Organic Chemistry II (Jan 24, 2020) PDF

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Organic Chemistry II (Jan 24, 2020)...

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Organic Chemistry II January 24, 2020 ______________________________________________________________________________ Recap: - We talked about basic interpretation of mass spec and found that the M+ peak is a useful indicator of the size of a molecular (molecular mass), and can sometimes be used to predict the molecular formula. - We also saw that an oddi mass can indicate the number of N - i.e., an odd number. ● In the molecule it means there will be 1 or, very uncommonly, 3 N. - (Also re: slide 15.11) We can identify the presence of Cl or Br in the structure. This comes from a particular pattern of the M+ and (M+2)+. This is indicative of a Clcontaining compound because of the natural abundance of Cl35 and Cl37. If you see this type of pattern where the M+ peak (furthest to the right in the spectrum) with an intensity of relative 3 to relative 1, then that tells you that there is a Cl atom in your unknown structure. ● This reflects the probability of, for example, say, a chlorobenzene...the probability that it contains a Cl37 (about 75% of the time) or a Cl 35 (about 25% of the time). So we’ll have a … about 4:1 in terms of peak height. ● Cl35: 77+35=112 (M+) ● Cl37: 77+37=114 (M+2)+ ● If you see this pattern in a mass spec for an unknown, you pretty much are guaranteed to have a Cl. ● E.g., (on board) bromobenzene ○ Mass: Br-79: 77+79=156 (M+) ○ Br-81 is equally abundant: 77+81=158 (M+2)+ ○ Unlike the chlorobenzene, this one will be in about a 1:! Ratio in terms of intensity. This is diagnostic of your organic molecule containing exactly one Br atom. - These are the three things to extract from a mass spec. 15.12 Analyzing the Fragments - You could explain a mass spec based on things below the mass of a molecular ion appearing in the spectrum. ● E.g., for ?? you can expect to see a series of series around M-57, M-43, M-29, and M-15. ● We won’t use this to solve the structure of molecules. It’s best to leave this kind of analysis to NMR spectroscopy. -

I will leave the rest of the slides on fragmentation for you to look at should you choose; this will not appear on the midterm or final.

15.13 High Resolution Mass Spec (HRMS) - Looking particularly at the M+ peak. ● Gives the mass of the molecular ion to 4 decimal places. ● Gives you the molecular formula of the compound. - Why is this useful? Protons/neutrons both weigh approximately 1 atomic mass unit, but they’re not exactly 1. They differ in the 3rd and 4th decimal places. Different nuclei/atoms are going to have different weights depending on whether they’re C, N or O because they contain different numbers of protons and neutrons. - C12 is exactly 12.0000 atomic mass units. - For the proton, for example, this is the exact mass of an H atom (re: slide 15-52). It weighs just a little over 1 atomic mass units, whereas C weight exactly 12, N weigh a little over ??, etc. - Because of these small differences, although these two compounds have the same molecular weight apparently. They differ slightly in the sense that this molecule formula will give a different decimal places. ● If you were to calculate for those 2 molecular formulas, the exact mass for that collection of atoms would be 84.0573 atomic mass units (amu). … - How would we use this in practice? ● I wouldn’t give you the whole spectrum because it doesn’t give you that much more information. ● If I give you a mass of, say, 120.0575, and I said this is the high-res mass you get from the mass spec of this unknown compound. You can use this to predict your molecular formula. ● Start by identifying approximately how many heteroatoms or heavy atoms you have in the structure. ● Take the base number - 120 - and assign it to molecular formula: ○ C10 would equal 120. ○ If we replace 1 C with a bunch of H, then we could have C9H12 = 120. ○ Of these two, for an organic compound anyway, the most likely one would be the latter, because it contains both C and H, and they’re both in reasonable amounts. So we can take this as one possibility to get a mass of 120. ○ Then you can also replace one of the C with an O and also come up with a reasonable molecule formula: C8HxO → C8H8O = mass of 120. ○ It can’t contain exactly 1 N, cause that would give an odd mass, we have an even mass (120). ○ Other possibilities: - C7H4O2 - C7H8N2

○ So, we have 4 reasonable guesses as to what our molecular formula would be based on the amu: - C9H12 - C8H8O - C7H4O2 - C7H8N2 ● Take the exact numbers from the chart and plug them in to this type of formula (re: slide 15-54). For each of the above molecule formulas, then, you’d generate an exact number to 4 decimal places. And it would be the same for each of them. 99% of the time, one will match the observed mass. ○ It turns out that C8H8O is what works out in this case. ○ You should use the exact mass of the most common isotope. - E.g., H1, N14, O16, etc. Spectroscopy Problems - These problems are similar to what you can expect on the midterm, with similar difficulty. - Questions on the exam will range from easy to tricky, but they will be tricky to similar extents that these examples are. - You’ll be given a sets of spectra may or may not have numbers pointed out on them. - You’re given a mass spec, and proton and a carbon NMR, and a ??. You’d be asked to determine the structure based on these sets of spectra. ● Multiplicities and integration values for peaks are provided. ● I’ll also tell you the integration values, but you should know how to do that. - This mass spec in this first example is wrong. The one on the last page is the correct one. ● All we want from this structure is the M+ peak value = 102. ● We can ignore the rest of the mass spec and just focus on the rest of the spectra to give us the structure of the final compound. - E.g. #1 ● Based on the mass (102) we know that: - There are likely no N. - Three are no Br or Cl atoms. ● Proton spectrum - They’re already labeled for you. Peak D C B A

Chemical Shift 4.2 2.4 1.3 1.1

Integration 2 2 3 3

Multiplicity quartet (q) q triplet (t) t

Assignment X-CH2-CH3 Y-CH2-CH3 CH3-CH2 CH3-CH2

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● If we don’t have the molecular formula like we have had in previous examples, this gives us an approximate number of H - in this case, about 10. ● The ?? gives a C count - in this case, 5 C. ● First ?? due to the standard, the last due to the solvent. ● So we end up with C5H10. ● We could add these up: 5 C will weigh about 60; 10 H will weigh about 10, giving a total of 70. ○ 102-70 = 32 ○ So we now have to find, in terms of the mass, 32 amu. That is likely going to be 2 times 16, which means 2 O. ● So if we had to guess, we could say that the molecular formula is C5H10O2. ● If we have this information, we might as well calculate the degree of unsaturation = 12-10/2 = 2/2 = 1. ○ 1 degree of unsaturation. ● We’ve already surmised from the C spectra that there is a carbonyl present. ● We can then work on the proton to give us the actual structure. ○ Back to the chart above to complete “assignment” column. ○ From this, it looks like we’ve so far accounted for 2 ethyl groups. So we probably have this in the structure (re: diagram on board). ● We know we have a carbonyl based on the C spectrum and our analysis of the molecular formula and degree of unsaturation. ● From this, in total, we have C5 (two ethyl groups and a carbonyl), H10, and O = C5H10O. There is one O missing - a common occurrence in these kinds of system, because no spectroscopy clearly shows ether-like linkages. If we include this in our series of fragments, then that accounts for the whole molecular formula/mass that we worked out. ● The only real possibility is this (re: diagram on board) - an ethyl prop??. ● If you double check the spectra, this would fit. E.g., #2 ● A bit more complicated; has some more complex information in it. ● This is the general strategy to use. ● You’re not given a mass spec, but you’re given a molecular formula: C10H12O2. ● First, figure out your degree of unsaturation. ○ Saturated hydrocarbon would be C10H22, and adjusted molecular formula would be C10H12, and degree of unsaturation is then equal to 22-12/2 = 5. - This is quite high for a degree of unsaturation. If you see anything above 4, you can bet there will be an aromatic ring in the structure. - There are two peaks down in the aromatic region. Two doublets that integrat for 4, giving a total of 4 H. This is disubstituted benzene in that one case. And the fact that you see a pair of doublets in the proton tells you that it’s probably para-substituted.

So, for sure we have this structure in the molecule. This corresponds to a degree of unsaturation of 4. - We know that the total is 5, so we have to account for one more. So we turn to the IR spectra. ● Nothing in the OH or ?? region. No triple bond at 2200. But there’s a strong triangular peak at 1711, so that pretty much means that you have a carbonyl in the structure, which would be equal to 1 degree of unsaturation. ○ We’ve now accounted for all degrees of unsaturation: we have a benzene ring; ?? sets of ?? protons, uncoupled to each other. ○ We can count the C - 5, then 3 in the ?? region, giving a total of 8 unique C. This suggests at least some symmetry in the structure. So we can now look forward to putting this thing together. ● Instead of doing the whole chart, we’ll look at just eh proton. We’ve already assigned these peaks to the aromatic ring. So we just need to assign A, B, and C. Peak A B C

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Chemical Shift

Integration

Multiplicity

Assignment CH3 CH2 CH3

● In summary, so far we have the presence of a paradisubstituted benzene; we have a carbonyl; we have an isolated CH2; and we have 2 isolated CH3. If you added all those fragments together, you’d come up with the correct formula. We just have to find a way to put it together that makes sense. - I will leave that up to you. - Take the formula and the fragments and see if you can come up with the right answer. Go through the rest of the examples and we will discuss any problems you have with them in a review session closer to the exam....