CHAPTER 2: Question 2

 

2D explanations

a) To assign the stereochemistry, first compare the two groups (Me and H) attached to the left side of the double bond, then compare the two groups (Et and H) attached to the right side of the double bond. For the two groups attached to the left side of the double bond, CIP rule 1 states that the two atoms directly attached to the alkene should first be compared, and the atom of higher atomic number given priority. In this example, these are the carbon and hydrogen atoms shown in red in the diagram below, and since carbon has atomic number 6 whilst hydrogen has atomic number 1, the carbon atom of the methyl group is the substituent of higher priority. Similarly, on the right side of the alkene the two atoms directly attached to the alkene are again carbon and hydrogen (shown in blue below), so the carbon atom of the ethyl substituent has the higher priority. Hence, the groups of higher priority on each end of the alkene are on opposite sides of the double bond, so the stereochemical descriptor is E.

b) First compare the two groups (Ph and H) attached to the left side of the double bond, then compare the two groups (Ph and H) attached to the right side of the double bond. For the two groups attached to the left side of the double bond, CIP rule 1 states that the two atoms directly attached to the alkene should first be compared, and the atom of higher atomic number given priority. In this example, these are the carbon and hydrogen atoms shown in red in the diagram below, so the carbon atom of the phenyl group is the substituent of higher priority. Similarly, on the right side of the alkene the phenyl substituent again has the higher priority. Hence, the groups of higher priority on each end of the alkene are on the same side of the double bond, so the stereochemical descriptor is Z.

c) First compare the two groups attached to the left side of the double bond, then compare the two groups attached to the right side of the double bond. For the two groups attached to the left side of the double bond, CIP rule 1 states that the two atoms directly attached to the alkene should first be compared, and the atom of higher atomic number given priority. In this example, these are the two carbon atoms shown in red in the diagram below. Since the first atoms are identical, rule 2 must be used and the atoms attached to the atoms shown in red are compared. In both cases, these atoms are a carbon and two hydrogen atoms, all of which are shown in blue below. Thus, using rule 2 again, the atoms attached to the blue atoms are compared. In the case of the propyl substituent, these are again a carbon and two hydrogens (shown in magenta), whilst in the case of the ethyl group they are three hydrogens (also shown in magenta). These atoms are compared, starting with the atoms of highest atomic number (ie C and H), and since carbon has a higher atomic number than hydrogen, the propyl substituent is the group of higher priority. On the right side of the alkene, the two substituents have the same atomic number, but different atomic masses. Thus, by CIP rule 5, the deuterium atom has a higher priority than the hydrogen atom. Hence, the groups of higher priority on each end of the alkene are on opposite sides of the double bond, so the stereochemical descriptor is E.

d) On the left side of the alkene, CIP rules 1 and 2 are applied successively along the two chains until a difference is found. The atoms where the first difference occurs have been shown in red below, and these are three carbon atoms in the upper substituent and two carbons and a hydrogen in the lower substituent. These are compared starting with the atoms of highest atomic number, so first a carbon atom in the upper chain is compared with a carbon atom in the lower chain and since these are identical, a second carbon atom in the upper chain is compared with the second carbon atom in the lower chain. Since there is again no difference, the final carbon atom in the upper chain is compared with the hydrogen atom in the lower chain, resulting in the upper chain having the higher priority.

On the right side of the alkene, the first atoms are again carbon so it is necessary to examine the next atoms along the chain. However, in this case a double bond is present, so CIP rule 3 is used to replace this by two single bonds both ending in an oxygen atom as shown below. This gives two oxygens and a hydrogen atom in the upper substituent which can be compared with the oxygen and two hydrogen atoms in the lower substituent, all these atoms being shown in blue. Again, the atoms are compared starting with the atoms of highest atomic number, these being an oxygen atom in both the upper and lower substituents. As there is no difference, the atoms of next highest priority are compared, and these are the second oxygen atom of the upper substituent and a hydrogen atom of the lower substituent. Thus the upper substituent has the higher priority, and the groups of higher priority on each end of the alkene are on the same side of the double bond, so the stereochemical descriptor is Z.


e) On the left side of the alkene, first use CIP rule 3 to replace all the 'double bonds' within the aromatic rings by two single bonds to carbon atoms as shown below. Then use CIP rules 1 and 2 until a point of difference is found. This occurs when the atoms shown in red in the diagram below are compared (this assumes that you have worked clockwise round the rings in both cases, though this is not essential). The nitrogen atom of the upper substituent is compared with one of the carbon atoms of the lower substituent, and since nitrogen has a higher atomic number than carbon, the upper substituent again has the higher priority. The same procedure is applied to the right side of the alkene, and results in the carbon and hydrogen atoms shown in blue being compared, so the upper substituent again has the higher priority. Hence, the groups of higher priority on each end of the alkene are on the same side of the double bond, so the stereochemical descriptor is Z.

f) On the left side of the alkene, use CIP rules 1 and 2 working both clockwise and anti-clockwise around the cyclohexane ring until a point of difference is found. This occurs when the atoms shown in red in the diagram below are compared, note that the arrowed carbon atom is part of both sets of atoms to be compared. The arrowed red carbon atom is first compared with itself, and since there is no difference, the carbon atom of the methyl group is compared with a hydrogen from the top side of the cyclohexane ring. Carbon has a higher atomic number than hydrogen, so the lower part of the cyclohexane ring has the higher priority. CIP rules 1 and 2 are also applied to the right side of the alkene, and results in the silicon and carbon atoms shown in blue being compared, so the silyl ether substituent has the higher priority. Hence, the groups of higher priority on each end of the alkene are on opposite sides of the double bond, so the stereochemical descriptor is E.

g) On the left side of the carbon-nitrogen double bond, first use CIP rule 3 to replace the double bonds in the aromatic ring by two single bonds as shown below, then use CIP rules 1 and 2 until a point of difference is found. This occurs when the atoms shown in red in the diagram below are compared, and since carbon has a higher atomic number than hydrogen, the phenyl group has the higher priority. On the right of the carbon-nitrogen double bond, the substituents are a lone pair of electrons and an oxygen atom. CIP Rule 4 says that alone pair of electrons always has the lowest priority, so the OH group has the higher priority. Hence, the groups of higher priority on each end of the carbon-nitrogen double bond are on opposite sides of the double bond, so the stereochemical descriptor is E.

h) On both sides of the nitrogen-nitrogen double bond, the substituents are a phenyl ring and a lone pair of electrons as shown below. CIP Rule 4 states that a lone pair of electrons always has the lowest priority, so in both cases the phenyl group is the substituent of higher priority. Hence, the groups of higher priority on each end of the nitrogen-nitrogen double bond are on opposite sides of the double bond, so the stereochemical descriptor is E.

i) On the left side of the carbon-nitrogen double bond, CIP rules 1 and 2 are applied until a difference is found. The atoms where the first difference occurs have been shown in red below, and these are three hydrogen atoms in the methyl group and a carbon and two hydrogen atoms in the lower substituent. These are compared starting with the atoms of highest atomic number, so the carbon atom in the lower substituent is compared with a hydrogen atom in the methyl group, resulting in the lower substituent being the group of higher priority. On the right of the carbon-nitrogen double bond, the substituents are a lone pair of electrons and a carbon atom. CIP Rule 4 says that alone pair of electrons always has the lowest priority, so the carbon chain has the higher priority. Hence, the groups of higher priority on each end of the carbon-nitrogen double bond are on the same side of the double bond, so the stereochemical descriptor is Z.

j) On the left side of the carbon-nitrogen double bond, first use CIP rule 3 to replace all the double bonds in the aromatic ring by two carbon-carbon single bonds, then apply CIP rules 1 and 2 until a difference is found. The atoms where the first difference occurs have been shown in red below, and these are three carbon atoms in the phenyl group and a carbon and two hydrogen atoms in the lower substituent. These are compared starting with the atoms of highest atomic number, so one of the carbon atoms of the phenyl group is compared with the carbon atom of the lower substituent. Since these are identical, the groups with the next highest atomic numbers are compared, and these are another carbon atom of the phenyl group and a hydrogen atom of the lower substituent. Since carbon has a higher atomic number than hydrogen, the phenyl substituent is the group of higher priority. On the right of the carbon-nitrogen double bond, the substituents are a lone pair of electrons and a carbon atom. CIP Rule 4 says that alone pair of electrons always has the lowest priority, so the carbon chain has the higher priority. Hence, the groups of higher priority on each end of the carbon-nitrogen double bond are on opposite sides of the double bond, so the stereochemical descriptor is E.

It is instructive to compare examples i) and j), since they show how as a result of the way in which the CIP rules work, a small difference in two very similar molecules can result in opposite stereochemical descriptors being used.


k) This molecule contains two double bonds which will be considered separately, starting with the double bond on the left. The atoms directly attached to the left end of this alkene are the carbon and hydrogen atoms shown in red below, so by CIP rule 1, the methyl group has the higher priority. On the right end of the alkene, the atoms attached to the alkene are again carbon and hydrogen (shown in blue below), so again the carbon substituent has the higher priority. Hence, the groups of higher priority on each end of the left alkene are on opposite sides of the double bond, so the stereochemical descriptor is E.

Exactly the same situation occurs for the alkene on the right side of the molecule, with the atoms to be compared being shown in magenta (on the left side of this alkene) and green (on the right side) below. Hence, the stereochemical descriptor is again E.


l) This molecule contains two double bonds which will be considered separately, starting with the double bond on the left. The atoms directly attached to the left end of this alkene are the carbon and hydrogen atoms shown in red below, so by CIP rule 1, the methyl group has the higher priority. On the right end of the alkene, the atoms attached to the alkene are again carbon and hydrogen (shown in blue below), so again the carbon substituent has the higher priority. Hence, the groups of higher priority on each end of the left alkene are on the same side of the double bond, so the stereochemical descriptor is Z.

For the alkene on the right side of the molecule, the atoms attached to the left end of the alkene are the carbon and hydrogen atoms shown in magenta below, so by CIP rule 1, the carbon substituent has the higher priority. On the right side of this alkene however, the first atoms are both carbon, so CIP rule 3 is applied to the C=O as shown below, then rule two is used, resulting in the three hydrogen atoms of the methyl group being compared with the three oxygen atoms of the ester, these atoms being shown in green below. Since oxygen has a higher atomic number than hydrogen, the ester substituent has the higher priority. Hence, the groups of higher priority on each end of the right alkene are on opposite sides of the double bond, so the stereochemical descriptor is E.


m) The six alkenes in this molecule will be considered in the order 1-6 as shown on the diagram below. For alkene 1, the two substituents on the left side of the alkene are an ethyl and a methyl group, the priority of which can be determined by a combination of CIP rules 1 and 2. The atoms where a difference occurs are the carbon and two hydrogen atoms of the ethyl substituent, and the three hydrogen atoms of the methyl group shown in red below. Thus the carbon atom is compared with one of the hydrogen atoms, and the ethyl group is found to have the higher priority. On the right side of alkene 1, the atoms attached to the alkene are a carbon and a hydrogen atom (shown in blue below), so CIP rule 1 is sufficient to assign the carbon substituent as the group of higher priority. Hence, the groups of higher priority on each end of alkene 1 are on opposite sides of the double bond, so the stereochemical descriptor is E.

Consider next alkene 2, on the left side of this alkene, both substituents start with a carbon atom, so first use CIP rule 3 to convert the C=C substituent into two C-C single bonds as shown below, then application of CIP rules 1 and 2 shows that two carbon atoms and a hydrogen atom in the upper substituent must be compared with one carbon and two hydrogen atoms in the lower substituent (atoms shown in red below). These are compared, starting with the atoms of highest atomic number, so one of the carbon atoms of the upper substituent is compared with the carbon atom of the lower substituent. Since these are identical, the second carbon atom of the upper substituent is compared with one of the hydrogen atoms of the lower substituent, resulting in the upper substituent having the higher priority. On the right side of this alkene, the atoms attached to the alkene are carbon and hydrogen (shown in blue below), so CIP rule 1 allows the upper substituent to be given the higher priority. Hence, the groups of higher priority on each end of alkene 2 are on the same side of the double bond, so the stereochemical descriptor is Z.

Considering alkene 3, on the left side of this alkene, the atoms attached to the alkene are carbon and hydrogen (shown in red below), so CIP rule 1 allows the lower substituent to be given the higher priority. On the left side, CIP rule 3 must be applied to both alkene 4 and alkene 5, giving the structure shown below. Then, application of CIP rules 1 and 2 results eventually in the atoms shown in blue being compared. Again, the first pair of atoms are both carbon atoms, but the second choice atoms are a carbon and a hydrogen atom, so the lower substituent has the higher priority. Hence, the groups of higher priority on each end of alkene 3 are on the same side of the double bond, so the stereochemical descriptor is Z.

Alkene 4 has no stereochemistry associated with it, since the two substituents attached to the right side of the alkene are both hydrogen atoms. For alkene 5, on both ends of the alkene, the attached atoms are a carbon and a hydrogen (shown in red and blue below), so in both cases CIP rule 1 allows the carbon containing substituent to be given the higher priority. Hence, the groups of higher priority on each end of alkene 5 are on opposite sides of the double bond, so the stereochemical descriptor is E.

Finally, for alkene 6, on the left side of the alkene the substituents start with a carbon and a hydrogen atom (shown in red), so CIP rule 1 says that the upper substituent has priority. To assign the priority of the groups on the right side of the alkene, first apply CIP rule 3 to the lower C=O as shown below, then use CIP rules 1 and 2, which result in the atoms shown in blue being compared. One of the oxygen atoms of the lower substituent is compared with the carbon atom of the upper substituent, and since oxygen has a higher atomic number than carbon, the lower substituent has the higher priority. Hence, the groups of higher priority on each end of alkene 6 are on opposite sides of the double bond, so the stereochemical descriptor is E.


n) The three alkenes in this molecule will be considered in the order 1-3 as shown on the diagram below. For alkene 1, on both ends of the alkene the atoms attached to the alkene are a carbon and a hydrogen atom (shown in red and blue below), and by CIP rule 1 the carbon atom has priority. Hence, the groups of higher priority on each end of alkene 1 are on opposite sides of the double bond, so the stereochemical descriptor is E. Similarly, for alkene 2, on both ends of the alkene the atoms attached to the alkene are a carbon and a hydrogen atom (the red carbon atom and magenta hydrogen on the left, and the green atoms on the right), and by CIP rule 1 the carbon atom has priority. Hence, the groups of higher priority on each end of alkene 2 are on the same side of the double bond, so the stereochemical descriptor is Z.

For alkene 3, on the left side of the alkene, first use CIP rule 3 to convert the triple bond into three C-N single bonds as shown below, then by CIP rules 1 and 2, then atoms shown in red need to be compared. The nitrogen of the upper substituent is first compared with one of the nitrogen atoms of the lower substituent, and as there is no difference, one of the hydrogen atoms of the upper substituent is compared with a second nitrogen atom in the lower substituent. Nitrogen has a higher atomic mass than hydrogen, so the lower substituent has the higher priority. On the right side of alkene 3, application of CIP rule 3 to the two alkenes followed by repeated application of CIP rules 1 and 2 fails to find any difference between the two substituents. At every stage, all of the atoms being compared are identical. The only difference between the two substituents is the stereochemistry of the alkene, so CIP rule 6 is used. This states that a substituent with a double bond in the Z-configuration has a higher priority than a substituent with the E-configuration. The configuration of the two alkenes was determined above, so the lower substituent has the higher priority. Hence, the groups of higher priority on each end of alkene 3 are on the same side of the double bond, so the stereochemical descriptor is Z.

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