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1

Two pulses are applied to your sample, with a distance $\tau$ between them, which gets incremented by $\Delta\tau$ in, lets say, 128 steps, starting at zero, or, more realistically (because you cant apply two pulses at the same time), at $\Delta\tau$. There is an FID after each first pulse, which is discarded, and one after the second pulse, which gets ...


1

There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $\alpha$ and $\beta$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that the state with $m=+\frac{1}{2}$ is referred to as the $\alpha$ state, ...


4

Thanks to the fine work by Karsten Theis the remaining task is much easier (well, it's actually possible; please read the answers by Karsten Theis which explains that the shifts reported in the OP are incorrectly referenced). The corrected chemical shifts (from his answer) are: $$\begin{align} &\delta\ 2.63 \text{ (triplet, 2 H)}\\ &\delta\ 3.83 \...


10

Dhruba Charan Dash published the problem in his analytical chemistry textbook (ISBN 9788120353008). This is the picture that goes with the question: Notice that the scale of the chemical shift increases with the strength of the magnetic field $B$. This means they are using the $\tau$ scale instead of the $\delta$ scale. The modern description of the signals ...


7

A general term used in all analytical chemistry is the lower limit of detection (LoD), which is generally defined as 3 times the standard deviation of the baseline/blank. So if you calculate the $\sigma$ standard deviation of the baseline in question, and you have a signal that is over $3 \sigma$, you can generally accept that as a peak. This comes down to ...


6

I'm writing this to show there is no solution. I'm hoping there is a flaw in my argument leading the way to a solution. The key question, as the OP already stated, is what the two 2H triplets at 6.2 and 7.4 ppm arise from. As they are the only multiplets in the spectrum, it makes sense to assume they couple with each other. If we had an image of the ...


2

It is not always a good idea to focus only on $\ce{^{13}C}$-$\mathrm{NMR}$ when you are doing structure elucidation on unknown compound using spectral data. Yes, $\ce{^{13}C}$-$\mathrm{NMR}$ gives you valuable info, so does FTIR to some extent. Yet, $\ce{^{1}H}$-$\mathrm{NMR}$ gives the most info about the compound as I depicted in following diagram: All ...


0

This is not a hard question when compared to other questions of same kind. The first easy clue is the molecular formula, which is already given. That make your life much easier. The given molecular formula for the compound is $\ce{C16H16O2}$, hence you can determine the degree of unsaturation of the molecule ($\ce{C_nH_nO_x}$): $$u= \frac{n \times 2 + 2 - m}...


2

I will ignore the IR as I always do and work only with the NMR. The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents. The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either ...


5

Structure elucidation from spectrocopic analysis needs some experience. I have feeling that you are novice to this field by your explanation of the structure. You cannot determine the molecular formula by the data given and, hence you cannot determine the degree of unsaturation of the molecule. Yet, based on $\ce{^1H}$-$\mathrm{NMR}$, you can propose there ...


4

My personal opinion is: screw IR, it does not give useful information – but NMR does. The first thing to check in the proton NMR is the integrals of your signals (I used a ruler on my PC screen – which interestingly enough seems to be good enough for this case!). You have clearly aromatic protons, one singlet at around $\pu{5 ppm}$ and one larger one at ...


3

Fraction vs Ratio If you have two species with the concentrations $c_1$ and $c_2$ and you want to know the fraction of species 2, $\chi_2$, you can to the following: $$c_\mathrm{total} = c_1 + c_2$$ $$\chi_2 = c_2 / c_\mathrm{total} = \frac{c_2}{c_1 + c_2} = \frac{1}{1 + \frac{c_1}{c_2}} = \frac{1}{1 + \mathrm{ratio}} $$ In the last part, we divided ...


3

Do you think this is the correct molecule? No, for three main reasons: As Ezze states in the comment, there are strong indications for phenyl groups in the NMR spectra. One phenyl group corresponds to 4 double bond equivalents, which is incompatible with your molecular formula (too many hydrogen atoms). The H-NMR has too few multiplets for your proposed ...


3

The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (...


5

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure. First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{...


4

Finding a structure with the correct formula that matches the chemical shifts reasonably well is not so difficult. The multiplet integrals, normalized such that the area corresponding to one $\ce{H}$ is $\approx 33$ suggests the following number of $\ce{H}$ (moving upfield): 1,1,2,2,3. The upfield singlet suggests an uncoupled terminal methyl group, so we ...


7

I like both answers provided before me where one has used exclusive use of internet to suggest structure by NMR spectrum, and the other has used thorough analysis of mass spectrum. Although these two are valuable techniques, I feel OP needs to analyse step by step analysis of spectra given to predict the structure since he/her seemingly in a graduate course, ...


10

I would probably also use the method Buck has suggested, but let’s say the NMR broke down or somebody is measuring a $\ce{^13C}$ of $\pu{2.5mg}$ meaning it will be blocked until tomorrow; in this case, we can still extract more information from the mass spectrum. In addition to the molecule peak at 122, you have: a chlorine-containing fragment $m/z=93$ a ...


10

Being an NMR fan myself I would inspect that NMR spectrum: The integrals suggest you have 11 $\ce{^1H}$ or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with $\pu{87 Da -11 Da}=\pu{76 Da}$ to explain. If you throw in an oxygen you ...


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