In the following reaction, what is the configuration of the products? Br₂ CH3OH

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Chapter22: Organic And Biological Molecules
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Problem 170CP: Alcohols are very useful starting materials for the production of many different compounds. The...
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**Understanding the Bromination Reaction of Alkenes**

**Reaction Description:**
In the following reaction, the starting compound is an alkene, more precisely 2-methyl-2-butene, which is depicted as follows:

```
     CH3
     |
CH3-C=CH-CH3
     |
     H
```

When this compound undergoes a reaction with bromine (Br₂) in the presence of methanol (CH₃OH), a bromination reaction occurs. The product is formed through the addition of Br₂ across the double bond (C=C).

**Procedure:**
1. Initially, Br₂ interacts with the electron-rich double bond to form a cyclic bromonium ion. 
2. Following this, a nucleophile (in this case, CH₃OH) attacks the more substituted carbon of the bromonium ion.
3. This results in the formation of a bromohydrin where the bromine (Br) and methoxy group (OCH₃) are added anti to each other, meaning they add on opposite sides of the previously existing double bond.

**Product Configuration:**
The final product configuration will show that the bromine and methoxy groups are on adjacent carbons and oriented trans to each other, ensuring the anti addition mechanism is fulfilled.

**Summary:**
The product of the reaction is a bromohydrin with anti addition across the double bond:
```
        Br
         |
CH3-C-CH-CH3
     |    |
    H3C OCH3
```

In this product:
- One of the Carbons that was part of the double bond now carries a Br atom.
- The adjacent Carbon carries a -OCH₃ group. 
- The Br and the -OCH₃ group are on opposite sides (trans arrangement).

**Conclusion:**
The described mechanism and resultant product are representative of typical electrophilic addition reactions seen with alkenes in the presence of halogens and alcohols.
Transcribed Image Text:**Understanding the Bromination Reaction of Alkenes** **Reaction Description:** In the following reaction, the starting compound is an alkene, more precisely 2-methyl-2-butene, which is depicted as follows: ``` CH3 | CH3-C=CH-CH3 | H ``` When this compound undergoes a reaction with bromine (Br₂) in the presence of methanol (CH₃OH), a bromination reaction occurs. The product is formed through the addition of Br₂ across the double bond (C=C). **Procedure:** 1. Initially, Br₂ interacts with the electron-rich double bond to form a cyclic bromonium ion. 2. Following this, a nucleophile (in this case, CH₃OH) attacks the more substituted carbon of the bromonium ion. 3. This results in the formation of a bromohydrin where the bromine (Br) and methoxy group (OCH₃) are added anti to each other, meaning they add on opposite sides of the previously existing double bond. **Product Configuration:** The final product configuration will show that the bromine and methoxy groups are on adjacent carbons and oriented trans to each other, ensuring the anti addition mechanism is fulfilled. **Summary:** The product of the reaction is a bromohydrin with anti addition across the double bond: ``` Br | CH3-C-CH-CH3 | | H3C OCH3 ``` In this product: - One of the Carbons that was part of the double bond now carries a Br atom. - The adjacent Carbon carries a -OCH₃ group. - The Br and the -OCH₃ group are on opposite sides (trans arrangement). **Conclusion:** The described mechanism and resultant product are representative of typical electrophilic addition reactions seen with alkenes in the presence of halogens and alcohols.
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