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Topic: Problem of the week - 20/02/2012  (Read 8407 times)

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Offline Borek

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Problem of the week - 20/02/2012
« on: February 20, 2012, 10:52:04 AM »
I am a little bit late today, so this is not my original question, but a borrowed one (I will give source next week). Besides, we will venture into a simple organic chemistry for a change.

Achiral organic compound A has a molar mass of 100 g/mol. From other trials we know that the compound is free of geometric isomerism and doesn't contain quaternary carbon. Complete combustion of 100 mg of the compound gives 372 mg of mixture of H2O and CO2. Gentle oxidation of the compound A with potassium dichromate gives compound B with molar mass of 98. Compound A doesn't react with bromine dissolved in carbon tetrachloride.

Give structural formulas of both compounds A and B.
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Offline Dan

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Re: Problem of the week - 20/02/2012
« Reply #1 on: February 20, 2012, 05:40:09 PM »
Nice little problem. I got it, but will leave it open.
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Offline XGen

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Re: Problem of the week - 20/02/2012
« Reply #2 on: February 22, 2012, 05:00:29 PM »
Here is what I have; please tell me if any steps have any error in reasoning.

The molar mass of CO2 is 44 g/mol and the molar mass of H2O is 18 g/mol.

We can then say that the sum of the mass of the products can be represented with the expression of 18x + 44y = 372. Solving for integer solutions, the only integer solutions for the pair (x, y) is (6, 6). Therefore, the products side of the equation has the form 6CO2 + 6 H2O, meaning that Compound A has 6 carbons and 12 hydrogens at the very least.

Two pieces of information verify my next step; in order for Compound A to be oxidized and lose 2 grams from its molar mass to form Compound B, two hydrogens must be lost. This can be done in the oxidation of an alcohol to either a ketone or an aldehyde. Therefore, there must be at least one oxygen.

One oxygen, with a molar mass of 16g, sums with 6 carbons (6 x 12 = 72) and 12 hydrogens (12 x 1 = 12) to 100g.

After determining that the formula of Compound A is C6H12O, the next step is to determine the structure. Assuming that there will be a hydroxide group to create the alcohol, effectively the formula is C6H12 in terms of structure. This means that the structure must be an alkene or a cycloalkane. If it is an achiral compound and does not react to Br2, then it can not be an alkene. Therefore, it must be a cycloalkane. If it is a cycloalkane, then there are a few options for its structure. However, only cyclohexanol fits the bill.

The structural formula for Compound A is (CH2)5CHOH, and the structural formula for Compound B is (CH2)5CO.

Offline Borek

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Re: Problem of the week - 20/02/2012
« Reply #3 on: February 23, 2012, 08:38:38 AM »
The structural formula for Compound A is (CH2)5CHOH, and the structural formula for Compound B is (CH2)5CO.

These are not structural formulas, and it would be better if you would give names of both compounds, but you are right. A is cyclohexanol and B is cyclohexanon.

Question taken from the first stage of the Polish 45th Chemistry Olympiad.
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Offline XGen

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Re: Problem of the week - 20/02/2012
« Reply #4 on: February 23, 2012, 08:50:36 PM »
I apologize, I seem to have misread D:

Offline Yggdrasil

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Re: Problem of the week - 20/02/2012
« Reply #5 on: February 25, 2012, 04:55:47 PM »
Couldn't cyclopentylmethanol/cyclopentylmethanal also work?  Since you specify gentle permanganate oxidation, that could mean that the primary alcohol cyclopentylmethanol is oxidized only to the aldehyde and not to the acid.

Offline Borek

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Re: Problem of the week - 20/02/2012
« Reply #6 on: February 25, 2012, 05:04:52 PM »
For the record: dichromate, not permanganate.

Original answer to the question states that gentle oxidation of the primary alcohol would end with an acid, so it has to be a secondary alcohol. Could be it depends on the meaning of "gentle".
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Offline XGen

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Re: Problem of the week - 20/02/2012
« Reply #7 on: February 25, 2012, 10:58:28 PM »
Original answer to the question states that gentle oxidation of the primary alcohol would end with an acid, so it has to be a secondary alcohol. Could be it depends on the meaning of "gentle".

With which piece of information can you conclude that gentle oxidation would end up with an acid?

Also, what is the use of bromine dissolved in carbon tetrachloride?

Offline Borek

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Re: Problem of the week - 20/02/2012
« Reply #8 on: February 26, 2012, 04:29:36 AM »
With which piece of information can you conclude that gentle oxidation would end up with an acid?

As I mentioned earlier - it may depend on the meaning of "gentle". My organic chemistry is pretty rusty, so I would prefer someone else to chime in, but as far as I remember oxidation of aldehyde to acid is not harder than the oxidation of the alcohol to aldehyde, so if the first step is observed, second would most likely follow in the same conditions. Secondary alcohol would not get oxidized further than to ketone - unless we are talking about brute force oxidation that ends just with water and carbon dioxide.

Quote
Also, what is the use of bromine dissolved in carbon tetrachloride?

Double bonds.
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