Leah here from Leah4Sci.com.
And in this video, I will show you the SN1 reaction, rate, and mechanism.
Before we go into the mechanism, what is SN1?
SN1 stands for Substitution Nucleophilic Unimolecular.
Substitution is when I have a molecule that has a substituent x and I simply substitute
it for substituent y.
Nucleophilic comes from the word nucleophile which breaks down into nucleus and loving.
In other words, a nucleophile is a molecule that is attracted to positive charge and since
the nucleus is positive, we can say that a nucleophile is positive-seeking.
And what type of molecule is attracted to positive?
It will have to be something that is negative or partially negative due to electronegativity
and lone pairs of electrons.
And finally, unimolecular means that the reaction takes place one molecule at a time or a first
order reaction.
Many students confuse this to think SN1 is one step.
But an SN1 reaction can actually be two or three steps so don't mix this up.
So let's take a look at the SN1 reaction.
If I have a molecule with a leaving group on the secondary or tertiary carbon, the leaving
group can grab the electrons that attach it to the carbon breaking away.
This leaves you with a carbocation or a carbon with a positive charge and a leaving group
in solution.
A weak nucleophile will be attracted to this positive charge and use its lone pair of electrons
to attack attaching itself to the carbon chain giving me a final product that has the nucleophile
attached where the leaving group used to be.
And this is where we get the idea of the substitution reaction.
Let's analyze this a bit further.
The first step of this reaction where the carbocation forms is a very slow step because
you're taking a carbon chain that starts out neutral and giving it a positive charge.
Now, even though carbon is not highly electronegative and can form a stable carbocation, we're still
talking about going from neutral to charge and that's not very happy.
Because the intermediate is not very happy, this will be a very slow step, so slow in
fact that this is the rate determining step of the reaction.
As soon as that positive charge forms, the nucleophile comes in to attack and that is
considered a very fast step of the reaction.
Now, let's discuss this in terms of kinetics.
The rate of any reaction is dependent on some constant K, which you don't have to worry
about in organic chemistry, times the concentration of any molecules involved in the rate determining
step.
Since the alkyl chain that forms the carbocation is the rate determining step, well, say that
the rate is simply K times the concentration of the alkyl chain.
Since the rate is only dependent on one molecule, this is considered a first order reaction
which gives you the 1 in unimolecular.
So what is a first order reaction?
It's a reaction whose rate will change dependent on the concentration of the alkyl chain.
This means, if you double the concentration of your starting carbon molecule, the reaction
rate will double.
If you triple it, the reaction rate will triple.
However, because the nucleophile is not part of that rate determining step, if you double
or triple the concentration of the nucleophile, you will see no change in the speed of your
reaction.
Now, let's analyze the SN1 reaction that happens when 2-bromo-2-methylpropane reacts with acid
in the presence of water.
There are four things you want to look at any time you study a reaction to determine
between substitution and elimination or the 1 in two-type reactions.
I have detailed videos that go into every analysis step which you can find in my website,
Leah4Sci.com/Substitution-Elimination.
The first thing to analyze is the alkyl chain.
Recall that we said, in order to have a one-type, in this case, an SN1 reaction, we have to
have the leaving group sitting on a carbon that is capable of holding a positive charge.
Tertiary and secondary carbons make stable carbocations.
And in this case, we have a tertiary carbon holding the leaving group.
Meaning, we will have a stable carbocation.
Bromine as a leaving group forms a large negative halogen in solution.
And due to its size, it's able to distribute that negative charge making it relatively
stable and therefore willing to leave.
Since we're looking at substitution reactions, we're not looking at the base as much as the
nucleophile.
But in this case, we don't appear to have one.
All we see is H+ in solution but H+ is merely a catalyst.
What we do see however is a solvent that has an electronegative atom capable of interacting
with this molecule resulting in a solvolysis reaction.
Solvolysis is a term that explains a solvent that also acts as the nucleophile or base
in solution.
Meaning, you're reacting the molecule directly with your solvent.
H2O as an attacking molecule does have an electronegative atom with lone pairs of electrons.
Because we're in a positive or acidic solution, we cannot form a negative oxygen because that
would be too strong of a base or nucleophile.
However, given that we're in a positive acidic solution, a neutral electronegative oxygen
while a weak nucleophile is still considered strong enough to attack.
And finally, we have our solvent.
Water is considered a very polar protic molecule, meaning that it has polarity that can stable
any intermediates that form in this reaction.
When the carbocation forms, the partially negative oxygen in water will stabilize that
positive charge and when bromine leaves and gets a negative charge in solution, that negative
will be surrounded by the partially positive hydrogen on the water molecule.
According to these four factors, everything here does point to a stable SN1 reaction.
So let's see what the mechanism looks like.
This reaction starts out when bromine grabs the electrons that connected to the carbon
chain and breaks away from the carbon atom.
A carbon used to have 4 bonds and now has only 3, leaving it deficient and giving it
a formal charge of +1.
Bromine now has 8 lone pairs of electrons giving it a formal charge of -1.
When working through the SN1 and E1 reactions, be sure that you always pay attention to your
charges and insure that you don't lose or gain anything along the way.
The first step in my reaction has a net charge of zero because it's a neutral molecule.
The next or intermediate step of the reaction has a positive charge on the carbon, a negative
charge on bromine.
Because positive and negative cancel out, once again, we still have a net charge of
zero.
Notice that bromine sits comfortably in solution surrounded by the partially positive hydrogens
of the water molecule.
I'm not going to show the oxygen surrounding the carbocations but do understand that it
happens.
Instead, we'll jump to the next step of this reaction where a water molecule will use its
lone pair of electrons to attack the positive carbon forming a bond between carbon and oxygen.
Since oxygen donated one of its lone pairs to the bond, it has just one lone pair left
giving it a formal charge of +1.
This is where you want to pay attention.
Even though oxygen is the atom holding the positive charge, you will never have another
oxygen come and attack that positive oxygen because the oxygen is still an electronegative
atom.
Instead, what you're going to see is the electronegative oxygen pulling some of the electron density
from the oxygen-hydrogen bond towards itself making the oxygen slightly less positive but
at the same time, making the hydrogen atoms partially positive.
Another water molecule that's in solution will use its lone pairs to grab one of the
hydrogen atoms off of our molecule giving the bonding electrons to the oxygen atom.
The final product will have an oxygen attached to the carbon chain with two lone pairs of
electrons and a neutral charge.
We can also show the side product of the hydronium with a positive charge as well as the Br-.
Once again, if we take in that charge, the molecule is neutral and therefore has a charge
of zero.
The hydronium is our asset catalyst regenerated with a positive charge and the bromide in
solution has a negative charge.
We have a conservation of charge because the hydronium and bromide cancel each other out.
A common question I receive from students is where does that second water molecule come
from?
When you're given a reaction, don't limit yourself to just one starting molecule and
just one solvent molecule.
Think about your most recent organic chemistry lab.
When you do a reaction at lab, you take an entire beaker full of your chemical to react.
That means you have thousands and thousands of molecules available for reaction and so
we're reacting this on a large scale.
When you dissolve in water, we're talking about a virtually infinite number of water
molecules constantly surrounding your reactant and waiting to attack.
So when you work through the mechanism and you need to find another water molecule to
react, remember that you have an infinite number surrounding your molecule waiting to
attack.
So go ahead and use as many as you need.
Be sure to join me in SN1 video 2 where I focus on concepts like stereochemistry carbocation
rearrangement and speed of reaction.
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