27

A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object: $$\omega = \sqrt{\frac{k}{m}}$$ We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with ...


11

You usually post interesting questions, which appear to be deceptively simple yet they are very challenging. Modern day data acquisition and signal processing is so complicated that it is almost like a black-box. It is amusing when I ask electrical engineers some signal processing questions, they don't know the answers and when I ask mathematicians it is too ...


10

According to Skoog, Analytical Chemistry: Using classical mechanics, assuming a diatomic molecule, the frequency of vibration $\nu$ may be described by $$\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} $$ where $k$ represents the force constant of this bond, and $\mu$ the reduced mass of the particles bond together, defined as $$\mu = \frac{m_1 \cdot m_2} {...


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 ...


9

If there is a noisy signal that decays away, such as the FID in an NMR experiment, the signal to noise ratio is larger at shorter times than at longer ones where the noise remains but the signal is weaker. Multiplying the FID by a decaying exponential, i.e. an apodising function, thus suppresses the signal where the noise is larger and so increases the ...


8

Yes it is possible. While no normal modes can be both IR and Raman active in molecules with a centre of symmetry, as mentioned in a comment molecules with silent modes exist. The easiest example is probably the HCH out-of-plane twisting of ethylene: Source: http://www.shodor.org/succeed-1.0/compchem/labs/vibrations/ethylene.jpg Due to the symmetry of ...


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, ...


6

Hamming apodization function is also known as a Hamming window. If you have data that looks like this: _________ | | | | | | | | _________| |___________ because your sensor only picks up data over a certain window, when you feed that to a FFT you get a pile ...


5

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is ...


5

You should first have a look at the gas phase spectrum of water to convince yourself that vibrational levels are indeed "discrete". The textbook treatment of vibrating molecules assumes that there is no molecule nearby and each and every unit is independent. Which state of matter is closest to these assumptions? The gas phase, the lower the pressure, the ...


5

If you look at the figures above, notice that the FID ends are "square." When this is Fourier transformed, this fast drop off shows up as high frequency components since sharp changes are equivalent to high frequencies. All the apodization functions that are used drop off to zero at the edges and eliminates this artifact. The various apodization shapes used ...


5

QUESTION: Assuming that the block quote is right and it is indeed "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape" why is this considered "safe" to do? Why aren't all instrumental effects incorporated ...


4

Disclaimer: I never used Gaussian, so I am just guessing about the program. Coriolis coupling: Coriolis coupling couples bending modes with rotation and there is no bending in diatomic molecules. There is no Coriolis coupling in Gaussian for HCl, because there can't be one. The relevant output could also be NaN, it is simply not defined. Alpha "matrix": ...


4

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$. Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. ...


4

Intensity is (in general) proportional to sample concentration, so yes, if you are performing IR in solution, then an overly dilute sample would have less intense peaks. But solution IR is somewhat uncommon; I suspect you are either doing IR on your sample in neat form (using an ATR stage) or in a pressed KBr plate? If the latter, you may not have used ...


3

I think all modern teaching labs use attenuated total reflectance (ATR)-FTIR systems... you place some crystals and then collect the spectrum right away. ATR is least sensitive technique because the IR penetrates only only a few nanometers of the sample. In good old school times, one would dry the sample-literally bone dry, then oven-dried KBr was ground ...


3

Solvent molecules can have important effects on the internal vibrations in a number of ways, resulting for instance in "collisional broadening" of the spectral lines. Significant broadening can happen also in a gas if collisions are frequent due to high pressure, in which case it is not surprisingly referred to as "pressure broadening". Collisions shorten ...


3

What I was trying to say was that firstly the molecular formula has to be determined correctly. From the mass spectrum of the given molecule, the molecular ion peak ($\ce{M+}$) was determined as $\frac{m}{z} = 142$, which corresponds to the molecular formula $\ce{C9H18O}$. Note that, it can also correspond to $\ce{C10H6O}$, but in your $^1H $ NMR spectra, a ...


3

For an undergraduate starts to learn IR spectroscopy, the stretching frequency of any $A-B$ chemical bond, $\over\nu$ (in $\pu{cm-1}$) can be calculated by using following equation: $${\bar\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}= \frac{1}{2\pi c}\sqrt{\frac{k(m_A+m_B)}{(m_Am_B}}$$ where $k=\text{the force constant of the bond}=\text{bond strength}$, $m_A=...


3

I think the mistake you're making is this conclusion: "Higher atomic charge. . . -> higher electron density -> more pi-backbonding". With higher (ie more positive or less negative) atomic charge, there is less electron density donation towards the ligand, because the electron density is pulled to the metal center more strongly. Thus, $\pi$ back-bonding is ...


3

The sharpness or broadness of a stretch in IR spectra depends on extent of Hydrogen bonding present in the molecule. Basically, if it undergoes immense intermolecular hydrogen bonding, the peaks tend to be broader and the lesser the hydrogen bonding becomes, the sharper the peaks get in the spectra. Now, if you recall the criteria needed for stronger ...


2

An elder way would be the via book examples, like «Organic Structures from Spectra» by Field, Sternhell, Kalman, with dozens and dozens of combined exercises in a format like the following: and currently available in the 5th edition by Wiley (e.g., here). There is a related project of «Interactive Spectroscopy Problems» by the University of Calgary (here), ...


2

I'm collecting ideas from the excellent comments and adding some of my own: The typical UV/Vis spectrum has a narrow range On the low energy side, it ends where the visible spectrum ends, at about 800 nm. On the high energy side, it ends where water and quartz start absorbing too much, at about 220 nm. Compared to an IR spectrum, it has a narrow range. ...


2

A more by the numbers approach is this. Gather some evidence before speculating about the molecular formula. Do a first pass to see what "jumps out" of each of the spectra. From the IR spectra: The bands below 3000 are C-H indicting that this is a totally unsaturated hydrocarbon. The strong peak at 1700 indicates that there is a carbonyl group in the ...


2

This doesn't make any sense to me, is there any good reason for this discrepancy? Provided you remember the convention you are using it shouldn't matter. You could argue that either way "makes sense". If you draw it the way that you suggest is logical, then you are right, the low energy configuration will see the dipole moment vector aligned along external (...


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 ...


1

Indeed that sentence should be taken as merely anticipating what the book will treat in following chapters. Nevertheless, to satisfy your impelling curiosity the statement trace a combination of facts that leads to the actual spectrum of a given molecule. Overtones IR absorption bands are those resulting from breaking of selection rules. Instead of having ...


1

The main working rule of a good FTIR spectrum is that everything must be bone dry! The absorbance in IR is additive just like Beer's law, i.e. your spectra is the sum of water traces in your sample as well as your suspected alcohol. There is no simple way to deconvolute it. Dry your product thoroughly in a vacuum dessicator for at least a day and re-run the ...


1

Let's take the above molecules, acetyl chloride and ethyl acetate, as examples of acyl compounds with electron-withdrawing and -donating groups, respectively. We know that Cl has a higher electronegativity than O as a halogen, so the electron density from the carbonyl carbon should be pulled toward the chlorine in acetyl chloride and pushed toward the ...


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