Leah here from Leah4Sci.com and in this video, I will show you how to analyze the alkyl chain
when determining between SN1, SN2, E1, and E2 reactions.
The first thing to analyze when determining between substitution and elimination reactions
is the degree of substitution of the carbon holding the leaving group.
And there are five types of carbons that you will see.
There is the methyl carbon, primary, secondary, tertiary, and quaternary.
A methyl carbon occurs when you have a methyl group which is CH3 and that's bound to a leaving
group which I will show as Lg in my videos.
For rest of the carbons, to determine their substitution, simply count the number of carbons
attached to them.
So to do a quick review, on this molecule here, we'll start by identifying primary carbons.
A primary carbon just has one other carbon attached to it.
That means every terminal carbon is going to be primary because they're only attached
to one other carbon.
If you have a leaving group attached to the primary carbon, that is considered a primary
leaving group.
A secondary carbon is a carbon that has two other groups attached to it.
For example, this carbon here is attached to one carbon on the left and one carbon on
the right.
So it's this one, these two, this one.
If we place a leaving group on a secondary carbon, this is considered a secondary leaving
group.
A tertiary carbon is a carbon that is attached to three other carbons.
And in this molecule, we have two tertiary carbons.
If I place a leaving group on the tertiary carbon, it's considered a tertiary leaving
group.
A quaternary carbon is a carbon that has four other carbons attached to it.
According to the octet rule, a carbon can only have 8 valence electrons and therefore
4 bonds.
So quaternary carbon cannot hold a leaving group however you will see it participating
in reactions if there's a potential for a methyl shift or carbocation rearrangement.
So how does this apply to reactions?
I like to start my analysis of the one-type reactions and this refers to SN1 and E1 which
both undergo a carbocation intermediate.
In order to analyze the carbocation intermediate, we have to talk about the stability.
A carbocation is a carbon that has only three groups attached to it that leaves a deficient
in its valence shell with a formal charge of +1.
While a carbocation in general is slow to form, you have different levels of stability
which will directly impact the rate of reaction.
You can refer to my carbocation video for a detailed analysis of its stability.
But as a quick review, a methyl carbocation is simply a carbon bound to 3 hydrogens with
a positive charge.
This is very unstable and will not form.
The next one you have is a primary carbocation and this is a terminal carbon that has a positive
charge.
This is also unstable and once again, will not form.
A secondary carbocation is a secondary carbon with a positive charge.
This one is considered moderately stable and will form.
And finally, a tertiary carbocation is a carbocation with 3 R groups attached.
This is the most stable of your standard carbocations and therefore the quickest to form.
But don't forget you pi systems including your allylic and benzylic carbocations.
If you have a molecule that has a positive charge with a double bond next to it, the
pi bond can resonate towards the positive charge allowing the carbons to share that
burden of charge and therefore making it more stable.
An allylic carbocation is going to be more stable than a tertiary carbocation and this
can occur as a primary allylic.
You can have secondary allylic carbocation.
You can even have a tertiary allylic carbocation.
In addition to gaining stability from the carbon surrounding that positive charge, you
have the added stability of the resonance allowing multiple carbons to share that burden
of charge.
And finally, you have your benzylic carbocations which are even more stable than the allylic
carbocations given that the positive charge can resonate into the ring allowing multiple
atoms to share that burden of charge.
Here we have a primary benzylic carbocation, here we have a secondary, and here we have
a tertiary benzylic carbocation.
Now remember, primary benzylic, even though it's primary is still a lot more stable than
a regular tertiary carbocation because we're not only looking at the primary position but
also the fact that the positive charge can go into the ring.
Once you identify that your molecule has the ability to support a positive charge, you
want to consider the E1 and the SN1 reaction as follows.
If we have an allylic or benzylic system, you know it's going to have a stable carbocation.
If we have a tertiary leaving group, that will be stable too.
And you want to think of the one-type reactions.
It gets a little tricky with a secondary carbocation given that it's somewhat slow to form and
you may have competing reactions between the one-type meaning, SN1, E1, and the two-type
meaning, SN2 and E2.
If you see a primary or a methyl leaving group, you can roll out your SN1 and E1 given that
these will not form stable carbocations and therefore the reaction will not occur.
Now, let's analyze the two-type reactions specifically SN2 and E2.
Unlike SN1 and E1, we can't classify the two together because the way they react or the
molecules that they prefer to attack can be different.
An SN2 reaction involves a nucleophile coming in to attack the carbon specifically holding
the leaving group.
So we're looking for a minimally substitute leaving group.
Therefore, for an SN2 reaction, your ideal molecule is going to have a leaving group
in the methyl position or in the primary position.
However, you can still have a reaction at the secondary position.
But it's not only the leaving group that you want to focus on.
You also want to see what's going on next to this molecule.
Let's compare these two molecules.
They both have the leaving group on a secondary carbon.
However, the molecule on the left is rather small with minimal groups on the side and
that means a nucleophile will have an easier time getting in.
However, the molecule on the right is considered very bulky and therefore sterically hindered.
If we compare the ability of a nucleophile to attack the molecule on the left compare
to attacking the molecule on the right, the bulkiness of the molecule on the right will
slow down that SN2 reaction.
In fact, depending on the situation, the molecule on the right might go for a one-type reaction
where the molecule on the left will go for a two-type reaction.
When dealing with E2, you have to analyze the molecule in a slightly different way.
E2 is beta elimination so we have to analyze not only the position of the leaving group
but also the availability of beta hydrogens.
This molecule has a leaving group on a tertiary carbon and we'll call that the alpha carbon.
However, the carbons next to it are considered the beta carbons and this will provide the
hydrogens that can be removed in the elimination reaction.
Therefore, despite the fact that the leaving group is tertiary which implies that we can
have a carbocation, the fact is we have plenty of beta hydrogens available.
And so, given proper conditions, we can have an E2 reaction.
That means we can have primary, secondary, or tertiary leaving groups experiencing an
E2 reaction.
However, an E2 reaction cannot take place when you have a methyl attached to a leaving
group and that's because even though the leaving group is on what we'll call the alpha carbon,
there is no beta carbon present to provide a beta hydrogen for elimination.
This can also happen with a larger molecule such as what we see here.
The carbon holding the leaving group is the alpha carbon.
However, the beta carbon is quaternary and therefore does not have a hydrogen that can
be removed for the elimination reaction.
Be sure to join me in the next video where I show you how to analyze the similarities
and differences between strong and weak nucleophiles and strong and weak bases for substitution
and elimination reactions.
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