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Topic: Problem of the week - 22/10/2012  (Read 14222 times)

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Offline Borek

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Problem of the week - 22/10/2012
« on: October 22, 2012, 09:58:34 AM »
100 g of 23% w/w solution of an aldehyde dissolved 19 g of the next aldehyde of the homologous series. 2 g of the solution reduces 4.35 g of silver from the solution of the ammoniacal silver nitrate.

Give molecular formulas of both aldehydes.
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Offline Schrödinger

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Re: Problem of the week - 22/10/2012
« Reply #1 on: October 22, 2012, 10:27:52 AM »
sbeznyqrulqr rgunany?
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Offline Borek

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Re: Problem of the week - 22/10/2012
« Reply #2 on: October 22, 2012, 10:40:41 AM »
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Offline Schrödinger

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Re: Problem of the week - 22/10/2012
« Reply #3 on: October 22, 2012, 11:08:16 AM »
rot13. To avoid publicly displaying the answer.

This happens to be my 1000th post! :D \m/
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Offline 123456789

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Re: Problem of the week - 22/10/2012
« Reply #4 on: October 22, 2012, 12:57:37 PM »
100 g of 23% w/w solution of an aldehyde dissolved 19 g of the next aldehyde of the homologous series. 2 g of the solution reduces 4.35 g of silver from the solution of the ammoniacal silver nitrate.

Give molecular formulas of both aldehydes.

by 2g of the solution, do you mean the original 23% solution, or the solution with the  further dissolved alderhyde?

Offline Borek

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Re: Problem of the week - 22/10/2012
« Reply #5 on: October 22, 2012, 01:04:08 PM »
Solution with both aldehydes.
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Offline 123456789

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Re: Problem of the week - 22/10/2012
« Reply #6 on: October 22, 2012, 01:20:42 PM »
(CHO)2 and CH3COCHO?

Offline Borek

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Re: Problem of the week - 22/10/2012
« Reply #7 on: October 29, 2012, 12:31:33 PM »
(CHO)2 and CH3COCHO?

I don't see how these are homologous aldehydes.

sbeznyqrulqr rgunany?

Decoded to "formaldehyde ethanal" - and that's the correct answer.

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Offline Rutherford

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Re: Problem of the week - 22/10/2012
« Reply #8 on: October 30, 2012, 10:08:08 AM »
I hope that it's not a problem that I write my calculations which are wrong, so someone can correct them.

Molar mass of the first aldehyde is x.
Molar mass of the second aldehyde is x+14 (because of -CH2-).
It is obvious that the first aldehyde is in excess, so his number of moles after the aldol reaction will be: n=23/x-19/(x+14).
The number of moles of the aldol is na=19/(x+14).
With silver, both the aldol and the remaining quantity of the first aldehyde react, when summed that is: 23/x.
In 2g of the solution, 4.35/108 mol of Ag reacted, in 119g of the solution 517.65/216 mol of Ag react. As two moles of Ag react per 1 mole of the aldehyde and aldol, 23/x=517.65/432 and x=19g/mol. What am I doing wrong?

Offline Schrödinger

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Re: Problem of the week - 22/10/2012
« Reply #9 on: October 30, 2012, 10:22:52 AM »
There isn't any aldol reaction, really. Just oxidation by Tollen's reagent.
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Offline Rutherford

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Re: Problem of the week - 22/10/2012
« Reply #10 on: October 30, 2012, 10:56:13 AM »
Oh, silly me ::)! The term "dissolved" confused me and I thought that it is a non-neutral solution. Without the aldol reaction, it is easier and I got that x=29.96g/mol which is HCHO. Thanks for the indication.

Offline Schrödinger

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Re: Problem of the week - 22/10/2012
« Reply #11 on: October 30, 2012, 11:04:39 AM »
Even if the conditions were somehow hypothetically aldol-conducive, there is still one flaw in your numericals : Aldol condensation needs an alpha-H right? The question doesn't state anywhere that the aldehydes contain alpha hydrogen. As a matter of fact, one of the answers methanal doesn't have one.
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Offline Rutherford

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Re: Problem of the week - 22/10/2012
« Reply #12 on: October 30, 2012, 11:58:44 AM »
If it was aldol, then it should be obvious that at least one of the aldehydes must have an α-H.
If HCHO is the first aldehyde (doesn't have an α-H), the second aldehyde must have an α-H.

Offline Schrödinger

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Re: Problem of the week - 22/10/2012
« Reply #13 on: October 30, 2012, 01:42:38 PM »
What if the first aldehyde is 2,2-dimethylpropanal? And the next one contains just one CH2 extra. It could be anywhere. Not necessarily alpha; like 2,2-dimethylbutanal.
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Offline Rutherford

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Re: Problem of the week - 22/10/2012
« Reply #14 on: October 30, 2012, 02:11:51 PM »
But only the molecular formula is needed: "Give molecular formulas of both aldehydes." so if I e.g. determine a molecular formula C4H9CHO, it won't mean the aldehyde you proposed, but any other that has α-H.

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