nm0690: okay looking [1.5] at the [0.3] thermodynamic stability of 
organometallic compounds [0.9] er what i'd like to do now is turn to [0.4] how 
you actually make [0.2] the things [0.8] in other words the f-, synthesis [1.2] 
now the first thing [0.5] er [1.0] to observe before we start actually on 
methods of synthesis [0.8] is that some of these compounds are extremely 
sensitive [0.5] to air and water [1.4] and the origin [0.4] of that sensitivity 
[0.9] is the metal-carbon bond [1.1] or [0.5] if they're very ionic of course 
the carbanion if you have a metal [1.5] directly bonded to carbon [1.1] which 
is w-, our definition of an organometallic compound [0.5] you either have a 
covalent bond [0.3] which must be the source of reactivity [0.7] or you have 
something approaching [0.7] M-plus [1.0] C-minus [0.4] which of course [0.8] is 
an ionic [0.5] er arrangement [0.2] now [1.4] i suppose it's true to say that 
no organometallic compound [0.7] is totally ionic i mean there's no 
organometallic equivalent should we say of caesium fluoride [0.9] but some 
organometallics let's say [1.0] er alkyls of [0.2] you know caesium and 
ribidium the very electropositive metals [0.8] and even of 
sodium and lithium [0.5] are very substantially [1.0] polar [0.4] they have a 
large polar [0.4] component [0.5] so if you regard them as ionic of course [1.
5] the carbanion [0.2] is the source of reactivity [0.7] carbanions [0.4] if 
even if you have them in organic chemistry are incredibly reactive they would 
react with oxygen with water with all sorts of things [0.9] if they're covalent 
[0.3] then it's the metal-carbon bond which is the source of the reactivity [1.
4] now it makes it much simpler when discussing the synthesis if we just 
consider [0.4] what precautions are necessary [0.6] so if materials are 
sensitive [2.9] to let's say air [1.3] and it's really by that we mean oxygen 
of course because [0.5] in the air [0.5] there's [0.2] one-fifth oxygen roughly 
[0.8] the remainder is mainly nitrogen which of course hardly reacts with 
anything but i'll come to the one thing it does react with [1.0] a little bit 
of C-O-two which will react with very reactive ones [0.5] and water [3.3] so if 
they're sensitive to air and water [0.8] and [0.2] let's remember by the way 
that [1.1] the air [0.7] is always wet [0.5] when you may think on a 
bright sunny warm summer's day it's pretty dry [0.6] but the air [0.2] 
certainly in this climate always contains substantial amounts of water [0.5] so 
if something is sensitive to water [1.3] you've got to keep out air [0.2] 
because otherwise it's just going to go off [0.5] now how do you do that what 
does that mean it means [0.7] that you must have an atmosphere [0.9] above your 
apparatus [1.4] of dry [1.7] nitrogen is the common one [0.2] it's cheap [0.5] 
it's available fairly pure [1.1] and sometimes much easier to use but much more 
expensive [0.7] you can use argon [1.3] experimentally argon is much easier to 
use for the very simple reason [0.4] it's much heavier [0.4] than air [0.5] 
nitrogen of course is roughly the same density as air [1.5] and so [0.5] if you 
want to keep something [0.2] in a [0.2] in a flask [0.6] that contains nitrogen 
[0.6] you have a problem you must keep it stoppered [0.5] or under vacuum or 
under a flow of nitrogen at all times [0.9] if you've got something that's air-
sensitive and you've got a [0.8] covered with argon of course [0.3] for quite 
some minutes argon [0.4] will stay [0.7] in the flask [0.4] 
because it's much heavier than air [0.2] so it's much easier to work with argon 
[0.5] but as i say it's much dearer [0.4] and so experimentally you know you [0.
2] generally only use argon for very [0.2] precious compounds [0.4] or 
extremely sensitive material [2.7] er and this also means then that all your 
solvents [1.5] must be degassed [2.4] the air removed which is normally done 
under vacuum [0.7] and of course [0.8] because of the water [0.6] they must be 
dry [0.7] and normally [0.5] if you're working with organometallic compounds [0.
7] for their own sake and you're not using them just as tests or as reagents 
but you want to study them [0.7] they're normally dried over sodium [0.6] 
potassium alloy [1.7] and the reason for that [0.4] is that it's much more 
reactive [0.4] than just using sodium and it's much more efficient at drying 
them [2.2] and [0.9] so [0.2] the other thing is it's so that's your atmosphere 
[0.5] those are your solvents and three the apparatus [3.1] must be either [1.
4] under vacuum [4.1] or if that's not [0.4] possible [0.6] it must be under [2.
0] a flow [1.5] of inert gas that's to say nitrogen or argon [2.4] so when i 
say for 
example [0.3] that a particular [0.3] kind of compound [0.5] is sensitive to 
air and water or is sensitive to one of them [0.6] those are the conditions 
that you need to use [2.1] okay [0.7] so how what methods do we use [0.5] now 
th-, obviously if you want to make a metal derivative [0.6] of an organic 
compound [0.8] the easiest way in principle by far [0.5] is to take metal [1.3] 
and some organic derivative of the compound that you want and that's normally 
[0.5] a halogen [0.6] so for example [0.3] let's [0.2] let's just take [0.2] 
one [0.3] the first method [0.5] from the metal [0.9] plus an organic compound 
[9.6] the classic metals for which this works are lithium and magnesium those 
are the best known [2.1] but of course it will work for the heavier alkali 
metals sodium potassium [0.4] s-, rubidium and caesium and so on [0.5] and i'll 
give you [0.7] details of which metals are useful [0.3] later on but those are 
the classic examples [0.8] so for example you could take [0.3] lithium metal [0.
8] plus [0.9] what should we say you could take er [0.2] ethyl bromide [1.8] 
and what you would get out of that [1.1] assuming [0.3] that you [1.4] used the 
conditions above you had a 
dry atmosphere dry solvents and all the rest of it [0.7] you would get [0.8] er 
[0.3] lithium bromide [1.3] plus [0.6] lithium ethyl [0.9] and i'm just using E-
T as an abbreviation [0.4] for C-two-H-five the ethyl group [4.7] lithium ethyl 
[0.7] is very sensitive to oxygen [0.6] and to water [1.7] and i should explain 
one other thing which perhaps i should have mentioned earlier but just let me 
come to it now [1.3] what is it [0.7] that lithium [0.2] ethyl [0.5] reacts 
with in water [0.6] it is of course the proton [0.4] if you take lithium ethyl 
[3.0] with water [1.3] what you will get [0.2] is lithium hydroxide [0.6] and 
ethane [2.4] that's what happens that's the decomposition reaction with water 
[1.0] and how does it work of course it works because [0.5] the water [0.2] 
gives you a very low concentration of protons [0.6] those protons cleave [0.4] 
the lithium to carbon bond in lithium ethyl [2.2] now [1.8] that means [0.7] 
that in general [0.9] any solvent which will give you a concentration of 
protons [0.4] is going to be bad news [0.8] for these organometallic compounds 
[0.3] and so [0.7] when you do this reaction [0.2] in a solvent the solvent [0.
7] must not provide [0.5] 
any concentration of protons otherwise you'll simply decompose the product as 
it's formed [0.7] so [0.4] we must use what we call an aprotic [1.0] solvent 
just one [3.5] aprotic without protons okay [0.2] of course it doesn't mean 
that there isn't hydrogen in the solvent it just means [0.4] that this solvent 
does not produce any appreciable concentration [0.6] of the proton in solution 
[0.6] so for example [0.4] you could use diethyl ether [1.0] you can use T-H-F 
[1.4] ethers are quite common [0.5] and they're v-, they're very good [1.4] in 
principle you could use an alkane you could use hexane [0.2] but i'm going to 
put that in brackets because [0.8] although you can get the reactions to go in 
hexane under industrial conditions with high temperatures [1.2] it's extremely 
difficult [0.6] to get that reaction to work in the laboratory at normal sort 
of temperatures [0.3] so in principle you can use hexane [0.7] but you [0.8] in 
the lab you'd be better off sticking with some kind of ethereal solvent [0.9] 
er [0.9] a protic solvent [3.7] which i should emphasize is no use here at all 
[1.4] so i'll put them in 
brackets but protic solvents would be things you know like ethanol [1.0] er 
methanol [1.4] any acid of course is highly protic by these standards and any 
amine [2.3] would be useless [0.3] we normally think of amines as being basic 
[0.3] because of the lone pair of electrons on the nitrogen [1.1] but an amine 
can produce a very low concentration of protons it's also acidic [0.9] in terms 
of [0.7] these highly sensitive [0.5] metal to carbon bonds [0.3] so you 
wouldn't [0.5] you couldn't use an amine [0.5] and of course equally [0.7] you 
couldn't use a thiol [0.5] no-, you probably wouldn't want to [0.3] but all of 
those that i've put in the square brackets there [0.7] provide tiny 
concentrations of protons in solution and they will simply decompose [0.4] your 
compounds so you cannot use [0.5] those [1.1] er materials [6.3] the other 
classic reaction which you may have done in the lab i don't know but you will 
certainly know about [0.6] where you can use a metal [0.2] is magnesium where 
you get a Grignard reagent [0.5] it's a slightly different reaction of course 
[0.6] because normally with a Grignard 
reagent [0.5] you simply start with ordinary magnesium and the halide [0.3] and 
you don't of course get the formation [0.2] of magnesium halide [0.3] it's an 
insertion reaction [0.3] of the metal [0.6] into [0.3] the carbon-halogen bond 
[0.4] so for example let's take just a very well known example [0.4] if you 
take magnesium [0.7] plus methyl iodide [1.7] you do this let's say under 
nitrogen [0.4] and you will do it [0.2] we'll say just to give you an example 
in diethyl ether as a typical aprotic solvent [0.6] you will produce methyl [0.
2] magnesium iodide [4.7] you don't get any magnesium iodide you just get [0.3] 
the [0.5] if you like the addition reaction the insertion of magnesium the 
metal [0.5] between carbon and halogen [2.1] and i write it of course with the 
lines here [0.3] just to emphasize that this is a genuine organometallic 
compound [0.3] it contains a bond [0.4] between carbon and magnesium [0.6] 
later on [0.5] we'll [0.4] discuss [0.7] what the actual structures [0.2] of 
these compounds are [0.2] because although i've written lithium ethyl up there 
as though it was simply [0.5] one lithium metal [0.3] bonded to one ethyl group 
[0.9] 
its structure actually turns out to be more complicated than that [0.3] for 
reasons that i will explain [0.5] in a few [0.4] minutes later on [0.5] but [1.
8] the point about it is they contain direct bonds between lithium and carbon 
[0.2] and so [0.3] they are organometallic compounds [2.9] now [0.2] er [4.2] 
oh no that's all right [0.2] that's okay [20.5] we will find that both of these 
reagents that's to say [0.7] oh we've lost them but both the Grignard reagents 
[0.7] and the lithium reagents are themselves [0.5] incredibly useful [0.5] for 
generating other organometallic compounds so just bear them in mind we'll come 
back to them many times [2.7] all right [1.4] now [1.5] so that brings us up to 
the next kind of general reaction for making organometallic compounds [0.5] and 
these we can call [0.7] exchange reactions [9.3] now what do i mean by exchange 
reactions er [0.5] the first kind [0.2] i'll talk about is where you exchange 
between two [0.6] different metals [0.6] so we'll call [1.1] this [0.7] metal-
metal exchange [5.9] and i think you'll see [0.2] why it's so called in a 
minute [2.1] let's just imagine [0.2] for a moment [0.7] that you want to make 
vinyl 
lithium [0.2] the compound [0.6] C-H-two [0.5] double-bond [0.6] C-H [0.8] 
lithium [1.0] let's assume that you want to make that [1.7] from what we've 
said so far what you might well try [0.2] would be [0.4] to take [0.3] vinyl [0.
4] chloride or vinyl bromide [0.4] and metallic lithium [0.4] it sounds like [0.
5] a good bet [1.3] there is a problem with that though [0.7] and the problem 
is simply it doesn't work [0.5] lithium does not react with vinyl chloride or 
vinyl bromide [0.4] there's a second problem of course [0.5] which in the 
laboratory you normally overcome which is that vinyl chloride and vinyl bromide 
are both highly carcinogenic [0.5] but assuming that you've got the c-, [0.2] 
proper apparatus for handling [0.2] dangerous compounds [0.5] even then [0.4] 
the thing the [0.2] the chemistry doesn't work if you had C-H-two-double-bond-C-
H-B-R [0.5] you find there's no reaction with lithium [0.6] and that's of 
course because you're trying [0.5] if you like [0.4] you've got an S-P-two 
carbon here [0.4] and you're trying effectively [0.6] to do a substitution [0.
6] of the halogen [0.3] by the metal and as you know [0.3] it's not easy to do 
substitutions at S-P-two 
carbon so you've got to have another method [0.7] and the method that works 
very well [0.5] is to take [0.8] the vinyl derivative [1.0] of another [1.1] er 
[0.7] metal [0.6] so in this case [0.3] you could take [0.4] tetravinyl tin [0.
4] C-H-two-C-H [0.9] four times [0.4] tin [0.6] i'll tell you how you make that 
later [0.2] but you can make that [0.3] without too much trouble [0.8] 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 [0.4] compound [0.9] 
and if you treat that with four molecules of phenyl lithium [1.2] just write P-
H to [0.3] abbreviate again for C-six-H-five the phenyl ring [1.0] then the 
reaction works very nicely [0.6] you get four [0.7] molecules of vinyl lithium 
[2.6] and the other product [0.3] is tin [0.8] tetraphenyl [0.4] so you've 
exchanged organic groups [0.2] between two metals [0.6] the metal lithium and 
the metal tin [0.3] hence the name metal-metal exchange [2.4] of course you 
still have to [0.5] carry this reaction out [0.4] under all these [0.3] 
conditions of an aprotic solvent [0.9] and in the absence of air [0.2] and with 
dry [0.3] solvent [0.4] because now [0.8] the lithium starting material and the 
lithium product [0.6] are both [0.2] sensitive [0.2] to water and oxygen so you 
have a double problem there [0.9] 
it turns out that [0.2] t-, organo-tin compounds are not sensitive either to 
oxygen or to water [0.4] but the lithium [0.2] derivative's what you're 
interested in [0.4] and if you don't take the precautions [0.3] you're going to 
be in trouble [2.4] why do i choose incidentally phenyl lithium i mean why 
don't i choose butyl lithium [0.8] the answer is [0.4] it's purely experimental 
[0.4] butyl lithium would work quite well [1.0] the problem then would be [0.5] 
i would be left [0.6] with butyl tin [0.5] which is a nice [0.2] soluble [0.3] 
organometallic reagent [0.5] just like vinyl lithium [0.9] but [0.4] 
tetraphenyl tin [0.6] is [0.4] almost totally insoluble in all the solvents 
that you can choose certainly in ethers [0.5] so it just precipitates out [0.5] 
and it leaves you [0.3] with a clean product you don't have to worry too much 
[0.4] about purifying the material [2.4] now there are many examples of this 
kind of exchange and we'll come across some of these [0.3] later on [0.4] but 
if i [0.4] 
say that a compound is made by metal-metal exchange [0.4] that is the kind of 
[0.5] er reaction we're talking about [2.4] okay [0.2] so [1.1] the next kind 
of exchange reaction is metal-hydrogen [8.1] 
nf1340: here's the attendance list [0.4] 
nm0690: oh [0.2] right thanks [1.4] [2.7] okay [0.7] i've been presented with 
the usual attendance list [0.8] er [0.9] could you oblige the powers that be by 
signing it and passing it on for me [0.2] thanks [2.1] okay [0.3] now then so 
we have metal-hydrogen [2.2] metal-hydrogen [0.9] is a very important reaction 
[1.8] and it's really based on the fact [1.9] that the organometallic compounds 
[0.2] let's say of lithium [1.4] are very sensitive as we've already seen [0.7] 
to anything that can give you protons [1.3] so [0.2] you can use that [2.0] to 
make [0.3] organometallic derivatives [2.8] if the hydrogen that you want to 
replace [1.7] has got some [0.4] little bit of acidity about it [0.9] so [0.3] 
let's take a very typical example [0.2] let's take [0.3] let's take a classic 
example first and then i'll go on and show you that it's actually quite 
extensive [1.1] er first of all [1.0] oh i've broken the board here but [1.1] 
oh well [2.3] let's take an 
example where you've got a fairly strong [0.2] carbon acid now [1.7] carbon-
hydrogen compounds are not generally acidic [0.5] so you need some special kind 
of structural feature [1.1] to make [0.2] a carbon acid which is what you need 
[0.4] one such compound is cyclopentadiene [0.6] which is [0.7] a five-membered 
ring of course [0.4] with [0.2] two double bonds in it now cyclopentadiene [1.
0] is peculiar [0.2] it has hydrogens of course all the way round [0.5] but two 
hydrogens [0.4] here [2.5] and i-, [0.7] its [1.0] unusual character is because 
[1.8] it's rather acidic for a hydrogen compound [0.6] it will give you [7.9] 
the anion [0.9] plus hydrogen [2.1] and let me just write that as hydrogen [0.
4] S for solvated because as you know H-plus doesn't exist in solution [0.5] 
but i can't write water obviously because we can't do this reaction [0.6] i-, 
in water because cyclopentadiene isn't soluble in water [0.3] and in any case 
the anion would be [0.3] decomposed by water [0.6] so this will be in some [0.
3] other solvent [0.2] that can sustain the ionization [0.2] maybe T-H-F [0.9] 
now why does this happen [1.1] well the answer is very simple [0.8] in the 
double bond 
of course i've got two pi electrons [1.7] and i've got two of them so that is 
four pi electrons [1.7] if i take one of these hydrogen bonds [0.9] to carbon 
[0.4] it obviously contains two electrons [0.7] if i remove the proton [0.7] 
over here [0.9] of course i take no electrons with me because the proton has no 
electrons [0.5] i leave those two electrons behind [0.6] so that gives me [0.2] 
six electrons [0.6] and of course as you know from benzene chemistry [1.1] if 
i've got six [0.2] pi electrons [0.5] i have an aromatic sextet [0.3] and just 
as with benzene that's very stabilizing [0.5] so for cyclopentadiene [0.5] the 
anion [0.9] is stabilized [0.4] by having [0.4] six pi electrons just as you 
find in benzene [0.5] and therefore [0.2] the reaction [0.8] goes [0.7] much 
more than it would with a typical [0.5] carbon-hydrogen compound with any 
alkane [0.6] now [0.2] when i say [1.0] it goes much more [0.4] i don't mean 
this is like you know acetic acid i mean [0.8] that th-, this is a tiny tiny 
concentration [1.3] but it's dynamic you you can look it up in the textbooks 
you can find the figures [0.3] i think for example [0.2] if i remember let me 
just 
give you a guess i think the P-K-A for this molecule [0.3] is something like [0.
4] fifteen [0.5] that means [1.4] that the concentration the molar 
concentration of hydrogen ions [0.6] will be ten-to-the-minus [0.3] fifteen [0.
5] water is ten-to-the-minus-seven [0.4] so you know [0.3] this is another [0.
2] ten [0.6] to-the- [0.2] eight [0.5] less acidic than water [0.6] but you 
will find [0.2] that that's useful enough [0.5] if you take [0.4] 
cyclopentadiene [2.6] er i'll write it out in full 'cause then you can see 
what's happening [0.7] if you take cyclopentadiene with sodium [1.5] what you 
get [0.5] is [0.6] er sodium cyclopentadienide [2.1] [cough] [4.8] and of 
course [0.3] we've lost a hydrogen [1.7] and i'll just write that as a half-H-
two because you know otherwise you've got to double everything up [0.6] so [0.
9] what i've done is i started with two hydrogens here [1.3] and i've replaced 
one of them [0.5] with sodium [0.9] now of course [0.2] this is [0.5] an ionic 
compound substantially [0.5] because the reason that it goes in the first place 
[0.3] is the stability [0.4] of the cyclopentadienide anion [0.5] and so the 
product [0.4] is [1.2] er a salt if you like [0.6] but it's in exchange of 
sodium 
for hydrogen because if you write it out in [0.3] a formula terms we start out 
with C-five-H-six [0.6] we react it with sodium [0.4] and the product is sodium 
[0.6] C-five-H-five [3.4] and so i've e-, substituted or exchanged [0.4] one 
hydrogen [0.5] for [0.5] a sodium [0.2] so it's a metal [0.2] -hydrogen 
exchange [0.3] that reaction in the textbooks is sometimes [0.4] referred to 
simply as metallation [5.1] some textbooks call it metallation i think that's a 
confusing term because [0.6] any reaction which puts a metal [0.5] in place of 
any other group be it a halogen or another metal [0.4] is if you like a 
metallation [0.3] you're adding metal to the compound [0.9] but you will find 
that some textbooks use the term metallation [0.4] so if you want to look it up 
you might have to look it up under the term metallation [2.3] er [0.6] let me 
give you [0.4] an example where it's [0.3] much more evidently an exchange [0.
9] suppose we take phenyl sodium [1.3] and we take toluene [6.7] now toluene of 
course is an even weaker acid I-E [0.4] it has [0.5] an even lower 
concentration of protons [0.5] than cyclopentadiene [0.6] but still [1.1] it is 
a [0.4] slightly 
stabilized anion C-H-two-minus [0.6] negative charge can again interact with 
the aromatic ring [0.3] so this is a [0.4] i don't know what the P-K-A for this 
is it must be minuscule [0.7] but nevertheless it's enough [0.2] in the 
presence [0.3] of a sodium [0.2] organometallic compound [0.4] the reaction 
goes [0.2] you get [1.6] exchange [0.2] of one of the acidic hydrogens [1.2] so 
that's benzyle sodium [0.7] and you exchange that hydrogen for the sodium [0.3] 
so the other product of course is if you like P-H-H [0.8] or benzene [4.5] so 
that's a typical metal-hydrogen exchange [7.4] so those are the the main [0.4] 
important type of exchange reactions [2.9] let's [0.7] er finish off by looking 
at another very important class [0.7] which are insertion reactions [9.9] now 
of course [1.0] i suppose in the number of [0.4] compounds made and the number 
of times it's been used [0.9] the synthesis of a Grignard reagent is [0.2] 
probably [0.7] the most widely-used insertion reaction in chemistry [0.7] so 
we'll put that down first [1.2] but [4.5] many textbooks [0.3] don't actually 
include it as an insertion reaction but of course formally 
it is [0.9] so if i just put down the general reaction R-X [0.2] plus [0.2] M-G 
[1.0] gives you [0.6] R-M-G-X [1.2] that is obviously [0.3] in a formal sense 
an insertion reaction so we should include it [5.2] a much more typical kind of 
insertion reaction [0.6] is a reaction [0.5] which involves insertion into [0.
7] a metal [1.5] carbon bond that's already been formed [0.5] so let's take for 
example [0.4] butyl lithium [3.1] which i'll write like that because that's the 
bond [0.5] into which [0.5] we're going to insert something [0.8] now if you 
insert something [0.5] we can use the example above the magnesium to show [1.1] 
the [0.2] the er [0.2] the idea [0.5] if i insert magnesium [0.5] into [0.6] 
the carbon-halogen bond of this [0.4] R-X group [2.3] the magnesium has got to 
be able to form [0.6] two bonds [0.8] because i've pushed it in between two 
elements and if the whole thing is going to stay together [0.3] the magnesium 
has got to bond to the R [0.2] and it's got to bond to the X [0.7] another way 
of looking at this of course if you want to [0.4] is to say [0.5] that [0.3] in 
terms of the metal [1.0] its oxidation state increases [0.2] by two [0.7] 
oxidation state of course 
simply tells you the number of electrons [0.4] that the metal is using to form 
bonds [0.6] in magnesium metal [0.5] its oxidation state is zero [0.2] it's not 
using any electrons to form [0.4] any [0.5] ordinary covalent bonds [1.0] when 
it goes to the Grignard reagent it's forming two bonds [0.3] so it goes from 
magnesium nought to magnesium two [1.1] and that shows you that the the thing 
that's inserted has got to be able to provide two electrons [0.3] so that you 
can form two bonds [0.5] in general [0.4] that means [0.3] that you've got to 
have some kind of unsaturated species [0.3] so for example suppose i was to 
take [1.2] let's say [0.5] an acetylene or an alkyne if you prefer [0.4] so 
we'll take R- [0.5] C [0.4] -triple-bond-C- [0.3] R [0.8] and it's the [0.3] pi 
bonds of the triple bond [0.4] that can give rise to the new [0.3] bonds [1.5] 
this can insert into the [0.4] carbon lithium bond here and what you would get 
would be R- [0.3] C- [0.7] C- [0.4] R [1.5] butyl [0.7] and lithium [1.9] and 
the insertion [0.7] you will notice [0.6] i chose a triple bond for a very s-, 
particular reason [0.8] and that is simply that [0.4] i can show with a triple 
bond [0.8] the 
geometry of [0.2] of the insertion [0.5] if i'd started with a double bond [0.
4] i would get an alkane here [0.3] and we'd have free rotation and we couldn't 
tell anything about it [0.6] but starting with a triple bond i'm left with a 
double bond [0.3] and so of course [0.8] i can have [0.2] the butyl and the 
lithium going cis about the double bond which is one [0.2] isomer [0.3] or i 
could have butyl here and lithium up here [0.3] the trans arrangement [0.3] 
which would also be stable [0.3] different isomer [0.5] but in fact they both 
go on the same side [0.3] and so the mechanism of that insertion [0.3] is 
always a cis one [10.2] so that is an insertion [0.9] that's [0.2] insertion 
into [0.5] a metal [0.7] to carbon bond [4.4] it's also possible to have 
insertion [0.6] into a metal [0.3] to hydrogen bond [2.8] have you in organic 
chemistry [0.5] done anything about hydroboration [2.4] you've heard of it have 
you [0.6] okay [0.6] hydroboro-, m-, boron isn't really a metal so in a sense 
it's not [0.3] relevant to organometallic chemistry [0.4] but this is exactly 
the same kind of reaction [0.8] 
so f-, let me just go over some familiar ground first and then [0.5] i'll show 
you how it can be used [0.7] to form [0.5] er metal-carbon bonds [2.6] if we 
start out with an olefin in this case [0.8] doesn't matter what olefin [0.7] 
i'll just ha-, we'll just have some general olefin [0.9] here [1.3] and i've 
got let's say borane [0.8] now [1.5] you know of course that we can't actually 
start with borane itself because [0.5] borane exists as a dimer B-two-H-six [0.
7] and that's rather inconvenient [0.4] er so let's say [0.2] we we'll start 
out [0.8] with the complex with T-H-F because [0.2] the T-H-F complex is stable 
[0.5] it doesn't dimerize [0.5] because you donate a pair of electrons from the 
oxygen [0.4] of the T-H-F [0.3] to the boron [0.3] and then the monomer is 
stable and that simplifies the r-, stoichiometry of the reaction [0.7] now [0.
2] what happens here is [0.8] we're going to insert into [1.2] the B-H [0.5] 
bond [1.2] so the first thing of course you will get [0.4] will be [0.4] er [0.
2] well 
we'll keep the T-H-F initially you'll get T-H-F [0.7] H-two [0.6] B [0.3] now 
[0.5] into the boron-hydrogen bond we're going to insert our olefin [0.3] so we 
will have [0.5] C [1.3] C [1.0] H [2.5] that's hydroboration [0.3] of course [1.
6] you still have in this molecule here two other boron-hydrogen bonds [0.2] 
and so [0.2] if you've got enough olefin [0.5] the reaction will proceed [1.8] 
in the presence of excess olefin [0.5] the reaction will proceed [0.4] until 
you get [0.4] boron [1.1] C-C depends what these groups are [1.8] H [1.6] three 
times [1.9] perhaps to be mathematically accurate i should put the hydrogen i 
guess inside the [2.1] the bracket [4.2] okay [0.3] so [0.5] that's an example 
you've come across before [0.2] an unsaturated species is inserted into the 
boron-hydrogen bond [0.4] and it's almost an organometallic example [0.5] but 
there are many other [5.7] there are many other metals well there are many 
metals [2.9] which [5.4] which undergo similar reactions [0.6] and [0.5] 
perhaps the most important of those is aluminium [0.9] because [1.2] the 
reaction between aluminium-hydrogen compounds [0.5] and unsaturated species [0.
6] is used in industry a great deal [0.5] to 
make aluminium alkyls [4.6] so for example if we just take [0.6] a typical 
example [1.2] er let's say we start out with [0.2] dibutyl [1.3] aluminium [0.
4] hydride [0.8] so that's a metal now aluminium [0.4] hydrogen bond [0.5] and 
we'll take ethene [0.5] just to make a nice simple example [2.8] the ethene 
inserts into the aluminium-hydrogen bond [0.5] and the product that you get is 
dibutyl [0.5] aluminium ethyl [6.8] i give you that example as i say it's used 
to make alu-, organo-aluminium compounds [0.5] and you probably know that [0.2] 
the chemical industry [0.7] makes [0.2] thousands of tons of aluminium alkyls 
[0.9] every year because they're used in the catalysts that [0.2] produce [0.4] 
polythene [0.5] polypropylene [0.6] and a whole range of polyolefin catalysts 
[0.9] er er [0.7] are derived [0.2] from organo-aluminium compounds [0.4] and 
it's quite [0.2] interesting that until you've had a bit of practice [0.8] it 
would be extremely [0.7] er difficult [0.5] to carry out this reaction in the 
laboratory [0.4] because a material like this [0.6] not only reacts with oxygen 
[0.5] if i was to have a bottle with this and take th-, [0.4] the stopper out 
of it [0.5] it 
simply bursts into flames it reacts with oxygen incredibly [0.9] readily [0.8] 
they're pyrophoric materials [0.8] take the stopper out and i'd just have a 
sheet of flame [0.4] from here [0.4] to the ceiling i have actually [0.3] not 
in one of these lectures but i have actually done it in a demonstration lecture 
[0.6] and by [0.7] gosh it i mean it [0.2] it certainly frightens the lecturer 
if it doesn't frighten the audience [0.5] 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 [0.7] er materials [0.6] but as i say industry [0.2] 
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 [0.4] but you know 
thousands of tons of these things are made every year for industrial use [1.6] 
okay [0.4] er the other metals just so that you have a note of it i don't 
expect you to learn this list [0.5] but the other metals that undergo this kind 
of reaction if i think about them [0.6] er [0.3] er tin [0.5] does it very 
readily [0.4] 
silicon does it [1.0] silicon is important because that reaction is used in the 
production of silicones [0.6] for floor polishes waterproofings and so on [0.3] 
germanium does it [0.7] zirconium does it [1.3] don't know if i [0.6] think of 
anything else er [3.3] well that i think that those are the most important ones 
[3.9] okay [1.1] and finally [3.3] this is really [0.5] er [0.3] another 
example of insertion into metal-hydrogen bonds [2.0] but usi-, this particular 
case of using [0.5] er diazomethane [8.8] 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 [0.7] if 
we were to take for example triphenyl [1.7] silane [2.1] again silicon isn't a 
true metal but you cou-, [0.2] you can put tin if you like it works just as 
well in fact i'll put tin because [0.9] it's meant to be a metal [1.3] so we've 
got tin [1.9] and if i treat that with diazomethane [1.3] which i'm sure you 
will have come across [0.2] in organic chemistry but you can shout out if [0.3] 
i'm wrong about that [1.1] diazomethane is a very useful reagent [0.3] because 
of course [0.2] it behaves as a source 
of methylene [1.2] because nitrogen is [0.4] eliminated and nitrogen is 
thermodynamically so stable [0.4] so the product from this [0.5] is [0.2] P-H-
three [0.5] 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 [0.7] and nitrogen [0.8] and really it's the 
formation of nitrogen [0.4] which is the driving force for that reaction [8.1] 
okay so let me just [0.7] since we the board has come full circle [1.9] let me 
just emphasize for you [2.7] the reactions where we'd have to be very careful 
to give you some feeling for this [1.6] a Grignard reagent [0.4] is decomposed 
[0.2] by [0.9] catalytic amounts of oxygen [0.7] so you ha-, if you're making 
this Grignard reagent [0.4] you'll have to work with pure dry nitrogen [0.4] 
pure dry ether and get everything spot on [3.2] lithium is the same [2.8] 
sodium is even more so and there's a lesson there [0.5] what what are we [0.6] 
er [0.5] actually [0.2] learning from this well [0.2] amongst other things [1.
1] sodium of course is one element lower in the periodic table than lithium [0.
7] so sodium is more electropositive [0.2] than lithium [0.6] makes the bond 
more 
polar [0.5] the species more reactive [0.2] and so [0.5] if you if you have to 
take precautions with lithium you have to take them [0.6] er doubled in spades 
[0.2] for sodium [0.3] so the same thing here if you want to make benzyl sodium 
[0.6] you must work with [0.2] pure dry nitrogen [0.3] dry solvent dry gases no 
air no water [1.6] same thing here this is lithium [0.5] er [1.9] well [0.2] 
here [0.6] borane actually is not sensitive to water oddly enough [0.5] and the 
reason for that [0.6] is quite simply [0.6] that boron and hydrogen have r-, 
roughly the same electronegativity [0.6] and so [0.4] the boron-hydrogen bond 
[1.0] is not very polar [0.8] and so it is not attacked by water but it is 
attacked by oxygen [0.5] so all of this would have to be done you could do it 
wet if you wanted [0.2] i mean it'd be a bit [0.5] peculiar but you could do it 
wet if you wanted [0.4] as long [0.3] as you excluded all [0.6] oxygen [0.2] 
from the reaction [1.4] aluminium of course is heavier [0.3] more 
electropositive [0.7] aluminium reagents as i've said catch fire spontaneously 
in air [0.5] they virtually explode with water [0.2] so you want to be careful 
about wetness 
there [1.9] er [2.9] tin [0.5] is group fourteen [0.9] and so [1.1] for reasons 
that i'll explain later [0.2] it [0.3] the organic compounds of tin [0.4] not 
sensitive to water [0.2] and they're not [0.2] sensitive [0.3] to oxygen [0.6] 
so [0.2] we can summarize the sensitivity [0.2] in the following way [5.1] er 
[2.0] oxygen and water [3.4] is all [0.7] group one [3.4] all [0.8] group two 
[3.6] and group [2.1] thirteen [1.2] er [1.2] from aluminium down [14.2] er 
oxygen alone [2.0] would be [0.4] er boron compounds generally [3.5] and in 
general hydrides [2.5] E-G [1.5] S-I-H [1.0] S-N-H [4.6] now we've got one 
minute left so i'll finish up with just explaining [2.2] the rationalization 
for that but we'll go on and discuss it in much greater detail next time [0.6] 
but just [2.4] to encapsulate these differences [1.8] all these that we're 
talking about are main group [0.5] metals [1.3] so they all want [0.2] or they 
all tend [0.4] for reasons that we can describe in detail [0.4] they all tend 
to have a complete octet of electrons [0.2] you know the stability is 
associated with a complete octet [1.9] now of 
course [0.8] group one only have one electron so a l-, a lithium methyl monomer 
only has two electrons not an octet [0.8] magnesium [0.6] Grignard reagent only 
have four electrons no octet [0.3] group thirteen only have six electrons [0.4] 
so these things are all [0.3] highly [0.7] reactive and unstable and their 
structures we shall see reflect those [0.9] if you go [0.5] boron is curious 
with water because it's non-polar [1.3] silicon and tin of course are in group 
fourteen [0.4] they have four electrons [0.2] they can form four covalent bonds 
which is an octet [0.3] so in general [0.2] their organometallic compounds are 
not sensitive to water or air [0.5] it's only that the hydrogen compounds [0.5] 
because it's hydrogen [0.5] are very [1.0] easily oxidized so they're sensitive 
to oxygen [0.5] okay [0.3] we'll [0.7] stop there