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Why is tert-butoxide often used in elimination reactions when it is not necessary?

Alkene reactions with sodium tert-butoxide and CHBr₃Why is ethyne and not ethene used for welding?Why does LiAlH(O-tert-Bu)3 not reduce acyl chlorides to alcohols?Why is tert-Butyl (tBu) often not listed as protecting group for alcohol?Rate of substitution and elimination reactions at high temperaturesTert-butoxide in SN2 reactionsWhy substitution and elimination reactions are favored respectively at lower and higher temperature?Why are elimination reactions in organic chemistry known as beta eliminations?Reactions of differents halohydrins with tert-butoxide

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I understand that tert-butoxide is used to favor the Hofmann product in elimination reactions due to its steric bulk (and thus inability to abstract a proton in a more substituted position); however, I have often seen proposed organic synthesis involving the use of tert-butoxide when it is seemingly irrelevant. For example, 1-bromobutane will undergo an elimination reaction with or without a sterically hindered base (such as tert-butoxide); however, anytime I see reactions involving elimination reaction with primary substrates, such as the bromo group in 1-bromobutane, tert-butoxide is used rather than ethoxide. Is tert-butoxide simply used to disfavor the formation of substitution products. Is its steric bulk enough to not allow it to act as a good nucleophile in substitution reactions?

organic-chemistry synthetic-chemistry

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asked 8 hours ago

ELI JONESELI JONES

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5

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I understand that tert-butoxide is used to favor the Hofmann product in elimination reactions due to its steric bulk (and thus inability to abstract a proton in a more substituted position); however, I have often seen proposed organic synthesis involving the use of tert-butoxide when it is seemingly irrelevant. For example, 1-bromobutane will undergo an elimination reaction with or without a sterically hindered base (such as tert-butoxide); however, anytime I see reactions involving elimination reaction with primary substrates, such as the bromo group in 1-bromobutane, tert-butoxide is used rather than ethoxide. Is tert-butoxide simply used to disfavor the formation of substitution products. Is its steric bulk enough to not allow it to act as a good nucleophile in substitution reactions?

organic-chemistry synthetic-chemistry

share|improve this question

asked 8 hours ago

ELI JONESELI JONES

35911 silver badge77 bronze badges

$endgroup$

add a comment | 

$begingroup$

I understand that tert-butoxide is used to favor the Hofmann product in elimination reactions due to its steric bulk (and thus inability to abstract a proton in a more substituted position); however, I have often seen proposed organic synthesis involving the use of tert-butoxide when it is seemingly irrelevant. For example, 1-bromobutane will undergo an elimination reaction with or without a sterically hindered base (such as tert-butoxide); however, anytime I see reactions involving elimination reaction with primary substrates, such as the bromo group in 1-bromobutane, tert-butoxide is used rather than ethoxide. Is tert-butoxide simply used to disfavor the formation of substitution products. Is its steric bulk enough to not allow it to act as a good nucleophile in substitution reactions?

organic-chemistry synthetic-chemistry

share|improve this question

asked 8 hours ago

ELI JONESELI JONES

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share|improve this question

asked 8 hours ago

ELI JONESELI JONES

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share|improve this question

asked 8 hours ago

ELI JONESELI JONES

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asked 8 hours ago

ELI JONESELI JONES

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asked 8 hours ago

ELI JONESELI JONES

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2 Answers
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5

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Several reasons:

1. t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.




2. It is easily available, like Raphaël insists.




3. It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $ceK+$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.




4. Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.

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answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

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Generally, it's just because there's a big bottle of $cet-BuOK$ lying around, so why bother using something else?

In the same range of ideas, I've seen publications where they used $cen-BuLi$ to deprotonate an alcohol; it's completely overkill, but was probably just more convenient in their case than $ceNaH$ or something else.

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answered 8 hours ago

RaphaëlRaphaël

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5

$begingroup$

Several reasons:

1. t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.




2. It is easily available, like Raphaël insists.




3. It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $ceK+$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.




4. Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.

share|improve this answer

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Several reasons:

1. t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.




2. It is easily available, like Raphaël insists.




3. It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $ceK+$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.




4. Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.

share|improve this answer

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

$endgroup$

add a comment | 

$begingroup$

Several reasons:

1. t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.




2. It is easily available, like Raphaël insists.




3. It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $ceK+$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.




4. Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.

share|improve this answer

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

answered 8 hours ago

William R. EbenezerWilliam R. Ebenezer

1,72011 gold badge55 silver badges2323 bronze badges

4

$begingroup$

Generally, it's just because there's a big bottle of $cet-BuOK$ lying around, so why bother using something else?

In the same range of ideas, I've seen publications where they used $cen-BuLi$ to deprotonate an alcohol; it's completely overkill, but was probably just more convenient in their case than $ceNaH$ or something else.

share|improve this answer

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Generally, it's just because there's a big bottle of $cet-BuOK$ lying around, so why bother using something else?

In the same range of ideas, I've seen publications where they used $cen-BuLi$ to deprotonate an alcohol; it's completely overkill, but was probably just more convenient in their case than $ceNaH$ or something else.

share|improve this answer

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges

$endgroup$

add a comment | 

$begingroup$

Generally, it's just because there's a big bottle of $cet-BuOK$ lying around, so why bother using something else?

In the same range of ideas, I've seen publications where they used $cen-BuLi$ to deprotonate an alcohol; it's completely overkill, but was probably just more convenient in their case than $ceNaH$ or something else.

share|improve this answer

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges

answered 8 hours ago

RaphaëlRaphaël

49133 silver badges88 bronze badges