How many stereoisomers are possible for each of the following structures? Draw them, and name each by the R-S and E-Z conventions
a) 4-methyl-2,5-heptadiene
b) 1,2,5-trichloro-3-hexene
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(Simplified Molecular-Input Line-Entry System)
CC=C(C)C=CC
4-methyl-2,5-heptadiene has two points of E/Z isomerism--the double bond--and one point for R/S isomerism-the central carbon.
When both bonds have the same isomerism the central carbon is symmetric.
When the bonds have different isomerism, the central carbon is chiral.
That's four stereoisomers.
C\C=C/C(C)\C=C/C
Z,Z-4-methyl-2,5-heptadiene
C/C=C/C(C)\C=C\C
E,E-4-methyl-2,5-heptadiene
C\C=C/[C@@H](C)\C=C\C
R,Z,E-4-methyl-2,5-heptadiene
C\C=C\[C@@H](C)\C=C/C
S,E,Z-4-methyl-2,5-heptadiene
a) To determine the number of stereoisomers for 4-methyl-2,5-heptadiene, we need to consider two factors: geometric isomerism (E-Z) and stereoisomerism (R-S).
First, let's draw the structure of 4-methyl-2,5-heptadiene:
H H
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H-C=C-C=C-H
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CH3
Now, let's consider geometric isomerism. Since there are two double bonds in the molecule, there are four possible geometric isomers: E-cis, E-trans, Z-cis, and Z-trans.
E-cis:
H H
| |
H-C=C-C=C-H
| |
CH3
E-trans:
H H
| |
H-C=C-C=C-H
|
CH3
Z-cis:
H H
| |
H-C=C-C=C-H
| |
CH3
Z-trans:
H H
| |
H-C=C-C=C-H
|
CH3
Now, let's consider stereoisomerism (R-S) for each geometric isomer.
For each geometric isomer (E-cis, E-trans, Z-cis, and Z-trans), there are two stereoisomers: R and S. Therefore, for each geometric isomer, there are a total of two stereoisomers.
Hence, for 4-methyl-2,5-heptadiene:
- E-cis has 2 stereoisomers (R and S)
- E-trans has 2 stereoisomers (R and S)
- Z-cis has 2 stereoisomers (R and S)
- Z-trans has 2 stereoisomers (R and S)
b) To determine the number of stereoisomers for 1,2,5-trichloro-3-hexene, we only need to consider stereoisomerism (R-S), since there is no geometric isomerism.
First, let's draw the structure of 1,2,5-trichloro-3-hexene:
Cl Cl
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Cl-C-C=C-C-H
|
H
Now, let's consider stereoisomerism (R-S).
For a molecule with two chiral centers (the carbons with the chlorine atoms), there are a total of 2^2 = 4 possible stereoisomers.
1) R,R:
Cl Cl
| |
Cl-C-C=C-C-H
|
H
2) R,S:
Cl Cl
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Cl-C-C=C-C-H
|
H
3) S,R:
Cl Cl
| |
Cl-C-C=C-C-H
|
H
4) S,S:
Cl Cl
| |
Cl-C-C=C-C-H
|
H
Hence, for 1,2,5-trichloro-3-hexene, there are a total of 4 stereoisomers.
To determine the number of stereoisomers for each structure, we first need to identify the presence of chiral centers or double bonds with restricted rotation.
a) For 4-methyl-2,5-heptadiene:
To identify chiral centers, look for carbon atoms that are bonded to four different groups. In this case, there are no chiral centers since each carbon atom is bonded to only three different groups (one is a hydrogen atom). However, this molecule does contain a double bond, which restricts rotation.
To determine the E-Z configuration, we need to assign priority to the groups bonded to each carbon of the double bond. Using CIP (Cahn-Ingold-Prelog) rules, we assign priority based on atomic number, with the highest atomic number having the highest priority.
In this case, both carbon atoms are bonded to a hydrogen atom, but they differ in their other substituents.
The priority of each substituent is as follows:
- Higher priority: substituents with higher atomic numbers (e.g., Cl > C > H)
- Lower priority: substituents with lower atomic numbers (e.g., H < Cl < C)
Once we have assigned priorities, we compare the two substituents on each carbon atom. If the higher-priority groups are on the same side of the double bond, it is designated as "Z" (from German "zusammen," meaning together). If they are on opposite sides, it is designated as "E" (from German "entgegen," meaning opposite).
Since we have two double bonds in 4-methyl-2,5-heptadiene, we need to analyze each individually, and then determine the number of possible combinations:
For the first double bond:
- The higher-priority substituents on each carbon are different, allowing for an E-Z configuration.
- Therefore, there are two stereoisomers possible for the first double bond.
For the second double bond:
- The higher-priority substituents on each carbon are the same, meaning there is no E-Z configuration.
- Therefore, there is only one stereoisomer possible for the second double bond.
Overall, for the structure of 4-methyl-2,5-heptadiene, there are two stereoisomers for the first double bond and one stereoisomer for the second double bond. So, there are two possible stereoisomers for the structure.
b) For 1,2,5-trichloro-3-hexene:
Similar to the previous example, we need to determine if there are any chiral centers or double bonds with restricted rotation.
In this case, there are no chiral centers since each carbon atom is bonded to only three different groups (two chlorine atoms and one hydrogen atom). However, this molecule does have a double bond, which restricts rotation.
To determine the E-Z configuration, we again assign priorities to the groups bonded to each carbon atom of the double bond. Using the same CIP rules, we compare the substituents on each side of the double bond.
For the double bond in 1,2,5-trichloro-3-hexene:
- The higher-priority substituents on each carbon are both chlorine atoms.
- Since they are identical, there is no E-Z configuration possible for this double bond.
Therefore, there is only one possible stereoisomer for the structure of 1,2,5-trichloro-3-hexene.
To summarize:
a) 4-methyl-2,5-heptadiene has two possible stereoisomers, one for the first double bond and one for the second double bond.
b) 1,2,5-trichloro-3-hexene has only one possible stereoisomer.