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Iodometric titration is sometimes used to find out how much lactose there is in milk. The overall oxidation reaction is written as follows.

$$\ce{C12H22O11 + I2 + 3NaOH ->[\ce{CuSO4}]C12H21O12Na + 2NaI + H2O}$$

How does this reaction proceed? What is the role of $\ce{CuSO4}$?

The solution is then acidified with $\ce{HCl}$. Excess $\ce{I2}$ is determined via

$$\ce{2Na2S2O3 + I2 ->2NaI + Na2S4O6}.$$

What I think might be going on (with references)

Using basics of inorganic chemistry, iodine reacts with cold sodium hydroxide to give sodium hypoiodate.

$$\ce{I2 + 2NaOH <=>NaIO + NaI + H2O}$$

$\ce{NaIO}$ is the main oxidiser of lactose.$^{[1]\ [a]}$

$$\ce{C12H22O11 +NaIO + NaOH ->C12H21O12Na + NaI + H2O }$$

I have no clue as to what the complete mechanism might be here.

Alternatively, $\ce{Cu^{2}^+}$can be reduced. The following picture$^{[2]}$ shows the general transitions for (semi)linear lactose.

Lactose oxidation

Sadly, neither reference mentions the other option, nor are there arrow-pushing mechanisms which is what I am after.

Sodium iodate is prone to decomposition even under room temperature, but the reaction is vigorous from $80 \ ^\circ \text{C}\ ^{[3]}$:

$$\ce{3NaIO->NaIO3 +2NaI}.$$

This is partly why the acid is added. To get an accurate reading of "unreacted" iodine we must somehow re-release $\ce{I2}$ from the ions $\ce{IO^-}$ and $\ce{IO_3^-}$.$^{[1]}$

$$\ce{NaIO + NaI + 2HCl -> I2 + 2NaCl + H2O\\ NaIO3 + 5NaI + 6HCl -> 3I2 + 6NaCl + 3H2O}$$

Thiosulphate is a widespread reagent that is subject to inaccuracies$^{[4]\ [5]}$ due to

$$\ce{Na2S2O3 + 4I2 + 10NaOH -> 2Na2SO4 + 8NaI + 5H2O}.$$

This reaction is less favoured in acidic conditions.$^{[1]}$

Clarification

I am asking for

  • arrow-pushing mechanisms that show the role of $\ce{NaIO}$ and/or $\ce{CuSO4}$ with cyclic lactose (if possible).

$^{[1]}$ A Study of the Method for Titrating Aldose Sugars with Standard Iodine and Alkali. G. M. Kline, S. F. Acree. Bureau of Standards Journal of Research. Link [Comment: the study actually deals with aldose sugars. The equation for cyclic lactose might thus be a stretch.]

$^{[a]}$ Agu-Tõnis Talvik. Organic chemistry. (1996). page 490 (as far as I know, not available in English)

$^{[2]}$ Dairy Chemistry and Biochemistry. Second Edition. P. F. Fox, T. Uniacke-Lowe, P. L. H. McSweeney, J. A. O'Malhony. page 62. Link

$^{[3]}$ Иодат натрия. Russian Wikipedia. Link

$^{[4]}$ (G. Topf). Zeit. anal. Chem., 26, p. 137; 1887

$^{[5]}$ (Bougault). Compt. rend., 164, p.; 949; 1917

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  • $\begingroup$ This also got me wandering... What is the go-to software for chemists who wish to draw reaction mechanisms on the computer? $\endgroup$
    – Linear Christmas
    Jun 17, 2016 at 18:16
  • $\begingroup$ @LinearChristmas: wondering*... :( $\endgroup$
    – Linear Christmas
    Jun 17, 2016 at 18:35
  • 3
    $\begingroup$ ChemDraw - if you can afford it or get it free from your institution. $\endgroup$
    – bon
    Jun 17, 2016 at 19:05
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    $\begingroup$ @bon OR ChemDoodle which is very good and not too expensive. $\endgroup$
    – matt_black
    Jun 17, 2016 at 19:25
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    $\begingroup$ There are a number of free software tools around for Linux. Didn’t find one I would gladly recommend yet, though. They all have their shortcomings somewhere … $\endgroup$
    – Jan
    Jun 19, 2016 at 12:01

1 Answer 1

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Using the basics of organic chemistry, solutions of iodine and sodium hydroxide retain enough free $\ce{I2}$ to do other things. I would disocunt $\ce{NaIO}$ as a significant component of this process.

What if instead, we had exactly what your picture shows (except the last few steps are prevented - we'll see how in a few lines). You are worried abut the cyclic form, but the cyclic form and the acyclic form are in equilibrium in aqeous solution, especially if acid or base is present. Note that in the reactions equations below, I am using the notation $\ce{R}$ to refer to parts of the molecule that are not involved.

Isomerization to the ene-diol

$$\ce{R-CH(OH)-CHO <=>[\ce{OH-}] R-C(OH)=CHOH}$$

Oxidation by copper(II)

$$\ce{R-C(OH)=CHOH + 2Cu^2+ +3OH- -> R-CH(OH)-CO2^- + 2Cu+ + 2H2O}$$

Now, if that copper(I) cations stumbles into a hydroxide anion, then it's all over since $\ce{Cu2O}$ is not soluble. However, it's iodine to the rescue! Iodine oxidizes $\ce{Cu+}$ back to $\ce{Cu^2+}$.

$$\ce{2Cu+ + I2 -> 2Cu^2+ + 2I-}$$

These last two equations add to give the one you quote (only I left out the sodium spectator cations and simplified the carbohydrate).

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  • $\begingroup$ Good answer! I do agree that there will be free iodine; and it can be iodine itself that reacts. However, so far I have only been able to find references for $IO^-$. Since posting, I found even another source that supports $IO^-$ over $I_2$ (edited post). Also, the question asked for the mechanism. So is this answer to say that the mechanism is not known, or not useful in understanding the process? (I am interested in the $Cu^{2+}$/$IO^-$ intermediary steps.) Definite +1 and thanks for pointing out the iodine oxidation of $Cu^+$ which I had completely overlooked. $\endgroup$
    – Linear Christmas
    Jun 19, 2016 at 17:45
  • $\begingroup$ If the terminal oxidant is $\ce{IO-}$, then its reaction with $\ce{Cu+}$ is $$\ce{2Cu+ + IO- + H2O -> 2Cu^2+ + I- + 2OH-}$$ However, the mechanism of a redox reaction is a series of one-electron transfers. $\endgroup$
    – Ben Norris
    Jun 20, 2016 at 11:25
  • $\begingroup$ Could $\ce{Cu(OH)2}$ also participate in the process instead of or in parallel to $\ce{Cu^2^+}$? This is also what a few sources suggest. $\endgroup$
    – Linear Christmas
    Jun 27, 2016 at 17:00

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