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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5852_Библиотеки_им_академика_М_И_Перельмана.pdf
Maleic acid
meso-Tartaric acid
Alk. KMnO
4
or OsO
4
COOH
C
C
COOH
H
H
OH
OH
HCCOOH
HCCOOH
COOH
H
HOOC
H
Fumaric acid
COOH
C
C
H
OH
OH
HHOOC
HCOOH
COOHH
(+) Tartaric acid
(−) Tartaric acid
Alk. KMnO
4
or OsO
4
+
(Racemic mixture)
C
C
OH
OH
C
C
O
O
Oxidation
or
H
H
HOOC
HOOC
m.p. 130°C
Benzene
Quinone
Maleic acid
C
C
C h a P T E r 3 Stereochemistry
d. Method of formation from cyclic compounds: The isomer obtained by the rupture of a ring must be
a cis-isomer. Suppose we oxidize benzene or quinone to an unsaturated dicarboxylic acid of (m.p.
130°C), which is identical with maleic acid (m.p. 130°C). From the structure of benzene or quinone,
it becomes clear that the two carboxylic groups must be on the same side. Therefore, maleic acid,
the isomer having m.p. 130°C, must be cis-isomer and the other isomer fumaric acid must be trans-
isomer.
171
3.17.3 Geometrical Isomerism of Carbon–Nitrogen (C 5 N) Double Bond and N 5 N Double Bond
G
eometrical isomerism is not conned to molecules having carbon–carbon double bonds, rather it exists in
other molecules which possess carbon–nitrogen and nitrogen–nitrogen double bonds. Both the carbon and
nitrogen atoms in oximes and azo compounds are sp2 hybridized as in ethene. Thus, the C 5 n or n 5 n
bond consists of one sigma (s) and one pi (p) bond. The third sp2 orbital of nitrogen is occupied by a lone
pair of electrons. Thus, the oximes and azo compounds exhibit geometrical isomerism as there is no free
rotation around the C 5 n or n 5 n bond.
syn-Acetaldoxime anti-Acetaldoxime
H
3
C
H
OH
C NC N
H
3
C
H
OH
CN
H
3
C
H
5C2 OH
H3C
H
5C2
OH
syn-Ethylmethyl ketoxime anti-Ethylmethyl ketoxime
CN
H5C
6
syn-Diphenyl diazonium anti-Diphenyl diazonium
NN
H
5C6
C6H
5
C6H
5
NN
and
Type I
transcis
CC
a
a
b
b
CC
a
b
b
a
transcis
and
a
x
b
a
CCCC
a
a
b
x
Type II
172
PharMaCEuTICal OrGanIC ChEMISTry
In nomenclature, the prexes syn and anti are used instead of cis and trans. In the syn-aldoxime, the
hydroxyl group on nitrogen and hydrogen on the carbon atom are on the same side, and in the anti-isomer
they are on opposite sides:
In the syn-ketoxime, the hydroxyl group on nitrogen and the rst named alkyl group on the carbon are on
the same side, and in the anti-isomer they are on opposite sides:
In the diazo compounds, the two groups attached to each nitrogen atom are on the same side and in the
anti-isomer, they are on opposite sides:
3.17.4 E and Z System of Nomenclature
We know that the geometrical isomerism is possible in structures of the following three types:
trans
cis
CC
ax
by
ay
bx
and
CC
Type III
Cl
I
HBr
12
C
C
HI
Cl Br
CC
1
2
C h a P T E r 3 Stereochemistry
In the rst two types in which at least one group is common both to doubly bonded carbon atoms, the
geometrical isomers are labelled as cis and trans depending upon whether the common groups are on
same or opposite side of the double bond. neither the cis–trans nor the syn–anti system of congurational
nomenclature is applicable to the structures of type III, where all the four substituents are different. Moreover,
these systems of nomenclature are often ambiguous because the conguration descriptions have not been
dened according to any general and clear set of rules. So an unambiguous system of congurational
nomenclature for all types of structures showing geometrical isomerism was developed in 1968. The system
is known as E–Z system of nomenclature and is based on the sequence rules of Cahn, Ingold and Prelog for
naming optical isomers on the R–S system discussed earlier. In order to specify the E or Z conguration to
an geometrical isomer, the following rules are followed:
1. For any double bond compound to be assigned the conguration, the two groups of each of the
double-bonded carbons are arranged according to the priority sequence rules as in R and S rules for
specifying conguration.
2. That isomer in which the two groups of higher priority (one each from the two doubly bonded
carbon) are on the same side of the double bond is assigned the Z conguration. Z stands for
the German word Zusemmern meaning ‘together’. If, on the other hand, the two groups of higher
priority are on opposite sides of double bond, the isomer is given the E conguration. E stands for
the German word Entgegen, meaning ‘against’.
173
Thus in structure,
priority. Similarly out of the groups I and Br on the carbon 2, I has priority over Br. The group of higher
priority on each of the carbons that are doubly bonded (Cl and I) are on the same side of the double bond
and therefore the structure gets the Z conguration. Based on the above arguments, the structure
above will get E conguration since Cl and I on the two doubly bonded carbons are on the opposite
side to each other.
Similarly the congurational symbols to the following structures may be assigned as shown below
each:
the two groups on carbon 1 are Cl and H, out of which Cl has higher
EZ
HCl
H
3
CH
C
C
HH
HOOC COOH
C
C
Z-Acetaldoxime E-Acetaldoxime
CN
H
3
C
OH
H
CN
H3C
HOH
2,4-Hexadiene (2Z, 4Z)2,4-Hexadiene (2E, 4Z)
H
3
C
1
23
45
CH
3
6
H
H
H
H
C
C
C
C
H3C
1
2 3
HH
45
CH
3
6
HH
C
C
C
C
CH
3
CH
3
HH
cis-1,2-Dimethylcyclohexane
H
CH
3
CH3H
trans-1,2-Dimethylcyclohexane
174
PharMaCEuTICal OrGanIC ChEMISTry
E and Z system is particularly useful for assigning the congurations to a pair of oximes. For
example,
If the organic compound contains two or more double bonds capable of showing geometrical
isomerism, E and Z notations are given to each double bond. For example,
3.17.5 Geometrical Isomerism in Alicyclic Compounds
We have observed that geometrical isomerism arises only when there is hindered rotation about a bond.
Such hindrance is not conned to double bonds only. For instance disubstituted derivatives of cyclopropane,
cyclobutane, cyclopentane and cyclohexane can also show cis–trans isomerism because they full the basic
condition of hindrance to rotation about a linkage between atoms as joined in a ring are not free to rotate
around a sigma bond.
For example, 1,2-dimethyl cyclohexane exists in two geometrical isomeric forms. If two methyl groups
are on opposite sides of the ring, they are trans and when they are on the same side they are cis. These
compounds are geometrical isomers of each other.
H
OH
H
OH
H
OH
OH
H
CH
3
H
CH
3
H
H
CH
3
CH
3
H
cis-1,2-Dimethyl
cyclopropane
trans-1,2-Dimethyl
cyclopropane
cis-1,2-
Cyclopentanediol
trans-1,2-
Cyclopentanediol
C h a P T E r 3 Stereochemistry
The following are other examples of geometrical isomers of cyclic compounds.
3.18 CONFORMATION OF ALKANES
175
Rotation around carbon–carbon single bond: Single covalent bond (s bond) present between two
3
carbon atoms is formed by the overlap of their sp
hybrid orbitals along the internuclear axis. The
electron distribution of the molecular orbital, thus formed, is cylindrically symmetrical around the
axis of the bond. Due to the axial symmetry of the molecular orbital, rotation around the C–C bond is
almost free. As a result of this rotation, alkanes can have different spatial arrangements, i.e. different
relative arrangement of their atoms in space.
Such different spatial arrangements of atoms or groups of atoms in a molecule that can be readily interconverted by rotation around C–C single bond are called conformers. They are rotational isomers and the
phenomenon is called conformational isomerism. The molecular geometry corresponding to a conformer
is known as conformation.
3.18.1 Conformation of Ethane
In the ethane molecule, if it is supposed that the position of one of the carbon atoms is kept xed and the
other is rotated about it, a large number of arrangements of the hydrogen of the carbon with respect to the
hydrogen of the other can be obtained. Out of the innite number of possible conformations of ethane, two
conformations represent the extremes, as shown in Fig. 3.8. These are as follows:
1. eclipsed conformation: In this conformation, the tetrahedrally attached three hydrogen atoms to
front carbon are exactly in front of those attached to the back carbon, i.e. hydrogen atoms of both
the carbon atoms are crowded together.
2. Staggered conformation: In this conformation, the tetrahedrally attached hydrogen atoms to two
carbon atoms are as far apart as possible. It is important to note the basic structure of the molecule
and the various bond lengths and bond angles remain the same in both the conformations. These are
called sawhorse method of representing the conformations (Fig. 3.8).
Newman’s style of representing the conformations of molecule is more popular. Thus, the two
conformations of ethane may be represented by newman’s projections as shown in Fig. 3.9.
C
H
H
H
H
H
C
H
Eclipsed
C
H
C
H
HH
H
H
Staggered
Cl
H
H
H
H
H
H
Eclipsed
Cl
H
H
H
H
H
H
Staggered
176
PharMaCEuTICal OrGanIC ChEMISTry
Figure 3.8 Eclipsed and staggered conformations of ethane.
In the Newman projection, the two carbons forming a bond are represented by two circles, one behind
the other, so that only the front carbon is seen. The hydrogen atoms attached to the front carbon are
depicted by C–h bonds from the centre of the circle. The C–h bonds of the back carbon are drawn from
the circumference of the circle.
One conformation of ethane gets converted into the other when rotated through an angle of 60°.
Relative stabilities of conformations of ethane
It must be pointed out here that rotation around the single bond is not completely free. If it were so, the
potential energy of different conformations should have been the same. But in actual practice, the potential
energy of the molecule changes somewhat with the rotation around C–C single bond. Thus the potential
energy of ethane molecule is minimum for staggered conformation and maximum for eclipsed conformation.
The energy difference between the two is 12.6 kJ mol21. In other words, staggered conformation is the
most stable conformation of ethane while eclipsed conformation is the least stable (Fig. 3.10).
The small energy difference between different conformations is due to the repulsive interactions between
the electron clouds of the C–h bonds attached to the central C–C bond. In the staggered conformation of
Figure 3.9 Newman’s projection for the conformations of ethane.
Cl
CH
3
H
H
H
H
H
Cl
H
CH
3
H
H
H
H
Cl
H
H
H
H
H
H
Staggered
Cl
H
H
H
H
H
H
12.6 kcal
per mole
Potential energy
Cl
H
H
H
H
H
H
C h a P T E r 3 Stereochemistry
177
Figure
3.10 Relative stabilities of conformations of ethane.
ethane, the electron clouds of these carbon hydrogen bonds are as far apart as possible. But in the eclipsed
conformation, the three C–H bonds of one carbon are closest to the three C–H bonds of the other carbon.
The repulsive interactions between the electron clouds in this position increase the energy of the molecule
and thus decrease its stability.
But the energy difference is not large enough to prevent rotation. Even at ordinary temperature, the
molecules possess sufcient thermal or kinetic energy to overcome the energy barrier through effective
collisions and thus conformations keep on changing from one form to the other. As such, it is not possible to
separate the different conformations of ethene.
3.18.2 Conformations of Propane
In case of propane (Ch3–Ch2–Ch3), rotation can take place about either of the two carbon–carbon bonds
(Fig. 3.11). The energy difference between the two extreme conformations, i.e. eclipsed and staggered, is
13.8 kJ mol21, which is nearly the same as in case of ethane. The most stable conformation is the staggered
conformation although it can freely change into the eclipsed conformation and vice versa.
Figure 3.11 Newman’s projections for the conformations of propane.
Eclipsed Staggered
H
C
1
C2C
3
C4HH
H
HHH
HHH
C2C3CH
3
CH
3
HHH
H
178
PharMaCEuTICal OrGanIC ChEMISTry
3.18.3 Conformations of n-Butane
If we consider the rotation around the single bond between two carbon atoms (C2 and C3) of n-butane, we
see that molecule is similar to ethane except that one hydrogen on each carbon has been replaced by a methyl
group:
The various possible conformations of n-butane are as follows:
1. Anticonformation (i): In this conformation, methyl groups and hydrogen linked to two carbon
atoms (C2 and C3) are at a maximum distance apart. It is assumed that the angle of rotation around
the C2–C3 bond is zero for this conformation.
2. eclipsed conformation: Starting from the anticonformation, if we rotate the C2–C3 single bond
3. Gauche or skew conguration: On rotating C2–C3 single bond by another 60°, we get the staggered
4. Fully eclipsed conformation: rotating the C2–C3 bond by another 60° gives rise to fully eclipsed
through 60°, we get the partially eclipsed form II. In this conformation, the methyl group attached
to one carbon atom is at the back of hydrogen atom (rather than the methyl group) attached to the
other carbon atoms.
conformation. III is also known as gauche or skew form. In this conformation, the various bonds
are farthest apart but the methyl groups make an angle of 60° with respect to each other.
conformation (IV). In this conformation, the methyl groups and hydrogen atoms attached to two
carbon atoms completely eclipse each other.
Further rotation of 60° gives rise to the gauche form (V) in which the methyl groups are again 60°
apart. On further rotating C2–C
rotation of C
by 60° and thus completing the rotation through 360°, we get antiform (I) again.
2–C3
bond by 60°, we get another eclipsed conformation (VI). Still further
3
The various conformations are shown in Fig. 3.12.
III III
IV VVI
C h a P T E r 3 Stereochemistry
179
Figure 3.12 Newman’s projections for the conformations of n-butane.
Relative stabilities of conformations of n-butane
The stability of conformations of n-butane is in the order:
Fully eclipsed < partially eclipsed < Gauche < anti
(IV) (II and VI) (III and V) (I)
1. In fully eclipsed conformation IV, there is maximum repulsion between the bonding electrons
and bulky methyl group. repulsion between bonding electrons produces torsional strain in the
conformation while the crowding together of the bulky methyl group causes steric or van der Walls
strain in the molecule. Due to torsional strain and steric strain, this conformation has maximum
energy and minimum stability.
2. In the partially eclipsed conformation (II and VI), the repulsive interaction between the bulky methyl
group and bonding electrons is less than that in fully eclipsed conformation (IV). Thus partially
eclipsed conformations are more stable than fully eclipsed conformation.
3. In gauche conformation (III and V), there are weak repulsive interactions between two methyl
groups as they are 60° apart. In other words, there is a slight van der Walls strain but no torsional
strain. Thus it is more stable than partially eclipsed conformation.
4. In anti-conformation (I), the repulsive interaction between the two methyl groups and bonding electrons
is minimum because the two methyl groups are maximum distance apart. Thus, this conformation is free
of torsional as well as steric strain. hence it is the most stable conformation.
Cl
CH
3
H
H
CH
3
H
H
Potential energy
Cl
H
H
H3C
CH
3
H
H
Cl
H
3
CH
3
H
CH
3
H
H
Cl
CH
3
H
H
CH
3
H
H
AntiAnti GaucheGauche
60° 120° 180° 240° 300° 360°
3.8 kJ/mol 3.8 kJ/mol
14.6 kJ/mol14.6 kJ/mol
18.4 kJ/mol
Eclipsed
Fully eclipsed
Eclipsed
Cl
CH
3
H
H
H3C
H
H
Cl
H
CH
3
H
H3C
H
H
Cl
H
H
H
3
C
H3C
H
H
Rotation
180
PharMaCEuTICal OrGanIC ChEMISTry
Figure 3.13 Energy changes during rotation around the carbon–carbon bond in n-butane.
5. As anticonformation and gauche conformations have staggered arrangements, they have maximum
6. The difference in energy contents between conformations I and IV is about 18.4 kJ mol21, between I
7. As the maximum energy difference between two conformations is only 18.4 kJ/mol, which is easily
energy and maximum stability (anticonformation is slightly more stable than gauche conformation).
These conformations are referred to as conformational isomers.
and II (or VI) is about 14.6 kJ mol21, and between I and III (or V) is only 3.8 kJ mol
provided by colliding molecules at room temperature. Therefore, the various conformations of
n-butane are interconvertible and hence cannot be isolated. In other words, rotation around carbon
single bond in n-butane is almost free. At any time, n-butane consists of an equilibrium mixture of
all the possible conformations, which contains the highest percentage of the anticonformation(s) and
least proportion of the fully eclipsed conformation (IV).
21
(Fig. 3.13).