Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: Heory on February 08, 2010, 06:19:32 AM
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I have no idea about these problems. Could anybody do me a favor? Thanks.
1. The molecule illustrated below can react through either Path A or Path B to form salt 1 or salt 2 . In both instances the carbonyl oxygen functions as the nucleophile in an intramolecular alkylation. What is the preferred reaction path for the transformation in question?
2. Explain the "abnormal" diastereoselection of the reaction.
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I think that the answer to question 1 can be solved by reviewing Baldwin's rules for ring closure:
www.indiana.edu/~msvlab/.../Baldwins%20Rules.pdf
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I think that the answer to question 1 can be solved by reviewing Baldwin's rules for ring closure:
www.indiana.edu/~msvlab/.../Baldwins%20Rules.pdf
I don't think it can be solved by Baldwin's rules. :)
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on first glance, I was going to say part a was neighboring group participation, but now I'm not so sure...
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azmanam I don't think neighboring group participation is too far off. Part one seems eerily like a nitrogen mustard (See http://en.wikipedia.org/wiki/Nitrogen_mustard and http://en.wikipedia.org/wiki/Sulfur_mustard). Since N is more nucleophilic than O, it would make sense to form the three membered ring with N, then have the carbonyl oxygen reopen the three membered N-containing ring.
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azmanam I don't think neighboring group participation is too far off. Part one seems eerily like a nitrogen mustard (See http://en.wikipedia.org/wiki/Nitrogen_mustard and http://en.wikipedia.org/wiki/Sulfur_mustard). Since N is more nucleophilic than O, it would make sense to form the three membered ring with N, then have the carbonyl oxygen reopen the three membered N-containing ring.
It does look like a nitrogen mustard but after all it's an amide, so the nucleophilicity of the nitrogen atom is much poorer than that of the one of a nitrogen mustard. I think the oxygen atom of the amide would act as a nucleophilic reagent directly without formation of an azirdine or an oxazolidone (another Br kicked out). I want to find out in which path in the TS the lone pair of the amide oxgen overlaps the anti C-Br sigma bond better, which would solved this problem, but I don't know how, since I don't know the bond lengths and the bond angles.
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For #1 Heory's last structure made me think of the preferred confirmation of esters due to the lone pair. Even if neighboring group participation does not play a huge role, I would think it could still offer a slight rate increase to favor path B
#2 is interesting, is there a given Lewis acid or is it general?
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LA is general
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Two things:
1) I don't think the answer comes from bond lengths and angle strain. Both the nitrogen and oxygen are sp2 so they have approximately the same angles. And I just don't feel that the small difference in bond lengths is going to have any significant influence here.
2) I think the N is more nucleophilic than you would expect. Sure an amide is less nucleophilic than an amine due to resonance. Here we have a carbamate though... so both the N and O are donating electron density into the carbonyl. I would argue therefore that the N is doing less resonance in a carbamate than it does in an amide, and thus the N is a fine nucleophile here. See Tetrahedron Letters 2001, 42, 1799 for an example of a mild N-alkylation of carbamates with alkyl halides.
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For #2 my initial thinking would be the Lewis acid is blocking the top face of the enone, so the nucleophile comes in on the same side as the other substituent.
Also, my earlier comment about ester geometry probably isn't going to be a big factor in this problem, and stewie has won me over to the anchimeric assistance argument.
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what class is this for and what chapter are you studying (and from what book)?
I know I've seen problem 1 before... but I just can't place it. It's driving me crazy. Maybe with some context, I'll remember and be able to help better.
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On thinking furthur I agree with you more than before. I've no access to the data base so could you show me a picture of the reactions in the literature?
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These problems are from lectures of Evans' group.
See the attachment.
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yes! that's where i'd seen it. Of course Evans doesn't give the answer. Whew. You have no idea how crazy it was making me. I had all my lecture notes from every grad class I'd ever taken, plus both volumes of Carey & Sundberg, plus March. I'll have to think about it more, now.
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I think that:
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Oxy, it seems by your drawing you meant to say that path A is preferred..?
I certainly agree that the geometry about the oxygen is the way you have drawn and tmartin sort of alluded to that stereoelectronic effect. However I see no reason why the geometry about the nitrogen has to be the way you've drawn. The N-H sigma bond can donated into the C=O pi star if placed in the other amide geometry.
I was tempted to answer similarly for question two, but I was taught in my synthesis course that OSiR3 groups are terrible at chelation. So I'm not 100% convinced yet.
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If the SiR3 were a benzyl or PMB I would believe that the beta-chelation could explain the stereochemistry. However, like Stewie said, I've been taught that SiR3 groups are bad for chelation. Now, if the Lewis acid Heory denotes is specifically Me2AlCl I would believe that there was chelation because it is particularly strong at enforcing beta-chelation with silyl groups.
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Most amide bonds in peptides have the trans conformation (the carbon chains on opposite sides of the peptide bond). I think the cis conformation is drawn by Oxy. so I'm not sure which pathway is preferred
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According to the reference given in Evans' notes, the lewis acid problem was discovered by Danishevsky.
The argument is this: As the nucleophile attacks, a sigma* orbital is formed. If the nucleophile adds from the same side as the OR group, then the sigma star is stabilized by the hyperconjugative effects of the CH bond on the gamma carbon. If the nuc adds from the anti face, the sigma* orbital is not stabilized once again because of the electronic effects of the gamma carbon (C-O bond).
Now, that is straight out of JOC 1991 387.
Maybe someone can explain to me why stabilizing a sigma* orbital actually favors bond formation and doesn't slow down the formation of that corresponding sigma bond.
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I think these problems have been solved.
For #1, I think Oxy is on the right track. C-N should be aniti to C=O for C-N σ* is a better σ*-acceptor than the N-H σ* for the carbonyl oxgen's lone pair. As a result, path A is favored. Mirro mentioned amide bonds in peptides have the trans conformation. In order to solve this paradox I want to say here CH2 in the peptide was substituted by O so steric repulsion is reduced?
As for #2, I am not able to download the document so I post it for help. Thank you philmont702a. C-H σ is a much better σ-donor than C-O σ so when the nucleophile adds syn to OSiR3 the TS would be stabilized by the hyperconjugating effect. Here electronic effect overrides steric effect.
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Oxy, it seems by your drawing you meant to say that path A is preferred..?
I certainly agree that the geometry about the oxygen is the way you have drawn and tmartin sort of alluded to that stereoelectronic effect. However I see no reason why the geometry about the nitrogen has to be the way you've drawn. The N-H sigma bond can donated into the C=O pi star if placed in the other amide geometry.
Yes, path A is prefered.
Nitrogen has only one lone pair, and it is conjugate with C=O double bond. . I think if the carbon chain is cis to the C=O double bond, it may be repulsed by the lone of oxygen atom. The N-H sigma bond can donate into the C-O pi star, but not effect as a lone pair and the energy of repulsion is larger. So it prefer trans conformation.
I was tempted to answer similarly for question two, but I was taught in my synthesis course that OSiR3 groups are terrible at chelation. So I'm not 100% convinced yet.
I haven't heard or read that OSiR3 groups are terrible at chelation before. I only read that LA catalyse the aldol reaction using silyl enol ether by chelation. OSiR3 of silyl enol ether can coordinate to metal, OSiR3 of another silyl ether can too, can't it?
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Oxy, as I know N-H sigma bond can donate only slightly into the C-O pi star. And also, I have seen examples in which OSiR3 coordinated to metals.
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An FYI, the Lewis acid is HgI2 (from the same class below, which gave the same article that philmont found).
I suppose I should clarify, I'm not saying that OSiR3 will not chelate metals. I'm saying bulky silyl groups have been shown to not chelate metals nearly as well as less sterically encumbered groups. A benzyl group would be much better for chelation than a silyl group (excluding maybe TMS, etc.) TIPS, TBS, and TBDPS decrease in chelate selectivity with size, this is from: JACS 1992, 114, 1778 using Me2Mg as a Lewis acid. I imagine it varies with Lewis acid as well.
The above article also discusses kinetic evidence for chelation and silyl groups certainly slow down the reaction... five member-ed chelates (alpha) are faster than 6 member-ed chelates (beta), etc. Also, beta chelation is not as strong as alpha chelation, so without proper choice of solvents, it can be hard to obtain good selectivity.
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I do not have first hand experience with carbamates, so I Googled "conformation of carbamate" and came across this: http://www.journalarchive.jst.go.jp/jnlpdf.php?cdjournal=bcsj1926&cdvol=44&noissue=11&startpage=3148&lang=en&from=jnlabstract from the Bull Chem Soc Jpn 1971, 44, 3148-3151. Here they use NMR, IR, etc to show that the geometry about the amide portion of the carbamate is almost exclusively the s-trans (ie, not how Oxy has it drawn). Interestingly enough, due to what is apparently called the "rabbit ear effect" the geometry about the ester portion is mostly the s-cis. I disagree that path A is preferred.
Also Oxy mentioned that the N has only one lone pair which is tied up in resonance and is sp2. But the oxygen of the ester part is also sp2 with a lone pair tied up in resonance....
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I cannot open the page. :(
Could anybody please explain what rabbit ear effect is (and post the document for #2 as an attachment)?
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Heory, you can read more about the rabbit ear effect in JACS 1968, 90, 7174.
Basically is says that if we have s-trans with respect to both the amide portion and ester portion of the carbamate, that the lone pairs on N can have an electronic repulsion with the lone pairs on O. However if you put the ester portion in the s-cis geometry, this repulsion is no longer present. This is the first time I've ever heard of it, so I am unfamiliar with it myself.
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stewie griffin, could you please post those documents? Thank you!
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Here is the document which stewie griffin posted.
In the image which illustrate the "rabbit ear" effect, it seem to be N and O are sp3-hybridization ???.
I think if both of them are sp2-hybridization, there is no room for the "rabbit ear" effect.
I have used Chem3D to calculate energy of methyl methylcarbamate in four conformation: s-trans-s-trans, s-cis-strans, s-trans-s-cis and s-cis-s-cis. The first conformation is belong to Me-N-C-O, and the latter is belong to N-C-O-Me. Well, energy is increased follow the order above. Acorrding to it, we can't say which path A or path B is preferred.
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Oxy, I too agree that the rabbit ear effect seems odd b/c it requires sp3 orbitals. But Eliel was quite a smart man, so I hesitate to rule out his idea and am willing to entertain it.
Heory, is there any chance that you will get the answer in class? I am quite interested now in what the "correct" answer is.
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I return to the idea of neighboring group participation. Heory said, nitrogen atom of amide group is poor nucleophile. Of course, but I think nitrogen's lone pair doesn't attack the C-Br bond, so no matter about nucleophilicity. Mechanism is described below.
So we can expect that path B is preferred ?
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Oxy, I like your initial idea better. Base is not present according to the problem. I think the major product depends on the confirmation of the carbamate.
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After stewie griffin posted the document and due to calculation of Chem3D, I'm not sure about the first idea. I'm try finding a more reasonable way. Base is not required if proton isn't lost. Hmm... Heory, could you get the answer?
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No I don't have the answer, nor would I have the chance to get it.
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:-\ I really got my hopes up waiting for the answer too... now we'll just have to argue these points over and over again like Groundhogs Day! :P
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I asked a friend for help :P. Here is his solution.
Due to inductive effect, the C-Br sigma* MO (b) has lower energy than the other one.
The lone pair (1) interact with C-O sigma* MO better than the other one and lower its energy.
So energy between lone pair (1) and C-Br sigma* MO (a) is larger than energy between lone pair (1) and C-Br sigma* MO (b).
Conclusion, path A is preferred.