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How can I prepare 1,2-dibromopropane from but-1-ene in a three-step reaction?
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Mark Goodwillie
How can I prepare 1,2-dibromopropane from but-1-ene in a three-step reaction?
To prepare [math] ext{1,2-dibromopropane}[/math] from [math] ext{1-butene}[/math], you have to CLEAVE a [math]C-C[/math] bond, i.e. form a 3-carbon chain from a 4-carbon chain; not an easy proposition… Are you sure the question did not specify [math] ext{propene}[/math] as the starting material?
To prepare [math]ext{1,2-dibromopropane}[/math] from [math]ext{1-butene}[/math], you have to CLEAVE a [math]C-C[/math] bond, i.e. form a 3-carbon chain from a 4-carbon chain; not an easy proposition… Are you sure the question did not specify [math]ext{propene}[/math] as the starting material?
My thought process at solving this multi-step reaction started as such, and note that I am working backwards through a lot of this—I will provide a forwards synthesis below:
I first compared the final product with the starting material, and noticed that the alcohol was attached to the same carbon as the cyclopentyl group. This lead me to think of a reaction that would result in the formation of a secondary alcohol. One method is attack by a cyclopentyl magnesium bromide Grignard reagent and propanal followed by a quenching with acid.
However, this means I must form the cyclopentyl magnesium bromide from cyclopentene.
To start, I considered how to prepare the Grignard, which is the addition of metallic magnesium to a haloalkane (halogenated cyclopentane) in this case. This means that I must now form 1-bromocyclopentane from cyclopentene*. There are many ways to do this (I first thought NBS), but I’m going to stick with something simple—hydrobromic acid reacting with cyclopentene. This allows for a monosubstitution of bromine antiperiplanar to the adding hydrogen (not that it really makes a difference here, but it’s good to consider stereochemistry!).
My thought process at solving this multi-step reaction started as such, and note that I am working backwards through a lot of this—I will provide a forwards synthesis below:
I first compared the final product with the starting material, and noticed that the alcohol was attached to the same carbon as the cyclopentyl group. This lead me to think of a reaction that would result in the formation of a secondary alcohol. One method is attack by a cyclopentyl magnesium bromide Grignard reagent and propanal followed by a quenching with acid.
However, this means I must form the cyclopentyl magnesium bromide from cyclopentene.
To start, I considered how to prepare the Grignard, which is the addition of metallic magnesium to a haloalkane (halogenated cyclopentane) in this case. This means that I must now form 1-bromocyclopentane from cyclopentene*. There are many ways to do this (I first thought NBS), but I’m going to stick with something simple—hydrobromic acid reacting with cyclopentene. This allows for a monosubstitution of bromine antiperiplanar to the adding hydrogen (not that it really makes a difference here, but it’s good to consider stereochemistry!).
To prepare [math] ext{1,2-dibromopropane}[/math] from [math] ext{1-butene}[/math], you have to CLEAVE a [math]C-C[/math] bond, i.e. form a 3-carbon chain from a 4-carbon chain; not an easy proposition… Are you sure the question did not specify [math] ext{propene}[/math] as the starting material?
[math]H_{2}C=CH-CH_{3}(g) + Br_{2}(l) stackrel{H^{+}}longrightarrow BrH_{2}C-C(Br)H-CH_{3}(g)[/math]
To prepare [math]ext{1,2-dibromopropane}[/math] from [math]ext{1-butene}[/math], you have to CLEAVE a [math]C-C[/math] bond, i.e. form a 3-carbon chain from a 4-carbon chain; not an easy proposition… Are you sure the question did not specify [math]ext{propene}[/math] as the starting material?
[math]H_{2}C=CH-CH_{3}(g) + Br_{2}(l) stackrel{H^{+}}longrightarrow BrH_{2}C-C(Br)H-CH_{3}(g)[/math]
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My thought process at solving this multi-step reaction started as such, and note that I am working backwards through a lot of this—I will provide a forwards synthesis below:
I first compared the final product with the starting material, and noticed that the alcohol was attached to the same carbon as the cyclopentyl group. This lead me to think of a reaction that would result in the formation of a secondary alcohol. One method is attack by a cyclopentyl magnesium bromide Grignard reagent and propanal followed by a quenching with acid.
However, this means I must form the cyclopentyl magnesium bromide from cyclopentene.
To start, I considered how to prepare the Grignard, which is the addition of metallic magnesium to a haloalkane (halogenated cyclopentane) in this case. This means that I must now form 1-bromocyclopentane from cyclopentene*. There are many ways to do this (I first thought NBS), but I’m going to stick with something simple—hydrobromic acid reacting with cyclopentene. This allows for a monosubstitution of bromine antiperiplanar to the adding hydrogen (not that it really makes a difference here, but it’s good to consider stereochemistry!).
Below I have a picture of my reaction diagram:
My thought process at solving this multi-step reaction started as such, and note that I am working backwards through a lot of this—I will provide a forwards synthesis below:
I first compared the final product with the starting material, and noticed that the alcohol was attached to the same carbon as the cyclopentyl group. This lead me to think of a reaction that would result in the formation of a secondary alcohol. One method is attack by a cyclopentyl magnesium bromide Grignard reagent and propanal followed by a quenching with acid.
However, this means I must form the cyclopentyl magnesium bromide from cyclopentene.
To start, I considered how to prepare the Grignard, which is the addition of metallic magnesium to a haloalkane (halogenated cyclopentane) in this case. This means that I must now form 1-bromocyclopentane from cyclopentene*. There are many ways to do this (I first thought NBS), but I’m going to stick with something simple—hydrobromic acid reacting with cyclopentene. This allows for a monosubstitution of bromine antiperiplanar to the adding hydrogen (not that it really makes a difference here, but it’s good to consider stereochemistry!).
Below I have a picture of my reaction diagram:
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