Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on February 22, 2013, 10:40:50 AM
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A weighed amоunt оf mixture of glycerol and butane-1,2-diol (mА=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 nMn = 0.14 mоl equivalents оf KMnО4 (1/5 KMnО4). Determine the mоlar cоmpоsitiоn оf mixture А.
The reaction is:
2RCHO+5MnO4-+6H+ :rarrow: 2RCOOH+5Mn2++3H2O.
nKMnО4=5*0.14=0.7mol. From the reaction, nRCHO=0.28mol. Number of moles of the aldehydes is twice as the number of moles of the diols. ndiols=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?
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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.
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OH-(CH2)2-OH :rarrow: 2HCHO
OH-(CH2)4-OH :rarrow: HCHO+C2H5CHO
Correct?
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OH-(CH2)2-OH :rarrow: 2HCHO
OH-(CH2)4-OH :rarrow: 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+ :rarrow: 2RCOOH + 5Mn2+ + 3H2O balanced?
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:-[
2RCHO+5MnO4-+6H+ :rarrow: 2RCOOH+5Mn2++3H2O.
nRCHO=0.28mol
CHOH(CH2OH)2 :rarrow: 3HCHO
CH2(OH)-CH(OH)-C2H5 :rarrow: HCHO+C2H5CHO
3x+2y=0.28
92x+90y=1.64
Again negative answer ???.
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OH-(CH2)2-OH :rarrow: 2HCHO
OH-(CH2)4-OH :rarrow: 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.
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What are the two products that are obtained in different ratio?
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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.
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Are those HCHO and HCOOH?
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Yes.
So what do you expect with glycerol?
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Actually, I think there is a problem:
which required nMn = 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.
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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.
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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:
nMn = 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
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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?
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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?
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Of course.
xCHOH(CH2OH)2 :rarrow: 2xHCHO+xHCOOH
yCH2(OH)-CH(OH)-C2H5 :rarrow: yHCHO+ yC2H5CHO
Now it is said that, that only aldehyde groups reacted with KMnO4, so HCOOH won't react.
5C2H5CHO+2MnO4-+6H+ :rarrow: 5C2H5COOH+2Mn2++3H2O
5HCHO+4MnO4-+12H+ :rarrow: 5CO2+4Mn2++9H2O
From the 1st reaction, 0.4y of the permanganate reacted and from the 2nd one, 1.6x+0.8y reacted, so it is:
1.6x+1.2y=0.028
92x+90y=1.64
x=1.536·10-2
y=2.53·10-3
Molar share of glycerol is therefore is 85.86%. Did I do this correct?
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My answer of 67% is wrong - I misread the question and thought it was butane-2,3-diol - sorry about that.
When I do the calculations for butane-1,2-diol and assume that formic acid is oxidised to CO2 by permanganate, I get 44% glycerol (mole ratio). The method is the same as yours, but I use 2x + 1.2y= 0.028 because I think the formic acid formed in the periodate cleavage of glycerol will react with permanganate.
If it is true that you can ignore the formic acid from the oxidative cleavage of glycerol and that formaldehyde is oxidised to CO2, then your answer is correct. However, I do not think you can do that - my opinion is below if you are interested:
If you are assuming that formic acid does not react with permanganate, then you should work on the assumption that formaldehyde is oxidised to formic acid (not CO2), because formaldehyde is oxidised to CO2 in two steps via formic acid:
Formaldehyde + oxidant :rarrow: Formic acid (eq 1)
Formic acid + oxidant :rarrow: CO2 (eq 2)
If we can ignore the formic acid produced in the reaction of periodate with glycerol, then we are saying that eq 2 does not happen. Therefore any formaldehyde from the periodate step is oxidised only to the formic acid level based on the first assumption. If we do this, then the equation is 0.8x + 0.8y = 0.028. Solving the simultaneous equations gives a nonsense answer.
I admit that the question does say that the "aldehyde groups were titrated...", but I think this is unrealistic. If permanganate oxidises formaldehyde to CO2, then it follows that any formic acid in the system will also affect the titration.
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Yes, it is unrealistic, but as I see ambiguities are the main part of IChO problems. Your way seems much more appropriate. Thanks for the help.