What would happen if sugar and cream of tartar react with each other? Would it create an ester?

$$\ce{KC4H5O6 + C12H22O11 -> KC16H21O14 + 3H2O}$$

Is this correct? If not, why?


Rather than "cream of tarter" I'm going to stick to the pure chemical potassium bitartrate and tartaric acid, and I'm going to speak from the point of view of a chemist about the difference between potassium bitartrate and tartaric acid (rather than a cook or winemaker).

First a simple answer to your questions and then the details. Simplistically speaking, when you add potassium bitartrate to an aqueous solution of sucrose you will get some hydrolysis (producing glucose and fructose) and some ester formation. The key word here, however, is some -- and the amount depends strongly on the reaction conditions. At room temperature, for example, you're likely to get essentially no reaction within, say a day. More details below.

Tartaric acid is a dicarboxylic acid (a weak acid). Each of its carboxyl groups (-COOH) can donate a proton (H+ ion), and when this happens it's "acting as an acid". This is important to your question because it is the H+ ion that acts as a catalyst to sucrose hydrolysis and not the tartaric acid itself. For this situation, chemists call it a specific acid catalyzed reaction. By contrast, if a specific molecule such as tartaric acid were involved in the catalytic step chemists would call that "general acid catalysis".

Acidity of molecules are gauged by assigning a pKa value. If there are 2 acidic groups in the molecule (as with Tartaric acid) then it has 2 pKa values. Adding KOH to a solution of Tartaric acid will give you potassium bitartrate. At 25C, the pKa for removing Tartaric acid's first H+ is about 3.0 and the pKa for removing potassium bitartrate's H+ is 4.3. This makes both of them what chemist's call "weak acids" because they don't release anywhere near as much H+ into a solution as a strong acid (like HCl).

The other reaction, ester formation, is also specific acid catalyzed and so would also occur as I mentioned, but to a small extent. Esters are formed in acidic solutions whenever carboxyl (-COOH) and alcohol (-OH) groups are present, as is the case here. Note that there are many -OH groups here, on both molecules.

To understand the extent of each of these reactions it's important to consider the role of water. Water is a reactant in the hydrolysis reaction and this tends to drive the reaction forward. By contrast, water is a product of the ester formation reaction. This is especially important for the ester reaction since it is highly reversible. In other words, whenever molecules collide to form an ester, it's much more likely that that ester will collide with the far more abundant water molecules and revert back to the starting materials, resulting in no net reaction.

Usually one thinks of these reactions happening somewhere near room temperature. Under these conditions neither of the above reactions happens to a great extent; they are both generally very slow. In order to push them forward you have to do things like increasing the temperature, adding more concentrated reaction materials or catalysts, or removing reaction products. Of course, in this case increasing the concentration of the reaction materials also increase the amount of catalyst (H+), but since they are such weak acids, adding HCl (or another strong acid) would be much more effective.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.