Write the mechanism for (E)-2-5-dimethyl-hex-3-ene and bromine?
Assign the R, S configuration for each of the products and determine the stereochemical relationship of each set of products.
I think the products will be trans but I am not quite sure how the mechanism work. Any help would be appreciated
To determine the mechanism for the reaction between (E)-2-5-dimethyl-hex-3-ene and bromine, you need to understand the process of bromination of alkenes.
Bromination of alkenes typically occurs via an electrophilic addition reaction, where the double bond of the alkene acts as a nucleophile attacking the electrophilic bromine molecule. The reaction takes place in two steps: the initiation step and the propagation step.
1. Initiation step:
In the presence of light or heat, bromine is activated by homolytic cleavage. This results in the formation of bromine radicals (Br·):
Br2 -> 2 Br·
2. Propagation step:
The first propagation step involves the bromine radical (Br·) attacking the double bond of the alkene, causing the bond to break and a new bond to form between bromine and one of the carbon atoms. This results in the formation of a bromonium ion intermediate:
Br· + CH3CH=CHCH2CH(CH3)CH2Br -> BrCH2CH(CH3)CH2CH(Br)+CH3CH2·
In the second propagation step, the bromide ion (Br-) acts as a nucleophile attacking the backside of the bromonium ion. This leads to the formation of the trans product, as the bulky methyl groups prefer to be in the anti-periplanar arrangement:
BrCH2CH(CH3)CH2CH(Br)+CH3CH2· -> BrCH2CH(CH3)CH2CH(Br)CH3CH2Br
The overall reaction can be summarized as:
(E)-2-5-dimethyl-hex-3-ene + Br2 -> BrCH2CH(CH3)CH2CH(Br)CH3CH2Br
Now, let's assign the R and S configurations for each of the products:
For the first bromine (Br) attached to the double bond, the two bromines are on opposite sides, resulting in a trans configuration.
For the second bromine (Br) attached to the carbon with the methyl group (CH3), the methyl group and the remaining two substituents (CH2CH(CH3)CH2CH(Br)CH3CH2) can be prioritized based on the Cahn-Ingold-Prelog (CIP) rules. Rearranging the atoms so that the lowest priority group (H) is in the back, the order of priority for the remaining groups is as follows:
1. Br (higher atomic number than C)
2. CH3 (higher atomic number than H)
3. CH2CH(CH3)CH2 (higher atomic number than CH3)
With the highest priority group (Br) in the front, the configuration is assigned as R.
For the third bromine (Br) attached to the terminal carbon, the configuration is assigned as S based on the CIP rules.
In summary, the stereochemical relationship of the products is as follows:
1. The bromines attached to the double bond are trans.
2. The first bromine is on the opposite side of the methyl group, and its configuration is trans.
3. The second bromine has an R configuration.
4. The third bromine has an S configuration.
Therefore, the products have a trans configuration overall.