Virtual Office Hours - CEM 252

Virtual Office Hours are your opportunity to ask a question that either can't wait until regular office hours, or when you are unable to make it to office hours. Any question you ask will be placed and answered on this page, usually in less than 24 hours. Your name will not be included on this page, but the question will not be answered unless you provide your name and it appears on the class list for CEM 252 at Michigan State University. If you include your email address, a text-only answer will be emailed to you as well as a text+ version being posted here.

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If your question requires structures, either name them or refer to structures on particular pages in the book.


Current Questions and Answers

Question: (asked 2/26/97) Could you correct the steps i've used to turn 2-pentanone into 2-phenyl-2-pentanol. It seems like an easy problem but a need a better feeling for these reactions.
1) NaOH, adds hydroxide and forms cation.
2) Bromobenzene adds the phenyl group.

Answer: Sodium hydroxide reacts with ketones and aldehydes to form enolates (negatively charged species, not positive). These enolates indicate that the alpha carbon, not the carbonyl carbon, are nucleophilic. So if the bromine on bromobenzene could be displaced by a nucleophile (which it can't) you would have 3-phenyl-2-pentanone as your product.

What you need is a reagent that will attack the carbonyl carbon rather than being attacked by the alpha carbon. Review some of the early reactions from chapter 17 and take another stab at this question.


Question: (asked 2/16/97) How does one go about determining the charge on a tripeptide, such as those involved in problem 24.21, at a certain pH?

Answer: Solve the problem the same way you approach a problem dealing with a single amino acid - you need to look up the pKa values of all ionizable groups to get started. The following dipeptide example should demonstrate the process.

Example: Glutamic Acid-Alanine at pH 5.0
Relevant pKa Values (from table 24.3):
Glutamic Acid alpha-amino group pKa = 9.47
Glutamic Acid sidechain pKa = 4.07
Glutamic Acid alpha carboxylic acid has been converted to amide - no longer an ionizable group
Alanine alpha-carboxylic acid group pKa = 2.35
Alanine alpha-amino group has been converted to amide - no longer an ionizable group

At very low pH, the amino group (glutamic acid) will have a positive charge, and the carboxylic acid groups will be neutral. The net charge on the molecules will be positive one. At pH values higher than 2.35, the alanine carboxylic acid will occur predominantly in the negatively charged form, giving molecules that are nearly neutral (depending on how close the pH is to the pI). At pH values higher than 4.07, the glutamic acid sidechain carboxylic acid will be mostly negatively charged, giving molecules that would migrate to the positive electrode at pH 5.0. See the diagram below for structures to match these descriptions.



Question: (asked 2/12/97) I can easily see that Chymotrypsin will cleave the first aromatic it encounters but how do you go about determining the positive species that Trypsin will cleave. Answer: Minor correction: chymotrypsin will cleave the amide bond on the carboxyl side of EVERY amino acid with an aromatic sidechain. This may result in cleavage in several positions in a peptide chain.

For trypsin, you need to assume that it is being used in a solution of neutral pH (since you weren't given a pH) and you need to identify the amino acids which will have a positively charged sidechain at pH=7. Those amino acids are lysine (sidechain pKa=12.48) and arginine (sidechain pKa=10.53). Trypsin will then cleave the amide bond on the carboxyl side of EVERY lysine and arginine.
Amino Acid Structures and pKa Values


Question: (asked 2/9/97) For problem 24.1 (page 1046), the answer guide lists lysine as having a positive net charge at pH 10.0. Is that correct? The alpha CO2- is always negatively charged at this pH, but one of the NH2 s, the alpha amino group, is neutral in 91% of the molecules and the other amino group (from the side chain) is positive in 78% of the molecules. Doesn't this molecule then have a negative net charge because of the 100% alpha CO2-?

Answer: You are correct, the pI of lysine is 9.74. At a pH above the pI the molecule will have a net negative, rather than positive charge.


Questions and Answers Previous to First Exam

Question: (asked 1/26/97) On the "More Spectroscopy Problems" handout, I have a question on #8. According to the HNMR, there are 3 sets of lines, and all are singlets. The one I am not understanding is the set at 7.3 (5H) ppm. The correct answer is given as 'J'. However, it seems as if the 5 hydrogens on the ring were treated as equivalent hydrogens, especially since the problem states that the line is a singlet. I do not see how these five hydrogens are equivalent. This reasoning seems to be in disagreement with the text. I would have thought the 5 H's would be broken down into the following: two sets of 2 equivalent H's and one single H.

Answer: The five aromatic hydrogens are not equivalent but have chemical shifts so close together that they appear as a broad singlet. It is very common for non-equivalent aromatic hydrogens to appear as a singlet. Professor Reusch mentioned this in class as well as another point that deserves reiteration. The second point to remember is that the splitting you calculate with the n+1 rule is a MAXIMUM splitting. You don't always experimentally observe all the splitting that you expect.

Question: (asked 1/23/97) what exactly is an allylic carbon and what does it have to do with fragmentation and forming of carbocations in mass spectrometry?

Answer: An allylic carbon is any carbon next to a C=C. If you form a carbocation at an allylic carbon it will be delocalized with the double bond ( +C-C=C <-> C=C-C+ ) giving a resonance-stabilized carbocation. Fragments will form more frequently if they are more stable than other possible fragmentations. So bonds to allylic carbons (except the bond to the C=C) will be very likely to break to give fragment ions.

Question: (asked 1/20/97) Page 490, Section 13.9 B - The book says that the more s-character a molecule (or atom) has, the more electronegative it is and therefore the more deshielded the molecule is. At the moment, I have forgotten some of my basic chemisrty knowledge - why does the amount of s-character determine its electronegativity and deshielding abilities?

Answer: An orbital with more s-character is more stable (because s orbitals lie closer to the nucleus than p orbitals do). So, let us compare carbon-hydrogen single bonds in which the carbon uses an sp3 (~25% s character), an sp2 (~33% s character) and an sp (~50% s character) and hydrogen in every case uses an s orbital. The electrons in that bond are RELATIVELY more stable close to the sp carbon than they are close to the sp2 carbon, which is more stable than having them close to the sp3 carbon. So, greater s-character in the orbital makes carbon somewhat more electronegative. If the electrons are close to the sp carbon than they are to the sp2 carbon in the analogous C-H bonds we are comparing, then the hydrogen attached to the sp carbon is more exposed and is deshielded relative to the carbon attached to the sp2 carbon.

Question: (asked 1/20/97) Page 492, Figures 13.10 and 13.11 - The top figure shows how an alkyne decreased chemical shift is formed and the bottom figure shows how an alkene induced chemical shift forms. In reading the text and the captions (of those 2 figures), I understand why one is induced chem.shift and one is decreased chem. shift. However, looking at the pictures that illustrate ths concept confuse me - I can't tell the difference between the 2 pictures. To me both have the arrows going in the same direction, the only difference I noticed was one of the molecules was laying horizontal while the other molecule was vertical.

Answer: You have noted the correct difference in the figures. What you want to look at in those two pictures is the direction the arrows point IN THE VICINITY OF THE PROTONS. The acetylene protons are located in a region where the induced magnetic field due to the electrons is pointing AGAINST the external magnetic field (the induced magnetic field arrows are point down in the diagram as they go in toward the center of the molecule). The vinyl protons are located in a region of the induced magnetic field where is is WITH the external magnetic field (the induced magnetic field arrows are point up in the diagram as the loop around the outside of the molecule).

Question: (asked 1/19/97) I have a question regarding problem 13.26, page 519. Shouldn't the peaks shown be multiplets? Doesn't spin spin splitting have any effect on the peaks of the structure (ethylene cyclohexane?)? The answer in the student study guide (page 382 doesn't mention splitting of the peaks.

Note: The compound name is methylene cyclohexane

Answer: A phenomenon mentioned in class about spin-spin splitting is that sometimes it occurs but the peaks are not split far enough apart to see in the spectrum. You will note that the answer book does refer to two of the peaks as 'broad', indicated that there is some splitting occuring that is not clear. The splitting that you observe is related to the relative orientation of the hydrogen atoms involved. The hydrogen atoms in a ring are commonly constrained in orientations where the spin-spin splitting can be too small to observe experimentally.

Question: (asked 1/18/97) In the first problem on the "Sample Problems for Mass Spectroscopy" sheet handed out in lecture (the same problem in the online practice for mass spec.), I am having problems trying to figure out the gas. I think that it may have a molecular formula of C2H4O, but I can see no structure that will allow for this answer. I thought that CH2CHOH would be a solution... Can any rearrangement be done on such a small molecule to come up with m/e 28 & 16, or is the problem less difficult than I am making it?

Answer: You are trying to make this problem more difficult than you need to. The first thing you should think of are some common gases that weigh as little as 44. (Consider what 44-28 is equal to for an idea of an element that might be present and do the same with the other fragment.)

Question: (asked 1/18/97) Why is the propargyl cation such a stable cation (structure given on page 463 of the text)? Is the formation of resonance structures a factor, or is there a better reason?


Answer: Yes, resonance is the best explanation for the stability of the propargyl cation. Resonance is a much stronger stabilizing factor than induction (such as the electron-releasing ability of an alkyl group). Keep in mind that the propargyl cation is a common fragmentation of alkynes, that doesn't mean that this cation is more stable than a resonance-stabilized tertiary cation, just that it is commonly a good fragmentation for alkynes.

Question: (asked 1/14/97) I am confused as to how to go about proposing structural formulas for cations. For example, how would I go about doing problem 12.16 on page 472?

Answer: Question 12.16 reads "Following is the mass spectrum of 3-methyl-2-butanol. The molecular ion m/e 88 does not appear in this spectrum. Propose structural formulas for the cations of m/e 45, 43, and 41."
To approach a problem like this you should begin by looking at the structure of the molecule. Determine where bonds could be broken to leave behind a relatively stable cation. Then determine the weights of those cations and you will usually find that you have answered the question. See the figure below for this procedure performed for 3-methyl-2-butanol.



Notice from looking at the potential fragmentation of the molecular ion we have already explained the structure of the m/e 45 peak - it is the resonance-stabilized primary carbocation generated when the bond labelled 3 in the diagram is broken. The m/e 43 peak can also be explained if a small percentage of molecular ions in which bond 3 breaks do so leaving the positive charge on the secondary carbon. This cation is not as stable as the resonance-stabilized cation, and the abundance of this fragment ion (as demonstrated by the height of its peak in the mass spectrum) is much smaller. The m/e 41 peak is not quite as easy - there is no straightforward way to get it from the molecular ion. If there is no simple fragmentation of the molecular ion, look at fragment ions of slightly higher m/e to see if they give you ideas. The 41 peak is due to the allyl cation probably by loss of H2 from the propyl cation as shown in the image below.





Last modified 2/26/97
Dr. Parrill
Department of Chemistry
Michigan State University

These pages may be downloaded and linked from other pages freely for academic and educational purposes. Questions, problems, and errors should besent to parrill@argus.cem.msu.edu.