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| Q&A
for 14C |
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Dr. H, I have a question on #14 on part B of the final exam: doesn't bistramide already experience resonance through the nitrogen's lone pair interaction with the carbon's pi-bonds prior to deprotonization? I realize that the conjugate base of the N-H group is a great deal more stable than the conjugate base of the O-H group, but isn't the N-H acid nearly as stable its conjugate base?
Miguel, I have no idea what you are asking. --DrH
Hi Dr. Hardinger, I know the quarter's officially over, but I just have one question from our final exam. On part A #18, why can't maltose be an answer? Can't maltose, just like glucose, be seen as being part of the entire molecule?
Lisa, Maltose could be circled, but the point distribution for the question prevents assignment of any credit for this answer. --DrH
For the previous question I submitted, I made a mistake! I meant part A #18.
Lisa, So that VOH readers don't have to hunt all over to find the question and correct reference together, please resubmit your question along with the corrected reference. Thanks. --DrH
Hi professor Hardinger, On partB#18 of our final exam, I was wondering why maltose can't be a choice? Can't you say a part of the molecule contains maltose since two glucose molecules are together?
Lisa, For Part B #18, maltose cannot be the answer because its molecular weight is not 132. --DrH
Hi Dr. H, for our final (Fall 2005 Final Part A) #4, I was wondering whether dipole-dipole would also be a correct answer. On the Noncovalent Molecular Forces CFQ #4, it says under dipole-dipole interactions that in addition to electronegativity, "high polarizability increase the magnitude of the bond dipole." I reasoned that dipole-dipole forces are influenced by BOTH electronegativity and polarizability, that in this case it was influenced more by polarizability. Is this OK? Thanks.
Siamak, If dipole-dipole forces dominated in this case, then the molecule with the largest dipole (CH3F) would have the strongest attraction (highest bp). That is not the case, so dipole-dipole is not an acceptable answer. --DrH
I was just wondering when our course grades will be available. Thanks
Lisa, The grades will be submitted to the Registrar probably early next week (Monday 12/19 or Tuesday 12/20). --DrH
Hi Dr. H! I'm a bit confused about some of the di/polysaccharide structures (particularly cellulose), because it seems that wherever I see it's structure, the two B-D-glucopyranose monosaccharide components are either flipped in different ways each time (like a ring flip) or simply oriented slightly differently. Do all these various structures for the molecule work? Thanks!
Donald, The glucose units may be rotated so that it is easier to see or draw the molecule. --DrH
Hi Dr H, Is there any way to tell whether the sugars present in the theme molecule are beta or alpha linked?
Miguel, Review the carbohydrates lecture supplement to see how to distinguish between alpha and beta configuration at the anomeric carbon. --DrH
On Spring 2004 Exam Part A, neither O, Se, nor S appear to be what the hydrogen is attached to (its attached to Carbon in all cases). Why then, is its size the S,O, and Se the determining factor in which base is the strongest?
Joseph, I assume you are refrring to question 9 on that exam? The question is about basicity, not acidity. Breaking a Se-H bond is a moot point. --DrH
Hi Prof. H! I was reviewing Acids and Bases when I came across this statement: "smaller atoms make for stronger bases, and larger atoms make for weaker bases." I thought that smaller atoms (atomic radius is small & EN is high) are weaker bases and larger atoms are stronger bases (since its EN is small, it wants to share e- instead of keeping it for itself). [CFQ: Acids and Bases, p253 #11 "atomic size"] Thanks ~ nhung
Nhung, "Smaller" in this context means "smaller atomic radius" not "smaller electronegativity." m--DrH
Professor, on the Fall 2004 Part B Exam, Question number 10, the answer key says the Keq < 1 but then goes on to explain that the equilibruim lies to the left. If the equilibrium goes to the left, towards the weaker base of aminopyrrrole, then shouldn't the Keq be > 1 ?
Eric, The answer is correct. If an equilibrium lies to the left then Keq < 1. --DrH
Hi Dr. H, On page 304 answer 3(a) isn't the pyrimidine structure III not structure II.
Justin, This issue has been addressed in previous VOH questions. Please scroll down for the answer. --DrH
Hi Dr. H, Thanks for being so prompt with your response to my previous question. I looked up the chart for elecronegativity in Atkins and Jones, which is how I arrived at the conclusion I did. On page 73, it's stated that the electronegativity of C is 2.6, N is 3.0 O is 3.4, and H is 2.2. Does this count as an error? Thanks again.
Margaret, It's not an error. There are different electronegativity scales. It appears that Atkins and Jones is using a slightly different one than Bruice. --DrH
Hi Dr.H! Is this a misprint of the text: [pg298, CFQ&PP #7 (answer)] "... the molecule's ability to hydrogen bond CANNOT be significantly decreased." Why would it be 'cannot'-- shouldn't it be 'can'? Aren't we trying to make it lipophilic (therefore decreasing the hydrogen bonding ability)? Thanks ~ Nhung.
Nhung, One oxygen atom does not provide very many opportunities for hydrogen bodning. Therefore changing one oxygen atom does not significantly decrease or increases the hydrogen bonding options for the molecule. --DrH
Hi Dr. Hardinger, Fall 2004 final exam partB 11(a): It doesn't specify the configuration at the anomeric carbon so shouldn't the OH be connected to the anomeric carbon by a squiggly line?
Lisa, The configuration is unspecified by the question, so there is no restriction on what you draw. You could draw alpha, beta, or a mixture. --DrH
Hello Dr. H, I am not sure if this is reported yet, but under CFQ & PP: Nucleotides and Nucleic Acids: pp #3a) the solution states that "I is a purine, because the heterocyclic portion contains two fused rings." (which is correct). The error is in the following statment of the same answer: where it states 'A pyrimidine, such as II, has only one ring.' Shouldn't the pyrimidine be III?
Harrison, The issue has been addressed in a previous VOH question. Please scroll down for the answer. --DrH
Hello Dr. H, I am not sure if this is reported yet, but under CFQ & PP: Nucleotides and Nucleic Acids: pp #3a) the solution states that "I is a purine, because the heterocyclic portion contains two fused rings." (which is correct). The error is in the following statment of the same answer: where it states 'A pyrimidine, such as II, has only one ring.' Shouldn't the pyrimidine be III?
Harrison, The issue has been addressed in previous VOH questions. Please scroll down for the answer. --DrH
Hi Dr. H, for Spring 2004 Final Part B #6c, would it have been OK to list two resonance contributors OTHER THAN the one in part a? (The impression that I got from the question was to list two other resonance contributors but the solution lists the structure from Part A as one of the two). Thanks.
Siamak, The question asks for the two most significant resonance contributors. The contributor given in part (a) is significant, and cannot be ignorned. --DrH
Hi Dr. H, for Lipids CFQ #3 solution, for phospholipid structure, I don't understand what the OR' attached to the phosphate represents. (last time I asked this question, you said that the solution clearly says that the R (not R') represents a fatty acid, but I am asking about R-prime). Hope this clarifies my question.
Siamak, For most phospholipids, R'=CH2CH2NH3+. --DrH
Hi Dr. H, When comparing BRNH3+ and PH4+ can you explain why BRNH3+ is not the stronger acid? (Spring 04, Part A, #8). Thanks very much.
Margaret, Phosphorous is larger than nitrogen, and size usually has more influence than inductive effects. --DrH
For Fall 2002 Final, #6f, could associative also be an answer since it will use associative mechanism to add more ligands? Thanks.
Siamak, Associative refers to a process, not a single structure. It is not a correct answer. --DrH
In the Fall 2002 Exam 3, question 6g, shouldn't it have ketopentose circled instead of ketohexose because fructose is a fice carbon sugar? Thanks Bindu Patel
Bindu, use the index in Bruice to verify the number of carbons in fructose. --DrH
For the Fall 2004 exam part b question 11B, how do you know in the fischer projection that at the top there will be a CH2OH and not a COH. Thanks.
Chloe, The question asks for a ketose (a ketone) not an aldose (aldehyde). --DrH
Good Evening Dr. H, I was wondering if we were expected to know the transition metal section for the final exam. Thanks.
Student, Every course topic might appear on the final, so the answer is yes. --Instructor
Hi Dr. H, On the Winter 2005 Final Exam Part B, #5 why is the most polar bond C-O? The electronegativity difference between C and O is 0.8, but is also 0.8 for N and H. Thanks.
Margaret, Either your electronegativities or your math is wrong. Correct electronegativites: C = 2.5, O = 3.5, N = 3.0, H = 2.1 --DrH
Hi Dr. H, I asked once before but I didn't understand your response: if asked to draw the skeleton for phospholipids, what does the OR' attached to the phosphate stand for? (The last time I asked you said that it clearly states that R (not R') = fatty acid but I'm asking about R-prime). One of the old exams puts a CH2CH2NH3+ in place of the R -- what type of group is this? Thank you.
Siamak, I do not know which question you are asking about, so I cannot give an answer. --DrH
I understand the relative strengths of the 4 main noncovalent bonds (ionic, dipole-dipole, H-bonding, and van der Waals), but i was wondering how the other noncovalent bonds (pi stacking, ion-dipole, and ion-pi) compare with these in strengh. Thank you.
Carolina, They are all much weaker than ionic, but not always stronger or weaker than the other forces. --DrH
Hi Dr. H ~ On the Fall 2001 Final, problem 2, shouldn't the COH bond be a little bit larger than 109.5 (as opposed to smaller) becuase the C is part of a ring and thus this attachment would be the biggest and push the other three attachments (two lone pairs and the hydrogen) farther away?
Bindu, It depends upon what assumptions are made concerning the magnitude of the group/hydrogen repulsion versus the magnitude of the repulsion between two lone pairs. --DrH
Hi! I was wondering for the fall 2003 final, #7, why is A aromatic? Doesn't Sulfur have two pairs of lone pairs (sp3)? Thanks.
Vivian, For sp2 sulfur only one of the lone pairs is in a position to participate in the aromatic sextet. The other lone pair is perpendicular to the pi electron cloud. --DrH
hi, Are we expected to know how to draw in detail a protein alpha helix or beta sheet? thanks!
Jamie, This issue was addressed in a previous VOH question. Please scroll down for the answer. (Same answer applies for both helix and sheet.) --DrH
Dr. H, on Spring 2004 Final Part B, on #13, was the Carbon that is labeled "anomeric carbon" really the anomeric carbon?
Siamak, Yes it really is the anomeric carbon. --DrH
Dr. Hardinger, when a question asks for the MOST IMPORTANT resonance contributor, should you always draw the resonance hybrid (since it is the most accurate) or would you draw a resonance contributor other than the hybrid? Thanks.
Siamak, The hybrid is not a resonance contributor, but instead is a weighted average of all the resonance contributors. Therefore drawing a reonance hybrid when the question asks for a resonance contributor will earn zero credit. --DrH
Dr. Hardinger, on Spring 2004 Final Part B #9c, when you ask for the two most important resonance contributors, would it have been OK to draw two resonance contributors OTHER than the structure in part A?
Siamak, The exam question has nothing to do with resonance, so I do not understand what you are asking. --DrH
professor, in the answer for question number of the nucleotide cfqs, you state that DNA and RNA are linked by "3', 5' phosphodiester group." isnt the phosphodiester group linked to 3',5' positions of the nucleoside, not the other way around since there is no such thing as 3', or 5' when it comes to phophoric acid. the wording seems to be a bit confusing.
Abdul, If 3',5' applies only to the nucleoside and not to the phosphate, then the context of the numbering makes it clear. --DrH
in the sprin 2004 part a final exam, On question 8, why is +PH4 the correct answer?
Jodi, Because phosphorus is larger than nitrogen. --DrH
hi professor, on the spring 2004 final exam part A, please explain question 9. On question 9, why is the answer not CH3Se- if in the order of importance atomic size is is more important than EN.
Jodi, Selenium large, so this makes CH3Se- a weaker base. --DrH
Dr. H, for #10 on Spring 2005 Final Part B, would it have been OK to put hydrogen bonding and dipole dipole? Do these count as two different noncovalent forces? Thanks.
Siamak, Other forces are more significant in this case. --DrH
Professor Hardinger, shouldn't 9a on Spring 05 Final Part B say: 12 - 2 + (2 x 2)+ e- from water? The solution has a 10 instead of 12 (which would make the e- from water = 6, not 4.
Siamak, This issue was addressed in a previous VOH question. Please scroll down for the answer. --DrH
Professor Hardinger, For Spring 2005 Final Part B #9b, why is Keq < 1? Since in the structure on the right, Zn has noble gas configuration, wouldn't the equilibrium lie to the right making Keq > 1? Thanks.
Siamak, You are correct. An updated exam key will be posted. You have been awarded one error bounty extra credit point. --DrH
Hi Dr. Hardinger, for Spring 2005 Final Part B #1, would there be an ester present (or is that CO2- considered something else)?
Siamak, RCO2- is a carboxylate ion. An ester requires carbon on both sides: C-CO2-C. --DrH
Hi Dr. H, In the Chem 14C thinkbook for the Nucleotides and Nucleic Acids Practice Problem #3a and #3c, the answer key states that "a pyrimidine such as II has only one ring" and that "I pairs with II". Isn't structure II a purine not a pyrimidine because the structure has two rings and four N-atoms? Also, because structures I and II are both purines (structure I is guanine and structure II is adenine) they cannot pair with eachother in a DNA structure. I believe that this might be an error because it was also written that "a pyrimidine, such as II has only one ring" when the previous page shows this structure as having two rings. Thank you for your time. Natalie
Natalie, You are correct. The error has been posted at the Known Typos web page. You have been awarded one error bounty extra credit point. --DrH
Hey proffessor, Im still struggling with the rare instances in which resonance can actually increase basicity by increasing the concentration of electrons. Is there a general rule to follow to be able to tell when this would happen?
Joseph, There is no simple general rule other than to look at the resonance hybrid. --DrH
Hi professor Hardinger, On the winter 2005 exam part B #3(b), why isn't the OCH3 oxygen sp2 hybridized? Wouldn't resonance make one of the oxygen's lone pair to move into a Pz orbital?
Lisa, You are correct. An updated exam key will be posted. You have been awarded one error bounty extra credit point. --DrH
Do we need to know how to draw cytosine, thymine, uracil, guanine, and adenine for the final? Also, do we need to know how to draw a alpha helix for protein?
William, As a life science major no molecule is more important to you than DNA. Therefore you should be intimately familiar with its structure. You should be able to draw evey base, every lone pair and every hydrogen bond. An alpha helix is a bit less important, so the expections of your familiarity with its structure are lower. --DrH
Will we be given a periodic table for the final exam?
Stacy, Yes, you will have a periodic table. --DrH
I have just a basic question; is it only considered hydrogen bonding if the hydrogen is covalently bonded to an oxygen or nitrogen? Thanks
Eric, Hydrogen bonding is not limited to N-H or O-H bonds, although these are the most common. For exxample, H-F can be a hydrogen bond donor. --DrH
Hi Dr. Hardinger, for Fall 2001 Midterm 3 #4, on reason #2, how could you know just from looking at the structure that it was not of biological origin? Thanks.
Siamak, See page 70 of the Thinkbook. --DrH
Dr. H, for Winter 2005 Final Part A #11, on the right side of the reaction, the solution has an arrow drawn from the C=O double bond to the S-H bond. Would it have been OK to first draw an arrow from the C=O double bond to the O, then an arrow from that O to the H? (I think someone asked something similar to this in a previous VOH question but I don't understand the answer). Thanks.
Siamak, It is not correct. Curved arrows are drawn in a way that minimizes the number of arrows used. Your curved arrows give electrons to the oxygen, and then take them away with the second arrow. The electrons want to avoid the middleman (the oxygen)...why pay retail whrn you can pay wholesale? --DrH
Hi Dr. H, for #3b on Fall 2002 Exam 3, it says that E is the least acidic hydrogen. But since E's conjugate base has resonance with the carboxylic acid, wouldn't it be stronger than B? Thanks.
Siamak, This is a case where electronegativity outweighs resonance. --DrH
Hi Dr.H! For PP of Carbohydrates (pg. 283) #10, I used the model kit to construct fructosefuranose (product of sucrose) and found that the structure is different than the solution given. Instead of having 'H' attached to C-4 on the left side, I have 'HO'... I'm not sure which is correct. Thanks for helping ~ Nhung
Nhung, The structure in the Thinkbook is accurate (I have just verified it), so the error probably lies in your model. --DrH
Hello Dr. Hardinger. Do errors found in the Klein book count for an error bounty point? If so, I found one on page 40. The resonance structure of 2.33 is wrong because it lacks a negative charge on the carbon that accepted a lone pair. (Answer is on page 232)
Cynthia, You are correct about the error, but it was added to the Known Typos web page some months ago. --DrH
In the Thinkbook on page 302 number 3c, the answer says I-II. I-III has 1 more (3 as opposed to 2) hydrogen bond though, doesn't it?
Jonathan, This issue has been addressed in previous VOH questions. Please scroll down for the answer. --DrH
Dr. H, I'm confused on the spring 2005, final exam part B number 9. The key says that the zinc atom has 10 valence shell electrons, but I think it has 12. So, I think the final answer should be n=1. Or am I missing something?
Stephanie, Zinc has 12 electrons, but due to a math error, the number of water molecules remains the same. An updated exam key will be posted. Thank you for bringing this to my attention. --DrH
Hi Dr. H! For LS of Carbohydrates (pg. 62), why is it 1,2'-alpha-glycoside instead of 1,1'-alpha-glycoside for the sucrose structure? Also, the answers given for parts p and q are switched (pg. 278, CFQ & PP: Carbohydrates). Thanks ~ Nhung
Nhung, The numbering starts at the end of the carbohydrate chain closest to the carbonyl (when acyclic). For an aldose in pyranose form this means the anomeric carbon is #1. For a ketose carbon #1 is CH2OH (when acyclic) and the carbonyl carbon (which becomes the anomeric carbon) is #2. --DrH
Dr. H, On Spring 2005, Final Exam Part A, for #17 part f), why is that end of the molecule considered an "unnatural" amino acid? I thought it was because if the amine group was on the right, the side chain had to be pointing away from the viewer. However, wouldn't any of the other amino acid groups be considered "unnatural?" Or is only that end amino acid considered "unnatural?"
Stephanie, In a natural amino acid, the side chain points out when the alpha carbon points up 9and the amino is on the left and the CO2H on the right). When the alpha carbon points down the side chain is back. Verify this by examining a model. --DrH
Hi Dr. H ~ Will you be having your normal OH on Monday at 10 AM? Thanks.
Bindu, I will be available 10-11 AM. --DrH
On problem #10 for Spring 2005 Final Exam Part A, why is the sulfur a stronger acid than the oxygen? Shouldn't the sulfur be less acidic because sulfur is less electronegative than oxygen and thus more willing to share electrons? Or is it a question of my misreading the question- "concerning acidity"?
Stephanie, Size has more influence than electronegativity. Sulfur is larger than oxygen. --DrH
For #9 of Spring 2004 Final Part A, doesn't it make more sense that F- be the stronger base because it is smaller (and atomic size outranks both electronegativity and inductive effect)? Thanks.
Siamak, As discussed in lecture and in the Acids & Bases Supplementary Reading, O and F have about the same size, so electronegativity is dominant. Oxygen is less electronegative than fluorine so CH3O- is a stronger base than F-. --DrH
Hi Dr. Hardinger, on page 282, #8, is the monosaccharide on the left a glucose even though the OH on the #2 carbon is pointing towards the right?
Timmy, Yes it is a glucose. Study a model to verify this. --DrH
Hi Dr. Hardinger, I'm confused about the how the phosphodiester should be drawn when drawing DNA or just a nucleotide. The past exams have alot of phosphodiesters and many of them are different (are bonded to the furanose ring with different oxygens, double bond is on different oxygens, sometimes there are two oxygens with formal negative charge and sometimes only one oxygen with formal negative). How do you tell which of these should be used? Are any of them acceptable? Thanks.
Siamak, Resonance. --DrH
Evening DrH. When drawing an acyclic form of monosaccharides in the Ficher projection, I know the OH is on the right, directly above the primary alcohol for it's D form. But in which direction do the other OH's and H's have to point?
Rosie, The position of the other OH groups varies between the various carbohydrates, as shown on page 60 of the Thinkbook. --DrH
Ok, so I know that delocaliztion of electrons in a molecule makes the atom a weaker base becasue it is less willing/able to share electrons. If that is the case, why in the thinkbook on pg 270 in the answer to #25 does it say "Alternately, one can say that protonation of the oxygen is easier because the charge in the product is more delocalized by resonance."? Wouldn't delocalizion make it harder to protonate? Thanks.
Jenny, "Resonance makes a weaker base" is a false generalization. Sometimes resonance makes for a stronger base. For problem 25 resonance of the base is not the issue, because it's the same base in each case. The conjugate acids, hwoever, are different so we need some way to analyze their relative stabilities. Resonance is the key. --DrH
For curved arrows, do we draw them starting from the atom or the negative charge? Thanks.
Jenny, The atom, and not the negative charge, is the source of the electrons. Therefore the curved arrow starts at the atom and not the charge. --DrH
Why is the nucleic acid primary structure a nucleoside sequence and not a nucleotide sequence? Thanks.
Jenny, They are the same thing, except for the phosphate linker. --DrH
I was doing sample problem 2 in the proton NMR lecture supplement and I'm not exactly sure what the final structure should look like (I got two possibilities). I was wondering what the answer was. Thanks. Bindu Patel
Bindu, The answer is CH3CH2CO2CH3. Note the methyl singlet at 3.7 ppm, which is more consistent with CH3O than with CH3 next to a carbonyl. --DrH
Hi Dr H, I was wondering if you could explain example 2 on the MS lec supp (pg 33). Why can you subtract 0.36 from the M+1 relative abundance. It says in the note that its because three nitrogens are unusual, but i don't understand the correlation. Thank you. Bindu Patel
Bindu, Check the natural abundance of nitrogen on page 30 of the Thinkbook. This should answer your question about 0.36. --DrH
Hi Professor, you said that we don't need to memorize structures, but we need to only recognize them right? or should we be able to draw all of them too?
Jan, There are some structures that your should definitely know in detail. These include (but are probably not limited to): glucose, ribose and DNA. In other cases you shoudl be able to draw the skeleton, such as steroids or phospholipids. Look at the Thinkbook and old exams for more examples. --DrH
Dr. Hardinger, did we cover molecular recognition?
Siamak, That has not been a Chem 14C topic for several years. --DrH
Hi Dr. H- On acid and base PP #37 why is carboxylic acid a weaker acid than hydronium ion? doesn't carboxylic acid have resonance so shouldn't that mean that its likely to share electrons so its a poor base which means its a strong acid?
Kathy, This is a case where formal charge carries more influence that resonance. --DrH
The thinkbook practice problems mention cases in which resonance can actually increase basicity instead of decreasing it. Is this sort of thing fair game for the final, or should we follow the general rule that resonance decreases basicity? If we are expected to know this, is there a general rule of thumb for whe resonance increases basicity?
Joseph, Anything in the Thinkbook or lecture is "fair game" for the final, including unusual cases where resonance increases basicity. --DrH
Hi Professor Hardinger, on Fall 2003 Final Part A #14, would another reason be that the side chain of natural amino acids are pointing TOWARDS the viewer (when the amine is drawn to the left, carboxylic acid to the right, and alpha carbon up)? Thanks.
Siamak, Sounds like a reasonable answer. --DrH
On Acid Base PP # 39, isnt resonance supposed to make it a poor base so equilibrium would lie to the left? I dont understand the thinkbook explanation. Can you explain it a little simpler and clearer PLEASE! Thank you.
Lawrence, As mentioned in lecture and the Acids & Bases supplementary reading, resonance ofetn (but not always) reduces basicity. This is a case in which resonance enhances basicity. --DrH
Hi Dr. H ~ On pg. 25 of the thinkbook, it says on the lecture supplement that cyclooctatettraene is actually tub shaped as opposed to planar becuase there is no aromaticity. However, even if there isn't aromaticity, shouldn't there be conjugation which would still force the structure to be planar? Or is it because the strain from being planar is too great for conjugation to outweigh it? Thanks. ~Bindu Patel
Bindu, It is not possible in this molecule to have just two of the four alkenes to be coplanar. The molecular structure is such that all four would lie in the same plane. This leads to a situation of high instability called antiaromaticity, which is described in the texttbook. --DrH
Dr. H- on practice problem number 3, part C of the DNA section, is the answer supposed to be molecules I and III form the most stable base pairs?
Anh, This issue has been addressed in previous VOH questions. Please scroll down for the answer. --DrH
Dr. Hardinger, For Spring 2005 Final Part A #1, is pi stacking present?
Siamak, The molecule is question does have pi stacking. --DrH
Hi Dr. Hardinger, I just wanted to make sure that #12 on Fall 2004 Final Part A is not applicable to us (listing the two monosaccharides in sucrose). Thanks.
Siamak, It is applicable. This fact can be found in the Thinkbook, so it could be asked on an exam. --DrH
For Fall 2004 Final Part A #2, I don't understand why the conjugate base from losing the benzene proton wouldn't have resonance. If it lost the proton, wouldn't the C have a negative charge which would be delocalized around the benzene ring? Thanks.
Siamak, The orbital which contains the C-H bond (which becomes a lone pair after deprotonation) is perpendicular to the p orbitals of the benzene ring, so there is no resonance interaction. Verify this with a model. --DrH
Hi professor, is there an error in the solution for practice problem 3.(c) on pg. 304 of the thinkbook? I thought that the answer should be the base pair I-III because it's between a pyrimidine and a purine (guanine and cytosine?) The answer in the book would have two pyrimidines forming a DNA base pair. Is that possible?
Lan-Anh, This PP was addressed in a previous VOH question. Please scroll down for the answer. --DrH
Professor, based on the reasoning you provided for Practice Problem number 3 on page 304 of the nucleotide section, shouldnt the answer be both I and II and I and III because both those pairings have an equal amount of hydrogen bonds?
Abdul, Given the points of attachment to the ribose/phosphate backbone of DNA, not every hydrogen bond may be able to form. This PP was addressed in previous VOH questions. Please scroll down for the answer. --DrH
Hi Dr. Hardinger. Is it important to know how draw the base pairs H-bonding to each other (for example, where to draw the hydrogen-bonding between guanine and cytosine from p.77 of the Thinkbook)? Thanks.
Siamak, As a life science major there are few molecules les important to you than DNA. Therefore you should be intimately familiar with its structure, including every base, lone pair, formal charge and hydrogen bond. So yes, you should be able to draw these hydrogen bonds. --DrH
Hi. In the OWLs for Carbohydrate #4, the beta-D-galactose form a trisacchide through 1,3' linkage. So, there can be other possible linkages besides 1,4' or 1,6' as in amylopectin, because i thought it was usually 1,4' linkage? If so, how do we know when the linkage is not 1,4'? Thanks.
Monika, The problem does not limit which carbons can be used for the galactose-galactose glycosidic linkage. You could make it 1,2 or 1,3 or 1,4 or 2,4 or whatever you like. --DrH
Professor Hardinger, can you explain why, on page 266 of the thinkbook, #16c, protonation of the amide would provede the molecule with more stability? Doesn't it have resonance even if the oxygeon on the ketone is not protonated? Thanks.
Timmy, The issue is not one of the stability of the amide, but rather the relative stability of the protonated amide versus the protonated ketone. The more stable product is formed more readily. The protonated ketone has several resonance contributors, but only one of these, (CH3)2C=OH+, is significant. The protonated amide has two significant resonance contributors. Using our general rule that molecules with a greater number of significant resonance contributors are more stable, the protonated amide is more stable than the protonated ketone. Therefore the amide accepts the proton more readily than the ketone. (For more on determining the relative importance of resonance contributors, review the resonance tutorials on my web site.) --DrH
Hi professor hardinger- on page 304 of the Nucleotides and Nucleic Acids practice problems solutions for number 3c, it says that I pairs with II but these are both purines, I thought that purines paired with pyrimidines? Is this a misprint? Any clarification would be appreciated.
Amy, This issue was addressed in a previous VOH question. Scroll donw a bit for the answer. --DrH
Dr. H, I don't understand how the acyclic D glucose and the chair conformation next to it (where is say "or") are the same thing (on Page 61 of the thinkbook), the chair has a carbonyl but the fisher projection does not, it has a CH2OH. Also, I thought that glucose is in the family of aldoses so the fisher projection on page 61 is confusing me. Thank you. Sarah
Sarah, As mentioned in lecture, there is an error on this page. The top carbon of the acyclic structure on the left should be an alkdehyde (CHO) not a primary alcohol (CH2OH). --DrH
Dr H, on the nucleotide practice problems number 3C p304, you said that the most stable DNA base pairs are formed between I and II. I dont understand why this is the case when I and III form more Hydrogen bonding, thus making it more stable. Thank you, Mahyar
Mahyar, This issue was addressed in the last VOH question. Scroll down for the answer. --DrH
hi professor-
On nucletides and nucleic acids PRACTICE PROBLEMS #3C (page 302 of the thinkbook), doesnt the base pair (I-III)
Min, You are correct. Note the purine-pyrimidine pairing, just like in DNA. The error will be posted at the Known Typos web page. You have been awarded one error bounty extra credit point. --DrH
Hi professor Hardinger, On pg 102 #3 (b) in the thinkbook,shouldn't the oxygen atoms that are part of the peptide bond be double bonded to the carbon atoms? The solutions posted on the web has oxygens in Glycine and Leucine with single bonds.
Lisa, They are double bonds. The problem may lie in the resolution of your computer screen, or the use of an old version of the pdf reader on your computer. Try printing out the page, and/or updating your pdf reader. --DrH
Hi Dr. H, for CFQ #3 on the Lipids section, it doesn't say what the R' on the phospholipid represents? Is it a long-chain alcohol?
Siamak, It clearly states that R = fatty acid. --DrH
Hi Professor Hardinger, for the Lipids CFQ's, question 1 says fatty acids have 12-20 Carbons but question 3 says they have 11-19 Carbons. Which do you prefer us to know?
Siamak, The answers are identical. In question 3 you are probably forgetting the carbon of the carboxylic acid. --DrH
Hi Dr H, I was just wondering what the difference between D-Glucopyranose and D-Glucose is. Does it make a difference which direction you draw the OH on the Primary Alcohol? Thanks, REbecca
Rebecca, Glucopyranose refer to glucose specifically in its pyranose form. Glucose is any kind of glucose, acyclic or cyclic. --DrH
Hi Dr. H, Can you please explain how to do the acid/base problems in which we are to predict if the equilibrium lies to the left or the right....for example pp#34 in the acids/ base questions. I do not understand the explanation where it says to pick the side with the weak acid/base combination....which one is meant to be weak the acid or the base? Thank you. Sarah
Sarah, Recall from general chemistry that an acid/base equilibrium favors the side with the weakest acid and the weakest base. (If the equilibrium does not have the weakest acid and the weakest base on the same side then it is not written correctly.) Refer to the acids & bases supplementary reading on my web site for more information. --DrH
Hello, what is the difference between triacylglycerol and triglyceride and what does it mean to triester something? thanks
Student, The answer you seek can be found using the index to the textbook. --Instructor
Hi Dr. H ~ What are some biological functions of fatty acids? Thanks. ~Bindu Patel
Bindu, Mostly as components of other lipids. --DrH
Hi Professor, When we talked about quatenary structure in lecture, we said that it has to do with the noncovalent forces mentioned earlier and their influence on the amino acids and peptides. However, PP 9 Amino acids says that quaternary structure deals with subunits. Are these subunits influenced by the noncovalent forces? Where do noncovalent forces come in? Thank you.
Panthea, Noncovalent forces hold the subunits together to form the quaternary structure of the protein. --DrH
Hi Professor, For PP 8a of Amino Acids, how can you tell that the two peptide chains are parallel or antiparallel? Wouldn't you say they are parallel because the Hydrogen bonds match up exactly between the O and H? Plus if they are antiparallel, how can they form a Beta sheet to stay rigid? Thanks.
Panthea, If both chains have the N-termini pointing in the same direction, they are said to be parallel. If they point in opposite directions they are called antiparallel. --DrH
Hi Professor, why is an Alpha amino acid is not the same as alpha and beta with regards to carbohydrates?
Panthea, Because they refer to completely different things. --DrH
Hi Professor, will we get all 3 hours to complete the final exam? Or will it just be equivalent to two 50 minute midterms?
Lan-Anh, As mentioned twice in lecture, you have the entire three hour time slot to work on the exam. --DrH
Please disregard my previous question; I understand it now.
Margaret, No problem. --DrH
Hello Professor, I am confused on how an amino acid can be hydrophillic nonacidic. Isn't that just the same thing as a hydrophillic basic side chain? And the reason it is hydrophillic is because of the polarity brought on by the base and its want for stability, right? Thank you.
Panthea, "Nonacidic" refers to a side chain that is neither acidic or basic. I agree that the term is misleading, but as far as I can tell, it is the standard descriptor. --DrH
Hi Dr. H, In the Carbohydrates PP, should number 5 have a double bond O instead of the CHO drawn at the top? Thanks.
Margaret, "CHO" is an abbreviation for aldehyde, just like Ph = benzene ring or CO2H = carboxylic acid. --DrH
Dr. H., On Final for Winter 2005, Part A, #11, for the molecule on the right, I drew the arrow from O- to the bond next to it, and then an arrow from the double bond to the O attached to the double bond, and then another arrow from the double bonded O to the H that is attached to the S. Is that acceptable? Your answer did not have the arrow from the double bond to the O that's attached to it.
Cam, A curved arrow never ends at a bond, because atoms do not form bonds with bonds. A curved arrow ends at an atom, or the space between two atoms. --DrH
Hi Dr. H, For the final exam, will we be given both parts of the exam at the same time, or can we only work on one part at a time (meaning we turn in part A before receiving part B). Thanks. -Bindu Patel
Bindu, As I mentioned twice in lecture you have the entire three hour time to work on both portions. You will receive parts A and B at the same time, stapled together. --DrH
Hi professor Hardinger, We know that NaCl has a high boiling point because of its ionic bond. But does ionic bond refer to the bond between Na and Cl in the same molecule or the bond between Na and Cl of another NaCl molecule? All the other molecules we discussed involves breaking the intermolecular bonds when boiling point is reached, but which bonds are broken in NaCl when its boiling point is reached?
Lisa, NaCl does not exist as separate NaCl molecules. Instead, it is a collection of Na+ and Cl- all sitting in the crystal lattice. In sme respects, the whole crystal can be considered as one molecule. --DrH
Hi Dr. H, In the conjugate acids and bases tutorial, for practice problem c, should the first species (H20-H) in the reactants have a positive charge rather than a negative charge. Thanks. Bindu
Bindu, It appears to me to have the correct positive formal charge. The picture may be distorted on some web browsers, giving the appearance of the wrong formal charge. --DrH
Hi Dr. H, Will we be required to draw Haworth projections on the final exam? Thanks.
Margaret, Probably not. Using models will certainly make this easier, however. --DrH
For #2 on the Acids/Bases OWLS, the solution for "Second Step" doesn't indicate that the electrons that formed the C-H bond become another C-C bond, making it a C=C double bond. Is this unimportant?
Siamak, An expanded solution has been posted. --DrH
Professor, I'm having trouble understanding the solution for question 16(c) of the Acids and Bases Practice Problems on page 266 of the Thinkbook. The solution discusses protonation of the oxygen atom but I'm unable to see this from the illustrations on page 256. Also, why are there alcohols present in the solution structures?
Albert, I do not understand that source of your confusion. Please ask me or an IA in person so we can sort it out. --DrH
Hello Professor, For the disaccharides with beta-linkages, I'm more comfortable drawing the monomers with the ring oxygen at the head of the chair (sorry, i don't know how else to describe it). As long I maintian the 1-4 linkage, is this acceptable?...or should i draw it like it's drawn in the lecture suppliment?
Mitra, As long as you draw the same molecule, it doesn't matter how you draw it. Subject, of course, to any restrictions imposed by the question that causes you to draw it. --DrH
Hi Professor, For the PP of Fundamentals, on page 113, problem 7d, there is a circle around an amide. When I did this problem, I wasn't sure if that was an amide or an aldehyde, because if you circle the same carbonyl, but instead of the N include the H on the other side of the carbonyl, isn't this similar to an aldehyde? Thanks.
Mansoor, It is similar to an aldehyde, but it is not an aldehyde. An aldehyde is characterized by a carbonyl group flanked with a hydrogen atom on one side and a carbon on the other. --DrH
Hi Professor, For problems where they ask us to draw the Fischer projection from a cyclic form, is there a way to know which direction the OH and H groups, respectively, should point in the Fischer projection, or do we only know that, for the D conformation, the OH directly above the CH2OH should point to the right? Thanks.
Mansoor, Build a model of the acyclic Fischer projection, then curl it into the cyclic form. Voila! --DrH
hey professor, i'm having trouble understanding why with all things the same, a molecule with Br instead of Cl will be a stronger base (weaker acid). Isn't Br on the same column so atomic size should govern rahter than electronegativity? so problem 40 on acidbase PP's part b) BrCH2CO2H shouldnt this be a weaker base than ClCH2CO2H?
Pei, Atomic size matter only for the atom that is sharing an electron pair to make the new bond with hydrogen. In this case, that atom is oxygen. Br and Cl are not at "the business end" of the base, only their inductive effects will matter. --DrH
hey professor, for problem 21 on acid bases PP's, part b) we are comparing HOCl and HOBr, but in the solutions it compares HOCl to HOF. In the case of HOBr, Br is greater in size so is weaker base and stronger acid, but compared to HOF, HOCl is the stronger acid right? So, does that count as an error? =)
Pei, Between H2O, HOCl and HOBr, HOCl is more acidic, The HOF/HoCl discussion covers the second part of the question that asks you to suggest a molecule that is more acidic than H2O, HOCl or HOBr. --DrH
Hi Professor, For CFQ for acids and bases #11, it says that hybridization of an atom can increase electronegativity because the electrons are closer to the nucleus so the molec is a weaker base, but doens't that contradict the earlier saying that higher charge concentrations increase the drive to share electrons which makes it a stonger base because it wants to make more bonds? I'm confused here. Thanks.
Panthea, I agree that the hybridization effect seems contradictory to other effects. I don't have an explanation that makes it fit with everything else. --DrH
Hi Dr. H., On page 258 in Thinkbook, #33, why is the pKa for carbonic acid is greater than pKa of acetic acid? I thought the OH on carbonic acid would cause carbonic acid more acidic due to inductive effect. Likewise, shouldn't the CH3 in acetic acid hinder the deprotonation of the H in acetic acid?
Cam, The OH has lone pairs and thus increases electron density at the carbonyl through resonance. Using our usual assumption that resonance outweighs inductive effects, the net effect of the HO group is to release electron density. --DrH
Hi Dr. H, In 5c of the solutions for the OWLS of Organic Acids and Bases, should guanidine have 3 resonance contributors? Thanks.
Margaret. There are other resonance contributors, but only the two that are shown are significant. --DrH
Hi professor Hardinger, On pg 258 #28 in the thinkbook, doesn't the deprotonated form of Nitromethane have 3 resonance contributors? If it only has 2 resonance contributors, Nitromethane also has 2 contributors, so what makes Nitromethane so acidic? Is it because the deprotonated form of nitromethane have charges shared by two oxygen atoms?
Lisa, There are other resonance contributors, but the question does not require that we draw all of them. --DrH
Hi professor, Please disregard my previous question! I should have read further down for the answers.
Lisa, No problem. We can all gain wonderful insight from the miracle of the printed word. --DrH
Hi professor Hardinger, On pg 263 #10(a) in the thinkbook, I drew two arrows for the step taken by arrow #2: one going from the double bond to the oxygen atom, and a second one from the lone pair on the oxygen to the hydrogen atom. The product molecule is the same, but is this wrong?
Lisa, Use the minimum number of curved arrows to show the bond changes, You have one extra arrow. You have given the oxygen a lone pair that used to be the pi bond, but the second arrow takes the lone pair away. Just use one arrow for this. --DrH
hello professor, I have a question regarding nomenclature of glucose. Is the cyclic form glucose called glucopyranose, or is glucopyranose just a synonym for glucose? the reason i ask is because in question 8 of the cfq, you distinguish between glucose alpha/beta glucopyranose(cyclic) and pure glucose (acyclic). Then in problem one of the practice problems, you dont make the same distinction, merely naming the structures alpha/beta glucose? is this a mistake, shouldng they be named alpha/beta glucopyranose? thank you
Abdul, "Glucose" refers to either the acyclic or cyclic forms. "Glucopyranose" refers to glucose in its pyanose (cyclic) form. The use of the terms depends upon the context. --DrH
Dr. H, Please disregard the previous question. I solved it by reading some supplementary material. My apologies.
Stephanie, No problem! --DrH
Dr. H, I'm confused by PP #6 on the carbohydrates. Can you clarify what you mean by "open form" in the solution? Does that mean in a chain form versus closed form- a ring? If that's the case, i'm confused why glucose is an aldohexose because a hexose is a 6 carbon chain and the structure indicates a ring?
Stephanie, "Open form" means the carbohydrate does not have a ring, like the structures on page 60 of the Thinkbook. --DrH
Dr. H, Regarding PP #39 acids and bases, I can understand that it's a result of the delta - charges and thus are more willing to share electrons. However, are we always supposed to examine resonance structures to determine their effect or is this case a special exception (thus not needing to follow this procedure for all resonance structures)? In other words, how do you know which molecules are exceptions and which are not? I just need a little clarification. Thanks.
Stephanie, We always need to consider resonance because it has such a strong influence when it is present. --DrH
Hi Dr. Hardinger, Why is electronegativity more important in controlling basicity than formal charge? I am confused because Formal charge actually has a full minus charge while electronegativity is a partial charge (delta + or - ). Shouldn't a lone pair (ie formal charge) be stronger? Thanks.
Panthea, Electronegativity is not always more influential than formal charge. Sometimes it is, somtimes it is not. This is why I labeled it as the "x factor" in lecture, and did not give it a specific place in the list of relative importance of factors that influence basicity. --DrH
Hi Professor, On pg 16 of the supplementary reading for acids and bases, there is a general rule about acidity and basicity with regards to the atom and what it is in. Why does the solution that the atom is in make a difference with whether or not the atom is going to be a base or an acid in a molecule? If the electronegativity of the atom sharing electrons with the Hydrogen is greater, isn't that molecule going to be a weaker base regardless of what type of solution (acid or base) that it is in? Thanks.
Panthea, Solvent influences acidity and basicity in a variety of ways not discussed in the reading. For example, hydrogen bonding can stabilize a base, and therefore influence the position of an acid-base equilibrium. This changes a pKa value. --DrH
Hi Professor, I just wanted to let you know that on pg 12 of your supp reading chapter for acids and bases, 3rd paragraph, last line, I think you meant to say "Their existence will greaty facilitate (not facility) your study of organic chemistry." That's all, just a suggestion.
Panthea, OK, thanks. I will fix that. --DrH
Professor Hardinger, Never mind my last question about PP#37 for Acids & Bases. The answer is given by PP#41.
Siamak, OK! --DrH
Hi Dr. Hardinger, I was just wondering about p. 8 of the extra reading on your website for acids and bases. How can you substitute the Keq and the pKa to find the what the total Keq is? Also why are you multiplying by the reciprocal Ka of NH4+ and how do you even know what that equation would be? Thanks for your help.
Panthea, I'm not sure I understand the source of your confusion. Please ask me or one of the IAs in person. --DrH
Hi Professor Hardinger, For Acids & Bases, on Practice Problem #37B, I don't understand why equilibrium would lie to the left. Since the base on the products side has resonance, wouldn't it be favored, pushing the reaction to the right?
Siamak, This is a case where formal charge has more influence than resonance. --DrH
Hi Professor, when discussing whether an anomeric carbon is in alpha or beta configuration, can we say that it is alpha because the substituent is axial, and beta if equatorial?
Sheela, This is not correct. In the most stable conformation of alpha-glucopyranose, all groups are equatorial except the anomeric OH. If the molecule suffers a chair-chair ring flip (like cyclohexane), then the anomeric OH becomes equatorial. Regardless of which chair the molecule is in, however, the anomeric OH and CH2OH groups are still trans, so the stereochemistry at the anomeric carbon is still alpha. --DrH
Hi Professor, I had a question on number 39 in the practice problems for acids and bases. How do you know that resonance increases electron density to nitrogen. I thought that resonance decreases electron density. thanks
Annie, I mentioned in lecture that resonance often but not always decreases electron density. This is a case where resonance increases it. Compare the magnitude of delta- charges in the resonance hybrids of acetate ion and the molecules in PP 39. --DrH
Hi Dr. H, I was a little confused on the PP #26 for introduction to the reactivity of organic molecules: acids and bases. On the second molecule from the left, why is the hydrogen bonded to the sulfur the most acidic? Why isn't the answer they hydrogen bonded to the oxygen? I thought electronegativity had more of an influence than atomic size.
Stephanie, The influence of atomic size is usually greater than electronegativity. For more on this review Acids and Bases CFQ 11. --DrH
Hi Professor, did you know that in CFQ Acids/Bases Problem 40, the acid whose pKa you asked us to find is actually on the table already?
Sheela, Yes I did know that. That particular example illustrates that the estimate can be close, but will not necessary give you the excact pKa. --DrH
hi professor, I dont understand practice problem #20E under acids and bases. The answer explains that the CH3 is a weak electron donor compared to hydrogen, but wouldn't this mean that oxygen in CH3CH2O will have a negative charge to a lesser extent than the oxygen in the OH, thus, making it a weaker base than OH (therfore a stronger acid)?
Preethika, Electron donor means it is releasing electron density to its neighbors, causing an increase in the neighbor's electron density. Relative to water, the oxygen atom of ethanol (CH3CH2OH) has more electron densite because the ethyl group is giving electron density to the rest of the molecule. A hydrogen atom is not an electron donor. --DrH
Hi Professor Hardinger, I think there's an error on pg 229 #30.d. in the solutions manual for Bruice: one of the NH2s should be changed to a CH3. Thanks you.
Lisa, You are correct. The error has been posted at the Known Typos web page, and you have been awarded one error bounty extra credit point. --DrH
Hi Professor, just making sure...for CFQ #38 in the Acids & Bases section, would H2SO2F2 be an acceptable answer? (basically, replacing each of the OH oxygens with a fluorine)
Sheela, That is not an acceptable answer. Halogens generally form one single bond only (example: H-F or HO-Cl) unless bonded to other halogens (example: IF5). So H-F-S-restofmolecule is not possible. --DrH
Hi Professor, for Problem 3 in Chapter 22 of Bruice, why are the answers for parts a and b L-glyceraldehyde? In a and b the OH group is on opposite sides, and yet their names are the same.
Sheela, Special rules apply to the manipulation of Fischer projections. Fopr example, a 90 degree rotation charges the stereochemistry (in this case between L and D). It may be easiest to use models that correspond to the Fischer projections. If you still have difficulty with this exercise after you use models, please ask again. --DrH
Hi professor Hardinger, I think I found an error on pg 45 of Bruice. On the very bottom of the page where relative acidities are ranked, I think NH4 should be NH3 instead because NH4 would yield a pka value of 9.4 and it wouldn't fit in the ranking. (it also lacks a positive charge)
Lisa, You are correct. The error has been added to the Known Typos web page, and you have earned an error bounty extra credit point. --DrH
Hi Professor, In regards to drawing carbohydrates, if it is only specified that we are to draw a aldohexose, how do we know the orientation of the OH's along the chain? Other than the fact that all naturally occuring (D) carbohydrates have the OH right about the CH2OH pointing to the right, is there any convention for the others? I am looking at PP #5 of Carbs. Based on how I have seen it before, numbering from top to bottom, I would have put the OH on the opposite sides of carbons 3 and 4. Thanks.
Mansoor, If the question requires only an aldohexose and has not other requirements, then the configuration of each H-C-OH stereocenter does not matter. (If your first grade teacher asks you to draw a car does it matter what color crayon you use?) --DrH
Hi professor Hardinger, On pg 189 of Atkins and Jones, 5.1(b)N2H4, the answer key only lists london forces and dipole-dipole forces. I was wondering why hydrogen bonding isn't included? (isn't there an N-H bond with lone pair of electrons on N that can interact with a hydrogen atom of another molecule?)Thank you.
Lisa, You are correct. Hydrogen bonding is also operating in the case of hydrazine (H2N-NH2). The error has been added to the Known Typos web page, and you have been awarded one error bounty extra credit point. --DrH
Hi professor, Number 10 of practice problems of noncovalent molecular forces, I don't understand part of the solution in the thinkbook. It says that due to inductive effect in HCCl3, hydrogen has a large delta +. But, it also mentions that HCCl3 is a weak hydrogen bond donor. Shouldn't it be a good hydrogen bond donor? Thanks.
Preethika, Although the hydrogen of chloroform (CHCl3) does have a delta+ charge, the charge is small. Since hydrogen bonding is an electrostatic force, the attraction is small when the charge is small. In other words, the hydrogen bond formed with chloroform is weak. --DrH
Hi Professor, You gave us an example today in class about how for the size of an atom you don't worry about the size of the total molecule. You gave us two molecules, CH3O- and (CH3)3CO-, and made the point that the stronger base was the larger one. I don't understand why. Resonance being a non-factor in both cases, and atomic size not differing, and having no electronegativity difference, we are left with the inductive effect. Since C is more electronegative than H, and one would assume that this would mean that the CH3O- was a stronger base, do we assume that, in this case, the CH3 groups are donating electron density through the inductive effect? If so, is this a general rule? Thanks.
Mansoor, You got it! Alkyl groups are (as you have discovered) electron-donors through an inductive effect. There are other issues involved in the case of CH3O- and (CH3)3CO- as well, but they are beyond the scope of the course. --DrH
Hi Dr. H., For Exercise "c." in the "Acids and Bases: Conjugate Acids and Bases" tutorial, the formal charge above the O atom in the H2Ö-H molecule is given as negative (-1). Since there are 3 bonds and 2 lone pair electrons, shouldn't the formal charge on the oxygen be positive (+1) instead?
Andrew, There is a positive charge at that oxygen. It may look strange due to your web browser, and because the graphics in that tutorial are rather old (need to be updated). --DrH
Hi Professor, For problem 29 of the CFQ for acids and bases, shouldn't the carbon that is connected to the oxygen on one side, and the carboxylic acid group on the other carry a formal charge of -1? Thanks.
Mansoor, I assume you mean PP 29 because there is not CFQ 29 for that material. I do not understand your question, but there is no formal charge error in the answer. --DrH
Hi Professor, For # 30 of the practice problems for Acids and Bases, H3O+ has a lower pKa than CH3CO2H even though the latter has resonance. Are we to assume that the formal positive charge on the hydronium offsets this resonance, because normally resonance is the most important thing that makes something a weak conjugate base, thereby giving it a strong acid. Thanks.
Mansoor, Resonance usually dominates over other factors such as formal charge, but this is one exception to that assumption. --DrH
Dear Dr. H: I was reviewing for the final the part with formal charges on your website, On practice problem 5 for formal charges worksheet, it states that the molecule is a carbon monoxide. Is this an error?
Greg, By "the part with formal charge" I assume you mean the formal charge tutorial? Practice problem 5 is about CO, carbon monoxide. --DrH
Also, on page 63 of the Thinkbook, when referring to cellulose in wood, you state that it is stabilized by "inter" molecular hydrogen bonding... shouldn't it be "intra" molecular hydrogen bonding, as stated on page 948 of the Bruice textbook? Chemically significant?
Rosie, Cellulose enjoys both intermolecular and intramolecular hydrogen bonding. --DrH
Hi Professor, Is there an error in the Fischer structure of acyclic D-glucose on page 61 of the Thinkbook? Shouldn't glucose have an aldehyde instead of a primary alochol as its uppermost functional group?
Roisie, You are correct. The "top" carbon of glucose is an aldehyde not a primary alcohol. The error has been posted at the Known Typos web page, and you have been awarded one error bounty extra credit point. --DrH
Professor Hardinger, I believe there is an error in the posted solutions for the Fall 2005 midterm II that we just took. Question two asks us circle the bond with the highest stretching energy, which I believe should be a Carbon/nitrogen triple bond, but the posted answer key states that it is C-C. Thanks.
Joey, An updated exam key was posted Sunday morning. --DrH
Hi Professor. For part c of the first question on the midterm, why can't we assume that the M peak is the base peak when we have said before that the most abundant ion, the M ion, is usually the base peak? Just as we have no reason to assume that our M is an exception, we similarly have no reason to assume that it is NOT an exception. Thanks.
Mansoor, The molecular ion is not always the most abundant ion. When comp[aring M, M+1 and M+2 for the purposes of determining molecular formula, we set M = 100%. This does not mean it is the most abundant ion in the entire mass spectrum. --DrH
For Exam 2 Fall 2005 (our exam) why does C-C have the highest stretching energy? Since it's in the fingerprint region, wouldn't it have one of the lowest energies?
Siamak, I do not understand your question. The answer key indicates that a carbon-nitrogen triple bond (not a carbon-carbon single bond) has the highest stretching energy. --DrH
For our exam 2 (Fall 2005) #6, shouldn't the first formula on the solution have 23 hydrogens, not 24?
Simak, I do not understand your question. The answer key for question 6 does not include any formulae with 24 hydrogens. --DrH
Hi Professor, after conversing with my friends about the second midterm, we feel one of the questions was worded ambiguously. The question stated that the proton H had an integration of 1 - did the "H" stand for a generic Hydrogen atom, or the atom labled H on the molecule?
Matthew, The H stood for proton H on molecule X. Given the context of the question and the comparison with similar questions on previous exams, I do not think this was ambiguous. --DrH
Dr H i have a question about the Spring 2004 second midterm, problem #10. Why is the integration for D and E 1.5? Since the given information establishes that the ration is .25:1H, and d has three H's shouldn't the integration be .75? Or are the D and E hydrogens considered equivalent thus giving us (6)(.25)=1.5? Thanks!
Chloe, Equivalent hydrogens have the same NMR signal, which includes integration. For example two equivalent methyl groups give a 6 hydrogen singlet, not two 3 hydrogen singlets. --DrH
For Fall 2003 Exam 2 #8, why is it that the carbonyl is only attributed to the ester? why not the carboxylic acid as well?
Siamak, The exam key will be annotated to clarify this point. Thank you for bringing it to my attention. --DrH
Professor, On page 42 of the think book, the Schematic NMR spectrum given says that downfield has "low magnetic field strength" and upfield has "high magnetic field strength". Should this be the opposite?
Michelle, It is correct as written. --DrH
Hi Professor H. On Fall 2002 Exam 2 for the last problem, why is CH=CH? Since the double bond is a free spacer, wouldn't the 1st CH be 4 lines, the 2nd CH is four lines and CH2 be triplet? In that case, they do not follow the splitting given on the test. Thanks
Nancy, "Free spacer" means the coupling can extend beyond this point, but a pi bond is not 100% free. Coupling through the pi bond can sometimes cause a decrease in coupling constants (J values), which in turn may alter the nature of the coupling patterns. In other words, the coupling may no longer be what is called "first order", and the simple n+1 rule may no longer apply. --DrH
Hi Professor, For the OWLS Proton NMR solutions, on 1b, the diagram is labeled incorrectly. THere are three types of H's that are nonequivalent, but the diagram only shows two. (It has Ha, and Hb, and in place of Hc it re-uses Hb). Just wanted to bring this to your attention. Thanks.
Mansoor, An updated version of the solutions which includes a correction of this error was posted a few days ago. --DrH
Hi Professor, With regards to Question #1 of Fall 2003 Exam 2, I am having a bit of trouble with the chemical formula. I've counted it many times, and get one carbon and two hydrogens lower that what the key says. Since then, two of my friends have also counted the same C28, H37, whereas the key says C29, H39, and both of these give the same DBE? Are we missing something, or is there a mistake? Thanks.
Mansoor, I have counted the atoms five times and each time I get the formula given in the answer key. Without watching you count I do not know what you are missing. --DrH
For Spring 2005 Exam 2 #10, the solution says that the best choice is 4.2. But using the method that the IA's taught us I get a different answer. 2.9 (for anything bound to oxygen) + 1.3 (for acyclic methyline) + 0.9 (for being bound to a C=C) = 5.1. Am I making a mistake by using these numbers? or is my procedure correct?
Siamak, Perhaps your calculation is in error. Or, this could be a case that the estimation procedure does not handle very well. As I mentioned in lecture, this is just an estimate, with large errors bars because it uses averages and assumptions that may not be broadly valid. --DrH
About my last question about Exam 2, Winter 2005 #12, I know that there is no reason why the ring would flip. But doesn't this make it so the hydrogens on E (axial and equitorial) would not be equivalent and would couple with each other?
Siamak, I think you misunderstand. The ring WILL flip. --DrH
For Exam 2, Winter 2005 #12: the solution says that e and f are both triplets (because they are each others neighbors and there is nothing else to couple with). But since that is a cyclohexane which wouldnn't easily do a ring flip, shouldn't the axial H and equitorial H be different for both e and f? (This would give each three neighbors to couple with and they would be a quartet.)
Siamak, There is no good reason why the ring would not readily flip. --DrH
Professor Hardinger, if you have a carbonyl, and the carbon is attached to hydrogen on one side and oxygen (bound to carbons) on the other, would zone 4 show the IR carbonyl peak associated with ester or the carbonyl peak associated with aldehyde? Regardless of where the carbonyl shows up, would you still get peaks at 2700 and 2900? The structure looks like this: H-C-OR (with carbon also doubled bonded to another O)
Siamak, An aldehyde has a hydrogen on one side of the carbonyl and a carbon group on the other. You have described as ester not an aldehyde. It is an ester of formic acid, HCO2H. --DrH
Hi Dr. H! I have a question about the solution to Solving Spectroscopy Problems PP#3 on page 229 of the thinkbook: It says that the M+2 peak would be greater than 100% if chlorine and/or sulfer were present in addition to bromine. Wouldn't the M+2 peak still be smaller than the M peak in this case, as M peak would have a greater intensity/abundance (still having the more common isotopes of Cl, S, and Br)? Sorry for the confusion.
Donald, The explanation is pointing out that Cl or S cannot be present in addition to Br. If the molecule had Br and Cl, M+2 would be ~133%. --DrH
Hi proffesor there was a previous question posted about OWLS NMR spec # 4b about why the answer key talks about Chlorine but the molecules in question doesnt have cholorine. The answer to part 4a seems fixed but the answer to 4b still talks about cl wheras the question is asking about the esters am i missing something?
Andrew, Updated solutions were posted, but I guess I sent them to the wrong place. I cannot update my web site from home (hwere I am now), so I will post the update tomorrow morning. In the mean time I sent a copy of the update to your email. Sorry for the hassle. --DrH
To clarify, I was referring to the very first problem.
Mansoor, So many questions get posted to VOH that it is not feasible to refer back and forth between posts to see what you are referring to. Therefore please resubmit your question along with the reference, so that we can all understand what you are asking about. --DrH
Hey Dr. H, for the Fall 2002 Exam 2 Question 8, why are there 8 sets of nonequivalent hydrogens intead of 6?
Julie, Without knowing what you have or have not labeled as equivalent, I would guess you didn't include two to different NH protons, or the three different benzene ring protons. --DrH
Hi Professor, With regards to Fall 2003 Exam 2, I am having a bit of trouble with the chemical formula. I've counted it many times, and get one carbon and two hydrogens lower that what the key says. Since then, two of my friends have also counted the same C28, H37, whereas the key says C29, H39, and both of these give the same DBE? Are we missing something, or is there a mistake? Thanks.
Mansoor, The fall 2003 exam 2 has more than one problem on it. Please resubmit your question with a reference to the specific problem you are asking about. --DrH
HI dr. H, i was just wondering if we should know the structure from the firt midterm. thank you, Rebecca
Rebecca, This structure is not applicable to the exam. --DrH
Dr. Hardinger, In many of your previous tests you ask for an 'estimate' of the intensity of the M+1 peaks and M+2 peaks. What exactly should this estimation include? Just the carbon abundance for M+1, or the carbon and the hydrogen abundance? Just the bromine, chlorine, and sulfur abundances for M+2? Would we get marked down if we solved it out exactly?
Eric, A more accurate estimate is always preferrable. For M+1 this includes all elements. For M+2 we are more concerned with an approximate value than an exact number, so it is sufficient to include just Cl, Br and S. --DrH
For midterm II, will we need to know how to calculate exact chemical shifts in PPM (for NMR), and if yes, how do we do it? (specifically for PP #2.C, methyl propionate and ethyl acetate)
Dave, You may need to do this. The procedure is a bit more than can be described in this little VOH answer space, but it is explained in an NMR OWLS problem. --DrH
I have a question regarding #1 of the Spring 2004 Midterm. In this question the molecular formula is C5H11O2NS and I am suppose to find the relative abundance of M+1. When I did the problem, I added the M+1 percentage for C, O, N, and S. I got 6.7%. In the answer key, the answer is 5.8% (carbon with nitrogen). I don't understand why the M+1 percentage of S and O are not added.
Wing, The question asks for an approximation. A more precise calculation would account for oxygen and sulfur, like you included. --DrH
Hi professor, on practice exam fall 2004 exam 2 Question number 18, Why would the addition of C CL2 to proton e's carbon create a quartet splitting pattern? I thought only Hydrogens could split other hydrogens?
Andrew, Chlorine atoms do not change the splitting. Replacing one of the two "e" hydrogens with a CH (of CHCl2) causes the increase in splitting. This may be clearer to you if you draw in all the hydrogens of the modified molecule. --DrH
Hi Professor, I had a question about the Combined Spectroscopy PP #6. There is a choice here between an alkyne and a nitrile, and when i drew it, I drew both. The explanation on pg. 235 says that the shift of 1.1 of the CH3's means they are attached to a C, which is fine, but since 1.1 is larger than the typical 0.9, i left the option open for their attachment to nitrogen. In this alternative setup, I have the alkyne next to the propyl group, bonded to a nitrogen which is attached to one lone pair, and the two methyls which are singlets. Basically, I wanted to know if, on an exam, either structure would be accepted, or if we would be expected to figure it out as done in the explanation.
Mansoor, It sounds like your answer agrees with the NMR splitting pattern but not the chemical shifts. An N-CH3 hydrogen has a typical chemical shift of 2-3 ppm, but the actual chemical shift for the CH3 is 1.1 ppm. --DrH
Hey Dr Hardinger, for the Fall 2004 Exam 2 on the very last problem: for the CH2 double bonded to another carbon, why can't the hydrogens of the CH2 couple with the hydrogens attached to the carbons to the left of oxygen (and vice versa), since pi bonds are free spaces, they should be able to couple and be doublets.
X. Ran, I do not understand your question. The molecule does not include a CH2 group doubl ebonded to another carbon. --DrH
For 4B of the NMR OWLS, the question asks what structure is the better fit for the solution doesn't seem to answer this. Also, I don't understand why you used the shift associated with Chlorine (there is no chlorine in the given structures).
Siamak, I changed the problem some time ago, but not the answer! Updated solutions have been posted with the correct answer for 4(b). You have been awarded one error bounty extra credit point. --DrH
Dr. H, For problem 11 in the NMR CFQs, can you please explain the methodology of determining the chemical shift of the hyrdogens labeled D. Thank you.
Chloe, The middle of the characteristic chemical shift range for an alcohol OH (Thinkbook page 43) was used. --DrH
For 4A on the NMR OWLS, why does the solution say that the estimated shift for a methine next to an oxygen is 4.65? The problem asks for the chemical shift of CH2 (methylene).
Siamak, The procedure is correct for a methine, but the problem is about a methylene (CH2) proton. The OWLS solution set has been updated, and you have been awarded one error bounty extra credit point. --DrH
For question #5 in the CFQs of the IR section, couldn't enone be added to this list since there would be absorptions in zone 4 to account for the carbonyl and zone 5 to account for the alkene?
Chloe, It could be, although enone is usually not considered to be a distinct functional group. If enone IS included, then aryl ketone must also be included. --DrH
Hi Dr. Hardinger, I just wanted to know whether it matters when analyzing an IR spectrum whether it matters to have a stretch of aromaticity in the 1950-1750 region if there is not simultaneously a benzene ring stretch in the 1600-1450 region.
Sheela, If the molecule has a benzene ring the spectrum will have peaks both in zone 4 (the aromatic overtones) as well as in zone 5. However, one of both of these may be obscured. Therefore we look for both in order to determine the presence or absence of a benzene ring with greater certainty. --DrH
Hey Dr Hardinger, for the solutions of OWLS: Proton NMR Spec, there is an error for #1 part c, the solution says that the hydrogens attached to the oxygens are equivalent to the two hydrogens attached to the middle carbon, the oxygen hydrogens should be Hc instead of Hb.
X. Ran, You are correct. The OH hydrogens should be labeled Hc not Hb. The posted solutions have been updated, and you have been awarded one error bounty extra credit point. --DrH
hi professor, why is it that for combined spectroscopy practice problem #5, nitrogen is not shown in the IR or NMR? Is it because it's not directly attached to hydrogens? thank you.
Preethika, A tertiary amine (R3N) does not have a characteristic stretching frequency that lies in our five zones. The presence of a tertiary amine in a molecule is determined by elimination of other possibilities. --DrH
Hi Dr. H., In some of the past midterm 2, you had a theme molecule. Will you be posting a theme molecule for our midterm like you did for midterm 1?
Cam, Friday's exam does not have a theme molecule. --DrH
Hi professor, Can an NMR signal specifying the presence of a benzene ring appear as a triplet if there are three hydrogens attached to each carbon on the ring and four other groups that are not hydrogens? One of the problems in the thinkbook dealt with a benzene ring with 2 hydrogens and the peak appeared as a doublet. In lecture we learned that monosubstituted benzene will appear as a multiplet since all the hydrogens participate in splitting. So, I guess my question is, when do we start observing the 'benzene ring' peak as a multiplet?
Lisa, If the benzene ring has just two hydrogens because the othe four positions have something other than hydrogen, then the benzene ring protons are doublets not triplets. (Unless they are equivalent in which case they do not couple.) --DrH
Professor-On page 208 of practice problem 1D, why are HC-HD quintets? From my analysis, I found that C gets its H's from B and D and D gets its H's from C and E. Where does the fourth H come from for each?
Amy, Hc axial and Hc equatorial are not equivalent, so they contribute to the splitting as well. --DrH
hi professor, for the midterm, is it ok if we just give the structure of the final molecule and not the formula? For an ex. in the thinkbook, problem 2 of solving spectroscopy probs, gave the answers in the form of structural formula. Do we have to do this? thanks
Preethika, That would depend upon what the exam question asks you to do. --DrH
Hi Dr.H, I have a question on page 203 of the thinkbook, step 4. I understand that 2XCH in CHCH2 , have two neighbor. I know we need two CH to statisfy the # of hydrogen, but wouldn't two of the CH have 4 neighbor? Or do we not count the neighbor for symmetry? Thanks. nancy
Nancy, I am not sure what you are asking. Perhaps you are forgetting that the all the hydrogens that correspond to one NMR signal must be equivalent? --DrH
Regarding my last question about Spring 2004 Exam 2 #10, I can see that for D and E, you put 1.5 instead of .75 because they are equivalent and you combined them. But if would it still be correct to put 0.75 in each box since combined they equal 1.5?
Siamak, 0.75 is not correct because the signal corresponds to six protons (D and E combined), not just 3 protons. --DrH
Hi Dr. H., Does the height of each band in H-NMR matter? What I mean is that does the height of the band actually tell us something, such as how many of a specific functional group are in the molecule?
Cam, Each NMR signal is actually (well, approximately) a very skinny triangle. It is the area under the peak that matters, not the height. --DrH
For Sping 2004 Exam 2, why does the answer to #10 say that D and E both have integration of 1.5? They seem to each have 3 protons (each of which is 0.25) so shouldn't each one of them say 0.75, not 1.5?
Siamak, Protons D and E are equivalent, so they are together in the same signal, and are integrated together. In this problem an integral area of 1.5 corresponds to six protons. --DrH
Hi Dr. H, I am not sure how to estimate chemical shifts. Is there a more systematic way of doing it?
Hellen, I do not understand your question. The method presented in the OWLS problem set for NMR is pretty systematic. If I knew of a better method I would have used it instead. --DrH
hi professor, for practice problem 5 in NMR, does the attached chlorine have any influence on the CH3 groups (a, b)?
Preethika, because of its high electronegativity, the chlorine atom causes hydrogens (a) and (b) to be somewhat further downfield than typical alkane methyl group hydrogens. --DrH
hi professor, answer to practice problem 3b in the thinkbook (in NMR section) discusses splitting patterns para, ortho and meta. Do we have to know what these are?
Preethika, It is not the most important detail of NMR spectroscopy, but any idea presented in the Thinkbook or lecture is possible for an exam. --DrH
Hello Professor, I would like to know if you will postpone the second midterm since we have not yet covered Solving Combined Spectroscopy Problems, which should have been covered on November 9. Thank You
Lisa, The exam will not be postponed. The Solving Spectroscopy Problems lecture presentation is mostly just practice. A significant portion has already been introduced in previous lectures. In addition, the NMR CFQ & PP gives you plenty of material to practice with NMR problems on your own. And I am aware of the short time between the lecture and exam, and the exam has been written accordingly. --DrH
Hi Professor Hardinger, there's an error in the Mass Spec/IR Spec OWLS. The solution to #8 says that there are no aryl or vinyl H's even though all three structures (including the right s