http://orgo.curvedarrow.com/punbb Curved Arrow Press Forum en http://backend.userland.com/rss http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=512#512 512@http://orgo.curvedarrow.com/punbb Topic: Identify benzyl methyl maleate or fumarate from the following data. Message: Yahoo!Answers in chemistry wrote:A 2.81g of an optically active diester A (containing only C, H and O atoms) was hydrolysed with 30 mL of 1M NaOH. Following the hydrolysis, the solution required 6 mL of 1M HCl solution to react with excess NaOH only. The organic products are found to contain methanol, an organic acid B and an alcohol C, C8H10O. Compound C gives a pale yellow precipitate with warm alkaline aqueous iodine.i) Calculate the MW of the diester A.ii) Draw structural formulae for the compounds A, B and C, explaining the above observationsiii) Compounds B and C exhibit stereoisomerism. Explain clearly, the type of stereoismerism exhibited by compound B and C, and draw displayed formulae of the isomers. A has a MW of 234. Of the 30 mmol of NaOH, 6 were in excess, since it was a diester, only half of the 24 were on a molar basis. 2,810 mg/12 mmol = 234 mg/mmol (or g/mol). Okay, lets add up the pieces, CH3O (31), C8H9O (121), 2 x C=O (56), + unknown = 234. The missing pieces = 26 or 2 x 13 (CH). The alcohol C has 4 degrees of unsaturation, therefore it probably is a benzene ring. That leaves C2H5O to be attached to it (C8H10O - C6H5). However, since it must contain a chirality center, it cannot be CH2CH2OH. That means C is an alpha methyl benzyl alcohol. The stereoisomers will be mirror images of the chirality center. It will give a positive result from the iodoform reaction (the pale yellow precipitate with alkaline iodine). The two carboxylic acid groups must be separated by two CH-groups. Therefore it must be maleic or fumaric acid (the cis and trans isomers, the two isomer forms).The structures A, B, and C could be as follows (with only a single isomer drawn of the cis, trans, R, and S isomers).           Sun, 09 May 2010 13:43:43 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=508#508 508@http://orgo.curvedarrow.com/punbb Topic: Identify 4-methylpentan-2-one from the IR and NMR spectra Message: Can you identify the unknown from the following NMR and IR spectra?This appears to be a student NMR. As such, it lacks the detail to enable easier interpretation. However, there is sufficient detail to identify the compound as it is a fairly simple structure. The integral is imprecise, but we can apply simple thinking here. It appears that there is a doublet(?) (2.3 ppm), singlet (2.1 ppm), and doublet (0.9 ppm). The compound is not an ester or ether with a CH-O-bond by the lack of a chemical shift greater than 3 ppm. The IR contains a C=O peak and the peaks at 2+ ppm are consistent with CH-groups attached to a C=O. However, it is difficult to think of an arrangement that will give a 1H doublet, a 2H singlet, and a 3H doublet. We should try an alternate by doubling the integral values. That will give a 2H doublet, a 4H singlet, and a 6H doublet. A four hydrogen singlet is not possible, but this singlet does not look clean. It looks as though a multiplet could be hidden within it. If we assume this to be the case, then a structure that would fit would be as follows. The 6H doublet is the isopropyl group, the two groups near 2 ppm are alpha to the C=O group. The missing CH-multiplet splits the CH2 and CH3 groups into doublets. This CH group broadens the singlet and increases the integral to 4H.  Mon, 26 Apr 2010 18:15:06 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=475#475 475@http://orgo.curvedarrow.com/punbb Topic: Identify butyl acetate IR and NMR spectra Message: Yahoo!Answers in chemistry wrote:I have an unknown organic compound which I extracted, purified, and tested in lab. I spent 6 hours with my data and the CRC handbook, only to get what my lab TA decided was "wrong but almost close." I would like some advice on determining this compound. Here is my data,It is a clear liquid.It smells like nail polish.It dissolved in the organic layer, not aqueous.It has a b.p. range of 124.5-129.5 degrees C.It is a neutral compound, neither an acid or base.C13 NMR revealed 6 peaks:170.5363.8230.3420.4118.7713.25HNMR revealed 5 peaks (Integration #, Splitting, Shifting):3, Triplet, 0.772, Sextet, 1.222, Quintet, 1.463, Singlet, 1.862, Triplet, 3.90IR Spectrometry revealed peaks (transmittance in cm-1):~3500, broad~2960, moderately broad1747, moderately sharp1460, sharp1360, sharp~1250, moderately sharp~1030, moderately sharp950, weak-sharp633, weak-sharp606, weak-sharpI determined the compound to be 2-Hexanone (Butyl methyl ketone), and, as I stated previously, I was told it is close but not right. I would appreciate ANY help you can give. I do not know the molecular formula, so I'm not sure if there are any halogens or if the oxygen I guess is actually a nitrogen. I would love any hints. My lab report isn't due until next Wednesday, but I am really frustrated by this and would love to get it finished. Please help. Thank you. Let me collect the minimum of facts. C13 NMR peak at 170 ppm, an ester, not a ketone, and it has six carbonsIR 1747 cm-1 consistent with an ester1H-NMR2 x CH3, 3 x CH2, therefore linear, no branching. 3H singlet at 1.85 ppm, therefore not connected to an oxygen atom. It must be a CH3C=O2H triplet at 3.90 ppm, connected to an oxygen atom and CH2 adjacentOf the six carbons, only five have hydrogen attached to themReally, do we need more to ID this? CH3CO2CH2CH2CH2CH3Some of the other data may not be helpful. The broad bands in the IR, I ignored as they wouldn't agree with an authentic sample. Wed, 23 Sep 2009 19:51:16 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=470#470 470@http://orgo.curvedarrow.com/punbb Topic: NMR of butyl acetate, unknown of C6H12O2 Message: Yahoo!Answers in chemistry wrote:Organic Chem help please?The H NMR spectrum of a compound with molecular formula C6H12O2 is shown here.Draw its structure. Identify the compound with MF = C6H12O2.IHD = (2x+2-y)/2IHD = (12+2-12)/2 = 1This problem may have had an IR associated with it to more easily identify a C=O bond being present. We could argue that it could have a ring with two oxygens placed at various points, but there are only two hydrogens attached to an oxygen (peak at 4.1 ppm). The integral is 2:3:2:2:3. Therefore one of the six carbons does not have a hydrogen attached. The others are CH3 and CH2 units. That makes it pretty simple. The peaks at 4.1 ppm has an adjacent CH2The peak at 2 ppm has not adjacent hydrogens as it is a singlet. The peak at 0.9 has an adjacent CH2. The compound is butyl acetate, CH3C(=O)OCH2CH2CH2CH3 Mon, 07 Sep 2009 10:22:15 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=469#469 469@http://orgo.curvedarrow.com/punbb Topic: NMR of 4-t-butyl-1-methylbenzene, unknown of C11H16 Message: Yahoo!Answers in chemistry wrote:The H NMR spectrum of a compound of formula C11H16 is shown here. Draw its structure. For a compound with MF = CxHyO, ignore the oxygen. IHD = [(2*x + 2)-y]/2IHD = [(2*11 + 2) - 16]/2IHD = 24 - 16 = 8/2 = 4Therefore there are 4 rings or double bonds. The peaks around 7 ppm indicate a benzene ring, therefore it represents one ring and three double bonds. From the integral, we can see there are 4 aromatic hydrogens, therefore the other peaks are 3Hs and 9Hs by measuring their heights (about 0.5 block per hydrogen). Since hydrogens must come in 1H, 2H, and 3H units, the peak at 2.3 ppm is a CH3 group and the peak at 1.3 must be 3xCH3 or a t-butyl group. Since there are four aromatic hydrogens, the benzene ring is di-substituted. The choices are 1,2-, 1,3-, or 1,4-substituted. 1,2- and 1,3-substitution would result in 4 aromatic signals, as no symmetry would be present. 1,4-substitution would result in two signals with agrees with this spectrum. Putting this all together will result in 1-(tert-butyl)-4-methylbenzene.                   Sun, 06 Sep 2009 23:57:07 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=428#428 428@http://orgo.curvedarrow.com/punbb Topic: NMR of 1-chlorobutane, the adjacency rule. Message: Here is 1-chlorobutane. You can see that the number of peaks depends on the number of adjacent hydrogens. You can examine the spectrum and see how the number of peaks in a signal depends on the total number of hydrogen atoms on the neighboring atoms.                   Tue, 17 Mar 2009 05:01:40 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=424#424 424@http://orgo.curvedarrow.com/punbb Topic: Identify 7-bromohept-4-ynal from IR and NMR spectra Message: Yahoo!Answers in chemistry 3/14/09 wrote:Open Question: Organic Chemistry Synthesis Products and IR help please!!!! Check my work?Which functional groups do you see?ketone, ester, ether, terminal alkyne (-CCH), internal alkyne (-CC-), benzene, alkene, nitrile, cycloalkane, alcohol, aldehyde I see: nitrile, alcohol, terminal alkyneI see: B? I suppose I should stop picking on this guy. He has provided fodder for my jibes and problems to post. Let's analyze what we have. The IR shows a CΞC at 2200 cm-1 and a C=O at 1700 cm-1. The MF C7H9BrO means we have 3 rings and/or double bonds. With the C=O, that leaves us with 2 rings or double bonds. The CΞC means no other rings or double bonds. The NMR shows a peak near 10 ppm, 1H, 2 adjacent hydrogens. If you look in a table, you will find this shift is consistent with an aldehyde. Therefore we have a CHOCH2-The rest of the NMR shows several CH2 groups by the integral. None has more than a single CH2 adjacent as they are all triplets. The highest field almost looks like a quartet, but I am guessing that since these are hand drawn, it is in error. This means all of the CH2 hydrogens are in isolated units of CH2CH2. Let's try to put this together. A CHO + CH2CH2 + CΞC + CH2CH2 + Br. That gives us D. Mon, 16 Mar 2009 16:54:32 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=423#423 423@http://orgo.curvedarrow.com/punbb Topic: Identify ethyl isobutyrate from IR and NMR spectra Message: Yahoo!Answers in chemistry wrote:Open Question: Organic Chemistry Synthesis Products and IR help please!!!! Check my work?Can someone please check my work? I am so confused about all of this stuff! :( Which functional groups do you see on the spectrum? alkyne, alkene, benzene, ester, ether, aldehyde, alcohol, nitrileI see: alkene, alcohol What is the compound?A? Obviously, the questioner is lost. A is neither an alkene or an alcohol, his answer above. Let's analyze what we have. From the IR, we have a C=O by the big peak around 1700 cm-1. There is a C-O peak, but it isn't necessary. From the NMR, we know C6H12 contains a ring or double bond. The C=O is that double bond. The peak at 4 ppm is a carbon connected to an oxygen and it is a CH2 from the integral. The adjacency rule says that if it has 4 peaks, then N + 1 means there are 3 hydrogens adjacent. That gives us a -OCH2CH3. It also means the peak at 1.3 ppm, 3H, triplet is the CH3.The peak at 1.0 ppm, 6H, doublet must be two CH3 groups and they are connect to a CH. The peak at 2.3 ppm, multiplet, 1H must be the CH. That gives us a -CH(CH3)2Okay, let's put this together, a CH3CH2O- plus a C=O plus a -CH(CH3)2. I get CH3CH2OCOCH(CH3)2, J. Mon, 16 Mar 2009 16:24:20 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=422#422 422@http://orgo.curvedarrow.com/punbb Topic: Yahoo NMR problems number of signals Message: Yahoo!Answers in chemistry 3/10/09 wrote:Open Question: Organic Chemistry Synthesis Products and IR help please!!!! Check my work?Can someone please check my work? I am so confused about all of this stuff! :(1.How many signals would be there be for each molecule on a H-NMR? So I get that signals means different groups of hydrogens. This is what I got:A:4 B:3 C:3 D:6 A-5, B-7, C-4, D-5This is the same guy that deletes his question and MY ANSWERS. Yahoo!Answers in chemistry wrote:A: 5 B: 7 C: 4 D: 4 ??? Huh? Did I get one wrong? Does he want me to say it is something else? FIVE! Mon, 16 Mar 2009 16:02:03 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=421#421 421@http://orgo.curvedarrow.com/punbb Topic: Yahoo NMR problems adjacency problem Message: Yahoo!Answers in chemistry 3/14/09 wrote:I need help with A and D.For A, I can't tell whether it's a nonet (so eight H neighbors+1= 9 peaks=nonet)or whether it's a quartet (just take one methyl group so 3+1= quarter= 4 peaks)or whether you just take one methyl group since the other one is in the same env't and then also take that CH2 group..so (3+2 then +1= sextet= six peaks)?For D: there's symmetry so do you look at only one CH2 group so 3 peaks (triplet) OR do you count both neighbors, so 4+1= pentet ??? Is this an acknowledgement of my answer? It's nine! Are you truly dense? Yahoo!Answers in chemistry 3/14/09 wrote:I need help with A and D.For A, I can't tell whether it's a nonet (so eight H neighbors+1= 9 peaks=nonet)or whether it's a quartet (just take one methyl group so 3+1= quarter= 4 peaks)or whether you just take one methyl group since the other one is in the same env't and then also take that CH2 group..so (3+2 then +1= sextet= six peaks)?For D: there's symmetry so do you look at only one CH2 group so 3 peaks (triplet) OR do you count both neighbors, so 4+1= pentet ??? What you also don't see if a problem with ethyl isobutyrate and its NMR. The NMR showed the COCH(CH3)2 peaks. If he is looking at his questions and the answers, there was an example spectrum for him to look at. Is it a quartet (NO) or a septet (yes)?I don't know if I had answered this prior question as it is only one of several. I may have skipped it, but as I said, this guy deletes his question and my answer, argh!Finally, I asked him for his email address and this is what I sent to him. You need to add up all of the adjacent hydrogens. A has 3 + 3 + 2 = 8 adjacent hydrogens. You know this all stems from the magnetic property of a proton. Thus a single proton adjacent to another adds to or subtracts from the applied field of the NMR instrument. That is why the measured signal is split into two peaks, one for the applied + the neighboring field and one for the applied - the neighboring field. That principle does not change with additional hydrogens. Mon, 16 Mar 2009 15:52:55 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=420#420 420@http://orgo.curvedarrow.com/punbb Topic: Yahoo NMR problems adjacency problem Message: Yahoo!Answers in chemistry 3/11/09 wrote:How many peaks are in the signal (you have to look at which H the arrow is pointing to) ?Please help. I am really, really confused.So I have to look at the differenty groups? Something to do with quartet, singlet, doublet, triplet, etc.??? I have no idea what I might have answered and as I said, he deleted it anyway. Mon, 16 Mar 2009 15:40:42 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=419#419 419@http://orgo.curvedarrow.com/punbb Topic: Yahoo NMR problems adjacency problem Message: I feel really anal on posting the following lengthy dialog. How could I be so stupid? Yahoo!Answers in chemistry 3/10/09 wrote:Open Question: Organic Chemistry Synthesis Products and IR help please!!!! Check my work?Can someone please check my work? I am so confused about all of this stuff! Please look at where the arrow is pointed (like which H it is pointed to) and I need to know:how many peaks are in the signal for each molecule. I have no idea how to go about solving this problem. How many peaks= singlet, doublet, triplet, quarter right? I know there is some sort of N+1 rule too.I just don't know. Because this guy deleted each question and my answers, I am only guessing at what my answers were. I believe I answered something like this.A: 3 + 3 + 2 + 1 = 9B: 2 + 1 = 3C: sD: 2 + 2 + 1 = 5 Mon, 16 Mar 2009 15:38:19 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=416#416 416@http://orgo.curvedarrow.com/punbb Topic: How to deduce a structure from IR, NMR and MS spectra, Example 1 Message: This is an example of how I deduce a structure from spectra. While I have read instructions or guide on how to do this in books, I realized that was not how I did them. In class, I simply illustrated how I did them. I have had it on my objectives to write out several examples of how to do this for some time, this is one of the first examples. Interestingly, this problem is probably more difficult compared to the kinds of problems I see in textbooks or posted by students. Open the following pdf file:Example problemIRNMR, NMR insetMSThe IUPAC name for the structure is (E)-3-(4-methoxyphenyl)-1-phenylprop-2-en-1-one (a chalcone derivative). Fri, 13 Mar 2009 23:33:24 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=380#380 380@http://orgo.curvedarrow.com/punbb Topic: How many C & H signals in 4-chloro-1,1,2,2,4-pentamethylcyclopentane? Message: Basically, I have found two methods for a problem like this, one is symmetry and the other is replacement. In the upper structure, there is a plane of symmetry that passes through the CH3-C-Cl atoms and bisects the C-C bond at the back of the ring. Every atom on either side is chemically equivalent due to this symmetry. I have used colors to show the identical features.The other method, is to replace each atom with another atom and if the two atoms or attachments give the same compound, then the atoms or attachements are identical. In the example, I have replaced one of the hydrogens with a fluorine atom. If the two structures are the same, the hydrogens are identical. While less obvious, if the hydrogens attached are identical, then the carbons to which they are attached also be identical.         I arbitrarily placed the first chlorine down. The hydrogens replaced by fluorine in the center set of structures are identical as you can see by the mirror image on the right after rotation. The others are not and their names are different. A cis or trans designation would be simpler to show this. While the symmetry method is easier in this instance, in others, this replacement method is also very useful. Tue, 16 Dec 2008 19:17:46 -0500 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=376#376 376@http://orgo.curvedarrow.com/punbb Topic: What does intensity mean in C13 NMR Spectra? Message: Yahoo Answers chemistry wrote:What does intensity mean in C13 NMR Spectra? I cannot figure out what the height (intensity) of the peaks means in terms of its structure. How I could use it to identify a molecules composition from the NMR spectrum? It is identical to an integral in 1H NMR, however due to the low abundance of 13C and a lower sensitivity, some tricks are performed in 13C NMR. The relaxation times are much longer for carbon than hydrogen and the longest for tertiary carbons. Therefore, a smaller RF pulse is used for 13C NMR and a much larger number of pulses are captured. Consequently, the signals for all of the carbons can be found, but the intensity of the signals is not proportional to their number. However, there are some generalizations one can make. Carbons with hydrogens attached will show up as a larger peak than a carbon without hydrogens. If you have two similar carbons, their intensity will generally be greater than another carbon.  A quantitative 13C NMR spectrum can be obtained, however a much longer pulse delay is used so each carbon can contribute equally and the total acquisition time is much longer. Sun, 30 Nov 2008 23:19:48 -0500