reaction mechanism of aldehyde with tollens reagent

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Fa123123
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can someone explain how silver is released when an aldehyde reacts with tollens reagent.

Would be great to show the reaction mechanism please
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Charlotte1303
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(Original post by Fa123123)
can someone explain how silver is released when an aldehyde reacts with tollens reagent.

Would be great to show the reaction mechanism please
no mechanism needed. the solution is silver nitrate in ammonia, it is colourless. The Ag is +1 in the solution. when reacting with aldehydes it produces Ag, it is reduced. the Ag produces the silver mirror as it is pure silver
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Kallisto
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(Original post by Fa123123)
can someone explain how silver is released when an aldehyde reacts with tollens reagent.

Would be great to show the reaction mechanism please
I will send you a link to the reaction, here.

Here are some explanations:

1.) The electrons in the aldehyde migrates when a reaction with a silver-ion starts.
2.) The silver-ion attracts one of those migrated electrons in the aldehyde and reacts to silver.
3.) The C-atom in aldehyde is positive charged now. That causes a nucleophilic reaction with negative charged OH-group.
4.) An H-atom is elimated and protonized. This proton is bond to the negative charged O-atom and makes another OH-group.
5.) The C-atom of the aldehyde is negative charged now. It donates an electron to a silver-ion and gets neutralized.
6.) An H-atom on an OH-group is eliminated, gets protonized and reacts with an OH-group to a water molecule.
7.) Another electron migration in carboxylic acid causes an elimination and protonation of an H-atom in COOH-group.
8.) This protonation leads to another water molecule and finally to the carboxylic acid.

The Tollen's test for the aldehyde is finished.
Last edited by Kallisto; 3 months ago
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Fa123123
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(Original post by Kallisto)
I will send you a link to the reaction, here.

Here are some explanations:

1.) The electrons in the aldehyde migrates when a reaction with a silver-ion starts.
2.) The silver-ion attracts one of those migrated electrons in the aldehyde and reacts to silver.
3.) The C-atom in aldehyde is positive charged now. That causes a nucleophilic reaction with negative charged OH-group.
4.) An H-atom is elimated and protonized. This proton is bond to the negative charged O-atom and makes another OH-group.
5.) The C-atom of the aldehyde is negative charged now. It donates an electron to a silver-ion and gets neutralized.
6.) An H-atom on an OH-group is eliminated, gets protonized and reacts with an OH-group to a water molecule.
7.) Another electron migration in carboxylic acid causes an elimination and protonation of an H-atom in COOH-group.
8.) This protonation leads to another water molecule and finally to the carboxylic acid.

The Tollen's test for the aldehyde is finished.
thank you!, could I say these are the steps for figuring out if the organic compound is an aldehyde or would this just be mechanism
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Kallisto
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(Original post by Fa123123)
thank you!, could I say these are the steps for figuring out if the organic compound is an aldehyde or would this just be mechanism
With no aldehyde in this tollen's test, this reaction and thus the whole mechanism around would not work, so both.
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