To solve the problem of the addition of \( \text{Br}_2 \) to (E)-but-2-ene, we will follow these steps:
### Step 1: Identify the structure of (E)-but-2-ene
(E)-but-2-ene has the following structure:
\[
\text{CH}_3 - \text{C}=\text{C} - \text{CH}_3
\]
In the (E) configuration, the two methyl groups (CH₃) are on opposite sides of the double bond.
### Step 2: Understand the addition reaction
When \( \text{Br}_2 \) is added to an alkene, it undergoes an electrophilic addition reaction. The double bond acts as a nucleophile and attacks a bromine molecule, leading to the formation of a cyclic bromonium ion intermediate.
### Step 3: Draw the bromonium ion intermediate
Upon the addition of \( \text{Br}_2 \), the double bond attacks one of the bromine atoms, forming a cyclic bromonium ion. The bromonium ion has a three-membered ring structure where one bromine atom is positively charged and is bonded to both carbons of the former double bond.
### Step 4: Nucleophilic attack by bromide ion
The bromide ion (Br⁻), which is generated from the dissociation of \( \text{Br}_2 \), can now attack the more substituted carbon of the bromonium ion. This attack occurs from the opposite side of the bromonium ion due to steric hindrance, leading to anti-addition.
### Step 5: Determine the products
The product of this reaction will be 2,3-dibromobutane. Since (E)-but-2-ene is symmetrical, the addition of bromine will yield two stereoisomers:
1. (2R,3S)-2,3-dibromobutane
2. (2S,3R)-2,3-dibromobutane
However, these two stereoisomers are actually meso compounds because they are superimposable on their mirror images.
### Step 6: Conclusion
The final product of the addition of \( \text{Br}_2 \) to (E)-but-2-ene is meso-2,3-dibromobutane, which can be represented as:
\[
\text{meso-2,3-dibromobutane}
\]
### Final Answer
The addition of \( \text{Br}_2 \) to (E)-but-2-ene gives meso-2,3-dibromobutane.
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