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Topic: Problem of the week - 13/05/2013  (Read 25617 times)

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Borek

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Problem of the week - 13/05/2013
« on: May 13, 2013, 05:01:04 AM »
Organic compounds X (molar mass 26 g/mol) and Y (60 g/mol) react in the presence of catalyst, yielding compound Z (86 g/mol). Z decolorizes bromine water, but it doesn't react with ammoniacal solution of disilver oxide. Left in acidic solution Z undergoes hydrolysis, and products of the reaction don't react with bromine water, but they do react with the ammoniacal solution of silver dioxide, producing silver mirror on the test tube surface (plus ammonium acetate). What are compounds X, Y and Z, and what are the reaction equations? Please use SMILES for answers.
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iamback

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Re: Problem of the week - 13/05/2013
« Reply #1 on: May 13, 2013, 07:46:04 AM »
Please also mention the level of question. Like I am a high school student, I am not able to solve it, even though it looked like I would be able to solve it. I spent half and hour on it, and  was only able to get some clues.

I have studied everything which the question mentions, but I think it's difficult.

Any hint.

Borek

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Re: Problem of the week - 13/05/2013
« Reply #2 on: May 13, 2013, 08:12:46 AM »
Definitely HS, can be difficult if you have not taken any organic chemistry yet.
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kriggy

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Re: Problem of the week - 13/05/2013
« Reply #3 on: May 13, 2013, 09:41:28 AM »
Man this one is good. I think I have it, so lets see:
compound X -
compound Y -
which together in catalyzed by acid
gives compound Z -

Which undergoes adition of Br2 (aq) on double bond

but esters arent reduced by Tollen´s reagent.
By hydrolysis we got acetic acid

and ethene-1-ol
which is unstable and by keto-enol tautomery isomerizes to acetaldehyde
which reduces Tollens reagent but doesnt react with bromine water.
__
btw HS? And I thought we had good chemistry teacher at grammar school. I couldnt solve it with only my HS chemistry.
« Last Edit: May 13, 2013, 10:54:22 AM by kriggy »

Borek

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Re: Problem of the week - 13/05/2013
« Reply #4 on: May 13, 2013, 09:52:46 AM »
HBr?
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kriggy

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Re: Problem of the week - 13/05/2013
« Reply #5 on: May 13, 2013, 10:10:01 AM »
HBr?
Bromine water?
I might live in ilusion that bromine water is highly diluted HBr..

edid: I used some google-fu and hopefuly corrected my mistake

opsomath

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Re: Problem of the week - 13/05/2013
« Reply #6 on: May 13, 2013, 10:49:19 AM »
Quote
but ketones arent reduced by Tollen´s reagent.

Kriggy, your answer is good, but that's an ester not a ketone.

Borek, I would point out only that IPA and n-propanol also weigh 60, and could conceivably be induced to add to ethylene with an appropriate catalyst.

kriggy

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Re: Problem of the week - 13/05/2013
« Reply #7 on: May 13, 2013, 10:53:57 AM »
Quote
but ketones arent reduced by Tollen´s reagent.

Kriggy, your answer is good, but that's an ester not a ketone.

Borek, I would point out only that IPA and n-propanol also weigh 60, and could conceivably be induced to add to ethylene with an appropriate catalyst.

youre right.. I missed that one too. thx
btw what would you get from n-propanol and IPA? Hept-1-en and 2-methylbutane?

Borek

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Re: Problem of the week - 13/05/2013
« Reply #8 on: May 20, 2013, 02:06:14 PM »
Kriggy got it - whole question is built around vinyl acetate.

Borek, I would point out only that IPA and n-propanol also weigh 60, and could conceivably be induced to add to ethylene with an appropriate catalyst.

Nitpicker
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kriggy

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Re: Problem of the week - 13/05/2013
« Reply #9 on: May 20, 2013, 04:49:28 PM »
BTW do you know mechanism of the esterification? Is this one correct?

I cant find anything wrong in it but seems weird..

opsomath

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Re: Problem of the week - 13/05/2013
« Reply #10 on: May 20, 2013, 05:10:55 PM »
Ok, your curved arrows are weird. Are you showing the triple bond electron pair going to the carbon on the right, like the arrows suggest?

Remember, curved arrows always start AT an electron pair and go TO an electron acceptor. Your second curved arrow is going from the positively charged carbon to the negatively charged oxygen, which is Very Bad. The correct way to draw this would have the arrow going from a lone pair on the oxygen, to the carbon. (Always draw electron pairs in when you're showing a bond forming or breaking, it helps keep you from screwing up.)

Briefly, the alkyne donates an electron pair to the acetic acid, pulling a proton off it. This gives a carbocation and a acetate anion. The anion is a nucleophile, the carbocation is an electrophile, so the anion donates an electron pair to the empty orbital on the carbocation and makes vinyl acetate.

kriggy

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Re: Problem of the week - 13/05/2013
« Reply #11 on: May 20, 2013, 06:05:04 PM »
Ok, your curved arrows are weird. Are you showing the triple bond electron pair going to the carbon on the right, like the arrows suggest?

Remember, curved arrows always start AT an electron pair and go TO an electron acceptor. Your second curved arrow is going from the positively charged carbon to the negatively charged oxygen, which is Very Bad. The correct way to draw this would have the arrow going from a lone pair on the oxygen, to the carbon. (Always draw electron pairs in when you're showing a bond forming or breaking, it helps keep you from screwing up.)

Briefly, the alkyne donates an electron pair to the acetic acid, pulling a proton off it. This gives a carbocation and a acetate anion. The anion is a nucleophile, the carbocation is an electrophile, so the anion donates an electron pair to the empty orbital on the carbocation and makes vinyl acetate.
didnt know about the arrows. thanks for explaining