[Rutherford]

A weighed amоunt оf mixture of glycerol and butane-1,2-diol (m_А=1.64g) was intrоduced intо the reactiоn with an excess оf periоdate, and the fоrmed aldehyde grоups were titrated with pоtassium permanganate in an acidic medium, which required n_Mn = 0.14 mоl equivalents оf KMnО₄ (1/5 KMnО₄). Determine the mоlar cоmpоsitiоn оf mixture А.

The reaction is:
2RCHO+5MnO₄^-+6H⁺ 2RCOOH+5Mn²⁺+3H₂O.
n_KMnО4=5*0.14=0.7mol. From the reaction, n_RCHO=0.28mol. Number of moles of the aldehydes is twice as the number of moles of the diols. n_diols=0.14mol
If I mark with x  and y the number of moles of glycerol and butane-1,2-diol respectively, then:
x+y=0.14
62x+90y=1.64g
When solved, a negative answer is obtained. Where am I wrong?

[Dan]

If I mark with x  and y the number of moles of glycerol and butane-1,2-diol respectively, then:
x+y=0.14

Not true.

Start by writing separate chemical equations for the reactions of periodate with glycerol and butane-1,2-diol.

[Rutherford]

OH-(CH₂)₂-OH 2HCHO
OH-(CH₂)₄-OH HCHO+C₂H₅CHO
Correct?

[sjb]

OH-(CH₂)₂-OH 2HCHO
OH-(CH₂)₄-OH HCHO+C₂H₅CHO
Correct?

Nope, check the formula for glycerol (and for butane-1,2-diol, though it's OK for the products).

Also, is 2RCHO + 5MnO₄^- + 6H⁺    2RCOOH + 5Mn²⁺ + 3H₂O balanced?

[Rutherford]

 
2RCHO+5MnO₄^-+6H⁺ 2RCOOH+5Mn²⁺+3H₂O.

n_RCHO=0.28mol
CHOH(CH₂OH)₂ 3HCHO
CH₂(OH)-CH(OH)-C₂H₅ HCHO+C₂H₅CHO

3x+2y=0.28
92x+90y=1.64
Again negative answer  .

[Dan]

OH-(CH₂)₂-OH 2HCHO
OH-(CH₂)₄-OH HCHO+C₂H₅CHO
Correct?

I don't think so - the glycerol one is not correct. Glycerol will cleave to give two different products in a 2:1 ratio.

Also, your permanganate equation is too general. Oxidation of formaldehyde with permanganate does not stop at formic acid.

[Rutherford]

What are the two products that are obtained in different ratio?

[Dan]

What are the two products that are obtained in different ratio?

That is a little puzzle for you to ponder!

Hint: 2-hydroxyacetaldehyde reacts with periodate to give two different products in a 1:1 ratio.

[Rutherford]

Are those HCHO and HCOOH?

[Dan]

Yes.

So what do you expect with glycerol?

[Dan]

Actually, I think there is a problem:

which required n_Mn = 0.14 mоl equivalents оf KMnО₄

I am confused by this.

1. Equivalents with respect to what?
2. Pemanganate is 5 electron oxidant, the oxidation of an aldehyde to an acid is a 2 electron oxidation. It follows that you need at least 0.4 mol equivalents of permanganate with respect to an aldehyde to oxidise it. If measured with respect to a 2,3-butanediol you would need 0.8 mol equiv. The numbers are even higher for glycerol.

Maybe I misunderstand the question, but it looks to me as though you will always get a negative solution with <0.8 equiv permanganate. By my calculations, I find that the equiv of permanganate must be between 0.8 and 2 for the question to make sense.

I think that the 0.14 mol equiv is a mistake. Are you sure it's not 1.4?
Did you post the question exactly as you have it in front of you?

I also do not understand why the mass of the mixture is given, as I've not used it.

[Rutherford]

It is the 16th IChO preparatory problem: http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf.
The text has mistakes (words are connected), so the number of moles could be a mistake, too.

[Dan]

Maybe I'm doing this wrong, but I can't see how to get a positive solution from those numbers.

Hopefully someone else will join in.

I think that this:

n_Mn = 0.14 mоl equivalents оf KMnО₄ (1/5 KMnО₄)

May just be an unnecessarily convoluted way of saying "0.028 mol KMnO₄". If that is true, I get a positive answer and I do have to use the mass information given.

I get this from:

equivalents оf KMnО₄ = 1/5 KMnО₄

so

0.14 mol equivalents оf KMnО₄ = 0.14*(1/5) KMnО₄ = 0.028 mol KMnО₄

But anyway:

Start by writing separate equations for the reactions of butane-2,3-diol and glycerol with periodate

[Rutherford]

Should I assume that HCHO is oxidized not only to HCOOH, but to CO₂?
It seems logical, so that way and using 0.028 mol of KMnО₄ I got that the molar share of glycerol is 85.86 %. Got the same?

[Dan]

Should I assume that HCHO is oxidized not only to HCOOH, but to CO₂?

I am working with that assumption - I think it is true.

It seems logical, so that way and using 0.028 mol of KMnО₄ I got that the molar share of glycerol is 85.86 %. Got the same?

I got 67%, but I could be wrong. Can you show your working?