字幕表 動画を再生する
>> Bond character is somewhere between that of a single bond
and that of a double bond.
In fact, it's sort of about 30% double bond and 70% single bond
that makeup this picture.
So as a result, you have slow rotation about this bond
and I'll say slow which means if I put a star next
to this ethyl group here, I'll just put sort of a star here
to remind us it's special, you do have a dynamic equilibrium
where you swap positions right.
So in this case, the start ethyl group assists to the carbonyl.
In this case, the start ethyl group is trans to the carbonyl
but this equilibrium is slow
on what we'll say the NMR time scale.
What I really want to do in today's talk is to give us more
of a feeling of what's slow, what's fast both
in time and also in energy.
So when I saw this is about 30% double bond character,
I want you to get a feeling for what that means
in kilocalories per mole and how I know that number.
Then I also want us to get a feeling for what happens
as we cross from the slow regime to the fast regime
so we'll get sort of one equation,
a couple of calibration points on energy
and time out of today's talk.
Now, if you think about this, let's take a simpler situation.
The case that people invariably use
for didactic purposes is something with singlets.
The case that sort of was the classic was dimethylformamide
just because it's easy to think about and easy to simulate.
So here with methyl groups, of course, you have singlets
and just like the ethyl groups for one of them is more
down field; the one that's CIS to the carbonyl is more
down field; the one that's trans
from the carbonyl is more up field.
You have the same with the methyl groups here.
I just want to, we'll call these HA and HB.
So you have some equilibrium here we can call this KA and KB.
In the case of a perfectly symmetrical molecule,
the rate constant is going to be the same
but if these were two groups say of disparate size
like a tert butyl group and a methyl group,
then the rate one way would be faster than the other
because you would have an equilibrium constant
that wasn't 1.
In other words, you'd have an equilibrium constant
where if you had a bulkier group,
say the bulkier group would spend more time here
and the less bulky group would spend more time here
or vice versa.
All right.
So let's just imagine a little thought experiment
for this situation.
So, the situation we just saw was one
where you have 2 singlets so this is just my old drawing
of an H1 NMR spectrum.
At some condition where you're slow, you're going
to see 2 sets of peaks.
For the same window if it were very fast, you'd see 1 peak
and I'm just drawing the same spectral window and sort
of plotting this out so I guess I'll kind of put these
on the scale, on the same scale.
So we can say this is fast and, of course, what's the best way
to take something that's slow
and make it fast in the laboratory?
>> Heat it up.
>> Heat it up.
So this is cold and this is hot.
All right somewhere between that point of cold
and hot you hit a middle point
which is called medium, yep, medium, okay.
You hit a middle point, medium, where you're
at what's called coalescence.
Now at coalescence what's happening is each
of these peaks is getting broader
and broader until they merge.
This is the uncertainty principle at work.
Remember we talked about line width and I said
that if you were able to measure the velocity
for infinitely long, in other words,
if there were never any spin flip, any relaxation,
any swapping of population between alpha and beta states,
your lines would be infinitely sharp,
but I said when we were talking
about the uncertainty principle your lines are typically
about a hertz wide or a little less than hertz wide
because your relaxation time is on the order
of a couple of seconds.
In other words, you cannot get your lines infinitely sharp
because you're only literally measuring the velocity
for finite amount of time.
As you heat things up,
what's happening is you're flipping faster and faster
and so your lines are broadening out.
So, what's really happening at coalescence
so let's just go back to the equation I presented
by the uncertainty principle, which is if you look
at your line, in theory there's some exact position of the line.
In other words in theory if you could make
that measurement infinitely,
your line would have this position
but what you're getting is signal out here
because you're not measuring that line with infinite time.
You're not able to because of relaxation
and so you have a certain width and that's the value
when we talked about the uncertainty principle we called
Delta nu, right?
Delta nu basically is the half line width at half height.
In other words, it is the level where you're sort
of within these arrow bars so you're plus or minus Delta nu
of that theoretical central value and we said
from the uncertainty principle Delta nu times the time that's
the lifetime so it's not really a half life it's almost
like a half life, it's an eth life, you know,
1 over 2.3 instead of 1 over 2.0.
Delta nu times time is equal to 1 over root 2 pi.
In other words, if you're able to make that measurement
for a second, then Delta nu is going to be .22 hertz
or the line width at half height,
the full line width is going to be .44 hertz.
If on the other hand you're only able to have
that lifetime be say a tenth of a second,
so I'll say t equals .1 second,
now we get to Delta nu is equal to 2.2 hertz.
In other words, now that line has become 4.4 hertz wide
at half height.
[ Pause ]
So what you were really seeing at coalescence
when I have this sketch of this broadened hump like this,
what you're really seeing is 2 fat lawrenciums [phonetic]
that are adding up underneath there.
So let me make this sort of with dotted lines
to show a fat lawrencium like so and if the separation
of the lines here, right, if this separation here
if we call this Delta nu lines for want of a better term,
in other words, the separation of the lines in hertz,
at this point at coalescence then now each
of these lines is fattened out so that its Delta nu,
not the Delta nu of lines, but this distance here,
this Delta nu is now, so if this is what we call Delta nu
of lines that this Delta nu is now half
of the Delta nu of lines.
Does that make sense?
In other words, each of the lines is broadened out so
that it's half width at half height is halfway across
and that is when your lines are going to be coalesced
where you're no longer going to see a distinct line
on the left, line on the right.
If they're broadened anymore, they're going
to be merged together
and eventually you just have a single peak and you're
at this situation here, but right now when they're broadened
out they're broadened out to a point
where they have merged together and so at that point Delta nu
of the lines the separation of the lines times tau,
which is now going to be our lifetime
at coalescence is equal to 2 over root 2 pi.
This is just the equation that we have over there except now
because we have the difference each
of these is fattened out halfway.
If we have 2 of them, it's going to now be 2.
So, what this boils down to then is a simple equation that tau
when you just work this out is equal to 0.450
over the Delta nu of the lines.
In other words, the lifetime at coalescence.
[ Writing on board ]
Is equal to .450 divided by the separation of the lines
at a lower temperature.
[ Writing on board ]
Does that make sense?
[Inaudible question] The .54 is simply what happens if I take
in my calculator 2 divided by root 2 divided by 3.1415
and then I put that in the numerator
and put the Delta nu lines in the denominator.
[Inaudible question] Well, you mean the 2?
Well, I'm saying because here our Delta nu is half
of the separation.
[Inaudible question] Within what you can measure it's exactly.
Let me show you.
Maybe the best way is for me
to show you how things look as you vary here.
So, basically if you go any, coming to, if it was more,
you'd start to pull in and you'd start to pull together.
If it's less, you'll see a dimple in the middle
and let me show you what this can best be pictured as
and this is just a simulation,
this is from a chapter undynamic NMR spectroscopy
in a book, let's see which book?
This may be a book on dynamic NMR spectroscopy.
So this is an old, just an old drawing of a simulation
of what you would expect and it's really based
on dimethylformamide and it's actually probably based
on why DMF on a 60 megahertz spectrometer
or something like this.
So their simulation is as follows,
and the reason I say it's a 60 megahertz spectrometer is the
lines in this simulation are Delta nu lines is equal
to 20 hertz.
In other words, out of 60 megahertz NMR spectrometer
that would be about 3/10 of a PPM, which is pretty reasonable.
Now, on a 500 megahertz spectrometer
that would be .04 PPM.
So anyway for their little simulation,
they're saying imagine that you have 2 lines, those 2 singlets,
and they're separated by 20 hertz.
Imagine that you have a T2,
that's a relaxation time, of .5 seconds.
In other words, imagine that the native lifetime
for this molecule due to relaxation was half a second.
In other words, that your lines are about .9 hertz width,
full width at half height, not half width
but full width at half height.
So that's how your normal spectrum would look.
Now, imagine that you start to heat this sample up so
that you have rotation between the 2.
So you have the 2 flipping back and forth.
So, imagine here, for example, that K for, you know,
our equilibrium this is our like dimethylformamide spectrum
where we can call this A and B or star.
[ Pause ]
Imagine now that our rate constant, oh, yeah,
imagine that our rate constant was 5.0 per second.
If your rate constant is 5 per second,
then your lifetime is 1 over K, right?
So your lifetime at this point is 200 milliseconds,
it's .2 seconds.
[ Pause ]
So your lines have met because they're not staying in the CIS
or trans state as long.
You can think of this as we started, here we're told
and here we're starting to heat the sample up.
Here they actually have a very slow K,
K is equal to in this case .1 per second.
In other words, it's a 10-second lifetime.
Do not swap at any appreciable rate.
As you heat up the sample in this simulation,
they go to K equals 5 and K equals 10 per second
so your lifetime is now .1 seconds and finally you get
to a point and you notice so here you are
and now your line width is still less
than half the distance between them.
So you still see this dimple here, this is it.
Now you can kind of see right
at this point now they are coalesced together.
So, right at K is equal to 44.4 per second they're now
coalesced together.
Then as you heat the sample up more, as you get hotter
and hotter, as you get faster and faster,
now we get to K equals 100
so now your lifetime is 10 milliseconds and then you get
to K equals 500 so your lifetime is now 2 milliseconds
and finally you get to K equals 10,000.
So now your lifetime is a tenth of a millisecond.
[ Pause ]
[Inaudible question] That's at relaxation time.
That's your native line width.
So what they're saying, of course,
is all lines have a native line width.
Let us pretend that we had a native line width
of about a hertz.
[Inaudible question] You could just as well have it be T1.
Either T1 or T2 is going to contribute to line width.
I don't know why they chose T2.
It's completely, it's completely arbitrary
because whether your magnetization is spreading
out in the X, Y plane so you're no longer able to get signal
or your magnetization is returning
to the Z axis you still have line width.
In the case of small molecules,
typically T1 is the predominant relaxation pathway in the case
of very large so strychnine, for example.
In the case of very large molecules like proteins,
T2 is the predominant relaxation mechanism and this depends
in part on how fast the molecules tumble
and how viscous the solvent is.
Other thoughts and questions?
[Inaudible question] It's flat, yeah, it's basically this is,
right, this is perfect coalescence.
[ Pause ]
All right so this is simulated data for sort
of a textbook example and now what I want
to do is show you a real example, show you how
to get a rate out of this and then we're going to translate
that into a free energy activation.
So, okay, so the case that I'll show you which is kind of cool,
this is just one I pulled from my own experience.
[ Pause ]
It's a sort of neat molecule because we're going to see
that there's actually 2 different things going on here.
So, the molecule is diethyl ortho toluamide
and I'll show you the spectrum of it here.
I have another handout for you.
[ Pause ]
Yeah, one of the great things about being
in graduate school if/else a lot of times you get
to observe stuff that's cool and beautiful and relates
to your classes and this just happened
to be something I noted back when I was in graduate school
and it's like, oh, this is cool; I'll keep this as an example.
There's just something I was doing
on the synthetic methods project just working out a method.
So I had my sample of diethyl toluamide,
diethyl ortho toluamide and I started to warm it up.
I noticed, I was curious because there was some broadness
to the peaks here.
So you had your two ethyl peaks, this is your methyl peak.
So these are your CH2s, these are your CH3s,
and I was just curious about what was going on.
So this was a sample in DMSO D6.
DMSO has a very high boiling point so you can heat it
up to a high temperature.
Deuterochloroform boils at I think 66 degrees.
So if you were to try to heat it up to 160 in NMR tube,
the NMR tube would if you're lucky just blow the top off the
tube if you're not lucky exploding the probe.
Either way you'd have a very angry department at UB
because you would trash the NMR spectrometer
and do serious damage.
So, as you warm it up, the CH2s coalesce
and 110 is really perfect
for the coalescence temperature of the CH2.
Now the CH2s are pretty far apart.
The CH2s are at this, this one is at 3.45.
This is a 300 megahertz NMR spectrometer
and the other one is at 3.03.
The methyls are a little closer together
so their Delta nu lines, the separation
of the lines is smaller so they actually will coalescence even
with smaller rotation.
So the further you are apart the faster you have to spin in order
to have the 2 lines coalescence into 1.
If 2 lines are very close together, you only have
to spin it slowly, only have
to have rotations slowly, to get coalescent.
If 2 lines are very far apart,
you have to have rotation very quickly.
So we're already coalesced.
I'll write coalesced here to indicate that it's
in the past tense and here
at 100 degrees we're not yet coalesced.
So somewhere here at about 105 I would say would have be coalesce
if I had bothered to do that experiment,
but let's for a moment focus on these 2 methyls and I want us
to figure out the rate here and then we're going to translate
that rate into an energy.
So for the CH2s the Delta nu lines, the separation
of the lines, is equal to 3.45 minus 3.03 times it's a 300
megahertz spectrometer so that's 126 hertz.
So, now literally it's plug and chug in this equation.
Tau, the lifetime at coalesce.
[ Writing on board ]
Which is 110 degrees for this particular set of resonances,
the lifetime at coalesce is just equal to .450 divided by 126
which is equal to 3.6 times 10 to the negative 3 seconds.
I didn't measure it at 105, but I think it's
about 105 is the coalesce temperature.
So for the CH3s, the CH3s were separated by 60 hertz and so
in that case tau so the lifetime at their coalesce 105 degrees
for them let's say is going to be about 7.5 times 10
to the next 3 seconds.
In other words, about 7.5 milliseconds.
To put it in terms of that other example,
at room temperature rotation
about this amide bond is slow on the NMR timescale.
In other words, it's on the order of let's say seconds
or hundreds of milliseconds so we see 1 ethyl peak,
we see another ethyl peak.
As we warm it up, it rotates faster and faster and faster
as it gets warmer and warmer and warmer.
By 105 degrees it's spinning
around with a lifetime of 7.5 milliseconds.
The methyls being close together have coalesced and we heated
up a little more to 110 degrees the methylenes being further
apart now they have it spinning at 3.6 millisecond lifetime.
The methylenes have coalesced and by the time I heated
up to 150 or 160 the peaks are now relatively sharp and usually
by that point it's pretty hard to get good shims
so we probably would see a quartet and a triplet there
if I could shim the spectrometer better.
[ Pause ]
All right I want to translate this lifetime
into a free energy.
So one of the take home messages here
from this example is below a millisecond let's say is fast
on the NMR time scale and, you know, above 10 milliseconds
or 100 milliseconds is slow on the NMR time scale.
Let's now see how that relates to free energy of activation.
So, what I want to do is translate our K
to Delta G double dagger [phonetic] and one came
from transition state theory you have the Eyring equation
which basically deals with the Boltzmann population
of molecules that are able to cross an energy barrier
and the Eyring equation is that the rate constant is equal
to kappa which is the transmission coefficient
which is generally taken
as 1 times the Boltzmann constant times the temperature
over Planck's constant times E
to the negative Delta G double dagger over RT
where you have the gas constant with temperature.
What this is, the way the Eyring equation is derived is basically
you're setting up an equilibrium between molecules
in the ground state molecules in the transition state
and then assuming that half the molecules go
over the transition state at that point and, of course,
K here at coalesce K is equal to 1 over tau so we're going
to use our 100 degrees K is equal to 1
over tau, 1 over our lifetime.
So what I want to do is figure out our free energy here.
So add 110 degrees Celsius if I now just plug
into this thing I get 1 over 3.6 times 10
to the negative 3 is equal to 1.35 times 10 to the 23rd,
10 to the negative 23rd and at 110 we're at 383 Kelvin
and we divide by Planck's Constant, 6.63 times 10
to the negative 34th and for the sake of taking on our math
for a moment I'll just keep this as E
to the negative Delta G double dagger over RT.
[ Pause ]
All right if I just continue to work through my equation,
I get 3.484 times 10 to the negative 11th equals e
to the negative Delta G double dagger over RT and if I work
through that's Delta G double dagger is equal
to negative 1.987 times 10
to the negative third times 383 times the natural log
of 3.484 times 10 to the negative 11th
and when all is said and done number I get is
18.3 kilocalories.
All right.
So what is that saying?
That's saying coming back to this point I raised
at the beginning of class what's the degree
of double bond character in an amide?
Well, you've got an 18 kilocalorie per mole barrier
to rotation.
If that were a single bond,
you'd have essentially no barrier to rotation.
If that were a pi bond, you could say that would be maybe,
you know, 60, 70 kilocalories per mole.
In other words, like in ethylene the pi bond is 60,
65 kilocalories per mole so I look at that and say oh,
that's about 30%, 25-30%, of a pi bond.
In other words, if I had ethylene, well, if I had,
if I had CIS 2 butene, no matter how hot I heated it
in the NMR spectrometer, I'd never see isomerization
between CIS and trans 2 butene, the energy barrier to rotate
about a real pi bond is so high you just don't get
that from thermal energy, but with this partial pi bond,
I have 18 kilocalorie per mole barrier.
All right.
There's something else that's really cool that's embedded
in this spectrum.
So take a look at the spectrum at 30 degrees Celsius
and you'll notice even at this point one
of our methylenes is a little board.
You see that?
There's actually, this is a cool molecule.
There are actually 2 dynamic processes
that are going on here.
So I figured at the time I was just curious but now, of course,
I'm using this as an example and it's a fun example
because it actually ties
into some cool concepts in stereochemistry.
So anyway I figured I want to cool the same down
and take a look at it.
Now, DSMO freezes just a little below room temperature
so you can't do super high temperature NMR in chloroform,
you can't do super low temperature NMR in DMSO.
There are chlorinated solvents you can use like 1, 1, 2,
2 dichloroethane can up to very high temperatures and down
to very low temperatures, but DSMO is common so I use that
and chloroform is common.
You can use methylene chloride if you needed to go
to a lower temperature.
So I took the NMR spectrum in chloroform
to see what the heck was going on.
It's really beautiful.
So, at room temperature, which happened to be
that day 22 Celsius now we were just not quite almost
at coalesce.
[ Pause ]
And you notice as you cool it down now we're at 10 degrees,
well, my Xerox didn't come out well.
By the time you're at 0 degrees you notice
that CH2 is resolving itself into 2 sets of peaks.
By the time you've gone down to negative 40, you can see these,
these happen to be doublet of quartets and by the time you're
down to negative 40 degrees Celsius you can see
that even the other methylene which started
as a quartet now has a more complex splitting pattern.
So this is really kind of cool.
So there's a second dynamic process
with a coalesce temperature that's just a hair
above 22 in chloroform.
Remember, it's a different solvent
so you have slightly different rate constants.
So we'll say that I didn't measure it exactly but we'll say
that coalesce is probably at about 25
in chloroform, 298 Kelvin.
So, what I want to do now is to play with this process figure
out what's going on and then look at the energies
that are involved and get us calibrated on energy.
[ Pause ]
All right.
So, this system happens to be way cool.
So, because you have these 2 substituents
and the methyl group,
your methyl group is not going to be coplanar.
In other words.
[ Pause ]
You have a situation where the tolumide ring is rotated
out of planarity from the amide group.
They are orthogonal or close to orthogonal to each other.
This is a situation where you have what's called
axial chirality.
Stereochemistry is cool.
It's the same thing you have in allene.
If you have 1, 3 dimethylallene, there are 2 antimeres of it.
So if you have simplistic example you can come up with.
[ Writing on board ]
This molecule is chiral.
All right so we have an equilibrium here of 2,
atropisomeric rotamers.
[ Writing on board ]
And these 2 atropisomeric rotamers are antimeres.
[ Writing on board ]
Which means your CH2s are diastereotropic.
[ Writing on board ]
Now the CH2 that's next
to the carbonyl has very low magnetic anisotropy so you see
that at very low temperature you do see something that's other
than a simple quartet, but this one the 2 protons, the pro R
and the pro S have a high degree of magnetic anisotropy.
The one that's on the same side as the ring.
So we actually have separation there
and in this case the Delta nu as I cooled it
down remember we could see the 2 different lines here
and so here one of these lines is at 3.94 and the other line is
at 3.30 and so the Delta nu lines then is going to now be
for those 2 lines for the methylenes it's going
to be 3.94 minus 3.30 times 300 is 192 hertz.
So our tau, the lifetime at coalesce, is 0.450
and that's divided by 192 is 2.3 times 10
to the negative 3 seconds at I don't know if it's 298 or 295K.
I said, I don't know.
That's close to coalesce.
If it's at 295K, I'll plug in Delta G double dagger is equal
to 13.7 kilocalories per mole.
All right so take a minute to think about this.
We've got 2 processes in this molecule.
One that's almost invisible at room temperature because, well,
that's invisible slightly above room temperature,
which is the rotation about this bond, and the other
that is the rotation about this bond.
The rotation about this bond has an 18 kilocalorie per mole,
18 point what did I saw 3 kilocalorie per mole barrier,
so it doesn't become fast until you heat the thing up hot,
like 110 degrees or 105 degrees.
The rotation about his bond is kind
of medium scale at room temperature.
You cool it down, it becomes slow.
You heat it up it becomes fast and is invisible.
So that's kind of cool.
Now I want to give you 1 caveat because every student,
I had to say student, every new person's first excuse
when they see something in the NMR that they don't understand
when they see 2 sets of peaks is say hinder rotation.
All right I'm going to make a very sweeping generalization.
The hindered rotations that you see are only going
to involve things where you have an SP2 atom connected
to an SP2 atom.
Chances are if it's an SP2 atom connected to an SP3 atom
or an SP3 atom to an SP3 atom it's going to be fast.
So hindered rotation generally only for SP2 to SP2.
So here we have an SP2 hybridized benzene connected
to an SP2 hybridized carbonyl.
When you have some extra stereic hindrance it's slow to rotate.
Here you have an SP2 connected
with partial double bond character,
you have slow rotation.
[Inaudible question] It's pretty darn flat.
It's actually that nitrogen really is SP2.
What I mean specifically is I can think
of no simple bonding situation
where you have SP3 atoms connected, single SP3 atoms,
where anything is slow without some SP2 intervention.
Cyclohexane ring flip where you have 2 sets of eclipsing,
10 kilocalories per mole, which is still fast
on the NMR time scale at room temperature.
Cyclohexane is coupled together.
That one if you cool to negative 80 degrees,
does become, can become slow.
Actually let me use this as a chance, did that sort
of answer your question?
>> Yeah, yeah.
>> Other questions?
>> How did that [inaudible] change?
>> You rotate, you spin so basically the benzene ring is
like this here with a methyl group pointing out
and it spins back and forth, but it has to bang in doing
so the methyl group has to bang past the carbonyl.
There's enough steric hindrance there
that it can't do it rapidly.
[Inaudible question] It's okay.
It's not going to be perfectly perpendicular.
It'll be at about a 60-degree angle and it'll nicely rock back
and forth but to cross to the other atropisomer.
That's where it's hard.
If you want to make a model of it, this is a great one
to use pinmol [phonetic].
You can easily make a model in pinmol
and you'll see how they sit and how they sit and how they bang.
[Inaudible question] It'll put it,
you still have axial chirality as long as you have a barrier.
[Inaudible question] Oh, did I, oh, ah.
Oh. My goodness.
Oh, thank you, yes, no I meant to have this going back, yeah.
There you go.
Okay, yeah, yeah.
Methyl back, methyl forward.
>> Does it matter though actually?
>> They are 2 antimeres.
Yeah, so. [Inaudible question] So, okay,
my thumb is the methyl group.
Methyl out, flip, rotate methyl back.
All right.
Thank you, thank you, thank you.
All right.
I want to show you, the last thing I want
to do is give you 2 take home messages and let me start
with the message and then I'll go
and show you my thought on this.
My thought is the take home message is the NMR time scale,
you know, I like to have simple things in my mind as ways
to keep things, is let's say less than 1 millisecond is fast,
about 1 to 10 milliseconds is intermediate and greater
than 10 milliseconds is slow.
These are obviously sweeping generalizations
because they're going to depend on separation of lines
and they're going to depend on field strength
of the spectrometer, but let me show you my thinking on this.
If we imagine a Delta nu of lines and it's 50 hertz,
I'm going to give us 2 scenarios.
Let's start with a scenario where we're 50 hertz
and let's say what that is in PPM
at 500 I'll just take 500 megahertz because that's sort
of a typical modern spectrometer so that's going to be .1 PPM.
So you see 2 lines at .1 PPM separation and the tau,
the lifetime at coalesce is .009 seconds.
In other words, it's 10 milliseconds
or K is equal to 111 per second.
So in other words, with 2 lines that are close together
if your process is occurring on the order of 1 millisecond,
it's going to be fast.
If it's occurring on the order of 100 milliseconds, it's slow.
If our separation of lines was 500 hertz,
that's pretty far apart.
That's 1 PPM but we saw half a PPM over in that example.
If it's 1 PPM, the lifetime at coalesce would be .009 seconds.
In other words, 111, 1,111 per second.
In other words, if the lines are further apart,
if it's spinning around, you know,
many times per millisecond, it's fast.
If it's spinning around once every 10 milliseconds,
it's slow.
So that's how I calibrate myself.
Let me give you my other calibration that I
like to keep in my head.
So the other calibration I like to keep
in my head is typical NMR energies.
[ Writing on board ]
Is going to be say 10 to 20 kilocalories per mole.
In other words, a process that's 15 kilocalories per mole kind
of teeters between slow and fast at room temperature.
A process that's 20 kilocalories per mole is slow
at room temperature but it's going to be fast
at a very hot temperature .
A process that's 10 kilocalories per mole will be slow
at very low temperature but fast at room temperature
and so let me just show you my thinking on this.
So if I take, if I go ahead and look at 1 over K
and I'm just going to make a little table of 1 over K
in seconds as a function of Delta G double dagger
and temperature and so this is me windowing myself there's your
take home message on top but let me show you my window
in my cell.
So, imagine we consider energies of 10, 15,
20 and 25 kilocalorie per mole Delta G double dagger
and then consider from the,
from the rate equation we consider temperatures
and I'm just going to window at 3 temperatures,
negative 50 degrees C, which is kind of cold, 298K, 25C,
which is kind of room temperature, and 373K,
100C which is kind of hot.
All right then if I simply calculate
from the calculate the rate that applies the lifetime
in seconds is 1 millisecond for a process
with a 10 kilocalorie barrier at negative 50.
That process has a lifetime of 3.5 times 10
to the negative 6 seconds at room temperature
and 9.3 times 10 to the negative 8th seconds at high temperature.
In other words for a 10 kilocalorie barrier process
at negative 50, we teeter between slow and fast.
So that's sort of our intermediate,
our coalesce temperature.
By this point we're fast and by this point we're very fast.
So 10 kilocalorie barrier cyclohexane ring flip slow
and cold.
Okay, if we go to 15,
108 seconds is the lifetime at negative 50.
In other words, it's slow at negative 50,
but that 15 kilocalorie barrier process becomes milliseconds,
16 milliseconds, at room temperature.
In other words, 15 kilocalories becomes intermediate
at room temperature.
That was like that rotation about the benzene bond.
It's fast by the time we're hot, it's 7.9 times 10
to the negative 5th seconds lifetime.
Twenty kilocalories per mole, 8.6 times 10
to the 6 lifetime that's forever.
Seventy-five at room temperature, 75 seconds is slow,
but you get to hot and it's 68 milliseconds and now we get
into the intermediate regime in high temperature.
That was like our amide bond rotation;
18.3 kilocalories we had to heat it up to 110.
Finally by the time we get to 24 kilocalories, 6.9 times 10
to the 11 seconds 3.5 times 10
to the 5th seconds and 57 seconds.
In other words, by the time we have a 25 kilocalorie per mole
barrier even at high temperature, you are still slow.
All right so that's where I window myself
and I say 10 degrees slow at low temperature, 15 degrees,
you know, intermediate at room temperature,
20 degrees intermediate at high temperature.
All right.
Our midterm will be next time.
We have an in-class part on Friday
and then an open book part on Saturday.
So a closed book part on Friday. ------------------------------858acd1bce6f--