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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

  • 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.

Let's see if we can learn a thing or

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B1 中級

Sn2反應 (Sn2 Reactions)

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    Chao Joanne 發佈於 2021 年 01 月 14 日
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