QUESTION:
Name all the following molecules. Using your knowledge from
previous chemistry courses, order the following molecules in order
of increasing boiling point.
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ANSWER
HINT
Question Section
(12) ANSWER:
Ethanethiol
2-mercaptoethanol
1,2-Ethanedithiol
Ethanedial
2-mercaptoethanal
ANSWER: Ethanethiol < Ethanedial < 2-mercaptoethanal < 1,2-Ethanedithiol < 2-mercaptoethanol
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ANSWER
HELP
Question Section
(13) QUESTION:
The following molecule is the commonly known analgesic (Drug that
reduces pain) and antipyretic (Drug that reduces fever),
Acetaminophen (Tylenol™). Name all functional groups within the
Molecule.
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ANSWER
HINT
Question Section
(13) ANSWER:
Amine
Ketone
Hydroxyl
Benzene
ANSWER: hydroxyl, ketone, amine, benzene
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ANSWER
HELP
Carboxylic Acids
CARBOXYLIC ACIDS are those molecules containing a carbon bonded
to a hydroxyl and bonded to a carbonyl group simultaneously. When
naming these molecules, ensure the addition of “oic” to the parent
name followed by “acid”.
The Carboxylic Acid
Functional Group
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Amines and Amides
AMINES are the derivatives of Ammonias in which the hydrogen atoms (protons)
have been replaced with alkyl groups (carbon) or aryl groups (benzene). As a
parent chain the addition of the suffix “amine” is used. As a substituent, “amino is
used. AMIDES are derivatives of carboxylic acids where the hydroxyl group has
been substituted for a nitrogen group. Amides will often include structure that
branch from the nitrogen group. The use of the suffix “amide” is used for as the
parent chain.
The Amine
Functional Group
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The Amide
Functional Group
Esters
ESTERS are those molecules where a carbonyl group and a ether groups are
situated on the same carbon. Naming esters involves using the suffix “oate”. You
will have two side chains, and thus will name them as follows. The side chain
whose derivative is a carboxylic acid will be considered the parent side chain, and
the side chain whose derivative is an alcohol will be a substituent.
The Alcohol
Derivative
The Carboxylic Acid
Derivative
The Ester Function
Group
Butyl Propanoate
H₂O
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Question Section
(14) QUESTION:
Name the molecule:
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ANSWER
HINT
Question Section
(14) ANSWER:
STEP 1: Parent chain is 3 Carbons long.
STEP 2: substituents ( 1 amino)
STEP 3: Amino at 2
ANSWER: 2-Aminopropanoic acid
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Alanine
ANSWER
HELP
Question Section
(15) QUESTION:
Viagra™ the popular erectile dysfunction drug contains many
interesting functional groups. Name all functional groups and state
the number of tertiary amine groups.
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ANSWER
HINT
Question Section
(15) ANSWER:
ANSWER: 3 Tertiary Amines,
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Sulfone, Amine, Amide, Benzene (alkane and alkene)
ANSWER
HELP
Question Section
(16) QUESTION:
Name the following Molecule:
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ANSWER
HINT
Question Section
(16) ANSWER:
ANSWER: (3 amino-5-hydroxy-4-mercapto-2-oxo)-propyl propanoate
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ANSWER
HELP
Others
Here are some other names you must be familiar with:
PEROXIDE
NITRILE
ARENE
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DO YOU REMEMBER
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Periodic Table Trends
EFFECTIVE NUCLEAR CHARGE
Although this is typically not a trend
discussed in class, it is very important in
understanding other periodic trends.
Is the average nuclear charge felt by an individual electron in an atom, taking into
consideration the “shielding” effect of inner-shell electrons. Since negatively
charged electrons are attracted to the positively charge protons in the nucleus,
and at the same time, repelled by other electrons in the atom, we use effective
nuclear charge to determine the positive charge on a specific electrons in a
specific orbital. A general approach to calculating nuclear charge is using this
equation where Z is the total number of protons in an atom and S is the number
of electrons in the inner-orbitals of the orbital at interest.
𝑍𝑒𝑓𝑓 = 𝑍 − 𝑆
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Periodic Table Trends
EFFECTIVE NUCLEAR CHARGE
-
Electron of interest
-
-
Electron of interest
-
-
Na
11+
-
-
-
-
-
And so we see that
sodium has a less
effective nuclear
charge than
chlorine.
-
-
𝒁𝒆𝒇𝒇 = 𝒁 − 𝑺 = 𝟏𝟏 − 𝟏𝟎 = 𝟏
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-
-
-
Cl
17+
-
-
-
-
-
-
-
𝒁𝒆𝒇𝒇 = 𝒁 − 𝑺 = 𝟏𝟕 − 𝟏𝟎 = 𝟕
Periodic Table Trends
ATOMIC RADII
Is the measure of the size of an element from it’s nucleus to its outer most cloud
of electrons. Typically, we see as we go from
left to right on the periodic table, Atomic
Radii DECREASES since the effective nuclear
charge increases and electrons in the outer
most orbital are more effectively attracted
to the positive nucleus. We also see that if
we go from top to bottom of the periodic
table Atomic Radii INCREASES since
electrons are filling new outer orbitals.
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Periodic Table Trends
IONIZATION ENERGY
The amount of energy required to remove an
electron from a gaseous atom (i.e. an atom
that is all by itself, not hooked up to others is
in a solid or liquid). Typically, the ionization
energy INCREASES from left to right on the
periodic table since the effective nuclear
charge increases, meaning that electrons are
more effectively attracted to the nucleus and
this will require more energy to be extracted.
The ionization energy DECREASES from top to
bottom of the periodic table since electrons in the distance from the nucleus to the
outermost shell is increased and thus the effective nuclear charge on an electron is
weakened due to the distance of attraction.
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Periodic Table Trends
ELECTRONEGATIVITY
Is the power of an atom to attract electrons to
Itself. Typically, as we go from left to right on
the periodic table we see that
electronegativity INCREASES because effective
nuclear charge increases and thus an atom’s
ability to effectively attract electrons is higher.
as we go from top to the bottom of the
periodic table we see that electronegativity
DECREASES since effective nuclear charge is
weakened due to the distance from the nucleus to the further most orbital.
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Question Section
(17) QUESTION:
14
𝐶,
a radioactive isotope, contains two extra neutrons. What effect
would this have on the atomic radii of carbon? Explain.
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ANSWER
HINT
Question Section
(17) ANSWER:
The addition of neutrons within an atom will have NO (by our definition of nuclear charge) effect on
the atomic radii. Since Atomic Radii is the distance from the nucleus to the outer most cloud of
electrons which is determined by the effective nuclear charge and its ability to effectively attract
electrons toward the nucleus, we notice that the addition of neutrons will not affect our nuclear
charge and thus 14𝐶 will exhibit an identical Atomic Radii.
NEXT QUESTION
ANSWER
HELP
Question Section
(18) QUESTION:
What is true about 𝑂2− :
i. 𝑂2− has a greater atomic radii than 𝑂
ii. 𝑂2− has a greater electronegativity than 𝑂
iii. 𝑂2− has a smaller ionization energy than 𝑂
a) i
b) ii
c) iii
d) i & ii
e) ii & iii
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ANSWER
HINT
Question Section
(18) ANSWER:
i)
This answer is TRUE. More electrons in a orbital with the same effective nuclear charge between
atoms implies that the electrons are not as tightly bound as they are when more are present.
ii)
This answer is FALSE. Ionic oxygen has already received two electrons and is stable. The addition
of any more electrons would imply the placement in a new orbital which would result in a
electron that has a 0 effective nuclear charge since all inner most electrons pair up with a proton.
iii) This answer is FALSE. Ionic oxygen is stable and thus the amount of energy to remove an electron
is more than if oxygen wasn’t stable.
ANSWER: A
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ANSWER
HELP
Question Section
(19) QUESTION:
Arrange the following Atoms in increasing Atomic Radii, if the inequality is
reversed, does this match the definition of increasing electronegativity?
(Phosphorus, Neon, chlorine, Arsenic)
a) Chlorine < Neon < Phosphorus < Arsenic
b) Arsenic < Phosphorus < Chlorine < Neon
c) Neon < Chlorine < Phosphorus < Arsenic
d) Arsenic < Phosphorus < Neon < Chlorine
e) Phosphorus < Neon < Chlorine < Arsenic
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ANSWER
HINT
Question Section
(19) ANSWER:
Following the Trend on the periodic table, we notice the appropriate increasing
order of atomic radii of these elements is Neon < Chlorine < Phosphorus < Arsenic.
If this inequality is switched, we have Neon > Chlorine > Phosphorus > Arsenic,
which is false in terms of increasing electronegativity as noble gases have a
electronegativity of 0.
ANSWER: C , False
NEXT QUESTION
ANSWER
HELP
Bond Type
There are three different types of bonds we will consider:
NONPOLAR COVALENT: The sharing of electrons equally, typically found in
identical atoms or atoms with slight electronegativity difference.
POLAR COVALENT: The sharing of electrons unequally due to the effective nuclear
charge on one atom having a higher effective pull of electrons than that of the
other atom.
IONIC: The taking of electrons from an atom whose effective nuclear charges are
on both sides of the “spectrum”, results in atoms being stable and similar to that
of their respected noble gas.
Determining bond types by electronegativity difference:
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.4
1.7
Determining Lewis Structure
When determining Lewis Structure of a molecule, we utilize the valence electrons number as the electrons available
to form bonds and lone pairs within a molecule. We will also use the octet rule which allows states that all atoms
must have 8 electrons around it.
Lets determine the Structure of the following Molecule (𝐶2 𝐻4 𝑂2 ) :
Valence Electrons
C = 2x4 = 8
O = 2x6 = 12
H = 1x4 = 4 -> 28 available electrons to form bonds and lone pairs
*** Keep in mind of the most common forms of each individual atom…Oxygen “typically” has two lone pairs and two
bonds, Carbon “typically” has 4 bonds. Hydrogen “typically” has 1 bond.
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Formal Charge
Formal Charge, is defined as the charge associated with an atom that does not exhibit the appropriate
number of valence electrons. In other terms, atoms situated in a molecule all have a standard valence
electron value according to their position in the periodic table. If for some reason the number of
electrons they are contributing within a molecule is different then their valence electron value, we
assign a Formal Charge according to the addition or loss of electrons contributing within a molecule for
the particular atom.
Oxygen has 6 valence
electrons, in this molecule,
oxygen has 3 pairs of lone
pairs (6 valence electrons) +
1 electron contributing in a
sigma bond. Thus, we assign
a Formal Charge of -1
Chlorine has 7 valence
electrons, in this molecule,
chlorine has 2 lone pairs (4
valence electrons) and 2
electrons contributing in
sigma bonds. Thus, we assign
a Formal Charge of +1.
In general, we can define Formal Charge as follows:
𝐹𝐶 = # 𝑜𝑓 𝑉𝑎𝑙𝑒𝑛𝑐𝑒 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 − # 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑖𝑛 𝐿𝑜𝑛𝑒 𝑃𝑎𝑖𝑟𝑠 −
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(# 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑖𝑛 𝐵𝑜𝑛𝑑𝑠)
2
Question Section
(20) QUESTION:
Determine the Formal Charge on Oxygen in the following Molecules:
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(20) ANSWER:
Oxygen has 6 valence
electrons, in this
molecule, oxygen has
2 pairs of lone pairs
(4 valence electrons)
+ 2 electron
contributing in a
sigma bond. Thus,
we assign a Formal
Charge of 0
The Carbon Atom in
this molecule breaks
this octet rule, under
standard conditions,
this molecule does
not exist. Though, if
we were to assign a
Formal Charge, it
would be +1
Oxygen has 6 valence
electrons, in this
molecule, oxygen has
3 pairs of lone pairs
(6 valence electrons)
+ 1 electron
contributing in a
sigma bond. Thus,
we assign a Formal
charge of -1.
Oxygen has 6 valence electrons, in this molecule, oxygen has 1
pairs of lone pairs (2 valence electrons) + 3 electrons
contributing in sigma bonds. Thus, we assign a Formal charge
of +1.
ANSWER:0 , +1(if considered a standard molecule), -1, +1
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(21) QUESTION:
Determine the Lewis Structure of the following Molecular Formulas:
𝐶1 𝐻2 𝑂2
𝑁𝑎1 𝐶4 𝐻3 𝑂2
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(21) ANSWER:
Solving these problems usually
requires multiple figures. But we
can use them to determine the
appropriate figure. Lets move
one of the hydrogen atoms to
the oxygen and make the other
oxygen a double bond.
𝐶1 𝐻2 𝑂2
Valence Electrons
C=4
=4
H = 2x1
=2
O = 2x6
= 12
= 18e⁻
C typically has 4 Bonds, H typically has 2 Bonds, O typically has 2 bonds and 2 lone pairs
Satisfies all
Requirements
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(21) ANSWER:
𝑁𝑎1 𝐶4 𝐻3 𝑂2 → 𝑁𝑎+ + 𝐶4 𝐻3 𝑂2 −
Valence Electrons
C = 4x4
= 16
H = 3x1
=3
O = 2x6
= 12
= 1e⁻
= 32e⁻
Electrons in outer most orbital
C = 8x4
= 32
H = 3x2
=6
O = 2x8
= 16
= 54e⁻
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Since there exists an additional
electron, we know that the
general rules for atoms cannot
be applied, thus we have a new
rule that determines the amount
of bonds in the molecule:
We can deduce the number of
bonds by (54-32)/2 = 11. Since
this results in bonds with other
molecules, lone pairs would be
included in the valence
electrons, which we have
subtracted.
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(21) ANSWER:
𝑁𝑎1 𝐶4 𝐻3 𝑂2 → 𝑁𝑎+ + 𝐶4 𝐻3 𝑂2 −
ANSWER
This molecules is missing the
negative charge. Lets moving the
hydrogen on the oxygen,
creating the negative charge and
place it on the last carbon.
This molecule breaks the octet
rule on carbon 4. Let remove the
double bond and make a triple
bond between carbon 2 and 3.
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ANSWER
This molecule satisfies all requirements!
Though notice the negative charge could
exist on the double bonded oxygen. Will
discuss soon.
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(22) QUESTION:
Epinephrine, the hormone used to aid with anaphylaxis, exhibits
multiple bond types. State the bonds types present within the
molecule and give an example of their location.
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(22) ANSWER:
Polar Covalent Bonds
Non-Polar Covalent
Bonds
ANSWER: Multiple Answers Possible, Note, no ionic bonds present!
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VSEPR
Valence Shell Electron Pair Repulsion (VSEPR) Theories main purpose is to show that “electron pairs” (i.e. bonds and
lone pairs) will exist within a molecule such that there location in space is to minimize repulsion with other atoms and
electron pairs. This theory has given us the following observed shapes within molecule and these structure have been
proven via other methods (i.e. X-ray Crystallography).
TETRAHEDRAL
all atoms
spaced 109.5°
(ex. Methane)
TRIGONAL PLANAR all
atoms spaced 120°
(ex. Boron Trifluoride)
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LINEAR
all atoms spaced 180°
(ex. Beryllium Difluoride)
TRIGONAL
PYRIMIDAL
atoms spaced 107°
(ex. Ammonia)
BENT
atoms spaced
104.5°
(ex. Water)
Atomic Orbitals
𝜓 2 = the probability of where electrons
exist in space, determined from 𝜓 , which is
the solved wave equation (wave function)
that takes into account the wave property
of electrons.
Some Principles About Placement of Electrons
AUFBAU PRINCIPLE: lowest-energy filled first.
PAULI EXCLUSION PRINCIPLE: each orbital can have a maximum of two electrons that have opposite spins.
HUND’S RULE: when dealing with degenerate orbitals, such as p orbitals, one electron is placed in each degenerate
orbital first.
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Hybridization
__ __ __
2P
__
2S
__
1S
This does not coincide with what we see. So, let’s say
during formation,
carbon undergoes a
HYBRIDIZE
“HYBRIDIZATON” of 2S and 2P orbitals, we will call
them sp³ hybridized orbitals as the are the average of
1 s orbital and 3 p orbitals.
ENERGY
ENERGY
If we look back at VSEPR, it tells us the positioning of atoms relative to each other, (i.e. Methane = 109°). If we look at
Orbitals we would expect methane to have angles of 90°…but this is not observed??? If we consider the electron
configuration of Carbon in methane we find:
__ __ __ __
SP³
__
1S
An average of 3 p orbitals and 1 s orbital, an sp³ orbital
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Hybridization
__ __ __
2P
__
2S
__
1S
HYBRIDIZE
ENERGY
ENERGY
METHANE
__ __ __ __
SP³
__
1S
H
H
H
H
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Hybridization
__ __ __
2P
__
2S
__
1S
HYBRIDIZE
__
2P
ENERGY
ENERGY
BORON TRIFLUORIDE
__ __ __
SP²
__
1S
F
F
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F
Hybridization
__ __ __
2P
__
2S
__
1S
HYBRIDIZE
__ __
2P
ENERGY
ENERGY
BERYLLIUM DIHYDRIDE
__ __
SP
__
1S
H
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H
Hybridization
SIGMA BONDS: are covalent bonds formed from the overlap of atomic orbitals
PI BONDS: are covalent bonds formed from the overlap of p atomic orbitals.
ENERGY
__
2P
__ __ __
SP²
__
1S
The Hybridized electron
configuration of Carbon in
ethylene (ethene). Notice
the 2P orbital, it requires
one more electron, which
it receives from the second
carbon
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Question Section
(23) QUESTION
Tetrahydrocannabinol, is the principal psychoactive constituent of the cannabis
plant. It’s chemical structure is very interesting! What is the number of pi bonds
and sigma bonds?
a)
b)
c)
d)
e)
𝜎
𝜎
𝜎
𝜎
𝜎
= 57
= 26
= 26
= 53
= 56
𝜋=4
𝜋=4
𝜋=3
𝜋=4
𝜋=4
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(23) ANSWER:
Count…All bonds (including the
OH bond and double bonds)
is considered a Sigma bond.
All Pi bonds are double bonds
Or Triple bonds (Note,
Though this molecule does not
Contain a triple bond, triple
Bonds account for two pi
Bonds).
ANSWER: A
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(24) QUESTION
When a double bond is formed between two atoms, one of the bonds is a sigma
bond and the other is a pi bond. The pi bond is created by the overlap of...
a)
b)
c)
d)
e)
sp
sp²
sp³
P orbitals
S orbitals
RESOURCE: http://chemistry.boisestate.edu/people/richardbanks/inorganic/mc111/vol10/mcquestions111j.htm
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(24) ANSWER:
Ethene, an example of
a double bond, is
formed from the
overlap of two sp²
orbitals.
ANSWER: B
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(25) QUESTION
What is the electron-pair geometry of the central oxygen atom of ozone (O3)?
a)
b)
c)
d)
e)
Linear
Trigonal planar
Bent
Tetrahedral
Trigonal Pyramidal
RESOURCE: http://chemistry.boisestate.edu/people/richardbanks/inorganic/mc111/vol10/mcquestions111j.htm
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(25) ANSWER:
Central Atom is Trigonal Planar
Lone Pair
ANSWER: B
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Resonance
Suppose we had a divided chamber of gas where one chamber the gas had a higher pressure than the other:
__ __ __ The hybridizing of a
positive carbon.
2P
__
HYBRIDIZE
2S
__
1S
__
2P
ENERGY
Suppose we were studying the orbital
diagram of the following molecule:
ENERGY
If the slit in the chamber were to be removed, What would happen?
__ __ __
SP²
__
1S
Notice the hybridization of the positively charge carbon, The P orbital is empty and all SP² orbitals have been filled by 2 bonds between 2
hydrogen atoms and a bond between a carbon atom.
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Resonance
Thus, the orbital diagram (if we only draw pi bond) would look like:
Note, the empty P orbital
C
C
C
If we consider the electron cloud over Carbon 1 to Carbon 2 as the chamber of high pressure and the electron cloud
from Carbon 2 to Carbon 3 as the chamber of low pressure, what should we expect?
Some sort of combination between the first structure and this structure:
This is what we will refer to as RESONANCE, and each of these structures
Will be called Resonating Structures.
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C
C
C
Resonance
The problem is with our method of drawing structures, Lewis Diagrams, but since these are so practical and so well
known, we must deal with learning resonance via these structures though they should be thought an average , not a
equilibrium, among electron pi clouds. Though, we still must abide by standard rules including the Octet Rule. So, we
will draw all Resonating Structures as Lewis Diagrams but think of them as an average of all structures combined:
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Rules for Resonance
• Only electrons can move between resonance structures.
• The position of the atoms does not change
• The only electrons that can participate in resonance are nonbonding electrons (lone pairs) and pi electrons (double or triple
bonds)
• Sigma bonds cannot participate in resonance (cannot break sigma
bonds)
• Cannot EXCEED the octet rule for elements in rows 1 and 2 (8
electrons max)
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Question Section
(26) QUESTION
Determine the Resonating Structures of the following molecule
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(26) ANSWER:
C
C
C
C
C
ANSWER:
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C
C
C
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(27) QUESTION
Determine the Resonating Structures of the following molecule:
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(27) ANSWER:
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(28) QUESTION
Determine the Resonating Structures of the following molecule:
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(28) ANSWER:
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(29) QUESTION
Determine the Major Contributor of this molecule:
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(29) ANSWER:
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(30) QUESTION
Considering all of the major resonance structures for the molecule below; the
number of major resonance contributors in which a negative charge resonates
onto an oxygen atom is?
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(30) ANSWER:
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Intermolecular Forces
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