spin-1/2-nuclei

Hi... Unfortunately, the variations you've reported are too much to say that your compound is consistent with data observed in literature. Changing the MHz from 500 to 400 should not effect the center of the peaks. Your signal to noise and resolution should improve, but the center of the peaks should not change, therefore they should be at the same ppm. Is it possible that you have a metal contaminant? Shimming should not change the ppm of your peaks, bad shimming can cause them to be mishapen, so you likely have a sample related problem. Another good way to check this is to see if the peaks for a standard sample are also off. If that is the case you might need to ask the NMR technician for help.

Hi, I think the other posts did a very good job of explaining how a catalyst works and you can easily apply that to organic reactions. Hopefully I'm understanding you correctly, but I think you might be asking if catalysts behave differently in organic reactions? If that is your question then it can't be generalized in this way. Catalysts are typically very specific for a certain type of reaction like there are specific catalysts for things like Suzuki couplings (a named reaction in organic chemistry) and there are catalysts for polymerization reactions (like Ziegler Natta). Both of those catalysts are different and cannot be compared in the method I believe you might be hoping to compare all catalysts. I guess a good way to demonstrate this would be to say that you cannot use the palladium catalyst in the suzuki reaction to do the polymerization reaction that requires the Ziegler Natta, they are almost never interchangeable. A lot of extremely hard work and research goes into designing reaction specific catalysts as result. Someone can have an entire career working on catalysts for one specific type of reaction. Sorry I couldn't be of more help.

Hi... I really don't think you should try this.. I certainly can't speak for other people, but I personally wouldn't feel comfortable telling someone how to make this. Chemistry can be fun but can also be very hazardous and proper training is required to do it safely. Without proper equipment and training you cannot do this safely moreover if you don't know the percent of the ammonium hydroxide you cannot get your equivalents right. This is an acid base reaction they can be very dangerous and this reaction will get very hot.

Hello, hopefully I'm understanding you correctly, if not let me know and I'll try to answer your question. Are you talking about the amine protons in the final product? Those are stable to base they have a pka of approximately 38 or so in the presence of the carboxylate because making dianions becomes more difficult. Here is a link to a good pka table. This is hands down the best pka table to have in my opinion. - http://docs.google.com/viewer?a=v&q=cache:UY6cBoMV640J:evans.harvard.edu/pdf/evans_pKa_table.pdf+evans+pka+table&hl=en&gl=us&pid=bl&srcid=ADGEESiNn7DVNlwVhRq6cGOxdfWQAHsjnCcQ6QZpJ4GvbQZB0Y7H_XIaxQXxllxBGCZ0d-Z3Dg0KCx7HlQq9zvrLmMkERPhGTn40Zp2ACSM2mXTTlFcjp2EA0jw6KUraN6GEmJ8pfcn5&sig=AHIEtbQ7OPFAjGYUC34QH5JZ5rH-m4vRlQ

Hi... I'll do my best to take a stab at this question. the Br2 will add trans across the double bond. So you will get a total of 4 products if your stereocenter at carbon 3 is racemic. You will get 2 products if it is not. The 2^n rule applies here, where n is the number of stereogenic centers... in this case because your reaction is facially selective you only get 4 possible products instead of 8. You will get 2 enantiomers and 2 diasteriomers here. The "absolute stereochemistry" will be different for each of the 4 molecules, however the relative stereochemistry for the bromines of carbons 1 and 2 will be the same for each of the 4 molecules. This is because the reaction is facially selective therefore you get only trans addition across the double bond. If the addition of Br2 was not facially selective then you would not have any relative stereochemistry. hopefully this was useful. edit: I should clarify that the relative stereochemistry is the same for all of the 4 possible products in the sense that we can know for sure that the bromines in all 4 products on carbons 1 and 2 will be trans to each other. What we can't know is if they added from the top face or the bottom face - therefore you get 2 products from addition to the top and 2 products from addition to the bottom - thus always trans to each other but we cannot know if the center formed is R or S... absolute stereochemistry is knowing that carbon 1 is S carbon 2 is R carbon 3 is S and so forth. Typically you must have a chiral reagent to introduce absolute stereochemistry into the product or you must have a way to separate each enantiomer and disasteriomer from each other. I should also add that because the center at carbon 3 is undefined - this is why you will get 2 products from addition to the top face and 2 products from addition to the bottom face. This is probably called something like Re face and Si face in your notes.. but if carbon 3 were defined you would just have 2 products 1 from the Re side and one from the Si side. So a helpful hint is to always remember to divide the possible number of products by 2 for each stereocenter that is already fixed and by 2 if you have a controlled reaction (i.e.) facially selective etc...

Hi lets see if we can walk through how to solve this. The first thing you want to do is look at all of your data and compare one diagnostic technique to the other, so lets see what that tells us. 1. You have no signal in the IR, which could be because you have no IR active functional groups (not likely) - or - the ones that are present have high level of symmetry. 2. you have only one signal in the proton and only one signal in the carbon - also suggesting a high level of symmetry. 3. you have a singlet at 3.7ppm in the proton and signal at 67ppm in the carbon - this suggests an oxygenated carbon with symmetrical hydrogens on it. 4. your mass spec data and the high level of symmetry suggests that you probably have more than 1 oxygenated carbon. This is because of the mw of 88. 5. Now lets look at your peaks in the mass spec. The loss of 30 from 88 and 58 suggests that you have a CH2O fragment. The loss of 28 suggests a CO fragment. 6. Now you have to ask yourself how many CH2 and O fragments can you get into 88. 7. Therefore the data suggests a dioxane. However, since this is a homework assignment I will not tell you which dioxane. You can figure this out from your data though. Remember that you have no splitting in the proton NMR, what does this tell you about the hydrogens and why? you can ask more questions if you still need help. hopefully this was useful.

Hello. I'm sorry I don't have a figure to link you to so I will describe it with words. hopefully this will be of some use to you. Step 1: hydroxide attacks one of the carbonyls - (it doesn't matter which because they are symmetrical) - and one of the bonds between the carbon of the carbonyl and the oxygen of the carbonyl moves to the oxygen to make a negatively charged oxygen. Step 2: one of the lone pairs on the negative oxygen swings down to reform the carbonyl and the bond between that carbon and the nitrogen goes to the nitrogen (breaking open the five membered ring) - and the lone pair of the nitrogen removes the hydrogen from the "tertiary alcohol" formed in this intermediate. *this isn't done by a water molecule because of pka(s) Step 3: the lone pair of the primary amide now attacks a the Br2 and a Br- is kicked out. Step 4: another hydroxide deprotonates the positively charged nitrogen. Step 5: the Hofmann rearrangement occurs: another hydroxide attacks the reamaining hydrogen on the nitrogen, that bond goes to the carbon bearing the amide on the benzene ring, the bond between the benzene carbon and the amide goes to form a double between the nitrogen and the carbon of the amide carbonyl. Bromide is kicked out as well during this process. Step 6: another hydroxide attacks the carbon of the isocyanate - (hopefully that is what it's called?) - and the bond between the carbon of the isocyanate and the nitrogen deprotonates a water molecule to give a hydrogen to the nitrogen. Step 7: a decarboxylation occurs: the lone pair of the nitrogen attacks the hydrogen of the "tertiary alcohol" of the isocyanate, the bond between the hydrogen and the oxygen goes to the carbon of the isocyante creating an additional CO double bond, and the bond between the carbon of the isocynate goes to the nitrogen to produce a new lone pair. step 8: you now have the primary amine and the carboxylate species in the 1 and 2 position of the benzene ring. H3O+ (will have to be added) - and is needed to protonate the carboxylate because water is not a strong enough acid, or you should draw your final product with just the carboxylate... hopefully this was helpful.