LDA and an aldehyde Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Selectivity in aldol condensation between pivaldehyde and acetoneWhy can you not make a compound with a quaternary alpha carbon using malonic ester?How to explain the different regioselectivity of ketones/imines reacting with LDA?nucleophilic attack in Acid chloride formationWhy is Fmoc base-labile and Moz acid-labile?Can a base like OH- attack aldehyde and ketone to create hydrate instead of enol?Can aldol reaction occur during α-halogenation of ketone?Synthesis of methylenecyclohexane from cyclohexylmethanolReduction of Carbonyl Compounds with LiAlH4Salt used in Perkin Reaction
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LDA and an aldehyde
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LDA and an aldehyde
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Selectivity in aldol condensation between pivaldehyde and acetoneWhy can you not make a compound with a quaternary alpha carbon using malonic ester?How to explain the different regioselectivity of ketones/imines reacting with LDA?nucleophilic attack in Acid chloride formationWhy is Fmoc base-labile and Moz acid-labile?Can a base like OH- attack aldehyde and ketone to create hydrate instead of enol?Can aldol reaction occur during α-halogenation of ketone?Synthesis of methylenecyclohexane from cyclohexylmethanolReduction of Carbonyl Compounds with LiAlH4Salt used in Perkin Reaction
$begingroup$
Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $ceC4H5O^-Li^+$.
I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $mathrmpK_mathrma$ because it is an $mathrmsp^2$ C and the C is slightly positive from the carbonyl.
However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $mathrmsp^3$ C next to the carbonyl.
Where would LDA preferentially attack an aldehyde and why?
Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $ceH+$ from the $mathrmsp^3$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $mathrmsp^3$ C-H for some reason?
organic-chemistry reaction-mechanism
edited 40 mins ago
andselisk
19.6k665128
asked 4 hours ago
MirteMirte
21919
$endgroup$
add a comment |
$begingroup$
Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $ceC4H5O^-Li^+$.
I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $mathrmpK_mathrma$ because it is an $mathrmsp^2$ C and the C is slightly positive from the carbonyl.
However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $mathrmsp^3$ C next to the carbonyl.
Where would LDA preferentially attack an aldehyde and why?
Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $ceH+$ from the $mathrmsp^3$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $mathrmsp^3$ C-H for some reason?
organic-chemistry reaction-mechanism
edited 40 mins ago
andselisk
19.6k665128
asked 4 hours ago
MirteMirte
21919
$endgroup$
$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago
add a comment |
$begingroup$
Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $ceC4H5O^-Li^+$.
I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $mathrmpK_mathrma$ because it is an $mathrmsp^2$ C and the C is slightly positive from the carbonyl.
However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $mathrmsp^3$ C next to the carbonyl.
Where would LDA preferentially attack an aldehyde and why?
Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $ceH+$ from the $mathrmsp^3$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $mathrmsp^3$ C-H for some reason?
organic-chemistry reaction-mechanism
edited 40 mins ago
andselisk
19.6k665128
asked 4 hours ago
MirteMirte
21919
$endgroup$
Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $ceC4H5O^-Li^+$.
I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $mathrmpK_mathrma$ because it is an $mathrmsp^2$ C and the C is slightly positive from the carbonyl.
However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $mathrmsp^3$ C next to the carbonyl.
Where would LDA preferentially attack an aldehyde and why?
Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $ceH+$ from the $mathrmsp^3$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $mathrmsp^3$ C-H for some reason?
organic-chemistry reaction-mechanism
organic-chemistry reaction-mechanism
edited 40 mins ago
andselisk
19.6k665128
asked 4 hours ago
MirteMirte
21919
edited 40 mins ago
andselisk
19.6k665128
asked 4 hours ago
MirteMirte
21919
edited 40 mins ago
andselisk
19.6k665128
edited 40 mins ago
andselisk
19.6k665128
edited 40 mins ago
andselisk
19.6k665128
19.6k665128
asked 4 hours ago
MirteMirte
21919
asked 4 hours ago
MirteMirte
21919
asked 4 hours ago
MirteMirte
21919
21919
$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago
add a comment |
$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago
$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $mathrmsp^2$ and $mathrmsp^3$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.
edited 39 mins ago
andselisk
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $mathrmsp^2$ and $mathrmsp^3$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.
edited 39 mins ago
andselisk
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
$endgroup$
add a comment |
$begingroup$
If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $mathrmsp^2$ and $mathrmsp^3$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.
edited 39 mins ago
andselisk
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
$endgroup$
add a comment |
$begingroup$
If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $mathrmsp^2$ and $mathrmsp^3$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.
edited 39 mins ago
andselisk
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
$endgroup$
If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $mathrmsp^2$ and $mathrmsp^3$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.
edited 39 mins ago
andselisk
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
edited 39 mins ago
andselisk
19.6k665128
edited 39 mins ago
andselisk
19.6k665128
edited 39 mins ago
andselisk
19.6k665128
19.6k665128
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
answered 3 hours ago
Oscar LanziOscar Lanzi
16.3k12749
16.3k12749
add a comment |
add a comment |
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$begingroup$
LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.
$endgroup$
– Waylander
4 hours ago
$begingroup$
While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.
$endgroup$
– Zhe
2 hours ago