Professor, will you still be at UCLA during spring quarter?
Yes, from April 23 onwards I'll be here.
Hello, i looked at the solutions of the first synthesis question for synthesis. I noticed that u used the OH group to substiture the Chlorine on the acyl chloride. Could we just use a fisher esterfication instead, adding an OH from both compounds?
Yes.
Can you post the directions for final exam pick up, is there a chance that we can come get it from you before you leave for England?
If you are around next Wednesday or later, you can pick your exam up directly from me. I will be gone from the 1st to the 16th of April, and again on the 20th April.
Can you tell me why it is that the enol is not the ketone resonant structure? On midterm #3 problem #1C, how many points are they worth each since there are sixteen points total for this problem?
I believe you are referring to the 1(a) on the final. Ketones and enols are examples of tautomers, not equilbrium structures. They are chemically distinct in their connectivity, and so can not be resonance structures by definition. Also, the fact that the enol form can be populated by catalysis suggests we are not dealing with resonance forms, whose "concentration" can not be influenced by chemical means. In problem 1 on midterm 3 each part (a–d) is worth 4 points.
Dr. Mascal, Iwas wondering, in comparing ketones and enols, we know that ketones are the favored form. However, with amines, when comparing enamines to imines, enamines are the more favored (stable) form. Why is that? Thanks, Amine
Remember that when we make enamines for the purpose of generating a carbon nucleophile from a ketone or aldehyde, we use secondary amines. This means that the choice is between the enamine form and the charged iminium species, and the enamine wins out. If however we use ammonia or a primary amine, there will be a substantial proportion of the imine tautomer. There is much more enamine present than there would be enol in the corresponding carbonyl, I think because the C=N bond is weaker than the C=O.
Hi Professor Mascal, When the time comes will you be posting the averages and the grade scales for the last midterm and final?
I post the mean, high and low scores, but I do not work out individual grade curves for the last two exams. The curves on the first and second midterms were only done to give you approximation of how you are doing so far. I will however post a total curve (i.e. out of 400 points) and the stats for the course.
Doctor Mascal, could you please post the answer key midterm 3 and final.
Yes, I hope to do that sometime Tuesday in my webspace.
Hi Professor, i forgot to bring a self-addressed,stamped manilla envelope for my final/MT #3, is it possible for me to drop one off to your office tomorrow?
Yes, no problem.
Professor, If a Claisen condensation is performed with an ester that has only one alpha-hydrogen, the last step of the reaction (which is really the driving force), where the methoxide anion is supposed to deprotenate an alpha-hydrogen, can't take place. Does this decrease the yield considerably? Thanks.
Yes, the equilibrium would lie substantially towards the starting esters in this case.
Professor, On page 567, near the bottom, the book demonstrates how a methyl ketone can be indirectly oxidized to a carboxylic acid through the haloform reaction. They use 5-Methyl-3-hexen-2-one which contains a double bond. Are they enforcing that the addition of Cl2 across the double bond is selectively avoided through the haloform reaction? Or are they only showing the major product? Thanks.
Under the conditions, the CH3 alpha to the carbonyl is much more reactive towards Cl2 than the double bond. Like with many chemical reactions, this selectivity is just a matter of relative rates of reaction.
Refering back to the Mid 2 review sheet. #8 asks to synthesize a cyclohexanone (with an attached carbn phenol group) from a cycloheptenone. YOu suggested a route where we break the alkene functional group and do a synthesis in the intramolecular aldol fashion followed by oxidation, nucleophilic attck and then decarboxylation to get the final product. I was wondering if this other route would be reasonable. If i started converted the alkene func group on the cycloheptenone into a glycol (OsO4) and did a pinacol rearrangement and then converted alcohol into a ketone. then react cyclohexanone with LDA and attack a Br-CH2-Ph. would that work??? thanks, mary
I'm not following your reasoning. Are you sure you have the correct product of the pinacol reaction? I get a cyclohexyl substituted methanal derivative when I do it, which doesn't seem to correspond to what you are suggesting. Have another look at the pinacol...
Prof. Mascal. Can you get a ring closure (like the Robinson Annulation) from the product of any Michael addition? Like if I take acetone, turn it into an enamine and do a Michael conjugate addition, can that product react with itself in an aldol reaction and form a ring (since intramolecular reactions are faster)? Or does the Michael "donor" have to be in a cyclic ring?
The Michael donor does not have to be a ring. The reaction you suggest is fine and would formally be a Robinson annulation.
Hi Prof Mascal, When you have 2-hexanone, and perform an enamine to the 2-hexanone, can you also get an enamine product that has a double bond between the 2nd and 3rd carbon (i.e. ch3ch2ch2ch=cNR2ch3)or is this considered the minor product compared to (i.e. ch3ch2ch2ch2cNR2=ch2) a double bond at the 1st and 2nd carbon. I'm asking if we perform a enamine for 2-hexanone, can we say that the major will always be the kinetic (in our case), the one with the double bond on the 1st and 2nd carbon. THanks
The enamine will not always be the kinetic one, especially when there is a group to which the double bond of the enamine can conjugate (like an aromatic ring, see p. 707 in the book, which gives 100% of the thermodynamic enamine). When enamines are made from aldehydes there is no ambiguity, but with non-symmetric ketones like 2-hexanone there would be a mixture of isomers, and perhaps it is best to avoid this situation altogether...
Hi professor, i wanted to know if their are any limitations to the uses of MgBr. For example, can the alkyl chains it is attached to have functional groups, or can it attach to a tert butyl, etc.
Grignard reagents can be prepared primary, secondary, or tertiary alkyl halides and also vinyl or aryl halides. You can have a C=C double bond in the presence of a Grignard reagent, but any electrophilic functional group (i.e. C=O, C=N, cyano, nitro, halides, acidic hydrogens, etc) can not be present. You should generally consider Grignard reagents as 'hydrocarbon' nucleophiles.
Dr. Mascal, Is it possible to generate an ylide with a terminal leaving group. ie Ph3P + BrCH2CH2Br -->(over BuLi) to give Ph3P+CH-CH2Br ? Also, can we have ylides with alkene groups in them as well? Thanx, Amine
If you try to make a phosphonium salt of a dihalide you will substitute both halogens, so this is not a good idea, but ylides with alkene groups in them are ok.
Hi Dr. Mascal. I was wondering if you could tell us how many synthesis questions you will put on the test. The reason why I'm asking is because it takes me a good 20-25 minutes to devise an answer for each of the practice problems on the review sheet, and I'm concerned because I think I'm going too slow.
It is my intention that you do not have more synthesis problems than you can do in three hours... I can not tell you the exact number, but I hope I have got the balance right. I also normally set the level of difficulty on the exam below that of the review sheets, so hopefully you can count on doing most of the synthesis questions on the exam more quickly than those you have been doing lately.
Is sulfur a better nucleophile than nitrogen or does it depend on the type of molecules.
Sulfur and nitrogen are fairly closely matched in terms of nucleophilicity. Both are capable of forming stable hypervalent salts (ammonium and sulfonium salts).
Can enamines directly attack the carbonyl C of an aldehyde or should the aldehyde be first converted to an acid chloride?
Yes, this reaction is possible but it requires a Lewis acid catalyst to work. I would also note that aldehydes are not 'converted' to acid chlorides directly; this involves a change in oxidation state.
Hi Professor, Can we have a triple bond on the Grignard, or do we have to start with the double bond and then make the triple bond?
If I understand your question correctly, you mean an MgBr at the terminus of an alkyne. If so, there is no reason to do this, because the CH of a terminal alkyne is acidic enough to be deprotonated directly. Alkenes and alkanes however can not be deprotonated like this and so need to be converted to Grignards or organolithiums to become nucleophiles.
Hi Professor, Just so I'm clear on this, on the exam do we need to prepare the ylides and Grignard reagents from the starting materials? Thanks.
Yes, this is generally a good idea, since Grignard and Wittig reagents are not usually available to buy. However, if you were to leave it out it would not be a major error...
Prof. Mascal: I have a few questions: What conditions are necessary for the Michael addition? I mean why can't any enol react with any conjugated molecule? How do we know when we need a (-CO2R) group in the synthesis? Thanks
Check the book and your notes on thermodynamic vs. kinetic conjugate addition. Enols themselves are not sufficiently reactive to do conjugate addition, and simple enolates are too reactive and add mainly in a 1,2 (i.e. kinetic) fashion. Carbon nucleophiles of intermediate reactivity, like beta dicarbonyl enolates, enamines, and organocuprates are the only ones which undergo this reaction cleanly.
Hello Professor Mascal, are you giving any of the review session on Saturday? Also, will you be available /or on campus on Saturday?
The review sessions are being run by the TAs. Although I will be on campus Saturday, I'm afraid I need this time to finish off writing/producing the exams, so I will not be generally available. I will however be here to answer any last minute questions from 8-12 on Monday.
When you say that we can begin with any kind of fragments, as long as it it the same or under the Carbon amount specified, do you also mean that we can have as many things attached to the carbons that we want (say -OH, =O, or anything we like)? THanks
Yes. It wouldn't do much good if the carbon fragment didn't also have functionality to work with...
Will you be giving out scratch paper for the exam(s)?
Yes. I did mention this in lecture... Please do not bring anything to the exam except your ID and something to write with.
are there gonna be any extra TA office hours today Thursday and Friday?
As mentioned in lecture, I operate an 'open door' policy the week before the final, which means you can come and see me without an appointment anytime up to noon on Monday the 19th. It's 'office hours' all week.
professor... i group of us was wondering if you can post up the answers to the practice problems either on the web or on the door of your office... we would really love to compare our answers/methods (usually 85% correct) to yours (100% correct)... many of us would appreciate it if you posted up the answers..thank you
Please see the response to another VOH question (which either appears above or below this one) regarding answers to the review sheet problems. I would like to be able to accommodate this request, but I'm afraid does more harm than good in the long term. Please feel 100% free however to seek advice from myself or the TAs - although this will not involve simply being given the answers!
Hello Dr. Mascal, I was wondering if you could point me towards some practice synthesis problems like the handout you gave us, except with solutions. Its somewhat easier to learn from the problems if I can not only compare my technique with another but also get some direction when I'm stuck. Thanks.
The reason I do not give solutions to these problems is that they can often short-circuit the learning process. It is too easy to give up and just work backwards from the solution, by which little is actually learned. I am not aware of any books, etc which have problems (and answers) at the level we are now working at in synthesis. There are some which are more advanced, but these would serve mainly to confuse you by the use of reactions you have not yet heard of. My recommendation is that you seek advice on the problems you have done on the review sheet, either from me, one of the TAs, or at the TA review sessions.
Professor, can Grignards react intramolecularly, if there is a carbonyl group present within the molecule?
In principle, yes. But Grignards are so reactive that they will not react exclusively in an intramolecular sense, i.e. you would expect the intermolecular product as well. For that reason, I do not recommend you propose such a reaction in a synthesis.
May I have Denise's email address? I can't find it on the web. Thank you.
Go to the department webpage (www.chem.ucla.edu) and click on 'faculty and staff directory' and you will find it. If you have questions, I think it would be preferable if you were to visit her personally rather than email her...
Can enamines attack esters the way they attack acid chlorides and acid anhydrides?
No, the reaction of enamines with esters is too slow to be of synthetic use.
in a synthesis, we are required to break the molecule into several smaller pieces however, how do we know which piece is allowed with same number of carbon? in another word, when the starting material is not given in a problem, how do we know which compound we are allowed to use from the store room or should be broken into simplier pieces? thank you!
If I tell you the limit is, for example, three carbons, then any starting material with three carbons or less is ok.
couple of questions: 1) do u want reagents/conditions on the practice midterm?... 2)will there be any forward synthesis problems on the final? 3) on the final day- will there be two separate exams- 3rd exam and cummulative exam- OR one test consisting of both materials... thank you prof.
1) Major reagents and conditions are always necessary for the 'forward' synthesis to be correct. What you do not need to worry about are details of solvents or workup conditions (e.g., adding water at the end of a reaction involving carbanions, etc). 2) I can not answer this directly, but if you've been listening in lecture and apply the principles of common sense, you should be able to arrive at the right conclusion. 3) Formally, they are two separate exams, but you have the full three hours to devote to both of them as you see fit.
Prof. Mascal, I have several questions that require only short answers. 1. Can a Grignard reagent attack any partially positive charge(s) other than the C of a C=O bond? Eg. CH3-X 2. Is Ph-MgBr a reasonable molecule?? 3. What effect, if any, will [H2+, Pt] have on C=O groups? 4.Can CH2=CHBr be attacked by a nu- ? 5.Can radical addition of HBr occur to Ph-CH3 resulting in Ph-Ch2Br? (I'm thinking similar to allylic addition of a halide)
1. Grignard reagents do not react cleanly with alkyl halides; they should normally only be used with C=O electrophiles or unsubstituted epoxide (i.e. ethylene oxide). 2. Yes. PhMgBr (and H2C=CHMgBr) are perfectly reasonable Grignard reagents. 3. C=O bonds can be reduced under forcing conditions with H2/Pt, but C=C bonds are reduced more quickly and can be selectively reduced in the presence of a C=O bond. 4. No, SN2 reactions do not occur at sp2 centers. 5. Yes, benzylic and allylic reactivity is very similar, and radical halogenation works in either case.
I was working on the practice midterm questions- on the final- do u want us to state the reagents as well as the conditions during the retro-synthesis? or just showing structure is enough... also- are there going to be any questions concerning straight forward synthesis? For example- synthesizing a target molecule JUST starting w/ actelyene?..
Reagents are usually not specified during retrosynthesis. The real purpose of retrosynthesis is to break the molecule up into appropriate nucleophiles (-) and electrophiles (+) which can be plugged into a sensible forward synthesis. In the forward synthesis however you should always include reagents and any important conditions.
Hi Dr. Mascal You are going to stay in your office until what time?
It varies, but I am usually here until about 7 pm. This evening (Tuesday) I have to leave a little earlier, but otherwise I am happy to make late appointments if this is what people prefer.
Professor...today is sunday, march 12th... I can't seem to access your page for an extra copy of the practice midterm #3...
The problem here was that the mainframe was rebooted a few days ago, but the associated web-server was not reset. The problem has now been fixed - apologies for any inconvenience.
Just a note that I was also unable to access your personal webspace today or yesterday (either by direct link or hyperlink through VOH).
The problem here was that the mainframe was rebooted a few days ago, but the associated web-server was not reset. The problem has now been fixed - apologies for any inconvenience.
Hi Dr. Mascal Do you have more examples on retrosynthesis, so, we can practice on TA review sessions? Like last quarter, we have the "Extra problems for TA review sessions worksheet." thanks
I think the 21 molecules on the sheet are as many as you will need. I do not anticipate being able to cover more than half of them myself, and hopefully the TA review sessions will be able to deal with the rest.
I still can't access your personal webspace using either Netscape or Internet Explorer. I was wondering if there was anyway for me to see the key for Midterm 2? Please let me know how I can access this information. Thank you.
I wish I knew why this is happening in your case. Anyway, please come and see me and I will give you a hardcopy of the midterm 2 key.
Just a couple of quick questions re: starting materials. Ylides, malonic esters, and Grignard reagents, for example, can also be synthesized from starting materials. Are these substances that can be bought ready made or would you have to prepare PPh3 with CH3-BR, for example, to make an ylide? Should we breakdown ylides, malonic esters, and Grignards in our synthesis or can we throw them in at will where needed?
Ylides and Grignards generally need to be made from the corresponding halide. Malonic esters however are commecially available and normally should not be further disconnected.
Professor, I noticed that fisher esterification is not part of the assigned reading...but part of the hw... i assume that you have taught this in your 130A class last quarter and we are expected to know it for the midterm/final...
It was taught in 30 last quarter. It does the same thing as diazomethane (which we have more recently covered), i.e. converts an acid to an ester. It is a useful thing to know, so it may be a good idea to review it.
Thanks for such a fast response, but when you say substituted carbonyl compound are you including an ester or malonic esters? Do these NOT interact with each other by the natural 1,3 and 1,5 relationships?
I'm not entirely sure what you are asking, but let me make an attempt at an answer anyway: Formally, the CO2R group can be called a substituent at the alpha carbon in malonate or acetoacetate type carbonyl compounds, but the C=O of this CO2R group already has a natural (1,3) functional relationship to the other C=O. Reactivity-wise, nothing has changed. When I talk about 'substituted carbonyl compounds', I mean substituents that will change (actually invert) the natural pattern of reactivity. So, for an unsubstituted (or even alpha CO2R substituted) carbonyl compound, the C=O carbon itself is labeled with a (+) as a synthon, because it is naturally electrophilic, whereas the 'alpha' position is labeled with a (-), since it can be nucleophilic. Putting a halogen in the alpha position inverts this reactivity, since the carbon must now be labeled as (+), because the halogen is a leaving group (i.e. the carbon has become electrophilic). I hope this is helpful.
Hello Prof. Mascal, I was wondering if you could restate the relationships (natural and unnatural) and the usual kind of reactions they coorespond with, such as the 1,5 = Michael. Thank you and have a nice weekend.
The natural functional relationships are 1,3 and 1,5. 1,3 corresponds to aldol/Claisen-type chemistry and 1,5 to conjugate (Michael) additions. They are referred to as natural because they represent the characteristic reactivity of carbonyl compounds with each other, rather than with other reagents (epoxides, etc) or with substituted carbonyl compounds (alpha halo, etc).
Hi Prof. Mascal, In the final, will there be sufficient time to analyze the structures, if we are to do retrosynthesis, since it takes a while to figure out the starting structures. What I'm asking is how many retrosynthesis problems will be on the midterm3 exam compared to carboxylic derivatives and enamines? Thanks
I haven't written the exam yet, so I'm afraid I am not able to comment. Remember you have 3 hours for midterm 3 and the final. That should be sufficient time to do some retrosynthesis...
Hi Dr. Mascal Could you suggest which chapters we can read or practice the problems for the last learning objective, synthesis? thanks
As I mentioned on Wednesday, I am putting together a review sheet with plenty of practice problems on it. You will get this on Friday, or you can download it now from my webspace if you can't wait for it to come out in hardcopy. There is really not much in the way of reading material in BF on synthesis, and indeed since we are not covering any new chemistry, there is not much to read about. This part of the course is designed to be practical, rather than purely academic, so I don't think reading would necessarily be a good use of your valuable study time...
Hi, Professor. You mentioned in class that we should avoid using protecting groups whenever possible. Why is that?
You should avoid protecting groups when you can simply because they add two otherwise unnecessary steps to a synthesis: putting them on and then taking them off.
Professor, In reacting the acetylene anion with a substituted ketone (such that the pH of the alpha hydrogens is less than 20), would an acid-base reaction or a nucleophilic reaction ensue? If the acetylene anion (or even Grignard reagents for that matter) were reacted with a carboxylic acid, the acid-base reaction would be observed exclusively right? What difference in pH between the base and the acid would be required to conclude that the acid-base reaction would be observed over the nucleophilic one? Thank you for your time.
When the pKa of the alpha C-H of a carbonyl electrophile is relatively high, like that of a ketone (> 20), then the dominant reaction with carbon nucleophiles/bases would be attack on the C=O, since alpha deprotonation is slower than nucleophilic attack, even for very strong nucleophiles/bases (like organolithiums). When the pKa slips down in the range of the dicarbonyl compounds (~10-12), deprotonation by strong nucelophiles/bases (alkyne anions, Grignards, etc) becomes a serious issue. For highly acidic carbonyl compounds (like carboxylic acids, pKa ~5), the acid-base reaction is predominant with C-nucleophiles of any sort.
Professor, Problem 16.20 deals with the preparation of Carboxylic Acids. I did not recognize part (b) as a reaction I had seen before. I even checked your Chem 30 syllabus and saw that no such reading was assigned on carboxylic acids. Am I mistaken? If so please let me know what page I can find further information on that particular reaction. Thank you.
The oxidation of an aldehyde to an acid using Tollen's reagent (Ag+ in ammonia) was covered in the organic redox part of the Chem 30 syllabus, Ch. 15 p. 568.
Professor Mascal, Just so we aren't surprised on the final, I was wondering what format we should expect, like the percentage or ratio of reactant/product questions (like our last midterm), mechanisms, and true/false or short answer problems? Thanks.
I have not yet written the final, so I am not able to comment. But I don't think the format of the exam should be the issue in any case; you either know the material or you don't, and even the most friendly question format is not going to change that... I can say that the final in Chem 30 had an average about 15 points higher than a typical midterm, so perhaps if you prepare yourself to answer questions, regardless of how the exam is organized, you'll do well.
I cannot access the key to midterm 2 on your webpage.
If you can access any of the other documents on the webspace, the key to midterm 2 should work as well. I have just tried it myself and it works. Come and see me if you continue to have problems and I will give you a hardcopy.
what does the average for the midterm look like it's going to be? Thanks.
I will give the statistics for midterm 2 on Friday.
When I first worked through #57b of chpt 18, I made an enol out of cyclopentanone and did a mechanism that lead to the same product. I realized that I needed an enamine reaction so I reworked the problemwith morpholine. If the problem doesn't explicitly state it wants an enamine mechanism, how would we know to use an enamine versus a Claisen or aldol? Or are both acceptable?
Certainly, there may be more than one approach to enol/enolate type reactions, i.e. with or without enamines. For 18.57b they say to use the enamine, and that one would not work by simple catalyzed enolization of the ketone. It might work with the specific lithium enolate or in a mixed aldol-Claisen type reaction, but I could see problems occuring there too. Diketones like the one in 18.57 suggest preparation by the enamine route, and likewise simple ketones or keto-esters suggest the Claisen, and alpha-beta unsaturated ketones suggest the aldol.
I have a question about the addition of the tertiary bromide to the dicarbonyl in teh presence of NaOEt. Since it is done in a step by step process, why is it not possible that the OEt rips of the H and forms EtOH and the tertiary bromide is then added through SN1? Thanks a lot.
The EtO(-) removes the proton between the two carbonyl groups of the keto ester to give EtOH and the acetoacetate anion; this is true. But SN1 only takes place in the presence of weak nucleophiles/bases. Since the anion is a fairly strong nucleophile/base, SN1 really can't happen here. The only reasonable reaction to suggest in this case with the teriary halide is E2.
professor Mascal, I saw on a news program that MTBE can leak from tankers go into the ground water and react with H2O. do you know how this ether would react with water to destroy the water table?
MTBE, being an ether, is not particularly reactive, so I think the issue is that it simply contaminates the water table. It was originally thought to be of low toxicity, but it is now suspected that chronic exposure (in drinking water or as a pollutant) can be harmful.
Hi Dr. Mascal I have a question on the third page of midter 2. Is it possible that eposide is formed form the reaction of crown ether and conc. HBr? thanks
The formation of an epoxide under these conditions is not really possible. It is true that 2-bromoethanol would be an intermediate product on the way to the 1,2-dibromoethane final product, and that this halohydrin can in principle cyclize to give an epoxide, but such a cyclization would not occur under acidic conditions. And even if some epoxide were formed, it would cleave again very quickly in conc. HBr...
Professor Mascal, For the clainsen condensation on page 1 ( problem 3), isn't MeOH, or acid, required as a second step to obtain the product you've written as the solution. That is, without alcohol to donate a proton then wouldn't the product contain an anion on the alpha carbon? Also, for the very last problem (EtMgBr), the book usually uses HCl, H20 as the second step in order to turn the O- into the final alcohol. YOu, however, did not include this second step and yet have the 0H rather than an O-. Why?
In the instructions to the question it says, "assume any necessary solvents, workup conditions, etc. are included." Normally, adding water at the end of a reaction is not considered another step of the reaction, but a "workup". If you put the anion in the cases you mention, that's fine too and your answer will get full credit.
Hi Prof. Regarding the same question as before about the "OMe" synthesis, could the "NaOMe" also deprotonate at the alpha-Hydrogen between the alcohol and ketone group, thus forming a "Me" bond in the middle. thanks
No. As discussed both in lecture and the book, alkylation alpha to a carbonyl requires strong base, and even if you had strong base in this case the OH would come off before anything else.
Hi Prof. I just looked at the key, and I think the last problem is incorrect. The 1,2 addition of the nucleophile, should it not also have an "OEt" instead of an "Et". Also the 4th box on page 3 of the midterm, counting left to right, how is the "OMe" added. Thanks
No mistake, a Grignard adds twice to an ester - first to give a ketone and again to give a tertiary alcohol. The fourth box on p. 3 is a Williamson ether synthesis.
Hi. Would you post answer key for midterm2?
Yes, it will be posted hopefully this afternoon (Friday) in the Chem 30/130A webspace. Watch for a VOH announcement for details.
Hi Prof. Mascal, do we have to know the mechanism for acetal formation for the midterm? thanks.
Acetals are a Chem 30 topic. While they have not been a focus of attention in Chem 130A, a background understanding of this, and indeed all Chem 30 material, is important for getting the most out of 130.
Hi Dr. Mascal- I came to office hours today after class (about 10:05) and couldn't find you. Will you have office hours tomorrow afternoon? Thanks.
About 50 people showed up to office hours yesterday and we had to go to a bigger room. You must have just missed us. I have office hours tomorrow between 3:30-4:30.
Hi, for problem 18.52, why does the double bond in the ring acts as a nucleophile in the Michael's Rxn? (I normally thought that double bonds are electrophiles!) Does it have to do with the nitrogen in the molecule?
18.52 is a question involving enamines. We have not covered enamines yet and any problems which involve enamines should not be attempted until after we have discussed them in lecture. Enamines will not be on the exam.
Hi Dr. Mascal I have two questions on review sheet: 5 and 7a. what do we need to write for the answer? thanks
Did I not cover 5 and 7a on Wednesday? Come and see me if you require further clairification.
Is Chap. 16 going to be on the midterm? Thax.
Decarboxylation of beta-keto acids is described in Ch. 16 (as well as Ch. 18). I did mention this during lecture. We will cover the rest of Ch. 16 after we finish Ch. 18 (also as mentioned during lecture).
Prof. Mascal, What is the cut off point for the second midterm from the book/syllabus? What pages and problems are we responsible for from "enolates/enamines" and "carboxylic acids and their derivatives?"
I have already answered this question at least twice. Please find the response to your question below.
Professor, I understand the usefulness of malonic esters in synthesizing monosubstituted and disubstituted acetic acids, but how would we go about synthesizing the malonic ester itself? Do we need to be concerned with this? Thanks.
No, don't worry about this. Malonic ester is sort of like acetone, methanol, etc; we don't make them, we buy them from a chemical company. Making it would involve a crossed-Claisen like reaction between an acetate ester and a dialkyl carbonate (i.e. a diester of carbonic acid, H2CO3).
For acid catalysis of alpha keto, why is this process more synthetically favored over the base promoted to alpha keto?? (the book refers to the addition of the second halogen on an acid catalysis as being slower or harder to add).
If I understand your question, I think you are referring to acid-catalyzed vs. base-catalyzed alpha halogenation. If so, the reason the acid-catalyzed reaction is favored is just as the book says: it can be stopped after the addition of one halogen, whereas the base-catalyzed reaction can not be controlled.
Hi, prof mascal. I have several questions. ON 15.28 there is a hydrolysis of a vinyl ether. How does the methyl group get released upon protonation of the Oxygen. For 15.49: How does the configuration change from an r-gylceraldyhyde to S-gylceraldehyde (how does the base catalysis cause it to change the configuration?) 15.56: When added the HCl to the alkene functional group, why isnt the Cl not added on the alpha carbon postion (because that position also exhibits a stable position because of the carbocation) is it not added there because of the acidity of that position???
1) The methyl group does not get 'released', the OMe group is substituted by an OH group. Push from the oxygen and protonate at carbon, then add H2O into the =OMe(+) function to get a hemiacetal. Review BF section 15.8 on acetal chemistry if you are having problems. 2) If the chiral center tautomerizes into a carbonyl group, all stereochemical information must be lost, so the number of R and S centers becomes random. 3) The HCl adds in normal Markovnikov fashion to the double bond in 15.56. Addition the wrong way around is further discouraged by the carbonyl group, which would destabilize the positive charge.
hi dr. mascal, what's the cutt-off for midterm 2 ? and was all of chapter 18 covered in lecture ?
Up to section 18.8 (conjugate addition), but not including enamines (section 18.5) or anything in the rest of Ch. 18 which involves enamines.
are we responsible for saponification, a few problems that you assigned in chp 18 requires saponification
Saponification (i.e. ester hydrolysis) should have been covered in Chem 30. If you are having some trouble with this I recommend you review it.
Dr. Mascal, In 18.27b, after the double bond undergoes hydrolysis, the OH collaspes onto the carbon and electrons are pushed into the alpha carbon. What is the motivation for this to occur? Does the alpha carbon have characteristics of a nucleophile?
18.27 is best described as the conjugate addition of water as a nucleophile to an alpha, beta-unsaturated ketone followed by a reverse aldol reaction. If you look at the aldol condensation as an equilibrium process (as you should), you can see that the reaction could be shifted backwards in the presence of large quantities of water. And all CH carbons alpha to a carbonyl have characteristics of a nucleophile; that is what the aldol reaction is all about...
I was wondering if we needed to know the things in the book that you did not lecture on... such as the clemmenson reduction, the wolff-kisner reduction, the enamines, and the acetoacetic ester systhesis?
I put up a transparency last week at the beginning of this learning objectives unit which explained how I had reorganized some of the material. Among other things, I said that p. 573-576 in Ch. 15 would be covered after Ch. 18. The acetoacetic ester synthesis was covered in Friday's lecture. It pays to go to class...
Professor, Regarding the Wittig reaction, you mentioned in class that the reaction also has stereochemistry involved (Z product being the kinetically favored one and the E product the thermodynamically favored one). I understand why the E product is the thermodynamically favored product, but what is going on (comparable to secondary orbital overlap from Diels-Alder) in the reaction that's making the Z product the kinetically favored one? I just want to be able to predict the stereochemistry of the product. Thank you.
There has been much speculation concerning why the Z is the kinetic product. It involves a fairly complex treatment of orbital symmetry arguments in the mechanism which is rather beyond the scope of Chem 130A. If you want more information about this I can suggest some books to consult. Otherwise, in this case, it's just a matter of memorization.
For midterm 2,will we be responsible for the sections in chapter 18 (and related problems) involving enamines, robinson annulation, and conjugate addition of Diorganocopper reagents? I don't think you've covered those in lecture. Thanks.
We have not yet covered enamines (section 18.5), so please don't concern yourselves with those yet. I did mention in lecture that dialkylcuprates could be used for conjugate addition (even though the reaction with these in not reversible), and the Robinson annulation is just a fancy term for the combination of two reactions you have already learned, i.e. the aldol and the Michael addition...
Mr. Mascal, for problem #25 in Ch. 11, is the CH3O-Na+ in part a. used in the rxn, or is it just a solvent used for the rxn to go forward? The book doesn't thoroughly explain what happens.
The book assumes you recognize CH3O(-) Na(+) as a strong nucleophile. It is a salt, like NaOH, so it obviously could not be used as a solvent... The solvent in this reaction is the CH3OH.
Hello, Dr. Mascal. I have a question about the grading of the first midterm. I don't recall if you ever told us the procedure we are to follow in asking for a re-grade of our exam. Are we to talk to our TA about this, or do we come directly to you?
Please see the syllabus.
Are we responsible for the base-promoted haloform rxn on p. 567?
Although we didn't address this subject in lecture, it was part of your assigned reading. I do not consider it a reaction of tremendous importance, but anything in the book could in principle end up on an exam...
Dr. Mascal....up to what section will the midterm cover?
Up to section 18.8 (conjugate addition), but not including enamines (section 18.5) or anything in the rest of Ch. 18 which concerns enamines.
What was the last subject covered that will apear on the test
The last subject covered (on Friday) which could appear on the test was Michael addition.
Hi Dr. Mascal could ester undergo keto-enol tautomerism like aldehyde and ketone? or ester requires different stands? thanks
Yes, esters (and other carboxylic acid derivatives) enolize like aldehydes and ketones, but the acid-catalyzed tautomerization and processes mediated thereby (condensations, etc.) are less efficient than for aldehydes and ketones.
Dr. Mascal, on page 548 at the top, what directly indicates the formation of either E or Z isomers?....and on page 564 problem 15.10 part b), why don't you get the formation of two ketones since two alcohols are present. The book indicates only formation of one ketone, why is that so ?
1) Z alkenes derive from reactions of carbonyl compounds with simple alkyl ylides, whereas the E isomers derive from reactions of carbonyl compounds with stablilzed ylides (with a neighboring C=O group, for example). 2) Although either of the –OH's can become a keto group, there is only one hydroxy ketone tautomer possible in 15.10b. Push the arrows around from either enol and you get the same molecule in the end - try it.
Could you restate the changes you made to the third learning objective?
I have posted a revised syllabus in the Chem 30/130A webspace. But for the record, I have moved enolates and enamines (Ch. 18) in front of carboxylic acids and their derivatives (Ch. 16). Additionally, I will cover Ch. 15 pp. 573-576 after Ch. 18 is finished.
Hi Dr. Mascal 1)What's the difference between relative mixtures and racemic mixtures? 2)In the lecture, you talked about both organocuprate and grignerd reagents can open the epoxides ring. However, there is side product as we use grignerd reagents. What side product could be formed? thanks
1) I'm not sure I've ever heard of a relative mixture. Perhaps you are referring to the relative stereochemistry between two or more chiral centers? If so, this concerns the relationship of one stereocenter to another in a molecule, usually in a racemic mixture. If you are talking about a single enantiomer, this relationship is better described as 'absolute sterechemistry'. 2) In principle, Grignard reagents should also open epoxide rings, but they are also very basic (unlike cuprates), which can cause deprotonation alpha to the ring giving an elimination-type product.
Hi Dr. Mascal 1)When we have nucleophilic ring opening reation on the epoxide, does it require strong nucleophile to open epoxide or it doesn't matter? 2)On p419, can we also call nucleophilic ring opeing reaction as "base-catalyzed reation" since RO- could be work as base and recycle at the end of net reaction? thanks
1) Good to moderate nucleophiles are required to open up epoxides under non-acidic conditions. Weak ones (like methanol, water, chloride, etc) normally won't do it. 2) Although catalytic base might promote the reaction, it wouldn't really be accurate to call it 'base-catalyzed' because the epoxide is not actually deprotonated by the base at any time, and the focus of the reaction is the epoxide itself, not the nucleophile.
Hello Dr. Mascal, Could you please post the changes that were made in the syllabus? Thank you.
Yes, I'll post a revised syllabus in the Chem 30/130A webspace tomorrow (Tuesday). Sorry for the delay.
Professor, Regarding problem 11.39, in the solution manual, the first suggested step is the treatment of 1-chloro-3-methyl-2-butene with -OH to convert it to an alcohol and simultaneous conversion of styrene to an epoxide. In the conversion of styrene, is there a reason why the peracid would attack, selectively, the alkene not in the ring? Thank you.
Yes. Aromatic rings are much more stable than isolated double bonds, and hence are only attacked by powerful electrophiles (peracids are weak electrophiles). You'll learn more about this in Chem 130B.
Hi Prof. Mascal, For problem 11.22a, the mechanism for the synthesis is not shown in the answer key, so I proposed the following (using the the answer key's H+/2CH3OH as reagents). 1) Protonation (H+) of the epoxide. 2) A CH3OH attacks anti to the epoxide, opening the ring... 3) A proton (-H+) gets dropped off. 4) the 2nd CH3OH protonates the OH group (in order to make it a good L.G.) 4) Finally, the left over (CH3O-), does a nucleophilic substitution with the (OH2+), dropping off (h2o), and resulting in the answer. Is this a correct mechanism? One last question on 11.22b, in the answer key, it shows CH3O-Na+ and CH3OH for the first reagents used, but won't using "CH3OH" just protonate to CH3OCH2CH2OH, instead of the book's CH3OCH2CH2O-Na+ ? Thanks
Yes, except for the fact that CH3OH is not an acid, so the -OH group would be protonated by a strong acid catalyst, and that CH3O(-) would not exist in acidic solution, so you need to attack the C–OH2(+) center with methanol (intermolecular dehydration). I would note that I took 11.22 out of the list of assigned problems (announced during lecture a few weeks ago...).
Hi Dr. Mascal I have a question on p.547. The paragraph two, it said because phosphines are "weak bases", treatment of a tertiary halide with a phosphine give largly an alkene by an "E2" pathway. However, on p.310 table 8.13, tertiary halide gives an alkene by an E2 pathway when there is "strong base." Why there is the difference? thanks
The main issue here is that SN2 is not possible at tertiary centers. E2 requires a good to moderate base (of course the stronger, the better, but weaker bases can also do it), so the point the book is making is that phosphines will not be alkylated by tertiary halides, but are more likely to cause an elimination if they do anything at all.
Dr. Mascal, In BF ch 11 number 21 we are asked to synthesize Dioxane. There is an intermediate product before dioxane in the mechanism that gets protenated with acid. This intermediate product has 2 hydroxyl groups and an ether functionality. How do we know which Oxygen gets protenated, and for that matter why both of the OH groups aren't protenated by the acid? Thanx, Amine
In the presence of acid, any O group can get protonated. Multiple protonation would be disfavored by the neighboring positive charges, but might occur to some minor extent. When you protonate the O in the middle, nothing really can happen (at least not at any sensible rate), and so the proton just comes back off again. But when one of the OH groups at the end is protonated, it invites an SN2 to take place, closing the dioxane ring. Thus, although they all get protonated from time to time, only protonation at the terminal O atoms leads to a reaction.
Mr. Mascal, are we responsible for knowing how to calculate the equilibrium constant (acids and bases chapter)? Since you've said that calculators are not allowed on exams, would it be on an upcoming exam? Or are we just supposed to know which direction the equilibrium flows?
Any questions on exams which require calculations will involve only simple math which can be done in your head or on paper. It is not impossible to write an equilibrium constant question which fits this description...
Hi Dr. Mascal I have a question regarding to mideterm 1. For the last filling box on the second page, since acetylide is strong base as well as good nucleophile, is it possible that there is any Sn1 product? thanks
Are you sure you mean SN1? This only really occurs in the presence of weak nucleophiles, not strong ones... The reaction in the last box of question 3 is best described as an E2.
Hello Dr. Mascal, Can crown ethers be used to introduce polar ions into the nonpolar inner core of a DNA molecule? If yes, how does it permeate the polar phosphate backbone? Does it get in through the major and minor grooves of the helix or would denaturing the DNA strand make this easier? If not, is there a way to coat this crown ether with polar functional groups?
I don't recall any examples of crown ethers being used to bring metal ions into DNA, but some well known chemotherapy drugs, such as cis-platin, do include metals. There are examples of crown ethers with polar functional groups on the periphery, as you suggest, which make the macrocycle more hydrophilic. If you want more information on macrocycle chemistry, I can give you some library references to consult.
In the reaction with the epoxide on page 418(middle), the nucleophile attacks the more substituted carbon, yet on page 419(top), the nucleophile attacks the less subsituted carbon. Can you please explain why this occurs? Thanx
The difference here is between acid-catalyzed and nucleophilic ring opening, where the regioselectivity derives from the stability of the carbocation intermediate (acid) or the steric hindrance to the approach of the nucleophile (nucleophilic). Review Monday's lecture notes for details.
can you list the type of questions that you are willing to provide answers for?
Yes: I am perfectly willing to address questions which can not be answered by simply reviewing lecture notes or reading the book.
What is the difference between an Intermolecular dehydration reaction and a Intramolecular dehydration molecular reaction? In 11.3 a, could ch3(ch2)3O-Na+ and Brch(ch3)2 react to give the desired product. In part (b) of this same reaction, could (R)-ch3ch2chch3O-Na+ amd BrCH2CH3 react to give the desired product?
Inter- vs intramolecular dehydration: see notes for my lecture on Jan. 24. 11.3a: the answer is no; see the scheme at the top of the page for an explanation. 11.3b: the only apparent difference between what you suggest and the book's approach is that you want to use ethyl bromide instead of ethyl iodide. That would work fine.
What's our class average in the midterm1? thanks
I should have that information by Friday. Sorry for the delay.
I apologize for my mistake. In haste I misread your response. No need for a second look at problem 11.31 and 11.32.
Ok. I only get these questions one at a time, so I attempted again to answer your previous question.
Professor, I asked a question on Sun, Feb 4 regarding 11.32 and 11.33. You managed to state some general truths about the topic but for some reason avoided answering the question at hand. Please review the question. Thank you.
You can't speak of 'absolute configurations' of alkenes, since they do not possess chirality. I assumed you were referring to the product epoxides, which are produced with absolute stereochem. I have noted in the lecture that the Sharpless reagent is able to distinguish the top and bottom faces of alkenes, so I am not sure where you are still having trouble with this. Do what the book says – draw the allyl alcohol with the OH group in the lower right-hand corner of a box and then attack with the reagent from the appropriate side.
could not access midterm #1 key
Perhaps you'd like to explain the problem in more detail... Could you not access the page, or could you not read the file? Reading the files requires the Abode Acrobat reader, which is a free program available from www.adobe.com. Come and see me if you continue to have difficulties.
Professor, Regarding problem 11.31 and 11.32, they make reference to a reaction that was discovered in the last 15 years that delivered O either to the bottom face of an alkene or the top. Are we to assume that, in doing the problems, the way the molecules are presented in the book is their absolute configuration in space? It seems arbitrary to just use bottom face and top face since the molecules don't stay in any definite orientation in space. Thank you.
This is the first example of an enantioselective reaction on a nonchiral substrate we come across in Chem 30/130A. The configurations are indeed absolute: A chiral reactant is able to discrimminate between the top and bottom faces of the alkene, and the resulting stereochemistry in the product is absolute, not relative.
Hello Professor Mascal, I have a question regarding to the midterm. For the last problem on the second page (alkyene reacting with NaNH2 and then a 3rd degree Br), after the alkyene react with NaNH2, can it still react with a 3rd degree Br? in the book, it says that this reaction is a SN2 reaction and I thought SN2 don't occur with a 3rd degree halide? Thank you
Please see the answer key posted in my webspace.
Hello Dr. Mascal, When do you think you will be able to post midterm 1 exam key? Have a great weekend.
The midterm 1 key is now available in my webspace. To access it, just go to VOH and click on the hyperlink associated to my name.
Hi Prof. Mascal, when determining if a reaction is under kinetic or thermodynamic control, would looking at the mechanism and observing a formation of an intermediate versus formation of a transition state categorize it as a thermodynamic controlled reaction?? Basically, how can you tell if a reaction is thermo or kinetic without only looking at the 1,2 vs 1,4 addition?? (do you look at the formation of intermediates?)
You can't tell for sure just by looking at the molecules or intermediates involved, you have to determine a potential energy profile for the reaction and consider the conditions under which the reaction takes place to know the extent of thermodynamic vs. kinetic control.
Hi, how many questions will there be on the exam?
Come on Friday morning and find out!
Hi Professor, 1. you covered some ether material in last friday's lecture and i remembered you saying the midterm will cover up to last friday's lecture. however, in one of the VOH questions, you said ether will not be on the midterm?? 2. in oxymercuration of alkene, why is reduction necessary? why can't you just hydrolyze the Hg?
1. I don't remember saying the midterm would cover all of last Friday's lecture, I'm pretty sure I said we would draw the line between alcohols and ethers. 2. Apparently, the Hg–C bond is too stable to just hydrolyze under normal conditions, so it has to be reduced.
Hello, in chapter 9, the purpose of the reactions of alcohols with SOCl2, PBr3, MsCl, TsCl is to convert alcohol into a better leaving group and to form a alkyl halide, correct?? If so, then couldn't the mechanism proceed with the conversion of -OH to good Leaving group by protonation be just as efficient?? When would one be preffered over the other??
In principle, you are right: both derivatization to the halide (or sulfonate) and protonation act to make the OH a leaving group. But remember that protonation must take place in acidic solution, and most nucleophiles will be protonated (and thus not work) under such conditions. For example, if you wanted to substitute an –OH for an –NH2, you would not be able to use NH3, since it is not acidic enough to protonate OH, and you couldn't run the reaction in the presence of better acid since the NH3 would be protonated to NH4(+) and thus no longer be nucleophilic. The conversion of ROH to RX is the only practical means to accomplish nucleophilic substitution via alcohols.
Hello Professor Mascal, refering to lecture on the reaction coordinate of conjugated dienes, How is it that the higher positive density is on the secondary carbocation and not the primary carbocation? You said that the 1,2 addition results in a secondary carbocation and that carbocation because of its positive density has high affinity for the nucleophile (X-). I thought that the positive density on the secondary carbon would be delocalized by other surrounding groups.
Think of it this way: The intermediate carbocation exits as two resonance structures, the primary allylic carbocation and the secondary allylic carbocation. But these two resonance structures do not contribute eqally to the overall resonance hybrid; the more stable secondary one is the major contributor and the primary one the minor contributor. This means when we do a summary of the electron density in the hybrid, there is a greater density of positive charge at the secondary position than at the primary position. The favorable electrostatic interaction between the more highly (+)-charged carbon and the nucleophile mean the Ea for this process is lower than for the alternative attack.
HI Prof Mascal, on the reactions for the dehydration of primary (non beta branching ) alcohols, one would go through an elimination (E2) mechanism. Upon dehydration of primary (non beta branching) alcohols in chapter 11(ethers) the reactant goes through a SN mechanism and forms an ether. How would you account for the difference in product formation when the conditions seem to be the same for both reactions to take place???
The issue here is concentration. If I want to dehydrate a primary alcohol, you will add it dropwise to a relatively large volume of acid, so the overall concentration of alcohol in solution at any one time is low. If you want an 'intermolecular dehydration' to give an ether, you will add concentrated acid to the alcohol, so that the concentration of alcohol during the reaction is high.
professor mascal, in reviewing the reading for chapter 6, it includes pages 219-224, and 219 is about Heats of Hydrogenation and Relative Stabilities of Alkenes. you didn't go over this in class, but is it going to show up on the midterm? thanks.
I did briefly discuss the resonance stabilization of conjugated dienes, but I did not go into detail by covering delta-H values. In principle, anything which appears in the reading could turn up on a midterm...
just to make sure....alcohols are on the midterm but not ethers?
That's correct.
hi prof. Mascal on question 10.24, the fisrt answer is "Br2 with heat" I remembered reading something about it during Chem 30, but i forgot how it's suppose to work. I don't remember what chapter where i can find my answer, so i turn to you. How does Br2 or Cl2 with heat work? how does it apply to functional groups? thanks
These are radical substitution reactions, see BF Ch. 7 to review.
1- if oh is present in an Alkene or Alkyne and is reacted with HX, will the OH be hydrated to H20, or will the alkene/alkyne be reduced? 2) IN the acid catalyzed hydrolosi of Benzaldeyde Cyanohydrogen, why would the triple bond be attacked but not the OH group? (546) 3) Does addition of pbr3 to an alchol involve inversion of configuration 4)If the addition of BH3 to an alkene is in solution with CH3C02H instead of Hydrogen peroxide and sodium hydroxide will Boron be displaced by H?
1) If a hydroxyl group is present in a molecule which also contains a multiple bond, you could expect both the OH and the multiple bond to react, depending on the conditions. 2) What can happen to the OH group under the circumstances? If it got protonated and left (which is not so easy, since it is next to the electron-withdrawing CN group), the cation would have nothing to do but react with H2O to give the alcohol again... 3) The reaction involves an SN2 of the –OPBr2 group by Br(-), so it should go with inversion, but a small amount of rearrangement product is also expected with 2? alcohols, which can complicate the overall picture. 3) Be careful here. The addition of BH3 and reaction with acetic acid or H2O2/OH- are separate steps. The BH3 is added first. If this is followed by acetic acid, the B will be replaced by an H. If it is followed by H2O2/OH-, the B will be replaced by OH.
Professor Mascal, do you think you can post practice midterm's answer key (just the ones you didn't go over in lecture today)? some of us have class right after the lecture AND can't attend OHs. thanks (even a very rough outline of how we should approach the question would be fine).
I really prefer not to do this. If you can't attend any office hour or discussion section between Wednesday morning and Thursday evening, make an appointment to see me anytime Thursday and I'll be happy to go over the answers with you.
Hi Prof. Mascal, I was wondering for the inorganic reagents, do we have to include everything (i.e. if we used BH3, would we have to write 1. BH3, 2. H202/NaOH), or would writing just BH3 be enough. thanks
You should definitely write out both steps if you are going from an alkene to the alcohol (or alkyne to carbonyl). Only if you are stopping at the organoborane should you just put BH3.
Hi Dr. Mascal, giving a Diene and a Dienophile, how would you determine if the product would have the Exo or Endo confirmation?. Is this determined by the relative position of the Substiuentrs in the double bond of the dienophile to the diene or by their relative position to the double bond formed in the product?? thanks
The 'endo' or 'exo' approach of the dienophile to the diene transmits the respective stereochemistry to the product. The position of the substituents relative to the bicyclo[2.2.1]heptane ring system can also be described as 'exo' or 'endo', and corresponds to the approach taken by the dienophile.
Will thiols be included in the midterm?
No. We have not yet covered thiol chemistry.
Dr. Mascal, on page 335, the mechanism in the middle, the textbook uses 1-butanol to illustrate conversion from a primary alcohol to a primary alkyl halide by Sn2 reaction. There is no carbocation formed in this reaction. But on page 342, 1-butanol is used again to form both E1 and E2 products (cis, trans 2-butane and 1-butane). My question is why wouldn't the same compound undergo migration of H in Sn1?
We formulate mechanisms to explain what we observe in the most plausible way. For the conversion of 1-butanol to the bromide, it is most reasonable to propose an SN2-like mechanism, by which we avoid the intermediacy of primary carbocation. In the dehydration of 1-butanol, we propose a concerted mechanism like the one shown on the bottom of p. 343 to give 1-butene, and also an H migration to give a secondary carbocation, which then loses H+ as expected to give the mixture of alkenes. In neither of these scenarios do we need to invoke a primary carbocation. I hope this explains what is happening, because I'm not sure what your question is asking, since there is no SN1 involved.
Is the degree of substitution about a carbon determined by the number of Hydrogens attached to it or the number of carbon? (9.41e). I noticed in the study manual that sometimes a concentrated form of sulfuric acid is used, why is this necassary? In 9.35(D), are we to assume that HBr is in excess, for more than one mole would be needed to replace both OH groups, as the solution manual has it. Why isn't there sterochemistry shown in 9.26 d How is D different from H, what is its purpose. Leo
1) The degree of substitution (primary, secondary...) is determined by the number of attached carbons. 2) For dehydrations of alcohols, for example, it is necessary to keep the concentration of H2O as low as possible to suppress the reverse reaction (hydration), so conc. H2SO4 (98%) may be used. 3) If you are not given the quantity, you should assume at least enough reagent is there to do the job, so yes, in 9.35(d) there should be at least 2 mols of HBr present. 4) In 9.26d there is only one chiral center in the molecule; no specific stereochemistry is involved here. 5) D (deuterium) is the same as H but containing a neutron in the nucleus. It is simply an isotope of hydrogen, just as the majority of other elements consist of a mixture of isotopes.
Dr. Mascal, You have not covered any of the thiol reactions in lecture but it is blended with alcohol reactions in the book, are we responsible for thiol reactions on the midterm? Thanks
We have not covered it yet, it follows epoxide chemistry. You are only responsible for the subjects we have so far covered.
Hi. Dr. Mascal Can we use hydrolysis to do the cleavage C-B, like p.383 the cleavage of C-Hg. Thanks
Yes, the C–B bond can also be cleaved by H3O+.
when/where is the review session for midterm 1? thanks
This Wednesday's lecture will be the review session.
when/where is the review session for midterm 1? thanks
This Wednesday's lecture will be the review session.
Hi Dr. Mascal On p.383, water is used to do the cleavage of the carbon-mercury bond. However, can we use carboxylic acid to do the cleavage, like on p.376 replacement of boron by hydrogen from acetic acid? thanks
Yes, but remember the Hg(2+) catalyzed hydration of an alkyne is carried out in acidic aqueous solution, so there is really no need to add acetic acid.
Is the midterm on friday going to cover up to ch. 22?
See today's VOH announcement. Chapter 22 is included.
Professor, I don't know if you mentioned this in class but- up to what section is the midterm going to cover? thanks...
See today's VOH announcement.
Hi Dr. Mascal 1)On p.209, the one paragraph above 6.4, it mentions "the more substituted one has a greater degree of partial positive charge and ..." However, we know that third degree carbocation, more substituted one, is stable because of the delocalization. That means desity of the positive charge on the more substituted one should be reduced. So, why the book still says more substituted has a greater degree of "positive charge?" 2)Which diels-alder reaction will be faster if I put "methyl group" or "tert-butyl group" on the diene? thanks
1) I think there's a little confusion here. Alkyl carbocations are not stabilized by delocalization, but by the inductive effects of increasing substitution. This means the majority of positive potential will reside on the carbon which can best stabilize it, i.e. the more highly substituted carbon. 2) It depends where you put the methyl vs. tert butyl group. A tert-butyl group at the 1- or 4-position of a 1,3-diene would cause steric problems and make the reaction slower.
Professor, Regarding 9.42, the mechanism that the book suggests doesn't consider the elimination reaction which eliminates the hydrogen on the #3 carbon (relative to the O in the 6 carbon ring). How can this reaction be so selective when there appears to be no factor governing the formation of the desired product? Thanks.
There is a factor: the main product will be the most stable alkene, i.e. the one in conjugation with the oxygen. Some of the alternative alkene might however also be observed as a minor product.
Professor, Regarding 9.41 (e), why would the -OH group selectively bind to the #3 carbon and not the #2 carbon. Aren't they equivalent in terms of saturation? Thanks.
Question 9.41e is problematic. You wouldn't expect much regioselectivity in the hydration of 2-pentene. I predict there would be a mixture of 2- and 3-pentanol, but the solutions manual seems to suggest only the 3-isomer would be formed. This is a mistake.
Professor, Problem 9.39 (f), deals with section 9.9, from which reading wasn't assigned. Is this a typo, or do you hold us responsible for material to which no reading is assigned? Thanks.
Section 9.9 was covered in Chem 30, and Chem 130A builds on the basis of the chemistry developed in 30. I recommend that you review this material if you need to.
Hi Dr. Mascal 1)Problem 9.36, should this undergo dehydration which the product is akene? How come the solution shows some secondary alcohol as products? 2)When rearrangement occurs, does it require energy to make -H or -CH3 migrate? thanks
1) You could expect some dehydration (E1) in this reaction, but the SN1 product would typically predominate. 2) Yes, it requires activation energy.
Hi Dr. Mascal 1)Can we place electron-releasing group on dienophile and electron-withdrwaing group on diene which is opposite from the example in the text? 2)What reaction will be faster if I place electron-releasing or electron-withdrawing group on diene? And why? Thanks
1) Yes, this works too. But for various reasons it is usually done with the electron withdrawing group on the dienophile and the electron-donating group on the diene. 2) The reaction is generally accelerated by putting electronically complementary functions on the reactants; I don't think it's any faster one way or the other.
Hello Prof. Mascal, I as well as many other students would like to know what material would be covered on the midterm this friday. Will we be responsible for the material up to the midterm (ie. next wednesday's lecture). THank you.
See today's VOH announcement.
Hi again, this question pertains to chapter 9 on alcohols. Problem 9.39e gives a cyclohexane with a double bonded methyl groupp attached and it asks to convert it to a cyclohexane with a methylchloride attached. At first glance I would approach this problem using a hydrohalogentation of the alkene to form the product, but the sol'ns manual says to convert alkene to alcohol first before adding a chlorine. Why would converting reactant to alcohol first be favored over the reaction of the alkene with H-Cl??
If you simply add HCl across this double bond, you will get the Markovnikov product, which is not the product you want in this case. So, the alkene is converted regioselectively to the primary alcohol first.
HI prof Mascal, In chapter 9 (alcohols and thiols), we were assigned reading that excluded the oxidations of primary, secondary, tertiary alcohols to form carboxylic acids, aldehydes etc. My question is, are we responsible for this material because the suggested problems that you assigned involved the knowledge of these oxidation strategies. (ie. using chromic acid or PCC)
Oxidation/reduction of alcohols was part of the Chem 30 syllabus. It would be a good idea to review any Chem 30 material you are not confident about, since many of the reactions learned in 30 will reappear in 130A (as in this case).
Professor, If a -SH bond is less polar than a -OH bond how is the -SH hydrogen more acidic? Doesn't the increased polarity in the alcohol show less electron density by the hydrogen, thereby making it easier to pull off?
There are two opposing issues in acidity/basicity: the ability to stabilize the negative charge after the loss of a proton, and the proton affinity of the anion, e.g. O(-) or S(-). Considering O, it is better able to stabilize a negative charge than N or C, so the O-H bond is more acidic than the C-H or the N-H. But the O with its 2s/2p outer shell has greater proton affinity than S, with its more diffuse 3s/3p outer shell. This makes the proton stick more strongly to O(-) than S(-), making S-H more acidic than O-H.
Professor Mascal, is there going to be a lot of nomenclature on the midterm? Thanks.
Nomenclature is the language of organic chemistry, and the naming of carbon chains and functional groups was covered in Chem 30. For that reason, I have not taken time to go over the nomenclature of alkynes, alcohols, etc. during lecture. Although I will not be asking you directly to name molecules on the exam, the name of a molecule might appear in a question, and it would be useful to know what the name refers to...
Hi Dr. Mascal On problem 10.22(e), I do know first we need (sia)2BH as our reagent. However, I don't understand why we need deuteroacetic acid as our second reagent? What is the reaction called when organoborane reacts with deuteroacetic acid? Thanks
When a borane reacts with acetic acid (or deuteroacetic acid), the boron gets replaced by the H (or the D) of the acid. The reaction involves the acetate O pushing electrons into the boron empty orbital, causing the weaker B–C bond to cleave, with the electrons going towards the carbon and picking up the H+ (or D+) from the acid. If you don't use deuteroacetic acid in 10.22(e), you won't have D in the product.
Hi Dr. Mascal I have a question that on p. 894 table 22.1 why ester can be either electron-releasing or withdrawing groups?
You have to consider which way the ester is attached to the diene/dieneophile: if it is bonded through the carbonyl carbon, it withdraws electrons, if it is bonded through the oxygen, it releases electrons. You should be able to draw resonance forms to demonstrate this.
Are rections with sulfonyl chlorides useful with all alcohols, or are the only useful with primary and secondary alcohols. Is HCL a poor nucleophile?
Sulfonate esters can be easily prepared from primary and secondary alcohols. Tertiary alcohols also react but the R group of the sulfonate has to be small (like methyl) or the reaction is very slow. HCl is not a good nucleophile, but it is usually 'mopped up' by adding a base (like pyridine) to the reaction between alcohols and sulfonyl chlorides, just to make sure it doesn't interfere.
In step 3 of the mechanism on page 346, why is tertiary carbocation with the positive charge placed on the carbon attached to the OH group more stable than the tertiary carbocation with the positive charge placed on the carbon with the methyl group?
For the reason it gives under the structures: it is a resonance-stabilized cation. Remember, resonance is much more important than inductive effects in stabilizing carbocations.
On page 347 2-methyl-12-propanediol is converted to 2-methylproanal. In order for 2-methylpropanal to form, a primary carbocation would have to form. Isn't the driving force for rearrangements to create a more stable carbocation intermediate. Why wouldn't the reaction take place in the presence of the tertiary carbocation?
In the pinacol rearrangement of 2-methyl-1,2-propanediol to 2-methylpropanal, a tertiary carbocation is the intermediate as shown in the text. The next step of the mechanism is concerted, with the lone pair of the oxygen pushing down and pushing the hydride out and over to the neighboring carbon. There is no need to do the hydride migration first to give a primary carbocation. I know the book seems to suggest this mechanistic approach on the previous page, but that causes obvious problems in this case. The mechanism I presented in lecture includes the concerted migration - this is the one to follow.
Dr. Mascal, Is it true that thermodynamic control is always under high temperature? thanks for your time
It depends entirely on the activation energy of the reaction: some reactions are under thermodynamic control at room temperature or even below, some have to be heated to achieve this.
Dr. Mascal, In BF ch10 problem 13c) It asks to produce CH3CH2CH2CtriplebondCD from CH3CH2CH2CtriplebondCH and it doesn't specify reagents. Could this product be made by treating the initial reactant 1-pentyne with Br2 to give you a dihalide and then follow this reaction with 2NaND2? Where I use a deuterated hydrogen in lieu of normal hydrogen. The text uses 1.NaNH2 followed by 2. D2O . Thanx
The answer in the solutions manual is correct: you deprotonate the terminal alkyne with a strong base then add D2O, which behaves much like H2O would in the same circumstances (i.e. re-"protonates" the carbon). The sequence of reactions you describe does not seem to get to the product you want. NaNH2(-) is a base, and using NaND2(-) does not put D in the product.
Hi Dr. Mascal 1)Few days ago, you answered to me that 1,2-dichloroethane becoming vinyl chloride is under the process of elimination. However, at the end of p.380, there is no base to eliminate H and Cl. So, how this reaction accomplish? 2)What is "secondary orbital overlap?"
1) The reaction is formally an elimination of HCl from 1,2-dichloroethane, regardless of the mechanism. In this case, heating at very high temperatures in the presence of a catalyst accomplishes the reaction without a base. The use of high temperatures and catalysts instead of reagents is typical in industrial chemistry. 2) We define the primary overlap as that which leads to the electronic reorganization which we call cycloaddition, and a secondary overlap as an orbital-orbital interaction which is energetically favorable, but which does not result in the formation of covalent bonds.
Hi Dr. Mascal 1)Few days ago, you answered to me that 1,2-dichloroethane becoming vinyl chloride is under the process of elimination. However, at the end of p.380, there is no base to eliminate H and Cl. So, how this reaction accomplish? 2)What is "secondary orbital overlap?"
1) The reaction is formally an elimination of HCl from 1,2-dichloroethane, regardless of the mechanism. In this case, heating at very high temperatures in the presence of a catalyst accomplishes the reaction without a base. The use of high temperatures and catalysts instead of reagents is typical in industrial chemistry. 2) We define the primary overlap as that which leads to the electronic reorganization which we call cycloaddition, and a secondary overlap as an orbital-orbital interaction which is energetically favorable, but which does not result in the formation of covalent bonds.
professor,do we have to know all the rxns and the mechanisms for making alcohol that u mentioned in the class?
All of those were reactions you should already know...
Regarding the mechanisms not discussed in class (the reduction by NaBH4 in oxymercuration and the oxydation by H2O2 in hydroboration), do we need to know them for the exam? If so, then can you post them, or tell us where we may find them in the book. Thank you.
We did not discuss in detail the NaBH4 reduction of the organomercury product of the oxymercuration because the mechanism involves free radicals and is not simple. Do not worry further about this reaction. We DID however discuss the mecanism of the oxidation of organoboranes with H2O2...
Professor, In the book on page 895, regarding the reaction of cyclopentadiene and methyl propenoate to form methyl bicyclo[2.2.1]hept-5-en-endo-2-carboxylate, isn't it true that in order to end up with the endo configuration, the dienophile would have to approach the diene with the substituent pointing towards the diene and not away from it (in order to get secondary orbital overlap)? In the book, it is drawn pointing away from the diene and I was wondering if this was done on purpose or if the drawing was arbitrary since the molecule doesn't have any definite orientation in space. And on an exam, would you want us to draw the dienophile in the needed orientation to produce the desired adduct (endo or exo) or would any orientation do? Thank you.
Good observation. In the example on p. 895, it doesn't matter whether the approach is endo or exo, since there is nothing in the diene to differentiate between the two possibilities. But formally, the transition state should be drawn with the dienophile approaching endo, not exo. In the book, I think the intention was just to show the two reactants, rather than inply the approach.
Hi Dr. Mascal 1) At the end of p. 380, the book talks about 1,2-dichloroethane loses HCl to form vinyl chloride by heating it. What type of reaction it is?(eg. elimination?) 2) On p. 383, the cleavage of carbon-mercury bond is used by water. How come it is different from oxymercuration when C-Hg bond is broken by adding NaBH4? 3) Do we need to know the "Wacker process" for the midterm 1 on the p. 384. Thanks
1. Yes, it is an elimination. 2. The vinyl-metal bond cleavage is strongly influenced by the enol function (or the ketone if rearrangement can occur while the metal is still there). You should be able to draw a mechanism where the ketone (or protonated enol) assists the hydrolysis (try it). The same is not possible for the simple alkyl-metal bond. 3. I know we did not specifically discuss this in class (there is not time to cover everything), but in principle, anything in the book could show up on the exam...
Professor, Regarding the Diels-Alder reaction, I understand that the relative stereochemistry of the dienophile is retained in the product (cis remains cis and trans remains trans). But, what about the absolute stereochemistry? Is there something that can be said about that? It seems that there shouldn't be a way of assigning absolute stereochemistry, because the diene can either dock on top of the dienophile or vice versa. Secondly, regarding exo and endo products...does the exo product form in miniscule amounts or does it not form at all? Thank you.
The stereochemical relationships developed in the Diels-Alder reaction are indeed relative, not absolute. Think of it this way: if you do not start with optically active materials in the reaction, you can not have optically active products. Regarding your second point, the endo product is normally the major product, but some exo product may also be observed. It really depends on the specific reactants involved.
Hello Dr. Mascal 1)When terminal alkynes react with borane, we need disiamylborane to prevent the second hydroboration step. Why terminal alkynes are used with different substituted borane from other alkynes like 3-hexyne? 2)Why vinylic carbocation is less stable than alkyl carbocation?
1) If I understand your question right, the reason for this is that the internal alkynes undergo a second hydroboration quite slowly, so for them the reaction can easily be stopped after the addition of one borane. Therefore, a bulky borane is not necessary. But terminal alkynes add a second borane too quickly to stop after just one, so a bulky borane is required to slow things down. 2) The degree of s-character in the orbital bearing the charge has some influence over the stability of charged carbon. The more s-character, the more stable a negative charge and the less stable a positive charge will be, and vice-versa for the degree of p-character. Thus the sp2 vinylic cation is expected to be less stable than the corresponding sp3 alkyl cation.
Do we need to the mechanism for the additions or is it alright to just know the reagents and the product?
Mechanisms are the key to understanding organic chemistry - you should definitely know them. Note that although there are a lot of different reactions, there are only a few types of mechanisms which appear in slightly different guises over and over again.
Hi Dr. Mascal I have some questions on BF p.371. 1)What is geminal dihalide? Did you mention in class or chem 30 before? 2)In the end of this page, it said addition of water is neccessary to complete the alkyne preparation. Why? Is it just a simple acid-base reaction? Thanks
1) See the right margin on page 371 for the definition of geminal. 2) Yes, water simply reprotonates the alkyne anion at the end of the reaction.
Hello- Problem #6.40a states that 4 cis-trans isomers are possible for the product. In general this is true, but for this reaction wouldn't only 2 be possible, since BH3 adds with syn-addition? Thanks
Yes, that's correct. Although four stereoisomers are possible for that compound, only two can be produced by the hydroboration (and of those only one is observed to any great extent).
One more thing, about the biosynthesis of terpenes for example,the synthesis of using pyrophospates. Will we have the structure provided or will we have to memorize it? And same question goes for the other organic terpenes (Farnesol,Vitamin A,cholesterol). And what resources will be provided for the exam? THANK YOU.
There is no need to memorize complex structures (farnesol, cholesterol...) for the exam. You will be provided with a periodic table, and this is all you will need. You should not bring anything to the exam except your I.D. and something to write with
Hi prof Mascal, I was looking over your chem 30 webpage,and it indicates that you have review sheets for each midterm. Will you be distributing review sheets for 130A and if you are,when will you be distributing them?
Yes, I hope to be able to do this. If so, review sheets will be distributed at lecture, perhaps Monday the 29th.
Professor, On page 380, in reacting ethyne (acetylene) with hydrochloric acid to produce chloroethene (vinyl chloride) wouldn't there be a primary vinylic carbocation intermediate? If not, then is this reaction concerted as opposed to the mechanism proposed on page 379 for the addition of hydrogen halides?
You are right, the reaction would appear to go through a primary vinylic carbocation intermediate. Although we avoid invoking primary carbocations where possible, in this case we have little choice. More involved treatments of this mechanism might be able to skirt the issue by invoking a three-center two electron proton-bridged intermediate, but I don't think you need to worry about that at this stage. Good question, though.
Professor, In general, if a molecule is symmetric (for example trans 4-Octene) and you wish to convert it to 4-Octanol, a simple acid-catalyzed hydration would do, wouldn't it? However, the answer book insists on using oxymercuration or hydroboration in such instances when it seems, to me, to be unnecessary. Please comment on the situation. Thank you.
In principle, the simple acid-catalyzed reaction should give the product. But stewing organic molecules up in concentrated aqueous acid is really only practical in a few cases where the molecule is otherwise very simple. Often, other functional groups will be affected by such harsh conditions. So, as a rule, when an alkene function needs to be hydrated, the milder approach (hydroboration, oxymercuration) is used.
Professor, On page 380, in reacting ethyne (acetylene) with hydrochloric acid to produce chloroethene (vinyl chloride) wouldn't there be a primary vinylic carbocation intermediate? If not, then is this reaction concerted as opposed to the mechanism proposed on page 379 for the addition of hydrogen halides?
You are right, the reaction would appear to go through a primary vinylic carbocation intermediate. Although we avoid invoking primary carbocations where possible, in this case we have little choice. More involved treatments of this mechanism might be able to skirt the issue by invoking a three-center two electron proton-bridged intermediate, but I don't think you need to worry about that at this stage. Good question, though.
Why does the 1,4 product predominate at equilibrium(once the energy level of the system is at a certain level)? Although the 1,4 product is more stable, the 1,2 addition product has a faster rate of reaction at both higher and lower tempratures. Does this mean that the 1,2 product will be in a higher concentration than the 1,4 product?
Under equilibrium conditions, the rate of reaction becomes less important than the thermodynamic stability of the products. The more stable product must predominate as given by the relationship G=-RTlnK(eq), where G is the Gibbs free energy (delta H - T delta S) and K(eq) is the equilibrium constant. The equilibrium concentration of each species thus relates to their energy (stability). Consult your general chemistry textbook for a more detailed discussion of this.
I just have a general question. If I didn't do too well in Chem 30, does that mean I am not going to do well in 130A? I am worried because you keep mentioning how we learned everything in 30. For me, some parts don't connect from 30 to 130a. I took 30 during summer 2000, maybe I might have forgotten something, but I get worried whenever you say all these things should be familiar. What do you suggest?
Chem 130A and 130B build on basics learned in Chem 30, and it is essential to have a good understanding of the fundamental reactions of organic chemistry (addition, elimination, substitution, rearrangement) in order to completely understand the more involved reactions we will study this quarter. Although 30 by definition must cover certain issues, each instructor will have his/her own agenda, and will choose to emphasize (or de-emphasize) certain subjects. If you did not have me in Chem 30, I strongly recommend you do the following: First, download the syllabus, exams and worksheets from my Chem 30 course last quarter and see if there is anything you don't recognize or need to review. Second, go to the union bookstore and get a copy of my LectureNotes from last quarter, which give the most direct indication of what was covered in 30 on a day to day basis. I am sorry if you feel uncomfortable with my references during lecture to the previous course, but I feel it is very important to relate new chemistry to reactions I know the students have already studied. I hope this is helpful.
Hi Dr. Mascal I have a question on p.208 BF. In the middle of the paragraph, it says "unlike bromine, mercury has no electron pair to donate..." How come mercury has no electron pair to donate since atomatic number of mercury is biger than bromine? thanks
The chemistry of metal ions is very unlike that of the nonmetals, and even though it seems Hg has lots of electrons to donate, experience shows that it does not really demonstrate nucleophilicity, thus it is assumed that Hg does not share its outer shell (6s, 5d) electrons in the same way Br might share its 4s and 4p electrons.
I was trying to find the mechanism for how the organomercury compound in oxymercuration undergoes reduction by NaBH4. In other words, how does Hg-OAc get replaced by H.
The mechanism involves radicals (which means that the reduction step is not stereoselective) and hence is not simple. Mercury is reduced to Hg(0) in the process, but I would have to check in a textbook on organometallic chemistry to be sure of the answer to this question.
Why are allenes less stable than alkynes? In example 10.4, 2-pentyne is suggested to have the same degree of substitution about the carbon double bond. One carbon of the double bond has an ethyl group attached and the other has a methyl group attached to it. Wouldn't the carbon with the ethyl group attached have a higher degree of substitution than the group with the methyl attached to it? On page 376, it says that treatment of a trialkenylborane with a carboxylic results in steroselective replacement of boron by hydrogen. Is there a difference if H2O2/NaOH instead of a carboxylic acid is used? Why is a carboxylic acid used here?
First question: it can be shown that allenes are less stable than the corresponding alkyne by thermochemistry, as noted in the book. The theoretical basis for this lower stability is complex and beyond the scope of this course. Second question: each carbon of the triple (not double) bond is attached to another carbon (alkyl group), which means the borane has little basis for discrimmination between the two positions. The ethyl is somewhat larger than the methyl, but the difference here is much less than that between an alkyl group and a hydrogen (i.e. an internal vs. terminal alkyne). So, little regioselectivity is expected. Third question: the reaction of the alkenylborane with an acid gives a different result (protonation) than reaction with H2O2/NaOH (oxidation). Both are stereoselective, but in the case of the oxidation an enol results, which tautomerizes to the carbonyl and thus loses its stereochemistry.
Professor, On page 377 in our text, 2-Butyne is converted to 2-Buten-2-ol and 2-Butanone.....my question is why isn't more than one borane bonded to 2-Butyne since it isn't a sterically hindered disubstituded borane?
Multiple reaction with borane is usually only a problem with terminal alkynes. 2-Butyne has an internal triple bond and so a sterically hindered borane is not required.
In question 6.21(E), OH and HgAcO are added to the same side of the double bond although anti-steroselectivity that is shown in the mechanisms. Is this due to there not being enough space for the HgAcO molecule to bond to the same side of the double bond as Methyl? If so, how would it be possibble for H2O to attack the same side of the molecule as HgACO?
The answer in the study guide is not meant to show stereochemistry. However, the addition is stereoselective (anti), but we didn't emphasize this in the lecture because in the subsequent reduction with NaBH4 you lose the stereochemistry (because it goes through a radical intermediate). Since oxymercuration can not really be used to synthesize alcohols stereoselectively, I don't think it's worth worrying about the stereochemistry of the organomercury intermediate.
What determines where isoprene