CHAPTER 2: Question 1

 

2D 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.

The two isomers are shown in the diagram above, and 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.


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.

The two isomers (A and B) are shown in the diagram above, and 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.

The two isomers (A and B) are shown in the diagram above, and 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.


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.

The two isomers are shown in the diagram above, and 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.

The two isomers (A and B) are shown in the diagram above, and 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.

The two isomers are shown in the diagram above, and 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. 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. 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. 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. 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.


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. 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. 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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