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Course: Organic chemistry > Unit 5
Lesson 5: Sn1 and Sn2- Identifying nucleophilic and electrophilic centers
- Curly arrow conventions in organic chemistry
- Intro to organic mechanisms
- Alkyl halide nomenclature and classification
- Sn1 mechanism: kinetics and substrate
- Sn1 mechanism: stereochemistry
- Carbocation stability and rearrangement introduction
- Carbocation rearrangement practice
- Sn1 mechanism: carbocation rearrangement
- Sn1 carbocation rearrangement (advanced)
- Sn2 mechanism: kinetics and substrate
- Sn2 mechanism: stereospecificity
- Sn1 and Sn2: leaving group
- Sn1 vs Sn2: Solvent effects
- Sn1 vs Sn2: Summary
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Sn2 mechanism: stereospecificity
How nucleophilic attack in Sn2 reaction results in inversion of configuration at carbon with leaving group.
Want to join the conversation?
- At8:30why Bromine was substituted not florien ?!(12 votes)
- It is simply a matter of good, better, best. Bromine and Fluorine are both good leaving groups, but Bromine is a better leaving group than Fluorine. Why? The next video explains. Stronger acids have more stable anionic forms, or conjugate bases. Bromine, further down the Halogen group, is more stable in its anionic form than Fluorine.(12 votes)
- Where did the Na go in the end?(9 votes)
- at1:33it is mentioned R&S configuration . what is it?(4 votes)
- The R,S system is a system of nomenclature devised by R.S. Cahn, C.K. Ingold and V. Prelog. The R,S system is used for naming enantiomers. For example, if we take into consideration a molecule, suppose 2-butanol, then it would have two enantiomers. If we name these two enantiomers by using only the IUPAC system, then both enantiomers would have the same name. According to the R,S system, one enantiomer of 2-butanol should be designated (R) - 2- butanol and the other should be (S)-2-butanol.
Hope this helps. Cheers!(2 votes)
- at8:27, does the sodium cation of sodium methoxide disassociates in the media?(4 votes)
- I'm a bit confused. In SN2 reaction, inversion takes place but it is not necessary that the configuration is inverted so then what gets inverted if not configuration?(3 votes)
- In my view, there is inversion of configuration, but you can't always say that R changes to S and S changes to R.(3 votes)
- Why is the first reaction SN2, but not SN1?(3 votes)
- The reaction shows there is inversion of stereochemistry and this only happens with SN2. SN1 results in a racemic product.(2 votes)
- If there is no stereocenter in the SN2 reaction, will there be a change in R/S priority when the nucleophile replaces the leaving group?(2 votes)
- That depends what you mean really. If there’s no stereocentre then it can’t “change” from R to S or S to R (because it wasn’t R or S in the first place!), but a SN2 reaction could create a stereocentre if for example that carbon now has 4 different groups bonded to it.(2 votes)
- at8:20, why did you just draw the nucleophile as the methoxide anion? Wouldn't the sodium also be bonded to the O-CH3, making it not charged at all?(2 votes)
- It would be assumed to be run with a solvent it is soluble in such as methanol, in which case it would be split up into Na+ and OCH3- just like ions in water.(2 votes)
- At10:00minutes, isn't the molecule on the left supposed to be R configuration?
Since 4th priority is in between 2nd and 3rd, from 1st priority to 3rd is S configuration (counterclockwise), shouldn't the molecule be flipped to be R configuration?(2 votes)- Moving from the 1st priority group through the 2d to the 3rd makes us travel in a counterclockwise direction which is S configuration. And from our perspective the lowest priority group is pointing away from us so we don't need to flip and so it remains S configuration.
Hope that helps.(2 votes)
- Around6:30you mention that our product is S-2butanol, would we obtain a racemic mixture from this reaction, or all S conformers?(2 votes)
- All S. After the inversion happens you can't get any R conformers(2 votes)
Video transcript
- [Narrator] In the last video,
we looked at the mechanism for the SN2 reaction. The hydroxide ion will
function as a nucleophile in this case and attack our electrophile. So right here at this carbon
and since the SN2 mechanism is concerted, the nucleophile
attacks the electrophile at the same time that
our leaving group leaves. So bromine leaves as the bromide anion, and the OH, the nucleophile, substitutes for our leaving group. So for our final product we now have an OH attached to our carbon chain. In this video we're going to
look at the stereo specificity of the SN2 reaction. And that just means
that the stereochemistry of the reactant determines
the stereochemistry of the product. For example if we look at our substrate, we know that this carbon
is a chiral center. And our bromine is on a wedge. It's coming out at us in space. So the configuration at
this chiral center is R. We look at our product,
this is that chiral center. But now we have our OH
going away from us in space and the configuration at
this chiral center is S. So the stereochemistry of
our product is determined by the stereochemistry of our reactant. And that's because of our SN2 mechanism. We observe inversion of configuration. So let me write that down here. So inversion of configuration. We're going from an R configuration here to an S configuration. And that means that the
nucleophile can only attack from the side that's
opposite of the leaving group and that's consistent
with our SN2 mechanism. One way of thinking about
this is your bromine is relatively large and
it has a large amount of electron density around it with all these loan pairs of electrons. And that will repel your
negatively charged nucleophile so your nucleophile has
to approach from the side opposite of your leaving group. And that requirement means that you get inversion of configuration. So next we're going to do
a video where I'm gonna use the model set to show
you how R2 bromobutane turns into S2 butanol, and
hopefully it'll be more obvious. Here is R2 bromonbutane and
I've made bromine yellow. And I've left the hydrogens
off the alkyl groups just to make it easier to see. So if I turn the model a little bit and we look at our chiral center, so we have tetrahedral
geometry around this carbon. There's a methyl group
coming out at us in space, a hydrogen going away from us in space. An ethyl group going down and our bromine is off to the right. So we get our nucleophile, our nucleophile is trying to attack this carbon. So the hydroxide ion has
to approach our substrate from the side that is
opposite of our leaving group. So next let's look at
the transition state. Notice that we still have
our methyl group coming out at us in space, our hydrogen
going away from us in space, our ethyl group going down and
our halogen off to the right. But now there's a partial bond that forms between this oxygen and this carbon. There's also a partial
bond between this halogen and this carbon. Now if we look at that
central carbon there and we look at the three
groups, the methyl, the hydrogen and the ethyl group, those 3 groups are all in the same plane. So if I rotate this a little bit, right, you can see they're all planar there. So the transition state has a bond forming at the same time a bond is breaking. If I keep this model of
the transition state up, we can compare it with the final product. So here's the final product, the alcohol. We still have the methyl
group coming out at us space. With the hydrogen going away
from us, our OH on the left and our ethyl group going down. So this is one way to
represent our product. But if I rotate this
model set a little bit, so we can see our carbon chain. So if I turn it, if I turn it this way, now our OH is going out at us in space. So that's another way to
view our final product. I can also turn this
model set a little bit so if I turn it over to view
the carbon chain this way, now the OH is going away from us. And that's how I drew it
in the original reaction. So let's draw out what
we saw in the video. So our hydroxide ion, our
nucleophile attacks this carbon. At the same time these electrons
come off onto the bromine. Let's draw out the transition state. So we're gonna put some brackets in here. And I'm gonna put in my OH, I'll have some lone
pairs of electrons here. And we saw that a partial
bond forms between this oxygen and this carbon here. In the video I pointed
out how the methyl group is still coming out at us in space, so our methyl group is coming out at us. Our hydrogen is going away
from us and our ethyl group is going down and those three things, the methyl, the hydrogen, the ethyl group, are in the same plane. And at the same time right,
we have a partial bond between our carbon and our leaving group, which is our bromine here. So let me put in these
electrons here on the bromine. So this is our transition state. And we designate the transition state with this little symbol up here. So let me draw that in. Now the hydroxide ion over
here was a full negative charge but if it's forming a partial
bond with this carbon, we would give it a partial negative so it's losing some of
its electron density as that bond is forming. And as the electrons are
coming off on to the bromine, to form the bromide anion,
we're gonna get a little more negative charge on that bromine. So I'm putting a partial negative here. So here's our transition
state where the nucleophile is in the process of forming a bond. At the same time the bond to
the leaving group is breaking. So finally let's draw our product here and just like I did in the video, right, I'm gonna point out how
our methyl group stays coming out at us here. So let me draw that on a wedge. So here's our methyl group. Our hydrogen is still going away from us. And our ethyl group is going down. So let me draw our ethyl group going down and now we have a bond to ROH. And we're back to tetrahedral geometry. So this carbon, right, our chiral center has tetrahedral geometry to start with. Here for our three groups
we have planar geometry. But for our final product we're back to tetrahedral geometry. And this is one way that we
drew our product in the video I showed moving the model set around and one of the view
points that we looked at was if your carbon chain looks like this, that OH group is coming
out at you in space. And that's the same thing
as if you turn the model set around where you would have
your carbon chain like this. Your OH is going away from you in space. So these are three different
ways to represent our product which is S2 butanol. Let's look at another SN2
reaction with stereochemistry. And let's figure out the product. So first I would say, okay this ion here is negatively charged so
that must be my nucleophile. And for my substrate,
for my alkyl halides, I know that chlorine is
withdrawing electron density from this carbon since chlorine
is more electronegative. Which means that carbon
is partially positive. So my nucleophile is going
to attack my electrophile. At the same time, these
electrons are gonna come off onto the chlorine to
form the chloride anion. So I know that's what
happens in my SN2 mechanism and now that we've looked
at stereochemistry, I know that the nucleophile has to attack from aside its opposite
of our leaving group and we get inversion of configuration. So when I draw my product here, if I'm drawing the
carbon chain the same way that we started with, since
this chlorine is on a dash going away from us in
space, we have to show the bond to our nucleophile
here coming out as a wedge. So I'm gonna put the SH here. So this will be our final product. For simple systems you'll see
inversion of configuration. If you're starting compound
as R, you will get S for your product and if you
start with S, you will get R. However it doesn't always
have to be that way. If we look at this reaction on the left is our starting compound. This is an SN2 reaction
and our nucleophile would be the methoxide anion. So let me go ahead and draw that in. So we have an oxygen, we have a CH3. We have a -1 formal charge on the oxygen. So our nucleophile is going to attack our partially positive carbon
which is right here. So our nucleophile
attacks our electrophile at the same time that we
get loss of a leaving group. And the bromide ion would be
the best leaving group here. So these electrons come
off onto the bromine to form the bromide ion and we form a bond between the oxygen and the carbon. So this lone pair of
electrons on the oxygen forms this bond and we get our product. If you look at the stereochemistry, let's first start on the left here so we know that this
carbon is a chiral center and if I'm assigning
priority to the four groups attached to that chiral center, bromine has the highest atomic number so that would be priority number one, followed by fluorine which
should be number two, followed by carbon here
which is number three and then hydrogen is the lowest
priority group number four which is going away from us. So we're going around in this direction which is counter clockwise and
so that's an S configuration to our starting compound. For our product again this carbon here, let me change colors, this
carbon is our chiral center and when we assign priority this time, now the fluorine has the
highest atomic number, followed by this oxygen here, followed by this carbon and then of course with the hydrogen. So when we go around in this direction, again the configuration is still S. So here's an example of an SN2 mechanism where our nucleophile has to
attack from the opposite side of our leaving group, so
we still get inversion in terms of the mechanism. But we don't get
inversion of configuration because of how we assign
priority to our groups. In this case when our bromine leaves, fluorine becomes the
highest priority group and that's the reason why
the configuration stays S.