Sn2 Reactions Sn2 Reactions
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- Let's see if we can learn a thing or
- 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
- 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
- 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
- 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.
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