Section Three pp. 1
Section Three: Introduction to Organic Structure, Isomerism, and Chirality
Reading:
Bettelheim, Brown, Campbell, and Farrell Chapters 10.1 – 10.4, 11.1 –
11.8, 12.1 – 12.4, 12.6, 15.1 – 15.5, 13.1.
Recommended Problems:
Chapter 10 #14, 15, 16, 21, 22, 26, 29, 31, 32, 43, 49, 50.
Chapter 11 #13, 14, 16, 17, 18, 19, 20, 24, 32, 33, 56, 57.
Chapter 12 #14, 15, 16, 17, 18, 23, 34, 36, 43, 62.
Chapter 15 #18, 19, 20, 21, 22, 23, 24, 26, 31, 32, 37, 38.
INTRODUCTION AND IDENTIFYING FUNCTIONAL GROUPS
Organic chemistry is the study of carbon-containing compounds and their properties. We
begin our discussion with the abbreviated notation for writing organic compounds.
Example: C5H12 (pentane)
Fully Expanded Formula:
Condensed Formula:
H H H H H
H C C C C C H
H H H H H
CH3CH2CH2CH2CH3 OR CH3(CH2)3CH3
Bond-Line Formula:
Problem #1: Consider the formula of the alkane shown below. Write the molecular
formula, condensed formula, and bond-line formula of the compound.
H H H H CH3
H C C C C C CH3
H H H H H
Problem #2: Write the bond-line formulas for the following Lewis dot structures.
H
H
H
H
C
C
C
C
H
H
H
H
A.
H
B.
H
(CH3)2CHCH2CO2H
H
C.
O
H
C
H
C
H
C
H
H
H
C
C
H
H
H
Section Three pp. 2
As we will see shortly, certain portions of organic structures can be identified which
undergo specific chemical transformations. We refer to atoms or groups of atoms of an
organic molecule or ion that undergo predictable chemical reactions as functional groups.
The same functional group, in whatever organic molecule it occurs, undergoes the same
type(s) of chemical reactions. Functional groups are responsible for the physical
properties of a compound. Moreover, the recognition of functional groups is increasingly
important in this area of chemistry, especially in medicine and pharmacology, where new
compounds are constantly synthesized and marketed globally. Consider the following
(“R” = alkyl group AND “X” = halide):
alkyne
alkene
amine
R
R O R
R X
alcohol
ether
alkyl halide
O
O
R NH2
R OH
R
H
O
R
ketone
aldehyde
R
O
O
OH
carboxylic acid
R
R
OR
NH2
amide
ester
Our discussion of alkenes and alkynes will ensue shortly. Let us first examine alcohols,
which contains an –OH (hydroxyl) group, and the amino group, or a nitrogen atom singly
bonded to carbon(s). Both alcohols and amines can be classified as primary (1º),
secondary (2º), or tertiary (3º), depending on the number of carbon atoms bonded to the
carbon bearing the –OH/-amino group, respectively. For example,
1o alcohol
2o alcohol
OH
1o amine
NH2
3o alcohol
OH
2o amine
H
N
OH
3o amine
N
Problem #3: Draw the bond-line formulas of all the alcohols with a molecular formula
C4H10O.
Problem #4: Draw the bond-line formulas of all the amines with a molecular formula
C3H9N.
Section Three pp. 3
Problem #5: Consider the following derivative of Neomethymycin, an antibiotic.
C
E
O
B
D
N(CH3)2
O
HO
O
HO
O
O
F
A
Label the functional groups identified with capital letters above.
Problem #6: Identify all the functional groups in the compounds shown below.
CH3
O
NH2
HO
H
HO
H
CH3
O
OH
L-DOPA (treatment of Parkinson's disease)
H2C
(R)-Carvone (mint taste)
O
O
OH
H3CO
N
H
O
HO
CH3
Capsaicin ("Hot" taste of peppers!)
O
Aspirin (acetylsalicylic acid)
NAMING ALKANES, ALKENES, AND ALKYNES
Organic compounds can be named using rules adopted by the International Union of Pure
and Applied Chemistry (IUPAC). We will begin with basic alkanes, alkenes, and
alkynes.
Section Three pp. 4
Rules for Naming Alkanes – tetrahedral geometry (109.5º bond angle)
1.
(CnH2n+2)
Select the longest continuous carbon chain for the parent name (with functional
group where applicable).
Number of Carbons
1
2
3
4
5
6
7
8
9
10
Prefix
MethEthPropButPentHexHeptOctNonDec-
The alkane is named by adding the suffix –ane to the Greek root for the number
of carbon atoms.
2.
Number the carbons in the chain, from either end, such that the substituents (i.e.
groups appearing off the parent chain) are given the lowest numbers possible or
closest to the functional group that has priority.
3.
The substituents are assigned the number of the carbon to which they are attached.
When naming these substituents, the –ane suffix is dropped and replaced with the
–yl suffix.
4.
The name of the compound is now composed of the name of the parent chain
preceded by the name and the number of the substituents, arranged in alphabetical
order.
A.
B.
C.
If a substituent occurs more than once, the prefixes “di-“, “tri-“, “tetra-“,
etc. are used to indicate how many times it occurs. These prefixes are
NOT included in the alphabetization of the substituents.
If a substituent occurs twice on the same carbon, the number of the
substituent is repeated.
Numbers are separated by commas, and numbers are separated from
names by hyphens. By convention, halogen substituents are placed before
alkyl substituents.
Section Three pp. 5
Alkenes – possess the structural formula CnH2n (contain one or more carboncarbon double bonds)
Rules for Naming Alkenes – trigonal planar geometry (120º bond angle)
1.
Select the longest continuous chain of carbons containing the double bond. This
is the parent structure and is assigned the name of the corresponding alkane with
the suffix changed from “-ane” to “-ene”.
2.
Number the chain so that the position of the double bond is designated by the
lowest possible number assigned to the first doubly-bonded carbon.
3.
The substituents are assigned the number of the carbon to which they are attached.
When naming these substituents, the –ane suffix is dropped and replaced with the
–yl suffix.
4.
The name of the compound is now composed of the name of the parent chain
preceded by the name and the number of the substituents, arranged in alphabetical
order.
A.
B.
5.
If a substituent occurs more than once, the prefixes “di-“, “tri-“, “tetra-“,
etc. are used to indicate how many times it occurs.
If a substituent occurs twice on the same carbon, the number of the
substituent is repeated.
If each carbon atom of the double bond is bonded to TWO different groups, then
cis-trans isomerism exists for the alkene. Consider the following two possibilities
for the aforementioned molecule:
trans isomer - larger portions of
the main chain lie on OPPOSITE
sides of the double bond (denoted E)
cis isomer - larger portions of
the main chain lie on the SAME
side of the double bond (denoted Z)
Section Three pp. 6
Alkynes – possess the structural formula CnH2n-2 (contain one or more carboncarbon triple bonds) – linear geometry (180º bond angle)
Rules for Naming Alkynes – same as alkenes, only use “-yne” suffix for parent structure
Notice the linear arrangement
of the triple-bonded carbon atoms
here; this linear geometry is also
represented by the bond-line
structure as depicted above!
When multiple functional groups are present, the highest priority functional group is on
the end of the name. Priorities progress according to the CIP Sequence Rules (CahnIngold-Prelog), named after the three chemists who developed them:
Alkene < alkyne < amine < alcohol < ketone < aldehyde < amide < alkyl halide < ester
< carboxylic acid
Example: Name the following structure.
OH
Problem #7: Consider each of the compounds shown/written below. Utilize the
appropriate IUPAC rules to properly write/name each compound.
A.
C.
cis-5-chloro-4-methyl-2-hexene
D.
2,2,4-trimethylpentane
F.
E.
OH
G.
B.
H
Section Three pp. 7
Besides forming chains, carbon atoms also form rings. The simplest of the cyclic
alkanes is cyclopropane, shown below. These cyclic alkanes follow the general formula
CnH2n.
= cyclopropane
Since the carbon atoms in cyclopropane form an equilateral triangle with 60° bond
angles, their orbitals do NOT overlap head-on as in normal alkanes. This results in
unusually weak, or strained, C-C bonds; therefore, the cyclopropane molecule is much
more reactive than straight-chain propane.
For simplicity, the cyclic alkanes are often represented by the following bond-line
structures:
Problem #8: Name each of the following cyclic alkanes.
A.
B.
CH2CH3
CH2CH2CH3
Now let’s make the molecules a bit more complex; try naming each of the following:
Problem #9: MORE CYCLIC STRUCTURES!
A.
B.
C.
D.
Finally, we will consider one more class of compounds, known as aromatic
hydrocarbons. The simplest of these is benzene (C6H6), which is shown below:
Benzene has a planar ring structure with a trigonal planar geometry on each carbon.
Section Three pp. 8
When two substituents occur on a benzene ring, three possibilities exist. We locate the
substitutents either by numbering the atoms of the ring or by using the following locators:
ortho (for 1,2-positions), meta (for 1,3-positions), and para (for 1,4-positions).
orthometaparaFor three or more substituents, you denote their locations by numbers.
Problem #10: Write structural formulas for each of the following compounds
A.
ethylbenzene
B.
m-bromochlorobenzene
C.
1,3,5-trifluorobenzene
D.
p-dimethylbenzene
Section Three pp. 9
PHYSICAL PROPERTIES
Alkanes have the lowest melting and boiling points of all the organic compounds.
The attractions between nonpolar alkanes in the solid and liquid states result from
London dispersion forces. In longer carbon chains, the greater number of electrons
produces more attractions between molecules, which results in higher melting and boiling
points. Consider the following:
CH4
bp = -164 ºC
CH3CH3
bp = -89 ºC
CH3CH2CH3
bp = -42 ºC
The boiling points of branched alkanes are generally LOWER than continuous alkanes
with the same number of carbon atoms. The branched chain alkanes tend to be more
compact, which reduces the amount of contact between the molecules. Moreover,
cycloalkanes have higher boiling points than the continuous-chain alkanes. Because
rotation of carbon bonds is restricted, cycloalkanes maintain a rigid structure.
bp = 10 ºC
<
<
<
bp = 28 ºC
bp = 36 ºC
<
bp = 37 ºC bp = 43 ºC
Problem #11: Identify the compound in each pair that has the HIGHER boiling point.
A.
pentane or heptane
B.
pentane or cyclopentane
C.
hexane or 2-methylpentane
Problem #12: Identify the compound in each pair that has the HIGHER boiling point.
A.
nonane or octane
B.
cyclohexane or cyclopentane
C.
pentane or 2,2-dimethylpropane
Alkanes are not soluble in water, but they are soluble in other nonpolar organic
compounds. All liquid and solid alkanes are less dense than water; higher molecular
Section Three pp. 10
weight alkanes possess greater densities than lower molecular weight alkanes. Because
alkanes are insoluble in water and less dense, they float on it.
ISOMERISM
Isomers
Stereoisomers
Structural
(or Constitutional)
Conformational
Configurational
Enantiomers
Diastereomers
Constitutional (Structural) Isomers
Molecules having the same molecular formula but a different arrangement of
atoms are called constitutional, or structural, isomers. These isomers are unique
compounds because of their structural differences, and they have different physical and
chemical properties.
To check whether you have accidentally made duplicate isomers, name them
using the IUPAC rules. All isomers MUST have different IUPAC names. If two names
are identical, then the structures are also identical! Consider the following isomers for
C4H10O:
OH
OH
O
OCH3
OH
O
Problem #13: Write all the constitutional isomers having the molecular formula C5H12
using bond-line notation, and name each different compound.
Section Three pp. 11
Problem #14: Write all the constitutional isomers having the molecular formula C6H14
using bond-line notation, and name each different compound.
Configurational Isomers
Now we will study the relationship that exists between molecules and their mirror
images, or stereoisomers called enantiomers. Enantiomers are very important to study in
chemistry because most molecules in the biological world exhibit this type of isomerism.
Moreover, generally one enantiomer of a molecule is preferred by enzymes in
physiological systems by binding the enantiomer to a specific binding site on the enzyme
surface (recall the lock-and-key model; for example, your house key cannot unlock your
car door).
I.
Enantiomer: same molecular formula, same connectivity, different spatial
arrangement, NONSUPERIMPOSABLE MIRROR IMAGE (said to be chiral)
Example: tetrahedral carbon with 4 different substituents
A
D
A
B
B
C
D
C
mirror
plane
An achiral molecule does not have 4 DIFFERENT substituents on it; such
a molecule is identical and does NOT exhibit enantiomerism. Consider
the following:
A
D
B
B
Section Three pp. 12
Problem #15: Draw the mirror image (or enantiomer) for each structure below.
A.
H
HS
H3C
CH2CH3
OH
B.
H3C
Cl
H
The chiral carbon (which contains four different substituents) is called a stereocenter.
R/S/ Assignments using Cahn-Ingold-Prelog Rules
CIP Priorities:
A.
B.
C.
Higher atomic number has higher priority (numbered #1)
Compare the first atom of each group on the stereocenter. If no
decision is possible because the atom is the same, them move outward
until a point of difference is encountered.
If a double or triple bond is encountered on an atom, count the
bonded atom multiple times (i.e. twice for a double bond or three
times for a triple bond) when prioritizing.
With these rules in mind, you are now ready to assign each stereocenter as R
or S:
A.
B.
C.
Determine the priorities of the 4 different groups on the stereocenter
using the CIP Rules above.
View the molecule with the lowest priority group BEHIND (on the
dash); label this group as #4. If the lowest priority group resides on
the wedge, then invert the assigned configuration.
Label the other three groups from highest (#1) to lowest (#3) priority.
If the motion of groups #1 – #4 is clockwise, then the configuration is
R. If the motion of groups #1 - #4 is counterclockwise, then the
configuration is S.
Section Three pp. 13
1
Cl
Priorities (high to low)
-Cl
-CH(CH3)2
-CH2CH2CH3
-H
Therefore, S
4H
(H3C)2HC
2
CH2CH2CH3
3
4
H
(C, H, H)
1
OCH3
3
2
Therefore, S
(first C has C, C, H attached)
(first C has C, H, H attached)
(should be on "dash"; if on "wedge",
then invert the assigned configuration)
-CH=CH2 is treated as
H
H
C
C
C
C
H
(C, C, H)
Problem #16: Re-examine the given molecules in Problem #15 above, and determine the
absolute configuration (R/S) at each stereocenter.
Problem #17: Explore the molecule below. Determine the absolute configuration at each
stereocenter, and then write ONE enantiomer for the given structure. What do you notice
about the absolute configuration at each stereocenter for the enantiomer that you drew?
H2N
O
H
OH
HO
H
Section Three pp. 14
To find the max. # of stereoisomers possible, calculate 2n, where n = # of stereocenters
Enantiomers have identical melting points, boiling points, solubilities, heats of
formation, standard free energies, and densitites. A pair of enantiomers is distinguished
by their optical activities (i.e. the way they rotate plane polarized light). That is, a pair of
enantiomers rotates plane polarized light by equal amounts in opposite directions. A plus
(+, dextrorotatory) or minus (-, levorotatory) sign is used with the name of a chiral
compound to indicate the sign of its optical rotation. There is NO simple relationship
between the sign of optical rotation and absolute configuration (R or S).
Finally, a racemic mixture is an equimolar mixture of two enantiomers. This
mixture, often designated as d, l (or +) does NOT rotate plane polarized light because the
rotation due to one enantiomer is canceled out by the other enantiomer.
II.
Diastereomers: same molecular formula, same connectivity, different spatial
arrangement, but NOT MIRROR IMAGE.
Problem #18: Consider the molecules shown below. How many total possible
stereoisomers exist for each? Draw all of them, label the absolute configurations at each
stereocenter, and determine if your drawn structure is an enantiomer or diastereomer of
the given compound.
O
H2N
H
Cl
H
B.
A.
OH
H
Br
HO
H
Section Three pp. 15
In our discussion of organic structure, we have shown that chiral compounds can possess
up to 2n stereoisomers. Insofar as these molecules can exist as a number of
stereoisomers, almost invariably only one stereoisomer is found in nature. Of course,
instances do occur in which more than one stereoisomer is found, but these isomers rarely
exist together in the same biological system.
Consider the drug Captopril, effective for the treatment of high blood pressure and
congestive heart failure. It is manufactured and sold as the (S, S)-stereoisomer.
O
N
SH
H3C
H
COOH
H
A large number of chiral drugs, however, are sold as racemic mixtures. As an example,
consider the (S)-enantiomer of Ibuprofen shown:
H3C
H
HO
O
This specific stereoisomer is active as a pain and fever reliever. Its (R)-enantiomer is
biologically inactive in our bodies. Recently, the U.S. Food and Drug Administration has
established more rigorous criteria for the testing and marketing of chiral drugs. As such,
many pharmaceutical companies have decided to develop only single enantiomers of new
chiral drugs.
Now consider Thalidomide shown below.
O
N
O
O
O
H
The story of Thalidomide underscores the importance of organic syntheses that produce
only ONE enantiomer of a chiral compound. Thalidomide was introduced as a sedative
Section Three pp. 16
in the late 1950s and was used to treat morning sickness. Soon thereafter, an abnormally
large number of birth defects was observed (shortened and/or completely absent limbs in
infants). In 1961, these abnormalities were shown to correlate with the ingestion of
Thalidomide during the first trimester of pregnancy. Further research indicated that,
while both enantiomers are active sedatives, only the (S)-enantiomer is teratogenic.
Today, there has been a resurgence of interest in Thalidomide because it has been found
to be useful in the treatment of leprosy and may also be useful in the treatment of AIDS.
Finally, another perspective that you can use to visualize chiral molecules involves
drawing a Fischer projection. In such a projection, the molecule is first arranged with the
horizontal bonds to its chiral center projecting ABOVE the plane of the page, and the
vertical bonds project BEHIND the page. The longest continuous carbon chain resides
along the vertical, beginning with the higher priority on top. Recall the Cahn-IngoldPrelog Sequence Rules when organizing your Fischer projection. Consider the following:
CH(CH3)2
Cl
recall:
H
(H3C)2HC
CH2CH2CH3
=
Cl
H
CH2CH2CH3
The molecule (as written on the left) must first be rotated so that the longest continuous
carbon chain does not reside on either the wedge or the dash. That is,
Cl
CH(CH3)2
H
(H3C)2HC
=
CH2CH2CH3
Cl
CH2CH2CH3
H
Hold -CH2CH2CH3 bond in place and rotate along other three bonds
Once the Fischer projection has been written, drawing an enantiomer (mirror image) is
just like before:
CH(CH3)2
CH(CH3)2
Cl
H
H
CH2CH2CH3
mirror
plane
Cl
CH2CH2CH3
Section Three pp. 17
Now let’s try converting a bond-line structure to Fischer projection for a molecule
containing multiple chiral centers!
H
CH2CH3
Cl
=
H
OH
CH3
Problem #19: Convert the molecules below to the appropriate Fischer projection, and
then construct one enantiomer for your Fischer projection.
O
OH
OH
A.
OH
H
OH
OH
CH2CH3
CH3
O
B.
OH
Cl
Cl
CH2CH3
Conformational Isomers (emphasis on cyclohexanes!)
Conformational isomers are a subcategory of stereoisomers that possess the same
molecular formula, same connectivity of atoms, but differ spatially due to internal bond
rotations. We will focus primarily on the cyclohexane structure, of which its most stable
conformation is the chair conformation, in which all bond angles are approximately
109.5º. Consider the following:
ax
ax
ax
e
e
2
e
ax
e
1
e
ax
ax
e
ax
e
e
e
ax
ax
ax
ax
ax = axial; e = equatorial
e
2
e
1
e
ax
Section Three pp. 18
Now you practice drawing the previous structures!
In a chair conformation, the outer bonds are arranged in two different orientations: six of
them are axial bonds, and the other six are equatorial bonds. Notice that three of the axial
bonds point up; the other three point down. Furthermore, the bonds alternate, first up
then down, as you move from one carbon to the next. Also notice that if the axial bond
on a carbon points upward, the equatorial bond on that carbon points downward, and vice
versa.
Ha
Ha
Hb
Hc
Hc
Hb
is best written as:
Hd
Ha
Hc
Hd
Hb
Hd
Each equatorial bond is oriented PARALLEL to two ring bonds on opposite sides of the
ring. An equilibrium exists between the first chair conformation and the second. The
two chair conformations are called “flipped”, but this does NOT mean a gymnastic-type
flip, but rather an internal bond rotation that causes the axial atoms to become equatorial,
and the equatorial atoms to become axial. Moreover, ring flips interconvert all equatorial
and axial groups. Consider the following example of a mono-substituted cyclohexane
(i.e. methylcyclohexane):
CH3
CH3
MORE STABLE
As shown in the equilibrium above, the structure on the right (equatorial methyl
substituent) is more stable and is therefore favored. This is due to 1,3-Diaxial
Section Three pp. 19
Interactions that exist when the methyl substitutent lies in the axial position. Examine
the structure on the left in more detail:
1,3-Diaxial Interactions
H
CH3
H
The crowding that exists between the axial methyl and the axial hydrogens on C-3 and
C-5 destabilizes this particular conformation. In general, the greater the size of the
substituent on the axial position, the more the opposite conformer is favored in order to
minimize these 1,3-Diaxial Interactions.
Problem #20: Draw the two chair conformations for each of the following molecules, and
select the more stable conformer.
CH3
CH3
B.
A.
CH3
C.
CH3
D.
Section Three pp. 20
REACTIONS OF ALKANES AND ALKENES
A.
Alkane and similar compounds react with oxygen to undergo combustion
(as presented in Section Two)
Example: Solid sucrose (C12H22O11) is burned in air.
B.
Alkenes react with hydrogen halides (Hydrohalogenation)
Note trend for carbocation stability: 3o C(+) > 2o C(+) > 1o C(+)
Example:
HBr
C.
Alkenes react with water in the presence of an acid catalyst (Hydration)
Example:
H2SO4
+ H2O
D.
Alkenes react with bromine and chlorine (Halogenation)
Example:
Cl2
CHCl3
E.
Alkenes react with hydrogen in the presence of a transition metal catalyst
such as Pt, Pd, Ru, or Ni (Hydrogenation)
Example:
+ H2
Ni
Section Three pp. 21
Problem #21: Predict the MAJOR product for the reactions shown below.
HBr
Br2
CH2Br2
+ H2
+ H2O
Ru
H2SO4