two about Sn2 reactions.

And you're going to see a lot of these in organic chemistry.

This sounds like a very fancy term.

But it really just stands for-- the S stands for

substitution.

The n stands for nucleophilic.

And then the 2, and this is probably maybe the least

obvious part of Sn2, the 2 comes from the fact that the

rate-determining step in this reaction

involves both reactants.

And I'm going to show you what we mean.

And it'll actually be probably more clear when we compare it

to Sn1 reactions.

So let me show you an Sn2 reaction.

Let's say we have bromomethane.

And let me draw it in three dimensions.

You have a hydrogen sticking out.

Maybe the bromine is right over there.

And then you have another hydrogen in the back.

And then you have a hydrogen that comes up just like that.

And here, the three dimensions won't matter that much.

This is not a chiral carbon.

There's no handedness here.

But in the future, we'll think about chiral carbons and what

might happen to them as they undergo Sn2 reactions.

So I have this bromomethane here.

And let's say we also have some hydroxide in

solution with it.

So it's a hydroxide, and this oxygen has

seven valence electrons.

You could imagine that the way that you get a hydroxide is if

you start off with water, that looks like this.

And maybe there's another water someplace else that

looks like that.

And maybe in one step, this water right here takes this

guy's hydrogen.

Remember, there's a partially positive charge on the

hydrogen, partially negative charge on the oxygen.

So maybe he gives one of his electrons to the hydrogen,

which will end up with a positive charge.

It'll be a hydronium cation.

And then this electron takes back that extra electron from

the hydrogen.

And so now, he's going to have seven valence electrons.

He's going to have this one.

That was at this end of the bond of the one that he's

taking back.

And then that's two, three, four, five, six, and then this

electron right there, seven.

So he's going to have one, two, three, four, five, six,

seven valence electrons.

And he's going to have a negative charge.

It's neutral when it's water.

When he takes an extra electron, it's going to have a

negative charge.

And you could imagine this is in a solution with water, that

water is our solvent.

So let me put a negative charge right here.

I can write minus 1 if I want.

And in this reaction, just so we know what's what, this

hydroxide molecule, or this hydroxide anion, is our

nucleophile.

It is the nucleophile right there.

And the best way to think about a nucleophile is it

likes other people's nucleuses.

It likes them because nucleuses are positive.

It has extra electrons.

In this case, it has a negative charge, so it is

attracted to other nucleuses.

Nucleuses of atoms are positive, so it is attracted

to electrons.

Phile, I'm not a Greek scholar, but it means to be

attracted or to like

something, so it likes nucleuses.

And if you want to compare it to acids or bases, you'd

classify this actually as a Lewis base, something that

gives electrons.

But we're not going to go into that right now.

This is a nucleophile.

It is attracted to nucleuses, to positive things.

It wants to give away electrons.

Now, what's going to happen is that this is going to give an

electron to this carbon.

This carbon's going to say, oh, I got an electron.

Let me give away an electron to somebody who probably wants

it really badly.

And in this case, it's going to be the bromine because

we've seen multiple times that the bromine is very

electronegative.

It already is starting to hog the electrons, and it would

like to take an electron altogether.

And so let me show you the reaction.

It's going to happen very quickly.

Or it's actually going to happen very quickly the way I

drew it, because it's really a one-step reaction.

So you're going to have one of these electrons right here

will attack the carbon.

Or I shouldn't say attack.

It sounds very aggressive.

It will be given to the carbon.

And then the carbon says, oh, I'm getting an electron from

this direction.

It actually has a slightly positive charge at this end,

because the bromine is hogging some of the electrons, so

they'll be attracted to each other.

They're going to bump into each other just the right way.

And then at the exact same time, this all has to happen

kind of simultaneously.

At the exact same time, this electron, sitting at that end

of the bond, is going to go to the bromine.

And so what is it going to look like when this reaction

has actually occurred?

So it's literally a one-step reaction.

So you have your carbon.

You have the hydrogen that is pointing up.

Now, let me color code this.

So that'll make it interesting.

So you have this hydrogen, that is

coming out of the screen.

And it's still just coming straight out of the screen,

but I'm going to draw it to the down right, because this

hydroxide's attacking from behind.

And this bromine, which we call the leaving group, or

actually, it's going to be a bromide anion once it leaves,

that's going to leave from the right.

Attack from the left, leave from the right,

from different sides.

So then this is that hydrogen that's sticking out.

The hydrogen in the back-- well, we know it's in the

back, so I'll just continue to draw it in the back like that.

And now this hydroxide is going to be

attached to the carbon.

So this carbon now gets one of those electrons, and it is

bonded now to the oxygen.

Let me draw the oxygen.

So you have the oxygen.

Let me see the best way that I could do this.

So you have the oxygen here bonded to a hydrogen.

And let me draw its valence electrons.

So it had one, two, three, four, five, six, and then it

had a seventh electron, but it just gave it away.

So its seventh electron is going to bond with this one.

They were already a pair over here.

Let me make it very clear what the color is.

This electron right over there is this electron.

But it was already paired with this guy, so when he gives it

away to the carbon, it forms this covalent bond right here.

So now you have this OH group, this hydroxyl group, off of

the carbon now.

And now the bromine, you had a negatively charged group here.

Now, what I've drawn so far, everything is neutral.

So something has to be negatively charged.

It's going to be the bromine because that's what getting

the electron.

So now the bromine, if I wanted to draw bromine's

valence shell, before it would have seven electrons, so one,

two, three, four, five, six, seven.

Now that it's taking that electron from carbon, it's

going to have eight: one, two, three, four, five, six, seven,

and let me color code it again.

So this electron right here, when it goes to the bromine,

is going to be that electron right there, and we're going

to have a negative charge.

And so here, in this situation, just to make sure,

this is the nucleophile.

The bromine right here, this is the leaving group.

And then the carbon, I guess the thing that is reacting,

that's getting substituted, where the bromine leaves it

and the hydroxide joins it, that's called the substrate.

So I could draw it like this.

This is the substrate.

And the whole reason why I did it, this was

not a chiral carbon.

But you could imagine, once we deal with chiral carbons, when

they undergo an Sn2 reaction like this, their chirality

will actually change.

So I'll leave you there right now.

That'll give you something to think about, Sn2 reactions.

And then in the next video, we'll think a

little bit about Sn1.

Oh, and before I leave you, I left you hanging on the 2.

I said this is called substitution.

You saw that the bromine was substituted by the hydroxide.

It's nucleophilic.

It involved this hydroxide, which likes nucleuses.

And then I said the rate-determining step involves

both reactants.

And here, it might not be completely clear, but the

rate-determining step was actually what's

going on right here.

And if I were to draw an intermediate, you could

actually draw an intermediate step here.

Let me copy and paste this down here.

So you could actually draw an intermediate step, so let me

copy the other one.

Let me copy this.

Copy and paste.

This all happens over one step, but if we wanted to show

what happens in between, for a small amount of time, you have

this transition state, where the carbon is going to have a

partial bond, so it has this hydrogen in blue.

Let me draw this hydrogen in blue sticking out.

And it has another hydrogen behind it.

I'll draw it like it's right behind it.

And then for some small fraction of time, it is bonded

to both the bromine-- so I'll draw it with a dotted line.

It is bonded to both the bromine, which now will have a

partial negative charge because it's starting to get

an electron, and it's also bonded to the hydroxide, which

also has a partial negative charge.

You can imagine this negative charge is now getting split

between the two.

This is a transition state.

And the notation for a transition state, you put

brackets around it, or parentheses, in some cultures,

they call them, and you write this little symbol up here.

This tells us this is a transition state.

This is not one of the semi-stable intermediates.

This is something that we're going through.

This is an Sn2 reaction because this transition state

is actually what determines the rate of the reaction.

This requires the highest energy state of this reaction.

This is what's going to-- well, I said it already--

determine its rate.

And in this transition state, this rate-determining step,

you have all of the reactants.

You have your hydroxide right there, and you also have your

bromomethane that we started with, so it

has two of the reactants.

That's why it's called an Sn2 reaction.