So let's talk about one of the simplest ways that they could react which would be to react
in an acid/base with a Bronsted-Lowry reaction.
So when a nucleophile and electrophile react together to exchange a proton, we call that
a Bronsted-Lowry reaction.
So this is an example that we used earlier from acids and bases.
You could see here I have a negative charge and I have a neutral substance.
If I were to figure out which one is the nucleophile, what would you say that is?
Well, the nucleophile was always the thing that was good at donating electrons, so I'd
say that would be this thing right here.
That species is my nucleophile, which means that invariably, the other thing needs to
be my electrophile even if it's difficult for me to see how it's electrophilic, but
it must be because the other thing is a better nucleophile.
So let's go ahead and say this is my electrophile here.
Now in this reaction I'd have to figure out now I know my nucleophile and my electrophile,
where does the arrow start from?
Remember that with mechanisms we always start from the nucleophile, so I would know that
I need to draw an arrow starting from this negative charge.
Now the question is where does it go to.
To figure out which atom it's going to go to because there's no positive charge directly
drawn.
If there was a positive charge, then I would just go ahead and attack that.
But the electrophile doesn't have a positive, so I'm going to have to use dipoles to figure
out what's the most positive atom on this molecule.
So I would say I've got a few different bonds.
I've got a carbon-sulfur bond.
I've also got a hydrogen-sulfur bond.
Which direction would those dipoles go?
They would both go towards the sulfur.
What that means is that eventually, my sulfur would have a partial negative and both of
these would have partial positives.
Now notice that I have a positive on a hydrogen.
That means that that's the same thing as me saying that I have an acidic hydrogen.
Why is that acidic?
Because it's going to be easily donated because of the fact that it already has a partial
positive charge, so it's looking for something negative that can attack it.
To finish off this arrow, since I have an acidic hydrogen, that's going to be my electrophile
right away.
I'm going to go ahead and attack the H. And you guys could have predicted that's what's
going to happen because I just told you we're doing a Bronsted-Lowry reaction, so this is
kind of a recap of the acid/base chapter.
So I go ahead and I grab that H. Now, what's the next question that I ask myself?
Do you guys remember?
Well, the next question is always are we done with the mechanism or do we need to keep drawing.
Do you think we're doing with the mechanism just the way it's drawn?
No, we're actually not.
The reason is because remember that that arrow that I just drew represents the sharing of
two new electrons, so this is two electrons that are now going to be attached to that
H. How many electrons does the H want to have?
In total, it only wants to have two electrons.
So now it would have four electrons if I donated this new lone pair.
So what that means is that we're going to follow that predictable rule, which is that
if I make a bond, I have to break a bond in order to preserve the octet of the hydrogen.
So obviously the only thing that I can break is this bond to the sulfur, so I would go
ahead and dump the electrons on to the sulfur and since this is Bronsted-Lowry I would draw
equilibrium arrows and what I would wind up getting is now I have my O that now has a
new single bond and that new single bond is to an H because it pulled that H off of the
acid.
Then I also have to draw my conjugate base.
Remember that.
And my conjugate base would just be the thing that now it doesn't have a hydrogen anymore,
so I would draw my ring structure and then I would draw an S. And then I would ask myself
how many electrons did the S have before.
It had 8.
Let me just draw them in.
It had a lone pair here and a lone pair there.
So this S would still have those blue electrons from before, but now it's going to have one
extra lone pair that came from the breaking bond.
So I'm going to go ahead and add that lone pair here.
And then I would use the formal charge rules to figure out what kind of charges these should
have.
So the oxygen should be neutral because oxygen wants to have 6 electrons and right now it
does have 6.
But the sulfur should have a negative charge because the sulfur wants to have 6 as well.
It's actually in the same column as oxygen, but it has 7 electrons.
So notice that this is a very predictable Bronsted-Lowry reaction because what I'm doing
is I'm reacting a nucleophile and electrophile and what I'm getting is an exchange of a hydrogen
and the exchange of a lone pair.
That's easy.
This is what we've already done in acids and base.
This is really just a recap.
Is that cool?