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Chapter 15
Organic Compounds and the
Atomic Properties of Carbon
15-1
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ORGANIC CHEMISTRY
Organic chemistry is the study of carbon chemistry and the
compounds of carbon.
Majority of chemical compounds on earth are organic.
The main elements involved in organic chemistry are C, H, O, N,
HC = hydrocarbon
Organic chemistry reminds us of plants and animals but is not
limited to such.
Natural medicines: penicillin, cortisone, streptomycin
Manmade medicines: novocaine, sulfa drugs, aspirin.
Natural textile fibers: nylon, Dacron, latex, rayon
Polymers: saran, Teflon, Styrofoam, plastics, polyethylene,
PBC’s
15-2
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FUNDAMENTAL ASPECTS OF
CARBON
CARBON WILL HAVE 4 COVALENT BONDS
NITROGEN WILL HAVE 3 COVALENT BONDS
OXYGEN WILL HAVE 2 COVALENT BONDS
HYDROGEN WILL HAVE 1 COVALENT BOND
REMEMBER TO MAKE SURE THESE ATOMS
ALWAYS HAVE THE ABOVE # OF BONDS
15-3
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The Structural Complexity of Organic Molecules
Reviewing the atomic structure and properties of carbon, we can get
an idea of why organic molecules can be complex.
Contributing factors include:
1. Electron configuration, electronegativity, covalent bonding
2. Bond properties, catenation, and molecular shape.
catenation - two atoms of the same element bound to each other
3. Molecular stability
•atomic size and bond strength
15-4
•available orbitals
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Chemical Diversity
Diversity in structure and behavior is due to interrelated factors:
1. Bonding to heteroatoms - See Figure 15.2
2. Electron density and reactivity
•C - C bond
EN = 0; therefore the C-C bond is nonpolar
and in general unreactive.
•C - H bond
EN ~ 0; therefore the C-H bond is nearly
nonpolar and fairly unreactive.
•C - O bond
EN = 1; therefore the C-O bond is polar
and reactive.
•bonds to other heteroatoms are usually large and therefore
•weak and reactive.
15-5
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SHAPE, GEOMETRY, STRUCTURE
Since 0rganic chemistry is limited to a small number of elements,
why are there so many molecules and compounds possible?
Structure, geometry is very important. Remember molecules exist
in 3-D and Chem RX’s occur because the approach is easiest
(requires less energy) - use molecules to show difficulty of approach
(steric hindrance)
Different geometry, shape or structure will give molecules different
physical or chemical properties.
Most common geometry for carbon compounds:
Linear, trigonal planar, tetrahedral, cyclo
Stability can be demonstrated by using models and feeling the amount
of stress needed to make the model.
15-6
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CLASSIFICATION OF HYDROCARBONS
HYDROCARBONS
ALIPHATIC HYDROCARBONS
SATURATED
HYDROCARBONS
AROMATIC HYDROCARBONS
(unsaturated hydrocarbons)
UNSATURATED
HYDROCARBONS
BENZENE AND
DERIVATIVES
alkenes
(CnH2n)
Alkanes
(CnH2n + 2)
15-7
Cycloalkanes
(CnH2n)
alkynes
(CnH2n - 2)
FUSED-RING
AROMATIC
HYDROCARBONS
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Figure 15.2
The chemical diversity of organic compounds.
4 carbons linked with single bonds, 1 oxygen and needed hydrogens.
CH3 CH2 CH2 CH2 OH
CH3 CH2 CH2 O CH3
CH3 CH2 CH CH3
OH
CH3
CH3 CH CH2 OH
CH3
O
H2C
CH2
H2C
OH
CH2
CH2
CH3 CH CH CH3
CH3 C CH3
OH
CH2 CH
CH3 CH O CH3
H2
C
H2C
CH2
O
CH
CH2
CH2
CH3
O
15-8
CH3
CH3 CH2 O CH2 CH3
H2
C
CH CH2OH
H2C
CH CH2OH
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Figure 15.2
The chemical diversity of organic compounds
continued
OH
C H
O
CH2 CH CH2 CH3
H2 C
CH CH3
O
CH3 CH2 CH2 C O
H2 C
C
CH3
O
H2C
CH CH3
CH3 CH2 C CH3
O
H
CH3
CH2
O
CH3
H2 C
CH3 CH C H
O
15-9
H2C
CH C
H
O
CH2
C
CH2
CH2
O
C
H2 C
CH CH3
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HYDROCARBONS
Carbon Skeletons and Hydrogen Skins
When determining the number of different skeletons, remember that
Each C can form a maximum of four single bonds, OR two
single and one double bond, OR one single and triple bond.
The arrangement of C atoms determines the skeleton, so a
straight chain and a bent chain represent the same skeleton.
Groups joined by single bonds can rotate, so a branch pointing
down is the same as one pointing up.
15-10
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Figure 15.3
ring
Some five-carbon skeletons
C
single bonds
C
C
double bond
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
15-11
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
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Figure 15.4
Adding the H-atom skin to the C-atom skeleton
H
C C
H
H
H
C C C
C C C
C
H
A C atom single-bonded
to one other atom gets
three H atoms.
H
A C atom single-bonded
to two other atoms gets
two H atoms.
C
C C C
H
C
C
15-12
C
C
C
C
A C atom single-bonded
to four other atom is
already fully bonded (no
H atoms).
A C atom single-bonded
to three other atoms gets
one H atom.
H
H
A double-bonded C atom
is treated as if it were
bonded to two other
atoms.
C
C
A double- and singlebonded C atom or a
triple-bonded C atom is
treated as if it were
bonded to three other
atoms.
H
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SAMPLE PROBLEM 15.1
PROBLEM:
Drawing Hydrocarbons
Draw structures that have different atom arrangements for
hydrocarbons with
(a) Six C atoms, no multiple bonds, and no rings
(b) Four C atoms, one double bond, and no rings
(c) Four C atoms, no multiple bonds, and one ring
PLAN:
Start with the longest chain and then draw shorter chains until
you are repeating structures.
SOLUTION:
(a) Six carbons, no rings
H H H H H H
H H H H H
H C C C C C C H H C C C C C H
H H H H H H
H H H
H
H C H
H
15-13
H H H H H
H C C C C C H
H H
H H
H C H
H
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SAMPLE PROBLEM 15.1
Drawing Hydrocarbons
continued
(a) continued
C
C
C C C C
C C C C
C
C
(b) Four carbons, one double bond
H H H H
H
H
H C C C C H
H C C C H
H H
H H H H
H
H C H
H C C C C H
H
15-14
H
H
(c) Four carbons, one ring
H
H C
H
H
C H
H C H
H C
H C
H
C H
H
H C
H
C H
H
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Table 15.1 Numerical Roots for Carbon Chains and Branches
PREFIX + ROOT + SUFFIX
Roots
15-15
Number of
C atoms
meth-
1
eth-
2
prop-
3
but-
4
pent-
5
hex-
6
hept-
7
oct-
8
non-
9
dec-
10
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15-16
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Figure 15.5
15-17
Ways of depicting formulas and models of an alkane
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Figure 15.6
Depicting cycloalkanes
H
H
C
H C
H
cyclopropane
15-18
C H
H
H C
C H
H C
C H
H
H
cyclobutane
H
H
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15-19
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Figure 15.7
15-20
Boiling points of the first 10 unbranched alkanes
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Figure 15.9
Two chiral molecules.
optical isomers of 3-methylhexane
15-21
optical isomers of alanine
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Figure 15.10
The rotation of plane-polarized light by an optically active substance
15-22
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Figure 15.11
15-23
The binding site of an enzyme
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15-24
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SAMPLE PROBLEM 15.2
PROBLEM:
Give the systematic name for each of the following, indicate
the chiral center in part (d), and draw two geometric isomers
for part (c).
CH3
(a)
Naming Alkanes, Alkenes, and Alkynes
(b)
CH3 C CH2 CH3
CH3
CH3
CH3 CH2 CH CH CH3
CH3
(d)
(c)
CH2
CH3 (e)
CH3
CH3 CH2 CH
CH2CH3
CH
CH2
CH3
CH3 CH2 CH C
CH CH3
CH3
PLAN:
15-25
For (a)-(c), find the longest, continuous chain and give it the base
name (root + suffix). Then number the chain so that the branches
occur on the lowest numbered carbons and name the branches with
the (root + yl). For (d) and (e) the main chain must contain the double
bond and the chain must be numbered such that the double bond
occurs on the lowest numbered carbon.
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SAMPLE PROBLEM 15.2
Naming Alkanes, Alkenes, and Alkynes
continued
SOLUTION:
methyl
methyl
CH3
(a)
butane
CH3 5CH2 4CH 3CH CH3
CH3 2C CH
CH3
1
3 2
6
4
CH3
CH3
(b)
2CH2
methyl
CH3
hexane
3,4-dimethylhexane
can be numbered in either direction
1
2,2-dimethylbutane
cyclopentane
(c)
3
4
5
2
1
CH3
methyl
methyl
CH2CH3
ethyl
1-ethyl-2-methylcyclopentane
(d)
pentene
CH3
CH3 CH2 CH CH CH2
5
4
3
2
1
1-pentene
3-methyl-1-pentene
chiral center
15-26
methyl
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SAMPLE PROBLEM 15.2
Naming Alkanes, Alkenes, and Alkynes
continued
methyl
CH3
(e)
6CH3 5CH2
CH
C
4
3
3-hexene
CH CH
1 3
2
CH3
methyl
3-hexene
CH3
H
C
6CH3 5CH2
methyl
4
CH
2
CH3
CH
1 3
methyl
cis-2,3-dimethyl-3-hexene
15-27
methyl
6CH3 5CH2
C
3
3-hexene
4
H
C
CH3
C
3
CH
2
CH3
CH
1 3
methyl
trans-2,3-dimethyl-3-hexene
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Figure 15.12 Representations of benzene
or
15-28
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15-29
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Types of Organic Reactions
An addition reaction occurs when an unsaturated reactant becomes a
saturated product:
X
Y
R CH CH R
+
X Y
R CH CH R
Elimination reactions are the opposite of addition; they occur when a
more saturated reactant becomes a less saturated product:
X
Y
R CH CH R
R CH CH R
+
X Y
A substitution reaction occurs when an atom (or group) from an added
reagent substitutes for one in the organic reactant:
R
15-30
C
X
+
Y
R
C Y
+
X
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SAMPLE PROBLEM 15.3:
PROBLEM:
Recognizing the Type of Organic Reaction
State whether each reaction is an addition, elimination, or
substitution:
CH3 CH CH2 + HBr
(a) CH3 CH2 CH2 Br
(b)
+
H2
O
(c)
CH3 C Br + CH3CH2OH
PLAN:
O
CH3 C OCH2CH3 + HBr
Look for changes in the number of atoms attached to carbon.
•More atoms bonded to C is an addition.
•Fewer atoms bonded to C is an elimination.
•Same number of atoms bonded to C is a substitution.
15-31
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SAMPLE PROBLEM 15.3:
Recognizing the Type of Organic Reaction
continued
SOLUTION:
(a) CH3 CH2 CH2 Br
CH3 CH CH2 + HBr
Elimination: there are fewer bonds to last two carbons.
(b)
+
H2
Addition: there are more bonds to the two carbons in the second structure.
O
(c)
CH3 C Br + CH3CH2OH
O
CH3 C OCH2CH3 + HBr
Substitution: the C-Br bond becomes a C-O bond and the number of bonds
to carbon remain the same.
15-32
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15-33
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15-34
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Figure 15.15 General structures of amines
the amine functional group
primary, 10, amine
15-35
C N
secondary, 20, amine
tertiary, 30, amine
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Figure 15.16
Some biomolecules with the amine functional group.
Lysine (10 amine)
amino acid found
in proteins
Epinephrine
(adrenaline; 20 amine)
neurotransmitter in
brain; hormone released
during stress
15-36
Adenine (10 amine)
component of
nucleic acids
Cocaine (30 amine)
brain stimulant;
widely abused drug
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SAMPLE PROBLEM 15.4:
PROBLEM:
Predicting the Reactions of Alcohols, Alkyl
Halides, and Amines
Determine the reaction type and predict the product(s) in the
following:
(a) CH3 CH2 CH2 I + NaOH
(b) CH3 CH2 Br + 2 H3C CH2 CH2 NH2
Cr2O72-
(c)
H3C CH CH3
OH
PLAN:
15-37
H2SO4
Check for functional groups and reagents, then for inorganics added.
In (a) the -OH will substitute in the alkyl halide; in (b) the amine and
alkyl halide will undergo a substitution of amine for halogen; in (c)
the inorganics form a strong oxidizing agent resulting in an
elimination.
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SAMPLE PROBLEM 15.4:
Predicting the Reactions of Alcohols, Alkyl
Halides, and Amines
continued
SOLUTION:
(a) Substitution - the products are CH3 CH2 CH2 OH + NaI
(b) Substitution - the products are
CH3 CH2 NHCH2CH2CH3
+ CH3CH2CH2NHBr
(c) Elimination - the product is H3C C CH3
O
15-38
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Figure 15.18
methanal
(formaldehyde)
used to make
resins in
plywood,
dishware,
countertops;
biological
preservative
15-39
Some common aldehydes and ketones.
ethanal
(acetaldehyde)
narcotic product of
ethanol
metabolism; used
to make perfume,
flavors, plastics,
other chemicals
benzaldehyde
artificial almond
flavoring
2-butanone
2-propanone
(methyl ethyl
(acetone) solvent
for fat, rubber, ketone) important
solvent
plastic, varnish,
lacquer; chemical
feedstock
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Figure 15.19
15-40
The carbonyl group.
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Figure 15.20
Some molecules with the carboxylic acid functional group.
methanoic acid
(formic acid)
an irritating
component of ant
and bee stings
benzoic acid
calorimetric
standard; used in
preserving food,
dyeing fabric,
curing tobacco
15-41
butanoic acid
(butyric acid)
odor of rancid butter;
suspected component
of monkey sex
attractant
octadecanoic acid
(stearic acid)
found in animal fats;
used in making
candles and soap
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Figure 15.24
An ester and an amide of other nonmetals.
glucose-6-phosphate
15-42
sulfanilamide
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SAMPLE PROBLEM 15.7:
PROBLEM:
Circle and name the functional groups in the following molecules:
O
(a)
Recognizing Functional Groups
C OH
O
O
(c)
(b)
OH
O C CH3
CH CH2 NH CH3
Cl
PLAN:
Use Table 15.5 to identify the functional groups.
SOLUTION:
carboxylic
O
acid
(a)
C OH O
O C CH3
ketone
(b)
(c)
alcohol
OH
O
haloalkane
CH CH2 NH CH3
Cl
ester
20 amine
alkene
15-43
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Figure 15.25
15-44
Steps in the free-radical polymerization of ethylene.
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15-45
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Figure 15.27
15-46
The structure of glucose in aqueous solution and the
formation of a disaccharide.