Leah here from Leah4Sci.com.
And in this video, we'll continue the SN1 reaction by analyzing a couple of mechanisms.
We'll start by analyzing the reaction of 1-iodo-1-methylxcyclohexane reacting with an acid catalyst in a methanol
solution.
For all substitution and elimination reactions, you want to analyze four things that will
help you determine which reaction takes place.
And the four things we'll look at are the alkyl chain, leaving group, attacking nucleophile
or base, and solvent.
I have a video discussing each of these topics which you can find on my website, Leah4Sci.com/Substitution-Elimination.
We'll keep this list on the side so that we can refer to it giving us room to work through
the reaction.
Starting with the alkyl chain, we're specifically looking at the carbon that's holding the leaving
group.
In this case, we have a tertiary carbon which can form a stable carbocation, something we
want in an SN1 reaction.
This means that I can have my one-type, meaning an SN1 or E1 reaction.
So we'll write tertiary carbocation one-type.
The next thing we look at is iodine.
Iodine is a very large halogen and when it leaves and forms a negative charge in solution,
we'll be able to distribute that charge over the large molecule making it a good leaving
group.
This doesn't necessarily tell me between a one or two-type reaction but rather tells
me that the reaction will be relatively quick compared to a reaction that has a bad leaving
group or leaving group that doesn't want to leave at all.
Next, we look at the attacking nucleophile or base.
But in this case, we don't appear to have one.
All we have is an acid catalyst dissolved in a methanol solution.
This is a case of solvolysis which means the solution itself is going to be your attacking
molecule.
So let's analyze methanol as a nucleophile given that we're only looking at the substitution
reaction.
Methanol dissolved in an acidic solution cannot form methoxide which is a negative oxygen.
Methanol as a nucleophile has lone pairs of electrons on an electronegative making it
weak nucleophile not capable of kicking out a leaving group but it is capable of attacking
any positive charge that would form in solution.
And lastly, we'll look at methanol as a solvent.
Since it has a hydrogen atom bound to oxygen, it's a polar protic solvent capable of stabilizing
any charges that form in solution.
Once again, this would stabilize an SN1 reaction.
So let's begin.
We have the carbocation formation step.
When iodine takes the electrons that bind the two carbons and breaks away in solution,
this gives me a positive charge on the tertiary carbon and a negative charge on the iodide
in solution.
When you do a quick charge analysis, we start with a neutral molecule.
In the intermediate step, we have a positive and a negative that cancels out.
That means that charges all accounted for.
We won't focus on the iodine anymore keeping in mind that it will be dissolved in methanol
and just off somewhere in solution.
Since this is a solvolysis reaction, one of the solvent molecules will reach out and attack
the carbocation with a lone pair of electrons on oxygen forming a bond between the oxygen
and the carbon.
Since oxygen used the lone pair to create that bond, it only has one lone pair remaining
and a formal charge of +1.
To get rid of that charge, we bring another methanol molecule from solution to reach out
with its oxygen atom, removed the hydrogen from our molecule and give oxygen back its
lone pair of electrons.
For our final product, we have an OCH3 attached to the carbon atom that used to hold the iodine.
We also have a protonated methanol which is the regeneration of our acid catalyst.
It appears that the final step has a net charge of +1 but don't forget the I- off somewhere
in solution which cancels out that positive charge giving me a conservation of charge
through every step of this reaction.
One thing we did not look at here is the stereochemistry of the starting and ending molecule and that's
because our starting molecule was achiral.
Now, let's see what happens when a chiral molecule undergoes an SN1 reaction.
Notice that on the right, we have a chlorine coming forward but don't forget the invisible
hydrogen going back.
The same thing with the methyl on the left, the methyl comes forward but the invisible
hydrogen goes to the back.
That means our starting molecule has two chiral carbons.
Before we start the reaction, let's quickly analyze the checklist to verify that indeed
we can have an SN1 reaction take place.
We'll start with the alkyl chain.
We're specifically focusing on the carbon that's holding the chlorine atom.
Since it's a secondary carbon, it will form a stable carbocation but not as stable as
the tertiary so we can potentially have competition between a one-type or a two-type reaction,
meaning SN2, E2 or SN1 and E1.
We'll write 1 and 2 to show that they're both possible.
The leaving group chlorine forms a stable enough negative ion in solution and so is
considered a decent leaving group.
Next, we analyze the attacking nucleophile or base but once again, we're dealing with
a solvolysis reaction where the solvent is the nucleophile.
Again, because we're focusing on substitution rather than elimination in this reaction.
H2O is a weak nucleophile and can only undergo a one-type reaction, meaning SN1 or E1.
And finally, H2O is a polar protic solvent which will happily stabilize any charges that
forms as intermediates of the reaction.
Let's go back to the initial comment where we mentioned that you can have a competition
between a one and two-type reaction.
Given that the leaving group is on the secondary carbon, we do have that competition.
However, the presence of an acid catalyst or H+ in solution tells me that we cannot
form any strong negative charges.
So even though we can have a two-type reaction with a secondary leaving group, water isn't
strong enough to undergo a two-type reaction, meaning it's not strong enough to kick out
chlorine.
And instead, because it's so weak, it will have to wait for chlorine to leave, wait for
their positive carbocation to form and then jump on the opportunity of that positive charge
and attack because it's a weak nucleophile.
And this is why you have to analyze all four components before you determine your reaction
because some may be ambiguous but when analyze together, you'll get a much clearer picture
of the reaction about to happen.
We got rid of the invisible hydrogens to make it easier to see and the reaction starts out
when chlorine grabs the electrons connecting it to carbon and breaks off.
In addition to the Cl- forming in solution, we also have a positive charge on the carbon
that used to hold the chlorine.
The starting carbon was sp3 hybridized and that is how we were able to have a chiral
carbon.
However, this is where you have to be careful.
When the chlorine broke away leaving a positive charge in its place, the hybridization changed
from sp3 to sp2.
While sp3 has a tetrahedral geometry, the sp2 carbon is going to be trigonal planar
or simply flat.
Given that the sp2 carbon is flat, we lost chirality giving me an achiral intermediate
for the carbon that underwent the SN1 reaction.
Since is carbon is now flat, for the next step of this reaction, water can attack the
carbon from the top or it can attack the carbon from the bottom potentially giving me two
different stereochemical products.
I drew arrows in different colors so that we can follow the progress of each one separately.
We'll start with the blue arrow that shows the water molecule attacking from above.
The water is now coming out of the page on the product but it does have a lone pair and
a positive charge which means we have to bring in another water molecule to use its electrons
and remove that hydrogen, giving oxygen back its electrons.
As you can see, the final product has 2 chiral carbons where the OH and the chlorine appear
to have to have the same stereochemistry.
However, if we look at the second attack that happened where the water molecule attacked
from below, we show the OH2 having added with dashes rather than wedges giving me the exact
opposite chirality of our initial product.
We don't have room to show the next step but know that a water molecule comes in, grabs
that hydrogen, so we'll show that being removed, and gives oxygen another lone pair of electrons.
And these are the two products of my reaction.
Notice that the methyl group has not changed its chirality because it did not take part
in this reaction.
However, the OH group while chiral is forward on one molecule and back on the other.
Since we have two chiral groups where one stayed the same and the other changed, the
two products are considered to be diastereomers of each other.
But if we want to understand what happened in this reaction, you can say that the product
underwent racemization, meaning that we have a racemic product that is 50% R and 50% S
for the carbon that underwent the SN1 reaction.
And this happened because the chirality is lost in the sp2 carbocation intermediate and
that means, if you start chiral and end chiral but undergo and achiral intermediate, you're
going to lose the specificity.
Meaning, you're going to lose any pure R or pure S that you started with and instead wind
up with a 50/50 mix.
Be sure to join me in SN1 part 3 where we analyze one final reaction that involves a
hydride shift and a carbocation rearrangement.
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