Leah here from leah4sci.com 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 leah4sci.com/redox 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 leah4sci.com/redox