It's time to do an SN1 mechanism all by yourself.
Just kidding.
I'm actually going to help you on this one, but the next one I need you to do by yourself.
Let's go ahead and start off by reading the question.
It says provide the mechanism and final products for the following reactions and we're going
to start off with (a) which is giving us some kind of alkyl halide and some other reagent.
So the first thing to do is before we even draw any arrows, let's just double check that
these reagents even want to do an SN1 mechanism.
The alkyl halide is what degree?
What type of alkyl halide?
Well, the carbon it's attached to appears to be attached to three other carbons.
So that would make it tertiary.
So this is going to be a tertiary alkyl halide.
Now my other reagent, my nucleophile, is methanol or methyl alcohol can also be drawn as CH3OH.
And this is neutral.
It doesn't have a charge.
It's not negative.
It's not positive.
It's neutral.
So I'm just going to put here in parenthesis, it's neutral.
What did we learn about neutral nucleophiles and highly substituted leaving groups?
They prefer to react in SN1.
So if we had to pick between the two, would this be SN2 or SN1?
It would make sense that this would be SN1 because of those two factors.
It's neutral and it's highly substituted, which is the opposite of what SN2 likes.
So now that makes sense.
So now that we've verified that this could be an SN1, let's just talk about the arrows.
What would be the first arrow of an SN1 mechanism?
What can we always say is going to be the first arrow?
The leaving group leaving.
The leaving group takes off by itself and leaves something behind.
Now remember that this is different than SN2 because in SN2 the negative attacks and does
a back-side attack in the first step.
But since I have a neutral nucleophile, it's waiting around for something else to happen,
which is that we're going to basically leave and create something in its place.
So notice that what we get after it leaves is that you have an F- because now the F has
extra electrons but you also have a carbon that's missing a bond because now this carbon
is missing one of its four bonds, so what we're going to get is a three-membered kind
of – or three branches coming off the middle with a positive charge.
So we're going to get a carbocation.
So now I've got a carbocation and I just finished my first step.
This step, if you guys recall, was all the slow step or the rate-determining step because
this is the hard one to do to make your carbocation so that you can, later on, react it with something.
Now there's two things we can react it with.
We could either react it with F- or we could react it with the neutral nucleophile that's
hanging out.
Now could F- go ahead and re-attack that carbon?
Sure, but that's just going to take me back to where I started so that reaction doesn't
really make sense for me to draw because that would just be going back to the original and
then it would dissociate again.
But what is interesting to me is that I have this methanol around and we know that now
that the methanol sees this strong electrophile, that strong cation, it's able to attack.
So that's the next step.
The next step is that your nucleophile attacks the carbocation that you made What's that
going to do?
Well, that's going to give me my same carbon structure, but now I just made a fourth bond.
That fourth bond is the O attaching to the positive.
So what is attached to that fourth bond?
Well, you have to go in order from the thing that attacked.
The thing that attacked was the oxygen.
That's what had the lone pairs.
What is the oxygen attached to?
Two things still.
It's definitely attached to the CH3 but it's also still attached to that H that it had.
And guys going back to our initial lessons, way at the beginning, about formal charges
we learned that oxygen with three bonds isn't happy.
An oxygen with three bonds and one lone pair because it used to have two but now one of
them got used for the bond, this is going to have a positive charge.
That's not very cool.
Ending on a positive usually isn't a good sign.
It usually means that you messed something up.
But for you guys you might be saying, “Johnny, how did I mess something up?
I just did what you told me.
This should be the last step.”
And you're right, but there is one exception.
There's one caveat that I didn't tell you yet which is this note.
What the note says is that – check this out.
This is important.
Substitution reactions, so any type of substitution that is using a neutral nucleophile.
Neutral meaning that it doesn't have a charge at the beginning.
Are going to require one additional step.
It's called deprotonation.
You're going to have to take a proton off in the last step.
So the way you can think about it is that an SN1 reaction, since SN1 reactions usually
have neutral nucleophiles, after SN1 you have to do a second step and that's to deprotonate.
Basically, if you're using neutral nucleophiles such as water or such as alcohol, etcetera,
nucleophiles that are neutral don't have a charge.
You always have to draw in that extra deprotonation step, meaning that we're not done yet.
We still have to deprotonate.
Let's go ahead and draw just a little thing that says deprotonate here and bring this
down to the bottom.
And we're almost done, I promise you.
Now we're going to deprotonate.
What are we going to deprotonate with?
There's actually two things that you can deprotonate with here.
You can deprotonate with F- and you can also deprotonate with CH3OH.
Both of these guys are around.
Both of them have electrons around that can grab the H and take that proton away.
Now what you're going to find in my videos guys is that I'm not very picky about which
reagent you use.
Sometimes I'm going to use the F- or the X, sometimes I'm going to use the alcohol and
I'm not going to be very picky about which one.
You know why?
Because it doesn't really matter.
In solution, these protons are constantly being exchanged between the solvents, so the
H, if you grab it with the F, it's just going to probably come off the F at some point and
then go to the methanol.
And if you grab it with the methanol, same thing.
It's eventually going to come off and jump onto the F and jump back.
So these protons are constantly jumping around so it doesn't really matter that much and
that's actually going to be a common theme for a lot of deprotonations in my course.
You may see me use one reagent or another and I don't want you to freak out because
your professor used a different one.
Your professor probably isn't going to matter and probably isn't going to be very consistent
about it either.
No obviously, your professor's word is more important than mine.
So if your professor always says that he wants you to use the alcohol, by all means, do that.
I'm just basically saying in the end of the day, it doesn't really matter.
All that matters is are we deprotonating or not.
That being said, I'm going to go ahead and use the alcohol here.
The way this would work is that my lone pair would grab the H. And if you make a bond,
you break a bond.
So I just made that bond, I have to break a bond.
I'm going to break this down to the O. And what we're going to get now is a structure
that looks like this, three bonds to carbon, OCH3 plus F- and plus CH3OH2+.
That methanol now has an extra H because I grabbed it off of the oxygen.
Guys, I'm done.
Really the main product that you should be able to produce for your professor is this
one.
This is the one – now this is an ether.
We just made an ether product.
This important because this is what we call the organic product.
The other stuff kind of just floats in solution, not a big deal.
But your professor really wants you to be able to draw that actual SN1 product.
Now guys, one more thing before we break for the next problem and then you're going to
do that one all on your own hopefully, remember that SN1 came with a certain type of chirality,
a certain type of stereochemistry, do you guys remember what that was?
Guys it was called racemic.
Remember that racemic means that you get 50% of one, 50% of another of two different stereoisomers.
So in this case, should I draw two products or should I just draw one?
Have I made a mistake to keep it at one or should I make a second one as well?
What do you think?
So guys, the answer is this is the answer you should not draw two.
In fact, if you draw two, you're going to look like a noob to your professor because
it's going to look like you don't really know what's going on.
The reason I say that is because this is not chiral.
This carbon here is not a chiral center.
It has three of the same group.
So there's not chance for it to be receiving.
So you'd only get one product here.
Now if this was a chiral center, if instead I had let's say this is a methyl and this
is an ethyl and this is a propyl, so these are all three different groups and this is
a chiral center, then absolutely you would draw a racemic product.
You would draw the OCH3 is on a wedge and then you'd say plus it's also on a dash.
And you'd get 50% of both.
But since this is not chiral, I only get one product.
So I got lucky here.
It's always easier when you don't have a chiral center because then you just have to draw
one.
But as you might see in the next example, I'm going to show you guys what's going to
happen when you do have a chiral center.
Well, you'll find out.
So go ahead and try to do the next practice problem all on your own.