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In the last video, we saw how the conjugate base
can be stabilized by induction.
And we also went a little bit into orbitals,
which will be the focus of this video.
So let's review the concept that we
talked about in the last video, where
the different hybridization states of carbon
have different electronegativities, which
can affect the stability of the conjugate base.
And so we talked about how for these three carboxylic acids,
as we move to the right, we get an increase
in the acidity of these molecules.
And so the propynoic acid molecule
on the right, this one, is the most acidic out of these three.
And we said that had to do with the different hybridization
states of the carbons in the molecule.
So, for example, these two carbons are sp3 hybridized.
These two carbons in the double bond are sp2 hybridized.
And in this triple bond here, both carbons are sp hybridized.
Sp hybridized carbons are the most electronegative out
of these three.
So even though it's the same element, carbon,
sp3 hybridized carbons are more electronegative than sp2.
Sp2 hybridized carbons are more electronegative
than sp3 hybridized.
And we can think about the effect
this has upon the stability of the conjugate
base when we realize that the more electronegative something
is, the more of an electron withdrawing effect
it will have.
And so if I think about donating the proton
for the propynoic acid molecule, leaving the negative charge
on this oxygen here-- so that's a negative charge.
Let me just go ahead and fix that really fast.
So we have a negative 1 formal charge on this oxygen.
That negative 1 formal charge is delocalized by resonance
over this portion of the molecule.
And since our sp hybridized carbons are relatively
electron withdrawing, we could think
about an inductive effect, where electrons
are pulled to the left here.
And that is going to delocalize this negative charge
even further.
And so this negative charge can be spread out
over this portion of molecule.
The more you delocalize the charge, the more stable
the conjugate base.
The more stable the conjugate base,
the more acidic the molecule is.
And so it all has to do with the different electronegativities.
Let's apply the same concept to the relative acidities
of these three molecules.
And so on the left, we have ethane.
And so ethane, we could say this is the acidic proton on ethane.
It has a PKA of approximately 50.
So ethane is not very acidic at all.
Ethene, or ethylene, let's say this is the acidic proton.
It has a PKA value of 44.
So it's getting a little bit more acidic.
And then finally, ethyne or acetylene,
let's say this is the acidic proton.
The PKA value decreases even more to approximately 25.
And so as you move to the right, once again, we get more acidic.
So we get an increase in the acidity.
And of course, this has to do with
the different hybridization states of these carbons.
So if we look at this carbon right here,
so only single bonds to this carbon.
So this carbon is sp3 hybridized.
If we look at this carbon right here, that's in a double bond.
So it is, therefore, sp2 hybridized.
And then finally, this carbon right here,
the one the acidic proton is attached to, is sp hybridized.
So let's go ahead and draw the conjugate bases
to these molecules, and let's analyze them
in terms of their stability.
So we'll start with the conjugate base to ethane.
And so ethane has six hydrogens, but we're
going to donate to one of those protons.
And so we would be left with five hydrogens and a lone pair
of electrons on this carbon, giving this carbon
a negative 1 formal charge.
So we form a carbanion.
This lone pair of electrons is going
to occupy an sp3 hybrid orbital.
An sp3 hybridized orbital formed from one
s orbital and three p orbitals.
Therefore, it has 25% s character and 75% p character.
Since p orbitals are larger, we could think about this
as being a relatively large orbital.
So I'm going to exaggerate for effect here.
So that lone pair of electrons on the carbon
occupies an sp3 hybridized orbital.
Let's go ahead and draw the conjugate base to ethene.
So I'm going to have my carbons double bonded to each other.
I'm going to have three hydrogens.
And if we lose the proton that I marked in red,
that's going to give a lone pair of electrons on this carbon,
giving this carbon a negative 1 formal charge.
That lone pair of electrons is going
to occupy an sp2 hybridized orbital.
An sp2 hybridized orbital came from one s orbital and two p
orbitals.
Therefore, it has 33% s character
and about approximately 67% p character.
And so since it has more s character,
s orbitals are smaller.
And so we could think about this hybridized orbital
as being a little bit smaller than the one
we drew a minute ago.
So I'll try to show this as being a little bit smaller
than the one on the left here.
And finally, acetylene.
So if we lose the acidic proton on acetylene,
and we draw the conjugate base, we
have the carbons triple bonded to each other.
And then we have a lone pair of electrons on this carbon.
And since that carbon is sp hybridized,
this will be a negative 1 formal charge for our carbanion.
And since this carbon with a negative formal charge
is sp hybridized, that lone pair of electrons
occupies an sp hybridized orbital.
An sp hybridized orbitals form from one
s orbital and one p orbital.
And therefore, the s character has increased to 50%,
so 50% s character.
Once again, s orbitals are smaller.
And so, therefore, this hybridized orbital
is going to be even smaller than the one we just drew.
So I'll see if I can exaggerate that even more, so
an even smaller orbital to occupy
that lone pair of electrons.
Since acetylene has the lowest value for the PKA--
so we look over here.
Acetylene has the lowest value for the PKA.
Acetylene is the most acidic out of these three.
And therefore, the conjugate base to acetylene
must be the most stable.
And so there are many different explanations
for why this conjugate base is the most stable.
I think the simplest is just to think about electronegativity.
So you have sp hybridized carbons
as being more electronegative.
And so if I go back up to the acetylene molecule,
this sp hybridized carbon is more electronegative
than, say, an sp2 hybridized carbon.
And so you could think about the electrons in this bond right
here moving closer to the carbon, so an inductive effect
like that.
And so since it's pulling the electrons more than sp2
or an sp hybridized carbon, it makes sense
that the conjugate base is more stable
than those other examples.
Another way to explain it would be
to think about the electrons in s orbitals
as being closer to the nucleus.
And therefore, electrons in s orbitals
feel the positive charge of the nucleus
more than electrons in p orbitals, which
have a higher probability of being further away.
And therefore, the higher the s character,
the more it can better stabilize the negative charge
of the conjugate base.
And so it's electrons are closer to the nucleus
and feel that positive charge more.
Yet another way to explain this would be using energy.
Since s orbitals are closer to the nucleus than p orbitals,
the electrons in s orbitals are lower in energy.
Therefore, electrons in an sp orbital are lower in energy
than electrons in an sp2 or an sp3 orbital.
Therefore, you can say electrons are the most stable
in an sp orbital.
So again, there are many different ways
to think about this.
For me, electronegativity is good enough.
But there are, of course, more advanced explanations.
The important thing is just to have
a general idea of the trends in acidity
when you look at the different hybridization
states of the carbon that the acidic proton is attached to.