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<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

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(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

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The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

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<recording dur="00:45:44" n="5543">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



<person id="nm0690" role="main speaker" n="n" sex="m"><p>nm0690, main speaker, non-student, male</p></person>

<person id="nf1340" role="participant" n="n" sex="f"><p>nf1340, participant, non-student, female</p></person>

<personGrp id="ss" role="audience" size="m"><p>ss, audience, medium group </p></personGrp>

<personGrp id="sl" role="all" size="m"><p>sl, all, medium group</p></personGrp>

<personGrp role="speakers" size="4"><p>number of speakers: 4</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Chemistry</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG2</item>

<item n="module">Organometallic chemistry</item>





<u who="nm0690"> okay looking <pause dur="1.5"/> at the <pause dur="0.3"/> thermodynamic stability of organometallic compounds <pause dur="0.9"/> # what i'd like to do now is turn to <pause dur="0.4"/> how you actually make <pause dur="0.2"/> the things <pause dur="0.8"/> in other words the <trunc>f</trunc> synthesis <pause dur="1.2"/> now the first thing <pause dur="0.5"/> # <pause dur="1.0"/> to observe before we start actually on methods of synthesis <pause dur="0.8"/> is that some of these compounds are extremely sensitive <pause dur="0.5"/> to air and water <pause dur="1.4"/> and the origin <pause dur="0.4"/> of that sensitivity <pause dur="0.9"/> is the metal-carbon bond <pause dur="1.1"/> or <pause dur="0.5"/> if they're very ionic of course the carbanion if you have a metal <pause dur="1.5"/> directly bonded <kinesic desc="writes on board" iterated="y" dur="1"/> to carbon <pause dur="1.1"/> which is <trunc>w</trunc> our definition of an organometallic compound <pause dur="0.5"/> you either have a covalent bond <pause dur="0.3"/> which must be the source of reactivity <pause dur="0.7"/> or you have something approaching <pause dur="0.7"/> M-plus <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="2"/> C-minus <pause dur="0.4"/> which of course <pause dur="0.8"/> is an ionic <pause dur="0.5"/> # arrangement <pause dur="0.2"/> now <pause dur="1.4"/> i suppose it's true to say that no organometallic compound <pause dur="0.7"/> is totally ionic i mean there's no organometallic equivalent should we say of caesium fluoride <pause dur="0.9"/> but some organometallics let's say <pause dur="1.0"/> # alkyls of <pause dur="0.2"/> you know caesium and ribidium the very electropositive metals <pause dur="0.8"/> and even of

sodium and lithium <pause dur="0.5"/> are very substantially <pause dur="1.0"/> polar <pause dur="0.4"/> they have a large polar <pause dur="0.4"/> component <pause dur="0.5"/> so if you regard them as ionic of course <pause dur="1.5"/> the carbanion <pause dur="0.2"/> is the source of reactivity <pause dur="0.7"/> carbanions <pause dur="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 <pause dur="0.9"/> if they're covalent <pause dur="0.3"/> then it's the metal-carbon bond which is the source of the reactivity <pause dur="1.4"/> now it makes it much simpler when discussing the synthesis if we just consider <pause dur="0.4"/> what precautions are necessary <pause dur="0.6"/> so if materials are sensitive <pause dur="2.9"/><kinesic desc="writes on board" iterated="y" dur="7"/> to let's say air <pause dur="1.3"/> and it's really by that we mean oxygen of course because <pause dur="0.5"/> in the air <pause dur="0.5"/> there's <pause dur="0.2"/> one-fifth oxygen roughly <pause dur="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 <pause dur="1.0"/> a little bit of C-O-two which will react with very reactive ones <pause dur="0.5"/> and water <pause dur="3.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> so if they're sensitive to air and water <pause dur="0.8"/> and <pause dur="0.2"/> let's remember by the way that <pause dur="1.1"/> the air <pause dur="0.7"/> is always wet <pause dur="0.5"/> when you may think on a

bright sunny warm summer's day it's pretty dry <pause dur="0.6"/> but the air <pause dur="0.2"/> certainly in this climate always contains substantial amounts of water <pause dur="0.5"/> so if something is sensitive to water <pause dur="1.3"/> you've got to keep out air <pause dur="0.2"/> because otherwise it's just going to go off <pause dur="0.5"/> now how do you do that what does that mean it means <pause dur="0.7"/> that you must have an atmosphere <pause dur="0.9"/> above your apparatus <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> of dry <pause dur="1.7"/> nitrogen is the common one <pause dur="0.2"/> it's cheap <pause dur="0.5"/> it's available fairly pure <pause dur="1.1"/> and sometimes much easier to use but much more expensive <pause dur="0.7"/> you can use argon <pause dur="1.3"/> experimentally argon is much easier to use for the very simple reason <pause dur="0.4"/> it's much heavier <pause dur="0.4"/> than air <pause dur="0.5"/> nitrogen of course is roughly the same density as air <pause dur="1.5"/> and so <pause dur="0.5"/> if you want to keep something <pause dur="0.2"/> in a <pause dur="0.2"/> in a flask <pause dur="0.6"/> that contains nitrogen <pause dur="0.6"/> you have a problem you must keep it stoppered <pause dur="0.5"/> or under vacuum or under a flow of nitrogen at all times <pause dur="0.9"/> if you've got something that's air-sensitive and you've got a <pause dur="0.8"/> covered with argon of course <pause dur="0.3"/> for quite some minutes argon <pause dur="0.4"/> will stay <pause dur="0.7"/> in the flask <pause dur="0.4"/>

because it's much heavier than air <pause dur="0.2"/> so it's much easier to work with argon <pause dur="0.5"/> but as i say it's much dearer <pause dur="0.4"/> and so experimentally you know you <pause dur="0.2"/> generally only use argon for very <pause dur="0.2"/> precious compounds <pause dur="0.4"/> or extremely sensitive material <pause dur="2.7"/> # and this also means then that all your solvents <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="6"/> must be degassed <pause dur="2.4"/> the air removed which is normally done under vacuum <pause dur="0.7"/> and of course <pause dur="0.8"/> because of the water <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="1"/> they must be dry <pause dur="0.7"/> and normally <pause dur="0.5"/> if you're working with organometallic compounds <pause dur="0.7"/> for their own sake and you're not using them just as tests or as reagents but you want to study them <pause dur="0.7"/> they're normally <kinesic desc="writes on board" iterated="y" dur="6"/> dried over sodium <pause dur="0.6"/> potassium alloy <pause dur="1.7"/> and the reason for that <pause dur="0.4"/> is that it's much more reactive <pause dur="0.4"/> than just using sodium and it's much more efficient at drying them <pause dur="2.2"/> and <pause dur="0.9"/> so <pause dur="0.2"/> the other thing is it's so that's your <kinesic desc="writes on board" iterated="y" dur="27"/> atmosphere <pause dur="0.5"/> those are your solvents and three the apparatus <pause dur="3.1"/> must be either <pause dur="1.4"/> under vacuum <pause dur="4.1"/> or if that's not <pause dur="0.4"/> possible <pause dur="0.6"/> it must be under <pause dur="2.0"/> a flow <pause dur="1.5"/> of inert gas that's to say nitrogen or argon <pause dur="2.4"/> so when i say for

example <pause dur="0.3"/> that a particular <pause dur="0.3"/> kind of compound <pause dur="0.5"/> is sensitive to air and water or is sensitive to one of them <pause dur="0.6"/> those are the conditions that you need to use <pause dur="2.1"/> okay <pause dur="0.7"/> so how what methods do we use <pause dur="0.5"/> now <trunc>th</trunc> obviously if you want to make a metal derivative <pause dur="0.6"/> of an organic compound <pause dur="0.8"/> the easiest way in principle by far <pause dur="0.5"/> is to take metal <pause dur="1.3"/> and some organic derivative of the compound that you want and that's normally <pause dur="0.5"/> a halogen <pause dur="0.6"/> so for example <pause dur="0.3"/> let's <pause dur="0.2"/> let's just take <pause dur="0.2"/> one <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="13"/> the first method <pause dur="0.5"/> from the metal <pause dur="0.9"/> plus an organic compound <pause dur="9.6"/> the classic metals for which this works are lithium and magnesium those are the <kinesic desc="writes on board" iterated="y" dur="10"/> best known <pause dur="2.1"/> but of course it will work for the heavier alkali metals sodium potassium <pause dur="0.4"/> <trunc>s</trunc> rubidium and caesium and so on <pause dur="0.5"/> and i'll give you <pause dur="0.7"/> details of which metals are useful <pause dur="0.3"/> later on but those are the classic examples <pause dur="0.8"/> so for example you could take <pause dur="0.3"/> lithium metal <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="1"/> plus <pause dur="0.9"/> what should we say you could take # <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="4"/> ethyl bromide <pause dur="1.8"/> and what you would get out of that <pause dur="1.1"/> assuming <pause dur="0.3"/> that you <pause dur="1.4"/> used the conditions above you had a

dry atmosphere dry solvents and all the rest of it <pause dur="0.7"/> you would get <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="7"/> # <pause dur="0.3"/> lithium bromide <pause dur="1.3"/> plus <pause dur="0.6"/> lithium ethyl <pause dur="0.9"/> and i'm just using E-T as an abbreviation <pause dur="0.4"/> for C-two-H-five the ethyl group <pause dur="4.7"/> lithium ethyl <pause dur="0.7"/> is very sensitive to oxygen <pause dur="0.6"/> and to water <pause dur="1.7"/> and i should explain one other thing which perhaps i should have mentioned earlier but just let me come to it now <pause dur="1.3"/> what is it <pause dur="0.7"/> that lithium <pause dur="0.2"/> ethyl <pause dur="0.5"/> reacts with in water <pause dur="0.6"/> it is of course the proton <pause dur="0.4"/> if you take lithium ethyl <pause dur="3.0"/><kinesic desc="writes on board" iterated="y" dur="12"/> with water <pause dur="1.3"/> what you will get <pause dur="0.2"/> is lithium hydroxide <pause dur="0.6"/> and ethane <pause dur="2.4"/> that's what happens that's the decomposition reaction with water <pause dur="1.0"/> and how does it work of course it works because <pause dur="0.5"/> the water <pause dur="0.2"/> gives you a very low concentration of protons <pause dur="0.6"/> those protons cleave <pause dur="0.4"/> the lithium to carbon bond in lithium ethyl <pause dur="2.2"/> now <pause dur="1.8"/> that means <pause dur="0.7"/> that in general <pause dur="0.9"/> any solvent which will give you a concentration of protons <pause dur="0.4"/> is going to be bad news <pause dur="0.8"/> for these organometallic compounds <pause dur="0.3"/> and so <pause dur="0.7"/> when you do this reaction <pause dur="0.2"/> in a solvent the solvent <pause dur="0.7"/> must not provide <pause dur="0.5"/>

any concentration of protons otherwise you'll simply decompose the product as it's formed <pause dur="0.7"/> so <pause dur="0.4"/> we must use what we call an <kinesic desc="writes on board" iterated="y" dur="6"/> aprotic <pause dur="1.0"/> solvent just one <pause dur="3.5"/> aprotic without protons okay <pause dur="0.2"/> of course it doesn't mean that there isn't hydrogen in the solvent it just means <pause dur="0.4"/> that this solvent does not produce any appreciable concentration <pause dur="0.6"/> of the proton in solution <pause dur="0.6"/> so for example <pause dur="0.4"/> you could use <kinesic desc="writes on board" iterated="y" dur="2"/> diethyl ether <pause dur="1.0"/> you can use T-H-F <pause dur="1.4"/> <kinesic desc="writes on board" iterated="y" dur="1"/> ethers are quite common <pause dur="0.5"/> and they're <trunc>v</trunc> they're very good <pause dur="1.4"/> in principle you could use an alkane you could use hexane <pause dur="0.2"/> but i'm going to put that in brackets because <pause dur="0.8"/> although you can get the reactions to go in hexane under industrial conditions with high temperatures <pause dur="1.2"/> it's extremely difficult <pause dur="0.6"/> to get that reaction to work in the laboratory at normal sort of temperatures <pause dur="0.3"/> so in principle you can use hexane <pause dur="0.7"/> but you <pause dur="0.8"/> in the lab you'd be better off sticking with some kind of ethereal solvent <pause dur="0.9"/> # <pause dur="0.9"/> a protic solvent <pause dur="3.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> which i should emphasize is no use here at all <pause dur="1.4"/> so i'll put them in

brackets but protic solvents would be things you know like <kinesic desc="writes on board" iterated="y" dur="12"/> ethanol <pause dur="1.0"/> # methanol <pause dur="1.4"/> any acid of course is highly protic by these standards and any amine <pause dur="2.3"/> would be useless <pause dur="0.3"/> we normally think of amines as being basic <pause dur="0.3"/> because of the lone pair of electrons on the nitrogen <pause dur="1.1"/> but an amine can produce a very low concentration of protons it's also acidic <pause dur="0.9"/> in terms of <pause dur="0.7"/> these highly sensitive <pause dur="0.5"/> metal to carbon bonds <pause dur="0.3"/> so you wouldn't <pause dur="0.5"/> you couldn't use an amine <pause dur="0.5"/> and of course equally <pause dur="0.7"/> you couldn't use a thiol <pause dur="0.5"/> <trunc>no</trunc> you probably wouldn't want to <pause dur="0.3"/> but all of those that i've put in the square brackets there <pause dur="0.7"/> provide tiny concentrations of protons in solution and they will simply decompose <pause dur="0.4"/> your compounds so you cannot use <pause dur="0.5"/> those <pause dur="1.1"/> # materials <pause dur="6.3"/> the other classic reaction which you may have done in the lab i don't know but you will certainly know about <pause dur="0.6"/> where you can use a metal <pause dur="0.2"/> is magnesium where you get a Grignard reagent <pause dur="0.5"/> it's a slightly different reaction of course <pause dur="0.6"/> because normally with a Grignard

reagent <pause dur="0.5"/> you simply start with ordinary magnesium and the halide <pause dur="0.3"/> and you don't of course get the formation <pause dur="0.2"/> of magnesium halide <pause dur="0.3"/> it's an insertion reaction <pause dur="0.3"/> of the metal <pause dur="0.6"/> into <pause dur="0.3"/> the carbon-halogen bond <pause dur="0.4"/> so for example let's take just a very well known example <pause dur="0.4"/> if you <kinesic desc="writes on board" iterated="y" dur="9"/> take magnesium <pause dur="0.7"/> plus methyl iodide <pause dur="1.7"/> you do this let's say under nitrogen <pause dur="0.4"/> and you will do it <pause dur="0.2"/> we'll say just to give you an example in diethyl ether as a typical aprotic solvent <pause dur="0.6"/> you will produce methyl <pause dur="0.2"/> magnesium iodide <kinesic desc="writes on board" iterated="y" dur="3"/><pause dur="4.7"/> you don't get any magnesium iodide you just get <pause dur="0.3"/> the <pause dur="0.5"/> if you like the addition reaction the insertion of magnesium the metal <pause dur="0.5"/> between carbon and halogen <pause dur="2.1"/> and i write it of course with the lines here <pause dur="0.3"/> just to emphasize that this is a genuine organometallic compound <pause dur="0.3"/> it contains a bond <pause dur="0.4"/> between carbon and magnesium <pause dur="0.6"/> later on <pause dur="0.5"/> we'll <pause dur="0.4"/> discuss <pause dur="0.7"/> what the actual structures <pause dur="0.2"/> of these compounds are <pause dur="0.2"/> because although i've written lithium ethyl up there as though it was simply <pause dur="0.5"/> one lithium metal <pause dur="0.3"/> bonded to one ethyl group <pause dur="0.9"/>

its structure actually turns out to be more complicated than that <pause dur="0.3"/> for reasons that i will explain <pause dur="0.5"/> in a few <pause dur="0.4"/> minutes later on <pause dur="0.5"/> but <pause dur="1.8"/> the point about it is they contain direct bonds between lithium and carbon <pause dur="0.2"/> and so <pause dur="0.3"/> they are organometallic compounds <pause dur="2.9"/> now <pause dur="0.2"/> # <pause dur="4.2"/> oh no that's all right <pause dur="0.2"/> that's okay <pause dur="20.5"/> we will find that both of these reagents that's to say <pause dur="0.7"/> oh we've lost them but both the Grignard reagents <pause dur="0.7"/> and the lithium reagents are themselves <pause dur="0.5"/> incredibly useful <pause dur="0.5"/> for generating other organometallic compounds so just bear them in mind we'll come back to them many times <pause dur="2.7"/> all right <pause dur="1.4"/> now <pause dur="1.5"/> so that brings us up to the next kind of general reaction for making organometallic compounds <pause dur="0.5"/> and these we can call <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="11"/> exchange reactions <pause dur="9.3"/> now what do i mean by exchange reactions # <pause dur="0.5"/> the first kind <pause dur="0.2"/> i'll talk about is where you exchange between two <pause dur="0.6"/> different metals <pause dur="0.6"/> so we'll call <pause dur="1.1"/> this <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="8"/> metal-metal exchange <pause dur="5.9"/> and i think you'll see <pause dur="0.2"/> why it's so called in a minute <pause dur="2.1"/> let's just imagine <pause dur="0.2"/> for a moment <pause dur="0.7"/> that you want to make vinyl

lithium <pause dur="0.2"/> the compound <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> C-H-two <pause dur="0.5"/> double-bond <pause dur="0.6"/> C-H <pause dur="0.8"/> lithium <pause dur="1.0"/> let's assume that you want to make that <pause dur="1.7"/> from what we've said so far what you might well try <pause dur="0.2"/> would be <pause dur="0.4"/> to take <pause dur="0.3"/> vinyl <pause dur="0.4"/> chloride or vinyl bromide <pause dur="0.4"/> and metallic lithium <pause dur="0.4"/> it sounds like <pause dur="0.5"/> a good bet <pause dur="1.3"/> there is a problem with that though <pause dur="0.7"/> and the problem is simply it doesn't work <pause dur="0.5"/> lithium does not react with vinyl <shift feature="voice" new="laugh"/>chloride<shift feature="voice" new="normal"/> or vinyl bromide <pause dur="0.4"/> there's a second problem of course <pause dur="0.5"/> which in the laboratory you normally overcome which is that vinyl chloride and vinyl bromide are both highly carcinogenic <pause dur="0.5"/> but assuming that you've got the <trunc>c</trunc> <pause dur="0.2"/> proper apparatus for handling <pause dur="0.2"/> dangerous compounds <pause dur="0.5"/> even then <pause dur="0.4"/> the thing the <pause dur="0.2"/> the chemistry doesn't work if you had C-H-two-double-bond-C-H-B-R <pause dur="0.5"/> you find there's no reaction with lithium <pause dur="0.6"/> and that's of course because you're trying <pause dur="0.5"/> if you like <pause dur="0.4"/> you've got an S-P-two carbon here <pause dur="0.4"/> and you're trying effectively <pause dur="0.6"/> to do a substitution <pause dur="0.6"/> of the halogen <pause dur="0.3"/> by the metal and as you know <pause dur="0.3"/> it's not easy to do substitutions at S-P-two

carbon so you've got to have another method <pause dur="0.7"/> and the method that works very well <pause dur="0.5"/> is to take <pause dur="0.8"/> the vinyl derivative <pause dur="1.0"/> of another <pause dur="1.1"/> # <pause dur="0.7"/> metal <pause dur="0.6"/> so in this case <pause dur="0.3"/> you could take <pause dur="0.4"/> tetravinyl tin <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> C-H-two-C-H <pause dur="0.9"/> four times <pause dur="0.4"/> tin <pause dur="0.6"/> i'll tell you how you make that later <pause dur="0.2"/> but you can make that <pause dur="0.3"/> without too much trouble <pause dur="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 <pause dur="0.4"/> compound <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="6"/> and if you treat that with four molecules of phenyl lithium <pause dur="1.2"/> just write P-H to <pause dur="0.3"/> abbreviate again for C-six-H-five the phenyl ring <pause dur="1.0"/> then the reaction works very nicely <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="9"/> you get four <pause dur="0.7"/> molecules of vinyl lithium <pause dur="2.6"/> and the other product <pause dur="0.3"/> is tin <pause dur="0.8"/> tetraphenyl <pause dur="0.4"/> so you've exchanged organic groups <pause dur="0.2"/> between two metals <pause dur="0.6"/> the metal lithium and the metal tin <pause dur="0.3"/> hence the name metal-metal exchange <pause dur="2.4"/> of course you still have to <pause dur="0.5"/> carry this reaction out <pause dur="0.4"/> under all these <pause dur="0.3"/> conditions of an aprotic solvent <pause dur="0.9"/> and in the absence of air <pause dur="0.2"/> and with dry <pause dur="0.3"/> solvent <pause dur="0.4"/> because now <pause dur="0.8"/> the lithium starting material and the lithium product <pause dur="0.6"/> are both <pause dur="0.2"/> sensitive <pause dur="0.2"/> to water and oxygen so you have a double problem there <pause dur="0.9"/>

it turns out that <pause dur="0.2"/> <trunc>t</trunc> organo-tin compounds are not sensitive either to oxygen or to water <pause dur="0.4"/> but the lithium <pause dur="0.2"/> derivative's what you're interested in <pause dur="0.4"/> and if you don't take the precautions <pause dur="0.3"/> you're going to be in trouble <pause dur="2.4"/> why do i choose incidentally phenyl lithium i mean why don't i choose butyl lithium <pause dur="0.8"/> the answer is <pause dur="0.4"/> it's purely experimental <pause dur="0.4"/> butyl lithium would work quite well <pause dur="1.0"/> the problem then would be <pause dur="0.5"/> i would be left <pause dur="0.6"/> with butyl tin <pause dur="0.5"/> which is a nice <pause dur="0.2"/> soluble <pause dur="0.3"/> organometallic reagent <pause dur="0.5"/> just like vinyl lithium <pause dur="0.9"/> but <pause dur="0.4"/> tetraphenyl tin <pause dur="0.6"/> is <pause dur="0.4"/> almost totally insoluble in all the solvents that you can choose certainly in ethers <pause dur="0.5"/> so it just precipitates out <pause dur="0.5"/> and it leaves you <pause dur="0.3"/> with a clean product you don't have to worry too much <pause dur="0.4"/> about purifying the material <pause dur="2.4"/> now there are many examples of this kind of exchange and we'll come across some of these <pause dur="0.3"/> later on <pause dur="0.4"/> but if i <pause dur="0.4"/>

say that a compound is made by metal-metal exchange <pause dur="0.4"/> that is the kind of <pause dur="0.5"/> # reaction we're talking about <pause dur="2.4"/> okay <pause dur="0.2"/> so <pause dur="1.1"/> the next kind of exchange reaction <kinesic desc="writes on board" iterated="y" dur="5"/> is metal-hydrogen </u><pause dur="8.1"/><event desc="woman enters room, gives the lecturer an attendance sheet to be passed around" iterated="n" n="nf1340"/> <u who="nf1340" trans="pause"> here's the attendance list </u><pause dur="0.4"/><u who="nm0690" trans="pause"> oh <pause dur="0.2"/> right thanks <pause dur="1.4"/> <gap reason="inaudible" extent="1 sec"/> <pause dur="2.7"/> okay <pause dur="0.7"/> i've been presented with the usual attendance list <pause dur="0.8"/> # <pause dur="0.9"/> could you oblige the powers that be by signing it and passing it on for me <pause dur="0.2"/> thanks <pause dur="2.1"/><event desc="passes around attendance sheet" iterated="n"/> okay <pause dur="0.3"/> now then so we have metal-hydrogen <pause dur="2.2"/> metal-hydrogen <pause dur="0.9"/> is a very important reaction <pause dur="1.8"/> and it's really based on the fact <pause dur="1.9"/> that the organometallic compounds <pause dur="0.2"/> let's say of lithium <pause dur="1.4"/> are very sensitive as we've already seen <pause dur="0.7"/> to anything that can give you protons <pause dur="1.3"/> so <pause dur="0.2"/> you can use that <pause dur="2.0"/> to make <pause dur="0.3"/> organometallic derivatives <pause dur="2.8"/> if the hydrogen that you want to replace <pause dur="1.7"/> has got some <pause dur="0.4"/> little bit of acidity about it <pause dur="0.9"/> so <pause dur="0.3"/> let's take a very typical example <pause dur="0.2"/> let's take <pause dur="0.3"/> let's take a classic example first and then i'll go on and show you that it's actually quite extensive <pause dur="1.1"/> # first of all <pause dur="1.0"/><event desc="breaks board" iterated="n"/> oh i've broken the board here but <pause dur="1.1"/> oh well <pause dur="2.3"/> let's take an

example where you've got a fairly strong <pause dur="0.2"/> carbon acid now <pause dur="1.7"/> carbon-hydrogen compounds are not generally acidic <pause dur="0.5"/> so you need some special kind of structural feature <pause dur="1.1"/> to make <pause dur="0.2"/> a carbon acid which is what you need <pause dur="0.4"/> one such <kinesic desc="writes on board" iterated="y" dur="5"/> compound is cyclopentadiene <pause dur="0.6"/> which is <pause dur="0.7"/> a five-membered ring of course <pause dur="0.4"/> with <pause dur="0.2"/> two double bonds in it now cyclopentadiene <pause dur="1.0"/> is peculiar <pause dur="0.2"/> it has hydrogens of course all the way <kinesic desc="writes on board" iterated="y" dur="1"/> round <pause dur="0.5"/> but two hydrogens <pause dur="0.4"/> here <pause dur="2.5"/> and <trunc>i</trunc> <pause dur="0.7"/> its <pause dur="1.0"/> unusual character is because <pause dur="1.8"/> it's rather acidic for a hydrogen compound <pause dur="0.6"/> it will give you <pause dur="7.9"/><kinesic desc="writes on board" iterated="y" dur="15"/> the anion <pause dur="0.9"/> plus hydrogen <pause dur="2.1"/> and let me just write that as hydrogen <pause dur="0.4"/> S for solvated because as you know H-plus doesn't exist in solution <pause dur="0.5"/> but i can't write water obviously because we can't do this reaction <pause dur="0.6"/> <trunc>i</trunc> in water because cyclopentadiene isn't soluble in water <pause dur="0.3"/> and in any case the anion would be <pause dur="0.3"/> decomposed by water <pause dur="0.6"/> so this will be in some <pause dur="0.3"/> other solvent <pause dur="0.2"/> that can sustain the ionization <pause dur="0.2"/> maybe T-H-F <pause dur="0.9"/> now why does this happen <pause dur="1.1"/> well the answer is very simple <pause dur="0.8"/> in the double bond

of course i've <kinesic desc="writes on board" iterated="y" dur="3"/> got two pi electrons <pause dur="1.7"/> and i've got two of them so that is four pi electrons <pause dur="1.7"/> if i take <kinesic desc="writes on board" iterated="y" dur="5"/> one of these hydrogen bonds <pause dur="0.9"/> to carbon <pause dur="0.4"/> it obviously contains two electrons <pause dur="0.7"/> if i remove the proton <pause dur="0.7"/> over here <pause dur="0.9"/> of course i take no electrons with me because the proton has no electrons <pause dur="0.5"/> i leave those two electrons behind <pause dur="0.6"/> so that gives me <pause dur="0.2"/> six electrons <pause dur="0.6"/> and of course as you know from benzene chemistry <pause dur="1.1"/> if i've got six <pause dur="0.2"/> pi electrons <pause dur="0.5"/> i have an aromatic sextet <pause dur="0.3"/> and just as with benzene that's very stabilizing <pause dur="0.5"/> so for cyclopentadiene <pause dur="0.5"/> the anion <pause dur="0.9"/> is stabilized <pause dur="0.4"/> by having <pause dur="0.4"/> six pi electrons just as you find in benzene <pause dur="0.5"/> and therefore <pause dur="0.2"/> the reaction <pause dur="0.8"/> goes <pause dur="0.7"/> much more than it would with a typical <pause dur="0.5"/> carbon-hydrogen compound with any alkane <pause dur="0.6"/> now <pause dur="0.2"/> when i say <pause dur="1.0"/> it goes much more <pause dur="0.4"/> i don't mean this is like you know acetic acid i mean <pause dur="0.8"/> that <trunc>th</trunc> this is a tiny tiny concentration <pause dur="1.3"/> but it's dynamic you you can look it up in the textbooks you can find the figures <pause dur="0.3"/> i think for example <pause dur="0.2"/> if i remember let me just

give you a guess i think the P-K-A for this molecule <pause dur="0.3"/> is something like <pause dur="0.4"/> fifteen <pause dur="0.5"/> that means <pause dur="1.4"/> that the concentration the molar concentration of hydrogen ions <pause dur="0.6"/> will be ten-to-the-minus <pause dur="0.3"/> fifteen <pause dur="0.5"/> water is ten-to-the-minus-seven <pause dur="0.4"/> so you know <pause dur="0.3"/> this is another <pause dur="0.2"/> ten<pause dur="0.6"/>to-the-<pause dur="0.2"/>eight <pause dur="0.5"/> less acidic than water <pause dur="0.6"/> but you will find <pause dur="0.2"/> that that's useful enough <pause dur="0.5"/> if you take <pause dur="0.4"/> cyclopentadiene <pause dur="2.6"/> # i'll write it out in full <kinesic desc="writes on board" iterated="y" dur="23"/> 'cause then you can see what's happening <pause dur="0.7"/> if you take cyclopentadiene with sodium <pause dur="1.5"/> what you get <pause dur="0.5"/> is <pause dur="0.6"/> # sodium cyclopentadienide <pause dur="2.1"/><vocal desc="cough" iterated="n"/><pause dur="4.8"/> and of course <pause dur="0.3"/> we've lost a hydrogen <pause dur="1.7"/> and i'll just write that as a half-H-two because you know otherwise you've got to double everything up <pause dur="0.6"/> so <pause dur="0.9"/> what i've done is i started with two hydrogens here <pause dur="1.3"/> and i've replaced one of them <pause dur="0.5"/> with sodium <pause dur="0.9"/> now of course <pause dur="0.2"/> this is <pause dur="0.5"/> an ionic compound substantially <pause dur="0.5"/> because the reason that it goes in the first place <pause dur="0.3"/> is the stability <pause dur="0.4"/> of the cyclopentadienide anion <pause dur="0.5"/> and so the product <pause dur="0.4"/> is <pause dur="1.2"/> # a salt if you like <pause dur="0.6"/> but it's in exchange of sodium

for hydrogen because if you write it out in <pause dur="0.3"/> a formula terms we start out <kinesic desc="writes on board" iterated="y" dur="11"/> with C-five-H-six <pause dur="0.6"/> we react it with sodium <pause dur="0.4"/> and the product is sodium <pause dur="0.6"/> C-five-H-five <pause dur="3.4"/> and so i've <trunc>e</trunc> substituted or exchanged <pause dur="0.4"/> one hydrogen <pause dur="0.5"/> for <pause dur="0.5"/> a sodium <pause dur="0.2"/> so it's a metal<pause dur="0.2"/>-hydrogen exchange <pause dur="0.3"/> that reaction in the textbooks is sometimes <pause dur="0.4"/> referred to <kinesic desc="writes on board" iterated="y" dur="5"/> simply as metallation <pause dur="5.1"/> some textbooks call it metallation i think that's a confusing term because <pause dur="0.6"/> any reaction which puts a metal <pause dur="0.5"/> in place of any other group be it a halogen or another metal <pause dur="0.4"/> is if you like a metallation <pause dur="0.3"/> you're adding metal to the compound <pause dur="0.9"/> but you will find that some textbooks use the term metallation <pause dur="0.4"/> so if you want to look it up you might have to look it up under the term metallation <pause dur="2.3"/> # <pause dur="0.6"/> let me give you <pause dur="0.4"/> an example where it's <pause dur="0.3"/> much more evidently an exchange <pause dur="0.9"/> suppose we take <kinesic desc="writes on board" iterated="y" dur="10"/> phenyl sodium <pause dur="1.3"/> and we take toluene <pause dur="6.7"/> now toluene of course is an even weaker acid I-E <pause dur="0.4"/> it has <pause dur="0.5"/> an even lower concentration of protons <pause dur="0.5"/> than cyclopentadiene <pause dur="0.6"/> but still <pause dur="1.1"/> it is a <pause dur="0.4"/> slightly

stabilized anion C-H-two-minus <pause dur="0.6"/> negative charge can again interact with the aromatic ring <pause dur="0.3"/> so this is a <pause dur="0.4"/> i don't know what the P-K-A for this is it must be minuscule <pause dur="0.7"/> but nevertheless it's enough <pause dur="0.2"/> in the presence <pause dur="0.3"/> of a sodium <pause dur="0.2"/> organometallic compound <pause dur="0.4"/> the reaction goes <pause dur="0.2"/> you get <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> exchange <pause dur="0.2"/> of one of the acidic hydrogens <pause dur="1.2"/> so that's benzyle sodium <pause dur="0.7"/> and you exchange that hydrogen for the sodium <pause dur="0.3"/> so the other product of course is if you like P-H-H <pause dur="0.8"/> or <kinesic desc="writes on board" iterated="y" dur="5"/> benzene <pause dur="4.5"/> so that's a typical metal-hydrogen exchange <pause dur="7.4"/> so those are the the main <pause dur="0.4"/> important type of exchange reactions <pause dur="2.9"/> let's <pause dur="0.7"/> # finish off by looking at another very important class <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="11"/> which are insertion reactions <pause dur="9.9"/> now of course <pause dur="1.0"/> i suppose in the number of <pause dur="0.4"/> compounds made and the number of times it's been used <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="4"/> the synthesis of a Grignard reagent is <pause dur="0.2"/> probably <pause dur="0.7"/> the most widely-used insertion reaction in chemistry <pause dur="0.7"/> so we'll put that down first <pause dur="1.2"/> but <pause dur="4.5"/> many textbooks <pause dur="0.3"/> don't actually include it as an insertion reaction but of course formally

it is <pause dur="0.9"/> so if i just put down the general reaction R-X<pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="7"/> plus <pause dur="0.2"/> M-G <pause dur="1.0"/> gives you <pause dur="0.6"/> R-M-G-X <pause dur="1.2"/> that is obviously <pause dur="0.3"/> in a formal sense an insertion reaction so we should include it <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="4"/> a much more typical kind of insertion reaction <pause dur="0.6"/> is a reaction <pause dur="0.5"/> which involves insertion into <pause dur="0.7"/> a metal<pause dur="1.5"/>carbon bond that's already been formed <pause dur="0.5"/> so let's take for example <pause dur="0.4"/> butyl lithium <pause dur="3.1"/><kinesic desc="writes on board" iterated="y" dur="4"/> which i'll write like that because that's the bond <pause dur="0.5"/> into which <pause dur="0.5"/> we're going to insert something <pause dur="0.8"/> now if you insert something <pause dur="0.5"/> we can use the example above the magnesium to show <pause dur="1.1"/> the <pause dur="0.2"/> the # <pause dur="0.2"/> the idea <pause dur="0.5"/> if i insert magnesium <pause dur="0.5"/> into <pause dur="0.6"/> the carbon-halogen bond of this <pause dur="0.4"/> R-X group <pause dur="2.3"/> the magnesium has got to be able to form <pause dur="0.6"/> two bonds <pause dur="0.8"/> because i've pushed it in between two elements and if the whole thing is going to stay together <pause dur="0.3"/> the magnesium has got to bond to the R <pause dur="0.2"/> and it's got to bond to the X <pause dur="0.7"/> another way of looking at this of course if you want to <pause dur="0.4"/> is to say <pause dur="0.5"/> that <pause dur="0.3"/> in terms of the metal <pause dur="1.0"/> its oxidation state increases <pause dur="0.2"/> by two <pause dur="0.7"/> oxidation state of course

simply tells you the number of electrons <pause dur="0.4"/> that the metal is using to form bonds <pause dur="0.6"/> in magnesium metal <pause dur="0.5"/> its oxidation state is zero <pause dur="0.2"/> it's not using any electrons to form <pause dur="0.4"/> any <pause dur="0.5"/> ordinary covalent bonds <pause dur="1.0"/> when it goes to the Grignard reagent it's forming two bonds <pause dur="0.3"/> so it goes from magnesium nought to magnesium two <pause dur="1.1"/> and that shows you that the the thing that's inserted has got to be able to provide two electrons <pause dur="0.3"/> so that you can form two bonds <pause dur="0.5"/> in general <pause dur="0.4"/> that means <pause dur="0.3"/> that you've got to have some kind of unsaturated species <pause dur="0.3"/> so for example suppose <kinesic desc="writes on board" iterated="y" dur="2"/> i was to take <pause dur="1.2"/> let's say <pause dur="0.5"/> an acetylene or an alkyne if you prefer <pause dur="0.4"/> so we'll take R-<pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="4"/>C<pause dur="0.4"/>-triple-bond-C-<pause dur="0.3"/>R <pause dur="0.8"/> and it's the <pause dur="0.3"/> pi bonds of the triple bond <pause dur="0.4"/> that can give rise to the new <pause dur="0.3"/> bonds <pause dur="1.5"/> this <kinesic desc="indicates point on board" iterated="n"/> can insert into the <pause dur="0.4"/> carbon lithium bond here and what you would get would be <kinesic desc="writes on board" iterated="y" dur="8"/> R-<pause dur="0.3"/>C-<pause dur="0.7"/>C-<pause dur="0.4"/>R <pause dur="1.5"/> butyl <pause dur="0.7"/> and lithium <pause dur="1.9"/> and the insertion <pause dur="0.7"/> you will notice <pause dur="0.6"/> i chose a triple bond for a very <trunc>s</trunc> particular reason <pause dur="0.8"/> and that is simply that <pause dur="0.4"/> i can show with a triple bond <pause dur="0.8"/> the

geometry of <pause dur="0.2"/> of the insertion <pause dur="0.5"/> if i'd started with a double bond <pause dur="0.4"/> i would get an alkane here <pause dur="0.3"/> and we'd have free rotation and we couldn't tell anything about it <pause dur="0.6"/> but starting with a triple bond i'm left with a double bond <pause dur="0.3"/> and so of course <pause dur="0.8"/> i can have <pause dur="0.2"/> the butyl and the lithium going cis about the double bond which is one <pause dur="0.2"/> isomer <pause dur="0.3"/> or i could have butyl here and lithium up here <pause dur="0.3"/> the trans arrangement <pause dur="0.3"/> which would also be stable <pause dur="0.3"/> different isomer <pause dur="0.5"/> but in fact they both go on the same side <pause dur="0.3"/> and so the <kinesic desc="writes on board" iterated="y" dur="1"/> mechanism of that insertion <pause dur="0.3"/> is always a cis one <pause dur="10.2"/> so that is an insertion <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="9"/> that's <pause dur="0.2"/> insertion into <pause dur="0.5"/> a metal <pause dur="0.7"/> to carbon bond <pause dur="4.4"/> <kinesic desc="writes on board" iterated="y" dur="8"/> it's also possible to have insertion <pause dur="0.6"/> into a metal <pause dur="0.3"/> to hydrogen bond <pause dur="2.8"/> have you in organic chemistry <pause dur="0.5"/> done anything about hydroboration <pause dur="2.4"/> you've heard of it have you <pause dur="0.6"/> okay <pause dur="0.6"/> <trunc>hydroboro</trunc> <trunc>m</trunc> boron isn't really a metal so in a sense it's not <pause dur="0.3"/> relevant to organometallic chemistry <pause dur="0.4"/> but this is exactly the same kind of reaction <pause dur="0.8"/>

so <trunc>f</trunc> let me just go over some familiar ground first and then <pause dur="0.5"/> i'll show you how it can be used <pause dur="0.7"/> to form <pause dur="0.5"/> # metal-carbon bonds <pause dur="2.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> if we start out with an olefin in this case <pause dur="0.8"/> doesn't matter what olefin <pause dur="0.7"/> i'll just <trunc>ha</trunc> we'll just have some general olefin <pause dur="0.9"/> here <pause dur="1.3"/> <kinesic desc="writes on board" iterated="y" dur="6"/> and i've got let's say borane <pause dur="0.8"/> now <pause dur="1.5"/> you know of course that we can't actually start with borane itself because <pause dur="0.5"/> borane exists as a dimer B-two-H-six <pause dur="0.7"/> and that's rather inconvenient <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="1"/> # so let's say <pause dur="0.2"/> we we'll <kinesic desc="writes on board" iterated="y" dur="3"/> start out <pause dur="0.8"/> with the complex with T-H-F because <pause dur="0.2"/> the T-H-F complex is stable <pause dur="0.5"/> it doesn't dimerize <pause dur="0.5"/> because you donate a pair of electrons from the oxygen <pause dur="0.4"/> of the T-H-F <pause dur="0.3"/> to the boron <pause dur="0.3"/> and then the monomer is stable and that simplifies the <trunc>r</trunc> stoichiometry of the reaction <pause dur="0.7"/> now <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="8"/> what happens here is <pause dur="0.8"/> we're going to insert into <pause dur="1.2"/> the B-H <pause dur="0.5"/> bond <pause dur="1.2"/> so the first thing of course you will get <pause dur="0.4"/> will be <pause dur="0.4"/> # <pause dur="0.2"/> well

we'll <kinesic desc="writes on board" iterated="y" dur="8"/> keep the T-H-F initially you'll get T-H-F <pause dur="0.7"/> H-two<pause dur="0.6"/>B <pause dur="0.3"/> now <pause dur="0.5"/> into the boron-hydrogen bond we're going to insert our olefin <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="7"/> so we will have <pause dur="0.5"/> C<pause dur="1.3"/>C<pause dur="1.0"/>H <pause dur="2.5"/> that's hydroboration <pause dur="0.3"/> of course <pause dur="1.6"/> you still have in this molecule here two other boron-hydrogen bonds <pause dur="0.2"/> and so <pause dur="0.2"/> if you've got enough olefin <pause dur="0.5"/> the reaction will proceed <pause dur="1.8"/><kinesic desc="writes on board" iterated="y" dur="4"/> in the presence of excess olefin <pause dur="0.5"/> the reaction will proceed <pause dur="0.4"/> until you get <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="10"/> boron <pause dur="1.1"/> C-C depends what these groups are <pause dur="1.8"/> H <pause dur="1.6"/> three times <pause dur="1.9"/> perhaps to be mathematically accurate i <kinesic desc="writes on board" iterated="y" dur="5"/> should put the hydrogen i guess inside the <pause dur="2.1"/> the bracket <pause dur="4.2"/> okay <pause dur="0.3"/> so <pause dur="0.5"/> that's an example you've come across before <pause dur="0.2"/> an unsaturated species is inserted into the boron-hydrogen bond <pause dur="0.4"/> and it's almost an organometallic example <pause dur="0.5"/> but there are many other <pause dur="5.7"/> there are many other metals well there are many metals <pause dur="2.9"/> which <pause dur="5.4"/> which undergo similar reactions <pause dur="0.6"/> and <pause dur="0.5"/> perhaps the most important of those is aluminium <pause dur="0.9"/> because <pause dur="1.2"/> the reaction between aluminium-hydrogen compounds <pause dur="0.5"/> and unsaturated species <pause dur="0.6"/> is used in industry a great deal <pause dur="0.5"/> to

make aluminium alkyls <pause dur="4.6"/> so for example if we just take <pause dur="0.6"/> a typical example <pause dur="1.2"/> # let's say we start out with <pause dur="0.2"/> dibutyl <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> aluminium <pause dur="0.4"/> hydride <pause dur="0.8"/> so that's a metal now aluminium <pause dur="0.4"/> hydrogen bond <pause dur="0.5"/> and we'll take ethene <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="5"/> just to make a nice simple example <pause dur="2.8"/> the ethene inserts into the aluminium-hydrogen bond <pause dur="0.5"/> and the product that <kinesic desc="writes on board" iterated="y" dur="7"/> you get is dibutyl <pause dur="0.5"/> aluminium ethyl <pause dur="6.8"/> i give you that example as i say it's used to make <trunc>alu</trunc> organo-aluminium compounds <pause dur="0.5"/> and you probably know that <pause dur="0.2"/> the chemical industry <pause dur="0.7"/> makes <pause dur="0.2"/> thousands of tons of aluminium alkyls <pause dur="0.9"/> every year because they're used in the catalysts that <pause dur="0.2"/> produce <pause dur="0.4"/> polythene <pause dur="0.5"/> polypropylene <pause dur="0.6"/> and a whole range of polyolefin catalysts <pause dur="0.9"/> # # <pause dur="0.7"/> are derived <pause dur="0.2"/> from organo-aluminium compounds <pause dur="0.4"/> and it's quite <pause dur="0.2"/> interesting that until you've had a bit of practice <pause dur="0.8"/> it would be extremely <pause dur="0.7"/> # difficult <pause dur="0.5"/> to carry out this reaction in the laboratory <pause dur="0.4"/> because a material like this <pause dur="0.6"/> not only reacts with oxygen <pause dur="0.5"/> if i was to have a bottle with this and take <trunc>th</trunc> <pause dur="0.4"/> the stopper out of it <pause dur="0.5"/> it

simply bursts into flames it reacts with oxygen incredibly <pause dur="0.9"/> readily <pause dur="0.8"/> they're pyrophoric materials <pause dur="0.8"/> take the stopper out and i'd just have a sheet of flame <pause dur="0.4"/> from here <pause dur="0.4"/> to the ceiling i have actually <pause dur="0.3"/> not in one of these lectures but i have actually done it in a demonstration lecture <pause dur="0.6"/> and by <pause dur="0.7"/> gosh it i mean it <pause dur="0.2"/> it certainly frightens the lecturer if it doesn't frighten the audience <pause dur="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 <pause dur="0.7"/> # materials <pause dur="0.6"/> but as i say industry <pause dur="0.2"/> has found ways i mean of course they <trunc>ha</trunc> they don't handle them in glass of course they handle them in stainless steel apparatus <pause dur="0.4"/> but you know thousands of tons of these things are made every year for industrial use <pause dur="1.6"/> okay <pause dur="0.4"/> # the other metals just so that you have a note of it i don't expect you to learn this list <pause dur="0.5"/> but the other metals that undergo this kind of reaction if i think about them <pause dur="0.6"/> # <pause dur="0.3"/> # tin <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="2"/> does it very readily <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="3"/>

silicon does it <pause dur="1.0"/> silicon is important because that reaction is used in the production of silicones <pause dur="0.6"/> for floor polishes waterproofings and so on <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="5"/> germanium does it <pause dur="0.7"/> zirconium does it <pause dur="1.3"/> don't know if i <pause dur="0.6"/> think of anything else # <pause dur="3.3"/> well that i think that those are the most important ones <pause dur="3.9"/> okay <pause dur="1.1"/> and finally <pause dur="3.3"/><kinesic desc="writes on board" iterated="y" dur="4"/> this is really <pause dur="0.5"/> # <pause dur="0.3"/> another example of insertion into metal-hydrogen bonds <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="13"/> but <trunc>usi</trunc> this particular case of using <pause dur="0.5"/> # diazomethane <pause dur="8.8"/> and it's just to show you that you <trunc>c</trunc> if you want to make methyls this is one very easy way you can do it <pause dur="0.7"/> if we were to take for example <kinesic desc="writes on board" iterated="y" dur="4"/> triphenyl<pause dur="1.7"/>silane <pause dur="2.1"/> again silicon isn't a true metal but you <trunc>cou</trunc> <pause dur="0.2"/> you can put tin if you like it works just as well in fact i'll put tin because <pause dur="0.9"/> it's meant to be a metal <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> so we've got tin <pause dur="1.9"/> and if i treat that with diazomethane <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="6"/> which i'm sure you will have come across <pause dur="0.2"/> in organic chemistry but you can shout out if <pause dur="0.3"/> i'm wrong about that <pause dur="1.1"/> diazomethane is a very useful reagent <pause dur="0.3"/> because of course <pause dur="0.2"/> it behaves as a source

of methylene <pause dur="1.2"/> because nitrogen is <pause dur="0.4"/> eliminated and nitrogen is thermodynamically so stable <pause dur="0.4"/> so the product from this <pause dur="0.5"/> is <pause dur="0.2"/> <kinesic desc="writes on board" iterated="y" dur="10"/> P-H-three<pause dur="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 <pause dur="0.7"/> and nitrogen <pause dur="0.8"/> and really it's the formation of nitrogen <pause dur="0.4"/> which is the driving force for that reaction <pause dur="8.1"/> okay so let me just <pause dur="0.7"/> since we the board has come full circle <pause dur="1.9"/> let me just emphasize for you <pause dur="2.7"/> the reactions where we'd have to be very careful to give you some feeling for this <pause dur="1.6"/> a Grignard reagent <pause dur="0.4"/> is decomposed <pause dur="0.2"/> by <pause dur="0.9"/> catalytic amounts of oxygen <pause dur="0.7"/> so you <trunc>ha</trunc> if you're making this Grignard reagent <pause dur="0.4"/> you'll have to work with pure dry nitrogen <pause dur="0.4"/> pure dry ether and get everything spot on <pause dur="3.2"/> lithium is the same <pause dur="2.8"/> sodium is even more so and there's a lesson there <pause dur="0.5"/> what what are we <pause dur="0.6"/> # <pause dur="0.5"/> actually <pause dur="0.2"/> learning from this well <pause dur="0.2"/> amongst other things <pause dur="1.1"/> sodium of course is one element lower in the periodic table than lithium <pause dur="0.7"/> so sodium is more electropositive <pause dur="0.2"/> than lithium <pause dur="0.6"/> makes the bond more

polar <pause dur="0.5"/> the species more reactive <pause dur="0.2"/> and so <pause dur="0.5"/> if you if you have to take precautions with lithium you have to take them <pause dur="0.6"/> # doubled in spades <pause dur="0.2"/> for sodium <pause dur="0.3"/> so the same thing here if you want to make benzyl sodium <pause dur="0.6"/> you must work with <pause dur="0.2"/> pure dry nitrogen <pause dur="0.3"/> dry solvent dry gases no air no water <pause dur="1.6"/> same thing here this is lithium <pause dur="0.5"/> # <pause dur="1.9"/> well <pause dur="0.2"/> here <pause dur="0.6"/> borane actually is not sensitive to water oddly enough <pause dur="0.5"/> and the reason for that <pause dur="0.6"/> is quite simply <pause dur="0.6"/> that boron and hydrogen have <trunc>r</trunc> roughly the same electronegativity <pause dur="0.6"/> and so <pause dur="0.4"/> the boron-hydrogen bond <pause dur="1.0"/> is not very polar <pause dur="0.8"/> and so it is not attacked by water but it is attacked by oxygen <pause dur="0.5"/> so all of this would have to be done you could do it wet if you wanted <pause dur="0.2"/> <unclear>i mean it'd</unclear> be a bit <pause dur="0.5"/> peculiar but you could do it wet if you wanted <pause dur="0.4"/> as long <pause dur="0.3"/> as you excluded all <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="1"/> oxygen <pause dur="0.2"/> from the reaction <pause dur="1.4"/> aluminium of course is heavier <pause dur="0.3"/> more electropositive <pause dur="0.7"/> aluminium reagents as i've said catch fire spontaneously in air <pause dur="0.5"/> they virtually explode with water <pause dur="0.2"/> so you want to be careful about wetness

there <pause dur="1.9"/> # <pause dur="2.9"/> tin <pause dur="0.5"/> is group fourteen <pause dur="0.9"/> and so <pause dur="1.1"/> for reasons that i'll explain later <pause dur="0.2"/> it <pause dur="0.3"/> the organic compounds of tin <pause dur="0.4"/> not sensitive to water <pause dur="0.2"/> and they're not <pause dur="0.2"/> sensitive <pause dur="0.3"/> to oxygen <pause dur="0.6"/> so <pause dur="0.2"/> we can summarize the sensitivity <pause dur="0.2"/> in the following way <pause dur="5.1"/><kinesic desc="writes on board" iterated="y" dur="5"/> # <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="28"/> oxygen and water <pause dur="3.4"/> is all <pause dur="0.7"/> group one <pause dur="3.4"/> all <pause dur="0.8"/> group two <pause dur="3.6"/> and group <pause dur="2.1"/> thirteen <pause dur="1.2"/> # <pause dur="1.2"/> from aluminium down <pause dur="14.2"/> # oxygen alone <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="23"/> would be <pause dur="0.4"/> # boron compounds generally <pause dur="3.5"/> and in general hydrides <pause dur="2.5"/> E-G <pause dur="1.5"/> S-I-H <pause dur="1.0"/> S-N-H <pause dur="4.6"/> now we've got one minute left so i'll finish up with just explaining <pause dur="2.2"/> the rationalization for that but we'll go on and discuss it in much greater detail next time <pause dur="0.6"/> but just <pause dur="2.4"/> to encapsulate these differences <pause dur="1.8"/> all these that we're talking about are main group <pause dur="0.5"/> metals <pause dur="1.3"/> so they all want <pause dur="0.2"/> or they all tend <pause dur="0.4"/> for reasons that we can describe in detail <pause dur="0.4"/> they all tend to have a complete octet of electrons <pause dur="0.2"/> you know the stability is associated with a complete octet <pause dur="1.9"/> now of

course <pause dur="0.8"/> group one only have one electron so a <trunc>l</trunc> a lithium methyl monomer only has two electrons not an octet <pause dur="0.8"/> magnesium <pause dur="0.6"/> Grignard reagent only have four electrons no octet <pause dur="0.3"/> group thirteen only have six electrons <pause dur="0.4"/> so these things are all <pause dur="0.3"/> highly <pause dur="0.7"/> reactive and unstable and their structures we shall see reflect those <pause dur="0.9"/> if you go <pause dur="0.5"/> boron is curious with water because it's non-polar <pause dur="1.3"/> silicon and tin of course are in group fourteen <pause dur="0.4"/> they have four electrons <pause dur="0.2"/> they can form four covalent bonds which is an octet <pause dur="0.3"/> so in general <pause dur="0.2"/> their organometallic compounds are not sensitive to water or air <pause dur="0.5"/> it's only that the hydrogen compounds <pause dur="0.5"/> because it's hydrogen <pause dur="0.5"/> are very <pause dur="1.0"/> easily oxidized so they're sensitive to oxygen <pause dur="0.5"/> okay <pause dur="0.3"/> we'll <pause dur="0.7"/> stop there