Sn2 Stereochemistry Sn2 Stereochemistry
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- 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.
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