Mechanism of the formation of peracetic acid Unicorn Meta Zoo #1: Why another podcast? ...
Mechanism of the formation of peracetic acid
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Mechanism of the formation of peracetic acid
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Wikipedia says that the equilibrium $$ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.
The first possibility is that $ce{CH3COOH}$ is protonated into $ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $ce{CH3C+O}$. Because the oxygen atom in $ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $ce{H2O2}$ attacks the carbon atom in $ce{MeCOOH}$, then the $ce{OH}$ in $ce{COOH}$ and
one of the $ce H$ in $ce {H2O2}$ leave.
Is the mechanism above right or not? If it is not, what's the correct one?
P.S. (this question does not answer my question)
organic-chemistry reaction-mechanism
asked 1 hour ago
Kemono ChenKemono Chen
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$begingroup$
Wikipedia says that the equilibrium $$ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.
The first possibility is that $ce{CH3COOH}$ is protonated into $ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $ce{CH3C+O}$. Because the oxygen atom in $ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $ce{H2O2}$ attacks the carbon atom in $ce{MeCOOH}$, then the $ce{OH}$ in $ce{COOH}$ and
one of the $ce H$ in $ce {H2O2}$ leave.
Is the mechanism above right or not? If it is not, what's the correct one?
P.S. (this question does not answer my question)
organic-chemistry reaction-mechanism
asked 1 hour ago
Kemono ChenKemono Chen
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Wikipedia says that the equilibrium $$ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.
The first possibility is that $ce{CH3COOH}$ is protonated into $ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $ce{CH3C+O}$. Because the oxygen atom in $ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $ce{H2O2}$ attacks the carbon atom in $ce{MeCOOH}$, then the $ce{OH}$ in $ce{COOH}$ and
one of the $ce H$ in $ce {H2O2}$ leave.
Is the mechanism above right or not? If it is not, what's the correct one?
P.S. (this question does not answer my question)
organic-chemistry reaction-mechanism
asked 1 hour ago
Kemono ChenKemono Chen
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Wikipedia says that the equilibrium $$ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.
The first possibility is that $ce{CH3COOH}$ is protonated into $ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $ce{CH3C+O}$. Because the oxygen atom in $ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $ce{H2O2}$ attacks the carbon atom in $ce{MeCOOH}$, then the $ce{OH}$ in $ce{COOH}$ and
one of the $ce H$ in $ce {H2O2}$ leave.
Is the mechanism above right or not? If it is not, what's the correct one?
P.S. (this question does not answer my question)
organic-chemistry reaction-mechanism
organic-chemistry reaction-mechanism
asked 1 hour ago
Kemono ChenKemono Chen
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kemono ChenKemono Chen
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kemono ChenKemono Chen
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kemono ChenKemono Chen
1134
asked 1 hour ago
Kemono ChenKemono Chen
1134
1134
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kemono Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.
answered 45 mins ago
ringoringo
20.3k559112
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1 Answer
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$begingroup$
You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.
answered 45 mins ago
ringoringo
20.3k559112
$endgroup$
add a comment |
$begingroup$
You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.
answered 45 mins ago
ringoringo
20.3k559112
$endgroup$
add a comment |
$begingroup$
You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.
answered 45 mins ago
ringoringo
20.3k559112
$endgroup$
You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.
answered 45 mins ago
ringoringo
20.3k559112
answered 45 mins ago
ringoringo
20.3k559112
answered 45 mins ago
ringoringo
20.3k559112
answered 45 mins ago
ringoringo
20.3k559112
20.3k559112
add a comment |
add a comment |
Kemono Chen is a new contributor. Be nice, and check out our Code of Conduct.
Kemono Chen is a new contributor. Be nice, and check out our Code of Conduct.
Kemono Chen is a new contributor. Be nice, and check out our Code of Conduct.
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