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Chapter 24
The Organic Chemistry of Carbon
In 1828, F. Wohler synthesized urea from ammonium chloride and
silver cyanate. The compound synthesized was identical to the
urea generated by animals. Previous to this time it was thought
that only life forms were capable of producing organic compounds.
Wohler demonstrated that organic compounds could be made from
inorganic compounds.
Important Terms
❏ organic compounds
❏ saturated hydrocarbons
❏ hybridization
❏ functional groups
❏ geometric isomers
❏ optical activity
❏ SN1 and SN2
❏ primary carbon cation
❏ carbanion
❏ Markovnikov's Rule
❏ nucleophilic
❏ ortho, meta, para
❏ alcohol
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
structural formulas
unsaturated hydrocarbons
structural isomers
IUPAC
cis and trans
chiral
alkene
secondary carbon cation
radical
electrophilic
aromatic
carboxylic acid
amine
The Saturated Hydrocarbons: Alkanes and Cycloalkanes
A hydrocarbon is a compound that contains only hydrogen and
carbon. An important aspect of carbon is its ability to form chains
and branches with other carbon atoms.
The different ways that carbon can bond, makes it important for us
to be able to distinguish between compounds. For example, given
the formula C4H10, there are two compounds with this same
formula.
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
H
H
C
H
H
H
H
Structural formulas indicate the actual bonding arrangement for
organic compounds. They indicate how and to what the atoms are
Chapter 24
bonded to. There are two ways of writing structural formulas, the
expanded or full structural formula, and the line or semi-condensed
formula. In the semi-condensed formula, he number of lines
represent the type of bonding between carbon atoms. A single line
between two atoms means a single bond, a double line refers to a
double bond while a triple line refers to a triple bond. These two
formula types are illustrated for ethane and propane. below. One
should be able to write both expanded and condensed structural
formulas.
Ring systems, like structural formulas, are generally abbreviated. A
hydrocarbon ring is represented by a polygon where each vertex
represents a carbon with suitable hydrogen to satisfy bonding
requirements. Consider cyclopropane. This compound has the
general formula of C3H6. It is classified as an cyclic alkane , it fits
the general formula of CnH2n. The general structure is a triangle
where each vertex represents a carbon atom bonded to two other
carbon atoms in the ring. Each carbon atom of the ring has two
hydrogens bonded to it to satisfy bonding requirements.
Expanded Structural, Condensed, and Molecular Formulas.
H
H
H
C
C
H
H H
H H H
H C
C
C
H
H
H
H
CH3--CH3 or H3C-CH3 or CH3CH3
CH3-CH2-CH3 or H3C-CH2-CH3 or CH3CH2CH3
C2H6
C3H8
The Organic Chemistry of Carbon
Cyclopropane can be represented using any of the following
structures.
H
H
CH 2
C
H
H
C
C
CH 2
CH 2
H
H
The compound cyclobutane, C4H8, is a four-membered ring
system which can be represented by a square.
H
H
H
C
C
H
CH 2
CH 2
H
C
C
H
CH 2
CH 2
H
H
Cyclopentane, C5H10, is a five-membered ring system and can be
represented by any of the following structures.
H
H
H
C
C
H
CH 2
CH 2
H
C
C
H
CH 2
CH 2
H
H
C
H
CH 2
H
Structural or Positional Isomers
Structural or positional isomers have the same number of atoms
but exhibit different bonding arrangements. This usually results in
compounds that have different physical properties and sometimes
different chemical properties as well.
Chapter 24
Example 24.1. Draw all of the structural isomers for C6H14.
For C6H14 the longest continuous chain contains six carbon atoms
and yields
CH3-CH2-CH2-CH2-CH2-CH3 Structure I
Now, write five carbons in a row and number the carbons. The
sixth carbon atom may be placed on carbons 2, 3, or 4.
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
C
C C C C
C
C
C
C
C
C
C C C C
C
C
C
Structure II
Structure III
Structure IV
Structures II and IV are identical since the -CH 3 group is in the
second position from the end. Remember that when counting the
longest continuous chain, you should start from one end, and
count, then start from the other end and count, to be sure
duplication does not occur.
6
C
C C C C C
1 2 3 4 5
Do not make the common error of thinking a carbon in the up or
down position near the end or beginning of a chain is not part of
the longest continuous chain. Number the longest chain based
upon the continuous connection of carbon atoms.
We can place the proper number of hydrogens around each carbon.
Writing four carbons in a chain, and adding two CH3 groups on
carbon number 2 gives structure IV. We can have a CH3 group on
carbon 2 and a CH3 group on carbon 3 to give structure V.
The Organic Chemistry of Carbon
C H3
C H3 C
CH2C H3
C H3
H
H
C H3 C
C
C H3
C H3 C H3
Structure IV
Structure V
There are a total of five structural isomers for C 6H14.
The Unsaturated Hydrocarbons: Alkenes and Alkynes
Single, double, and triple bonds have been found between carbon
atoms. The table below indicates the relationship among bonding,
structure, and hybridization schemes.
Geometry and Hybridization Schemes for Carbon.
Bonds
Geometry
Hybridization
CH3-CH3
tetrahedral
sp3
CH2=CH2
planar
sp2
CH CH
linear
sp
Angles (HCC)
109.5°
120.0°
180.0°
Alkanes are saturated hydrocarbons with the general formula of
CnH2n+2 and each carbon atom has sp 3 hybridization with
tetrahedral site symmetry around each carbon; that is, a H-C-C
angle of 109.5°. Alkenes are unsaturated hydrocarbons that have
the general formula CnH2n. This type of compound contains a
double bond between carbon atoms, has trigonal planar site
geometry, and the carbon atoms of the double bond have sp2
hybridization. The approximate angle around the carbon atoms
involved in the double bond is 120°. Alkynes have the general
formula, CnH2n-2, are unsaturated hydrocarbons with a carbon
carbon triple bond, sp hybridization for the carbon atoms of the
triple bonds, linear site geometry, and an angle of 180°.
Example 24.2. Identify the following compounds as being
saturated or unsaturated hydrocarbons and also identify the class of
the hydrocarbons (alkanes, alkenes, or alkynes):
a) CH3-CH2-CH3
b) CH3-CH=CH-CH3 c) CH3-C
C-CH3
Chapter 24
Compound a has all single bonds as well as single carbon
hydrogen bonds. Hence, it is a saturated hydrocarbon, an alkane.
This compound also has the molecular formula of C3H8 which is
of the form CnH2n+2.
Compound b contains a double carbon-carbon bond and therefore
this compound is unsaturated. The presence of the double bond
indicates that the compound is an alkene. It has the molecular
formula of C4H8 which coincides with the general formula of an
alkene, CnH2n.
Compound c has a triple carbon carbon bond. It is an unsaturated
hydrocarbon and falls into the classification of an alkyne. The
molecular formula for this compound C4H6 agrees with that of an
alkyne, CnH2n-2.
Example 24.3. Indicate the hybridization and site geometry for
each carbon atoms in the following structures:
O
C H3 CH 2 C H
H2C C CH CH3
O
C H3 CH 2 C H
2
1
3
Carbon atom number 3 has four single bonds to it. It has sp3
hybridization and tetrahedral site symmetry. The same is true for
carbon 2. Carbon 1 has sp2 hybridization and planar site
symmetry. It has one double bond and two single bonds.
H2C C CH CH3
1 2 3 4
Carbon atom 1 has one carbon-carbon double bond and two CH
single bonds. It exhibits sp2 hybridization and has trigonal planar
site symmetry. The same situation is found for carbon number 3.
Carbon number 4 has sp3 hybridization. It has four single bonds
and exhibits tetrahedral site symmetry. Carbon atom 2 is a special
case for it has sp hybridization. This arises because the p orbitals
on carbon number 2 that are used in the p bonding to carbons 1 and
3 are perpendicular to each other.
The Organic Chemistry of Carbon
Geometric Isomers are isomers that have very similar chemical
properties but different physical properties. This arises because the
compounds have the same basic framework for their carbon
structures. The requirements for geometric isomers are fairly
straight forward:
1. Compound must have restricted rotation.
a. Double bond
b. Ring structure
2. No two groups attached to the same carbon atom of the
double bond may be identical.
Examples of geometric isomers (structures I and II) are given
below for 2-butene.
I
C H3
C
II
C H3
C H3
C
C
H
H
C
H
cis-2-butene
III
C H3
H
C
C H3
C H3
Structure III is not a geometric isomer for it contains identical CH3
groups on the same carbon atom of the double bond. It also
contains two H's on the other carbon atom of the double bond.
Geometric isomers are often referred to as being cis- or trans-. Cis
- means on the same side of the double bond while trans- means on
the opposite side of the double bond. Note that in the case of the
cis- isomer you pass through the double bond and arrive on the
same side of the double bond plane.
Example 24.4. Indicate if the following ring systems are examples
of geometric isomers. If so indicate if they are of the cis - or transtype.
a)
Cl
Cl
Cl
c)
b)
Cl
C
H
2-methyl-1-propene
trans-2-butene
Cl
H
Cl
Structure a has two chlorine groups on the same carbon atom.
Thus it is not a geometric isomer. It is named as 1,1dichlorocyclopropane.
Chapter 24
Structure b has the two chlorine atoms on opposite sides of the ring
therefore it is a trans- geometric isomer. It is named as trans-1,2dichlorocyclopropane.
Structure c has the two chlorine atoms on the same side of the ring.
This is a cis geometric isomer. It is named as cis-1,2dichlorocyclopropane.
Some Reactions of Alkanes and Alkenes
Example 24.5. Indicate a simple chemical test that would
distinguish between an unsaturated hydrocarbon and a saturated
hydrocarbon. Use the following general structures to illustrate
your answer.
Unsaturated hydrocarbons undergo electrophilic addition reactions
with bromine. This reaction is usually performed in the dark to
avoid photochemical substitution of bromine. The disappearance
of the dark red color associated with bromine is indicative of a
positive reaction.
Unsaturated hydrocarbon
Br Br
R
H
dark
C
+ Br 2
C
R C C H
R
H
red color
colorless
R
R
colorless product
Saturated hydrocarbon
R
H
H
C
C
R H
colorless
dark
H +
Br 2
red color
no reaction
The Organic Chemistry of Carbon
Example 24.6. Identify the major product (I or II) of the following
electrophilic addition of HBr across the carbon carbon double
bond.
C H3
C H3
C
C H3
C
+
HBr
C H3
H
Br
H
C
C
H
or
C H3
H
Br
C
C
H
C H3 C H
3
II
C H3 C H3
I
Markovnikov's rule states that the major product in an electrophilic
addition reaction of H-X across an unsymmetric double bond will
be the one where hydrogen adds to the carbon atom of the double
bond that already contains the most hydrogens directly bonded to
it.
C H3
C
CH
3
CH
3
C
+
H
HBr
C H3
+
C
H
H
C
H
CH CH
3
3
3° carbonium ion
or
C H3
C
+
C
H
C H3 C H
3
2° carbonium ion
Structure I is the major product. Structure II is the minor product.
The Aromatic Hydrocarbons and Their Derivatives
Benzene has the shape of a hexagonal ring with the molecular
formula of C6H6 and all of its carbon carbon distance the same
(0.139) nm compared to single carbon carbon bond distance of
0.154 nm and a double bond distance of 0.133 nm). Benzene does
not undergo electrophilic addition nor does it readily add hydrogen
or bromine. This stability is attributed to its unique bonding
system that involves the overlap of p orbitals which form a
continuous p electron cloud of electron density around the ring.
The resonance contributing structures of benzene can be written in
two forms
Chapter 24
H
H
H
C
C
C
C
C
H
H
H
C
C
H
C
H
H
C
C
C
C
H
H
H
A resonance hybrid structure of benzene is represented
as a full circle inside of the hexagonal ring to represent
the overlap of electron density. This is a short-hand
notation for benzene that is commonly used.
Compounds with benzene-like structures are classified
as aromatic hydrocarbons. Some common examples of aromatics
are naphthalene, anthracene, and substituted benzenes.
X
benzene
naphthalene
anthracene
substituted benzene
Substituted benzenes are formed by electrophilic substitution
reactions. Such reactions require the presence of a Lewis catalyst.
H
E
Lewis Acid
+ H-Nu
+ E-Nu
Some typical mono- and di- substituted benzenes are given below:
CH3
toluene
chlorobenzene
OH
NH2
Cl
aniline
phenol
The Organic Chemistry of Carbon
Example 24.7. Draw the structures for ortho-, meta-, and paraisomers of dichlorobenzene.
Cl
Cl
Cl
Cl
Cl
ortho-dichlorobenzene
meta-dichlorobenzene
Cl
para-dichlorobenzene
The terminology of ortho-, meta-, and para- is used only for
disubstituted benzenes.
Functional Groups
Specific arrangements of elements that act as reactive sites within a
molecule are called functional groups. Classification of reactions
that occur in organic chemistry are often by functional groups. For
example, the structure R-OH is a general structure for an alcohol.
R-O-R' is a general structure for ethers, where R represents any
attached system. One should be able to recognize and name the
common functional groups. Knowledge of the general structure of
common functional groups is a key concept in chemistry.
Example 24.8. Identify the functional group present in each of the
following structures
O
a) C H3 CH 2 N C H3
H
b) C H3 CH 2 C H
c) C H3CH 2CH 2 O C H3
a) The N-H linkage is attached to two carbon atoms and one
hydrogen atom. This is the structure for a secondary amine. The
name of this compound is ethylmethyl amine.
Chapter 24
b) The carbonyl functional group with a hydrogen atom attached
to it identifies this compound as an aldehyde. The IUPAC name
for this compound is propanal.
O
R C H
c) This compound has the general structure R-O-R, it is an ether.
The name for this compound is propyl methyl ether.
Organic Nomenclature
The variety of functional groups and the early development of
organic chemistry have led to an interesting situation for naming
organic compounds. Many organic compounds are identified by
their common name. The International Union of Pure and Applied
Chemistry (IUPAC) has rules for naming organic compounds and
there is a systematic method.
To begin with, we first determine the longest continuous carbon
chain and then determine the root name. The root name is derived
from the name of the compound having that number of carbons in
its longest carbon chain. You must memorize the names of the
first twelve hydrocarbons. The following table provides the
chemical formula, IUPAC name, the root name and the number of
carbon atoms in the chain.
IUPAC Names and Roots.
Chemical Formula
IUPAC Name
CH4
methane
CH3-CH3
ethane
CH3-CH2-CH3
propane
CH3-(CH2)2-CH3
butane
CH3-(CH2)3-CH3
pentane
CH3-(CH2)4-CH3
hexane
CH3-(CH2)5-CH3
heptane
CH3-(CH2)6-CH3
octane
CH3-(CH2)7-CH3
nonane
CH3-(CH2)8-CH3
decane
CH3-(CH2)9-CH3
undecane
CH3-(CH2)10-CH3
dodecane
Root
meth
eth
prop
but
pent
hex
hept
oct
non
dec
undec
dodec
Atoms
1
2
3
4
5
6
7
8
9
10
11
12
Alkyl (Al-keel) groups are formed by removing a hydrogen from
the alkane; for example, removing a hydrogen from methane H3C-
The Organic Chemistry of Carbon
H yields the methyl group H3C-. The alkyl group is named after
its corresponding hydrocarbon by removing the suffix ending ane
and adding the ending yl. Some of the more common alkyl groups
are listed on the next page.
Simple Alkyl Groups.
______________________________________________________
Group
Name
Group
Name
______________________________________________________
CH3-
CH3-CH2-
ethyl
CH3-CH2-CH2- n-propyl
CH3-CH2-CH2--CH2-
n-butyl
CH3-CH|
CH3
isopropyl
CH3-CH-CH2|
CH3
isobutyl
t-butyl
CH3-CH-CH2-CH3
|
sec-butyl
CH3
|
CH3-C|
CH3
methyl
F- fluoro
Cl- chloro Br- bromo
I- iodo
______________________________________________________
Example 24.9. Give the IUPAC name to the following chemical
compounds:
a) CH3-CH-CH2-CH3
|
CH2-CH2-CH3
c) CH3-CBr2-CHBr-CH3
b) CH3-CH-CH2-CH3
|
CH2Cl-CH-CH3
Chapter 24
For compound a, the longest carbon chain contains six carbon
atoms and it is numbered to give the methyl group the lowest
possible index number.
H
H 2
1
6
3
4
5
CH
C
H
C
CH
C
H
C H3 C
3
2 C H3
2
3
CH 2 CH 2 C H3
CH 2 CH 2 C H3
1
2
5
6
3
4
3-methylhexane
rather than
4-methylhexane
For compound b, the numbering scheme is selected to give the
lowest set of index numbers.
3
4
1
2
C
H
C
H
C
H C H3
C H3 C H
3
3
Cl CH 2 CH C H3
Cl CH 2 CH C H3
4
3
1
2
1-chloro-2,3-dimethylbutane rather than 4-chloro-2,3dimethylbutane
For compound c, you need to number the carbons so that the
lowest set of index numbers for the substituents is obtained.
1
2
3
C H3 CBr 2 CHBr
4
C H3
2,2,3-tribromobutane
3
2
4
C H3 CBr 2 CHBr
1
C H3
rather than 2,3,3-tribromobutane
Example 24.10. Give the appropriate chemical structure for the
following compounds:
a) 2,2-dichloropentane
b) 1,2-dibromo-2-methylpropane
c) 1,1,2-trichloroethane
d) 2,3-dimethylhexane
For compound a, the parent hydrocarbon is a five-membered
carbon chain (pent) that is saturated (all single bonds) (ane).
Writing this carbon sequence out and numbering the carbons yields
The Organic Chemistry of Carbon
Cl
1 2 3 4 5
|
C-C-C-C-C Inserting the chlorines results in C-C-C-C-C and
completing the structure
|
Cl
Cl
|
with hydrogen gives CH3-C-CH2-CH2-CH3
|
Cl
For compound b, 1,2-dibromo-2-methylpropane, the parent
hydrocarbon is a three-membered chain (prop-) that is saturated
(ane). Writing this carbon sequence out, numbering, inserting the
substituents, and then the hydrogen yields
C-C-C
Br Br
| |
C-C-C
|
CH3
Br Br H
|
| |
H-C- C- C-H
|
|
|
H CH3 H
For compound c, the parent hydrocarbon is a two-membered chain
(eth) that is saturated (ane). Writing the carbon sequence out,
numbering, inserting the substituents, and then adding the
hydrogen results in the following structure
1
C
2
C
Cl
Cl
C
C
Cl
H
Cl
Cl
C
C H
or
C HCl CH 2 Cl
2
Cl H
For compound d, 2,3-dimethylhexane, the parent hydrocarbon is a
six-membered chain (hex) that is saturated (ane). Writing this
carbon sequence out, numbering, inserting the substituents, and
then adding the hydrogen yields
Chapter 24
1 2 3 4 5 6
C C C C C C
C H3
C C C C C C
C H3 C H3
C H3 CH CH CH 2 CH 2 C H3
C H3
Alcohols: Some Reactions, Isomers, and Boiling Point
Information
The number of organic compounds is in the billions and these
compounds are usually placed in specific categories based upon the
type of functional group present. Once you are able to classify
reactants according to functional groups, you are able to learn
reaction chemistry of specific functional groups.
Example 24.11. Determine the number of structural isomers for
C4H10O. Please draw only alcohols and ethers. There are four
structural isomers with the alcohol functional group.
H
C H3CH 2CH 2CH
2 OH
OH
H
C H3C CH 2C H3
OH
C H3C CH 2 OH
C H3
C H3
C C H3
C H3
For the ethers there are three structural isomers.
H
C H3
O
CH CH C H3
2
2
C H3CH 2 O CH 2C H3
C H 3C
O
C H3
C H3
There are a total of seven structural isomers for C4H10O that are
ethers or alcohols.
Substitution Reactions
SN1 refers to a unimolecular nucleophilic substitution reaction
while SN2 refers to a bimolecular nucleophilic substitution
reaction. Both SN1-type and SN2 -type reactions are usually
determined by experimentation. An SN1 reaction rate depends
only upon the concentration of the substrate while an SN2 reaction
rate depends upon the concentration of the substrate (organic
reactant) and the nucleophile (Lewis base). Remember that
mechanisms are the intermediate steps in a reaction. The rate of
chemical reaction is determined by the slow step in the mechanism.
The Organic Chemistry of Carbon
For the unimolecular nucleophilic substitution reaction between
(CH3)2CHI + OH- ---> (CH3)2CHOH + Ithe formation of a carbon cation intermediate is the rate controlling
step. Experimental evidence indicates that SN1-type reactions are
favored by the formation of tertiary carbon cations over secondary
carbon cations which in turn are more favorable than primary
carbon cations. Carbon cations are identical to carbonium ions.
C H3
C H3 C
C H3
C H3 C +
I
slow
H
C H3
OH
+
C H3 C OH + I
I
fast
H
H
For the bimolecular nucleophilic substitution reaction between
methyl iodide and hydroxide ion, H3CI + OH- ---> H3COH + Ithe formation of a five-membered coordinate intermediate is the
rate-controlling step. Experimental evidence indicates that S N2
-type reactions are favored by carbon compounds that would form
primary or secondary carbon cations rather than tertiary carbon
cations.
H
I
H
H
C
H
+
OH
I
H
H
C
I
OH
slow
H
transition state
+
H
C
OH
fast
H
This reaction involves a collision between two species, the methyl
iodide molecule and the hydroxide ion. Carbocations or
carbonium ions are positively charged ions that exist for a short
period of time and are known as intermediates. You cannot bottle
and store reaction intermediates. The general representation of the
three different types of carbonium ions are shown below
Chapter 24
R
R
R
H
C+
R
H
primary carbon cation
R
C+
H
secondary carbon cation
R
tertiary carbon cation
3°
2°
1°
C+
R refers to any type of attached group. Two hydrogens and one R
group indicates a primary structure, one hydrogen and two R
groups indicates a secondary structure, and three R groups
indicates a tertiary structure. The type of intermediate structure
formed in a reaction often governs the nature of the products
formed. If the intermediate has a negative charge it is a carbanion.
If the intermediate does not have a charge, it is neutral, and is a
free radical.
Example 24.12. Identify the following structures as being
primary, secondary, or tertiary carbon cations, carbanions, or
carbon free radicals.
C H3
a) C H3
CH 2 C +
H
C H3 C H
3
C H3
b) C H3 C
H
c)
C H3 C
H
C+
CH
3
Structure a has a positive charge. It is a carbon cation. The carbon
with the charge has two groups attached to it and only one
hydrogen. It is a secondary carbon cation.
Structure b has a negative charge. It is a carbanion. The carbon
with the charge has two groups attached to it and only one
hydrogen atom.
Structure c has a positive charge. It is a carbon cation. The carbon
with the charge has three groups attached to it. It is a tertiary
carbon cation.
Oxidation and Reduction Reactions
Oxidation and reduction reactions were encountered in Chapters 2
and 19. In the field of organic chemistry, oxidation is usually
associated with the addition of oxygen, the removal of hydrogen or
The Organic Chemistry of Carbon
the conversion of alcohols to carbonyls. Reduction refers to the
addition of hydrogen, the removal of oxygen or the conversion of
carbonyls to alcohols. Primary alcohols are oxidized to aldehydes
or acids. Secondary alcohols are oxidized to ketones. Tertiary
alcohols do not undergo mild oxidation.
Example 24.13. Predict the products of the following oxidation
and reduction reactions
OH
a)
H + K 2Cr2O7
C H3 C
H SO
2 4
∆
H
OH
b)
CH
C H3 C
3
K 2Cr2O7
H SO
2 4
∆
H
OH
c)
C H3 C
C H3
K 2Cr2O7
H SO
2 4
∆
C H3
O
d)
C H3 C
H +
H2
∆
Reactions a, b , and d are oxidation-reduction reactions. Tertiary
alcohols do not undergo mild oxidation. Therefore, no products
are formed in reaction c.
Carboxylic Acids and Their Derivatives
Carboxylic acids are weak organic acids that have the carboxylic
functional group. The lower molecular weight carboxylic acids
are very water soluble while the heavier acids are water soluble.
The acidity of these acids is general represented by the following
equilibrium system.
O
O
+
R C O + H 3O
R C OH + H 2O
Chapter 24
Recall that the acid dissociation constant for such an equilibrium
system is
O
H 3O+
C H3 C O
Ka =
O
C H3 C OH
Ka values for organic acids are usually less than 10-3. Meaning
that at equilibrium most of species in solution are the unreacted
organic acid molecules. There are relatively few ions present.
Organic acids react with bases to produce a salt and water,
however, since it is a weak acid, an equilibrium system is
established.
O
O
+
R C O Na + H 2O
R C OH + NaOH
You should know the following four types of reactions involving
carboxylic acids:
Ionization (Carboxylic acid + water ∆ carboxylate ions +
hydronium ion)
O
O
+
R C O + H 3O
R C OH + H 2O
Neutralization (Carboxylic acid + base ∆ salt + water)
O
O
+
R C O Na + H 2O
R C OH + NaOH
Amide Formation (Carboxylic acid + amine ----> amide +
water)
O
H
H
O
∆
R C N
+ H 2O
R C OH + H N
R
R
The Organic Chemistry of Carbon
Acid Anhydride Formation (Carboxylic acid + Carboxylic acid
-----> acid anydride + water)
+
O
O
O
Η
O
R C OH + R C OH
R C O C R + H 2O
∆
+
Grignard Reagents
Grignard reactions involve the attack of an alkyl halide towards a
carbonyl group. Grignard reactions produces alcohols from
aldehydes and ketones.
Example 24.14. Predict the major product A and B in the
following Grignard reaction.
O
C H3
C
H+
ether
H
B
A
+ C H3 MgBr
H O
2
Addition of a Grignard reagent across a carbonyl bond of an
aldehyde results in the formation of a secondary alcohol because
Grignard reagents tend to follow the general scheme for
nucleophilic addition across a carbon oxygen double bond.
O MgBr
O
C H3
C
H + C H3 MgBr
ether
H
C H3
C
H
C H3
OH
+
H O
2
C H3
C
C H3 + HO MgBr
H
Optical Activity and Optical Isomers
Optical isomers exhibit identical chemical and physical properties
but differ from each other in the direction that they rotate plane
polarized light. The criteria for recognition of optical isomers is
the compound is nonsuperimposable upon its mirror image.
Optical isomers that are nonsuperimposable mirror images of each
other are referred to as enantiomers. Below are a pair of
enantiomers.
Chapter 24
mirror
A
A
E
B
E
B
D
D
In about 99 % of the encountered cases for optical isomers, we find
that carbon has four different groups bonded to it. A carbon atom
that has four different groups bonded to it is often referred to as
being a chiral or an asymmetric center. Carbon atoms with two
or more identical groups bonded to it are referred to as being
achiral or symmetric centers. Only chiral or asymmetric centers
are associated with optical activity; the rotation of plane polarized
light.
Example 24.15. Indicate the chiral centers in each molecule
below.
Cl
a)
C H3
C
OH OH
H
b) H
C
C H
Br OH
Br
OH OH OH
c) C H3 C
C C
C H3
Cl Cl Cl
a) The starred carbon atom is a chiral center. It has four different
groups bonded to it and has a nonsuperimposable mirror image.
Cl
a)
C H3
C H
*
Br
OH OH
b)
H
C C H
*
Br OH
OH OH OH
c) C H3 C C C C H3
*
*
Cl Cl Cl
b) The starred carbon atom is a chiral center. It has four different
groups bonded to it . This compound is optically active.
c) Compound c has two chiral centers which are indicated by the
starred carbon atoms. The central carbon atom is not a chiral
center because it has two identical groups bonded to it; that is,
"CH 3-C(OH)Cl-".
Chapter 25 Polymers: Synthetic and Natural
Polymers are large molecules formed from small molecules.
Polymers can be long straight chains or can be highly
branched. Examples of synthetic polymers include nylon,
polyester and polypropylene. Examples of natural polymers
include proteins, carbohydrates and deoxyribonucleic acids.
Important Terms
❏ polymer
❏ branched polymer
❏ addition polymer
❏ thermoplastic
❏ thermoplastic
❏ peptide
❏ monosaccharides
❏
❏
❏
❏
❏
❏
❏
linear polymer
crosslinked polymer
condensation polymer
proteins
amino acids
carbohydrates
disaccharides
Addition Polymers
Addition polymers are formed by addition of small
molecules called monomers into long chains. There is little or
no crosslinking between chains. These polymers can flow
easily when heated and can be molded into a variety of shapes;
thus we use the term thermoplastic to describe these
materials. Addition polymers can be recognized by the
repeating unit which always has the same formula as the
monomer from which the polymer is formed. An example is
the monomer ethylene CH2=CH2, which when it forms the
polymer polyethylene uses -(CH2CH2)- as the repeating unit.
Addition polymers can be made from free-radical
polymerization, ionic polymerization and coordination
polymerization. Free-radical polymerization proceeds by a
chain-reaction mechanism. Ionic polymerization proceeds
through ionic intermediates. Coordination polymerization
proceeds through a transition metal catalyst.
Chapter 25
277
Example 25.1. Given the following monomers, write a
plausible structure for the polymer:
a) CH=CH2
b) CH=CH
|
|
|
Cl
Cl
CH3
a) -CH-CH2-CH-CH2|
|
Cl
Cl
b) -CH-CH-CH-CH|
| |
|
Cl CH3 Cl CH3
Condensation Polymers
When condensation polymers are formed, a small molecule
such as water condenses out. The repeating unit in a
condensation polymer is smaller than the monomer. Once
formed, condensation polymers contain extensive
crosslinking and attempts to change the shape of this polymer
will be unsuccessful. These materials are called thermoset
plastics.
Peptides and Proteins
Proteins are formed from the polymerization of monomers
called amino acids. Amino acids contain both an amino (NH2) group and a carboxylic acid (-CO2H) group. There are 20
common amino acids used to synthesize proteins. Your
textbook will provide the structures of these amino acids.
Proteins are formed by the reaction of the -CO2H end of one
amino acid with the -NH2 end of another to form an amide.
This -CONH- amide bond is known as a peptide bond. As
more and more amino acids link together, a protein is formed.
By convention, proteins are listed from the N-terminal amino
acid to the C-terminal amino acid end. The order in which
amino acids are linked together gives rise to the specific
characteristics of the protein. When the amino acid of
aspartene is linked to phenylalanine, in the ASP-PHE order
the artificial sweetener aspartame if produced. If these
278
Polymers: Synthetic and Natural
amino acids are linked in the PHE-ASP order the peptide
formed is not sweet.
Example 25.2. Draw the structure for the dipeptide GLY-TYR
From your textbook, locate the structures of GLY and TYR.
Draw the structures of GLY and TYR. Link these two amino
acids by a peptide bond.
H
N
H
H
O
H
H
O
C
C
N
C
C
H
C
H
H
OH
OH
25.11 The Structure of Proteins
Protein structures can be described on several different
levels. The primary structure is the sequence of amino acids
listed one at time in order of appearance. The secondary
structure is how the protein folds back on itself, two of these
secondary structures are α-helix and β-pleated sheets. The
tertiary structure of proteins is produced by the interaction of
the amino acid side chains. In some proteins a quaternary
structure can be described between the individual polypeptide
side chains.
Carbohydrates: The Disaccharides and
Polysaccharides
Carbohydrates are the primary food source for living systems,
plants and animals. Carbohydrates include simple sugars
Chapter 25
279
such as glucose, fructose and sucrose as well as polymers of
these sugars such as starch, glycogen and cellulose. Three
important classes of carbohydrates include monosaccharides,
disaccharides and polysaccharides.
Monosaccharides are divided into two groups the aldoses or
ketoses, depending upon if they contain an aldehyde or
ketone functional group.
The dissacharides are formed by condensing a pair of
monosaccharides. Maltose, lactose and sucrose are examples
of disaccharides. Common polysaccharides include starch
and glycogen. Poly- saccharides serve two primary
functions, that of storage of energy and as mechanical
structure for cells.