Tollens Reagent Silver Mirror Test for Aldehydes

Tollens Reagent Silver Mirror Test for Aldehydes

Leah here from and in this video
we’ll look at the Tollens test or the silver mirror test for aldehydes including the reaction
and mechanism, compared to general chemistry, organic chemistry lab can be a little bit
boring because everything is typically colorless but if the reaction can create a visual change
we can use it to our advantage. In the case of the Tollens test or the silver
mirror test, we use the color change to test for the presence of aldehydes. In the previous redox videos which you can
find at we look at what happens to a molecule that is oxidized or
reduced, for example, if I have a secondary alcohol, we say that it’s oxidized if KMn04
turns it into a ketone. If I react the carboxylic acid with lithium
aluminum hydride, we say it’s reduced when it’s turned into a primary alcohol. Notice their focus is specifically on what
happens to the organic molecule where we tend to ignore the reagent itself now this is typically
okay in organic chemistry especially because your professors are just looking for the product
but don’t forget that redox reactions aren’t just reduction or oxidation, their antagonistic
reactions where when something get oxidized something else has to get reduced in that
same reaction. If oxidation is the laws of electrons and
reduction is the gain of electrons, where did the electrons come from and where did
they go? Well some of the reagent and solutions that
we don’t care about unless there is a color change when that reagent itself changes and
that’s what happens in the Tollens test. The Tollens reagent is a diamine silver one
complex in the presence of a base which you’ll typically see as OH- and this is prepared
in a tricky multistep process that will leave for you lab professor. What we care about is the reaction. If I react an aldehyde with a Tollens reagent
typically written as Ag(CNH3)2OH Followed by an Acid workup, the product is going to
be a carboxylic acid. Not exciting but the side product is silver
solid which will coat the inside of your beaker giving you a shiny silver mirror something
that’s very very visible to the naked eye and that is why it’s called the silver mirror
test. So how does this work? If you look on the periodic table, you’ll
see that silver or Ag is a transition metal, this means it can exist as multiple oxidation
states, we have silver solid with an oxidation number of 0, we can also have +1, +2, and
+3. What we care about is the +1 to 0 transformation. Silver and the Tollens reagent has an oxidation
number of +1, notice that ammonia is neutral which means the positive charge comes from
the Ag1 and when the aldehyde is oxidized, the silver is reduced in the following half
reaction. We have AG+ grabs one electron to give us
Ag0, Ag+ is Aqueous, it dissolves in solution and therefore colorless, Ag0 is a solid precipitate
in a high enough concentration will be visible as a shiny metallic silver. So what is the Tollens test? We’re checking to see if we have an aldehyde
or not. So you’re running a series of experiments
for your unknown and you know that you have a carbonyl present but you don’t know what
is the form of the carbonyl. Is it a carboxylic acid, is it an aldehyde
or is it a ketone? If I react a carboxylic acid with the Tollens
reagent I will get no reaction. for carboxylic acid is fully oxidized. If I react an aldehyde with a Tollens reagent
I will get silver solid as one of the products which is seen as a shiny silver mirror confirming
the presence of the aldehyde and if I react the ketone with the Tollens reagent once again,
I will get no reaction because the ketone cannot be oxidized further under regular conditions. I point this out to say it’s not the presence
of the carboxylic acid that gives us a positive test, rather the formation of the carboxylic
acid which in the process forms a solid silver that gives us the positive test. Let’s take a look at the mechanism and see
how this happens, Keeping in mind that most students do not have to know this mechanism
in under graduate chemistry so ask your professor before you take the time to memorize it. This is a tricky mechanism involving radicals
so make sure you pay close attention to the atoms and electrons specifically where each
color starts and ends so that you can make sense of what’s going on. Before starting the reaction, I want to remind
you that carbonyls have resonance where the pi bond can collapse up to the oxygen giving
us a less stable structure that has 3 lone pairs on the oxygen with a negative on the
oxygen and a positive on the carbon. The reaction starts out when one of the electrons
from oxygens lone pair is donated to the positive silver. Notice I drew this with a single headed or
fish hook arrow because radicals imply the movement of a single electron rather than
the typical electron pair that you’re used to seeing. This gives me an oxygen that now has 2 green
lone pairs, 1 lone red electron as a radical and still a carbocation on the carbon. As a side product, we have Ag which is now
neutral and therefore is a solid that precipitates out of solution since this takes place in
base, the OH- is attracted to the positive carbon and attacks, forming a bond between
oxygen and carbon. This yields the molecule that has the purple
OH bound to the former carbonyl carbon, the green hydrogen as well as the green oxygen
which still have 2 lone pairs and one lone radical. For every step you want to pay attention to
the sum of the charges because if the charge is off you can catch if you miss something. In this step, we have OH- and C+ which is
a net 0, this molecule is neutral which is also net 0 but it has a radical and that makes
it unstable which makes it very reactive. We’ll show the electrons binding carbon to
hydrogen in blue because for the next step, the hydrogen moves up to the oxygen but since
oxygen already has one electron we just need one of the hydrogen’s electrons to form
that bond, leaving that second electron to remain behind on the carbon as a radical. This gives me a carbon that now has the purple
OH bound, the green oxygen with it’s 2 green electron pairs and the bond to hydrogen which
is made up of 1 red electron from our previous radical and 1 blue electron that the hydrogen
brought with it, we also have another blue electron as a radical sitting on a carbon
atom. That radical is unstable and we have plenty
of Ag floating around in solution and so the carbon will donate it’s electron to the Ag+
which causes yet another Ag solid to precipitate out of solution. We also have a molecule that has a purple
OH, the green and blue OH and now a carbon with a carbocation because it doesn’t have
any electrons other than the 3 bonds. If you think back to your carbonyl chapter
you’ll remember that a carbon bound to 2 oxygens is not very stable especially when both of
the oxygens are greedy and want to donate their electrons to the carbocation. If the electrons from oxygen are pulled towards
the carbocation, the hydrogen that’s bound becomes very partially positive as the electrons
are pulled away from it. OH- and solution is going to take advantage
of this fact by reaching out for and grabbing one of the hydrogen atoms, collapsing the
electrons back on to the oxygen. This gives me a temporary intermediate that
has 1 OH and another O- sitting on a carbocation which will immediately collapse down as a
resonant structure to give us a carbon that is now double bound to the purple oxygen and
single bound to the green and blue OH. If you look at this molecule it’s just a twisted
carboxylic acid so lets redraw it. Same 2 carbon chain, same purple double bound
oxygen, same blue and green single bound oxygen, at this point we want to say that we’re done,
we have a carboxylic acid for our product except for one little detail and that is the
fact that we’re in a basic solution. A basic solution which is swimming with OH-
is no going to tolerate an acidic proton and as soon as that acidic proton show us it will
dissociate which will show as an OH- grabbing that proton giving oxygen back it’s electrons
because acids do not play nicely in basic solutions, they get neutralized as quickly
as possible. This gives me a carboxylate anion with a negative
charge which we neutralized by adding H3O+ into solution. When we wrote out the reagent we showed acid
as step 2 because if we put the acid in on step 1 it would neutralize the OH- and then
no reaction would happen but once the reaction is done and we have the carboxylate anion,
adding the H+ simply allows us to have oxygen, grab a hydrogen, reforming the carboxylic
acid and giving us our final product, a product that honestly we don’t care about. What we do care about are the 2 silver molecules
that precipitated out of solution keeping in mind that in your beaker you have thousands
and thousands of molecules so that the volume of silver coming out is enough to coat the
inside of your beaker. You’ll see this reaction again in biochem
when you are doing a test for sugars that have an aldehyde group for example, glucose. Cyclic glucose despite being stable as is,
exists in equilibrium with a linear form and notice the linear form has an aldehyde. If I react this with a Tollens reagent which
we’ll just show as the 2Ag+ and OH- followed by an acid workup, our product once again
will be a carboxylic acid as well as 2 solid silver molecules precipitating out of solution. You can find this entire video series along
with my redox practice quiz and cheatsheet by visiting my website


  1. Hey leah can you please make a video to explain the different reactions that take place in presence of KMnO4. There's acidic, basic, oxidative.. kinda confusing!

  2. Mam, please make a vedio about how we can selectively reduce or oxidise something among several functional groups

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