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pslct004

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<title>Rearrangements</title></titleStmt>

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<idno>pslct004</idno>

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

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

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

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

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

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The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any

way</p>

<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

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

British Academy and the Arts and Humanities Research Board. </p></availability>

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<date>16/02/2000</date><equipment><p>audio</p></equipment>

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

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<item n="partlevel">UG1</item>

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

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<u who="nm0691"> <event desc="looks through transparencies" iterated="y" dur="11"/> right when i <pause dur="1.0"/> finally find <pause dur="1.8"/> the relevant slides <pause dur="0.2"/> we can get going <pause dur="4.2"/> okay # <pause dur="1.3"/> we were talking are we <pause dur="1.5"/> sort of <pause dur="0.2"/> # the <pause dur="1.6"/> this lecture's being recorded for <pause dur="0.2"/> the benefit of <shift feature="voice" new="laugh"/> mankind <pause dur="0.4"/> so <pause dur="0.2"/> # <pause dur="0.7"/> that's why <shift feature="voice" new="normal"/> there's an additional member of the audience <pause dur="0.5"/> the <pause dur="3.0"/> form round that's being passed around you should sign which will tell us <pause dur="0.3"/> which members of the audience that should be here aren't here <pause dur="0.5"/> # <pause dur="1.2"/> was everyone here <trunc>b</trunc> <pause dur="0.4"/> present when i <pause dur="1.0"/> talked about the exam structure <pause dur="1.5"/> i think most of you were if anyone <trunc>ha</trunc> wasn't present when i discussed the exam structure <pause dur="0.3"/> can you see me afterwards <pause dur="0.3"/> 'cause we haven't got a huge amount of time now <pause dur="0.4"/> # <pause dur="4.1"/> right <pause dur="10.2"/> let me just <pause dur="5.4"/> recap <pause dur="0.2"/> we were talking about rearrangements <pause dur="0.6"/> the migrating group going trans and it occurred to me that you hadn't followed it <pause dur="0.4"/> wholly <pause dur="0.4"/> # <pause dur="1.2"/> wholly <pause dur="0.4"/> understood what i was saying <pause dur="0.5"/> let's look at this <pause dur="1.0"/> because this is in your tutorial <pause dur="0.9"/> okay <vocal desc="laughter" n="sl" iterated="y" dur="1"/> <pause dur="6.0"/><kinesic desc="writes on board" iterated="y" dur="23"/> cyclohexanol <pause dur="1.1"/> if you treat it with <pause dur="0.2"/> # in in in acid conditions <pause dur="2.0"/> dehydrating conditions <pause dur="0.4"/> you <pause dur="2.6"/> can get <pause dur="5.5"/> you

will get the formation of a <sic corr="carbocation">carbocathion</sic> <pause dur="0.6"/> but <pause dur="0.2"/> in actual fact <pause dur="14.1"/><kinesic desc="writes on board" iterated="y" dur="14"/> what reacts in cyclohexanol <pause dur="0.2"/> in acidic solution <pause dur="0.3"/> is <pause dur="0.4"/> the <pause dur="1.5"/> two groups that are trans antiperiplanar to each other <pause dur="2.3"/> so the O-H has got to be pointing down the hydrogen's got to be pointing up <pause dur="0.5"/> and <pause dur="0.2"/> it does that and you get an elimination reaction <pause dur="8.9"/><kinesic desc="writes on board" iterated="y" dur="9"/> right <pause dur="0.5"/> so <pause dur="0.4"/> elimination requires you remember from <pause dur="0.3"/> Dr <gap reason="name" extent="1 word"/>'s course or perhaps you don't <pause dur="1.0"/><vocal desc="laughter" n="ss" iterated="y" dur="1"/> # <pause dur="0.5"/> there's <pause dur="0.5"/> was <pause dur="0.8"/> elimination <pause dur="0.4"/> the groups have to be trans antiperiplanar <pause dur="0.8"/> now <pause dur="1.3"/> in the case of this weird and wonderful decalin system that i showed you <pause dur="0.5"/> # <pause dur="16.8"/><kinesic desc="writes on board" iterated="y" dur="14"/> we protonate that <pause dur="0.7"/> now i'm going to save time drawing it <pause dur="1.8"/><kinesic desc="writes on board" iterated="y" dur="2"/> i'm going to protonate it up there <pause dur="3.8"/> now <pause dur="0.2"/> can we get elimination <pause dur="2.3"/> well the answer is no because <pause dur="0.2"/> <trunc>n</trunc> we need to have a hydrogen <pause dur="1.1"/> obviously we're going to eliminate H-two-O <pause dur="0.6"/> we need to have a hydrogen in an antiperiplanar <pause dur="0.2"/> arrangement <pause dur="0.7"/> and that means that <pause dur="0.7"/> that in this case <pause dur="0.6"/> the ring

can't flip <pause dur="0.4"/> as you probably recall <pause dur="0.7"/> decalin rings <pause dur="0.7"/> fused rings in general have very very little <pause dur="0.3"/> conformational freedom <pause dur="0.3"/> so this ring is fixed <pause dur="0.2"/> and it means it can't flip such that the O-H is pointing down in an axial position <pause dur="0.7"/> and so in this case <pause dur="0.2"/> you don't get elimination <pause dur="0.3"/> what you get rather is a migration <pause dur="1.0"/> so you form a cation <pause dur="0.2"/> and you get a group <pause dur="0.2"/> migrating <pause dur="0.3"/> this group here <pause dur="1.2"/><kinesic desc="indicates point on board" iterated="n"/> for example <pause dur="0.2"/> will migrate <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> and the O-H will fall off <pause dur="0.8"/> in the normal way <pause dur="0.2"/> and <pause dur="16.9"/> and the product we eventually get is the eliminated product <pause dur="0.3"/> but now it's in actual fact <pause dur="7.0"/><kinesic desc="writes on board" iterated="y" dur="6"/> that compound <pause dur="4.3"/> it seems a bit mysterious really but it's not <pause dur="0.6"/> all it all that we're saying is that the groups have to be <pause dur="0.4"/> trans antiperiplanar <pause dur="0.2"/> to each other <pause dur="0.4"/> in order to <pause dur="0.5"/> be able to <pause dur="0.2"/> # <pause dur="0.4"/> migrate <pause dur="5.2"/> okay <pause dur="1.8"/> in general rearrangements really come into their own in cyclic systems <pause dur="0.4"/> the cyclic systems are where you can get some good transformations <pause dur="0.3"/> in particular <pause dur="0.3"/> # <pause dur="0.3"/> you can form <pause dur="0.2"/> five and six member rings and

you remember last week we were discussing the pinacol arrangement <pause dur="0.6"/> and we had <pause dur="0.2"/> two options with a <pause dur="0.2"/> with a one-two-cyclohexanediol <pause dur="2.2"/> depending on the <pause dur="0.3"/> # <pause dur="0.2"/> configuration of the diol at the start whether the O-Hs were trans or cis <pause dur="35.7"/><kinesic desc="writes on board" iterated="y" dur="24"/> now <pause dur="0.4"/> this is a good one <pause dur="0.4"/> here <pause dur="0.7"/> and this is also <pause dur="0.2"/> in your tutorial <pause dur="1.3"/> it's also <pause dur="0.2"/> if <pause dur="0.6"/> i don't normally recommend colleagues' books <shift feature="voice" new="laugh"/>only <pause dur="0.2"/> only my own should i happen to have one <pause dur="0.3"/> # <vocal desc="laughter" n="sl" iterated="y" dur="1"/> <pause dur="0.5"/> but <pause dur="0.3"/><kinesic desc="holds up book" iterated="n"/> Professor <gap reason="name" extent="1 word"/>'s little <pause dur="0.8"/> monogram on <pause dur="0.2"/> polar rearrangements <pause dur="0.3"/> has all the answers in <pause dur="0.2"/> because that's what i used to write the course before he came here <pause dur="0.3"/> even <pause dur="0.5"/> and so <pause dur="0.2"/> # <pause dur="0.5"/> it will have all the answers in <pause dur="0.8"/> it's <pause dur="0.6"/> probably not necessary to buy these things we did arrange to have all these <pause dur="0.4"/> Oxford primers in the library in the short loan collection <pause dur="0.5"/> so if you go over there you should be able to get hold of it <pause dur="1.3"/> but <pause dur="0.3"/> this is # <pause dur="0.4"/> a particularly good one because it <trunc>b</trunc> both of them give the same product <pause dur="4.2"/><kinesic desc="writes on board" iterated="y" dur="13"/> oh-oh <pause dur="0.2"/> which i've have

to draw out properly <pause dur="8.9"/> this spirocyclic compound <pause dur="17.7"/> perhaps i'll put in the # <pause dur="0.5"/> the wedge line <pause dur="0.4"/> here <pause dur="3.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> from the hash line at the back <pause dur="0.6"/> just to emphasize you see these are spiro compounds <pause dur="0.2"/> those <pause dur="0.3"/> spiro systems are are not necessarily very easy <pause dur="0.2"/> to form <pause dur="19.2"/><kinesic desc="writes on board" iterated="y" dur="9"/> but you often find <pause dur="0.5"/> that rearrangements can give you that sort of <pause dur="0.2"/> unusual <pause dur="0.4"/> # <pause dur="0.2"/> cyclic arrangement <pause dur="1.1"/> the other <pause dur="0.2"/> the other type of reaction where you get some really strange <pause dur="0.3"/> cyclic compounds are <pause dur="0.3"/> photochemical reactions <pause dur="0.7"/> but <pause dur="0.3"/> it's outside of the scope <pause dur="0.9"/> of this <pause dur="0.6"/> series of lectures to talk about that <pause dur="9.3"/><event desc="looks through transparencies" iterated="y" dur="11"/> now <pause dur="1.2"/> i'm going to put up <pause dur="3.7"/> some alternatives here i'll just # <pause dur="0.8"/><kinesic desc="turns on overhead projector showing transparency" iterated="n"/> try and <pause dur="4.6"/> because i # <pause dur="2.2"/><event desc="shuts curtains" iterated="y" dur="6"/> i'll have to shut the curtains i think <pause dur="3.1"/> because i'm so kind to you and gave you the printed handout you don't need to copy this down <pause dur="1.4"/> it's also on the web <pause dur="0.2"/> and some of that's in colour <pause dur="0.4"/> and that was brought to my attention yesterday that i'd <pause dur="0.3"/> i'd write on this sheet things are blue <pause dur="0.2"/>

when of course they're black and white <pause dur="0.5"/> but i couldn't afford colour photocopying to <pause dur="0.4"/> give to you all <pause dur="0.7"/> here we have some variations on a theme as it were <pause dur="5.4"/> and these are <pause dur="0.2"/> all <pause dur="1.4"/> what i would classify as semi-pinacol rearrangements <pause dur="16.6"/><kinesic desc="writes on board" iterated="y" dur="11"/> right <pause dur="0.2"/> the driving force is always <pause dur="0.2"/> formation of a carbocation <pause dur="0.3"/> with a positive charge <pause dur="0.2"/> adjacent to the lone pair on the oxygens <pause dur="0.7"/> that's what causes <trunc>u</trunc> us to get this pinacol arrangement <pause dur="1.8"/> that's what's happening here indeed <pause dur="0.2"/> one of those O-Hs is being lost <pause dur="0.2"/> and then <pause dur="0.2"/> a group is migrating <pause dur="1.3"/> i haven't put up the mechanism because i'm sure <pause dur="0.2"/> you will all be capable of doing that yourself now <pause dur="2.7"/> i hope so anyway <pause dur="0.6"/> right <pause dur="2.0"/><kinesic desc="indicates point on board" iterated="n"/> these are <pause dur="1.3"/> diols <pause dur="0.2"/> and and the # the classic pinacol arrangement is <pause dur="0.2"/> the rearrangement of a one-two-diol <pause dur="1.7"/> we can generate <trunc>w</trunc> there's a ready source of diols <pause dur="0.2"/> we can generate <pause dur="0.3"/> diols from <pause dur="0.3"/> <trunc>v</trunc> via the dimerization of <pause dur="0.4"/> # <pause dur="0.8"/> ketones <pause dur="1.2"/> dimerization of ketones <pause dur="0.2"/> using <pause dur="0.2"/> mediated by <pause dur="0.2"/> <trunc>n</trunc> normally magnesium <pause dur="0.4"/> magnesium and mercury or something

like that <pause dur="0.4"/> would be <pause dur="0.2"/> suitable <pause dur="0.2"/> but we can also generate diols from <pause dur="0.2"/> alkene plus osmium tetroxide <pause dur="0.5"/> should we so desire <pause dur="0.2"/> and that that way we can actually get <pause dur="0.8"/> a <trunc>f</trunc> # a handle on some unsymmetric <pause dur="1.0"/> diols <pause dur="1.5"/> there are <pause dur="1.9"/> other ways of producing diols of course <pause dur="1.2"/> osmium tetroxide always in cyclic systems gives you cis <pause dur="0.6"/> diols if you want trans diols <pause dur="0.3"/> then <pause dur="0.9"/> you can <trunc>s</trunc> <pause dur="0.3"/> also generate them from <pause dur="0.2"/> you can generate them from alkenes by <pause dur="1.7"/> epoxidization followed by <pause dur="0.4"/> # nucleophilic attack and ring opening with base <pause dur="1.3"/> so we we got a handle on diols there's there's plenty of them around <pause dur="0.2"/> so synthetically this has got <pause dur="0.6"/> enormous utility <pause dur="2.2"/>

if you want to make that <pause dur="0.7"/> type of compound <pause dur="1.9"/> and <pause dur="0.2"/> and <pause dur="0.4"/> really the gist of what i'm i'm trying to get across <pause dur="0.2"/> is that <pause dur="0.8"/> in <trunc>a</trunc> in aliphatic systems i mean <pause dur="0.2"/> that's all very well <trunc>ver</trunc> very interesting <pause dur="0.2"/> but in cyclic systems you can often use this to change <pause dur="0.2"/> the way in which the ring looks <pause dur="1.7"/> now <pause dur="0.4"/> we're not restricted to diols <pause dur="1.7"/> we are <pause dur="0.5"/> we have this range of other systems <pause dur="0.2"/> which generate <pause dur="0.5"/> in the last case a similar <pause dur="0.2"/> in the other cases the same <pause dur="0.3"/> carbocation intermediate <pause dur="0.5"/> here <pause dur="1.8"/> so <pause dur="1.8"/> where we have an amine there <pause dur="1.3"/> we <pause dur="0.6"/> treat with nitrous acid <pause dur="0.4"/> and we diazotize <pause dur="0.2"/> we then lose the nitrogen <pause dur="0.2"/> as gas <pause dur="0.3"/> it's entropically driven <pause dur="0.5"/> it's a very rapid reaction you remember <pause dur="0.6"/> in in <pause dur="0.5"/> aliphatic systems at least <pause dur="0.2"/> so we lose the nitrogen <pause dur="0.5"/> and <pause dur="0.7"/> we <pause dur="0.7"/> generate <pause dur="0.2"/> the carbocation <pause dur="1.0"/> similarly <pause dur="0.2"/> we can ring open an epoxide <pause dur="0.4"/> with a Lewis acid <pause dur="3.4"/> lithium iodide for example <pause dur="0.9"/> ring open the expoxide <pause dur="0.2"/> and <pause dur="0.2"/> we will <pause dur="0.6"/> somehow magically <pause dur="0.7"/>

presumably from solvent <pause dur="0.3"/> pick up a proton <pause dur="0.4"/> and <pause dur="0.2"/> we will <trunc>l</trunc> <pause dur="0.4"/> get once again <pause dur="1.0"/> a <pause dur="2.0"/> cation <pause dur="1.3"/> the same cation <pause dur="1.0"/> once again a Lewis acid and a <pause dur="1.2"/> alcohol halide <pause dur="3.4"/> the <pause dur="0.3"/> the silver <trunc>m</trunc> mops up the chloride silver <trunc>c</trunc> silver chloride being insoluble <pause dur="0.3"/> disappears from <pause dur="1.9"/> the <pause dur="0.4"/> the phase <trunc>th</trunc> from <trunc>th</trunc> from from availability as it were as a precipitate <pause dur="0.4"/> and <pause dur="0.2"/> we get this cation <pause dur="1.9"/> here <pause dur="0.2"/> proteination <pause dur="1.0"/> gives us the most stable <pause dur="0.2"/> driven by <trunc>m</trunc> Markovnikov's rules <pause dur="0.4"/> gives us the most stable cation <pause dur="3.4"/> right <pause dur="1.0"/> none of those cations exist very long # <pause dur="0.2"/> # <pause dur="0.3"/> they're all similiar cations <pause dur="0.6"/> but <pause dur="0.2"/> they don't exist very long because a group's going to migrate <pause dur="0.2"/> one of these groups is going to migrate <pause dur="1.1"/> depending on the migratory aptitude <pause dur="1.1"/> and <pause dur="2.0"/> we're going to get <pause dur="0.2"/> a cation there <pause dur="1.8"/> on this carbon <pause dur="0.2"/> which is adjacent to the lone pair on the oxygen <pause dur="0.2"/> which then <pause dur="0.2"/> delocalizes pumps electrons into the system <pause dur="0.2"/> and gives us a relatively stable cation <pause dur="1.6"/> okay is that <pause dur="0.4"/> does that make sense <pause dur="0.3"/> is everyone happy with that <pause dur="1.4"/> fine <pause dur="0.5"/> so these are all <pause dur="0.4"/> variations on a theme semi-pinacol type

rearrangements <pause dur="0.4"/> and <pause dur="3.7"/><kinesic desc="changes transparency" iterated="y" dur="2"/> i have got <pause dur="0.2"/> a few more examples here <pause dur="3.8"/> and we we we we have to <pause dur="0.3"/> sort of rush through them i'm afraid <pause dur="0.2"/> once again <event desc="looks through handout" iterated="y" dur="4"/> this should be <pause dur="0.4"/> i'll just check <pause dur="0.6"/> yes it's in your handout <pause dur="1.7"/> so <pause dur="0.8"/> <trunc>i</trunc> probably <trunc>i</trunc> <pause dur="0.2"/> more likely to be correct <pause dur="0.5"/> if you <pause dur="0.2"/> just leave it <pause dur="0.4"/> <trunc>with</trunc> without copying it down <pause dur="0.4"/> once again <pause dur="0.2"/> these these are all <pause dur="0.4"/> semi-pinacol rearrangements <pause dur="0.3"/> nitrous acid <pause dur="2.0"/> diazonium cation <pause dur="1.7"/> once it falls off <pause dur="0.2"/> and we then get <pause dur="0.2"/> this group <pause dur="0.4"/> migrating round <pause dur="0.2"/> note <pause dur="0.2"/> it's the trans <pause dur="1.8"/> antiperiplanar arrangement that allows migration <pause dur="4.4"/> because <pause dur="0.3"/> <trunc>yo</trunc> <pause dur="0.2"/> <trunc>th</trunc> <pause dur="0.7"/> stress the antiperiplanar because of course <pause dur="0.5"/> these are <pause dur="2.4"/> actually in a trans arrangement aren't they <pause dur="0.7"/> but they're not <pause dur="0.2"/> in a suitable arrangement for migration <pause dur="1.8"/> so it's got to be <pause dur="0.2"/> this group here <pause dur="0.9"/> <kinesic desc="writes on board" iterated="y" dur="1"/> and that group there <pause dur="0.2"/> <kinesic desc="writes on board" iterated="y" dur="1"/> which are <pause dur="2.6"/> going to be <pause dur="0.2"/> are going to be the relevant # <trunc>s</trunc> have the relevant stereochemistry <pause dur="0.5"/> and <pause dur="1.2"/> of course we then <pause dur="0.4"/> get <pause dur="0.3"/> loss of a proton from the rearranged cation to give us a ketone <pause dur="2.3"/> <kinesic desc="writes on board" iterated="y" dur="1"/>

so we've made converted a <pause dur="2.1"/> six member ring to a five member ring <pause dur="4.3"/> here <pause dur="1.1"/> we get two <pause dur="0.8"/> different stereochemistry <pause dur="2.0"/> in this case we've got <pause dur="0.2"/> <trunc>s</trunc> the O-H and the N-H-two-cis <pause dur="2.3"/> and <pause dur="1.1"/> now <pause dur="0.3"/> we have the possibility <pause dur="0.3"/> of <pause dur="0.3"/> two <pause dur="1.0"/> conformations <pause dur="0.4"/> mediating the <pause dur="0.3"/> the process <pause dur="0.4"/> here <pause dur="0.7"/> we can have <pause dur="0.2"/> this one formations of a cation <pause dur="0.3"/> migrating group trans to the leaving group <pause dur="0.3"/> note <pause dur="0.2"/> the <pause dur="0.2"/> the the <trunc>f</trunc> the fact is and i <pause dur="1.1"/> must admit <pause dur="0.2"/> <trunc>t</trunc> <pause dur="0.3"/> confess to being slightly surprised or even baffled by this <pause dur="0.4"/> that <pause dur="0.2"/> that the <pause dur="0.7"/> the azo group falls off and it appears this group does migrate selectively <pause dur="0.2"/> even though <pause dur="0.2"/> we don't get any of this neighbouring group participation <pause dur="0.2"/> and <pause dur="0.2"/> and compare the aliphatic example <pause dur="0.3"/> based on # <pause dur="0.3"/> neopentyl cation that i gave you last week <pause dur="1.2"/> where <pause dur="0.2"/> where one got really a quite a mixture of products so <pause dur="0.2"/> so even though <pause dur="0.2"/> this is going to fall off very very quickly <pause dur="0.2"/> you'd still get this this trans group migrating <pause dur="2.7"/> and we get <pause dur="0.2"/> the same <pause dur="0.2"/> similar product to <pause dur="0.3"/> previously the same product <pause dur="0.2"/> but <pause dur="0.2"/> we've got this other

conformational arrangement available here <pause dur="1.6"/> and <pause dur="0.2"/> in this case <pause dur="0.9"/> we now get <pause dur="1.5"/> migration and <pause dur="0.5"/> it's easier for the <pause dur="0.2"/> hydride <pause dur="0.2"/> to migrate <pause dur="0.5"/> because it's trans <pause dur="0.4"/> to <pause dur="0.2"/> the leaving group <pause dur="1.1"/> and so that's what happens and now we get a six member ring <pause dur="1.7"/> there <pause dur="0.7"/> of course <pause dur="0.2"/> if we get ring-flipping <pause dur="0.2"/> then we still are left with <pause dur="0.3"/> the <pause dur="2.1"/> the ring <pause dur="0.2"/> methylenes <pause dur="0.2"/> being trans to the leaving group <pause dur="0.2"/> so <pause dur="0.3"/> ring-flipping doesn't really matter there <pause dur="4.3"/> right <pause dur="0.2"/> now <pause dur="0.2"/> i have <pause dur="0.9"/> a huge range of other examples <pause dur="3.2"/> so <pause dur="1:19.6"/><kinesic desc="writes on board" iterated="y" dur="54"/> right <pause dur="2.5"/> ring expansions it's a good way <pause dur="0.2"/> to make to get ring expansions or contractions to give us # <pause dur="0.4"/> # <pause dur="0.5"/> by using rearranged <pause dur="0.3"/> materials <pause dur="0.4"/> here <pause dur="0.8"/><kinesic desc="indicates point on board" iterated="n"/> we start with cyclo-<pause dur="0.3"/>hexanone <pause dur="4.0"/> we then <pause dur="0.5"/> react it with <pause dur="0.4"/> <trunc>diazome</trunc> with sorry with <trunc>nitro</trunc> nitromethane <pause dur="2.1"/> in base <pause dur="1.3"/> nitromethane of course <pause dur="0.3"/> is <pause dur="0.5"/> # is somewhat acidic <pause dur="0.2"/> so in base we form this <pause dur="6.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> highly delocalized anion <pause dur="4.3"/> have to be very careful with diazomethane in base and don't let it dry out <pause dur="0.3"/> of course <pause dur="0.8"/> because <pause dur="0.6"/> one interesting when i for my sins used to do <pause dur="0.7"/> E-S-R spectroscopy <pause dur="0.3"/> i # <pause dur="0.4"/> we used to

use <pause dur="0.9"/> nitromethane as a <trunc>s</trunc> as a spin trap <pause dur="0.2"/> and if you shake the flask containing the nitromethane in the base <pause dur="0.4"/> you see a huge amount of radicals being produced in your E-S-R spectrometer <pause dur="0.5"/> from goodness knows what source <pause dur="0.3"/> but to me <pause dur="0.2"/> the production of radicals <trunc>spontaneous</trunc> # <pause dur="0.5"/> <trunc>spontaneous</trunc> <pause dur="0.7"/> spontaneously <pause dur="0.2"/> indicates <pause dur="0.2"/> something alarming could be happening <pause dur="0.3"/> and certainly is it says on the bottle <pause dur="0.3"/> that <pause dur="0.2"/> it is quite explosive that <pause dur="1.5"/> but <pause dur="0.5"/> that is a good # # a good nucleophile and it will react <pause dur="0.2"/> with <pause dur="0.7"/> cyclohexanone there to give us <pause dur="0.8"/> that alcohol <pause dur="0.2"/> which then <pause dur="0.5"/> we reduce <pause dur="3.9"/> # <pause dur="1.1"/> i really don't have to remind you the difficulties of making amines <pause dur="0.4"/> aliphatic amines you can't do it simply by <pause dur="0.4"/> # <pause dur="0.9"/> just # just whacking in ammonia and and doing a nucleophilic substitution <pause dur="0.4"/> type reaction <pause dur="0.3"/> because there's <pause dur="0.2"/> you get all these <pause dur="0.4"/> # quarternarized ammonium <pause dur="0.2"/> ions and that sort of thing <pause dur="0.3"/> so <pause dur="0.2"/> you <pause dur="0.2"/> this is a good way to do to <pause dur="0.2"/> to get an amine there <pause dur="0.3"/> we then <pause dur="0.8"/> we have that <pause dur="0.3"/> we then reduce it <pause dur="1.9"/> it would seem hydrogenation is the method of

choice here <pause dur="0.3"/> and <pause dur="0.3"/> then <pause dur="0.2"/> we treat <pause dur="0.2"/> with <pause dur="0.4"/> nitrous acid <pause dur="7.5"/> <kinesic desc="writes on board" iterated="y" dur="3"/> to give us <pause dur="29.0"/> <kinesic desc="writes on board" iterated="y" dur="19"/> the <trunc>n</trunc> the the <pause dur="0.9"/> <trunc>nit</trunc> <pause dur="0.2"/> the N-two falls off <pause dur="0.2"/> the cation the # the <pause dur="0.9"/> methylene group from the ring migrates and gives us the <pause dur="0.3"/> the ring-expanded product <pause dur="0.4"/> so we've got a seven member ring <pause dur="2.1"/> now alternatively <pause dur="10.6"/> <kinesic desc="writes on board" iterated="y" dur="9"/> we can use <pause dur="0.3"/> diazomethane <pause dur="1:01.1"/><kinesic desc="writes on board" iterated="y" dur="40"/> right <pause dur="0.9"/> now <pause dur="10.6"/> we can <pause dur="1.9"/> get <pause dur="0.5"/> a migration <pause dur="0.2"/> but in actual fact we get two migrations happening <pause dur="0.3"/> we get one path <pause dur="0.3"/> gives us <pause dur="25.4"/><kinesic desc="writes on board" iterated="y" dur="20"/> gives us cycloheptanone <pause dur="1.0"/> the other path <pause dur="0.6"/> is <pause dur="17.7"/><kinesic desc="writes on board" iterated="y" dur="21"/> gives us that <pause dur="3.0"/> that <pause dur="0.3"/> epoxide ring <pause dur="0.4"/> in a <pause dur="0.2"/> spirocyclic system again <pause dur="11.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> of course <pause dur="0.3"/> you can then treat that <pause dur="0.5"/> with lithium iodide <pause dur="1.5"/> as we mentioned earlier <pause dur="0.3"/> and <pause dur="0.2"/> produce <pause dur="0.3"/> the <pause dur="0.9"/> cyclohexanone <pause dur="29.1"/>

<kinesic desc="writes on board" iterated="y" dur="5"/> i guess the trick is somewhere in the <pause dur="0.2"/> the more basic conditions you use <pause dur="0.3"/> in the <pause dur="0.5"/> # <pause dur="0.2"/> diazomethane synthesis <pause dur="8.5"/> it seems to fall apart into <pause dur="0.4"/> sort of more <pause dur="0.2"/> more readily i guess would be <pause dur="0.3"/> would be the answer there's more <pause dur="0.7"/> <trunc>i</trunc> <pause dur="0.6"/> it's <trunc>n</trunc> there's not so much thermodynamic control of the product <pause dur="2.5"/> but <pause dur="0.8"/> the important point is <pause dur="0.3"/> that the root <pause dur="0.2"/> at which you do this <pause dur="0.3"/> does <pause dur="0.2"/> affect what products you get <pause dur="2.3"/> even though nominally you're getting the same intermediate there <pause dur="0.6"/> but of course the intermediates are being formed <pause dur="0.4"/> in a different way in one <pause dur="0.2"/> you <trunc>f</trunc> you generate the N-two-plus <pause dur="0.2"/> in highly acidic media <pause dur="0.3"/> in the other <pause dur="0.2"/> i <pause dur="1.6"/> it is it it is actually according to <pause dur="0.3"/> to my <pause dur="0.3"/> notes done in in acid but i guess it's not as <pause dur="0.4"/> quite as acidic as as previously <pause dur="12.5"/> another example <pause dur="36.3"/> <kinesic desc="writes on board" iterated="y" dur="29"/> right do you get that occurring <pause dur="1.6"/> no <pause dur="1.9"/> what you get is <pause dur="2.7"/><kinesic desc="writes on board" iterated="y" dur="3"/> that <pause dur="1.5"/> exclusively <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="1"/> gives us that exclusively <pause dur="9.0"/> right <pause dur="2.7"/> i <pause dur="0.2"/> think we should leave the semi-pinacol

rearrangement there <pause dur="3.4"/> and move on <pause dur="0.3"/> rapidly <pause dur="0.5"/> to <pause dur="2.2"/> the <pause dur="1.1"/> Beckmann rearrangement and <trunc>rel</trunc> <pause dur="0.2"/> related analogues <pause dur="24.7"/><kinesic desc="writes on board" iterated="y" dur="14"/> right <pause dur="0.6"/> i hope you're all happy with that <pause dur="0.2"/> idea of the semi-pinacol rearrangement <pause dur="0.3"/> we've seen <pause dur="0.3"/> the pinacol rearrangement <pause dur="0.4"/> guided <pause dur="0.3"/> governed by the formation of this <pause dur="0.3"/> # <pause dur="0.3"/> highly stable cation adjacent to the oxygen <pause dur="0.3"/> we've seen that we can modify that <pause dur="0.3"/> using these semi-pinacol <pause dur="0.8"/> type system <pause dur="0.2"/> where <pause dur="1.2"/> the classic the best example is where we have an amine <pause dur="0.3"/> on the the adjacent group carbon rather than <pause dur="0.3"/> # an O-H <pause dur="0.2"/> and then we get rid of that by diazotization <pause dur="0.7"/> and you can see <pause dur="0.3"/> how <pause dur="0.2"/> one can use that <pause dur="0.2"/> particularly to form <pause dur="0.2"/> rings that wouldn't <pause dur="0.7"/> that are larger to form <pause dur="0.2"/> ring-expanded systems <pause dur="1.3"/> or <pause dur="0.9"/> ring-contracted systems <pause dur="0.7"/> depending on the precise conditions you use <pause dur="0.5"/> but you can change <pause dur="0.2"/> the ring size <pause dur="1.9"/> it's actually very useful to be able to form a <trunc>cyclo</trunc> <pause dur="0.2"/> # <pause dur="1.0"/> heptanone <pause dur="0.6"/> or a # seven member ring <pause dur="0.3"/> it's a good it's a good skill to <trunc>le</trunc> to have in our armoury as <pause dur="0.3"/> as synthetic

chemists <pause dur="0.8"/> okay so it's useful to be able to know <pause dur="0.3"/> that these types of reactions lead us to seven member rings <pause dur="0.8"/> albeit with some <pause dur="0.2"/> complications <pause dur="0.3"/> particularly with the <pause dur="0.3"/> the diazomethane <pause dur="0.8"/> the diazomethane is a <pause dur="0.2"/> is a good is a really good reagent 'cause you can buy it from Aldrich <pause dur="0.4"/> and so <pause dur="0.2"/> that means that you don't have to go through several steps <pause dur="0.3"/> and <pause dur="0.2"/> it means that that that it much <pause dur="0.2"/> quicker than <pause dur="0.5"/> the <trunc>ste</trunc> the the the <trunc>a</trunc> the alternative methods i've shown you which <pause dur="0.2"/> require you to synthesize <pause dur="0.2"/> you know <pause dur="0.3"/> where does that come from <pause dur="1.1"/> well <pause dur="0.5"/> goodness only knows <pause dur="0.9"/> look at the look at the the <trunc>m</trunc> the the <pause dur="0.7"/> two steps you have to get to even to get to the amine <pause dur="0.3"/> in the <pause dur="0.3"/> first synthesis <pause dur="0.9"/> right <pause dur="0.3"/> so <pause dur="0.2"/> bit of a downer that you get a side product but you can convert the side product <pause dur="1.8"/> synthetic utility <pause dur="0.2"/> of <pause dur="0.3"/> semi-pinacol rearrangements <pause dur="0.3"/> and pinacol rearrangements <pause dur="0.6"/> it's <pause dur="0.3"/> particularly <pause dur="0.2"/> to my mind <pause dur="0.2"/> where you want to form larger rings <pause dur="0.9"/> okay <pause dur="1.0"/> right so <trunc>summarizi</trunc> i've summarize that as i <pause dur="0.4"/> always do after i've started the

title of the next bit <pause dur="2.1"/> the Beckmann rearrangement <pause dur="1.6"/> like i say <pause dur="0.2"/> i <trunc>d</trunc> <pause dur="0.7"/> i don't like all these named reactions but <pause dur="0.7"/> it is <pause dur="0.5"/> helpful to have a title for a section <pause dur="0.3"/> and <pause dur="1.3"/> this is the rearrangement <pause dur="11.1"/><kinesic desc="writes on board" iterated="y" dur="7"/> now to my ignorant <pause dur="0.4"/> <trunc>poi</trunc> # <pause dur="0.2"/> point of view <pause dur="0.5"/> the Beckmann rearrangement's the only <pause dur="0.2"/> sort of <pause dur="0.2"/> really big <trunc>rac</trunc> reaction of oximes <pause dur="0.3"/> of course <pause dur="0.4"/> in actual fact i will be <pause dur="0.2"/> hauled over the coals if i said that in <pause dur="0.2"/> the presence of <trunc>so</trunc> genuine synthetic chemists because they <pause dur="0.3"/> have all sorts of <pause dur="0.3"/> reactions and there's all sorts of chemistry of oximes that goes on <pause dur="0.3"/> but this is <pause dur="0.2"/> this is this is the biggie <pause dur="0.3"/> # <pause dur="6.3"/> perhaps i'm bias because this is the way that nylon is formed <pause dur="0.4"/> for example <pause dur="1.6"/> now <pause dur="0.2"/> what is the Beckmann rearrangement i'm just reminding myself <pause dur="0.5"/> ah yes <pause dur="12.8"/><kinesic desc="writes on board" iterated="y" dur="13"/> an oxime <pause dur="1.4"/> there is an oxime <pause dur="0.2"/> now you can get an oxime <pause dur="0.2"/> from <pause dur="0.8"/> the reaction of a ketone <pause dur="0.2"/> with <pause dur="0.4"/> a <pause dur="3.4"/> <trunc>hydro</trunc> with <sic corr="hydroxylamine">hydroxyl xylamine</sic> <pause dur="1.6"/> ketone plus hydroxylamine <pause dur="0.2"/> goes to <pause dur="0.7"/> oxime <pause dur="0.4"/> so we have <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="21"/> we would have our <pause dur="11.2"/> # <pause dur="11.2"/>

normally we use hydroxylamine hydrochloride and buffer it with # <pause dur="0.2"/> sodium acetate <pause dur="0.8"/> just like your first year practical <pause dur="0.5"/> you remember <pause dur="0.7"/> the one that doesn't work <pause dur="0.8"/> well <pause dur="0.3"/> <shift feature="voice" new="laugh"/> no <vocal desc="laughter" n="sl" iterated="y" dur="1"/><shift feature="voice" new="normal"/><pause dur="0.9"/> but <pause dur="0.3"/> well perhaps there are more of them that <pause dur="0.4"/> don't work but the one where you make the semi-carbazone <pause dur="0.5"/> and you were adding in buckets of sodium acetate and buckets of <pause dur="0.4"/> # semi-carbazone hydrochloride it's <trunc>s</trunc> very similar <pause dur="1.2"/> so you buffer it with sodium acetate <pause dur="1.0"/> easy to produce oximes <pause dur="3.0"/> now <pause dur="0.9"/> if you've got an oxime with a specific stereochemistry <pause dur="0.4"/> then <pause dur="1.1"/> like that <pause dur="0.3"/> then that's that's quite important <pause dur="0.3"/> because <pause dur="0.3"/> in fact the stereochemistry will <pause dur="0.3"/> follow through in the process <pause dur="1.0"/> what happens now we protonate we go to <pause dur="8.9"/> <kinesic desc="writes on board" iterated="y" dur="8"/> to that <pause dur="1.6"/> make that a good leaving group and that's going to fall off <pause dur="3.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> and as it falls off <pause dur="0.2"/> the group <pause dur="0.3"/> trans to the leaving group migrates <pause dur="34.9"/> <kinesic desc="writes on board" iterated="y" dur="32"/> so we're left with that horrible intermediate <pause dur="0.2"/> which then picks up water <pause dur="6.9"/><kinesic desc="writes on board" iterated="y" dur="7"/>

or i should say picks up hydroxide from water <pause dur="7.6"/><kinesic desc="writes on board" iterated="y" dur="6"/> to give us that <pause dur="1.1"/> which then <pause dur="0.2"/> tautomerizes <pause dur="21.8"/><kinesic desc="writes on board" iterated="y" dur="18"/> so the net result is <pause dur="0.2"/> we've generated an amide <pause dur="2.6"/> from <pause dur="1.2"/> our <pause dur="7.8"/> from from our oxime <pause dur="20.1"/><kinesic desc="writes on board" iterated="y" dur="6"/>

right <pause dur="0.3"/> now <pause dur="0.3"/> pay attention <pause dur="1.2"/> the group trans to the leaving group migrates <pause dur="1.9"/> if you get some <pause dur="0.2"/> of a mixture of material <pause dur="0.2"/> which doesn't contain <pause dur="0.4"/> just one group migrating <pause dur="0.2"/> it means either you started off with something of mixed stereochemistry <pause dur="0.7"/> or <pause dur="0.3"/> the system <pause dur="0.2"/> has <pause dur="0.2"/> somehow <pause dur="0.3"/> # <pause dur="0.6"/> isomerized <pause dur="0.2"/> in situ <pause dur="2.3"/> now this one <kinesic desc="writes on board" iterated="y" dur="14"/> for example <pause dur="0.9"/> let's show you what i mean <pause dur="10.8"/> oh sorry what am i doing <pause dur="0.8"/> C <pause dur="0.6"/> <kinesic desc="writes on board" iterated="y" dur="8"/> <gap reason="inaudible" extent="1 word"/> <pause dur="7.5"/> you get exclusively the phenyl group migrating <pause dur="3.9"/> but <pause dur="0.4"/> certainly <pause dur="0.2"/> there are <pause dur="0.6"/> there is some evidence to suggest <pause dur="0.4"/> that <pause dur="40.5"/><kinesic desc="writes on board" iterated="y" dur="40"/> you <trunc>w</trunc> <pause dur="0.2"/> you might you get a trace of that <pause dur="1.0"/> where both groups are aliphatic particularly according to <pause dur="0.4"/> # March <pause dur="1.4"/> March being a particularly useful <pause dur="0.3"/> textbook for <pause dur="0.3"/> sort

of detailed synthetic information <pause dur="1.0"/> so <pause dur="0.8"/> there but what's happening there well it's not possible that that <pause dur="0.6"/> that group can migrate it's not trans <pause dur="0.3"/> what happens is <pause dur="0.2"/> that you get <pause dur="2.0"/> <kinesic desc="writes on board" iterated="y" dur="3"/> H-plus <pause dur="0.2"/> and you can hydrolize the oxime <pause dur="0.3"/> back to the ketone and the hydroxylamine <pause dur="0.9"/> and then <pause dur="0.2"/> you <pause dur="0.6"/> reform it <pause dur="27.3"/> <kinesic desc="writes on board" iterated="y" dur="22"/> you reform it as the other isomer <pause dur="0.2"/> in actual fact you don't need to even go that far <pause dur="2.2"/> you don't need to go that far as soon as you protonate it then you you you <trunc>y</trunc> <trunc>y</trunc> if you protonate across here <pause dur="0.4"/> then you've got the option of the bond rotating <pause dur="0.2"/> of course <pause dur="0.2"/> but you're aware <pause dur="0.2"/> because you did this practical in the first year <pause dur="0.2"/> of the danger of adding H-plus <pause dur="0.4"/> to <pause dur="0.6"/> these sorts of <trunc>derivati</trunc> derivatives <pause dur="0.7"/> because <pause dur="0.2"/> they are it is a reversible reaction <pause dur="0.7"/> and i guess <pause dur="0.3"/> here <pause dur="0.4"/> the <pause dur="1.8"/> the exception to the rule <pause dur="0.2"/> is only because <pause dur="0.2"/> somehow you're getting some isomerization in the presence of the acid <pause dur="3.8"/> but <pause dur="0.4"/> on the whole <pause dur="1.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> the trans migration is a <pause dur="0.2"/> is a golden

rule <pause dur="0.9"/> so we can we can we can stick to that <pause dur="0.4"/> now <pause dur="11.4"/> you could do you could do other things <pause dur="0.2"/> for example you can use <pause dur="13.2"/><kinesic desc="writes on board" iterated="y" dur="17"/> <trunc>s</trunc> sorry P-C-L-five <pause dur="2.8"/> you use P-C-L-five <pause dur="0.2"/> in which case <pause dur="0.2"/> you get <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="12"/> one of these inorganic <pause dur="10.8"/> you get that which is a really <pause dur="0.3"/> whizzo leaving group as it were <pause dur="0.2"/> because of course it's going to form the phosphorous-oxygen double bond <pause dur="1.4"/> or you can <pause dur="0.2"/> make <pause dur="0.2"/> the tosylate <pause dur="3.4"/> or you can use <kinesic desc="writes on board" iterated="y" dur="32"/> boron trifluoride <pause dur="43.3"/> <kinesic desc="writes on board" iterated="y" dur="11"/> <trunc>perit</trunc> <trunc>t</trunc> # toluene <trunc>sulf</trunc> sulfinyl chloride <pause dur="1.3"/> forms a tosylate which also is a good leaving group as you remember <pause dur="0.4"/> the S-O-three-minus <pause dur="0.2"/> because <pause dur="0.2"/> para-toluenesulfonic acid <pause dur="0.2"/> is extremely acidic <pause dur="1.1"/> now the last point i wanted to <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> make on the # <pause dur="8.3"/><kinesic desc="writes on board" iterated="y" dur="7"/> sorry on the Beckmann rearrangement is <pause dur="0.2"/> nylon <pause dur="37.6"/> <kinesic desc="writes on board" iterated="y" dur="37"/> right <pause dur="1.5"/> epsilon caprolactam <pause dur="0.5"/> seven member ring <pause dur="1.3"/> formed <pause dur="0.2"/> from this oxime <pause dur="0.2"/> the oxime is of course

produced from <pause dur="19.5"/><kinesic desc="writes on board" iterated="y" dur="16"/> from that <pause dur="0.8"/> then <pause dur="0.3"/> we treat that with heat and # <pause dur="0.5"/> it can be acidic or basic conditions <pause dur="1.0"/> but largely heat <pause dur="27.4"/><kinesic desc="writes on board" iterated="y" dur="27"/> and that polymerizes <pause dur="1.2"/> to form nylon-six <pause dur="1.1"/> the six is <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> the number of carbons there in the monomer repeat unit <pause dur="2.2"/> nylon-six-six is where you have two monomers and that's the one that you <pause dur="0.5"/> may have seen the rope trick <pause dur="0.7"/> where you <trunc>ra</trunc> react adipoyl chloride <pause dur="0.3"/> with <pause dur="0.3"/> <trunc>he</trunc> hexane diamine <pause dur="1.8"/> but <pause dur="0.3"/> this is a commercial production of nylon <pause dur="0.2"/> epsilon caprolactam is <pause dur="0.2"/> a seven member ring <pause dur="0.2"/> formed <pause dur="0.3"/> from <pause dur="0.3"/> cyclohexanone <pause dur="0.4"/> and <pause dur="0.3"/> in nineteen-seventy-five this was a compound that was in the news <pause dur="1.0"/> when i was <pause dur="0.2"/> relatively young <pause dur="0.4"/> was <shift feature="voice" new="laugh"/> most of you weren't even born i suspect <shift feature="voice" new="normal"/><pause dur="0.4"/> # <pause dur="0.4"/> because it was the <trunc>appr</trunc> <pause dur="0.4"/> epsilon caprolactam plant at Flixborough that blew up <pause dur="0.4"/> creating <pause dur="0.7"/> certain amount of local rearrangement as it were <pause dur="1.6"/><vocal desc="laughter" iterated="y" dur="1"/> okay <pause dur="0.2"/> well i think we better finish there thank you for your attention

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