CHAPTER 2: Question 1
3D explanations
a) In order for an alkene to exhibit cis-trans isomerism,
the two substituents a and b attached to each carbon of the alkene unit
must be different as shown below. In this case, the two substituents attached
to the carbon atom on the left of the alkene are a hydrogen and an ethyl
group, and the two substituents attached to the carbon atom on the right
of the alkene are a hydrogen and a methyl group. Thus, in both cases the
two substituents are different and the compound will exhibit cis-trans
isomerism. Note that it is totally irrelevant that there is a hydrogen
atom attached to each carbon of the alkene, it is only important that the
two substituents attached to a particular carbon atom of the alkene be
different. Three dimensional rotatable structures corresponding to the
two isomers are shown below. Note that no matter how these two structures
are rotated, they cannot be made superimposable, i.e. they really do represent
different compounds and not just different ways of drawing the same structure.
The two isomers can be classified using either the cis / trans
or the E / Z nomenclature system. The cis-isomer is
the isomer in which the two hydrogen atoms are on the same side of the
alkene, whilst the trans-isomer is the isomer in which the two hydrogen
atoms are on opposite sides of the alkene. To classify the two isomers
using the E / Z nomenclature, the atoms directly attached
to each end of the alkene are compared. There is a hydrogen atom and a
carbon atom attached to the left carbon of the alkene bond, these are shown
in red in the diagram above. Using rule 1 of the CIP rules, the atom of
higher atomic number has the higher precedence, so C > H. Similarly, there
is also a hydrogen atom and a carbon atom attached to the right carbon
of the alkene bond, these are shown in blue in the diagram above. Using
rule 1 of the CIP rules, the atom of higher atomic number has the higher
precedence, so again C > H. In the cis-isomer of the alkene, the
groups on the left and right side of the alkene which have the highest
precedence (the red and blue carbon atoms) are on the same side of the
alkene, so this is the Z-isomer, whilst for the trans isomer of
the alkene, the two groups which have highest precedence are on opposite
sides of the alkene so this is the E-isomer.
b) In order for an alkene to exhibit cis-trans
isomerism, the two substituents attached to each carbon of the alkene unit
must be different. In this case, the two substituents attached to the carbon
atom on the right of the alkene are both hydrogen atoms, so the compound
will not exhibit cis-trans isomerism. A three dimensional rotatable structure
is included below, and this can be rotated to match any other structure
drawn for this compound.
c) In order for an alkene to exhibit cis-trans isomerism,
the two substituents attached to each carbon of the alkene unit must be
different. In this case, the two substituents attached to the carbon atom
on the left of the alkene are a methyl and an ethyl group, and the two
substituents attached to the carbon atom on the right of the alkene are
a hydrogen and a methyl group. Thus, in both cases the two substituents
are different and the compound will exhibit cis-trans isomerism. Note that
it is totally irrelevant that there is a methyl group attached to each
carbon of the alkene, it is only important that the two substituents attached
to a particular carbon atom of the alkene be different. Three dimensional
rotatable structures corresponding to the two isomers are shown below.
Note that no matter how these two structures are rotated, they cannot be
made superimposable, i.e. they really do represent different compounds
and not just different ways of drawing the same structure.
The two isomers (A and B) can be classified using the E / Z
nomenclature system. Note that the cis / trans nomenclature
can only be used for disubstituted alkenes and so is not appropriate in
this case. To classify the two isomers using the E / Z nomenclature,
the atoms directly attached to each end of the alkene are compared. For
the substituents attached to the carbon atom at the right end of the alkene,
the situation is exactly as in part a), so the carbon of the methyl group
has precedence. The two substituents attached to the left carbon of the
alkene bond both start with a carbon atom shown in red in the diagram above.
Hence, the precedence of the groups cannot be determined from rule 1 of
the CIP rules. Therefore, rule 2 is invoked and the next atoms along each
substituent are examined. For the ethyl group these are a carbon and two
hydrogen atoms (shown in blue), whilst for the methyl group only three
hydrogen atoms (also shown in blue) are present. The blue atoms in each
chain are then compared, starting with the atoms of highest atomic number.
These are C and H, and using rule 1 C > H so the ethyl group has a higher
priority than the methyl group. Hence, in isomer A the groups on the left
and right side of the alkene which have the highest precedence are on the
same side of the alkene, so this is the Z-isomer. For isomer B,
the two groups which have
highest precedence are on opposite sides of the alkene so this is the
E-isomer.
d) The two substituents attached to the carbon atom
on the left of the alkene are a methyl and an ethyl group, as are the two
substituents attached to the carbon atom on the right of the alkene. Thus,
in both cases the two substituents are different and the compound will
exhibit cis-trans isomerism. Note that it is totally irrelevant that there
is both a methyl and an ethyl group attached to each carbon of the alkene,
it is only important that the two substituents attached to a particular
carbon atom of the alkene be different. Three dimensional rotatable structures
corresponding to the two isomers are shown below. Note that no matter how
these two structures are rotated, they cannot be made superimposable, i.e.
they really do represent different compounds and not just different ways
of drawing the same structure.
The two isomers (A and B) can be classified using the E / Z
nomenclature system. To classify the two isomers using the E / Z nomenclature,
the atoms directly attached to each end of the alkene are compared. In
both cases, the atoms of an ethyl and methyl group are to be compared,
and the procedure is exactly as described in part b). Hence, in isomer
A the groups on the left and right side of the alkene which have the highest
precedence are on the same side of the alkene, so this is the Z-isomer.
For isomer B, the two groups which have highest precedence are on opposite
sides of the alkene so this is the E-isomer.
e) In principal, cis-trans isomerism
can occur in imines, since there is a bond (the C=N bond) about which rotation
cannot occur, and the lone pair of electrons attached to the nitrogen atom
provides the second substituent attached to the nitrogen end of the double
bond as shown below. However, in this case the two substituents attached
to the carbon end of the double bond are identical (both methyl groups),
so no cis-trans isomerism is possible. A rotatable structure of this compound
is shown below, and this can be rotated to match any other structure drawn
for this compound.
f) The two substituents attached to the carbon
atom of the C=N are a hydrogen atom and a methyl group, and there are also
two different groups (an OH group and a lone pair of electrons) attached
to the nitrogen atom of the C=N. Thus, in both cases the two substituents
are different and the compound will exhibit cis-trans isomerism. Three
dimensional rotatable structures corresponding to the two isomers are shown
below. Note that no matter how these two structures are rotated, they cannot
be made superimposable, i.e. they really do represent different compounds
and not just different ways of drawing the same structure.
The two isomers can be classified using either the cis / trans
or the E / Z nomenclature system. The cis-isomer is
the compound in which the two non-hydrogen atoms attached to the C=N are
on the same face of the imine bond, whilst the trans-isomer is the
compound in which the two non-hydrogen atoms are on opposite faces of the
double bond. To classify the two isomers using the E / Z
nomenclature, the atoms directly attached to each end of the C=N are compared.
In the case of the carbon end of the double bond, these are a carbon and
a hydrogen atom, and the carbon atom has precedence (rule 1). In the case
of the substituents attached to the nitrogen atom, the oxygen atom is compared
with the lone pair of electrons. Rule 4 states that a lone pair always
has the lowest precedence, so the oxygen atom has the higher precedence.
Hence, in the cis-isomer the groups on the left and right end of
the imine which have the highest precedence are on the same side of the
alkene, so this is the Z-isomer. For the trans-isomer, the two groups
which have highest precedence are on opposite sides of the imine so this
is the E-isomer.
g) The two substituents attached to the carbon
atom of the C=N are a propyl group and a butyl group, and there are also
two different groups (a methyl group and a lone pair of electrons) attached
to the nitrogen atom of the C=N. Thus, in both cases the two substituents
are different and the compound will exhibit cis-trans isomerism. Three
dimensional rotatable structures corresponding to the two isomers are shown
below. Note that no matter how these two structures are rotated, they cannot
be made superimposable, i.e. they really do represent different compounds
and not just different ways of drawing the same structure.
The two isomers (A and B) can be classified using the E / Z
nomenclature system. To classify the two isomers using the E
/ Z nomenclature, the atoms directly attached to each end of the
C=N are compared. In the case of the carbon end of the double bond, these
are both carbon atoms (shown in red), so rule 2 is invoked and the next
atoms along the chains are compared. In both case, these are a carbon and
two hydrogen atoms (shown in blue), so it is necessary to examine the next
atoms along the chain. Once again, in both cases the atoms are a carbon
atom and two hydrogen atoms (shown in magenta) so it is necessary to look
still further along the chain. At the next level however, a difference
is found since in the propyl substituent the next three atoms are all hydrogens
(shown in green), whilst in the butyl substituent the atoms are a carbon
and two hydrogens (shown in green). Comparing the green atoms of highest
atomic number gives C > H, so the butyl substituent has a higher priority
than the propyl group. In the case of the substituents attached to the
nitrogen atom, the carbon atom is compared with the lone pair of electrons.
Rule 4 states that a lone pair always has the lowest precedence, so the
carbon atom has the higher precedence. Hence, in isomer A the groups on
the left and right end of the imine which have the highest precedence are
on the same side of the alkene, so this is the Z-isomer. For isomer
B, the two groups which have highest precedence are on opposite sides of
the imine so this is the E-isomer.
h) The two substituents attached to both ends
of the N=N are a chlorine atom and a lone pair of electrons. Thus, in both
cases the two substituents are different and the compound will exhibit
cis-trans isomerism. Three dimensional rotatable structures corresponding
to the two isomers are shown below. Note that no matter how these two structures
are rotated, they cannot be made superimposable, i.e. they really do represent
different compounds and not just different ways of drawing the same structure.
The two isomers can be classified using either the cis / trans
or the E / Z nomenclature system. The cis-isomer is
the compound in which the two chlorine atoms are on the same face of the
N=N double bond, whilst the trans-isomer is the isomer in which
the chlorine atoms are on opposite faces of the N-=N bond. To classify
the two isomers using the E / Z nomenclature, the atoms directly
attached to each end of the N=N are compared, and in both cases these are
a chlorine atom and a lone pair of electrons. Rule 4 states that a lone
pair always has the lowest precedence, so the chlorine atom has the higher
precedence. Hence, in the cis-isomer the groups on the left and
right end of N=N which have the highest precedence are on the same side
of the alkene, so this is the Z-isomer. For the trans-isomer,
the two groups which have highest precedence are on opposite sides of the
N=N so this is the E-isomer.
i) This compound exhibits cis-trans isomerism
since the two halogens can both be on the same face of the six membered
ring (ie. both drawn with wedges or hashed bonds) or on opposite faces
(one wedge and one hashed bond) as shown below. Three dimensional rotatable
structures corresponding to the two isomers are also shown below. Note
that no matter how these two structures are rotated, they cannot be made
superimposable, i.e. they really do represent different compounds and not
just different ways of drawing the same structure. The two isomers are
classified using the cis / trans nomenclature, with the isomer
with the two halogens on the same face of the ring being the cis-isomer
and the other isomer being the trans-isomer. Note, the E
/ Z nomenclature cannot be used for systems such as this.
j) This compound exhibits cis-trans isomerism
since the two ethyl groups (and the two methyl groups) can both be on the
same face of the five membered ring or on opposite faces as shown below.
Three dimensional rotatable structures corresponding to the two isomers
are also shown below. Note that no matter how these two structures are
rotated, they cannot be made superimposable, i.e. they really do represent
different compounds and not just different ways of drawing the same structure.
The two isomers are classified using the cis / trans nomenclature,
with the isomer with the two ethyl groups on one face of the ring and the
two methyl groups on the other face being the cis-isomer and the
other isomer being the trans-isomer.
k) This compound exhibits cis-trans isomerism
since the two methyl groups can both be on the same face of the three membered
ring or on opposite faces as shown below. Three dimensional rotatable structures
corresponding to the two isomers are also shown below. Note that no matter
how these two structures are rotated, they cannot be made superimposable,
i.e. they really do represent different compounds and not just different
ways of drawing the same structure. The two isomers are classified using
the cis / trans nomenclature, with the isomer with the two
methyl groups on one face of the ring) being the cis-isomer and
the other structure being the trans-isomer. Note that the two chlorine
substituents on the third carbon atom of the ring are irrelevant to the
cis-trans isomerism.
l) This complex has a square planar geometry,
and so exhibits cis-trans isomerism since the two bromine atoms (and the
two triphenylphosphine groups) can adopt positions cis or trans to one
another as shown below. Three dimensional rotatable structures corresponding
to the two isomers are also shown below in which the triphenylphosphine
groups are represented by purple spheres. Note that no matter how these
two structures are rotated, they cannot be made superimposable, i.e. they
really do represent different compounds and not just different ways of
drawing the same structure. The isomers are classified using the cis
/ trans nomenclature system with the cis-isomer being the
compound with identical substituents at approximately 90o to
one another and the trans-isomer being the compound with identical
substituents at approximately 180o to one another.
m) This complex has a square planar geometry,
but since three of the substituents are identical, there is only one possible
structure and the compound does not exhibit cis-trans isomerism. A three
dimensional rotatable structure of this complex is shown belowin which
the triphenylphosphine groups are represented by purple spheres, and can
be rotated to be superimposable with any other structure drawn for this
complex.
n) This complex is octahedral, and does exhibit
cis-trans isomerism since the three identical substituents can be all located
on one face of the molecule, or on different faces of the molecule as shown
below. Three dimensional rotatable structures corresponding to the two
isomers are also shown below. Note that no matter how these two structures
are rotated, they cannot be made superimposable, i.e. they really do represent
different compounds and not just different ways of drawing the same structure.
The two isomers are classified using the fac / mer nomenclature
system with the fac-isomer being the isomer in which three identical
groups form one face of the octahedron, and the other isomer being the
mer-isomer.
o) This complex is octahedral, and does exhibit
cis-trans isomerism since the two chlorine atoms can be at approximately
90o or 180o to one another. Three dimensional rotatable
structures corresponding to the two isomers are also shown below. Note
that no matter how these two structures are rotated, they cannot be made
superimposable, i.e. they really do represent different compounds and not
just different ways of drawing the same structure. The two isomers are
classified using the cis / trans nomenclature, with the isomer
with the chlorine atoms at approximately 90o to one another
being the cis-isomer and the compound with the chlorine atoms at
approximately 180o to one another being the trans-isomer
as shown below.
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