Leah here from leah4sci.com and in this video
we'll continue our discussion of common Organic
Chemistry Mechanism Patterns.
In part 1, we looked at the Nucleophilic attack
and Loss of a leaving group.
If you missed that, you can find the entire
series on my website leah4sci.com/mechanism.
The third common mechanism pattern is the
Proton Transfer.
As the name implies, we have a proton and
we transfer it.
Just think of the game hot potato where the
proton is that potato but everybody is grabbing
it instead of trying to pass it off.
For example in an SN1 reaction that starts
with butanol we can't do a reaction because
we can't get rid of the leaving group.
OH is a terrible leaving group as is but if
we bribe the oxygen and add a proton we'll
turn a bad leaving group into a good one.
How does this happen?
With an acid catalyst that has plenty of protons
floating around.
Regardless of the source we'll represent the
acid catalyst as an H+ in solution.
The proton transfer happens when a greedy
negative or electronegative atom in this case
Oxygen uses a lone pair of electrons to grab
that proton.
We show the arrows starting at those electrons
and ending on the proton that we're grabbing.
As a result, Oxygen still has its initial
green proton, it still has one lone electron
pair but it now has a second bond going to
the black proton that it just grabbed out
of solution where the bond is formed by the
very electron that reached out and did that
attack.
But it doesn't just have to be an oxygen or
a big molecule that does a proton transfer.
Protons are moved around so much in solution
we can even show a proton transfer in solution.
In General Chemistry you learned about the
autoionization of water where you had for
example two molecules of H2O will be in equilibrium
with H3O+ and OH-.
For this mechanism, we see exactly how that
happens.
Let's show the first water molecule in blue
with two lone pairs of electrons, and the
second water molecule purple.
In this case any hydrogen can be the proton
in question because on a water molecule, oxygen
is partially negative and hydrogens are all
partially positive.
Oxygen will use one of its lone pairs of electrons
to reach out towards a partially positive
proton and grab it.
We show the arrow starting at Oxygen and reaching
towards that hydrogen atom.
Because we're not showing a simplified version
of grabbing it out of solution, we need a
second arrow.
Hydrogen only has one S electron.
If it makes a bond it fills that S orbital
but it cannot form two bonds because that
would require four electrons.
When Oxygen is reaching for Hydrogen it's
forming a bond so Hydrogen has to let go of
the initial bond to make sure it only ever
has the one bond.
We show this as a second arrow starting at
the electrons from that existing Hydrogen
bond and collapsing onto the atom that will
ultimately wind up with those electrons.
So this arrow here, that second arrow, shows
that bond breaking away from hydrogen to free
up hydrogen to move on to the blue oxygen
atom.
As a result the blue water molecule is still
Oxygen bound to two hydrogens with one blue
lone pair.
The second lone pair is now represented as
a bond between the oxygen and the purple hydrogen
that it grabbed from the other molecule.
The remaining fragment of the purple water
molecule is now a purple oxygen bound to a
hydrogen.
The two initial lone pairs from water and
a third lone pair from the hydrogen oxygen
bond breaking onto the Oxygen atom.
A quick formal charge will show us a positive
blue oxygen and a negative purple oxygen as
we expected from here with the H3O+ and the
OH-.
Last but not the least, we have the Rearrangement
arrow mechanism.
As the name implies, something is rearranging,
something is moving around.
You'll see this early on when you're studying
carbocation rearrangement, for example a hydride
shift or a methyl shift.
If you're not familiar with this, make sure
to study my tutorial link below.
For example, say we have a secondary carbocation
sitting directly near a tertiary carbon.
After you understand the rearrangement, recognize
this trick.
A secondary carbocation sitting near a tertiary
carbon will undergo a hydride shift.
Why?
Because a tertiary carbocation that forms
when the shift occurs is much more stable
than a secondary carbocation which is what
we have right now.
We show this mechanism by starting the arrow
at the electrons holding hydrogen onto the
carbon.
For rearrangements, I like to circle the atom
just to show what's moving but the mechanism
itself is an arrow starting at those electrons
and going towards the new carbon, in this
case the carbocation where that hydrogen will
now be bound.
As a result we have the same carbon skeleton
but then hydrogen atom is now sitting at the
position of the former carbocation.
The carbon that lost hydrogen is now deficient
and has the formal charge of plus one.
Never draw the arrow as the positive charge
moving because as I explained in the tutorial
below, the carbocation is simply a symptom
of hunger for electrons and when that hydrogen
shifts this carbon now becomes hungry and
this carbon now has the carbocation deficiency.
If we look at a similar structure but this
time the carbocation is near a Quaternary
carbon, there is no hydrogen to move, we're
going to have to move something else.
And the trick here is simple.
A secondary carbocation near a Quaternary
carbon is going to give you an alkyl shift.
The Alkyl shift refers to the alkyl group
or the carbon portion that's moving and you
always go for the smallest one, in this case
that'll be a methyl, sometimes you'll see
an ethyl, sometimes you'll even see a ring
expansion.
Once again for Rearrangement I like to circle
the methyl group but the mechanism itself
is an arrow starting at the electrons that
form the bond between the group you are about
to shift and the quaternary carbon.
And the arrow will point towards the carbocation
because that's where the electrons with the
entire group is moving as a whole.
As a result the methyl group is now attached
to the secondary carbon or the former carbocation.
And the carbon that lost the methyl which
was Quaternary is now tertiary and has a carbocation
has a positive charge to show that deficiency.
Why? because once again, a tertiary carbocation
is more stable than a secondary carbocation.
Now that we've gone through the four patterns,
I want you to practice these a few times and
then join me in part 3 where we look at this
mechanism in detail and identify the pattern
for each mechanism arrow along the way to
get to the product.
You can find the entire series on my website,
leah4sci.com/mechanism.
