Draw the structure of:
(E)-1,2-dibromo-3-isopropyl-2-hexene
My approach: I know how to answer this question in terms of basic nomeclature but I have no idea what (E) means in this.
This compound is a standard exam stumper.
http://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism#E.2FZ_notation
E means opposite, E is another way of saying trans and Z would be cis..
In organic chemistry, the (E)- isomer designation refers to the "E configuration," which stands for "entgegen" and indicates that the highest-priority substituents are on opposite sides of a double bond. On the other hand, the (Z)- configuration, which stands for "zusammen," indicates that the highest-priority substituents are on the same side of a double bond.
Now, let's draw the structure of (E)-1,2-dibromo-3-isopropyl-2-hexene step by step:
Step 1: Start with a hexane backbone, which consists of six carbon atoms in a straight chain.
H H H H H H
| | | | | |
H-C-C-C-C-C-C-H
Step 2: Add a double bond between carbon atoms 2 and 3 by removing two hydrogens from carbon atoms 2 and 3.
H H Br Br H H
| | | | | |
H-C-C=C-C-C-C-H
Step 3: Next, we need to add the isopropyl group, which means substituting a branched carbon chain on carbon atom 3.
H H Br Br H H H
| | | | | | |
H-C-C=C-C-C-C-C-C-H
|
(CH3)2CH-
Step 4: Finally, add the two bromine atoms on carbon atoms 1 and 2, which completes the structure of (E)-1,2-dibromo-3-isopropyl-2-hexene.
H Br Br Br H H H
| | | | | | | |
H-C-C=C-C-C-C-C-C-H
|
(CH3)2CH-
Note: Even though the structure does not display the stereochemistry explicitly in this format, since it is named as (E)-1,2-dibromo-3-isopropyl-2-hexene, it implies that the bromine atoms are on opposite sides of the double bond.
To draw the structure of (E)-1,2-dibromo-3-isopropyl-2-hexene, we need to understand the concept of E/Z isomerism in organic chemistry.
E/Z isomerism, also known as cis/trans isomerism, refers to the different spatial arrangements of substituents around a double bond. In an E isomer, the highest priority substituents are on opposite sides of the double bond, while in a Z isomer, the highest priority substituents are on the same side. Priority is determined by the Cahn-Ingold-Prelog (CIP) rules, which assign priorities based on atomic number.
In order to assign E/Z configuration, we need to determine the priority of the substituents on both ends of the double bond. The highest atomic number substituent has the highest priority. If multiple substituents have the same atom, we look at the next atom in the substituent until a difference is found.
Now, let's apply this to (E)-1,2-dibromo-3-isopropyl-2-hexene:
First, identify the position of the double bond, which is between the 2nd and 3rd carbons from the right end.
Next, let's determine the priority of the substituents on both ends of the double bond. The priority is based on the atomic number of the atoms attached to the double bond carbons. In this case, we have 1,2-dibromo and 3-isopropyl as substituents.
For the 1,2-dibromo substituent, compare the atoms directly attached to the carbons of the double bond. In this case, both carbons have bromine atoms attached, so we move to the next atoms. For the first carbon, we have a bromine atom and a hydrogen atom. For the second carbon, we have a bromine atom and an isopropyl group. Since the isopropyl group has a higher atomic number compared to hydrogen, the second carbon has a higher priority.
For the 3-isopropyl substituent, we compare the atoms directly attached to the third carbon of the double bond. In this case, we have a hydrogen atom, a carbon atom, and two additional carbon atoms in the isopropyl group. The two additional carbon atoms have a higher atomic number compared to hydrogen, so the 3-isopropyl substituent has a higher priority.
With this information, we can now determine the E/Z configuration. Since the higher priority substituents (2nd carbon and 3-isopropyl) are on opposite sides of the double bond, the compound is an E isomer.
Now, to draw the structure of (E)-1,2-dibromo-3-isopropyl-2-hexene, follow this step-by-step procedure:
1. Draw a six-carbon chain.
2. Identify the second and third carbons from the right end as the double bond carbons.
3. Add two bromine atoms to the second carbon.
4. Add an isopropyl group (CH3-CH(CH3)-CH3) to the third carbon.
5. Complete the structure by adding hydrogens to the remaining carbons as needed to fulfill their valency.
The final structure should look like this:
Br
|
H3C-C-CH3
|
H3C-C=C-CH2-CH2-CH3
|
Br
Note: The carbon numbering may differ depending on where you start numbering the chain.