16

When looking at IR spectra of hydrogen-bonding groups, always check how IR spectra was recorded. It appears that NIST spectra is recorded using gas phase, while the first one used liquid film. In liquid film $\ce{NH2}$ groups form a network of hydrogen bonds, resulting in shape distortion and widening of $\ce{NH2}$-related bands. Same is true for carboxylic ...


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 ...


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

Honestly I have doubt about your textbook version of IR spectrum. However, I can suggest by the experience I have gained that the shapes of the peaks depends on the method you have used to obtain the spectra such as ATR, smear, absorbed in inert surface, etc., thus cannot pinpoint which one is incorrect. Yet, you can compare an one with other related amines. ...


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

The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double ...


4

Yes, you are right. If you have $n$ components in a mixture each with a spectrum $S_1$, $S_2$,...,$S_n$, the observed spectrum $S_{obsvd}$ would be the sum of $S_1$+ $S_2$+...+$S_n$. In FTIR you either measure the transmittance or the absorbance. The term intensity is more or less used when you deal with molecular or atomic emission of light (fluorescence ...


4

I think your main question is related to the difference? Have you heard a famous poem, The Blind Men and the Elephant? It is a story, I'd rather copy from Wikipedia It is a story of a group of blind men who have never come across an elephant before and who learn and conceptualize what the elephant is like by touching it. Each blind man feels a different ...


4

As EdV said, when you have aqueous samples, Raman spectroscopy is the way to go. The problem is not the container (which should not be glass etc),the problem is water. Its absorption will hide the absorption of everything. If you ever open an FTIR instruments, you will see that there is nothing which is made of glass or quartz because they absorb too ...


3

There is a great difference between sensitivity, stability and accuracy. Neurons have great sensitivity to stimuli, i.e., detecting minuscule amounts, but neurons are not very accurate, i.e., they don't repeatably show the same response to a given stimulus, and are subject to synaptic fatigue. Given a large number of neurons in a very tightly controlled ...


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

Looks like it has an IR spectra to me: Source: https://webbook.nist.gov/cgi/cbook.cgi?ID=C74862&Type=IR-SPEC&Index=1


1

The table below illustrates a possible screen, values computed displayed are rounded to four decimals. The suggest is to use the observations prior the intended irradiation ($x_1, x_2$) separate from the observations after the intended irradiation ($y_1, y_2$). Per wavelength compute for $x$ and $y$ the arithmetical mean value determine the standard ...


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 ...


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