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.