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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?)

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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 Jun 12 at 22:14
  • $\begingroup$ How extended of an extended period of boiling did Cocktail Codex mean? $\endgroup$ – RonJohn Jun 13 at 14:46
  • $\begingroup$ The hangover bits may come from the fact that sweet drinks make you drink them more. $\endgroup$ – vasin1987 Jun 13 at 16:37
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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 50% 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.

Figure_1_from_Wolfenden_paper

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 53 days.

edits

The above has been edited to fix sloppy math errors, as noted by @gremin and @John Bentin in the comments. Thank you both.

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    $\begingroup$ 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 Jun 13 at 0:26
  • $\begingroup$ Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation? $\endgroup$ – Tomek Jun 13 at 4:09
  • $\begingroup$ It's just exponential kinetics. $\frac{dS}{dt} = -kS$, so $S = S_0 e^{-kt}$. If we wait $\frac{1}{k}$ time units, then we plug in $t = \frac{1}{k}$ to the formula, and get $S = S_0 e^{-1}$. $e^{-1} = \frac{1}{e} = \frac{1}{2.71828...}=0.368$. That's how much is left, so $1-0.368=0.63=63\%$ is left. Actually $1 - \frac{1}{e}$ if you want to be precise. $\endgroup$ – Curt F. Jun 13 at 5:01
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    $\begingroup$ Half-life isn't the same as lifetime ($1/k$ in the kinetic equation). $T_{1/2} = k \times ln 2$. Shouldn't it be 50% in this sentence "Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C"? The numbers from the graph are ok. $\endgroup$ – Gremlin Jun 13 at 11:03
  • $\begingroup$ I don't see you final statement (in bold). If it takes 11.5 days to reach a sucrose-residue ratio of e^{-1}, then (neglecting back-reaction) 34.5 days would take it to e^{-3}, or about 5%. To reach 1% would require 11.5 ln 100 or about 53 days. $\endgroup$ – John Bentin Jun 13 at 12:22

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