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| Q&A
for 14D |
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Dr. H, The exam key isn't up yet. Is there a problem with the chem server again? I've tested the site from several locations and none of them see the exam key.
Mark, The exam keys were posted on Sunday morning, December 18. --DrH
Hi Dr. Hardinger, This is a last minute question, but I was reviewing the Enolates, Enols,and Enamines section and for number 5(a) on page 234 of the thinkbook, I couldn't get the Keq to equal 10^-9.5 as listed. My answer was equivalent to 10^9.5. Is this an error? Thanks.
Jessica, The equilibrium favors the weakest acid, which is the ester. Therefore the equilibrium lies to the left, and Keq < 1. Remember this when figuring out which pKa gets subtracted from which other pKa. --DrH
Hey Prof. On Exam Part B for Spring 2003, for #5b, we're changing the solvent to a protic one. I remember learning that in a protic solvent, larger nucleophiles are preferred for Sn2 reactions. Since the nucleophile is quite large, wont its nucleophilicty be enhanced, thus making the reaction faster? Thanks.
Faysal, The atomic size effect on nucleophilicity applies just to the atom that shares electrons with the electrophile, not the entire molecule. --DrH
Hey Dr. H, In you rlecture supplement for electrophilic substitution, it says NO2 is a withdrawing group - a meta director. But, below that on page 18, it shows nitration causing ortho and para additions. Is this a mistake or is this an exception?
Sev, Directing effects are caused by the group already attached to the ring, not the incoming electrophile. --DrH
In fall 2004 final exam part A #3a, wouldn't the superior resonance stabilization provided by the F atoms, which are in the same row as carbon, make nucleophilic attack on the carbonyl diflouride slower?
Bridget, That is correct. A clarification has been added to the exam key, and you have been awarded one error bounty extra credit point. --DrH
In SUMMER 2005 final exam part B #8, why wouldn't the major elimination product be adjacent to the ester, which would contribute a (minor) resonance contributor, wouldn't it? (sorry, wrong exam).
Bridget, The resonance contributor provided by the "ether" oxygen of the ester is small. In addition, this elimination product is a less-highly substituted alkene. --DrH
i know this is kind of late, but the winter 05 final part a key is not working. can you fix it?
Kevin, I do not know what you eamn by "not working," but it works fine for me on my computer. --DrH
In Spring 2005 final exam part B #8, why wouldn't the major elimination product be adjacent to the ester, which would contribute a (minor) resonance contributor, wouldn't it?
Bridget, I don't understand your question, as the molecule under discussion does not contain an ester. --DrH
Dr H, For this quarter's Exam 1, 18c), can you explain why the reaction rate would have not much change? It seems like it should be slower since Cl- is a worser leaving group and a primary carbocation is formed. The carbocation could rearrange to a secondary with resonance, but can't the original carbocation also undergo this rearranging with a methyl group?
Chris, The acceptable answer depends upon what you wrote for the mechanism. The shift to primary chloride makes SN1 slower but the effect on Sn2 is unclear. --DrH
I was wondering if LiAlH4 will only remove the carbonyl oxygen in cases where the carbonyl is an amine, or if there are other times it will do this. Thanks.
Bridget. Of the carbonyl functional groups we have talked about, all but one is reduced to an alcohol by LiAlH4. Amides are the only C=O functional group have discussed which becomes an amine. --DrH
Professor, page 203 #15, after the nucleophilic attack on the carbonyl carbon, can we used H2SO4 to protonate the oxygen anion, or would that be incorrect?
Elina, H2SO4 is not present. Note the reactants include aqueous H2SO4, so all H2SO4 has been converted to H3O+. --DrH
Hi again, here is my resubmitted question: For Winter 2005 Exam Part A, #1c, since the reaction takes place in ethanol, why is the O- on the right hand side not protonated? I understand that the O- form allows for significant resonance stabilzation, but you indicate in the directions to assume that everything is in excess, so instinctively I would have protonated it. Also, what determines if (CH3)3CS- only attacks one carbonyl vs. attacking both of them? Again, since its in excess i figured it would attack both. Thanks.
Faysal, A carboxylate ion is a weak base (RCO2H pKa about 5) and ethanol a weak acid (pKa 16), so at one molar concentration the equilbrium favors RCO2- and ethanol by Keq = 10-11. The ethanol would have to be at an impossibly high concentration to shift the equilibrium to favor RCO2H. --DrH
Hi professor, on page 205 the answer to problem #5 when showing the faster reaction to enolate formation... what happened to the flourine atoms? Is this an error?
Elina, Check the Known Typos list. --DrH
Dr H. I don't know if you consider this of chemical/scientific importance, but to my knowledge, bacteria have cell "membranes," not cell walls ( which plants have) as stated in PP # 2 in the problem bio. Just trying for a bounty point. Thanks.
Ashkan, Do a google search on bacteria cell wall. --DrH
Hello, for Spring 2003 Exam 3 # 3, in the addition mechanism you protonate the O- at the end, however in the substitution mechanism you don't. Why the difference? The solvents are the same for both. Thanks!
Igot, It is because a carboxylate (RCO2-) is a much weaker base than an alkoxide (RO-). --DrH
Dr. H on practice exam spring 2005 part A..number 7...could you tell me why the fourth mechanism, the oxygen of the carbonyl is protonated instead of the ester that leaves the group?...I thought that you would want to protonate the ester to make it a better leaving group...is that wrong?
Nancy, I don't understand your question. The carbonyl of the ester is, in fact, receiving the proton. --DrH
Dr. H, On practice exam spring 2005 solutions for part B..number 19....you use LDA and wrote it as NiPr2...but in the thinkbook you refer to it as N(iPr)2...is the one used on the exam incorrect?
Nancy, The two abbreviations are interchangeable, although most students prefer -N(iPr)2. --DrH
Hi Dr. H, just for clarification. Pg. 185 #12b, why can't we do a methyl group transfer to form a tertiary radical? I know that hydrogen atom transfers are more common than methyl transfers but isn't a highly substituted radical(a tertiary vs secondary in this case) more stable?
Elina, The product derived from methyl transfer might be produced in a tiny amount, but as you pointed out, hydrogen transfers are much more common than methyl transfers, regardless of radical stability. --DrH
Hey Prof. For Winter 2005 Exam Part A, #1b, since the reaction takes place in ethanol, why is the O- on the right hand side not protonated? I understand that the O- form allows for significant resonance stabilzation, but you indicate in the directions to assume that everything is in excess, so instinctively I would have protonated it. Also, what determines if (CH3)3CS- only attacks one carbonyl vs. attacking both of them? Again, since its in excess i figured it would attack both. Thanks.
Faysal, I do not understand your question. It does not seem to be relevant to quiestion 1(b) on the winter 2005 part A final. --DrH
Prof, Winter 2005 Part A exam, #1a. Does the hydroxide perform nucleophilic substitution on the carbonyl on the right or does it simply deprotonate the existing OH? I know you just ask for the product but i'm curious about the mechanism. Thanks.
Faysal, Proton transfers are generally faster than nucleophile/electrophile steps, so RCO2H is deprotonated by HO-. --DrH
Professor, under #1 of the Carbonyls CFQ it lists acetonitrile/nitriles under functional groups that contain a carbonyl. I don't understand why if it doesn't contain C-O pi bond.
Char, See page 20 of the Thinkbook. --DrH
Hi Dr.H On Winter 2005 Final Part A #6, I'm confused about why the ketone enolate is more stable. Why doesn't the oxygen in ester in this case contribute to the resonance?
Ginette, The "ether" oxygen of an ester destabilizes the enolate through electron donation. Review page 31 of the Thinkbook. --DrH
Hello Dr.H, is this an error? Page 168 #12, the directions ask to show the mechanisms for the MAJOR products formed in the reactions from #11 (previous problem). Well in the answer, pg 176, the mechanism is not shown for the MAJOR product(which should be in the ortho position). The mechanism provided shows a para product, which is NOT the major product.
Elina, The mechanisms for the ortho products are the same as the para products, differing only in the position that the electrophile attacks the benzene ring. --DrH
Just a question about Spring 2005 Final Part A #6 part C. The product has an amine group still attached. Is this because Nitrogen is more basic than oxygen and that water leaves before amine?
Keun, It has to do with the presence of aluminum cations (from the LiAlH4), and how they alter leaving group properties. You are not responsible for this mechanism, but feel free to ask me or an IA in person for more details. --DrH
Dr. H, In Friedel Crafts alkylation and acylation using AlCl3 and a group with a Cl attatched to one end, does AlCl4 act as the base to remove the hydrogen from the ring to restore aromaticity or is another base required for this reaction to occur? Thank you.
Zak, AlCl4- deprotonates the arenium ion. Alternately, AlCl4- can ionize into AlCl3 amd Cl-, then the Cl- deprotonates the arenium ion. --DrH
Hi Professor, For the enolates OWLS problem, on solution 2b, can the other hydrogen that is next to the carbonyl be deprotonated to form a pi bond instead? Wouldn't this be more favorable because the enolate has more substituents bonded to it, making it more stable?
Sandhya, LDA has a preference to remove the less sterically hindered proton, unless the more hindered proton is much more acidic. It is true that in this case the more hinered proton leads to a more stable enolate, but no much more stable as to overcome the steric preference. --DrH
On page 41 of the Thinkbook, under example drugs that target receptors, I believe that there is an error. Famotidine targets "histamine" receptors not Histidine receptors. Histamine is derived from the amino acid Histidine (via histidine decarboxylase), but the receptor should be for the Histamine.
Jennifer, You are correct. The error has been added to the Known Typos web page, and you have been awarded one error bounty extra credit point. --DrH
Dr. H., Sorry to bother you with another definition technicality, but for Winter 2004 Final Part A, #11 (a) the answer key says that a subtrate mimic resembles "the natural substrate of an enzyme or receptor." However, receptors do not bind to substrates. Instead, they interact with signal molecules. Doesn't "substrate" refer to the target molecule of an enzyme only? Thanks.
Kimberly, You SHOULD be fussy about definitions, as their role is to very specifically define the is/is not of a concept. You are correct about the error. AN updated exam key has been posted, and your correct fussiness has earned you an error bounty extra credit point. --DrH
Professor, In the Spring 2005 final, part A, question 12, you state in the answer key that the selected product is more highly conjugated than the other possible product, as a reason for that particular reaction being faster. However, isn't it true that product stability does not affect the rate of a reaction, as the reactants don't, so to speak, "know" what product is going to be formed? It seems to me that the reason given is more of a reason to explain how far to the right the equilibrium lies than to explain how fast the reaction proceeds. Thanks for your help.
Kunal, For a single mechanism step, the reaction does "know" what will happen, because the transition state has both reactant and product character. However, the reaction does not "know" beyond a single mechanism step, unless everything is in equilibrium. Then the reaction "knows" about all the reactants, transition states, intermediates and products. In this case everything is in equilibrium, although the equilibrium strongly favors the final product over anything else. --DrH
Hi, for Fall 2004 Final Part A #4 the question says "...it reacts with water to form CO2, and another molecule of water." But the answer has -OH and CO2, you wouldn't protonate the -OH to make it water? Thanks!
Igor, Normally we don't bother to write mechanisms for inorganic reactions, such as HO- + H3O+ ---> 2 H2O. --DrH
Prof, for winter 2004 final exam part A #7, what is the role of the methanol in the reaction? Just a solvent? Because it doesnt seem to take part in the mechanism. Thanks.
Faysal, It is just the solvent. --DrH
For Summer 2005 Final Exam Part A #8d, why do we need to include H30+? Why is it necessary?
Katherine, 1,3-dicarbonyl compounds are easily deprotonated, so in a reaction with strong base they are converted to their enolates. This cannot be stopped, even if only a small amount of base is used. See page 35 of the Thinkbook. --DrH
Hi Dr. H On Winter 2005 Final Part B #1, I'm confused with the effect of fluorine. If fluorine is high electronegative, doesn't it tend to pull electrons toward itself, therefore destabilizing the carboxylic acid?
Ginette, Fluorine's inductive effect stabilizes the carboxylate and destabilizes the carboxylic acid. Both of these effects shift the proton transfer equilibrium away from the acid and towards the conjugate base. In other words, the inductive effect of fluorine makes it easier for RCO2H to become RCO2-. --DRH
Hi Dr. H On Final Exam Part B Summer 2005 #28, I don't understand how H20 can be used to attack the C=N bond. Doesn't the aromatic ring make this bond more stable? Would it be wrong if I protonated it first?
Ginette, That sounds reasonable, depending upon what you do with the rest of the mechanism. Note the comment at the start of that answer in the solutions. --DrH
Hi Dr. H., I was studying CFQ&PP: Carbonyl Chemistry - Fundamentals, and in problem #9 from the PP, you state, "Solution 9(b), page 207: In the last line of the answer the fastest reaction is H (not E). " But for 9(a) you show the mechanism for reaction I . Is this correct? Thanks. -Jeison
Jeison, Check the Known Typos web page. --DrH
Hi Professor, For the solution to 3f on p. 225, in the second to last mechanism step, why does the nitrogen deprotonate the H3O+ instead of the carbonyl product?
Sandhya, It is acceptable to use CH3NH2 instead of water as the base that causes the tetrahedral intermediate decomposition in the last step. --DrH
For Spring 2005 Final Part B solution for #19, aren't there supposed to be parentheses around (iPr) so that LDA is -N(iPr)2 instead of -NiPr2?
Katherine, N(iPr)2 and NiPr2 are equivalent, although I can see why the latter is less clear. I will post a clarified exam key. --DrH
Dr. H., Can you clarify what suicide inhibitors bind to (enzymes and receptors, or just enzymes). The reason for the confusion is that in the Lecture supplement (page 42) and Summer 2005 Final Part A #14, it only refers to covalent bonding with an enzyme. However, in CFQ #1 (o)(page 250) and Spring 2005 Final Part A #10 (a), it refers to both enzymes and receptors. Which is correct (and does this qualify for an error bounty)? Thanks!
Kimberly, The term "suicide inhibitor" is reserved for enyzmes. There are few if any pharmaceuticals that work by covalently binding to a receptor. Therefore remove "receptor" form solution 1(o) on page 250 of the Thinkbook and the exam key. The error has been added to the Known Typos list, and a corrected exam key posted. You have been awarded one error bounty extra credit point. --DrH
Dr. H., For Summer 2005 Final Part A, #8 (d), would the reactants 1.) LDA, benzaladehyde, and 2.) hryronium be an acceptable answer? Thanks.
Kimberly, Since I have no idea what "hryronium" is I cannot comment on your answer. --DrH
i was going over the OWLS for Carbonyl Chemistry: Survey of Reactions and Mechanisms and i have a question about the answer for 2(f). why does the answer show only one OH group on the ring. shouldn't there be one on each side of the ring. if not, then what happens to the other oxygen.
Michael, LiAlH4 converts an amide to an amine. The oxygen gets removed by LiAlH4. You are not responsible for the mechanism. --DrH
Dr. H., For CFQ #14 on page 238, H30+ is indicated as a reactant in the reaction. However, in the mechanim on page 239, the second step shows water as the nucleophile. Is this an error? If not, is H20 used because iodide takes a proton from hydronium, forming water, which then functions as the proton shuttle? If this is true, does this step need to be shown (i.e. H30+ + I- --> H20 + HI)? Thanks for clarifying.
Kimberly, The presence of H3O+ implies the presence of water as well. --DrH
Dr. H On practice exam winter 2004-Part A-#7..how were we suppose to know that the -OH was suppose to act as a nucleophile instead of a base that would create an enolate isntead of the nucleophilic attack shown?...in general..how do we know to make a enolate or nucleophilic attack when in this situation?
Nancy, The question asks you to provide a hydrolysis product. Hydrolysis is usually a substitution reaction with water, not an enolate reaction. --DrH
Dr. H On practice exam winter 2004-Part A-#7..how were we suppose to know that the -OH was suppose to act as a nucleophile instead of a base that would create an enolate isntead of the nucleophilic attack shown?...in general..how do we know to make a enolate or nucleophilic attack when in this situation?
Nancy, The question asks you to provide a hydrolysis product. Hydrolysis is usually a substitution reaction with water, not an enolate reaction. --DrH
Dr. H On practice exam winter 2004--#7..how were we suppose to know that the -OH was suppose to act as a nucleophile instead of a base that would create an enolate isntead of the nucleophilic attack shown?...in general..how do we know to make a enolate or nucleophilic attack when in this situation?
Nancy, There are four exams for the winter 2004. Please resubmit your question along with the exam number. --DrH
Dr. on practice exam part A of Winter 2005,--on #1-a---I dont understand why it is that the -OH attacks the carbonyl with amid istead of the other ketone..because isnt the resonce resistance stronger at the amide?
Nancy, I do not understand your question. The molecule has an amide and a carboxylic acid (which is converted to a carboxylate with HO-), but it does not have a ketone. --DrH
Dr. H, I was reviewing the lecture supplement for Elimination reactions and noticed that at the bottom of page 12 under "Other E2 examples" you show a reaction with 2-dibromobutane that results in a Z isomer of an alkene. Shouldn't the E isomer form and the carbon group and Bromine groups be on opposite sides of the double bond? Thank you.
Zak, Build a model and examine the conformation necessary to have the beta hydrogen anti-periplanar to the leaving group. As this moves from starting material to product, it forces the two methyl groups to give the stereochemistry shown. --DrH
Hi, for Final part B Summer 2005 Q#25, why wouldn't you first protonate the oxygen of the carbonyl? I believe you said when a strong acid is present, we should protonate first. And aren't we not allowed to have negative charges in an acidic solution? Thanks!
Igor, An aldehyde is sufficiently electrophilic that it does not need protonation prior to nucleophilic attack. The mechanism shown in the answer key is accurate. An alternate mechanism that begins with carbonyl protonation is also acceptable. --DrH
Dr H. For CFQ Solution #7 for Carbonyl Chemistry - Fundamentals... the question asks for the factor that controls the mechanistic fate of the tetrahedral intermediate that bears the negatively charged oxygen atom. If the two fates are to lose the LG or gain a H+ on the negative O, I don't see how the solution answers the question. What is the factor that causes it to decide? Is it whether the LG is a good one?
Ashkan, The question does not ask you to decide which tetrahedral intermediate fate happens first. It just asks you to decide what causes the tetrahedral intermediate to react the way it does. --DrH
Dr. H, on page 28 of the 14D Thinkbook, I carried out the reaction of peptide hydrolysis and got the right product except for the H3N+ --{ . Should the product not be H2N --{ as in page 27 since you can only protonate the nitrogen once to make it a good leaving group? Is this an error?
Antonio, RNH2 + H3O+ ---> RNH3+ (The acid in in excess) --DrH
Hi Dr. H For the molecule CH3OH in general, when is it strong enough to protonate an oxygen? It seems that there are mechanism where it is used, other mechanisms when it's not.
Ginette, CH3OH is a pretty weak acid (pKa 15.7), so it can only protonate a strongly basic oxygen...an oxygen that has a formal negative charge such as HO- or CH3O-. --DrH
Hey Prof, for Spring 2003 Exam 3, why does the LiAlH4 protonate each oxygen in #5a, but in #5c the NaBH4 only protonates 1 oxygen? Is there an inherent difference in the nucleophilicity of these two compounds? Thanks.
Hey Faysal, I do not understand your question. LiAlH4 is not an acid (in fact it's a pretty good base) so it doesn't protonate any oxygen under any circumstance. --DrH
Dear Professor Hardinger, On the PP#20e for Eliminations on pg. 138, the major product of the SN1 reaction is not shown. (a 1-methyl cyclohexan-(1)-ol) Is this a mistake? There is no reason why the major product of the E1 mechanism should be more prevalent than the SN1 product, is there? Thank you!
Anita, That is because any alcohol products would be dehydrated to alkenes under these reaction conditions. --DrH
Hey Prof, in the answer key for Spring 2003 Exam 3, #5b, the OH in the ring is missing a formal positive charge. Is this an error?
Faysal, I do not understand your question. The starting material is correct as drawn (no formal charge). The product does not have a ring. --DrH
Hi Dr. H, On Spring 2005 Final Part B #7, can you explain why the hydrogen next to Cl was deprotonated? Does it depend on the electronegativity of Cl and I, or whether they're good/bad leaving groups?
Ginette, Iodide is a better leaving group than chloride. In addition, the hydrogen in question is slightly more acidic due to the inductive effect of chlorine. --DrH
Hi Dr. H, on Winter Exam 2004 Part B Solutions problem 16) shouldn't the sulfur trioxide be protonated by the H2SO4 first so that it becomes more electron deficient? Then you'll have the -OSO3H as the proton bus to deprotonate in the second step of the reaction to regain aromaticity. Otherwise without deprotonating H2S04, how would you have -OSO3H for the second step.
Linda, Sulfur trioxide is sufficiently electrophilic to attack the benzene ring without the need to be protonated to make SO3H. This is especially true in this case because the benzene ring has an activating substituent. --DrH
Dear Professor Hardinger, I think a statement I ran across in Bruice is a false statement (or at least poorly worded). It says on page 422, "If the reactant is a secondary or a tertiary alkyl halide, it may undergo either SN2/E2 or SN1/E1 reactions, depending on the reaction conditions." If a reactant is secondary, then SN2 may occur, depending on other conditions, but isn't it true that a tertiary alkyl halide does not undergo SN2, no exceptions? Thank you!
Anita, It's not precisely true to say that a tertiary halide cannot undergo SN2, but rather its SN1 reaction is so much faster that it never gets a chance to do SN2. I agree that the statement in the text is a bit confusing. --DrH
Dr. H, ON the practice exam part A of winter 2004...on #7, you have the carbonyl go under a nucleophilic attack with OH as the nucleophile....what I was wondering is why wasnt OH used as a base to for enolate created istead of a nucleophilic attack?...
Nancy, The question asks for a hydrolysis reaction. For a carbonyl compound, hydrolysis usually entails nucleophilic carbonyl substitution. Enolate formation can compete, but there is nothing for the enolate to do (an amide cannot be attacked by an enolate), so it just gets protonated. --DrH
Dear Professor, For CFQ #10 on page 103, the solution says that the stereochemistry of the enantiomer products of an SN1 reaction are produced in an equal amount. My notes from lecture and Bruice both say that the inverted enantiomer is more prevalent than the one with retained stereochemistry even though both are formed because the leaving group sometimes gets in the way of frontside attack. So which one is true?
Anita, Either can be true depending upon the reaction conditions. Complete racemization is uncommon, however. Most of the time you get an unequal mixture favoring retention, as you described. --DrH
Hi prof Hardinger: When we're comparing Elimination rxns, if OH is set up as a LG, but can potentially be protonated to water and leave, do we still consider it a bad LG and write off E1/SN1? Or do we take into account the OH2+ and say its a moderate LG.
Teabruin, You're describing an impossible situation. Eliminating HO- requires strong base (and conjugated product). Converting HO leaving group to H2O leaving group requires strong acid like H3O+ or H2SO4. Strong acid and strong base neutralize each other, so it is not possible to have strong acid and strong base together in the same reaction. The mechanism choices you ask about could never occur together. --DrH
Hi Dr. H For Fall 2004 Final Exam Part A #7 step 2 of the mechanism, does it matter whether oxygen is protonated first or NH2 is deprotonated first? How do you determine which happens first?
Ginette, Does not matter as long as you do not have two atoms or groups with the same + or - charge attached to the same atom. Causes too much repulsion. For example - O--C--O - is bad. --DrH
Hi Dr. H, on fall 2004 Final Exam Part A problem 7, even though it is not necessary to protonate the oxygen in this reaction would it be wrong if Enz-NH3+ was used to protonate the oxygen since I think it's a strong acid due to the + charge it has more desire to donate its proton?
Linda, Enz-NH3+ is not a strong acid. Compare NH4+, pka 10..a weak acid. ENz-NH3+ is not sufficiently acidic to protonate a carbonyl. Carbonyl protonation requires H2SO4, H3O+, ROH2+, etc...strong acids. --DrH
i see that in the winter 2004 Final Exam - Part A, there is chance for extra credit based on the drug discovery and development section. i was just wondering if you had included a similar extra credit section in this quarter's final.
Manish, No extra credit of that type this time. --DrH
in final exam of spring 2003, the anwser key shows that the anwser for 1.a. with the chlorine in the para position. shouldn't it be in the ortho position.
Manish, In lecture I mentioned that there is a delicate balance between steric and probability effects, that could cause the major product to be either ortho or para. I have updated the exam key to reflect this. --DrH
Hi Dr. H, on Final Exam 2003 question 1b) solutions, is the reason why there possibly is NR because NO2 is a deactivator thus making the ring less nucleophilic therefore not being strong enough to attack the C(CH3)3? Thanks.
Linda, Friedel-Crafts alkylation and acylation reactions usually fail if the benzene ring has one or more electron-withdrawing groups, such as nitro. Your analysis is correct. --DrH
For Winter 2004 Final Exam Part B #19, in the third mechanism step, can we react the BrCPhCH3 with OH- instead of (CH3)3CO-? What is the purpose of (CH3)3CO-?
Katherine, The (1) (2) notation tells us that (CH3)3CO- is not present at the same time as HO-, so their use is not interchangeable. In addition, H2O2/(CH3)3CO- will not work in the second phase of the hydroboration/oxidation reaction. (CH3)3CO- serves as a base in an E2 reaction. --DrH
For Winter 2005 Final Part B, for the solutions for #6, the final product is missing two methyl groups on the carbon that is adjacent to the ring.
Katherine, You are correct. An updated exam key has been posted, and you have been awarded one error bounty extra credit point. --DrH
Dr. H, OWLS: Carbonyl Chemistry - Fundamentals Answer Key, Problem 5b asks us to complete the mechanism. The first step you've drawn is a nucleophilic attack at an oxygen to remove a hydrogen atom. Weren't the three fates supposed to be nucleophilic attack at a carbon, electrophilic attack at oxygen or enolate formation using a Hydrogen bounded to an alpha carbon? Is this a mistake? Or are we allowed to modify molecules before we decide which of the three fates to use? Thanks.
Vikram, I never said the first step must occur at the carbonyl group. --DrH
Dear Professor Hardinger, for CFQ#9d for Enols, can the answer also be as an enol? (since conjugation stabilizes this form, I thought it can occur too)
Anita, The alkylated product cannot form a very stable enol. If you mean form the enol from the starting beta-ketoester, then what is the purpose of the butyl bromide? --DrH
Dear Professor, On the final exam, are we going to apply concepts from the entire course such as SN2 vs. E2 reactions on the part of the exam that is just the new stuff?
Anita, As I mentioned twice in lecture, half of the final exam is cumulative. Concepts from any part of the course might be there. --DrH
Dear Professor, For PP#9b on page 246, could an E2 reaction with the (CH3)2CHI occur as well?
Anita, There might be some E2, but SN2 is major. --DrH
For Spring 2005 Final Part A #2, is it correct if we deprotoante the OH2+ with water in the second mechanism step before we use the oxoanion to kick out the leaving group? Thank you.
Katherine, Remember that intramolecular reactions are generally faster than intermolecular reactions. Your proton transfer is not as likely as the given mechanism. --DrH
Hey Prof, I just want to clarify something. You told a student we are not responsible for CFQ 12 or 13, but CFQ 11 deals with the steps of drug discovery so I just want to make sure we aren't responsible for that either since it wasn't covered in lecture. Thanks.
Faysal, I forgot to include CFQ #11 in the "not responsible" list. --DrH
I noticed that on some of the previous exams, the pharm. (drug discovery) material that we didn't cover was given as extra credit? should i look this material over or skip it entirely? Thanks.
Elya, Skip it. No extra credit this time. --DrH
Hi Dr. H, on Winter 2002 Exam 3 solutions, problem 8, insteading of being protonated, why wouldn't the oxyanion kick off the NH as a leaving group? I thought the ring strain and the - charge delocalization would help NH to leave without being protonated. Plus it would be a intramolecular rxn while protonating the oxyanion is intermolecular. Thanks.
Linda, That could happen, but a nitranion is a really poor leaving group. --DrH
Hi Dr. H, on Winter 2002 Exam 3 Solutions, Problem #1, the illustration for addition to a pi bond shows that the para position is attacked because of steric reasons. My question is, why wouldn't ortho be favored since it leads to a tertiary radical with resonace? I wouldn't think that the sterics from the methyl group would be large enough to cause para attack perference. Thanks.
Linda, I mentioned in lecture that there were many such inconsistencies throughout old exam keys. This is one of them that I have yet to fix. --DrH
Hey Prof, for Enolates OWLS solutions to #2d, on the second line of the mechanism, is it necessary to protonate the OH for it to leave? I thought we could just deprotonate the alpha carbon and the resulting conjugation could kick out the OH? Thanks.
Faysal, This point is covered in the explanation at the end of that mechanism. --DrH
Dear Professor Hardinger, for CFQ#13 on page 238, I know that the dimethyl amine has been corrected to be (CH3)2NH, but there is another error in the usage of dimethyl amine in the mechanism. In the third step of the mechanism where the amine group in the tetrahedral is being deprotonated, the solution shows (CH3)2NH2 deprotonating the amine group. (CH3)2NH should be deprotonating the amine group instead of (CH3)2NH2+ because it has a lone pair to attack with. Is this right?
Anita, The description of the error at the Known Typos web page covers all instances of extra N-H bonds in the mechanism. --DrH
Dear Professor, For CFQ#12 on page 238, in the second step of the base catalyzed reaction, shouldn't there be an arrow going from the lone pair on the carbon to form the C-C double bond and to break the C-O double bond? Thank you!
Anita, This issue is covered under the previous VOH question. Scroll down. --DrH
Dear Professor, For CFQ #12 on page 238, shouldn't there be an arrow showing the movement of electrons from the double bond to the +OH group in the second step of the acid catalyzed reaction?
Anita, This is not a mechanism step, because the structure are resonance contributors. It is not necessary to include curved arrow that show the hypothetical electron redistribution for resonance, but they can be shown if you need them. --DrH
For Summer 2005 Final Exam Part B #26, the answer key has the carboxylic acid protonated, but since there is a strong base (-OCH3) and the carboxylic acid is a strong acid, the answer should have the carboxylic acid deprotonated. Isn't this right?
Keun, Any strong base present would be protonated by the strong acid (H3O+). Even weaker bases such as RCO2- are also protonated by H3O+. --DrH
For Summer 2005 Final Exam Part B #20, I don't see how the product provided in the answer key is formed. I thought a bromine radical would be the answer for this question. Could you explain how the product in the answer key is formed?
Keun, The product is formed by free-radical halogenation, just like the given product. Bromine radical is not a reaction product (you can't put it in a bottle). It is a reactive intermediate, just like a carbocation. --DrH
There is an error in Summer 2005 Final Exam Part A #7. The product in the answer key for this question is totally off. The product contains an ester when the question doesn't even have an ester. The arrow should lead to a carbonyl but the product is not a carbonyl. The product is just totally wrong. And the methyl group on the five membered carbon ring is missing also. Do I get an extra bounty point?
Keun, Yes, the mechanism isn't quite right for the problem. An updated mechanism will be posted. You have been awarded one error bounty extra credit point. --DrH
Hey Prof, for OWLS page 71 #5a, would it be possible to just use water instead of NaOH? Thanks.
Faysal, Water is not a strong enough nucleophile to attack an ester. If it was, then all your phospholipids would hydrolyze and your cell membranes fall apart. --DrH
Hi Dr. H, On the Fall 2004 Final Part B #11, the requirement for the base in an E2 reaction is listed as having to be moderate or better. Doesn't the base have to be strong for E2 to proceed,as CH3O- is in this case? Thanks, Kinata
Kinata, Moderate bases will cause E2 if the leaving group is very very good, or if the product gains conjugation. --DrH
Dr. Hardinger, For OWLS, #4 on page 70, why doesn't the H30+ attack PhNH2? This would then form PhNH3+, which would then protonate PhCHO. Isn't PhNH2 more basic than PhCHO? Thanks,Taera
Taera, Some does, but it is reversible. It alters the kinetics, but not the mechanism. --DrH
Dr. Hardinger, For OWLS problem #3a on page 70, isn't the CH3OH supposed to protanate the alkoxide (O-) instead of H20 because in the reaction mechanism, the carboxylic acid does not necessarily need to be deprotonated before the negatively charged oxygen gets protonated...or does it? If this is the case then water wouldn't even be made to protonate the alkoxide. If this is not the case then isn't there still a lot more CH3OH present in the solution than H20? Using your analogy- isn't it like reaching for the cell phone in your pocket instead of searching for a pay phone?
Taera, CH3OH (pKa 15.5) and H2O (pKa 15.7) are essentially equal strength acids, so acidity does not differentiate. CH3OH is the solvent so there is probably much more CH3OH than water, so it may be more accurate to use CH3OH as the proton source instead of water. I guess this is worth an error bounty point... An update key will be posted. --DrH
Professor, For the answer for #3 on page 233, you say that the enolate for methyl acetate only has 2 resonance contributors, however the lecture supplement page 31 shows that it has 3 contributors. I realize that the contributor with 3 formal charges is less significant, but isn't it still taken into account (for example when comparing the relative stabilities of ketone and ester enolates)? By that same token, I counted more than 3 resonance contributors for the enolate for methyl acetoacetate (some with 3 formal charges), so would these not be considered in this comparison either becaause they aren't AS significant of contributors? Thanks!
Laura, The answer discusses significant resonance contributors, of which an ester enolate only has two. You could rephrase the answer to include the less important ester resonance contributor. --DrH
I think I found an error in FAll 2004 Final Exam Part A #6 !!!!! It states that sulfuric acid provides a proton that eventually ends up on the carbonyl "carbon". But the correct answer should be that the proton eventually ends up on the carbonyl "oxygen". Is this an error?
Keun, You are correct; the proton goes to the oxygen, not the carbon, of the carbonyl. An updated exam key will be posted. You have been awarded one error bounty extra credit point. --DrH
I think Fall 2004 Final Exam Part A #4 is missing a step of mechanism or the final product should be water instead of a hydroxide ion. The question states "once carbonic acid is formed, it reacts with water to form CO2 and another molecule of water", but the answer for this question has CO2 and Hydroxide ion as products.
Keun, We normally don't both with mechanism steps that do not include carbon, especially if they are just proton transfers. For example it is not necessary to show the curved arrows for H2SO4 + H2O ---> H3O+. --DrH
Dr. Hardinger, for the relative rates of carbonyl substitution reactions, is thioester a faster reaction, or anyhydride?
Taera, Either could be, depending upon what factors you assume to be most important. --DrH
Prof. H, I know we didn't finish all of the pharmaceuticals stuff in lecture, so could you specify which of the CFQ's and PP's we are responsible for? Thank you
Katia, You are not responsible for CFQ 12 or 13, or PP 7. --DrH
Professor H, on problem 12b) on p.241 of the thinkbook, in the answer, you state that "the enone of reaction J is more highly conjugated than the enone of reaction K...but isn't K more conjugated than J because it has the benzene rings on each end? Also, K being more conjugated would fall in better with the fact that it doesn't need heat for dehydration whereas J does...is this a mistake or am I missing something? Thank you!
Katia, Check the Known Typos web page for more on this issue. --DrH
Hi Dr. Hardinger, I've come upon similar situations many times but this is the first time I remember to ask. On page 255 of the thinkbook, you show the mechanism for the hydrolysis of B-lactam. The fifth step in the mechanism shows OH donating e- to form pi bond and N+ leaving group taking an e-. When I first drew this mechanism, I deprotonated the OH with H2O, then drew an arrow from the OH hydrogen to the HO-C bond, then drew an arrow from the C-N bond to N+, all in one mechanism step. I was wondering if this is an acceptable way to illustrate the mechanism. Or...because the OH must donate an e- and N+ must leave first before the OH can be deprotonated? Thanks, Mike
Michael, It really depends upon the exact case, but it sounds like your mechanism (for the case in question) is ok. VOH is not the best place to discuss curved arrows, so for a clarification please bring it to me or an IA in person. --DrH
Prof. H, on the exam, will we be told when a base is creating an enolate versus when it is acting as a nucleophile and attacking the carbonyl...I know that an enolate is formed with LDA but there are many problems in the thinkbook where an -OH forms an enolate. If we will not be told, how do we know which mechanism will take place? Thanks!
katia, That is for you to figure out. In the absence of a product to guide your thinking, remember that proton transfers are generally faster than nucleophile/electrophile reactions, so that a carbonyl plus strong base will form enolate more often than it will suffer nucleophilic attack at the carbonyl. --DrH
Dear Professor, I was wondering why we use CH3O- to deprotonate the OH group in PP #3d on page 224 instead of using the -OH which is more basic. Is it wrong to use the -OH? Thank you!
Anita, The basicity difference between CH3O- and HO- is tiny. CH3OH pka 15.5; H2O pKa 15.7. So it does not matter which is used for this task. --DrH
Dear Professor, I was wondering why we don't use HO3SO- to deprotonate the OCH3H+ group in the tetrahedral intermediate in PP#3a on page 223. Would it be wrong to deprotonate with HO3SO-?
Anita, CH3OH is a stronger base than -OSO3H. This is because CH3OH does not lose resonance when it accepts a proton but -OSO3H does. --DrH
Dear Professor, Is NRH2 a stronger base than the retinal in CFQ#2 on page 214? If so, why doesn't the H3O+ protonate the NRH2 before the NRH2 protonates the carbonyl as the first step of the reaction? Should the RNH2 and H3O+ be numbered so that H3O+ is not added until the reaction with the NRH2 is complete?
Anita, Retinal is an aldehyde and therefore a terrible base. An amine is easily a stronger base than an aldehyde. --DrH
Dear Professor Hardinger, In CFQ#2a on page 214, the solution uses H2O to deprotonate the positively charged amine group on the tetrahedral intermediate in the second step of the mechanism...I thought that NRH2 would deprotonate the amine group because NRH2 is a stronger base than water. (H30+ has a pKa < 0 while NRH3+ has a pKa of ~10...this makes the conjugate base of NRH3+ more basic than the conjugate base of H3O+) A stronger base will deprotonate instead of the weaker base. Is this correct?
Anita, Good question! The amine can be used as well. --DrH
Dear Professor Hardinger, I am having trouble understanding why anhydrides are one of the most reactive of the carbonyls. The way I see it, the anhydride's leaving group, (the OCO2-)is stabilized by resonance and therefore able to accomodate an extra electron pair, but I don't see why the leaving group would even have a tendency to leave when the anhydride has much more resonance stabilization as is. This resonance makes a molecule more resistant to nucleophilic attack...so how is an anhydride so reactive?
Anita, The resonance stabilization of an anhydride is less than that of an ester. In an anhydride, the oxygen next to the carbonyl must share with two carbonyls, whereas the same oxygen in an ester only shares with one. Sharing with two carbonyl means less stabilization for each carbonyl, so an anhydride does not have as much resonance stabilization as an ester. --DrH
Dear Professor, For the lecture supplement on page 27, it says that any base is a catalyst. In order to be a catalyst, doesn't the base have to speed up a reaction because AND also be regenerated in the reaction? I thought bases are not regenerated in carbonyl reactions (i.e. hydrolysis) and therefore are not considered catalysts. Am I wrong or is the lecture supplement wrong? Thank you.
Anita, You statement concerning the requirements for catalysis are correct. If you work out the mechanism for peptide hydrolysis with HO- catalysis (on the same page) you will see how the HO- is regenerated. --DrH
Sorry Dr. H, disregard my last question. I figured out that carboxylic acids undergo substitution but not addition reactions.
Chris, No problem. --DrH
On pg 227, PP#5d for Carbonyl Chem Survey, why can't you protonate the carbonyl and react further by an addition process that would leave a C(OH)3 structure?
Chris, The equilibrium RCO2H + H2O <===> RC(OH)3 favors the carboxylic acid because the carboxylic acid has resonance whereas RC(OH)3 does not. See PP 6(c) on page 220 for a related example. --DrH
Hi Dr. H, on Pg 241 problem 9d) if the base NaOCH3 was replaced with LDA, would the same hydrogen be removed? Or will there be steric hinderance from the ester so LDA would not be able to get in and deprotonate that hydrogen and instead deprotonate the other hydrogen on the ther side of the carbonyl that's directly attached to the ring? Sorry if this is confusing. Thanks
Linda, A good question! LDA has a strong preference to remove a less hindered hydrogen, but the hydrogen between the two carbonyls is significantly more acidic. Because of the large pKa difference, I suspect LDA would remove more of the more acidic hydrogen than the less hindered hydrogen. --DrH
Hi Dr. H, on Pg. 240 problem 8a) you chose that product as the product of the reaction because you still have base left. And in the solution on Pg.245, the structure that occurs before the final product (the one with two carbonyls) would not be as stable as the final product you've shown because it has more conjugation correct? Thus since we still had base in solution, it readily deprotonated the acidic proton. Thanks
Linda, I don't understand your question. You're trying to compare the stability of a neutral molecule with an enolate, and that is not a meaningful comparison. --DrH
Hi professor, I found an error in the thinkbook on page 98 (solution for problem 34 part B), the stereochemistry of the product is incorrect. I think the answer should have H as a wedge and the methyl group as a dash. I did not see this error mentioned on the "KNOWN ERRORS" page.
Elina, There is no error here. Verify the accuracy of the answer by examininh molecule models. (Inversion can be shown by switching any two groups. In this case, the nucleophile and aromatic moiety have been switched.) --DrH
on pg 248 in for question 1.(c) which asks for the definition of antimetabolite the CFQ anwser for it (on pg 249) and the definition in the lecture supplement (pg 47) are a bit different. i was just wondering if both definitions were acceptable/correct. thanks.
Manish, The best definition would include the important points from both. Antimetabolite: A drug that works against the normal metabolic processes of a microorganism, but not against the normal metabolic processes of the host. --DrH
Dr. H, on page 247 of the thinkbook in the solution to PP 13b you show the formation of the 3 membered ring after the ejection of the ethoxide group, but don't show the free OCH3 coming and removing a proton off the compound again the way it did in the mechanism shown in the solution to PP 13a. Shouldn't the methoxide remove a proton and the H3O+ restore the carbonyl?
Zak, Molecule P has so much strain that as soon as it forms it opens back up. The subsequent deprotonation is a moot point because the molecule never survives to be deprotonated. --DrH
Dr. H, On page 245-246, the solution to PP #9 a,b you show the procuct as a an addition of an alkyl group to one side of the carbonyl. However, if the carbonyl is reformed at the end of the reaction and the LDA is still present, why wouldn't alkyl groups be added to the other side of the carbonyl as well? You have also done this same single sided addition in the answer to CFQ #9c on page 236. Thanks.
Zak, We usually only use one mol of LDA for each mol of carbonyl compound. --DrH
Dr. H. Can you give an example of a Primary Alkyl Halide. It's the exception to the E2 before Sn2 Rule. Thanks, Ashkan
Ashkan, Lecture supplement page 5. --DrH
Dr. H, On page 236, solution to CFQ #9 b, you show that the alcohol is protonated before leaving. I thought that an alcohol can leave if it results in significant conjugation (as this does). Why then is the group protonated first, why can't it just leave on its own? Thank you.
Zak, HO- can leave to make an alkene only when that alkene is conjugated with a carbonyl, and only if it is eliminated by strong base such as HO-. --DrH
I have a question about Spring 2005 Final Part A #6 part C. The answer has a structure with N(CH3)2 and I found this very weird because doesn't N(CH3)2 get substituted by the hydride ion from ALH4? I was expecting an OH in place of this N(CH3)2. Is this an error? or is there a reason why OH leaves instead of N(CH3)2?
Keun, The LiAlH4/amide mechanism is a bit weird because of the way aluminum changes the leaving group. You are not responsible for the mechanism, but you should know the product. --DrH
Dr H, Dr.H, On question 9B of owls for carbonyl surveys, why is it that the mechanism chooses to make another carbonyl, istead of using the lone pairs of one of the OH to attack the carbocation that would be formed when the other OH left?(where the carbonyl reforms)
Nancy, I am not sure what you are asking, but it might be a resonance issue. --DrH
Professor Hardinger, I don't understand why the most stable enolate is the more highly substitued enolate. For example, for CFQ 2 on pg 231, I would assume that because you are forming a carbanion, you would not want it to be around the 2 electron donating methyl groups which would increase the magnitude of negative charge, making it more difficult to deprotonate. Thanks, Taera
Taera, Think of an enolate as an alkene with an O- substituent. Then the same rules that apply to alkene stability apply to enolate stability. --DrH
I have a question about the answer for Owls: Enolates, enols and enamnines #2 part b. Why doesn't LDA deprotonate the carbon that is on the right side of the carbonyl carbon that will result in a more substituted enolate? The answer on the website shows that the hydrogen on the carbon that is left of the carbony carbon is deprotonated by LDA which result in a less substituted enolate. I thought that more substituted enolate is more stable and thus forms the major product. Is this an error?
Keun, Because of its size, LDA removes less sterically hindered protons more readily. This overrides enolate stability unless there is a huge difference in the enolates (not just one alkyl group) or unless the reaction is set up to allow enolate equilibriation (we have not discussed this special circumstance and therefore will assume it is not possible.) Come visit me or an IA for more detail on this issue. --DrH
Hey Dr. H. In the solution to 2c of Enol OWLS, hydroxide makes the final deprotonation to create the final product. Why is this not -OCH3 since it's being produced in the rxn and is a stronger base, I believe, having an e- density donationg methly group to increase its basicity over hyroxide? It seems like the stronger base would more readily do the deprotonating. Thanks.
Jason, The deprotonation could be carried out by either HO- or CH3O-. Your prediction about the relative basicity of HO- and CH3O- is reasonable. --DrH
Hey Prof. In the thinkbook on page 228 solution to #6b, the last line says that the ketone or its acetal has no resonance to lose. However, in the actual problem on page 219, there is a ketone and a ketal, not an acetal. Is this an error? Thanks.
Faysal, This is not an error. A ketal is a kind of acetal that is derived from a ketone. --DrH
For PP #3f on p. 218, on the how come the lone pair on the N in the final product is not protonated by H3O+. In other words, why doesn't the reaction continue with the protonation of the N?
Kmyu@ucla.edu, I do not understand your question. The products given on page 218 and the mechanism on page 225 show that the nitrogen is protonated. --DrH
On pg. 219 for PP #5f, why doesn't intermediate C7H12O2 (which is formed after CO2 reacts with the grignard reagent)react with another moleucule of the grignard reagent?
Katherine, Carboxylate ions (RCO2-) have very significant resonance stabilization (takes a very powerful nucleophile to disrupt it). In addition, the formal negative charge repels most nucleophiles. The only nucleophile we have encountered that is powerful enough to add to a carboxylate ion is LiAlH4. --DrH
hi professor, I have a question about midterm 2, the last exam we took...for #6, why isn't the answer the fourth choice? Why is the benzene not part of the product?
Sandhya, The carbocation derived from the alkene will accept a nucleophile. Br- is a much better nucleophilie choice than benzene, because attack on benzene requires loss of aromaticity. --DrH
Hey Prof, Thinkbook page 222 #1e, I know it's wrong to have the OCH3- attack a protonated carbonyl, but is it ok to have it act as a nucleophile and attack the carbonyl and form a tetrahedral intermediate after the deprotonation? Thanks.
Faysal, Sounds ok to me. --DrH
Hey prof, thinkbook pg. 216. At the top you say that with strong acid present we should protonate the carbonyl first. Yet in the problem 4 right underneath it the nucleophile attacks first. Could either process work?
Faysal, When using LiAlH4, the strong acid is added only after the LiAlH4 has run out of things to do. The strong acid is not present at the same time as LiAlH4. (Strong acid + LiAlH4 ---> H2 + lots of heat + sparks ---> kaboom.) --DrH
Hey Dr. H. In the solution to PP 13b on page 247, the mechanism written for product M shows the oxyanion being protonated by HOCH3, and not H3O+. You can argue that HOCH3 is in solution so it's possible, but then my question is why can't the oxyanion in the solution of part a) do the same. It's protonated by H3O+ instead, which makes sense because it's the next reagent. If the oxyanion in the mech of part b can be protonated by HOCH3, why not the oxyanion in the mech of part a)? Thanks
Jason, Good question! Methanol is a weak acid (pKa 15.5), so it can only protonate strong bases. RO- or an enolate without extra conjugation is sufficiently basic. The dienolate of part (a) is a weaker base, and requires a stronger acid for protonation. CH3OH is not a strong enough acid for part (a) but H3O+ is. --DrH
I was wondering why in pg 237 9(d), the (-)oxygen in the last mechanism step could not attack a hydrogen to form a alcohol instead of going back to an ester- wouldn't the alcohol leave the molecule with conjegation? in general why do claisen condensation rxns go back to esters?
Manish, The enol tautomer of the product can also be formed. Read more about this in part (c) of the same question. --DrH
Professor Hardinger, If CH3OH was placed in solution with any carbonyl group and a strong base, would the base attack the carbonyl derivative in every situation before instead of the methanol? Or does it depend on the carbonyl derivative? Thanks, Taera
Taera, It is not possible to give a precise answer to your question because to contains too many generalities. For example, the base may or may not deprotonate the CH3OH or carbonyl compound to any significant extent. Not all carbonyl compounds are suceptible to nucleophilic attack by CH3OH or even CH3O-. Etc. --DrH
Professor Hardinger, For PP 5 (f) in Survey of Rxns, on pg 219, why can't the rxn continue and get attacked by another CO2? Thanks! Taera
Taera, I do not understand your question. Are you asking for RCO2- to attack CO2 and make RCO2CO2-? --DrH
Professor Hardinger, I was wondering is there is a typo on Pg 30 of the thinkbook- It says the beta carbon is nucleophilic, but I thought the alpha carbon is shown to be nucleophilic in the resonance hybrid. thanks! Taera
Taera, The carbon bearing the X group is the alpha carbon in this case, so it is the beta carbon that is the site of nucleophilicity. --DrH
sorry professor, disregard my question it is on the known typos page..
Lena, No problem! --DrH
Professor, on pg. 236, for answer #9b of the CFQs for Carbonyl Chemistry, Enolates, Enols, and Enamines, you show the leaving arrows directed for the OCH3 group to leave, yet in the next step it turns out it was the methyl group that leaves. how is this possible??
Lena, Check the Known Typos list for a comment on this. --DrH
Hi Professor, Monday you said Erin will have OH next Thursday, which is also what’s posted on the announcements and today you said Wednesday, which is correct???
Melody, As stated in lecture and on the web site, Erin's extra office hours for the final are on Thursday. (Perhaps you are looking at the announcement for Erin's extra office hours for the second midterm?) --DrH
Hi Dr. H On page 247 solution 13(a) step 5, I was wondering how do I know when to let the double bond attack a hydrogen, and when to protonate the oxygen?
Ginette, In this case both can happen because the enol and keto forms are of roughly equal stability. If the enol wasn't very stable (like in the case of acetone) then only C-protonation is considered. This issue is discussed in solution 9(c) on page 237. --DrH
Hi Dr. H, I was wondering for CFQ Enolate 9(b) on page 232, the solution shows that the OH must be protonated first in order to leave. But since the leaving of OH would form a conjugated system, can the OH leave without protonation?
Ginette, In this case it appears the reaction was stopped before dehydration could occur. The formula C10H20O2 corresponds to the hydroxyketone, which was then dehydrated with acid. This does not mean the hydroxyketone cannot be dehydrated with base, but rather, in this case, it was stopped before that could happen. --DrH
Hello Professor H., You can disregard my last question about #2 page 233 because I realized those are equilibrium arrows, not resonance arrows.
Ryan, No problem. --DrH
Hello Professor H., for solution #2 on page 233, is the first resonance contributor shown (the less stable enolate) actually in equilibrium with the more stable enolate? I noticed that it is not included in the resonance hybrid and can't be formed by pushing electrons in the more stable enolate but there seems to be an equilibruim arrow between the two.
Ryan, They are in equilibrium. The less stable enolate is in equilibrium with the ketone (via proton transfers). The ketone is in equilibrium with the more stable enolate (via proton transfers). Therefore the two enolates are in equilibrium (via proton transfers). --DrH
Hi Professor. On Thinkbook page 241 #12, it states that reaction J must be heated to lose a molecule of water, however it never loses a molecule of water -- the hydroxide is the group that leaves. Is this an error? Thanks.
Faysal, The OH leaving group plus the alpha-hydrogen removed by the base constitute a molecule of water. --DrH
Another question regarding if we have to memorize the structures of molecules. page 251 of the thinkbook describes how there are three molecules required for the synthesis of folic acid, p-aminobenzoic acid being one of them. The purpose of this question is to state that sulfa drugs can mimic paba and inhibit dihydropteroate synthase, one of the enzymes necessary for folic acid synthesis. So is it safe to just memorize the structure of Paba and sulfa drugs?.. I am just overwhelmed by the number of molecules introduced in this section and very worried if we are responsible for the structures of all of these molecules.
Keun, Read the answer to your previous question...scroll down. --DrH
I see that the organic chemistry Pharmaceuticals section introduces us to many many many molecules.. do we have to know the structures of all the drugs in the lecture suplement? To be more specific, do we have to memorize the molecular structure for Ethidium bromide, andriamycin, Quinine, Calicheamicin (DNA cutter), Amphotericin B, Thalidomide, sulfanilide.......there are so many that I cannot name then all. Or is it fast to assume that we are only responsible for the structures of drugs and other molecules in CFQ and PP?
Keun, As I mentioned twice in lecture it is not necessary to memorize all of the drug structures. --DrH
Hey Professor, on thinkbook pg. 238 solution to CFQ 13, the H2SO4 under the reaction arrow is missing the label (cat). Isn't it an error not to include the catalyst label? Because otherwise it would be interpreted as a solvent...the H2S04 would attack the carbonyl, and the entire mechanism changes. Thanks.
Faysal, It is not necessary to include the "cat" label. --DrH
On page 41 of the Thinkbook it says that Enzymes catalyze reactions; rate enhancement up to 10^28. But on page 250 of the Thinkbook it says that Enzymes can cause reaction rate enhancement of as much as 10^16. Which one is correct? and is this an error?
Keun, Various sources report different numbers. 10^28 is the largest reliable number that I have seen. Therefore the number on Thinkbook page 250 needs to be updated. The error has been added to the Known Typos list, and you have been awarded one error bounty extra credit point. --DrH
Dear Professor Hardinger, On page 695 in Bruice, for the illustrations that show the tetrahedral intermediate in acid-catalyzed ester hydrolysis vs. the tetrahedral intermediate in uncatalyzed ester hydrolysis, shouldn't the protonated tetrahedral intermediate's leaving group be indicated by only one arrow? The book has two of the two-barbed arrows, indicating that 4 electrons would be departing from the carbon-CH3OH bond, when there are only 2 electrons leaving. Is there an error in the book?
Anita, Yes there is an extra curved arrow. The error has been added to the Known Typos list. You have been awarded one error bounty extra credit point. --DrH
Dear Professor, for PP#15 of Carbonyl Fundamentals, can we protonate the oxygen on the carbonyl group first to increase the delta + on the carbonyl carbon before having the nucleophilic attack there? or is the order of the mechanism as given in the solution the only correct way? Thank you!
Anita, Yes. --DrH
Professor Hardinger, For PP#9 of Carbonyl Fundamentals I think that reaction I would be faster than reaction H because the reactant in reaction 8 has 3 resonance contributors. Therefore, for the carboxylate to leave in the reaction, wouldn't that be energetically expensive since neither of the products formed have such strong resonance stabilization as the reactant? I feel that the resonance in reaction H makes the leaving group weaker, and not stronger because with the departure of the carboxylate the resonance would be disrupted. Thus, looking at the next most important factor, I think that reaction I would be the fastest since sulfur is larger than oxygen and nitrogen and thus more able to accomodate the electrons. Do you think I have valid reasoning?
Anita, As implied by the answer, there are several factors governing these reactions. Which reaction you choose as faster will depend upon what assumptions you make about the relative importance of the various factors. Your reasoning seems valid. --DrH
Professor Hardinger, I was just wondering, if we had to choose between OCH3 and OCH2CH3 as leaving groups, which one would be a better leaving group and why? Thank you!
Anita, They are equal. --DrH
Professor Hardinger, for PP#3 for Carbonyl Fundamentals, in the part where we form an enolate, would it also be correct to speed up the reaction by having a CH3 group instead of a CH(CH3)2 group because there will be less steric hindrance as well as more hydrogens to reaction with the water?
Anita, I do not understand your question. Carbonyl Fundamentals PP 3 (Thinkbook page 201) does not require you to make the reaction faster. --DrH
Professor Hardinger, In regards to my question about electrophile attack on an oxygen by water for PP #3 on pg 201, I think that this attack would not occur since it was stated both in the Thinkbook and in lecture that only strong acids can protonate oxygen. Therefore, I think that this process would result in No Reaction since H2O is the only reactant we are asked to use. Since water is not a strong acid, isn't the solution to PP#3 contradicting what the Thinkbook and what was said in lecture? Moreover, would it be correct in an exam if we used water (granted it was the only "acid" given) for the electrophilic attack of oxygen? Thank you for your time! -Taera
Taera, It is possible that water can protonate a carbonyl, but the equilibrium favors the deprotonated side by a huge amount (probably 10 to the -22 or so). Therefore the given answer is possible but not very probable. It's not really an error, but more of a poor example. If you wrote "NR" on an exam you would be correct. (I wrote myself a note to rewrite this problem in the next version of the Thinkbook.) --DrH
Whoah, ignore my last questions, the solution for part b answered my question. Didn't read it thoroughly enough. Sorry bout that.
Alan, No problem. --DrH
professor, for owls carbonyl chem. survey of rxns and mechanisms on pg. 70 of the think book, for the mechanism of question 3a, why in the answer didn't you use ch3oh as the means by which the cleaved R0- was protonated and then have the resulting -och3 deprotonate the carboxylic acid instead of deprogtonating w. oh- and then protonating with water. isn't water too weak of an acid?
Lena, CH3OH (pKa 15.5) and water (pKa 15.7) are virtually equal strength acids, so it doesn't really matter which one is used to protonate the RO-. --DrH
professor, on the bottom of owls carbonyl survey solutions pg. 2 you draw the product with a hydroxyl group when it should be a methoxy substituent (OCH3 instead of OH)
Lena, You are correct about the error. Updated solutions for this OWLS problem set were posted a few days ago. --DrH
Dr.H, On question 9B of owls for carbonyl surveys, what is ROH refering to that is used as the proton bus? Also, why is it that the mechanism chooses to make another carbonyl, istead of using the lone pairs of one of the OH to attack the carbocation that would be formed when the other OH left?(where the carbonyl reforms)
Nancy, In this case ROH is HOCH2CH2OH, the only alcohol present at the start of the reaction mechanism. --DrH
Hello, for Enolates CFQ #7 pg232, I am getting confused on the wording in the answer. It says that if the Keq is small, then some other electrophile must be added. But then it also says if Keq is large, some other electrophile must be added. So another electrophile must be added either way? Thanks!
Igor, I do not see the inconsistency here. Please ask me or an IA in person about this. --DrH
Hi: For Enolates/Enols/Enamines PP #13(a) (page 241), I know the loss of a Hydrogen happens to the left C=O, but why does it happen on the left adjacent side of the C=0 and not on its other adjacent side? Thanks
Alan, The enolate derived from "the right side of the ketone" leads to the cyclopropanone issue covered in part b of that same question. --DrH
Hi Prof Hardinger: I understand the idea that OR- can only leave when 1)there's an oxyanion 2)when it leads to conjugation. But on problem 11 of Enolate PP (page 246), you say that the rate of reaction I is zero b/c OH can't leave; but could the OH group be protonated to H2O+ and then leave? ALSO: On page 236 (#9b), hydroxide can leave b/c it leads to conjugated system, but you show it getting protonated first before leaving. Does it matter which way we draw it?
An An, Protonation of an OH requires strong acid, which is not present in this reaction. (Any KOH would neutralize it!) Therefore protonation of an OH is not possible in this case. --DrH
In solution 8d, page 245, OH- is shown as the base. I haven't ever seen the reactant Ba(OH)2 before and so didn't know that OH- would be disassociated from Ba. (Or maybe there's a solvent missing?) What should we know about Ba(OH)2 for the final?
Zach, Start by looking at a periodic table. Note the barium (Ba) is an alkaline earth metal, so in compounds it will be Ba2+. Note the parallel with calcium and magnesium, elements that are in the same family (column) of the periodic table. --DrH
For PP #8d on page 240, how do we know that H20 is present in the reaction? In the solution to the problem, H20 is shown protonating the O- of the tetrahedral intermediate, but how do we know that H20 is present?
Katherine, Water is generated in the first mechanism step, when HO- deprotonates acetone to give acetone enolate. --DrH
Dr. H, On page 227 #5-G.You explained that HCN would be used as the proton bus,,So why then did you use H20 insted of HCN in the mechanism as the proton source?
Nancy, Check the Know Typos web page for a comment on this mechanism. --DrH
Dr. H, I think there is a mistake on page 238 solutions for #13cfq. On the third step of the mechanism you deprotonate the hydrogen on the nitrogen...however, you use (CH3)2NH2--a molecule that should have a positive charge on it and should not be used to deprotonate anyhing with no available lonepairs on nitrogen. I think you meant to use (CH3)2NH..which does have a lone pair to be used for deprotonation. Thanks
Nancy, Check the Known Typos web page for a comment on this mechanism. --DrH
Hey Dr. H. In solution b) to Survey of Carbonyl Chemistry OWLS Q3, the mechanism leads the molecule to possessing two negative formal charges, then protonates both to obtain the product. My mechanim was very similar, except I protaned the first oxyanion before proceding to protonate the carbonly a second time with the AlH4. I figure the molecule would prefer to take care of its anxieties before going on to be protonated a second time. Does it matter, or can the steps just be written in a slightly different order? Thanks
Jason, This issue has been addressed in a previous VOH question. Scroll down for the answer. --DrH
Dr. Hardinger, I don't understand why on the known typos page you say that H should be the fastest reaction for PP 39 on pg 202. Based on resonance contribution, shouldn't the thioester be the most reactive? Thanks! Taera
Taera, The anhydride (reaction H) is judged to be faster because its leaving group (CH3CO2-) has resonance stabilization whereas the leaving group for the thioester reaction (CH3S-) does not. --DrH
Dr. Hardinger, In your Lecture supplements on pg 21, you say that electrohile atack at oxygen requires a strong acid. But for PP #3 on pg 201 you have water as the electrophile that attacks the oxygen. Is this possible or do we use it just because that is what it says to use in the question? Thanks! Taera
Taera, The curevd arrows are correct, but water is not a strong enough acid to protonate a carbonyl to any greate extent. This is not really an error, but more of a bad example that I meant to fix when I updated the Thinkbook and forgot to do so. --DrH
Dr. Hardinger CFQ #4 on page 198 for enolate formation you have only one hydrogen in the reaction and assumably 2 methyl substituents, but the when hydroxide ion attacks the reactant the products contain 2 hydrogens instead of 2 methyl substituents. Is this an error? Thanks! Taera
Taera, This is not an error, It is drawn this way to clarify the nature of the structure, including the number of hydrogens, lone pairs, etc. --DrH
Dr. H, Regarding Survery of Reactions and Mechanisms CFQs, why is it that the OH is protonated to form +OH2 in #2a (pg. 214) and then leaves the molecule, but the OH in #5c is not protonated (pg. 216)? Furthermore, is there a general rule to know when an OH will be protonated and leave and when it won't be?
Hello 603203011, As mentioned many times in lecture, RO- can be ejected from an oxyanionic tetrahedral intermediate with prior protonation of the leaving group. If there is no negative charge then the RO must be protonated first. --DrH
Hello Professor H., for #5(c) on page 226, would it be wrong to only show the first nucleophilic attack in the mechanism (resulting in the aldehyde)and would you tell us that AlH4 is in excess on a test? Thanks.
Ryan, Yes it would be wrong, for the reason discussed in lecture. An aldehyde is more electrophilic than an ester, so AlH4- will react with the aldehyde before it reacts with more of the starting ester. It is not possible to stop the reaction at the aldehyde stage, even with a limited amount of LiAlH4. --DrH
You can disregard my prior question. I didn't realize the solution addresses my question. Sorry about that!
Zach, No problem! --DrH
For CFQ 9c, page 232 (Enols, Enolates, Enamines), could the third mechanism step happen another way: Instead of the electrons from the H removed going to the ring C-C=O bond, could they go to the substituent C-C=O bond? That would result in an oxyanion on the non-ring oxygen, which could then deprotonate the HOCH3 formed already (or even the acid added in the next step). The product would be an alcohol subsituent conjugated with the ring C=O with a double bond to the substituted ring C.
Zach, The enolate structure used is the most significant resonance contributor. Think of it as an alkene with an O- attachment: more substituents = more stability. As to the final product, it is ok to draw the enol tautomer of the beta-diketone. For most beta-diketones, the enol and keto tautomers are of comparable stability. --DrH
Hi Dr. H. On pg 214, #3 c) in the structure given, is it missing a Ph group at the end of the carboxylic acid group? Since the question does specifically asked about the 2nd step of the mechanism, I would have thought it was necessary to draw resonance structures relating to the given mechanism being examined rather than a general portrayal.
Thuy, You are correct. The error has been added to the Known Typos list, and you have been awarded one error bounty extra credit point. --DrH
Hi Professor, On page 236 for CFQ # 9c, the solution mechanism shows the OCH3 leaving, however on the next step, the OCH3 is still there. Shouldn't it have been CH3 instead of OCH3?
Emily, This issue was addressed in a previous VOH question. Please scroll down for the answer. --DrH
Hi Professor, on thinkbook page 224 solution to problem #3d it says that the reaction doesn't stop at the carboxylic acid and one of the reasons you list is that hydroxide is a strong base. However, in the mechanism you show the methoxide (not the hydroxide) deprotonating the OH, so shouldn't the reason explain the function of the methoxide's basicity, or instead, show hydroxide doing the deprotonation? Thanks.
Faysal, The discussion should focus on methoxide (not hydroxide). Methoxide is a strong base, just like hydroxide. The error will be added to the Known Typos list. You have been awarded one error bounty extra credit point. --DrH
Hi Dr. H, for the solutions to OWLS Survey of Reactions and Mechanism problem 3b), the third step indicates nucleophilic attack occuring, but my question is, shouldn't the oxygen bearing the negative charge be protonated first? Because in your fourth structure, you have a molecule with 2 formal negative charges which shouldn't be favorable. Can you clarify this? Thanks
Linda, Note the (1) LiAlH4 (2) acid notation, so the acid isn't added until LiAlH4 is completely done reacting. Also, it's ok to have two negative charges if they are far apart. Having to atoms with negative charges bonded to the same atom is a problem. --DrH
Dr. H, Although this is not chemically significant, I wanted to bring your attention to an error on page 243. It states that "enolate O is less stable than enolate L." I think you are referring to "enolate P"
Kimberly, Thanks for bringing this to my attention. --DrH
Dr. H., About the dimethyl amine used in CFQ #13 (page 238), I think (CH3)2NH2 is used by mistake later in the mechanism, too. To deprotonate the nitrogen on the tetrahedral intermediate, shouldn't (CH3)2NH be used?
Kimberly, There are several similar errors in the same mechanism. --DrH
Dr. H., For CFQ #13 of Enolates, Enols and Enamines (page 238), I think the formula for dimethyl amine has an extra hydrogen. The mechanism shows (CH3)2NH2 deprotonating sulfuric acid to become (CH3)2NH2+. Shouldn't dimethyl amine just be (CH3)2NH, as written in the general reaction shown above the paragraph?
Kimberly, You are correct. The error will be added to the Known Typos list. You have been awarded one error bounty extra credit point. --DrH
Hi Dr. Hardinger, for PP # 10 Carbonyl Chemistry - Enolates, Enols and Enamines (pg 241), the step between product E and product F protonates an O-. However, there is no proton shuttle for this step. Is it ok to assume that H2O is still present from the first step or should we specify the presence of H2O? Thanks.
Kevin, Water is still present, although the reaction does not make that point clear. --DrH
Hi Dr. H, for PP #12b of Enolates, Enols and Enamines, I think there is an error. The solution says that the enone in reaction J is more conjugated than the enone of reaction K and thus has a lower activation energy for the formation of the enone. However, the enone of reaction K has more conjugation due to the phenyl groups and thus the formation of the enone of reaction K does not require heat.
Jill, "J" and "K" are switched in this answer. The error will be added to the Known Typos list. You have been awarded one error bounty extra credit point. --DrH
In PP#3a: survey of reactions and mechanisms, it says at the end(solution) that the last 2 steps can be combined. In the mechanism, the OH group is being deprotonated by HOCH3, but in the combined step, it is being deprotonated by H2O. Shouldn't it be deprontonated by HOCH3 instead since water is not present?
Judy, This issue was addressed in a previous VOH question. Please scroll down for the answer. --DrH
Dr. H., For CFQ #9(c) of Enolates, Enols and Enamines (page 236), the third step of the mechanism shows O minus ejecting OCH3 from the tetrahedral intermediate. However, the fourth step (and subsequent steps) shows OCH3 still attached. Should the OCH3 be the CH3 instead (which is not ejected from the tetrahedral intermediate)? Furthermore, the final product should not be a beta-ketoester, right? Please clarify. Thanks.
Kimberly, This issue was addressed in a previous VOH question. Please scroll down for the answer. --DrH
Hi Dr. Hardinger, for the solution to PP #3 Carbonyl Chemistry - Enolates, Enols and Enamines (pg 243), the last sentence says N has the least acidic conjugate acid therefore it has the highest pKa and P has the least acidic conjugate acid therefore it has the lowest pKa. Should this be P has the most acidic conjugate acid therefore it has the lowest pKa? Also, the answer references enolate L, but L is in problem 13b. Should this be enolate P instead? Thanks.
Kevin, You are correct; molecule P has the most acidic conjugate acid and the lowest pKa. The error will be added to the Known Typos web page. You have been awarded one error bounty extra credit point. --DrH
Hi Dr H, for CFQ #9c of Enolates, Enols and Enamines, I think that there is an error in the solution. In the third mechanism step, the OCH3 leaves, but in the fourth step, the OCH3 is still in the molecule.
Jill, You are correct. The product is a beta-diketone not a beta-ketoester. The error will be added to the Known Typos web page. You have been awarded one error bounty extra credit point. --DrH
Dr. H. OWLS - Carbonyl Chemistry: Survey of Reactions and Mechanisms, your answer for problem number 2, part E shows the final product formed being an alcohol. However when looking at the mechanisms, you've formed an ether on the final step. Which one is the correct product? The OCH3 or the OH?
Vikram, You are correct. Updated solutions will be posted. You have been awarded one error bounty extra credit point. --DrH
I think there is an error on page 221 PP Survey #1 part (a) (the first diagram) When the hydroxide ion attacks the ester, the product should have OH attached to the molecule, not OH2.
Keun, See the Known Typos web page for a correction. --DrH
For Question #4 CFQ Carbonyl Chem-Survey of Reactions and Mechanism page 213, I think NaBH4 and CH3CH2OH should be numbered 1. NaBH4 and 2. CH3CH2OH to indicate that the acid is not added until the hydride ion is added to the ketone.
Keun, CH3CH2OH is the solvent for the reaction. NaBH4 does react with CH3CH2OH, but NaBH4 attacks the ketone much more quickly. Thus NaBH4 and CH3CH2OH can be together without preventing the desired reduction of the ketone. LiAlH4, on the other hand, is a much stronger base than NaBH4, LiAlH4 would deprotonate a protic solvent such as ethanol more quickly than it attacks a carbonyl. Therefore when using LiAlH4, the proton source must be added in a second step, after all of the carbonyl is gone. CFQ 4 on page 213 is a good example of this. --DrH
I think there is an arrow missing from #3 part d on page 215 CFQ Carbony Chem-Survey of Reactions and Mechanisms. In the first step (protonation of methanol) I think there is should be an arrow indicating the electron movement from the bond between o-H (part of H2SO4) to Oxygen.
Keun, Check the Known Typos web page for the correction. --DrH
The solution to problem 9b of the Survey of Reactions and Mechanisms OWLS shows a lone pair from an ethylene glycol oxygen displacing H20 as a leaving group. (Mechanism step 5 I think.) Could the +OH2 group leave on its own, leaving a secondary carbocation with resonance stabilization from the nearby oxygen? If so, could the terminal OH of the ethylene glycol then attack the carbocation to form the (protonated) ketal?
Zach, The mechanism as shown combines the resonance and ionization steps. It is also acceptable to show them as separate steps. --DrH
Carbony Chemistry-Survey of Reactions and Mechanisms page 214 #2 part (a) I think there is a arrow missing for the 6th step. I think there should be an arrow pointing from the lone pairs of nitrogen to the carbocation to form the double bond
Keun, It is not necessary to include the