How long would it take for sucrose to undergo hydrolysis in boiling water?How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols?How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent?How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose?How many total absorptions would appear on proton -NMR spectrum for this molecule?How do I turn room temperature water into boiling water for drinking using an chemical/additive?How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis?How long does it take to replicate Friedrich Wohler's synthesis of urea?How long can solid DPPH powder last for in terms of shelf-life before its free-radical status is quenched by the atmosphere?How would 2-(chloromethyl)-1-methylpyrrolidine recyclize into 3-chloro-1-methylpiperidine with only water?How can I preserve chlorophyll for long periods of time?
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How long would it take for sucrose to undergo hydrolysis in boiling water?
How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols?How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent?How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose?How many total absorptions would appear on proton -NMR spectrum for this molecule?How do I turn room temperature water into boiling water for drinking using an chemical/additive?How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis?How long does it take to replicate Friedrich Wohler's synthesis of urea?How long can solid DPPH powder last for in terms of shelf-life before its free-radical status is quenched by the atmosphere?How would 2-(chloromethyl)-1-methylpyrrolidine recyclize into 3-chloro-1-methylpiperidine with only water?How can I preserve chlorophyll for long periods of time?
$begingroup$
I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:
Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).
Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?
(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)
organic-chemistry food-chemistry carbohydrates hydrolysis
edited 8 hours ago
andselisk♦
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek 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 |
$begingroup$
I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:
Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).
Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?
(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)
organic-chemistry food-chemistry carbohydrates hydrolysis
edited 8 hours ago
andselisk♦
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago
add a comment |
$begingroup$
I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:
Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).
Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?
(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)
organic-chemistry food-chemistry carbohydrates hydrolysis
edited 8 hours ago
andselisk♦
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:
Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).
Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?
(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)
organic-chemistry food-chemistry carbohydrates hydrolysis
organic-chemistry food-chemistry carbohydrates hydrolysis
edited 8 hours ago
andselisk♦
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
andselisk♦
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
andselisk♦
21.4k774143
edited 8 hours ago
andselisk♦
21.4k774143
edited 8 hours ago
andselisk♦
21.4k774143
21.4k774143
asked 8 hours ago
TomekTomek
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
TomekTomek
1113
asked 8 hours ago
TomekTomek
1113
1113
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tomek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago
add a comment |
$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago
$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago
$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C. That value comes from thermodynamic extrapolation of experiments done at higher temperatures. (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after. Figure 1 from the paper has the data.
The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^-6$ per second. That means that 63% degradation would take $10^6$ seconds, or about 11.5 days. Getting 99% degradation would take about 34.5 days.
answered 5 hours ago
Curt F.Curt F.
15.9k13789
$endgroup$
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C. That value comes from thermodynamic extrapolation of experiments done at higher temperatures. (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after. Figure 1 from the paper has the data.
The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^-6$ per second. That means that 63% degradation would take $10^6$ seconds, or about 11.5 days. Getting 99% degradation would take about 34.5 days.
answered 5 hours ago
Curt F.Curt F.
15.9k13789
$endgroup$
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
add a comment |
$begingroup$
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C. That value comes from thermodynamic extrapolation of experiments done at higher temperatures. (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after. Figure 1 from the paper has the data.
The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^-6$ per second. That means that 63% degradation would take $10^6$ seconds, or about 11.5 days. Getting 99% degradation would take about 34.5 days.
answered 5 hours ago
Curt F.Curt F.
15.9k13789
$endgroup$
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
add a comment |
$begingroup$
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C. That value comes from thermodynamic extrapolation of experiments done at higher temperatures. (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after. Figure 1 from the paper has the data.
The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^-6$ per second. That means that 63% degradation would take $10^6$ seconds, or about 11.5 days. Getting 99% degradation would take about 34.5 days.
answered 5 hours ago
Curt F.Curt F.
15.9k13789
$endgroup$
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C. That value comes from thermodynamic extrapolation of experiments done at higher temperatures. (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after. Figure 1 from the paper has the data.
The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^-6$ per second. That means that 63% degradation would take $10^6$ seconds, or about 11.5 days. Getting 99% degradation would take about 34.5 days.
answered 5 hours ago
Curt F.Curt F.
15.9k13789
edited 5 hours ago
answered 5 hours ago
Curt F.Curt F.
15.9k13789
answered 5 hours ago
Curt F.Curt F.
15.9k13789
answered 5 hours ago
Curt F.Curt F.
15.9k13789
15.9k13789
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
add a comment |
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.
$endgroup$
– Andrew
5 hours ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?
$endgroup$
– Tomek
1 hour ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
$begingroup$
It's just exponential kinetics. $fracdSdt = -kS$, so $S = S_0 e^-kt$. If we wait $frac1k$ time units, then we plug in $t = frac1k$ to the formula, and get $S = S_0 e^-1$. $e^-1 = frac1e = frac12.71828...=0.368$. That's how much is left, so $1-0.368=0.63=63%$ is left. Actually $1 - frac1e$ if you want to be precise.
$endgroup$
– Curt F.
37 mins ago
add a comment |
Tomek is a new contributor. Be nice, and check out our Code of Conduct.
Tomek is a new contributor. Be nice, and check out our Code of Conduct.
Tomek is a new contributor. Be nice, and check out our Code of Conduct.
Tomek is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.
$endgroup$
– Karl
7 hours ago