In this video, not only are we going to pay attention to how
these two molecules will react, but we'll also pay
attention to the
stereochemistry of the end product.
To do that, we have to pay attention to the
stereochemistry of the things that we're beginning with.
And this is just regular hydroxide anion, we've seen
this show before.
This is going to act as the nucleophile.
It's not the strongest nucleophile.
it's not iodide, but it's a pretty good nucleophile.
So we have a good nucleophile here.
And then over here, we have a carbon.
It's not a methyl carbon.
It's not bonded to a bunch of hydrogens.
That would be a super fast Sn2 reaction.
This guy could attack it.
And it's not even a primary, it's a secondary.
But we said if we have a pretty good nucleophile, this
guy's going to want to give his electrons enough to that,
that this will happen.
It won't be the fastest reaction, but we can get an
Sn2 reaction, especially if the solvent is the right type
of solvent.
And we'll talk more about this in future videos on what type
of solvents are good for Sn2 reactions versus Sn1.
So we're going to have an Sn1 reaction.
We're going to have this situation where one of the
electrons from the nucleophile will be given to this
secondary carbon.
It's bonded to two other carbons.
So it'll be given to the secondary carbon, and then
that will allow this bromine to swipe the carbon that it
was a bonded with.
And we know bromine is pretty electronegative, so it's going
to do that.
It's going to swipe that electron.
And then at the end, you're going to end up with a
bromine, or I should say bromide now that it's an
anion, a bromide anion.
It swiped that electron.
Before bromine had one, two, three, four, five, six, seven
valence electrons, and now it's going to swipe this one
over here, so now it'll have eight: one, two, three, four,
five, six, seven, eight.
Put a negative charge, it just gained an electron.
And then the other stuff will look like this.
Let me draw that carbon in the center, that's just like this.
And notice, bromine is leaving from the right, and the
hydroxide is attacking, or it's giving its electron from
the left, or it's bonding from the left.
So what's it going to look?
Well, We have this ethyl group coming straight out
forward, CH2, CH3.
We have the hydrogen that is still going to
be behind the carbon.
So I could draw it like this.
The bromine is no longer there.
You still have this methyl group above
it, so that is CH3.
And now this hydroxide has attacked right in the same
direction that the bromine is leaving in, but from the
opposite side.
So that's leaving to the right, this guy's entering
from the left, so then the hydroxide will be
bonded like this: O-H.
Now we have just described a classic Sn2 reaction.
And I always like to emphasize the Sn2, because my brain
always confuses it with Sn1, so I like to remind myself,
this is Sn2 because it is substitution with a
nucleophile.
That's the hydroxide in this case.
And that number two says the rate-determining step, and
there is only one step in this reaction, involves both of the
reactants, and this definitely is involving both of the
reactants, so this is a classic Sn2.
The point of this video is to see what happened to the
stereochemistry of this molecule.
Obviously, it's a different molecule on this side, but it
had switched from left to right, or I guess we should
say that it switched from R to S, or S to R, or did it stay
in the same direction?
So let's name this one using our naming mechanisms. I guess
you could call it the R-S naming scheme.
And then we'll name this one, and then we'll see if we
switched our handedness.
So first of all, let's just name the base molecule.
This is good review.
So what's our longest carbon chain?
We have one, two, three, four carbons.
And actually, I probably shouldn't have named it that
way, because this bromo group is closer to that end.
So I should number it starting from here.
We have one, two, three, four carbons.
So methyl is one carbon.
Ethyl is two carbons.
Propyl is three carbons, or prop- is three, so it's but-,
But- is four carbons, so it's going to but-.
It's all single bond, so it's going to be a butane.
And on the number two carbon, we have a bromo group, so it's
2-bromobutane.
And this is a chiral carbon, so we have to describe it's
handedness.
And we know it's a chiral carbon, because everything
it's bonded to is different.
It's bonded to two carbons, but this is
a whole ethyl group.
There's two carbons here.
There's only one over there, and obviously, it's a
hydrogen, and that's a bromine.
So what's it's handedness?
And to do that, you want to put the group that has the
lowest atomic number in the back.
And we know hydrogen has the lowest atomic number, so it's
already in the back.
So that's convenient.
We don't have to do any visualization of how you
rotate it, so you put it in the back.
So that's already in the back.
So what we have in the front is this guy,
this guy, this guy.
And what you do is you start with the guy with the highest
atomic number, and that would be bromine right there.
And then you go in a circle in the direction of the next
highest atomic number.
Carbon and carbon is a tie.
The tie breaker is that this carbon is
bonded to another carbon.
Well, this carbon is only bonded to hydrogens.
So you're going to go in this direction right here.
You're going to go clockwise.
Let me do that in another color.
You are going to go clockwise.
And the way I like to think about is, on top, we're going
to the right.
We're going clockwise, so this is R.
S is to the left for sinister, R for the right.
That's the way I think of it.
So this is R-2-bromobutane.
So it's handedness is R.
Now, what's this thing over here?
Well, first of all, once again we have four carbons.
We have one, two, three, four carbons, so
but- will be our prefix.
But since the group that we are attached to is an OH
group, it's a hydroxide group, this is actually an alcohol.
So this a little bit of practice of naming alcohols.
So we still have but- as our prefix.
But we wouldn't call it butane.
Since it has an OH there, we'll call it butanol.
And maybe I'll write that in blue to show the -anol comes
from the OH.
Butanol, or maybe we could say the -ol comes from-- maybe
that's even better.
Let me just write butanol.
That's a different color.
But- -an- and then the -ol comes from the fact that we
have the OH there.
We're dealing with an alcohol.
And it's attached to the number two carbon, so we'll
call it 2-butanol.
So we went from having 2-bromobutane to 2-butanol,
but what is the handedness of this?
So same exercise, you put the lowest atomic
number in the back.
That's the hydrogen.
It's already in the back.
And then you start with the atom or the group that has the
highest atomic number of the things that are bonded to this
chiral carbon.
It's still chiral.
All of these things are different.
You start with whatever has the highest atomic number.
This oxygen has the highest atomic number.
And you go in the direction of the next highest. This carbon
and that carbon are tied, but this one is also attached to
other carbons.
Well, this is attached only to hydrogens, so we go in this
direction, or counterclockwise, or in the
sinister direction, or the leftward direction.
So this is S, S for sinister.
S-2-butanol.
So the whole point of this video is to show you that when
you have an Sn2 reaction on a chiral carbon, that it changes
the carbon's stereochemistry.
It changes its handedness.
It went from being right-handed to left-handed.
It changes its chirality.
You can even imagine a situation where a bromine
comes and attacks, and this bromine leaves.
And then you would actually have the same 2-bromobutane,
but it would actually switch its handedness.
This is super important, because this matters if you
care whether or not you want to flip the
stereochemistry or not.
What we're going to see in the next video, or maybe the video
after that, is that in Sn1 reactions, not only do you not
flip your stereochemistry, that you kind of lose-- if all
of your molecules were one-handedness, your end
product is going to be a mix of the two handedness when you
do an Sn1 reaction.
But Sn2, you flip the handedness.
You saw here.
We started with R-2-bromobutane.
The Sn2 reaction happened.
We ended up with S.
It makes complete sense.
You view this as coming from the back, so you view this as
kind of an umbrella.
This was kind of an upturned umbrella.
This was the handle.
And when this comes in, he becomes the handle, and the
umbrella flips in that direction.
So that's one way to think about it.
So visually, it also should make a little bit of sense.