nm0690: okay looking at the thermodynamic stability of organometallic compounds 
er what i'd like to do now is turn to how you actually make the things in other 
words the f-, synthesis now the first thing er to observe before we start 
actually on methods of synthesis is that some of these compounds are extremely 
sensitive to air and water and the origin of that sensitivity is the metal-
carbon bond or if they're very ionic of course the carbanion if you have a 
metal directly bonded to carbon which is w-, our definition of an 
organometallic compound you either have a covalent bond which must be the 
source of reactivity or you have something approaching M-plus C-minus which of 
course is an ionic er arrangement now i suppose it's true to say that no 
organometallic compound is totally ionic i mean there's no organometallic 
equivalent should we say of caesium fluoride but some organometallics let's say 
er alkyls of you know caesium and ribidium the very electropositive metals and 
even of 
sodium and lithium are very substantially polar they have a large polar 
component so if you regard them as ionic of course the carbanion is the source 
of reactivity carbanions if even if you have them in organic chemistry are 
incredibly reactive they would react with oxygen with water with all sorts of 
things if they're covalent then it's the metal-carbon bond which is the source 
of the reactivity now it makes it much simpler when discussing the synthesis if 
we just consider what precautions are necessary so if materials are sensitive 
to let's say air and it's really by that we mean oxygen of course because in 
the air there's one-fifth oxygen roughly the remainder is mainly nitrogen which 
of course hardly reacts with anything but i'll come to the one thing it does 
react with a little bit of C-O-two which will react with very reactive ones and 
water so if they're sensitive to air and water and let's remember by the way 
that the air is always wet when you may think on a 
bright sunny warm summer's day it's pretty dry but the air certainly in this 
climate always contains substantial amounts of water so if something is 
sensitive to water you've got to keep out air because otherwise it's just going 
to go off now how do you do that what does that mean it means that you must 
have an atmosphere above your apparatus of dry nitrogen is the common one it's 
cheap it's available fairly pure and sometimes much easier to use but much more 
expensive you can use argon experimentally argon is much easier to use for the 
very simple reason it's much heavier than air nitrogen of course is roughly the 
same density as air and so if you want to keep something in a in a flask that 
contains nitrogen you have a problem you must keep it stoppered or under vacuum 
or under a flow of nitrogen at all times if you've got something that's air-
sensitive and you've got a covered with argon of course for quite some minutes 
argon will stay in the flask 
because it's much heavier than air so it's much easier to work with argon but 
as i say it's much dearer and so experimentally you know you generally only use 
argon for very precious compounds or extremely sensitive material er and this 
also means then that all your solvents must be degassed the air removed which 
is normally done under vacuum and of course because of the water they must be 
dry and normally if you're working with organometallic compounds for their own 
sake and you're not using them just as tests or as reagents but you want to 
study them they're normally dried over sodium potassium alloy and the reason 
for that is that it's much more reactive than just using sodium and it's much 
more efficient at drying them and so the other thing is it's so that's your 
atmosphere those are your solvents and three the apparatus must be either under 
vacuum or if that's not possible it must be under a flow of inert gas that's to 
say nitrogen or argon so when i say for 
example that a particular kind of compound is sensitive to air and water or is 
sensitive to one of them those are the conditions that you need to use okay so 
how what methods do we use now th-, obviously if you want to make a metal 
derivative of an organic compound the easiest way in principle by far is to 
take metal and some organic derivative of the compound that you want and that's 
normally a halogen so for example let's let's just take one the first method 
from the metal plus an organic compound the classic metals for which this works 
are lithium and magnesium those are the best known but of course it will work 
for the heavier alkali metals sodium potassium s-, rubidium and caesium and so 
on and i'll give you details of which metals are useful later on but those are 
the classic examples so for example you could take lithium metal plus what 
should we say you could take er ethyl bromide and what you would get out of 
that assuming that you used the conditions above you had a 
dry atmosphere dry solvents and all the rest of it you would get er lithium 
bromide plus lithium ethyl and i'm just using E-T as an abbreviation for C-two-
H-five the ethyl group lithium ethyl is very sensitive to oxygen and to water 
and i should explain one other thing which perhaps i should have mentioned 
earlier but just let me come to it now what is it that lithium ethyl reacts 
with in water it is of course the proton if you take lithium ethyl with water 
what you will get is lithium hydroxide and ethane that's what happens that's 
the decomposition reaction with water and how does it work of course it works 
because the water gives you a very low concentration of protons those protons 
cleave the lithium to carbon bond in lithium ethyl now that means that in 
general any solvent which will give you a concentration of protons is going to 
be bad news for these organometallic compounds and so when you do this reaction 
in a solvent the solvent must not provide 
any concentration of protons otherwise you'll simply decompose the product as 
it's formed so we must use what we call an aprotic solvent just one aprotic 
without protons okay of course it doesn't mean that there isn't hydrogen in the 
solvent it just means that this solvent does not produce any appreciable 
concentration of the proton in solution so for example you could use diethyl 
ether you can use T-H-F ethers are quite common and they're v-, they're very 
good in principle you could use an alkane you could use hexane but i'm going to 
put that in brackets because although you can get the reactions to go in hexane 
under industrial conditions with high temperatures it's extremely difficult to 
get that reaction to work in the laboratory at normal sort of temperatures so 
in principle you can use hexane but you in the lab you'd be better off sticking 
with some kind of ethereal solvent er a protic solvent which i should emphasize 
is no use here at all so i'll put them in 
brackets but protic solvents would be things you know like ethanol er methanol 
any acid of course is highly protic by these standards and any amine would be 
useless we normally think of amines as being basic because of the lone pair of 
electrons on the nitrogen but an amine can produce a very low concentration of 
protons it's also acidic in terms of these highly sensitive metal to carbon 
bonds so you wouldn't you couldn't use an amine and of course equally you 
couldn't use a thiol no-, you probably wouldn't want to but all of those that 
i've put in the square brackets there provide tiny concentrations of protons in 
solution and they will simply decompose your compounds so you cannot use those 
er materials the other classic reaction which you may have done in the lab i 
don't know but you will certainly know about where you can use a metal is 
magnesium where you get a Grignard reagent it's a slightly different reaction 
of course because normally with a Grignard 
reagent you simply start with ordinary magnesium and the halide and you don't 
of course get the formation of magnesium halide it's an insertion reaction of 
the metal into the carbon-halogen bond so for example let's take just a very 
well known example if you take magnesium plus methyl iodide you do this let's 
say under nitrogen and you will do it we'll say just to give you an example in 
diethyl ether as a typical aprotic solvent you will produce methyl magnesium 
iodide you don't get any magnesium iodide you just get the if you like the 
addition reaction the insertion of magnesium the metal between carbon and 
halogen and i write it of course with the lines here just to emphasize that 
this is a genuine organometallic compound it contains a bond between carbon and 
magnesium later on we'll discuss what the actual structures of these compounds 
are because although i've written lithium ethyl up there as though it was 
simply one lithium metal bonded to one ethyl group 
its structure actually turns out to be more complicated than that for reasons 
that i will explain in a few minutes later on but the point about it is they 
contain direct bonds between lithium and carbon and so they are organometallic 
compounds now er oh no that's all right that's okay we will find that both of 
these reagents that's to say oh we've lost them but both the Grignard reagents 
and the lithium reagents are themselves incredibly useful for generating other 
organometallic compounds so just bear them in mind we'll come back to them many 
times all right now so that brings us up to the next kind of general reaction 
for making organometallic compounds and these we can call exchange reactions 
now what do i mean by exchange reactions er the first kind i'll talk about is 
where you exchange between two different metals so we'll call this metal-metal 
exchange and i think you'll see why it's so called in a minute let's just 
imagine for a moment that you want to make vinyl 
lithium the compound C-H-two double-bond C-H lithium let's assume that you want 
to make that from what we've said so far what you might well try would be to 
take vinyl chloride or vinyl bromide and metallic lithium it sounds like a good 
bet there is a problem with that though and the problem is simply it doesn't 
work lithium does not react with vinyl chloride or vinyl bromide there's a 
second problem of course which in the laboratory you normally overcome which is 
that vinyl chloride and vinyl bromide are both highly carcinogenic but assuming 
that you've got the c-, proper apparatus for handling dangerous compounds even 
then the thing the the chemistry doesn't work if you had C-H-two-double-bond-C-
H-B-R you find there's no reaction with lithium and that's of course because 
you're trying if you like you've got an S-P-two carbon here and you're trying 
effectively to do a substitution of the halogen by the metal and as you know 
it's not easy to do substitutions at S-P-two 
carbon so you've got to have another method and the method that works very well 
is to take the vinyl derivative of another er metal so in this case you could 
take tetravinyl tin C-H-two-C-H four times tin i'll tell you how you make that 
later but you can make that without too much trouble there are four vinyl 
groups of course 'cause tin is in group fourteen it's four valence electrons so 
this is a perfectly normal tin four compound and if you treat that with four 
molecules of phenyl lithium just write P-H to abbreviate again for C-six-H-five 
the phenyl ring then the reaction works very nicely you get four molecules of 
vinyl lithium and the other product is tin tetraphenyl so you've exchanged 
organic groups between two metals the metal lithium and the metal tin hence the 
name metal-metal exchange of course you still have to carry this reaction out 
under all these conditions of an aprotic solvent and in the absence of air and 
with dry solvent because now the lithium starting material and the lithium 
product are both sensitive to water and oxygen so you have a double problem 
there 
it turns out that t-, organo-tin compounds are not sensitive either to oxygen 
or to water but the lithium derivative's what you're interested in and if you 
don't take the precautions you're going to be in trouble why do i choose 
incidentally phenyl lithium i mean why don't i choose butyl lithium the answer 
is it's purely experimental butyl lithium would work quite well the problem 
then would be i would be left with butyl tin which is a nice soluble 
organometallic reagent just like vinyl lithium but tetraphenyl tin is almost 
totally insoluble in all the solvents that you can choose certainly in ethers 
so it just precipitates out and it leaves you with a clean product you don't 
have to worry too much about purifying the material now there are many examples 
of this kind of exchange and we'll come across some of these later on but if i 
say that a compound is made by metal-metal exchange that is the kind of er 
reaction we're talking about okay so the next kind of exchange reaction is 
metal-hydrogen 
nf1340: here's the attendance list 
nm0690: oh right thanks okay i've been presented with the usual attendance list 
er could you oblige the powers that be by signing it and passing it on for me 
thanks okay now then so we have metal-hydrogen metal-hydrogen is a very 
important reaction and it's really based on the fact that the organometallic 
compounds let's say of lithium are very sensitive as we've already seen to 
anything that can give you protons so you can use that to make organometallic 
derivatives if the hydrogen that you want to replace has got some little bit of 
acidity about it so let's take a very typical example let's take let's take a 
classic example first and then i'll go on and show you that it's actually quite 
extensive er first of all oh i've broken the board here but oh well let's take 
an 
example where you've got a fairly strong carbon acid now carbon-hydrogen 
compounds are not generally acidic so you need some special kind of structural 
feature to make a carbon acid which is what you need one such compound is 
cyclopentadiene which is a five-membered ring of course with two double bonds 
in it now cyclopentadiene is peculiar it has hydrogens of course all the way 
round but two hydrogens here and i-, its unusual character is because it's 
rather acidic for a hydrogen compound it will give you the anion plus hydrogen 
and let me just write that as hydrogen S for solvated because as you know H-
plus doesn't exist in solution but i can't write water obviously because we 
can't do this reaction i-, in water because cyclopentadiene isn't soluble in 
water and in any case the anion would be decomposed by water so this will be in 
some other solvent that can sustain the ionization maybe T-H-F now why does 
this happen well the answer is very simple in the double bond 
of course i've got two pi electrons and i've got two of them so that is four pi 
electrons if i take one of these hydrogen bonds to carbon it obviously contains 
two electrons if i remove the proton over here of course i take no electrons 
with me because the proton has no electrons i leave those two electrons behind 
so that gives me six electrons and of course as you know from benzene chemistry 
if i've got six pi electrons i have an aromatic sextet and just as with benzene 
that's very stabilizing so for cyclopentadiene the anion is stabilized by 
having six pi electrons just as you find in benzene and therefore the reaction 
goes much more than it would with a typical carbon-hydrogen compound with any 
alkane now when i say it goes much more i don't mean this is like you know 
acetic acid i mean that th-, this is a tiny tiny concentration but it's dynamic 
you you can look it up in the textbooks you can find the figures i think for 
example if i remember let me just 
give you a guess i think the P-K-A for this molecule is something like fifteen 
that means that the concentration the molar concentration of hydrogen ions will 
be ten-to-the-minus fifteen water is ten-to-the-minus-seven so you know this is 
another ten to-the- eight less acidic than water but you will find that that's 
useful enough if you take cyclopentadiene er i'll write it out in full 'cause 
then you can see what's happening if you take cyclopentadiene with sodium what 
you get is er sodium cyclopentadienide [cough] and of course we've lost a 
hydrogen and i'll just write that as a half-H-two because you know otherwise 
you've got to double everything up so what i've done is i started with two 
hydrogens here and i've replaced one of them with sodium now of course this is 
an ionic compound substantially because the reason that it goes in the first 
place is the stability of the cyclopentadienide anion and so the product is er 
a salt if you like but it's in exchange of sodium 
for hydrogen because if you write it out in a formula terms we start out with C-
five-H-six we react it with sodium and the product is sodium C-five-H-five and 
so i've e-, substituted or exchanged one hydrogen for a sodium so it's a metal -
hydrogen exchange that reaction in the textbooks is sometimes referred to 
simply as metallation some textbooks call it metallation i think that's a 
confusing term because any reaction which puts a metal in place of any other 
group be it a halogen or another metal is if you like a metallation you're 
adding metal to the compound but you will find that some textbooks use the term 
metallation so if you want to look it up you might have to look it up under the 
term metallation er let me give you an example where it's much more evidently 
an exchange suppose we take phenyl sodium and we take toluene now toluene of 
course is an even weaker acid I-E it has an even lower concentration of protons 
than cyclopentadiene but still it is a slightly 
stabilized anion C-H-two-minus negative charge can again interact with the 
aromatic ring so this is a i don't know what the P-K-A for this is it must be 
minuscule but nevertheless it's enough in the presence of a sodium 
organometallic compound the reaction goes you get exchange of one of the acidic 
hydrogens so that's benzyle sodium and you exchange that hydrogen for the 
sodium so the other product of course is if you like P-H-H or benzene so that's 
a typical metal-hydrogen exchange so those are the the main important type of 
exchange reactions let's er finish off by looking at another very important 
class which are insertion reactions now of course i suppose in the number of 
compounds made and the number of times it's been used the synthesis of a 
Grignard reagent is probably the most widely-used insertion reaction in 
chemistry so we'll put that down first but many textbooks don't actually 
include it as an insertion reaction but of course formally 
it is so if i just put down the general reaction R-X plus M-G gives you R-M-G-X 
that is obviously in a formal sense an insertion reaction so we should include 
it a much more typical kind of insertion reaction is a reaction which involves 
insertion into a metal carbon bond that's already been formed so let's take for 
example butyl lithium which i'll write like that because that's the bond into 
which we're going to insert something now if you insert something we can use 
the example above the magnesium to show the the er the idea if i insert 
magnesium into the carbon-halogen bond of this R-X group the magnesium has got 
to be able to form two bonds because i've pushed it in between two elements and 
if the whole thing is going to stay together the magnesium has got to bond to 
the R and it's got to bond to the X another way of looking at this of course if 
you want to is to say that in terms of the metal its oxidation state increases 
by two oxidation state of course 
simply tells you the number of electrons that the metal is using to form bonds 
in magnesium metal its oxidation state is zero it's not using any electrons to 
form any ordinary covalent bonds when it goes to the Grignard reagent it's 
forming two bonds so it goes from magnesium nought to magnesium two and that 
shows you that the the thing that's inserted has got to be able to provide two 
electrons so that you can form two bonds in general that means that you've got 
to have some kind of unsaturated species so for example suppose i was to take 
let's say an acetylene or an alkyne if you prefer so we'll take R- C -triple-
bond-C- R and it's the pi bonds of the triple bond that can give rise to the 
new bonds this can insert into the carbon lithium bond here and what you would 
get would be R- C- C- R butyl and lithium and the insertion you will notice i 
chose a triple bond for a very s-, particular reason and that is simply that i 
can show with a triple bond the 
geometry of of the insertion if i'd started with a double bond i would get an 
alkane here and we'd have free rotation and we couldn't tell anything about it 
but starting with a triple bond i'm left with a double bond and so of course i 
can have the butyl and the lithium going cis about the double bond which is one 
isomer or i could have butyl here and lithium up here the trans arrangement 
which would also be stable different isomer but in fact they both go on the 
same side and so the mechanism of that insertion is always a cis one so that is 
an insertion that's insertion into a metal to carbon bond it's also possible to 
have insertion into a metal to hydrogen bond have you in organic chemistry done 
anything about hydroboration you've heard of it have you okay hydroboro-, m-, 
boron isn't really a metal so in a sense it's not relevant to organometallic 
chemistry but this is exactly the same kind of reaction 
so f-, let me just go over some familiar ground first and then i'll show you 
how it can be used to form er metal-carbon bonds if we start out with an olefin 
in this case doesn't matter what olefin i'll just ha-, we'll just have some 
general olefin here and i've got let's say borane now you know of course that 
we can't actually start with borane itself because borane exists as a dimer B-
two-H-six and that's rather inconvenient er so let's say we we'll start out 
with the complex with T-H-F because the T-H-F complex is stable it doesn't 
dimerize because you donate a pair of electrons from the oxygen of the T-H-F to 
the boron and then the monomer is stable and that simplifies the r-, 
stoichiometry of the reaction now what happens here is we're going to insert 
into the B-H bond so the first thing of course you will get will be er well 
we'll keep the T-H-F initially you'll get T-H-F H-two B now into the boron-
hydrogen bond we're going to insert our olefin so we will have C C H that's 
hydroboration of course you still have in this molecule here two other boron-
hydrogen bonds and so if you've got enough olefin the reaction will proceed in 
the presence of excess olefin the reaction will proceed until you get boron C-C 
depends what these groups are H three times perhaps to be mathematically 
accurate i should put the hydrogen i guess inside the the bracket okay so 
that's an example you've come across before an unsaturated species is inserted 
into the boron-hydrogen bond and it's almost an organometallic example but 
there are many other there are many other metals well there are many metals 
which which undergo similar reactions and perhaps the most important of those 
is aluminium because the reaction between aluminium-hydrogen compounds and 
unsaturated species is used in industry a great deal to 
make aluminium alkyls so for example if we just take a typical example er let's 
say we start out with dibutyl aluminium hydride so that's a metal now aluminium 
hydrogen bond and we'll take ethene just to make a nice simple example the 
ethene inserts into the aluminium-hydrogen bond and the product that you get is 
dibutyl aluminium ethyl i give you that example as i say it's used to make alu-,
organo-aluminium compounds and you probably know that the chemical industry 
makes thousands of tons of aluminium alkyls every year because they're used in 
the catalysts that produce polythene polypropylene and a whole range of 
polyolefin catalysts er er are derived from organo-aluminium compounds and it's 
quite interesting that until you've had a bit of practice it would be extremely 
er difficult to carry out this reaction in the laboratory because a material 
like this not only reacts with oxygen if i was to have a bottle with this and 
take th-, the stopper out of it it 
simply bursts into flames it reacts with oxygen incredibly readily they're 
pyrophoric materials take the stopper out and i'd just have a sheet of flame 
from here to the ceiling i have actually not in one of these lectures but i 
have actually done it in a demonstration lecture and by gosh it i mean it it 
certainly frightens the lecturer if it doesn't frighten the audience they stand 
here you know you just take the stopper out and the next thing is you're hidden 
behind a sheet of flame they're very violent er materials but as i say industry 
has found ways i mean of course they ha-, they don't handle them in glass of 
course they handle them in stainless steel apparatus but you know thousands of 
tons of these things are made every year for industrial use okay er the other 
metals just so that you have a note of it i don't expect you to learn this list 
but the other metals that undergo this kind of reaction if i think about them 
er er tin does it very readily 
silicon does it silicon is important because that reaction is used in the 
production of silicones for floor polishes waterproofings and so on germanium 
does it zirconium does it don't know if i think of anything else er well that i 
think that those are the most important ones okay and finally this is really er 
another example of insertion into metal-hydrogen bonds but usi-, this 
particular case of using er diazomethane and it's just to show you that you c-, 
if you want to make methyls this is one very easy way you can do it if we were 
to take for example triphenyl silane again silicon isn't a true metal but you 
cou-, you can put tin if you like it works just as well in fact i'll put tin 
because it's meant to be a metal so we've got tin and if i treat that with 
diazomethane which i'm sure you will have come across in organic chemistry but 
you can shout out if i'm wrong about that diazomethane is a very useful reagent 
because of course it behaves as a source 
of methylene because nitrogen is eliminated and nitrogen is thermodynamically 
so stable so the product from this is P-H-three S-N well i can write it if you 
like as C-H-two-H but i mean it's just tin methyl of course and nitrogen and 
really it's the formation of nitrogen which is the driving force for that 
reaction okay so let me just since we the board has come full circle let me 
just emphasize for you the reactions where we'd have to be very careful to give 
you some feeling for this a Grignard reagent is decomposed by catalytic amounts 
of oxygen so you ha-, if you're making this Grignard reagent you'll have to 
work with pure dry nitrogen pure dry ether and get everything spot on lithium 
is the same sodium is even more so and there's a lesson there what what are we 
er actually learning from this well amongst other things sodium of course is 
one element lower in the periodic table than lithium so sodium is more 
electropositive than lithium makes the bond more 
polar the species more reactive and so if you if you have to take precautions 
with lithium you have to take them er doubled in spades for sodium so the same 
thing here if you want to make benzyl sodium you must work with pure dry 
nitrogen dry solvent dry gases no air no water same thing here this is lithium 
er well here borane actually is not sensitive to water oddly enough and the 
reason for that is quite simply that boron and hydrogen have r-, roughly the 
same electronegativity and so the boron-hydrogen bond is not very polar and so 
it is not attacked by water but it is attacked by oxygen so all of this would 
have to be done you could do it wet if you wanted i mean it'd be a bit peculiar 
but you could do it wet if you wanted as long as you excluded all oxygen from 
the reaction aluminium of course is heavier more electropositive aluminium 
reagents as i've said catch fire spontaneously in air they virtually explode 
with water so you want to be careful about wetness 
there er tin is group fourteen and so for reasons that i'll explain later it 
the organic compounds of tin not sensitive to water and they're not sensitive 
to oxygen so we can summarize the sensitivity in the following way er oxygen 
and water is all group one all group two and group thirteen er from aluminium 
down er oxygen alone would be er boron compounds generally and in general 
hydrides E-G S-I-H S-N-H now we've got one minute left so i'll finish up with 
just explaining the rationalization for that but we'll go on and discuss it in 
much greater detail next time but just to encapsulate these differences all 
these that we're talking about are main group metals so they all want or they 
all tend for reasons that we can describe in detail they all tend to have a 
complete octet of electrons you know the stability is associated with a 
complete octet now of 
course group one only have one electron so a l-, a lithium methyl monomer only 
has two electrons not an octet magnesium Grignard reagent only have four 
electrons no octet group thirteen only have six electrons so these things are 
all highly reactive and unstable and their structures we shall see reflect 
those if you go boron is curious with water because it's non-polar silicon and 
tin of course are in group fourteen they have four electrons they can form four 
covalent bonds which is an octet so in general their organometallic compounds 
are not sensitive to water or air it's only that the hydrogen compounds because 
it's hydrogen are very easily oxidized so they're sensitive to oxygen okay 
we'll stop there