Lecture Stoechiometric Organometallics 11 - 19.11.13
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| Teil | 11 | |
| Anzahl der Teile | 22 | |
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| Identifikatoren | 10.5446/37202 (DOI) | |
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00:00
GastrinSpurenelementFunktionelle GruppeAdvanced glycosylation end productsKonformationsisomerieÜbergangszustandHydroxylgruppeKohlenstofffaserAromatizitätWursthülleChemische ReaktionIsomerChemischer ProzessKohlenstoffgruppeWasserHydrolysatStereoselektivitätSäureAdditionsverbindungenUmlagerungChromosomenkondensationInselKörpertemperaturMannoseGesundheitsstörungDeltaAldehydeMultiproteinkomplexCarbonylverbindungenMetallAldolreaktionMenschenversuchClaisen-UmlagerungRacemisierungHydroxyaldehydeVorlesung/Konferenz
05:14
WursthülleSterische HinderungAlkoholische LösungWasserstoffChemischer ProzessFunktionelle GruppeÜbergangszustandReaktivitätChemische ReaktionAktivierungsenergieMischenIonenbindungFlüssigkeitsfilmSubstituentThermoformenKonformationsisomerieVorlesung/Konferenz
09:39
SubstituentTurnover <Physiologie>PhenylgruppeCupcakeNebenproduktStereoselektivitätWasserstoffHydrolysatAldehydeBenzodiazepineLithiumVorlesung/Konferenz
11:19
CobaltoxideKohlenstofffaserLithiumHydrolysatFremdstoffFunktionelle GruppeKoordinationszahlFettÜbergangszustandWursthülleTieftemperaturtechnikPropylgruppeFärbenSenseOrganische ChemieStereoselektivitätWasserstoffChemische StrukturElektronentransferEthylgruppeSterische HinderungRacemisierungSäureAdditionsverbindungenIsobutylgruppeAlkylierungChelatbildungHydrocarboxylierungAlkohole <tertiär->BenzaldehydNatriumEnoleSubstrat <Chemie>Vorlesung/Konferenz
18:14
KaliumhydroxidÜbergangszustandAlphaspektroskopieHydrolysatFunktionelle GruppeSäureOxidschichtEnergiearmes LebensmittelStereoselektivitätCarboxylateChemische ReaktionWursthülleMagnesiumGesundheitsstörungWerkzeugstahlWasserstoffPhenylgruppePhenylacetatKonjugateAlkohole <tertiär->Grignard-ReaktionAlkylierungCarbonylverbindungenRacemisierungEnoleIsomerBenzaldehydVorlesung/Konferenz
25:08
KaliumhydroxidWasserstoffGenexpressionFunktionelle GruppeWursthülleÜbergangszustandSubstitutionsreaktionAbleitung <Bioelektrizität>SenseEnergiearmes LebensmittelFärbenQuerprofilDoppelbindungKohlenstofffaserGletscherzungeChemische ReaktionPhenylgruppeMagnesiumSetzen <Verfahrenstechnik>EnoleAldehydeBenzodiazepineVorlesung/Konferenz
31:45
KaliumhydroxidChemische ReaktionFunktionelle GruppeEtherKörpertemperaturStereoselektivitätPrimärelementBisacodylLithiumCyclohexanÜbergangszustandSterische HinderungFärbenSystemische Therapie <Pharmakologie>Gangart <Erzlagerstätte>GesundheitsstörungBorChlorideCobaltoxideAldehydeEnoleIonenbindungBenzodiazepineOrganisches LösungsmittelRacemisierungEthylaminCyclopentanonAldolreaktionVorlesung/Konferenz
38:11
Metallmatrix-VerbundwerkstoffMedroxyprogesteronBlätterteigKaliumhydroxidMischenOrganokatalyseWursthülleKetoneHydrocarboxylierungSystemische Therapie <Pharmakologie>ÜbergangszustandChlorideGesundheitsstörungSäureCarbonateChemische ReaktionAldehydeBorTitanchlorideAldolreaktionVinyletherSensePufferlösungEnoleStereoselektivitätOrganische ChemieAmineIsomerMolekülProteinogene AminosäurenFunktionelle GruppePropionaldehydThermoformenMetallorganische ChemieEthanPyridinEnamineProlinMähdrescherSiliciumChromosomenkondensationKörpertemperaturAmine <primär->AromatizitätBukett <Wein>FließgrenzeHydroxyaldehydeTrifluormethansulfonsäureChiralität <Chemie>EtomidatSubstrat <Chemie>IonenbindungClaisen-UmlagerungHomocysteinBindegewebeSetzen <Verfahrenstechnik>Aktivität <Konzentration>SilylierungLevomethadonWasserstoffPantoprazolHeterodimereAdditionsverbindungenKohlenstofffaserGenDeprotonierungFormaldehydChemische EigenschaftBleitetraethylElektronische ZigaretteFärbenMilchproduktAktivierung <Chemie>Vorlesung/Konferenz
47:37
TillitStereoselektivitätMetallOrganokatalyseAldolreaktionElektronentransferBorMetallorganische ChemieWeinfehlerFunktionelle GruppeOxideChemischer ProzessUmkristallisationChemische ForschungAdvanced glycosylation end productsChemische StrukturKohlenstofffaserThermoformenStickstoffatomOxazolidinBenzodiazepineAldehydeHydrocarboxylierungFließgrenzeChiralität <Chemie>StöchiometrieCarbonylverbindungenAuxiliarEnoleChemische ReaktionPhenylgruppeLactitolVorlesung/Konferenz
53:19
StickstoffatomGoldCarbonateEnergiearmes LebensmittelElektronische ZigaretteKohlenstofffaserAllylalkoholAluminiumhydridReduktionsmittelBleierzUmlagerungAlkohole <tertiär->WasserLithiumChemischer ProzessChlorideAktivierung <Chemie>WursthülleMähdrescherIonenbindungThioesterChemische ReaktionMolekülKohlenstoffgruppeGesundheitsstörungStereoselektivitätSchutzgruppeInterferonFunktionelle GruppeEthylgruppeSäureEsterAuxiliarEnantiomereKettenlänge <Makromolekül>EnoleTrimethylsilylgruppeSiliciumtetrachloridAldehydeClaisen-UmlagerungValinAldolreaktionAcroleinBaseCarbonylverbindungenHydrolysatChiralität <Chemie>OxazolidinVorlesung/Konferenz
01:00:25
UmlagerungHydrolysatSystemische Therapie <Pharmakologie>StereoinduktionFunktionelle GruppeMannoseClaisen-UmlagerungAlkohole <tertiär->Vorlesung/Konferenz
01:02:33
Computeranimation
Transkript: Englisch(automatisch erzeugt)
00:00
So welcome to lesson 11. In the preceding lesson we all learned that the Ireland Claisen rearrangement proceeds via a six-membered aromatic transition
00:22
state which explains the stereoselectivity we have observed. Today's subject is stereoselectivity in aldol addition reactions and we will see that the stereoselectivity also in this case is explained by six-membered aromatic transition states. In this case these
00:47
are called Zimmerman-Truxler transition states. So again also six-membered
01:10
aromatic transition states. Well aldol addition reactions for instance an
01:23
aldehyde reacts with an enol. M is a metal. Here we have in focus a cis enolate
01:49
or Z-enolate and this will result after hydrolysis. So first reaction lower
02:05
temperature, secondly first reaction hydrolysis. This will give predominantly this better hydroxy-carbonite group and this reaction generates two
02:37
stereogenic center and if we start with the Z-enolate we will get
02:45
predominantly the Zn diastereoisomer. Of course racemic. So predominantly Zn but racemic. It is clear that we have to pay attention with the
03:09
hydrolysis process because under more vigorous reaction conditions for the hydrolysis we would observe then the aldol condensation process. So
03:29
we should avoid eliminating water which would then lead us to the
03:44
niles and our stereogenic centers are simply gone. So avoid this. How could we explain the observed stereo selectivity? Well as I said six-membered
04:05
transition state. Forgetting to the six-membered transition state the Lewis acidic metal. It's a metal cation essentially. Complexates at the
04:25
carbonyl group thus increasing the electrophilicity of that delta positive carbon. The reactive conformation should be drawn like this.
05:03
So the metal coordinates at the carbonyl group and now we can see we have
05:21
those six centers which participate in the reaction. One, two, three, four, five, six. So let's put it like this and then you will reach this situation and
06:30
in between we have the transition state where we would simply write those bonds in dotted lines. Well as I said this is the preferred conformation. Of
07:04
that we get a racemic mixture at the end. Now we have to think about well what are other conformers that could in principle form a reactive
07:22
intermediate. Well so in a reactive intermediate that could go into that transition state. Well simply change the R2 prime and that hydrogen. Let's see how that would look like. So in the following transition state we would
08:21
have R2 prime in an axial position just as this one and those two groups well are of course would have a repulsive steric interaction. This is
08:43
unfavorable and this means that the equilibrium between this conformer and this one is predominantly on this side and in addition the activation barrier
09:07
for the transition state should be lower in this case which finally explain the stereoselective outcome of the process. So let's have a look at
09:39
special examples where we know what the substituents are, generally are. Well
09:48
okay here a phenyl group with starting with benz aldehyde and lithium enolates with R, one substituent R. So first starting the reaction at
10:07
minus 78 degrees again careful hydrolysis as pointed out already
10:25
predominantly the Zun product should occur as explained above but with the
10:42
trans, with the anti product as the byproduct. So if substituent R is
11:01
very small a hydrogen well we don't get any stereoselectivity. We can imagine if R, so this here is just a hydrogen we don't have to bother with
11:25
steric repulsion very much and we don't get a good selectivity, no selectivity in this case. A bit larger an ethyl group then we have a selectivity
11:40
of 90 to 10 with an isopropyl group it stays essentially the same 90 to 10 and with tertiary butyl group we increase the selectivity to about 99 to
12:07
1. So this is then excellent selectivity. Another example where we reach 100 percent almost 100 percent diastereoselectivity should follow
12:27
now and this case it has a special reason an additional chelate effect starting with this substrate with TMS protected alcohol here LDA THF minus 78
12:52
degrees will give this enolate the lithium chelated by the two oxygen
13:11
atoms so that enolate. Adding benzaldehyde at this low temperature
13:41
then the Lewis still Lewis acidic lithium will coordinate in equilibrium to the carbonite group of the benzaldehyde. So final group here looks a
14:39
bit crowdy with that large group in axial position but it's only that
14:46
hydrogen in the one three diaxial position and the lithium is very well coordinated. So a chelate effect goes through that Zimmerman
15:04
Traxler transition state. Now the lithium is bound to this oxygen from
16:09
the carbonyl exchange to become the alkylate and the lithium is still chelated by the other two oxygens. Well to translate now this structure in
16:31
a stereo well to translate it in another drawing and well it makes sense to follow the rules of Kahn-Ingold-Prelog and determine the
16:45
absolute configuration of the stereogenic centers in this drawing. Then we will have an R here and S configuration at this position and
17:02
after hydrolysis we will get to the final product this one. So but of
17:42
course racemic so we should put on those stars here. In literature I found
18:02
that one can oxidase oxidize a setup like that with sodium pair your date to get to the acid. So by oxidation you can get rid of this part however I
18:37
careful with that information I'm not sure that one does not have to protect
18:45
that alcohol also secondary alcohols could be oxidized under reaction conditions. Let's change to an interesting special case it's known
19:14
under the name Ivanov reaction an instructive example starting with
19:33
phenyl acetic acid and treating that with two equivalents of a simple alkyl
19:49
Grignard. The first equivalent will of course react here the most acidic position we get the magnesium carboxylate and the second equivalent is
20:09
basic enough to deprotonate that alpha position. While the alpha position is still activated by the carbonyl group but also activated because of a
20:23
conjugation with a phenyl group. So we have the carbox the enolate of a
20:45
carboxylate in this case. So and it's clear there are no cis-trans or anti isomers of this enolate. Do we nevertheless observe stereoselectivity
21:16
for instance if a reaction with benzaldehyde indeed we do after
21:51
hydrolysis we will have anti and soon diastereoisomer of course racemic
22:04
and for selectivity is not that excellent but 76 to 24 so approximately three to one. How do we explain the selectivity in this case?
22:29
Again Zimmerman-Traxler transition states we could have as transition
23:20
state either this one and it's mirror image or well okay we could also
23:36
change the phenyl group and that hydrogen but as we worked out earlier
23:45
the phenyl group should remain in the equatorial position. It will become clearer with a Newman projection which will follow. Essentially we should
24:01
have a look at this configuration or a second one where the phenyl group and
24:21
that hydrogen change position based on the selectivity preferentially anti
25:03
observed in this case you have to assume that this situation is favored compared to that one. Now the question is why is this the case? Well one tends
25:25
to argue the phenyl group this phenyl group here in essentially an axial position of the of a chair like transition state is disfavored but why
25:41
should it be disfavored? Axial position generally is disfavored because of 1,3-diaxial interaction. Well there is no other axial substitute so this
26:03
can't be the argument here. In this case it makes sense to draw a Newman projection of the transition state where we have a look at to the axis of this
26:32
carbon and that carbon. Okay so Newman projection hydrogen this carbon is that
26:49
one here we have a carbon with a C-C double bond that carbon is located
27:18
behind this one and well the magnesium is here and if you want to
27:40
draw the transition state of course we have to draw that as dotted lines
27:46
and there should be the dotted line between the carbon in front and the carbon behind this would be the transition state. So let's draw the
28:13
transition state once again for this situation phenyl group here
28:25
hydrogen there so so what could be the explanation why this transition
29:02
state is disfavored compared to this one. Well now we can see that here in this area it is more crowded compared to this one. Well and you should know
29:30
the special expression for a situation like that it is the so-called gauche interaction during the transition state of the state. Here we have an
29:47
unfavorable gauche interaction well in this case with those two phenyl groups
30:05
also an unfavorable gauche interaction but this situation simply is worse than that that's the reason why this is preferred rather simple. Okay so the
30:23
message is in this case the stereo differentiation has nothing to do with 1-3-diaxial interaction gauche interaction we have to keep in mind. Next example once again with benz aldehyde it's obviously a
30:58
test aldehyde for this type of reactions but now a boron enolate
31:16
with two n-butyl groups. This setup results in the zen product selectively
31:41
of course racemic but true here well it's let's count one two three four
32:06
five six seven yeah it's it's right of course racemic zen and the better
32:21
than 97% diastereoselectivity so this is quite remarkable compared to what we are used to with lithium enolates. Generally the diastereoselectivity
32:44
of the aldol addition processes with boron enolates it is much better than with lithium enolates. The reason for that simply is that the
33:00
oxygen boron bond is much shorter well it's about 1.5 angstrom whereas with lithium enolates we are in the range of one of 2.0 angstroms. So if you
33:23
then compare the transition state the system has to run through the transition state with boron enolates is more compact and therefore those steric interactions and gauche interactions and so on you know simply work better
33:45
with a more compact system. Moreover the boron enolates it is very well tested how to get to the zen and anti enolates. For instance starting with
34:22
this is simply pentanone typical reaction conditions for the zenolate are
34:58
the corresponding triflate applied then diisopropyl ethyl amine also known as
35:26
eunuch space reaction in ether at minus 78 degrees 30 minutes and we get
35:42
the zenolate with high selectivity 97 percent. Next step reaction with
36:04
benz aldehyde at the same temperature was tested to give 99 percent zen of
36:24
course always racemic. Let's change the reagents a bit to cyclohexyl groups
36:44
at the boron and not the triflate but the chloride. A less sterically hindered amine triaphyl amine same solvent diaphyl ether same temperature a bit
37:11
shorter reaction time 10 minutes and astonishingly we get predominantly the
37:26
E enolate with better than 99 percent selectivity. Reaction with benz aldehyde
37:45
consequently leads to the racemic anti product with better 97 percent. So how
38:13
can we explain the difference without going into details most important
38:21
difference between those two reaction conditions is that we have the triflate here and the chloride there. Because of the triflate this boron is
38:41
much more Lewis acidic. In this case the boron will attack that carbonyl before that ketone is deprotonated. So in this case we have this situation as an
39:08
reactive intermediate this is deprotonated and this indeed makes the
39:35
difference you see it in the outcome. Here this is far less Lewis acidic
39:42
therefore the triaphyl amine generally has to deprotonate first with carbonyl and then the boron attacks. I think this is essentially the difference of those two systems. Well what we can also learn from literature
40:04
is if we somehow have wrong enolate we can find reaction conditions for trans isomerization of the enolates. Well in that case we will get the
40:29
thermodynamic equilibrium. This can be done with buffered systems for instance pyridine in combination with an acid or also it's a literature
40:49
treating it with simply pyridine at higher temperatures. So there are some possibilities for the equilibrium. Lutidine triflic acid at 70 degrees
41:09
was introduced I think by the Evans group. I hope you remember a special
41:24
case of aldol addition reaction we already talked about. Mukayama aldol addition. For Mukayama aldol addition we have
41:51
silyl enol ethers as substrates for instance with TMS enolate. Or we talked
42:05
about Mukayama aldol in connection with Ireland Claisen rearrangements I think. So and the aldehyde. Generally with Mukayama aldol the
42:23
electrophilicity of the aldehyde has to be triggered to be increased by a Lewis acid. Could be titanium tetrachloride for instance. So we have a Lewis acid coordinated at the carbon ion group. So also in this case we will
42:50
get some stereoselectivity. But I think it's important to know that in Mukayama aldol we don't have that Zimmerman-Trexler transition state. We
43:05
don't have a closed aromatic transition state because the silicon is not a Lewis acid good enough to coordinate at the carbon ion group. And as we have seen
43:23
we need a stronger Lewis acid to get the activation done. Okay. Nevertheless we can achieve high diastereoselectivity and even some Anansio selectivity if we apply chiral Lewis acids. And we have chiral
43:52
information already in the system. So just these remarks for the Mukayama
44:05
aldol addition. Then a special case is aldol addition by organocatalysis.
44:28
This has per definition nothing to do with stoichiometric organometallics. But we have we have to have a closer look to competing systems competing with
44:47
organometallics. So what can we achieve with organocatalysis? Catalysis with small organic molecules. Let's take propionic aldehyde and the most
45:14
typical organocatalyst proline we need 10 mole percent of this amino acid.
45:38
Reaction performed in dimethyl form amide at plus four degrees will lead
45:51
to the formation of this homo aldol addition product. You get an 80% yield
46:11
4 to 1 anti to 1. And moreover since we have chiral information in that
46:25
system. A 99% enosomeric excess. How does this work? That secondary amine
46:41
forms an enamine by condensation with the aldehyde. With a hydrogen
47:12
bridging bond the second equivalent of the aldehyde is coordinated to the
47:24
system. And this is activated in this sense. And the chiral information is
47:42
transferred. So that anti-syn selectivity is certainly not as high as with the boron enolates but the 99% enosomeric excess. This is convincing.
48:06
So one could ask of course why do we need this organometallic aldol addition where we now have organocatalysis as a powerful instrument.
48:27
So there are some methods that still have much importance and will it also have in future. The best known is Evans enolate chemistry which is a method for first diastereoselective
49:10
setups and enancy selective at the same time with stoichiometric organometallic chemistry. So let's see. The carbonyl moiety which has to be transferred in the aldol addition process.
49:48
We have this chiral oxazolidine as a chiral auxiliary. That means that part of the
50:24
structure transfers chiral information and afterwards you have just to break that nitrogen carbonyl group and hopefully we can recover the chiral auxiliary. So working with boron
50:45
enolates synthesizing Z enolate and let it react with simple aldehyde will give this product
51:53
diastereoselectivity 99.6%. This is a general trend for instance with benz aldehyde
52:10
this translates to about 250 to 1 with the benz aldehyde. We will have a phenyl group here
52:24
at this position. Then we have even 500 to 1 and a 95% yield. So an excellent reaction
52:46
and moreover what is very important is these products are normally crystalline and they easily crystallize. So one recrystallization normally purifies it even further that you have
53:09
in the product 100% diastereoselectivity and it's then enantiomerically pure because
53:26
the chiral auxiliary that you put in should be enantiomerically pure. Hydrolizing this nitrogen carbon bond should then lead to the pure enantiomer of that acid.
53:52
Actually this is not so easy. How do you hydrolyze this bond with not eliminating water
54:00
here? But you could do other things. You could put in a protective group here. You could reduce that carbonyl group to an alcohol. So hydrolyzation would then because it's an would lead to the aldehyde. You could then use this aldehyde again for a second
54:32
aldol addition process with evans enolate. Then you have already four stereogenic centers
54:41
in one row. This is an excellent method for building up rather complicated target molecules. Very reliable. So and where do you get this chiral auxiliary from? Well of course
55:07
from chiral pool you should certainly know. In this case it is valine reduction with
55:34
lithium aluminum hydride gives the chiral amino alcohol and treating that with diethyl carbonate
55:54
under suitable reaction conditions will give you the oxazolidine. Last example for today
56:22
a combination of evans enolate aldol addition and Ireland Claisen rearrangement.
57:00
Let's see. So the evans boron enolate treated with acrylic aldehyde. Fortunately instead of
57:41
Michael addition process we get one two addition having an allylic alcohol
58:18
at the end of this chain. More than 99 percent diastereoselectivity reduction with
58:34
lithium aluminum hydride will lead to this alcohol and unselectively pure. Base and
59:05
and silicon chloride will introduce the protecting group preferentially to the primary to the primary alcohol since this one is sterically less in it.
59:38
Then we can isolate the secondary alcohol
01:00:12
With LDA minus 78 degrees trimethylsilyl chloride present, we will get the ester enolate.
01:01:03
So, and the Claisen rearrangement after hydrolysis will lead us to this final product.
01:01:36
We obviously have a 1-5 relationship between two stereogenic centers.
01:01:53
Obviously a nice example for a remote chiral induction.
01:02:01
And we have lots of functional groups which we could do various things. Reducing that to an alcohol or oxidizing that to an aldehyde, making use of that olefin. Certainly an interesting system.
01:02:25
Well, enough for today. Thank you for listening. See you next week.
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