All right guys, so now we're going to jump into one of the most important mechanisms
in all of organic chemistry and it's a mechanism that you're never allowed to forget.
What I teach you in the next 20 minutes, it's going to stick because you're going to need
it for your – obviously for orgo, for your graduate exams and even in graduate school.
If you're planning on going to graduate school for anything pre-health, you're still going
to need to know this reaction and that's called the SN2 mechanism.
Let's dive right into it.
What the – if I were to just give it a tagline and just say in one sentence what an SN2 mechanism
is, what it is is that a negatively charged nucleophile – hopefully, all of that is
words you should be comfortable with – negatively charged nucleophile reacts with an accessible
leaving group.
Now, leaving group you should know what it is.
Accessible, maybe you're a little confused, but we'll define it.
To produce substitution – you know what that is – in one step.
Let's go ahead.
I want to just get right into it.
Let's just draw this mechanism out.
So I have this nucleophile that I'm just generally putting as Nu-.
There's a lot of different nucleophiles out there.
It doesn't really matter the identity right now.
Now I have to figure out – I'm reacting it with an alkyl halide.
What did I say alkyl halides were good at?
Leaving.
What that means is I have to figure out what is the electrophilic part of this molecule
because this is my electrophile and what is it going to look like after it reacts.
So how do I find out which part's electrophilic?
Does it have a positive charge already on it?
No.
So I'm going to have to draw the dipole.
What does the dipole look like?
Well, remember that halogens pretty much always pull away from whatever they're attached to,
so I would have – my only major dipole is pulling away from the carbon.
What I would have is a negative here, a partial positive there.
Where's my nucleophile going to want to attack?
It's going to want to attack the carbon.
So the electrophilic part is not the X, it's the carbon.
So I'm going to start off my arrow from my nucleophile and I know I'm going to attack
that carbon.
But now actually we have a choice.
Because what we have is a distinct set of sides.
So let's think about it this way.
This is my carbon.
And in the past, I haven't really worried about exactly how I draw my arrows because
I haven't been very picky.
But, if you think about it, there's actually two different sides to this carbon.
Let's say that the X side, the one with the halogen, is called the front side.
So the X has three lone pairs, one, two, three.
That would be what I would consider the front side.
So I'm just going to write here that's the front.
And the back side would be everything that's on the other side over here.
And the back side, what it's going to have is just a hydrogen and then some alkyl groups.
So this would be a methyl group and an ethyl group.
Which of these two sides, front or back, do you think is going to be the easiest for my
nucleophile to approach?
Let's think about it this way.
We know that it wants to hit the carbon.
So no matter what it's going for the carbon.
But all I'm asking is is it going to try to go from the front side or the back side.
It turns out that the front side is a really bad option.
Why?
Because the nucleophile, remember it already has extra electrons.
It's got extra electrons it's trying to get rid of.
In order to go through the front side, it would need to pass through a bunch of electron
clouds from the hallogen.
Do you think that's going to be very easy to do?
It's actually going to be almost impossible.
It doesn't happen.
Those electrons are going to repel each other like crazy.
So front side attack is actually impossible.
It's never going to happen.
What that means is that this is going to lead us to one of the most inappropriate phrases
in all of science and that is back-side attack.
So as messed up as that sounds, alkyl halides are totally down with it.
Back-side attack is something that they're all about.
And we're going to be doing this every day for the rest of organic chemistry.
So I hope you guys are cool with that.
Got to get used to it pretty quick.
So back-side attack is the way to go because it's the way that's basically less hindered.
It's going to be a lot easier for those electrons to pass through the back side where there
are not as many electrons as the front.
So now what we need to do is we have to draw the transition state of what this is going
to look like because let me just ask you guys.
Am I done with this mechanism?
Do I need to draw any more arrows or am I done?
No, we should draw some more arrows.
Why?
Because remember that this is going to be a nucleophile and electrophile that does not
have an empty orbital.
Notice that there's no empty orbital here.
This carbon already has four bonds.
So this carbon already has four bonds if I make a new one that's five.
So if I'm making this bond, I'm going to have to break a bond.
And you guys already know what I'm going to break.
I'm going to break the halide off.
I'm going to break the halogen off.
So that means that I'm making a bond and I'm breaking a bond at the same time.
This is going to lead to something called a transition state.
A transition state is just a high-energy phase of the reaction that is very, very short lived.
What it means is that it never even really happens – what I'm trying to say is that
it cannot be isolated.
It's a very high energy thing that must happen because we know that it must go from one state
to another, but if I tried to just isolate it in a test tube, I would never be able to
isolate transition states.
Let's go ahead and draw what it would look like.
It would be a carbon.
And it would be attached to three things for sure that are just single bonds.
It would be attached to a methyl group.
That's the one in the front.
A hydrogen in the back and then an ethyl group.
And I'll just put the ethyl group facing down because I'm going to need all the space I
can get.
Now this is the interesting part.
We just said that I'm making a bond and I'm breaking a bond and it's all happening in
one step.
That means that I'm going to have to draw partial bonds.
That means that my nucleophile is partially making a bond to that carbon.
And my halide is partially breaking a bond to that carbon.
On top of that, now this carbon has too many bonds.
It has five instead of four, so I'm going to have to put partial negatives on these
atoms.
What that means is that remember that carbon wants to have four bonds, it now has five,
so in this transition state, it's extremely unstable.
Carbon does not like to have this many bonds.
I have to indicate that it has one too many by putting negatives that are distributed.
So there we go.
That's our transition state.
If you ever see this double dagger, that means transition state.
Like I said, this is something that it must exist for like a nanosecond, but it's not
something you could isolate.
After this reaction is done happening, after it's all completed, all happens at one time,
now I have to figure out what my products are going to look like.
Well, what I'm going to have now is that I'm going to have my nucleophile.
But now my nucleophile is going to be attached to – have a single bond.
I'll draw it in blue because that indicates the arrow that was just made, to that red
carbon.
And what is that red carbon going to be attached to?
The same three things it was attached to before.
So it's going to have that methyl group, it's going to have that hydrogen.
And not the ethyl group because the nucleophile came from the back, my ethyl group is getting
pushed towards the front.
That's going to be important.
Let's just hold on to that thought.
On top of that, is there anything else that we need to draw?
Our leaving group.
So our leaving group is going to get X-.
So first of all, how can I tell that a substitution reaction just took place?
Well, I can tell because things substituted.
Before I had a carbon with an X, now I have a carbon with a nucleophile.
Before I had a nucleophile with a negative, now I have an X with a negative.
See how everything perfectly swapped.
So this is definitely substitution.