Starting from 1,3 -butadiene, how would you prepare 2,3-dibromobutane?

Short Answer

Expert verified
To prepare 2,3-dibromobutane from 1,3-butadiene, perform an electrophilic addition reaction using bromine (Br2). The reaction can be represented as: \( CH_2=CH-CH=CH_2 + Br_2 \longrightarrow CH_2Br-CHBr-CH_2-CH_3 \). The final product has an anti stereochemistry, with the bromine atoms located trans to each other.

Step by step solution

01

Identify the compound structures

First, familiarize yourself with the structures of the compounds involved in the reaction: - 1,3-butadiene: CH₂=CH-CH=CH₂ - 2,3-dibromobutane: CH₂Br-CHBr-CH₂-CH₃ Take note that the butadiene molecule contains two double bonds in a conjugated diene structure, while the product, 2,3-dibromobutane, has two bromine atoms added to adjacent carbon atoms.
02

Recognize the reaction type

In order to synthesize 2,3-dibromobutane from 1,3-butadiene, an electrophilic addition reaction is needed. The most common electrophilic addition reaction for alkenes and conjugated dienes is the halogenation. In this case, we will use bromine (Br2) as a halogen to achieve the desired product.
03

Add bromine to the diene

As explained in the previous step, we need to add bromine (Br2) to 1,3-butadiene. This occurs through an electrophilic addition reaction mechanism. The bromine molecule acts as an electrophile, attacking one of the double bonds of the diene and forms a bromonium ion intermediate. The reaction can be represented as: \( \: \: CH_2=CH-CH=CH_2 \: + \: Br_2 \longrightarrow \: CH_2Br-CHBr-CH_2-CH_3 \)
04

Observe stereochemistry

The electrophilic addition of bromine to the conjugated diene forms a vicinal dibromoalkane. In this case, bromine adds to 2nd and 3rd carbon atoms within the molecule, and the stereochemistry of the product is anti, meaning that the two bromine atoms will be located trans to each other in the final product. In summary, by performing an electrophilic bromination reaction on 1,3-butadiene, we can successfully prepare 2,3-dibromobutane. The mechanism involves the formation of a bromonium ion intermediate, and the stereochemistry of the product is anti.

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