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PROFESSOR: OK, so kinetics we're continuing with today.
We talked on Wednesday about first order kinetics and we'll
do a brief review of some of that.
And we're going to talk about second order kinetics.
Today we're going to come back up and talk about chemical
equilibrium, which is something I love to do to
review things that we've talked about before.
And we're going to start in on reaction mechanisms. So, I
thought I would mention, some of you have may have seen the
activity in the Infinite Corridor, but today is world
AIDS day, and today a lot of the compounds that are being
used to treat HIV were actually designed based on
making inhibitors to enzymes.
And so, to design those pharmaceuticals, people had to
understand the reaction mechanism of the enzyme, and
enzymes, of course, are catalysts in the body.
So, knowledge of what medical individuals needed to know to
design these inhibitors to treat HIV are actually a lot
from this unit that we're going to be talking about.
So, we'll be talking about reaction mechanisms, and we're
also going to be talking about enzyme catalysis, which were
key points in being able to come up with some of the
current treatments against HIV.
All right, so just a little review from last Wednesday.
We talked about first order half life.
We talked about first order kinetics, we came up with an
integrated first order rate law, and we also
talked about half life.
And you told me last time that an example of first order half
life is radioactivity, which we're going to be talking
about today.
So, just a little review from last time, you have your first
order integrated rate law, and the half life is defined as
the time it takes for half of the original material to go
away, and half life is abbreviated t 1/2, that's the
symbol for half life, so the time for half of the original
material to go away.
If you plug original material divided by 2 in there, then
the original material a to the o for original, drops out and
you come up with this equation of the natural log of 1/2
equals minus k t 1/2, and k is, of
course, our rate constant.
And so, then we can take the natural log of a 1/2 and we
get a value.
Rearranging that, you get half life equals 0 .
6 9 3 1 over k.
And so you told me last time for this plot for first order
half life, each half life, half of the original
material goes away.
So, one example of a first order half life process is
radioactive decay.
And the reason why this is a first order process is because
the decay of the nucleus is independent of the number of
surrounding nuclei that has been decayed.
So it's independent of the original starting
concentration, and since it's independent, notice that's a
blank in your notes, since it's independent, then that
makes it a first order process.
So, we can apply first order integrated rate laws to
radioactive decay.
So here were some of the equations we had last time.
We had, this is a different expression of the first order
rate law where the material, concentration material of
material a, at some particular time equals the original
concentration, e to the minus k t, where you have your rate
constant and the time that has elapsed, and we also just
talked about first order half lifes with this equation here.
So, we can use those same equations, but often you don't
see it in terms of concentration of a, you
usually see these expressions in terms of either the number
of nuclei or a different a, which is a for activity.
So instead of concentration, we're talking about the number
of nuclei that have decay or capital N.
So, we can write this same expression, but now using N
instead of concentration of a, same thing, number of nuclei
equal the original number of nuclei, e, to the minus k t,
where k is our rate constant, or in this case decay
constant, and t is time.
So, with chemical kinetics, we're usually talking about
the change in concentration of things over time, but with
nuclear kinetics, we're talking about the number of
decay events, the number of nuclei that have decayed.
And so, here with nuclear kinetics, we measure these
events using a Geiger counter.
So this can measure radiation, and I'm going to come around
and just check the room.
And so, the gasket's ionized and then you
hear different clicks.
See if you can hear the clicks as I come around.
So let's just see if we have any problems over here.
Oh, maybe a little bit.
No, this is fine.
There's always a little bit of radioactivity, it's all fine.
So, this is a Geiger counter, which will measure nuclear
kinetics, it will measure radioactivity.
And my lab has this particular one, because we use x-rays in
our experiments.
OK, we'll leave this on low just to check things
out as we go along.
All right, so we do have a term, A, that we talk
about in this unit.
Instead of concentration of A, it's activity.
And so, activity here, sort of the decay rate, is also called
activity, capital A, and so this is equal to the change in
the number of nuclei or our decay constant times the
number of nuclei.
And people will often talk about the activity of
particular radioactive compounds.
So, because activity is proportional to the number of
nuclei, you can also take this expression and write it as
this expression.
So you can have either the number of nuclei equal the
original number of nuclei, e to the k t, or you can do it
in terms of activity -- that the activity at some time is
equal to the original activity, e to the minus k t.
So, all of these equations can be re-written in this way.
So, let's talk a minute about units.
All right, so the activity for units, the new activity is Bq,
Becquerel. and that, actually is named
after a French person.
Henry was his first name, and my French
pronunciation is not very good.
This is the current unit.
It's equal to one radioactive disintegration per second.
The older unit, which you may be familiar with, is called
the Curie, and that is 3 .
7 times 10 to the 10 disintegrations per second.
Does anyone want to guess, the Curie unit, who
that was named after?
STUDENT: Marie Curie?
PROFESSOR: No, it was named after her husband actually,
Pierre Curie.
And I actually always assume, because Marie is actually more
famous than her husband.
But she, Marie Curie won two Nobel prizes, so we shouldn't
feel too sorry for her.
Her husband shared the first Nobel Prize with her in, I
think it was 1903, but then in 1906 he was killed in a road
accident, run over by something that was crossing
the street.
So he did not share the second Nobel Prize, because by the
time that came around about 1911, he had passed away.
So, at first, we had the Curie, but then that turned
out to be a really big number.
And so, when you were talking about sort of safe units for
workers to be exposed to, if they were being exposed to
things 10 to the 10, that really isn't very healthy.
So they wanted to have sort of a much smaller unit.
And so I guess that Marie Curie at that point talked
about how her husband would feel about having the Curie
not being the standard unit, but I guess she was OK with
it, because if we had kept that same unit, then people
would have been using it and it would of had to of been a
really, really small number, because it was sort of picked
to be set up to something that was too large, and she didn't
want her husband's name apparently associated with a
sort of an infinitesimally small quantity of something.
So, the Curie was sort of done away with.
And Henry Becquerel, who was one of the people who
discovered radioactivity and shared that first Nobel Prize,
had the unit named after him.
And I always ask the freshman chemistry class that as they
go through MIT, if they ever discover a unit that is named
after a female scientist, to please come
back and let me know.
This was the one I thought was named after a female
scientist, but as it turns out, it was
actually Pierre Curie.
So, if you hear of any, please let me
know for future reference.
So, the current unit you'll be using is Bq here for
radioactivity.
So, you're not responsible for knowing all the different
types of radioactivity.
When you're working problems, you can always get this
information.
I'll just mention that a number of different kinds of
radioactivity, some involve a mass change, some do not
involve a mass change.
So, alpha decay, this isn't actually in your notes,
there's a reference to where the table is.
You're not responsible for memorizing it, so I didn't put
it in the notes.
An alpha decay is equivalent to a helium 4 nucleus, so you
lose two protons, two neutrons, so
that's a big mass change.
Whereas say a beta decay involves a loss of electrons,
so there's no mass change associated with that.
So just to be aware that there are these differences in
different types of radiation.
There's also really big differences in terms of half
lives of radioactive isotopes, and again, this information
would be given to you on a test or a problem-set, so you
don't have to memorize it.
So, this table is similar to one in your book, and the
point here is just how different half lifes can be.
So the abbreviation a here is year, d is day.
So you see some of these half lifes are in multiple years,
some of them are days, so there are big differences in
terms of the half life of some of these radioactive isotopes.
Some of them stay around for a really, really,
really, really long time.
So, I thought I would share with you a poem about half
lifes today.
And this was written by a former graduate student at
MIT, Mala Radhakrishnan, and she is now a professor at
Wellesley college right here in Wellesley, Massachusetts.
So, her poem entitled "Days of our half lives," is from her
collection of chemistry poetry, "Chemistry for the
Couch Potato." And this particular poem involves the
uranium 238 decay series.
So, here we go.
"Days of our half lives.
My dearest love, I writing you to tell you all
that I've been through.
I've changed my whole identity, but loved, I can not
pretend to be.
When I was uranium 238, you were on my case to start
losing weight.
For 5 billion years I'd hoped and I prayed, and finally I
had an alpha decay.
Two protons, two neutrons went right out the door, and now I
was thorium 234.
But my nucleus was still unfit for your eyes, not positive
enough for it's large size.
But this time my half life was really not very long, because
my will to change was quite strong.
It took just a month, not even a millennium, to beta decay
into protactinium.
But you still rejected me right off the bat,
protactinium, who's heard of that?
So, beta decay, I did much more to become uranium 234.
Myself again, but a new isotope, you still weren't
satisfied, but I still had hope.
Three alpha decays, it was hard, but I stayed on through
thorium, through radium, and then radon.
I thought I would finally please you, my mass was a
healthy 222, but you said, although I like your mass, I
do not want to be with a noble gas.
You had a point, I wasn't reactive, so in order to
please you, I stayed proactive.
A few days later I found you and said, two more alpha
decays and now I am lead.
You shook your head, you were not too keen on my
mass number of 214.
I had a bad experience with that mass before, an unstable
acitone walked right out the door.
So in order to change, I went away, but all I could do was
just beta decay.
My hopes and my dreams started to go under, because beta
decays don't change a mass number.
To bismuth, then polonium, I hoped and I beckoned, my half
life was 164 micro seconds.
And then finally I alpha decayed and then I was lead
with the prize worthy mass of 210.
I've got to admit I was getting quite tired, my
patience with you had nearly expired.
You were more demanding than any I dated, and much of my
energy had already been liberated.
But you still weren't happy, but you had a fix, I really
like the number of 206, So I waited for years until the
day, which began with another beta decay, and then one more,
and finally in the end I alpha-ed to
lead 206, my friend.
To change any further I wouldn't be able, no longer
active, but happily stable.
It took me billions of years to do, and look how I've
changed and all just for you.
And wait, what did you say?
You've gotten so old that I'd rather be with a
young lass of gold?
Well, I give up, we're through, my pumpkin.
Shouldn't all my effort be counting for something?
Well, you won't be able to rule me any more, because I'm
leaving you, not for one atom, but four.
That's right, when you were away defusing, I met some
chlorines that I found quite amusing.
So we're going to form lead c l 4, and you won't be hearing
from me any more.
See, over the years I've grown quite wise, I've learned that
love is about compromise.
You still have half of your half lives to live, so now you
go out there, it's your turn to give."
And that is "The days of our half lives." So, Mala takes
great effort to make sure that all her poetry not only
rhymes, but it is chemically correct.
So, it's a good way to review material to read "Chemistry
from the Couch Potato."
All right, so let's do an example now and think about
how things will change over time.
So we have an example, we want to know the original activity,
and the activity after 17 years
of a sample of plutonium.
So let's take a look at how we'll do this problem.
So first, given the information up there, the
first thing we want to do is find the
original number of nuclei.
So first, capital N o, the original number of nuclei.
So, we're given information about grams, so we have 0 .
5 grams. And now if we want to know the number of nuclei,
what's the first thing I have to do?
Convert from grams to what?
STUDENT: Moles.
PROFESSOR: Moles, right.
And here we want to use the molecular mass that's given to
us in the form of that isotope.
So here, we are given information about 239, and so
that's the number we want to use in our conversions.
So we can convert that over, but that's going to give us
moles, so how do we go from moles to molecules?
Avagadro's number, 6.022 times 10 to the 23.
This time we're going to talk about it in terms of nuclei
per mole, and so that's going to give us 1.3 times
10 to the 21 nuclei.
OK, so now we know the original number of nuclei.
The next thing we're going to want to do is find k.
And k is our rate constant for decay or our decay constant.
So what do we know about k for a first order process?
We know the equation for what?
For first order half life, right.
So that's o .
6 9 3 1 over t 1/2.
And in this problem we were given, the half life, and
often you will be given the half life or you can look it
up, and so we can put that in, so we have 0 .
6 9 3 1 over 7 .
6 times 10 to the 11 seconds.
And we can calculate our constant, which is 9 .
1 times 10 to the minus 13 per second.
So now we were asked to find the original activity and the
activity after 17 years.
So, first we'll find the original activity, and the
original activity is going to be equal to our rate constant
times our original number of nuclei.
So we've just solved for both of these, so we can plug these
in, so we had 9 .
1 times 10 the minus 13 per second times the number of
nuclei, 1 .
3 times 10 to the 21 nuclei, equals 1 .
2 times 10 to the 9, and what are the units here?
It's just like a hum, it's hard to understand.
STUDENT: Nuclei per second.
PROFESSOR: Nuclei per second, which is the same as what?
That's equal to something else.
Yup, so that's the same as the Becquerel or the Bq, so it's
defined as nuclei per second, or number of disintegrations
per second.
All right, so let's do the last one.
OK, so now after 17 years, so now we can say that the
activity at some time is equal to the original activity, e to
the minus k t, and we can put in the activity that we just
found, which is 1 .
2 times 10 to the 9 Bq, times e to the minus k, which is 9 .
1 times 10 to the minus 13 per second times 17 years, which
in seconds is 5 .
4 times 10 to the 8 second.
So here, we want to make sure that our units are going to
cancel, and this is where people often run into
problems. They'll plug in 17 years, and then a rate
constant, which was calculated in seconds, and things will
not cancel appropriately.
So make sure that you get your units consistent so that your
seconds are going to cancel.
And so this term, if we do the math out here with the number
significant figures, we find that that equals 1.2
times 10 to 9 Bq.
That term is insignificant in our problem.
So, the original radioactivity and the activity after 17
years are the same in terms of the significant figures.
And so, I choose this problem to emphasize a problem that we
have, and that is radioactive waste.
It takes a very long time for some compounds to decay.
And so you have to think about storing radioactive waste, and
think about a container that will outlast
that radioactive waste.
And how do you know that the container is going to outlast
the radioactive waste.
You can't really do an experiment because the time
involved in doing the experiment, anyone who designs
the container won't be alive by the time you're concerned
about whether the container is going to be stable or not.
So taking radioactivity is an issue.
You heard some in the presidential campaign about
whether both candidates believe in nuclear energy or
not, and I think that both of them said, it needs to be
considered, we need to have everything on the table.
If we're going to have a real uniform energy policy, we need
to think about everything.
So, issues of radioactive waste and how to handle
radioactivity safely are going to come back as being current,
important topics.
And so these may be topics that you will, in your
lifetime, have to deal with, either as a scientist trying
to come up with new technologies, or as a citizen
deciding whether having a radioactive plant in your
hometown is a good idea or not.
A lot of people are happy about nuclear energy, as long
as the power plants are nowhere located near them.
But, these are things that you'll have to face, and you
probably will be voting on this in the future, if not
dealing with it directly.
So that's how you do a problem in this.
All right, so let's talk about a medical use of
radioactivity.
Radioactivity can definitely be our friend, as well as
something to be concerned about.
And I think I mentioned this in the first day of class, one
of the ways that the Chemistry Department has moved to being
the number one ranked Chemistry Department of U.S.
News and World Report over the years, is a little extra money
that came in from the work of -- a patent from Professor
Alan Davidson that we were able to do some pretty
exciting things with that money over the years.
So I always like to mention all the great money-making
discoveries that occurred using 511-1 material, and this
is another example.
So, he used an isotope of technetium, and it's being
used organ scanning, bone scans, it's one of the leading
ones for heart imaging.
It's also been used recently in breast cancer.
It's estimated 7 million uses annually in the U.S. And so,
this was patented as cardiolite, and it's really
just very simple chemistry.
So you're using a d block metal, an isotope of a d block
metal, which has your exciting d orbitals.
And what did he do, he made a coordination complex with that
metal, an isotope of it, and he found ligands, cyanide
ligands, those are pretty common ligands.
You've seen a lot of coordination complexes with
cyanide ligands, and he tried different ligands to get the
desired properties of stability and solubility, and
that's all it was.
So he used some knowledge of radioactivity, knowledge of
inorganic chemistry -- he was an inorganic chemist, he's
retired now.
And simple coordination chemistry, and made an
enormous amount of money for MIT, and particularly, the
Chemistry Department, and also, this has
saved a lot of lives.
So, imaging is something that chemists
do a lot of, actually.
Not just imaging for cancer or imaging of organs, but also
imaging of live cells to try to understand how the cell
works when it's healthy.
And so recently, Professor Alice Ting, in the Chemistry
Department, received an NIH pioneer award, NIH is National
Institutes of Health, and started giving these pioneer
awards for people coming up with very innovative ideas,
the kind of innovative ideas that most people would not
want to fund, because there's a good chance it might not
work, but if it did, it would be spectacular.
So she received one of these awards for trying to develop
technology to image protein-protein interactions
in living cells, which is something that people would
really, really love to be able to do.
And so she is involved in developing technology.
So developing of imaging tools is something that a lot of
chemists do, it's a very popular area in chemistry.
And if it's something that you're interested in, there's
definitely a lot of people around that you could think
about working with for a UROP position.
OK, so that is first order.
And now let's go on and talk about second order
integrated rate laws.
And we're going to have a little derivation for you, I
always like to warn people that it's coming, because all
of a sudden equations are coming in and out, and you
just want to know where these equations are coming from.
So, as we talked about last time, this is an expression
for rate law.
You have your rate constant, your concentration of
something, a, and it's raised to a coefficient, and here
that coefficient is 2, indicating it's a
second order process.
So if there's nothing up there, that's 1.
And then 2, and again, the order of the reaction can be
positive, negative, it can be integers, it can be fractions.
But this is second order, so we have 2.
Now, as we did with the first order expression, we're going
to separate our concentration terms and our time terms. So
we're going to bring our concentration term over to one
side, another concentration term here, and we're going to
have our rate constant and our time term on the other side.
And now we're going to integrate, because it is an
integrated rate law.
So we can integrate from the original concentration of a to
the concentration of a at some time, t, and then we'll also
integrate from zero time to that time, t,
on the other side.
Now, I'm going to take this expression and just bring it
up to the top of the page, so that's the exact same
expression, nothing has happened.
And now we're going to solve that integral.
So we can solve that integral, and if you want to look at
these -- the back of your textbook has all of these
conversions, if you want to look at them.
So, we're going to solve that integral, now we have minus
parentheses 1 over the concentration of a at some
time, t, minus 1 over the original concentration
equals minus k t.
We can get rid of some of these minus signs.
So we're going to bring the concentration of time, t, over
on this side, we have our k t, and now we have this other
term, the original concentration term is on the
other side, and this is expressed in a certain way
that gives you the equation for a straight line.
And again, kinetics, you need experimental data for
kinetics, and so when you measure your data, you plot
your data, and so there's a lot of equations for straight
lines that you have, because it's all about trying to plot
data, and figure out what the order is experimentally.
So, here's an equation for a straight line, and we can plot
this, and you can tell me what the intercept of this line is.
OK, let's just take 10 more seconds.
Yup, so all of you know how to analyze the equation for a
straight line.
So, here we have 1 over the original concentration, and
then our slope is equal to what here? k.
So, you can plot your data as your concentration of a
changes with time.
You can plot the data, and if the data, if it's plotted as 1
over the concentration of a versus time and it gives you a
straight line, that's consistent with it being a
second order process.
So, in terms of second order half life, we talked about
first order half life, and for any half life, it's just the
time it takes for half of the original material to go away.
So we can rewrite this and take a to the t, and
substitute in the original concentration divided by 2,
and then we can have t have a special name t 1/2, so that's
the half life.
And now we can just simplify this expression.
We can bring the 2 up here, and now we can combine our
concentration terms on the side.
So we take 2, we bring this over, minus 1.
And that simplifies 1 over the concentration of the original
material here.
And now we can solve for it in terms of the half life.
So the half life for a second order process equals 1 over k
times the original concentration of the material.
So, a second order half life depends on the starting
concentration.
So that's very different from a first order half life
process where concentration term cancels out entirely.
So for a first order process, the concentration of the
original material does not affect the half life, or for
radioactive decay, the original number of nuclei --
it's independent of how many nuclei were
around at the time.
But for second order process, the starting concentration
does matter.
So again, chemistry is experimental.
And so what you would be doing in a lab, you would be trying
to figure out what the order of the reaction is, and so you
could try out your data.
You say I don't know if it's first or second order, so for
a first order plot, you're going to be plotting the
natural log of your concentrations versus time,
and if second order, you plot 1 over the concentration
versus time.
And so you could plot your data and see that oh, look at
this, it fits a straight line really well if I plot it as 1
over the concentration.
If I plot it as the natural log versus time, the data
doesn't fit a straight line at all.
So this is not a first order process, this is much more
consistent with a second order process.
So again, figuring out where something is first order or
second order is done experimentally.
All right.
Now we're going to talk about kinetics
and equilibrium constants.
So, I always get very excited, as you know, when we come back
to equilibrium constants, so am always very happy at this
time in the course when we can relate kinetics and
equilibrium constants.
So, at equilibrium, another way to think about what's
happening at equilibrium, is that the rate of the forward
reaction and the rate of the reverse reaction are equal to
each other.
So, we can now talk about big letter K, which is our
equilibrium constant again, and our little letter k's,
which our rate constant.
So the equilibrium constant for a chemical reaction a plus
b equals c plus d is going to be equal to what, what do I
put on the top?
Concentration of? c, and concentration of d, right.
So our products, and at the bottom we put our
concentration of our reactants, or a and b.
Now, we can also think about this reaction in terms of
little rate constants.
So we have small letter k 1 on top, and small letter k to the
minus 1 on the bottom.
So, the forward reaction, the rate of forward reaction is
going to be equal to k 1 times the concentration of a and the
concentration of b.
And on the bottom, our rate is going to be equal to the
little rate constant, so for the reverse reaction it's the
reverse rate constant, k minus 1, and in the reverse
direction, our reactants are the products for the forward
direction, or c and d.
So, here we have these rates, and at equilibrium, those
rates are going to be equal.
So at equilibrium, little k 1, a times b is going to be equal
to little k minus 1 times c times d.
And at equilibrium we had c d over a b is equal to then k 1
over k minus 1, so if we just rearrange this expression and
move the rate constants to one side and concentration terms
to the other side, this expression is the same as this
expression, and we also know what this expression is equal
to, which is our big K. So, therefore, our equilibrium
constant equals the rate constant for the forward
reaction over the rate constant
for the reverse direction.
And so here is an expression that compares equilibrium
constants with rate constants.
So now, let's think about what is true about this.
So our equilibrium constant, then, is the ratio or the
forward rate over the reverse rate for
these elementary reactions.
And if we think about rate constants in kinetics terms,
if k is greater than 1, if there are more products than
reactants at equilibrium, what's true about k
1 and k minus 1?
Is k 1 greater than or less than k minus 1?
Right.
So the forward rate constant is greater than the reverse
rate constant.
And if K, big equilibrium constant, K is less than 1, if
at equilibrium there are more reactants than products, what
is true about this relationship?
It would be less than.
So, you can think about equilibrium constants now in
terms of rate constants, which we'll be doing a lot on
Wednesday, too.
All right.
So let me introduce you to a couple of more terms in the
last few minutes.
So reactions don't usually occur in one step, but occur
in a series of steps.
Each step is called an elementary reaction.
So, the overall reaction, the order and the rate law, can be
derived from the stoichiometry for an overall reaction, you
can't use the stoichiometry, but for an elementary
reaction you can.
So, for an elementary reaction, say one step in the
reaction mechanism, that step occurs exactly as written so
you can use stoichiometry.
And that's going to be handy in coming up with mechanisms.
So, let's just look at one example very briefly.
So here we have the decomposition of ozone, which
is another environmental issue that you will be faced with in
your lifetime.
So this is the overall reaction, and you can't use
the stoiciomery to figure out the order of the reaction, but
if you divide it up into elementary reaction steps,
then you can use the stoiciomery to write the rate
law for each step of that reaction.
So, the first step here is a unimolecular step.
You have one thing going to two things, and molecularity
is the number of reactant molecules the come together to
form product.
So, unimolecular you just have one thing that's forming some
kind of product.
What do you think it's called if you have two
things forming a product?
Bimolecular.
These are good little one or two point questions on a test,
they're not very hard, they should be pretty
easy to think about.
So, we have unimolecular, examples would be some kind of
decomposition or radioactive decay.
Bimolecular, two reactants coming
together to form products.
And termolecular is three reactants coming together to
form a product, and that's rare.
And you can remember that it's rare if you think about how
you would hold three tennis balls in your hand and have
them all come together at the same time to form product,
that is a difficult thing to do.
Two things coming together is easy,
bimolecular is very common.
Termolecular not very common, that they'd all come together
at the same time to form a product.
All right, so we can write rate laws for each step here.
For the first step here, the rate would be equal to a k --
I don't have the k written up here, but there's always going
to be a little k over the arrow.
So, k times the reactant would be here.
For bimolecular, again, assume a k over there.
What's that rate going to be equal to?
OK, let's just take 10 more seconds since
class is almost over.
Yup, so it's this one right down here.
So we have the rate is equal to k
times these two reactants.
You can sum up the steps and get the overall reactants.
Notice that o is an intermediate, it's formed
here, decayed here.
It goes away, and so o doesn't appear in the overall
expression.
So we're going to talk a lot about reaction
intermediates next time.
And also, remember that you can't prove reaction
mechanisms to be correct, they're just consistent with
the data that you have.