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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 ...
For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment: $$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$ in that the intensity $I \... 14 I am told that carbon dioxide is IR inactive. You're right, that's not true. Since carbon dioxide is linear it has$3n-5 = 4$vibrations and they are pictured below. The symmetric stretch does not result in a change (of the initially zero dipole moment), so it is ir-inactive. The asymmetric stretch does result in a change in dipole moment so it is ir-... 12 The$\ce{C=C}$stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment. If we examine the$\ce{C=C}$stretch in cis- and trans-2-butene we find that image source cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we ... 12 The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride. Source: Introduction to Spectroscopy, Pavia and Lampman You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=... 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 This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant. Let's start with IR (not my strong point) The first and second IR ... 10 No, you cannot say that larger energies will necessarily give larger frequencies. Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry)$\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates ... 10 The reference should really be to the Boltzmann distribution where the population is proportional to$\exp(-E_n/(k_\mathrm BT))$where$E_n$is the energy of the$n^{th}$energy level. This shows that the population is greatest when$n$is small. Also you need to know that absorption is proportional to the population difference between any two levels. As ... 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 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 ... 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 ... 9 The 'long chain band' described in your textbook (Pavia) is actually a vibrational mode in which the terminal methyl groups are 'rocking'. This rocking mode is observed in open chain alkanes, -(CH2)n-, where the number of methylene units is greater than 4. Polyethylene is the 'classic' examples, the IR spectra and corresponding vibrational modes are ... 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 In general there are 'fundamental' lines at frequency v_1, v_2 etc. Weak overtone transitions can occur for any fundamental vibration at frequency nv_1, nv_2 etc. where n=2, 3,.. and so on for each vibration v_i. Combination bands can also be present and involve exciting two (or more) vibrations and have frequency nv_1+mv_2, where n and m are ... 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 Have a look at typical IR bands (other source) and identify them. Have a look at you spectrum, identify your residual solvent signal. Have a look at the typical \ce{{}^{1}H-NMR} shifts table and identify the protons. What is the summary of your first conclusions? How many protons? Calculate the coupling constants via$$J=\Delta\delta_{\mathrm{ppm}}\... 7 In basic terms, for a molecule to absorb radiation there has to be an oscillating dipole being produced. This can occur by nuclear motion (vibrations, rotations) or electronic motion to produce electronically excited states. You ask about IR radiation, which does not have sufficient energy to produce electronic excitation, so this is ignored in the text ... 7 If we assume the simplyfied description, i.e. atoms being tethered together by springs, the spring constant allows describe where (energy wise, along the scale of wavenumbers, the abscissa in the specta plot) the absorption becomes observable. This however does not say a iota if the absorption band is strong, leading to a low remnant transmittance (the ... 7 To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different). IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in ... 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 Just to add on to NotEvans's answer: The stretching frequencies of compounds can theoretically be predicted by group theory and symmetry considerations - some introduction can be found here. In general, for "simple" molecules such as$\ce{CO2}$($D_{\infty\mathrm{h}}$) and$\ce{XeF4}$($D_{4\mathrm{h}}$), the bonds do not stretch individually, but rather ... 6 There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment. Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger than the typical size of a molecule ($\sim 1\mathrm{nm}$). Hence, to a very good approximation, the (time-varying) electric field of the IR radiation is ... 6 Given the general agreement surrounding my comment on the question and the similar abbreviations and descriptions in the chart on the page linked by Tyberius in his comment, the meanings appear to be: strong moderate (or medium) weak Notations such as m-w and w-s presumably indicate a peak intensity that is variable in the indicated range, depending on the ... 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 are almost certainly looking at bands corresponding to functional groups in the humic acid fraction of your sample. The band at 1750 - 1820 cm$^{-1}$is likely the$\ce{C=O}$stretching mode in$\ce{COOH}$groups in humic acid, but this band is also used to differentiate the fulvic fraction from the humic fraction (see quoted text below). The band at ... 5 Generally, the width of IR absorption peaks depends on the environment of the target molecule. Intermolecular forces alter vibrational modes. If intermolecular attractions are fairly strong (like the hydrogen bonding present in liquid ethanol), the number and strength of interactions can vary greatly from molecule to molecule, causing the vibrational modes ... 5 Your question implies that hydrogen bonds give rise to discrete peaks in the vibrational spectrum. It's better to say that hydrogen bonding affects the position of already existing peaks, in particular those vibrational modes coming from the$\ce{X-H}$. A hydrogen bonding network may also broaden low-frequency peaks coming from librations and hindered ... 5 The data given is this question has changed sufficiently from the original post that it warrants a new answer to the latest (and much more reliable) data, although should be taken in conjunction with Martin's response, and Long's original response. Let's first look at the IR data: IR Spectrum: A much better spectrum, and, compared to the original spectrum, ... 5 In spectroscopy (and spectrophotometry), the position of a peak/trough is defined by the location of the tip of the peak or trough. This is why you see such precise values given for their locations. The ranges you refer to (e.g.,$3200$to$3750\pu{ cm^{-1}}$for the$\ce{-OH}\$ stretch) are the ranges of wavenumber values where the tip of the relevant peak/...