Download presentation source

Chapter 15
Organic Compounds and the Atomic
Properties of Carbon
15-1
Organic Compounds and the Atomic Properties of Carbon
15.1 The special nature of carbon and the characteristics of
organic molecules
15.2 The structures and classes of hydrocarbons
15.3 Some important classes of organic reactions
15.4 Properties and reactivities of common functional groups
15.5 The monomer-polymer theme I: Synthetic macromolecules
15.6 The monomer-polymer theme II: Biological macromolecules
15-2
The position of carbon in the periodic table
Figure 15.1
15-3
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
H2
C
15-4
H2C
CH2
O
CH
CH2
CH2
CH3
O
CH3 CH CH CH3
CH3 C CH3
OH
CH2 CH
CH3 CH O CH3
CH2
Figure 15.2
CH3
CH3 CH2 O CH2 CH3
H2
C
CH CH2OH
H2C
CH CH2OH
Figure 15.2 (continued)
OH
C H
O
CH2 CH CH2 CH3
H2 C
CH CH3
O
CH3 CH2 CH2 C O
H2C
C
CH3
O
H2C
CH CH3
CH3 CH2 C CH3
O
H
CH3
CH2
O
CH3
H2C
CH3 CH C H
O
15-5
H2C
CH C
H
O
CH2
C
CH2
CH2
O
C
H2C
CH CH3
HYDROCARBONS
Carbon skeletons and hydrogen skins
When determining the number of different skeletons, remember that:
Each carbon forms a maximum of four single bonds, OR two
single and one double bond, OR one single and triple
bond.
 The arrangement of carbon 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.

15-6
Some five-carbon skeletons
single bonds
Figure 15.3
15-7
double bonds
ring structures
Adding a H-atom skin to a carbon skeleton
Figure 15.4
15-8
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 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
15-9
H H H H H
H C C C C C H
H H
H H
H C H
H
SAMPLE PROBLEM 15.1 (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
15-10
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
Table 15.1
Numerical Roots for Carbon Chains and Branches
roots
15-11
number of
carbon atoms
meth-
1
eth-
2
prop-
3
but-
4
pent-
5
hex-
6
hept-
7
oct-
8
non-
9
dec-
10
15-12
Ways to depict formulas and models of an alkane
Figure 15.5
15-13
Depicting cycloalkanes
H
H
H
C
H C
cyclopropane
Figure 15.6
15-14
H
C H
H
cyclobutane
H
H C
C H
H C
C H
H
H
Depicting cycloalkanes
H
H
cyclopentane
Figure 15.6
15-15
H
H C
H
H C
H C
H
C
C
H
H
C
C
H
H
H
cyclohexane
H
H
C
H
C H
C H
C
H
H
H H
15-16
Boiling points of the first 10 unbranched alkanes
Figure 15.7
15-17
An analogy for
optical isomers
Figure 15.8
15-18
Two chiral molecules
optical isomers of 3-methylhexane
Figure 15.9
15-19
optical isomers of alanine
The rotation of plane-polarized light by an optically active substance
Figure 15.10
15-20
The binding site
of an enzyme
Figure 15.11
15-21
15-22
The initial chemical event in vision
Figure B15.1
15-23
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 CH2 CH CH CH3
CH3
(d)
(c)
CH3
CH2
CH3 (e)
CH3
CH3 CH2 CH
CH2CH3
CH
CH2
CH3
CH3 CH2 CH C
CH CH3
CH3
PLAN:
15-24
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.
SAMPLE PROBLEM 15.2
(continued)
methyl
SOLUTION:
methyl
CH3
(a)
CH3
(b)
butane
CH3 5CH2 4CH 3CH
6
CH3 2C CH
CH3
3 2
1
2CH2
4
CH3
CH3
1
cyclopentane
3
4
5
2
1
CH3 methyl
CH2CH3
ethyl
1-ethyl-2-methylcyclopentane
methyl
(d)
pentene
CH3
CH3 CH2 CH CH CH2
5
4
3
2
1
1-pentene
3-methyl-1-pentene
chiral center
15-25
methyl
hexane
3,4-dimethylhexane
(can be numbered in either direction)
methyl
2,2-dimethylbutane
(c)
CH3
SAMPLE PROBLEM 15.2
(continued)
methyl
CH3
(e)
6CH3 5CH2
CH
C
4
3
3-hexene
CH CH
1 3
2
CH3
methyl
3-hexene
H
CH3
C
6CH3 5CH2
methyl
4
C
3
CH CH
1 3
2
CH3
methyl
cis-2,3-dimethyl-3-hexene
15-26
3-hexene
methyl
6CH3 5CH2
4
H
C
CH3
C
3
CH CH
1 3
2
CH3
methyl
trans-2,3-dimethyl-3-hexene
Representations of benzene
or
Figure 15.12
15-27
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 C X
15-28
+
Y
R C Y
+
X
A color test for
C=C bonds
C C
Figure 15.13
15-29
+ Br2
Br
C C
Br
15-30
SAMPLE PROBLEM 15.3
PROBLEM:
Recognizing the type of organic reaction
State whether each reaction is an addition, elimination, or
substitution:
(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.



15-31
CH3 CH CH2 + HBr
More atoms bonded to carbon is an addition.
Fewer atoms bonded to carbon is an elimination.
Same number of atoms bonded to carbon is a substitution.
SAMPLE PROBLEM 15.3
(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 remains the same.
15-32
Some molecules with the alcohol functional group
Figure 15.14
15-33
15-34
15-35
General structures of amines
the amine functional group
primary (1o) amine
Figure 15.15
15-36
secondary( 2o) amine
C N
tertiary (3o) amine
Some biomolecules with the amine functional group
lysine (1o amine)
(amino acid found
in proteins)
epinephrine
(adrenaline; 2o amine)
(neurotransmitter in
brain; hormone released
during stress)
Figure 15.16
15-37
adenine (1o amine)
(component of
nucleic acids)
cocaine (3o amine)
(brain stimulant;
widely abused drug)
Structure of a cationic detergent
CH
H3C N CH3
Cl
H2C
CH2
H2C
CH2
benzylcetyldimethyl-
H2C
ammonium chloride
H2C
CH2
CH2
H2C
CH2
H2C
CH2
H2C
CH2
Figure 15.17
H2C
CH3
15-38
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:
15-39
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.
SAMPLE PROBLEM 15.4
(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
15-40
Some common aldehydes and ketones
methanal
(formaldehyde)
used to make
resins in
plywood,
dishware,
countertops;
biological
preservative
ethanal
(acetaldehyde)
narcotic product
of ethanol
metabolism;
used to make
perfume,
flavors, plastics,
other chemicals
2-propanone
benzaldehyde
(acetone) solvent
artificial
for fat, rubber,
almond
plastic, varnish,
flavoring
lacquer;
chemical
feedstock
2-butanone
(methyl ethyl
ketone)
important
solvent
Figure 15.18
15-41
The carbonyl group
Figure 15.19
15-42
SAMPLE PROBLEM 15.5
PROBLEM:
Predicting the steps in a reaction sequence
Fill in the blanks in the following reaction sequence:
Br
OH-
CH3 CH2 CH CH3
Cr2O72-
CH3Li
H2O
H2SO4
PLAN: Look at the functional groups and reagents to determine the type of
reaction.
SOLUTION:
Br
OH-
OH
O
Cr2 O7 2-
CH3 CH2 CH CH3
CH3 CH2 CH CH3
2-bromobutane
2-butanol
CH3 CH2 C
H2 SO4
2-butanone
OH
CH3 CH2 C
CH3 Li
CH3
CH3
2-methyl-2-butanol
15-43
CH3
H2 O
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)
Figure 15.20
15-44
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)
Some lipid molecules with the ester functional group
cetyl palmitate
(the most
common lipid in
whale blubber)
lecithin
(phospholipid found in all cell membranes)
15-45
Figure 15.21
tristearin
(typical dietary fat used as an
energy store in animals)
Which reactant contributes which group to the ester?
General esterification reaction
O
R C OH +
+
H OR'
18
R C OH + R'
OH
18
R C OH + R'
18
R C O R' + H O H
O
O
15-46
R C O R' + H O H
O
O
Figure 15.22
H
O
18
OH
R C O R' + H O H
Some molecules with the amide functional group
acetaminophen
(active ingredient in
nonaspirin pain relievers;
used to make dyes and
photographic chemicals)
N,N-dimethylmethanamide
(dimethylformamide)
(major organic solvent;
used in production of
synthetic fibers)
Figure 15.23
15-47
lysergic acid diethylamide
(LSD-25)
(a potent hallucinogen)
SAMPLE PROBLEM 15.6
PROBLEM:
Predicting the reactions of the carboxylic
acid family
Predict the product(s) of the following reactions:
O
OH
(a) CH3 CH2 CH2 C OH
CH3
+ CH3 CH CH3
O
(b) CH3 CH CH2 CH2 C NH CH2 CH3
PLAN:
H+
NaOH
H2O
(a) An acid and an alcohol undergo a condensation reaction to form
an ester.
(b) An amide, in the presence of base and water, is hydrolyzed.
O
CH3
SOLUTION:
(a) CH3 CH2 CH2 C O CH CH3
CH3
O
(b) CH3 CH CH2 CH2 C O- + Na+ + NH2 CH2 CH3
15-48
The formation of
carboxylic,
phosphoric, and
sulfuric acid
anhydrides
Figure 15.24
15-49
An ester and an amide of other non-metals
glucose-6-phosphate
Figure 15.25
15-50
sulfanilamide
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
2o amine
alkene
15-51
A summary of the interconversions among the major
organic functional groups
Figure 15.26
15-52
Steps in the free-radical polymerization of ethylene
Figure 15.27
15-53
15-54
15-55
The formation of nylon-66
Figure 15.28
15-56
The structure of glucose in aqueous solution and the
formation of a disaccharide
15-57
Figure 15.29
The common amino acids
15-58
Figure 15.30
A portion of a polypeptide chain
Figure 15.31
15-59
Forces that
maintain
protein
structure
Figure 15.32
15-60
Shapes of
fibrous
proteins
Figure 15.33
collagen
15-61
silk
fibroin
Mononucleotide monomers and their linkage
portion of DNA
polynucleotide chain
mononucleotide of
ribonucleic acid (RNA)
Figure 15.34
15-62
mononucleotide of
deoxyribonucleic acid
(DNA)
The double helix of DNA
1.08 nm
Figure 15.35
15-63
1.08 nm
Key stages in protein synthesis
Figure 15.36
15-64
Key stages of DNA replication
15-65
Figure 15.37