26

A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I’ll highlight a few. The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would ...


24

Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain ...


24

Our group uses both methods a lot so here are some examples why you would use X-ray, in addition to NMR, in organic synthesis. Your compound isn't soluble enough: Colleagues produce very large aromatic systems for organic electronics which are basically insoluble in all solvents. With a good NMR machine (600MHz and up) it might still be enough to get ...


19

Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth et al. Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz et al., and these included some rather nice low temperature 19F and 13C NMR studies in both solution and solid state (and ...


18

You are both right and wrong. At low temperatures, the NMR spectrum will indeed be as you predicted: the $\ce{H}$ from $\ce{OH}$ will produce a quartet and likewise the $\ce{H}$ from $\ce{CH_3}$ will produce a doublet. Note that at low temperatures the formation of $\ce{H}$ bonds is favored, hence leading to more stable components and significantly ...


16

You get a better resolution in the indirect dimension with an HSQC compared to an HMQC. The HSQC signal is a singlet while the HMQC is usually a multiplet as you see certain couplings on it. So generally you should prefer the HSQC over the HMQC. One advantage of the HMQC is that it uses significantly fewer pulses than the HSQC. This means that errors in the ...


16

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one ...


16

Coming from natural product chemistry, of course the structure elucidation by NMR is the most commonly used method, especially in isolation. NMR requires only a little substance dissolved in whichever deuterated solvent you have on your shelf and is rapidly set up. The only downside is that for very small amounts of sample you will be blocking the strongest ...


15

Let me begin by thanking Martin and long for supplying the key reference (J. Am. Chem. Soc. 1996, 118 (34), 7995–8005). Importantly, long also provided the isomer ratio I was asking for; according to him, the ratio was estimated using the numbers at bottom of page 7997 (with the 0.002 added). The free energy differences between the isomers of ...


15

TMS has 12 protons which are all equivalent and four carbons, which are also all equivalent. This means that it gives a single, strong signal in the spectrum, which turns out to be outside the range of most other signals, especially from organic compounds. Although the chemical shift scales are still zeroed at the TMS peak, most spectra are now calibrated ...


15

In 1H NMR, spin-spin coupling leading to multiplet structure most commonly arises due to coupling to other 1H nuclei. This is most often taught in terms of neighbouring 1H nuclei having "up" and "down" spins, both of which lead to the nucleus of interest experiencing a slightly different magnetic field, and hence having a slightly different chemical shift. ...


14

You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons. If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are ...


14

Many $\ce{-OH}$ and $\ce{-NH2}$ protons do have characteristic shifts. However, their characteristic ranges tend to be much broader than those of $\ce{C-H}$ protons. For example alcohol chemical shifts ($\delta$) range from roughly 1-5 ppm while pheonlics range from 4-8ppm and carboxylic acid protons are from 10-13 ppm or so. The ranges are 3-5 ppm in ...


14

It looks as if the NMR of morpholine is an AA′XX′ spectrum (the chemical shift difference is 0.80 ppm, or 320 Hz on your spectrometer, two orders of magnitude larger than the coupling constant). Unlike linear AA′XX′ systems where the bonds can rotate, in the morpholine it very much has a fixed conformation. In this fixed conformation, the protons are ...


13

When you perform 1H-NMR spectroscopy in solution, there are many cases where you want to use deuterated solvent, so that signals coming from solvent hydrogen nuclei don’t interfere with the signal from your target molecule. Hence the frequent use of deuterated water, deuterated acetone, deuterated methanol, and deuterated chloroform. Aprotic solvents are ...


13

Johnson and Bovey first proposed a theory describing the ring current effect,[1] and in 1972 Haigh and Mallion published their often-cited paper on ring current shielding.[2] A later review by these authors in 1979 describes the mathematical prediction of ring currents very well.[3] You may often see this graphic describing ring current contributions to ...


13

$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent. As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin > 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the ...


13

Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact ...


13

I will provide a full quantum mechanical explanation here.[1] Warning: rather MathJax heavy. Hopefully, this lends some insight into how the diagrams that long and porphyrin posted come about. Finding the states between which transitions occur The Hamiltonian for two coupled spins is $$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \frac{2\pi ...


13

There is a minimal volume you need to fill into the NMR tube to be able to get a good shim, this is around 500 microliters with regular 5mm tubes. Not using a solvent would require a lot more of your actual sample just to fill the tube. In many cases you might not even have enough of your product to fill an NMR tube. The solvent does have some effect on the ...


13

The are a number of important factors that contribute to the shielding of a nucleus. Chemical shifts arise due to differences in the local magnetic field in the different environments within a molecule, driven by overall electron density about that nucleus. That is to say, the effective magnetic field experienced by a nucleus is a function of the shielding ...


12

I might suggest that you write a short program to simulate the spectra. The problem with using real data is the amount of points (upwards of 32k) that make publishing CSV files of raw spectra unreasonable. Below is a quick attempt to give you what you are looking for. I apologize for those who do not use Mathematica; however I tried an Excel version and ...


12

In addition to mad scientists answer I would like to explain what the quantum mechanical phenomenon is that leads to a J-Coupling. The interesting part about the J-coupling for chemists is that this coupling is a proof for the presence of an electron bond between the two coupling nuclei. The coupling is based on two principles: a Fermi contact between the ...


12

I had some free time and resources at hand and did some highly accurate quantum chemical calculations for this system. I took it as a chance to compute some benchmarks and play a little with density functional theory. Originally I also wanted to compute transition barriers, but based on the previous findings I had to omit that, since I currently cannot ...


12

A little bit about me - I manage a busy research NMR Facility at a leading international tertiary institution. I deal with questions like this not infrequently, and there are some very important things you need to understand about implants and magnetic fields. I have all users of our Facility sign a waiver prior to access to indicate whether they have ...


12

Yes, with a but. First the yes: they do appear, albeit at slightly different chemical shifts as, of course, the environment around a proton in $\ce{-CH2\!\;\!-}$ is slightly different to the electronic environment around the proton in $\ce{-NH2}$. The $\ce{N-H}$, $\ce{O-H}$, and $\ce{C-H}$ bonds all have slightly different electronegativies. Now for the ...


12

I know that you asked for an intuitive explanation, but I'm afraid that if one wants to go beyond ron's comment, the technical aspects are somewhat necessary. It isn't incredibly important to understand the maths, but I have included it as I want to enable interested readers to go further, and it provides a basis for the last section, so I make no apology ...


11

The chemical shift gives you information about how well shielded the nuclei are from the magnetic field. A proton at higher chemical shift values is deshielded, so the aromatic protons are obviously less shielded than aliphatic protons. One effect that causes deshielding is the presense of electronegative atoms that draw electrons away from other atoms and ...


11

The first important point to note is that magnetically equivalent nuclei do in fact couple to each other, however no splitting is observed in the spectrum. The second point is that chemically equivalent, but magnetically non-equivalent, nuclei couple to each other, and this coupling is observable in the NMR spectrum. Spin coupling comes from a magnetic ...


11

In the 1D-proton experiment a hard 90° pulse is applied to the sample before the FID is acquired over some time, t. This set of events is known as the ‘pulse sequence’, and the resulting data is in the time domain, and needs to subjected to a Fourier transform in order to get a 1D spectrum in the frequency domain. 2D experiments are measured much in the ...


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