text:
How to identify hydrogen bonds and other non-covalent interactions from structure considerations?
Chemistry is governed by a wide range of interactions, from ionic and covalent bonding, or other types of strong interactions, towards weaker types of bonding, attraction, or repulsion, that typically occur between molecules. The latter are often referred to as chemical, molecular, or non-covalent interactions. Popular examples include ion-dipole interactions, hydrogen bonds, and London dispersion forces.
These non-covalent interactions play very important roles in DNA, protein-drug interactions, catalyst-substrate interactions, self-assembly, chemical reactions, among others.
How can I identify and analyse hydrogen bonds and other non-covalent interactions, possibly from experimental and calculated structures?

It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system.
Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on electronegativities, then infer from these how they might arrange. I will work an example based on the interactions between adenine and thymine later.[1]
With the advent of the information age, tools that every chemist has at her or his disposal have become more sophisticated. We have access to many digital resources like databases,[2] or publication servers[3] to retrieve a vast amount of information. Molecular modelling,[4] or even more sophisticated quantum chemical calculations,[5] have become more important; and free tools are available for everyone to use.
With that being said, here are some points that might help you identifying hydrogen bonds and other non-covalent interactions. For that purpose, let's have a look at our example molecules:

Immediately we can formulate a couple of assumptions based on the schematic representation. In adenine there are five nitrogen atoms, which have a higher electronegativity than carbon. Negative partial charges will therefore be located at mostly at these. Hydrogens in organic compounds usually carry positive partial charges; albeit $\ce{C-H}$ tend to be a lot less polarised than $\ce{N-H}$ bonds. Whenever a hydrogen is involved in a bond, that bond can potentially act as a hydrogen-bond donor (see below).
Similar observations can be made for thymine. Here we have two terminal oxygen atoms, which will carry a negative partial charge, since they have one of the highest electronegativities. These are often able to accept hydrogen bonds. On the other hand we also have $\ce{N-H}$ bonds, which can act as hydrogen bond donors.
Charges & Electrostatic Potential Surfaces
For many molecules structures are readily available. If not, some molecular editors give you the possibility to use implemented force fields to optimise built (guessed) structures. Based on those you can already do a few analyses. One tool that is quite powerful for various tasks is Avogadro, it let's you read crystal structures, perform basic calculations and much more. If you are just playing around, this is a really good choice.
For example, I have imported the crystal structure of adenine into Avogadro, optimised it, and calculated the electrostatic potential. Or after extracting Cartesian coordinates, Molden let's you easily calculate the charges.[6]

Hydrogen Bonds
Many molecular editors try to guess hydrogen bonds based on their implemented cutoff values. That certainly is very helpful, but not everything can be automatised in this way. And especially weaker interactions won't be found. One has to go a bit deeper then.
As a nice and concise example I have picked an intermolecular 2:1 complex between adenine and thymine, for which the crystal structure is available.[1]

There are two principle structural parameters to decide about hydrogen bonds:  (a) The distance of the hydrogen $\ce{H}$ and the hydrogen-bond acceptor $\ce{Y}$ is significantly shorter than the sum of their respective van-der-Waals radii, $d(\ce{XH\bond{...}Y})<r_\mathrm{vdW}(\ce{H})+r_\mathrm{vdW}(\ce{Y})$.[7]
(b) The angle around the hydrogen is nearly linear, $\angle(\ce{XH\bond{...}Y}) \approx 180^\circ$. For weakly polarised $\ce{XH}$ bonds, isotropic dispersion forces become more important (while the directional electrostatic and covalent contributions decrease), therefore the angle becomes more flexible.

We easily see that the bond angles are close to what we expect for hydrogen bonds. I have reproduced a few values from Batsanov's paper below, with the caveat that the value for hydrogen strongly varies depending on the chemical environment from $\pu{110 - 161 pm}$, so I used the classic from Bondi.[7c] Since all the distance are around $\pu{200 pm}$, they are well below the threshold we set earlier.
$$
\begin{array}{lr}
\text{Element }\ce{Y} & r_\mathrm{vdW}(\ce{Y})/\pu{pm}\\\hline
\ce{H}   & \approx 120\\
\ce{C}   &  196 \\
\ce{N}   &  179 \\
\ce{O}   &  171 \\\hline
\end{array}\hspace{2ex}
\begin{array}{lr}\\
\text{H-Bond }\ce{XH\bond{...}Y} & \sum r_\mathrm{vdW}(\ce{Y},\ce{H})/\pu{pm}\\\hline
\ce{CH}   &  316 \\
\ce{NH}   &  299 \\
\ce{OH}   &  291 \\\hline
\end{array}
$$
A quite interesting approach of revealing non-covalent interactions was presented by Johnson et. al., and the corresponding program is easy to use and only requires Cartesian coordinates.[8]

The surfaces between the molecules represent these interactions, where green represents weak interactions, typically found for dispersion. Blue represents stronger interactions, typically found for hydrogen-bonds. Red displays repulsive forces, typically found within ring or cage systems.
If you have access to quantum chemical software, then you can obtain this plot also for wave function files .wfn.
Another possibility is to analyse the electron density in terms of the quantum theory of atoms in molecules (QTAIM).[9] For this you do need a wave function file. The analysis, however, is straight forward and will yield a bond path or not. If there is a bond path, we can estimate the strengths of these bond with the methodology developed by Espinosa et. al.. According to this the bondstrength is approximately half the value of the potential energy density at the bond critical point.
$$E_\mathrm{H-Bond} = \frac{1}{2}V(r_{\mathrm{BCP}[\ce{XH\bond{...}Y}]})$$
I have performed such a calculation on the DF-B97D3/def2-TZVPP level of theory with Gaussian 09. The optimised geometry will be at the end.

\begin{array}{lr}
\text{H-Bond} & E_\mathrm{H-Bond}/\pu{kJ mol-1}\\\hline
\mathrm{N(37)H \cdots  O(1)}   &  46.6\\
\mathrm{N(5)H  \cdots  N(36)}  &  38.5\\
\mathrm{C(41)H \cdots  O(2)}   &   3.2\\
\mathrm{N(20)H \cdots  O(2)}   &  50.5\\
\mathrm{N(3)H  \cdots  N(24)}  &  29.9\\
\end{array}
A general warning shall be applied to the above. Absolute values of these are only approximate, but fall within the range of what is expected. A very nice side effect of this methodology is, that it can be applied to intramolecular hydrogen bonds, too.
Concluding remarks
Dispersive interactions and hydrogen bonds become more and more important in rational reaction design. Be it for understanding of molecular structure of biomolecules, or as a guiding principle for catalyst-substrate interactions.
With further development of computer technology, it should become more accessible to everyone. I hope this post demonstrates that gaining more insight can actually be quite easy (and free).

Notes and References

(a) Based on the structure from S. Chandrasekhar, T. R. Naik, S. K. Nayak, T. N. Row, Bioorg. Med. Chem. Lett. 2010, 20 (12), 3530-3533. DOI: 10.1016/j.bmcl.2010.04.131 PMID:
20493694 CSD: 739016
(b) Adenine, CSID: 185
(c) S. Mahapatra, S. K. Nayak, S. J. Prathapa, T. N. Guru Row, Cryst. Growth Des. 2008, 8 (4), 1223–1225. DOI: 10.1021/cg700743w, CSD: 652573
(d) Thymine, CSID: 1103
(e) G. Portalone, L. Bencivenni, M. Colapietro, A. Pieretti, F. Ramondo, Acta Chemica Scand. 1999, 53, 57-68. DOI: 10.3891/acta.chem.scand.53-0057, CSD: 136916
(a) The Cambridge Structural Database (CSD), https://www.ccdc.cam.ac.uk/
(b) Crystallography Open Database (COD), http://www.crystallography.net/cod/
(c) Computational Chemistry Comparison and Benchmark DataBase, http://cccbdb.nist.gov/ (Only for 1799 small molecules and atoms)
(d) Handbook of Chemistry and Physics, http://hbcponline.com/faces/contents/ContentsSearch.xhtml
(e) ...
(a) SciFinder, https://www.cas.org/products/scifinder
(b) Google Scholar, https://scholar.google.de/
(c) Web of Science, (formerly known as Web of Knowledge) http://www.webofknowledge.com/
(d) ...
(a) MolCalc, http://molcalc.org/
(b) Pitt Quantum Repository, https://pqr.pitt.edu/ (At the time of writing it was dead.) Github: pittquantum
(c) Many open source molecular editors include the possibility to use force field calculations. For example: Avogadro, molden
(d) For more on molecular modelling in the open source domain see S. Pirhadi, J. Sunseri, D. R. Koes, J. Mol. Graph. Model. 2016, 69, 127-143. An updated online version of this catalog can be found at https://opensourcemolecularmodeling.github.io.
(a) For an extensive, but not necessarily complete, list of quantum chemistry software see Wikipedia.
(b) For the purpose of this demonstration I will be using the proprietary software Gaussian.
(c) To view crystal structures, Mercury can be obtained (for free) from the Cambridge Crystallographic Data Centre (CCDC), which also hosts CSD.
https://www.ccdc.cam.ac.uk/solutions/csd-system/components/mercury/
(a) Tutorial for Avogadro (b) Tutorial for molden
(a) A concise (and as far as I can tell newest) list of van-der-Waals radii of many elements can be found in S. S, Batsanov, Inorg. Mat. 2001, 37, 871-885. DOI: 10.1023/A:1011625728803 (mirrored pdf)
(b) A list of van-der-Waals radii can also be found on Wikipedia
(c) A. Bondi, J. Phys. Chem. 1964, 68, 441-451. doi: 10.1021/j100785a001
(a) The original publication: E. R. Johnson, S Keinan, P. Mori-Sánche§, J. Contreras-García, A. J. Cohen, W. Yang, J. Am. Chem. Soc. 2010, 132, 6498-6506. DOI: 10.1021/ja100936w
(b) The presentation of the program: J. Contreras-Garcia, E. R. Johnson, S. Keinan, R. Chaudret, J-P. Piquemal, D. N. Beratan, W. Yang, J. Chem. Theory Comput. 2011, 7, 625-632. DOI: 10.1021/ct100641a
(c) Download the code: http://www.lct.jussieu.fr/pagesperso/contrera/nciplot.html
(d) You'll also need VMD (Visual Molecular Dynamics) from the University of Illinois
(a) A very brief introduction can be found on Wikipedia. The corresponding book:
Bader, Richard (1994). Atoms in Molecules: A Quantum Theory. USA: Oxford University Press. ISBN 978-0-19-855865-1. (publisher)
(b) Multiwfn - A Multifunctional Wavefunction Analyzer; http://sobereva.com/multiwfn/ corresponding paper: T. Lu, F. Chen, J. Comput. Chem. 2012, 33, 580-592. DOI: 10.1002/jcc.22885
(c) Startup script (and examples) for Linux version: https://github.com/polyluxus/runMultiwfn.bash (shameless self-plug)
(d) E. Espinosa, E. Molins and C. Lecomte, Chem. Phys. Lett., 1998, 285, 170–173.

Appendix
Optimised Structure of the adenine-thymine 2:1 complex calculated at DF-B97D3/def2-TZVPP in Gaussian 09 Rev. E.01

  45
E(RB97D3/def2TZVPP/W06) = -1388.51095169
O         19.03780       11.79565        1.63996
O         14.74303       13.36808        1.71115
N         15.07940       11.09762        1.64043
H         14.04669       10.93403        1.64128
N         16.88238       12.55083        1.67499
H         17.23771       13.53695        1.70257
C         17.83067       11.52964        1.63864
C         17.29332       10.17534        1.60048
C         15.51719       12.40225        1.67775
C         15.94232       10.03610        1.60357
H         15.46489        9.06076        1.57654
C         18.24710        9.02139        1.56030
H         18.89656        9.08263        0.68030
H         18.90669        9.02997        2.43483
H         17.71085        8.06866        1.53480
N          8.28060       10.09397        1.65692
H          7.68055        9.28386        1.63653
N          8.91986       12.24943        1.71837
N         10.48101        9.01368        1.60709
N         11.91467       12.94472        1.71757
H         12.92612       13.11883        1.71578
H         11.26613       13.71385        1.74609
C         10.03193       11.42408        1.68464
N         12.26612       10.63929        1.64378
C         11.41986       11.70069        1.68284
C          9.65939       10.07380        1.64587
C          7.89700       11.42267        1.70074
H          6.85459       11.71169        1.71748
C         11.75955        9.39245        1.60917
H         12.49648        8.59107        1.57896
N         21.22700       16.46826        1.76840
N         17.31372       17.43807        1.81488
H         16.80939       18.31038        1.84233
N         19.61652       18.27309        1.82815
C         18.69253       17.30751        1.80467
N         17.71219       15.24141        1.74956
N         20.64990       14.21957        1.70686
H         19.98577       13.44513        1.68592
H         21.63800       14.02512        1.69520
C         20.27365       15.50797        1.74532
C         16.77892       16.17095        1.78070
H         15.71578       15.97320        1.77996
C         18.91965       15.92447        1.76371
C         20.85479       17.75435        1.80728
H         21.67003       18.47621        1.82413

Doesn't NCIPLOT (ref. 8) need the wave-function to be provided in the input? You wrote it "only requires cartesian coordinates". The first link for ref. 8 has a typo: "doi.rog" -> "doi.org".

@diogom NCIPLOT can do the calculation based on standard densities, and it can do more. Also I converted your answer to a comment. I'll take care of the typo when I'm back at the computer tomorrow. Thanks for pointing it out.

Multiwfn can do the NCIPLOT part as well

@ph13 Woahhhh. I can't believe it... probably hidden behin option 134 or so...

Besides the computational methods clearly described by @Martin, you can identify hydrogen bonds using a number of experimental methods, many are indirect such as melting and boiling points, more specific are changes in absorption/fluorescence  and changes in  ir and raman spectra. Microwave spectra can be used in the gas phase.
The best methods are inelastic neutron scattering, which provides a way of measuring the potential energy surface of protons in hydrogen bonds, and x-ray and neutron diffraction. Solid state NMR can also be used.
The best of these is neutron diffraction because the H atoms  can be located  with similar accuracy to, for instance, C and N atoms. But these are difficult experiments and require using a national facility. X-ray diffraction is routinely available in many labs but the H atoms hardly scatter x-rays as they have so few electrons compared to C or N and so the position of the H atoms usually has to be inferred from the known positions of other atoms.  This is not as arbitrary as it might seem as high quality x-ray structures are routinely produced nowadays.

The question actually assumes most of what you have described as a prerequisite; the way in which you get to the structure is of no concern to it at all. Even if you have done all what you have described, you are still left with actually identifying the NCI.


text:
Classification/Nomenclature of aroma and flavour compounds - terpenes, aldehydes, and alcohols, etc
I'm collating different types of aroma compounds but the nomenclature and classification structure is confusing. How do you classify flavour compounds that can belong to different groups (e.g. terpenoids and aldehydes)?
Citral (lemony smell), for example, is a mixture of neral and geranial. Citral seems to be generally referred to as a terpene, but is a mixture of aldehydes. This previous post refers to it as an aldehyde: What happens when deodorants or perfumes expire , but is generally referred to as a terepenoid.
Similarly, Citronellal is a terpenoid, but is it not also an aldehyde?
Is there a way to prioritise the classifications of aroma/flavour compounds - i.e. does a terepenoid structure take priority over presence of an aldehyde or alcohol group?

You touched a critical point as the situation is exactly as you described.  Terpenoids have a nomenclature that's not really systematic and likely originated long time ago  without answer from someone who can put some order, if this is very important for you mention me and I can look for old lectures notes that I have somewhere. ...

@Alchimista - yes please, that would be a great start for me to begin to make sense of it!

Perhaps by classification/nomenclature you are referring to odour descriptors. If I remember correctly I have already covered a similar question about possibility of defining smell, so I will extract relevant parts here:
Odour descriptors and odour profiles
Odour descriptors are the words that come to mind when smelling a substance.
The more generally understood the words are, the more useful they are as descriptors. In practice, it is easy for any observer, after a little training, to use the standard descriptors of fragrance chemistry. Example of descriptors include musky, camphoraceous  etc
Smelling chemical groups
A fact that has, in our opinion, received too little attention from olfaction researchers is the ability of humans to detect the presence of functional groups with great reliability.
The case of thiols ($\ce{-SH}$) is familiar, but groups ($\ce{NO2}$), aldehydes ($\ce{C=O(H)}$), can be reliably identified once the odour character the functional group character confers is known. When nitriles are used as chemically stable replacement for aldehydes, they impart a metallic character to any smell: cumin nitrile smells like metallic cumin (cuminaldehyde), citronellyl nitrile smells like metallic lemongrass (citronellal), and nonadienylnitrile smells like metallic cucumber (nonadienal). Oximes give a green-camphoraceous character, isonitriles a flat metallic character of great power and unpleasantness, nitro groups a sweet-ethereal character,etc.
Here are some odour categories and their representative molecules, chosen to illustrate
structural diversity:
Musk

Musk odour descriptors might be “smooth clean, sweet and powdery”. The
  molecules that possess this odour character are exceptionally diverse
  in structure. Macrocyclic musks contain a 15-18 carbon cycle closed
  either by a carbonyl or by a lactone and smell similar but fresher and
  more natural, often with fruity overtones (cyclopentadecanolide,
  ambrettolide). Nitro musks, discovered originally as a byproduct of
  explosives chemistry, smell sweeter and are reminiscent of
  old-fashioned barbershop smells.

Representatives from five chemical classes which yield musk odors. 1 androst-16-en-3a-
ol, a steroid musk. 2: ambrettolide, a macrocyclic musk. 3: Musk Bauer, a nitro musk. 4: Tonalid, a
tetralin musk. 4: Traseolide, a indane musk.
Ambergris

Originally derived from concretions spat out by whales and aged in the
  sun, ambergris odorants smell nothing like natural ambergris tincture,
  which has a weak animalic marine smell. The smell of ambergris
  odorants was once aptly described to us by a chemist-perfumer as
  “glorified isopropanol”. Ambergris odourants provide an interesting
  combination of very closely related smells with widely different
  structures: amberketal, timberol, karanal and cedramber

Two ambergris odorants, timberol (left) and cedramber (right)
Bitter almonds
This easily-recognized category is interesting  because it includes a small molecule (HCN) which, however, is perceived by a large fraction of observers  to smell as metallic not almond-like to. Benzaldehyde, nitrobenzene,trans-2-hexenal (but see above) are good examples.

Green

Cut grass, fresh green bean notes with a sharp, almost aggressive
  feel. Diverse compounds possess this descriptor, ranging from classic
  grassy notes of cis-3-hexenol and ligustral, to the cucumber peel of
  nonadienal and the bell-pepper green note of some pyrazines.

Some examples of green odorants. Clockwise from top left cis-3-hexenol, ligustral, nonadienal and ethylmethoxypyrazine.
Camphoraceous
Camphor, cyclooctane, cineole are good
examples of camphoraceous smells, and smell rather similar to each other.

Three camphoraceous odorants: 1,8 cineole, camphor and cyclooctane
None of the above

Many other categories such as musty, spicy, aldehydic, lactonic,
  indolic, marine, etc exist, each with subdivisions. It must be
  emphasized that the odor categories above are merely convenient
  descriptors and only cover a very small fraction of odor “space”. In
  fact, particularly when one steps out of perfumery materials proper
  into smells noticed by chemists in the course of organic and inorganic
  syntheses the most frequent descriptor appears to be sui generis, i.e.
  a smell associated with nothing in particular

The complexity of structure-odour relations, and the fact that the three dimensional structure of the receptor site is unknown, make it very difficult to apply conventional quantitative structure activity relationships. For detailed SAR and theories please refer to above linked post.
References
Structure-odour relations: a modern perspective: Luca Turin et al

I am in line with common noses if I tend to smell a forest/mushroom note (not musk, something like but more fungi/mushroom plus a "gas" note ) almost all the time that a C triple bonds is around and the molecule is not very volatile?  It might be a kind of musk as for some fish eggs have that note,  and I was relating hormones to that. But I see that steroids are "musk" even without triple bond.

@Alchimista Many naturally occurring volatile chemicals that are detected through the sense of smell contain unsaturated (double or triple) carbon-carbon bonds, shape-based theories assume that the molecule assumes a conformation close to that of the macrocyclic/linear musk when binding to the receptor , hence the similarity in odor. Also given that the resolution of the human olfactory system is infinite there is always a chance of an overlap and the cut-off point is
very sharp.

At first you can collect a certain sufficient number of well known flavours or perfumes and, in case they are mixtures, identify the major components.
Than you start to locate each of them in a category or class. Forget looking for functions like alcohols, or ketones and as such. The reason is clear already from your question.
An exception can be esters. Esters in my opinion can be a class as for they occur simply as they are in many fruits and spices.
A class will surely be that of terpenes and terpenoids, etc.

I said "Etc" but right now, writing, no other possible broad classes come to my mind except aromatics.***
Contrary to the believe of many, aromatic have/had indeed connection with aroma (spice in latin) and a somewhat nice perfume (up to taste. Benzene has a nice and particular odour, at least to me and surely to those that realised that several good smelling compounds had the benzene motif in them). In other words aromatics to indicate molecules containg the benzene ring preceded the concept of aromaticity as intended in a Kekulè or QM sense.

Adapted from my notes:
Isoprenoids are unsaponificable lipids formally constituted by isoprene units.
Note that isoprene is not a biological intermediate and is not the monomer from which terpenoids are formed in nature.
(Even isoprene such as natural rubber do not derive from isoprene polymerisation).
The biological precursor parent term of these compounds is mevalonic acid
https://en.m.wikipedia.org/wiki/Mevalonic_acid
Isoprenoids are divided in:

terpenes
camphors
steroids
carotenoids and some vitamins.

Often they are not exact multiple of isoprene but rather possess functionalization.
Terpenes and camphors are often perfumed or peculiarly and strong smelling compounds extracted from plants (roots, flowers, etc.).
Few terpenes can be found in animals as well.
On the contrary, steroids are ubiquitous in animals but rarely found in vegetalia.
TERPENOIDS
Camphors and terpenes go under the name of terpenoids. Their distinction is however not sharp. Generally enough camphors are solid while terpenes are liquids prone to resinification and decomposition if exposed to air.
A well known essence is that of trementine. Already known to the ancient Egyptians is obtained from pine trees family. The resin from the trees when distilled in vapour stream gives a mixture of volatiles alpha- and beta-pinene, trementine which is used as solvent, and a solid C20 residue named colophony.
(Interesting note.  Alpha-pinene occurs in its (+) form in trees from North America, while its (-) form is in European trees).
The isoprene unit C5H8 was discovered in 1918. Its structure was clarified two years later. The  distinction between HH and HT polymers was made much later in 1953.
The classification is by

acyclic OR cyclic nature of the system
number of terpenic units (5, 10, 15 ,...)

https://en.m.wikipedia.org/wiki/Terpene
As in my notes, the link contains several example of cyclic and acyclic monoterpenes, diterpenes, and so on.
I want to mention Ascaridole as it is one of the rare natural peroxides.
https://en.m.wikipedia.org/wiki/Ascaridole
Other example worth mentioning are retinal, squalene, beta-carotene, as well as that chlorophyll contains terpenic parts.
Amongst polyterpenes a notable example is ubiquinone, involved in the last step of cellular respiration.
My notes have a passage pointing out the importance of terpenoids as flavours and perfurmes as well as the importance of their structures in mechanicists and theoretical aspects (Mirvan Wagner rearrangements of carbocations and stereochemistry).
Getting more strict on nomenclature:
As for a number of natural substances are around since long time, most have common names that are still in use (sometimes and in niche environments alongside wrong chemical structures!). As not only with terpenoids, these common names must be memorised.  Unfortunately the task is complicated by the fact that the same molecule can have different names.
Confusion arises even in systematic approach as for two numeration systems are in use, a common or traditional and the IUPAC one.
Here the notes refer to decaline numeration as examples.
Following there are numerous examples of terpenes with interesting properties but wrong names still recurring in the literature. The correct or acceptable versions is given alongside. To resume this part of the notes is a task that I cannot effort here. Unfortunately my note are in my mother tongue and I cannot post so much photographs of the xerocopies (IUPAC) attached to them.
This is why I like technology. Without it, no Chemistry SE and no way to link to https://goldbook.iupac.org/T06279.html and links therein.
Likely this resume has the only merit to show that a clear classification is complicated in itself.
I try to suggest what I would have done as a student, without trying to invent a new classification or so. If I would have to do it now and professionaly, than I would surely have input on what exactly looking for to improve an already in use classification.
Let me think Apple. I Google to find . ...... googling right now. .....
http://www.suttonelms.org.uk/apple102.html
.....  a nice link from which at least we know that several esters are responsible for the fruity part of apple flavour. I definitely choose the wrong fruit as each of the ester is described by a prevalent note of a different fruit.
It is a good starting point. The esters category of your classification contains already a good half dozen of esters.
Let me think of ginger, and I do the same as before. ...
http://www.compoundchem.com/2014/11/27/ginger/
And we now have example of terpenoid to be classified according to cyclic, number of isoprene units and details.
For instance Zingiberene is a monocyclic sesquiterpene.
You can also see molecules which could go to a class Phenols, indeed aromatic, as gingerol.
And so on.
***The above occurrence of a phenol makes me think that phenols could be a class in your classification. Not really flavours in a candy oriented sense but surely responsible for stringent and or bitter taste in green tea, wines, apples, the ginger from which the idea came from, etc.
Quite a lot of work, but I do not see how else to proceed on reasonable chemical bases.
Another possibility, inspired by the "by function (bulk support or energy stock)" classification  of polysaccharides, could be that of classifying molecules for their predominant taste, like bitter, sweet, sour, etc.
This makes sense too, and it can be even more functional depending on the interest of you as compiler. If I would have to start my own production of charms and candies, I would perhaps proceed this way. Perhaps expanding the 5 tastes (umami?) in something more detailed or restricted (fruity, pungent etc.). This can be even more chemically chaotic. But the categories will be self explanatory.
Edit just after posting. In line with the final part of my answer (a "flavour carried" based classification) is the answer by @flurbiprofen. I invite you to refer to it for such a functional classification based on gustative and or olfactory factors.

As of now, this is very cumbersome to follow. I would recommend improving formatting, and adding a common, coherently structured and connected theme. It's difficult to learn from lists of facts, especially when they seem unrelated at first glance.

For the moment I typed enough.  I will shrink it perhaps to convey the message. A classification based on chemical analogy or one based on flavour and taste. I am on a mobile and I will never invest on much formatting.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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text:
nomenclature of organic compounds?
Is it necessary for a chain of compound to have a functional group or they can be named without a functional group?I know that the longest continous chain must be selected for naming ,but I found a quote which was saying that the longest chain with afunctional group ,"if present" must be chosen to  be named it confused me about the necessity of functional group?

What would you think will happen if you try to name a compound without any functional groups (BTW, yes, such compounds do exist)? Will the world go down with error 500?

I personally not aware of such compounds and I think it would be difficult to name a compound without a functional group

On the contrary, this is by far the easiest case, hence all textbooks start with it. Are you reading yours from the middle?

No ,this topic is not included in my textbook ,I find a quote in my mcqs book

What it means is that, if you have a compound like this:

You must name it as

4-propylnonan-3-ol

taking the chain having $\ce{-OH}$ instead of

4-(1'-hydroxypropyl)nonane

To add to this answer (+1): if a compound does not have any functional group, just don't worry about its absence, and take the longest possible carbon chain.

The alternative name "4-(1'-hydroxypropyl)nonane" is not correct. Even if you would not express the $\ce{-OH}$ group as a suffix, it would still be "3-hydroxy-4-propylnonane".

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How would ceramic coating inside the reactor affect heat transfer?
With the ceramic coating, would I be able to get its internal temperature raise upto 520.
I would like to get a reactor fabricated for plastic pyrolysis. The temperature would maintained around 520C. The material processed would be Waste Polypropylene and Polyethylene.
Pressure would be slightly less than atmospheric pressure outside the vessel.
The conditions inside the vessel will be mildly acidic.
There are quite a number of alloys which seems good. (SS316, SS314)
How would ceramic coating inside the shell affect heat transfer?

I think you want a "cold shell" design. Plain old carbon steel shell ( eg. A 516 ) with a refractory lining. The shell won't be hot , maybe 200 C. You may need both erosion resistant refractory and insulating refractory.

Fluor , KBR, etc would likely be happy to design something for you.


text:
Is it possible to calculate the momentum of an electron precisely?
In a national level, undergraduate examination (TIFR-2018) following question was asked:

Which of the following is not correct with regard to the uncertainty principle?

An electron in an atom cannot be defined by a well-defined orbit.
The momentum of an electron cannot be measured exactly (correct answer)
A harmonic oscillator possesses a zero-point energy
Measurement of one variable in an atomic system can affect subsequent measurements of certain other variables

In my view answer should be (4). My explanation:
To calculate an electron's momentum precisely, it must have a totally undeterminable position! Since our apparatus have some definite size, you have to confine it in some definite space.
$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
Since we want to make $\Delta p$ to approach zero, $\Delta x$ must approach infinity, which is not possible. So we can't measure the electron's precise momentum. So option (2) shouldn't be the answer.
Moreover, as far as I know, Heisenberg's uncertainty principle restricts us from measuring two quantities(which doesn't commute each other) simultaneously. We can measure them subsequently (one after another). So option (4) should be the answer.
P.S. I know that it is not necessary that we will get the same measurement on repeating the experiment, we can get any of the eigenvalues of that observable. But this is not related to Heisenberg's uncertainty principle. It's related to observer's effect.

As orthocresol's answer suggests in his remarks on option 2, the uncertainty in position can approach infinity. We can see that this must be true by looking at the case of the free particle, which has a momentum eigenfunction as its wavefunction.

@Tyberius Just nitpicking: I'd say "can have" instead of "has" - after all, the wavefunction of the free particle can well be a superposition of momentum eigenstates. The momentum eigenstates $\psi_i(x)$ are the stationary states which satisfy the time-independent Schrodinger equation Hφ = Eφ, but any linear combination $$\Psi(x,t) = \sum_i c_i\psi_i(x)\exp(-\mathrm i E_it/\hbar)$$ will satisfy the time-dependent Schrodinger equation $\mathrm i\hbar (\partial \Psi/\partial t) = -(\hbar^2/2m)(\partial^2\Psi/\partial x^2)$. Of course, this does not affect your actual argument in any way.

To preface this answer I suggest reading a proper QM text which should address many of your issues. Griffiths' Introduction to Quantum Mechanics served me well but you may prefer any of the other excellent texts out there.
Option 4
This depends on your interpretation of quantum mechanics, but in the usual Copenhagen interpretation, the measurement of an observable causes the wavefunction of the system to collapse into the corresponding eigenstate. To give a more concrete example, if we have energy eigenstates $|1\rangle, |2\rangle, |3\rangle, \cdots$ with corresponding eigenvalues $E_1, E_2, E_3, \cdots$ then the wavefunction may be written as a linear combination of energy eigenstates. [This is guaranteed to be true because the Hamiltonian is a hermitian operator, and hermitian operators have complete sets of eigenstates.]
$$|\psi\rangle = c_1|1\rangle + c_2|2\rangle + c_3|3\rangle + \cdots$$
When you measure the energy of this system, the chance of getting the result $E_i$ is equal to $|c_i|^2$. Let's say you measure the energy to be $E_1$. Immediately after the measurement, the wavefunction of the system collapses to $|1\rangle$:
$$|\psi\rangle \overset{\text{collapse}}{\longrightarrow} |1\rangle$$
Now let's say you want to measure some other observable $A$. Again, you can express the original wavefunction $|\psi\rangle$ as a linear combination of eigenstates of $A$.
$$|\psi\rangle = a_1|A_1\rangle + a_2|A_2\rangle + a_3|A_3\rangle + \cdots$$
and the probability of measuring the eigenvalue $A_i$ would be $|a_i|^2$. However, the new wavefunction after collapse will have different coefficients in this expansion:
$$|1\rangle = a_1'|A_1\rangle + a_2'|A_2\rangle + a_3'|A_3\rangle + \cdots; \qquad a_i \neq a_i' \text{ (in general)}$$
and hence the probability of measuring the eigenvalue $A_i$ is now $|a_i'|^2$, which is in general, not the same as $|a_i|^2$.
In the absolute simplest case, we can consider measuring the energy again immediately after measuring the energy and obtaining the value $E_1$. Since the measurement caused the wavefunction of the system to collapse into $|1\rangle$, a second measurement done immediately after this will inevitably return the same value $E_1$ again.
Hence option 4 is true, which means it is not the answer you are looking for. Measurement of an observable causes wavefunction collapse, which in turn causes the probabilities of measuring other eigenvalues to change.

Option 2
First, we note that for a measurement of an observable to have no uncertainty, the system must be in an eigenstate of the relevant operator. This follows from the definition of uncertainty
$$\sigma_A^2 = \left\langle (A - \langle A\rangle)^2 \right\rangle$$
If the wavefunction of an electron is an eigenstate of the momentum operator $\hat{p}$, then a measurement of the momentum will always return the same value. In real life, if we carry out this measurement, will there be some uncertainty due to the precision of the machine? Yes, of course. Is it practical to set up an electron in a momentum eigenstate? Probably not. (By the way, your electron cannot really be constrained within your apparatus, because of quantum tunnelling. That is, unless the walls of your apparatus are an infinitely thick infinite potential energy barrier.)
But, that question is not asking you how you carry something out in the lab. The question is asking purely about theoretical considerations - and there is no fundamental theoretical problem with having an electron in a momentum eigenstate and measuring its momentum to arbitrary precision without changing its state. You are correct that its position will be indeterminate, but that is not a problem in the context of this question.
For interest, an "ideal measurement" of momentum is described in Shankar's Principles of Quantum Mechanics, 2nd ed. (p 123):

Consider a particle in a momentum eigenstate $|p\rangle$. The postulate tells us that if the momentum in this state is measured we are assured a result $p$, and that the state will be the same after the measurement (since $|\psi\rangle = |p\rangle$ is already an eigenstate of the operator $P$ in question). One way to measure the momentum of the particle is by Compton scattering, in which a photon of definite momentum bounces off the particle.
Let us assume the particle is forced to move along the $x$-axis and that we send in a right-moving photon of energy $\hbar\omega$ that bounces off the particle and returns as a left-moving photon of energy $\hbar\omega'$. (How do we know what the photon energies are? We assume we have atoms that are known to emit and absorb photons of any given energy.) Using momentum and energy conservation:
$$\begin{align}
cp' &= cp + \hbar(\omega + \omega') \\
E' &= E + \hbar(\omega - \omega')
\end{align}$$
it is now possible from this data to reconstruct the initial and final momenta of the particle:
$$\begin{align}
cp &= -\frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
cp' &= \frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
\end{align}$$
Since the photon always loses energy to the particle (as is clear in the particle rest frame) $\omega' < \omega$ and by sending $\omega \to 0$, we can make the change in momentum $p' - p$ arbitrarily small.

I have a little doubt regarding option 4 ! Question is which of the following is not correct regarding the "Heisenberg's Uncertainty principle" , how is it connected to uncertainty principle.

I would agree that it is not directly related to the uncertainty principle.

thanks for your answer. I have a personal question , Whenever I read your answers I found them really insightful. Can you please share your method of study? Which journals, books and video lecture do you prefer ? Say when you started some topic how did you read it first time ? I am an undergraduate student. How to increase my grasp on chemistry ? Many times I come across some amazing things like SF6 doesn't have d orbital involvement , I never found such thing from any of my teachers. I want to read things correctly.

@Orthocresrol can you suggest me some really insightful chemistry books , on inorganic, organic as well as physical chemistry.If you have some problem in writing here, I can share my email id too.

@AdityaShrivastav I'm flattered, but there's no particular way to study. All I will say is to (1) read widely (2) always question what you are reading. Textbooks make many simplifications (and sometimes mistakes) and it is best to always go back to primary literature, if you can understand it (I couldn't until ~3rd year of undergrad). Even primary literature can be wrong, but that is another matter - you will need a lot of specialist knowledge to question it seriously. (3) https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry

and lastly, being resourceful is a skill that will get you further than any one book will - because that's what will let you find the correct books or articles. I can't teach you how to do that - it comes with practice, imo.


text:
Can Cl(g) cause quick rusting?
I'm actually digging in something that comes from Syria...
How does Chlorine cause metal to get rusted?
It's full of pictures of rockets (allegedly filled with chlorine gas) that are all rusted. In my mind i find it pretty obvious that a strong acid like HCl can cause rust, but what's actually the precise mechanism?
Does HCL even matters oj simple Cl(g) can cause rusting?
PS: consider all the reactions taking place in open air
REPHRASED:
Is it possible for Cl(g) to cause some metal to get rusted in a matter of minutes?
What are the chemical reactions that make it possible/impossible?

Chlorine and HCl are quite different things, though I guess both cause rapid rusting.

I know, but since i knew Cl + H2O made HCL in our lungs and "corroded" them, i thought something similar may happen.

It may happen, but this is a relatively unimportant pathway, both in our lungs and on metal containers. Chlorine is toxic and corrosive in its own right, not because of HCl.

I mentioned HCL cause (studying medicine) i knew the mechanism Cl causes damage to our body: CL(g) + H2O(from our tissues) = HCL (that causes damage). Do you think that CL2 liberated in the air can cause rusting by itself?

Yes, I am pretty confident it can.

Neither hydrogen chloride ($\ce{HCl}(g)$) nor hydrochloric acid ($\ce{HCl}(aq)$) are capable of oxidising iron to the ferric state ($\ce{Fe^3+}$) but in the presence of oxygen (air) the ferrous ions initially produced can then be further oxidised to ferric ions:
$$\ce{4 Fe^2+ + O2 + 2 H2O -> 4 Fe^3+ +4 OH-}$$
For reasons explained here, that process is slow in acidic conditions, though.
Dry chlorine does oxidise iron to $\ce{Fe(III)}$ but at room temperature this is likely to be a very slow process.
However, due to the equilibrium:
$$\ce{Cl2 + H2O <=> HCl + HOCl}$$
moist chlorine is likely to be far more aggressive towards most metals including iron, because $\ce{HOCl}$ is a very strong oxidiser, easily capable of oxidising iron (or $\ce{Fe(II)}$) to $\ce{Fe(III)}$. Due to $\ce{Fe(OH)3}$'s extreme insolubility, the hydrated ferric oxide, i.e. rust, would then precipitate.

So, dry Cl would cause rusting but quite slowly. Instead the addition of water produces HOCl that is a stronger oxidizer that speeds the process up. Then we have also HCl "producing" Fe(II) that with Oxygen is further oxidized to Fe(III), thus more rust. Now, are these three processes simultaneous or one excludes others?

Also, when you say "moist", does it mean in aqueous solution achievable with something like condensation of water on cold metal or you mean "dripping wet"? THANKS

Probably all three together, depending on circumstances. Moist here means something like water vapour containing gas or air.


text:
can a thermodynamically unfavorable reaction be made favorable by a catalyst?
The question i am asking about is pertaining simply to chemical use  of a catalyst. It drives a reaction along a different path without taking part chemically. I know this fact. Is it possible so like this?

No. That would be against the energy conservation law.

Catalysts don't change thermodynamics.

The catalyst increases the rates of both the forward and reverse reactions equally, so the equilibrium point does not change.

Also a catalyst most definitely takes part in the reaction chemically.

Chemically unfavourable means non-spontaneous which means Gibbs' free energy change ($\Delta G$) is positive, i.e. $\Delta G > 0$.
But we know that $\Delta G = \Delta H - T \Delta S$. Now $\Delta H$ and $\Delta S$ (enthalpy and entropy change, respectively) are state functions. So $\Delta G$ is also a state function.
Catalysts can only change the pathway of a reaction (in most of the cases, by lowering the activation energy of the reaction ($E_\pu{a}$)) but after using the catalysts also, the energy states of the reactants and products don't alter. So $\Delta H$ and $\Delta S$ remain the same as uncatalysed reaction. So $\Delta G$ also doesn't change at all. So if the reaction is non-spontaneous, it remains like that.
That means, the thermodynamically unfavourable reaction can't be made favourable by introducing a catalyst.
But changing the temperature to a very high value can make $\Delta G < 0$, so the reaction can be made favourable by abruptly increasing temperature.


text:
Radical halogenation regioselectivity
What is the drawn product drawn here the major product? I thought the radical (intermediate) is more stabilized at a primary carbon than at a tertiary one.

Carbon-centered radicals are electron deficient species in general--you have 7 electrons instead of the desired 8. This means that to first order, you can gauge the stability of a series of radicals the same way as you would for a series of carbocations. In this case, the tertiary cation and the tertiary radical would be the most stable in their respective series. So, no, definitely don't make a primary radical when you can make a tertiary radical.

Ah thank you, that makes sense intuitively. I think I got confused because I was reading this website ( https://www.masterorganicchemistry.com/2013/04/12/a-fourth-alkene-addition-pattern-free-radical-addition/ ) on free radicals and in one of the examples, it says addition of a bromine radical has anti-markovnikov selectivity. Do you know why this might be?

The link you provide has to do with the addition of HBr in the presence of peroxides (Kharasch reaction), not the broomination of an alkane.


text:
Rationalise the Relative Reactivity of Borane and DIBAL as Reductants
In Clayden's Organic Chemistry (Second Edition) it is written that borane is useful in chemoselectively reducing the most nucleophilic carbonyl to its corresponding alcohol. It is later said that DIBAL reacts similarly and is useful in reducing esters to aldehydes, but that they will not proceed to the alcohol.
How are these differences in reactivity rationalised? I tried to draw analogies from the comparison of $\ce{LiAlH4}$ and $\ce{NaBH4}$ but since $\ce{LiAlH4}$ is the stronger reductant, I would have expected DIBAL to be the more reactive species!

DIBAL will reduce esters to alcohols. In fact with certain substrates it can be sufficiently challenging to stop at the aldehyde that the easiest option is to DIBAL all the way to the alcohol and then oxidise back up to the desired aldehyde (Swern, DMP, TPAP etc)

Then why does it seem to so often be hailed as the answer to this exact problem. From MasterOrganicChemistry.com, for example, ''Like Lindlar’s catalyst, DIBAL is most notable for what it does not do. It reduces esters, but not to alcohols – it stops at the aldehyde stage".

Because undergrad organic chemistry is full of "lies". Sad but true... DIBAL+Swern is a really common sequence and you'll see it in many total syntheses. Lindlar is pretty good but the fully hydrogenated alkane is often a byproduct and if you leave it stirring too long, you will only get the alkane. Also, borane - a Lewis acid - should react with the most nucleophilic carbonyl, not least (typo, I guess?)

Yes - edited to "most"!
But, presumably it is easier to obtain the aldehyde using DIBAL than many other reagents (the lie can't be that extensive, can it?). In which case, it is still apparently more tameable than borane and I'm still struggling to explain that...

The challenge, of course, is that aldehydes are more electrophilic than esters. Your hope here is that you can trap something at the same oxidation state as the aldehyde to prevent further reaction. I guess, with DIBAL, in some cases, the aluminum oxygen adduct may be stable enough to prevent further reaction. Alternatively, you can go with the Weinreb amide strategy, which captures the aldehyde equivalent as an imine.

You are right. My suspicion is that it depends on the conditions (temperature, solvent) and the stability of the tetrahedral intermediate (your penultimate structure in both schemes). I think there is some possibility of aggregation with the aluminium adduct so it may not necessarily be sufficient to look at a monomeric species. Anything beyond this would be pure speculation from me.


text:
Strongest hydrogen bond between neutral molecules?
It is known that very strong hydrogen bonds can occur in charged systems, such as $\ce{[F-}\ce{-H+-F-]}$, while hydrogen bonds in neutral systems are much weaker in comparison.
What is the strongest known hydrogen bond within the following specifications?

both donor and acceptor monomers are neutral, and remain neutral upon association (no ionic bridges);
heterodimers are allowed (e.g. hydrogen fluoride and ammonia);
must be a single-site hydrogen bond - excluding dimers stabilized by multiple hydrogen bonds such as nucleobase pairs.

Well, you should change "possible" to known, otherwise it's speculation.

Also this: https://chemistry.stackexchange.com/questions/47346/do-symmetric-hydrogen-bonds-in-neutral-molecules-exist is highly relevant here.

This sounds like it could be a good target for design golf...

It probably is $\ce{HF}$ and $\ce{NH3}$. It's hard to get a better donor and acceptor pair than that. It's always possible one could find some very obscure system where a hydrogen becomes quite positively charged, but for small molecules I'd be surprised if something beats $\ce{HF}$ and $\ce{NH3}$.


text:
pH calculation of HK2PO4
I am struggling with a calculation I should know how to do, but I feel like I may be missing crucial information. Here's the outline:
I dissolve $\pu{6.0 g}$ of $\ce{HK2PO4}$ in $\pu{20 mL}$ of water. I then add $\pu{8.0 mL}$ of $\pu{4 M}$ $\ce{HCl}$ and dilute to $\pu{30.0 mL}$. I take $\pu{25.0 mL}$ of the resulting solution and adjust the pH to 3.5 with $\pu{1 M}$ acetic acid.
What volume of acetic acid is theoretically necessary to reach pH 3.5?
Experimentally, I attempted this and the initial pH of the $\pu{25.0 mL}$ aliquot was 5.71. Technically, I should have been able to reduce the pH to 3.5 using less than $\pu{15.0 mL}$ of $\pu{1 M}$ acetic acid. However, I observed buffering effects around pH 4.0 and the total volume required was $\sim\pu{50.0 mL}$ of $\pu{1 N}$ acetic acid.

I don't think this calculation is as straight forward as I thought. Since this is a polyprotic acid I have to perform a speciation calculation. This results in a linear system of two mass balance equations. Please correct if I am wrong.

Thanks Vinicius Godim! That was a very thorough answer and extremely helpful. Thank you for taking the time to help me.

In complex analytical chemistry problems like this, it's useful to consider a stoichiometric matrix of the reaction network.
                               | H2PO4- | H3O+ | H3PO4 | CH3COOH | HPO4-2 | PO4-3
------------------------------ | ------ | ---- | ----- | ------- | ------ | -----
1st phosporic aciddissociation | 1      | 1    | -1    | 0       | 0      | 0
2nd phosporic dissociation     | -1     | 1    | 0     | 0       | 1      | 0
3rd phosporic dissociation     | 0      | 1    | 0     | 0       | -1     | 1
Acetic acid ionization         | 0      | 1    | 0     | -1      | 0      | 0

Each score represent the stoichiometric coefficient of that species in the given reaction, with a negative sign when it's a reactant and a positive value otherwise. I am not using $\ce{KH2PO4}$ or $\ce{HCl}$ as species because they are always fully dissociated. It's better to think of them as $\ce{H2PO4-}$ and $\ce{H3O+}$ respectively in regards to their molar concentration.
We can approach the problem by consider a "mixing $\to$ reaction" two-step process. Because equilibrium is achieved regardless of the path chosen, we can do this freely, that is, consider the "initial" concentration as though we mixed all parts together with no reaction occurring.
The mixing step mass balance will be $C_{\textrm{initial}} \times V_{\textrm{pre-mixing}} / V_{\textrm{solution}}$. Here's the full table:
                               | Solvent | H2PO4-       | H3O+         | H3PO4 | CH3COOH      | HPO4-2 | PO4-3 | CH3COO- | Total
------------------------------ | ------- | ------------ | ------------ | ----- | ------------ | ------ | ----- | ------- | ------------
Initial Concentration (mol/L)  | 0       | 2,2044883383 | 4            | 0     | 1            | 0      | 0     | 0       |
Initial volume (L)             | 0,03    | 0,02         | 0,008        | 0     | 0,5614062187 | 0      | 0     | 0       | 0,6194062187
Solution concentration (mol/L) | 0       | 0,0711806976 | 0,0516623809 | 0     | 0,9063619346 | 0      | 0     | 0

Since we are basically solving a nonlinear problem, I took the liberty to assign a random value to the initial acetic acid volume, which we will later determine by minimizing an objective function.
Let's call the stoichimetric matrix $A$, and the solution concentration (post-mixing, pre-reaction) vector $C_{\textrm{post-mix}}$. Let's define the extent of reaction vector $\delta$ as the array of values representing the amount of mols (per unit volume) produced or consumed by each reaction.
Reaction                   | Extent (mol/L)
-------------------------- | --------------------
1st phosporic dissociation | 0,0034158968
2nd phosporic dissociation | 1,46105223300036E-05
3rd phosporic dissociation | -2,218252003941E-14
Acetic acid ionization     | -0,0547767774

Once again, we do not know these values a priori. They will be optimized later on. The amounts of each reactant that is produced or consumed ($C_{\textrm{reacted}}$) is then given by the matrix product
$C_{\textrm{reacted}}^T = A^T \delta$
The equilibrium concentration is then the post-mixing concentration plus the reacted concentration,
$C_{\mathrm{eq}} = C_{\text{post-mix}} + C_{\textrm{reacted}}$
Now consider the vector of potentials of the equilibrium constants fpr each reaction ($\mathrm pK = -\log(K)$),
Reaction                       | pK
------------------------------ | -------------
1st phosporic aciddissociation | 2,1611509093
2nd phosporic dissociation     | 7,2076083105
3rd phosporic dissociation     | 12,3187587626
Acetic acid ionization         | 4,7447274949

For the equilibrium step, equaling the equilibrium expression to the $K$ of each reaction is the same as imposing the following matrix product (think of why?)
$A (\mathrm pC_{\textrm{eq}})^T = \mathrm pK$
Where the vector $\mathrm pC$ is calculed by taking the negative log of the species concentrations vector $C_{\mathrm{eq}}$. One of these values is the $\mathrm pH$, which is given by the problem.
We are then left with the task of finding the initial acetic acid volume by solving:
Find $V_\ce{CH3COOH}$ and $\delta$ (remember, this is a vector) that minimize the residuals in $A (\mathrm pC_{\mathrm{eq}})^T = \mathrm pK$ and $\mathrm pC_{\mathrm{eq}} \Bigg|_{\text{species}=\ce{H3O+}} = (\mathrm pH)_{\text{given}}$.
I used libreoffice-nlpsolver to yield
$$V_{\ce{CH3COOH}} = 0.5614062187\ \mathrm L$$
But, wait, this is the volume of acetic acid we'd employ to react with the entire solution volume. But the problem asks about only $25\ \mathrm{ml}$, it is thus just a matter of scaling the proportion using the total volume ($V_\mathrm t = 0{,}6194062187$),
$$V = 0{,}5614062187 L \times \frac{0{,}025 L}{0{,}6194062187} = 0{,}022659048 L \ \mathrm L$$
Or about $22{,}7\ \mathrm{ml}$
I purposely did not take into account the self ionization of water, because of the low pH in question. But it could have been done nonetheless. The only changes would be the need to include $\ce{OH-}$ in the stoichiometric matrix and take $\mathrm pW$ into action.

Please add the missing units.


text:
Pericyclic Reaction
Does anyone know what PhMe is used for in this reaction (how it contributes to the mechanism?)? I thought pericyclic reactions only needed heat? Thanks!

Heard about solvents?

Is there any reason for that particular one?

Again @Zhe, have you thought about converting to an answer?

PhMe is methylbenzene, better known as toluene.
Toluene is a very common nonpolar, organic solvent. As a simple aromatic solvent, it is preferred over benzene, which is known to be carcinogenic.
In your example, it's a good solvent because it (a) dissolves both reactants quite easily and (b) has a high boiling point (about $110\ ^{\circ}\mathrm{C}$). The higher boiling point allows the reaction to be heated to a higher temperature, which helps the reaction go faster.


text:
production of HCl I have a long question
SO chlorine gas bubbled through water produces HCl acid. Presumably the chlorine ions split open the Hydrogen off the H2O molecule so what happens to the OH- radical produced why doesnt that reduce the pH of that solution seeking out further created protons in solution turning them back into water again changing the ph from say 2 back to 7

Under intense light, some of what you describe may happen. But actually, $\ce{Cl2}$ will disproportionate to $\ce{Cl-}$, $\ce{H+}$ and $\ce{HOCl}$.

Thanks for contributing to be fair this was a strange fever dream idea running through my head at 4am so I appreciate sense thanks again

When dissolved in water, chlorine gives an equilibrium mixture of chlorine, hypochlorous acid ($\ce{HOCl}$), and hydrochloric acid ($\ce{HCl}$):
$$\ce{Cl2 + H2O <=> HOCl + HCl}$$
In acidic solution, the major species are $\ce{Cl2 and HOCl}$ whereas in alkaline solution, effectively only $\ce{ClO-}$ (hypochlorite ion) is present (this is chlorine bleach). Very small concentrations of $\ce{ClO2-, ClO3-, and ClO4-}$ are also found. (source: Wikipedia)

Please don't simply copy text from Wikipedia, as it might be considered plagiarism. Could you add a link to the specific article you took the text from?

https://en.wikipedia.org/wiki/Water_chlorination

I thought giving the reference disproved plagiarism.


text:
Addition Reaction
Why does the following addition reaction use radicals rather than using the normal transfer of hydrogen (seen in last picture)? I would think that because the reaction has no light to indicate homolytic cleavage to begin the free-radical process, it would not proceed this way.

The $\ce{O-O}$ bond in peroxides is weak $(\Delta H^{\circ} = 150~\pu{kJ} = 36~\pu{kcal})$. So, slight heating can break the peroxy bond. And radicals are very reactive species.

Thank you. The first image is the question (which is asking us to draw the mechanism) while the second image is the answer. However, I believe the first image does not indicate that there is a peroxide in the reaction. Is there any way to know that there is a peroxide in the reaction just from looking at the first image?

See this.

impure is there

@ApoorvPotnis Re: "\ce{O-O} bond in peroxides" as Sarah Smith noted, I don't see why you're bringing in peroxides into the question. Yes, the solution mentions diethyl ether peroxide (to which you later linked) but there doesn't seem to be any immediate connection to the question. Quoting Sarah Smith again, "Is there any way to know that there is a peroxide in the reaction just from looking at the first image?" Please consider elaborating. Thank you!

@Gaurang What I want to suggest is that  diethyl ether gets converted diethyl ether peroxide (impure $\ce{Et2O}$) and then as peroxide bond is weak, it can be broken easily and the reaction can proceed via free radical mechanism.

The point of the "impure ether" is to imply that the hydroperoxide is present and that the radical Kharasch Reaction (anti-Markovnikov) is faster than the ionic, Markovnikov process. When Kharasch used peroxide-free ether, the ionic route prevailed.


text:
Reaction mechanism for reduction of alkyl halide to alkane

Alkyl halides are reduced by zinc and dil.  $\ce{HCl}$ into alkanes
   $$\ce{ CH3Cl + H2 ->[Zn][dil\text{.} HCl]CH4 + HCl}$$

What I don't understand is the underlying mechanism behind this reaction,  is it free radical or nucleophilic substitution.  If it is the latter, then why isn't $\ce{CH3-CH3}$ a product?
I would appreciate if someone could explain the entire mechanism behind this reaction.

I was taught the reactants are alkyl halide and nascent hydrogen, along with zinc in a copper couple and acidic medium. Just an observation though.

Related: https://chemistry.stackexchange.com/q/66124/31775

It's neither. If I had to guess, I would propose oxidative addition of zinc to form $\ce{R-Zn-Cl}$ as an intermediate, followed by hydrogenolysis.

The action of HCl on Zn results in the formation of hydrogen.
Next, there is some surface chemistry where the diffusion of atomic hydrogen (or hydrogen radical) occurs with select metals. See, for example, this 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, who cites the "ingress of atomic hydrogen into the metal".
Then, the action of a hydrogen radical on an alkyl halide:
$\ce{.H + CH3Cl -> HCl + .CH3}$  (Source on radical reactions)
$\ce{.H + .CH3 -> CH4}$
I suspect more reactions are possible resulting in minor products, for example:
$\ce{.CH3 + .CH3 -> C2H6}$
$\ce{.H + CH4 -> H2 + .CH3}$
......
The final products include HCl, CH4 and other minor products.

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text:
Why is it called "bicarbonate"?
I'm helping to teach some kids about the baking soda/vinegar reaction, and was hoping to point out how the chemical names are just a direct translation of the chemical formula.
However I was surprised to find there is no "2" in the formula for "sodium bicarbonate", $\ce{NaHCO3}$.
Can someone explain where the "bi" comes in?

From Wikipedia

The prefix bi in bicarbonate comes from an outdated naming system and is based on the observation that there is twice as much carbonate ($\ce{CO3}$) per sodium in sodium bicarbonate ($\ce{NaHCO3}$) as there is in sodium carbonate ($\ce{Na2CO3}$).

The IUPAC name for it agrees with its chemical formula: sodium hydrogen carbonate.


text:
Current in a Concentration Cell
I would like to know if there is a simple function that gives the current as a function of time in a concentration cell. The current is measured in the wire connecting the two terminals of the cell. The two half cells are isolated i.e. not connected by a salt bridge and the electrolyte is a liquid.
I am looking at a very simple case without involving the factors that affect the current and the internal resistance of the cell.
From my workings, I've got $\displaystyle I=C\ln\left( \frac { A-q }{ B+q }  \right) $ or more specifically
$\displaystyle \frac { dq }{ dt } =\frac { RT }{ nrF }\ln\left( \frac { VnF{ C }_\pu{ cathode }-q }{ VnF{ C }_\pu{ anode }+q }  \right) $
where $R$ represents the gas constant
$T$ represents the absolute temperature
$n$ represents the number of moles of electrons exchanged
$r$ represents the resistance of the wire
$F$ represents the Faraday's constant
$V$ represents the Volume of the solution (assuming both electrolytes' to be same)
$C$ represents the initial concentration of the electrolyte in each half cell.
Now I've tried integrating this function using WolframAlpha and then differentiating it to obtain the current explicitly as a function of time. Seems this function is too complicated for Wolfram to evaluate.(I've used the expressions with A,B,C)
Is my derived expression correct?
Also, shouldn't a simple dynamic process have a simpler equation?
Thanks.

Yes, your derived expression looks correct to me.
As to the closed form for the integral, WolframAlpha is most likely unable to find one because it is simply not there among the elementary and special functions. This has nothing to do with the equation being simple. Of course it is simple; nature "solves" it quite easily. It is just that the nature's definition of "simple" does not necessarily coincide with ours. Helium atom is simple, so good luck trying to solve the Schrodinger equation for it. Would you argue an atom is not actually that simple? Well, your concentration cell is made up of quite a few atoms.
So it goes.


text:
Relative rate of solvolysis

I underestimated this question and just compared the carbocation stability of all the compounds (after removal of chloride via $S_N1$) and got the order D>C>B>A... but my answer is incorrect, what am I missing out?

Can't we bend the ring and bring that O closer, so as to have some intramolecular stabilization?

Is the answer A>B>D>C , if it is then I'll provide my reasoning . If it's not then  I'll wait for the question to get answered.

@AvatarShiny I don't know the correct order but C has the fastest solvolysis rate

@IvanNeretin why would we want to bring the O close that would make its -I more intense, making the carbocation further unstable (except if we bring the O at $\alpha$ wrt -Cl group...so it can be stabilized by +M)

Bringing O closer does not make its -I more intense. -I spreads along sigma bonds, roughly speaking.

yh that's true...I meant bringing O closer would make -I more intense for the carbocation(as Inductive effects diminish with distance, right) until its at $\alpha$ position to the carbocation

-I does not care about distance. What it cares about is the number of bonds between O and (+).

That's exactly what I've been trying to say...more number of bonds =more distance(not always I know)

double sigh @IvanNeretin This OP is very resistant to any hints. @ OP just read about neighbouring group participation.

@Mithoron OP? what does that mean?

It's you - original poster. BTW I have a feeling you should start you read thoroughly some good organic chem. textbook, a bit advanced one maybe. NGP (anchimeric assistance or whatever to call it) is significantly more tricky then inductive effect.

@Mithoron you have no idea how many books I've read about organic chemistry lol...I've read Jerry march, Paula bruice, peter sykes, solomons, Morrison boyd....but I just don't seem to remember...this question isn't a very tough one I know...I think

@Mithoron I guess  i'll be deleting this question and i'll find a way out on my own...now this just got on my self-respect

Well, I don't think you should delete, while "homeworky" a question about NGP should be valuable for the site IMO. You have put reasoning, but it's rather over your head. Even I can't tell outright what's the answer. You can answer yourself if you find out, though.

@Mithoron I got the answer right after you said you should read more books lol...now i'll answer my own question.

Can you tell the source of this question and the answer too(if it is provided)?

@ApoorvPotnis I got this question in my test

The rate of solvolysis is highest in (C) as a 6 membered ring is formed which is more stable than 4 & 5 membered rings.
And (B) more readily formed than (D) because the NGP ability of oxygen is more than that of an alkene(due to high charge density of oxygen).
So the correct order is C>B>D>A.

Well this question was easy and I have no idea why I couldn't possibly answer it earlier...thank you everyone for still paying attention

And here I thought to tell you to be careful with answering. This is not how NGP works; neighbouring groups stabilize cation by binding to it. Read your textbook again.


text:
Simplest way to convert glucose into water and carbon dioxide other than burning?
What is the simplest pathway from glucose to $\ce{H2O}$ and $\ce{CO2}$, such that

all reactions happen in aqueous solution with pH between 5 and 8,
temperature is in range between 0 and 35 °C,
no (expensive) enzymes are required,
the reaction is completed in around 6 hours, and
no toxic or harmful intermediates are formed and no toxic or harmful catalysts are used?

I realize that one possible answer to this would be, in biochemistry,
glycolysis → pyruvate decarboxylation → citric acid cycle → oxidative phosphorylation;
this is a fairly complex reaction cascade, however. Is there a way chemistry could simplify this in comparable (more or less physiological) conditions?
Aside from a possible (retro-)synthetic approach, there might be a catalytic option. I've found a publication by Degering and Upson* that describes iron pyrophosphate as a catalyst, but the reaction times are far higher (around 3 weeks), the temperature is relatively high and the reaction appears to be relatively messy in terms of products.
* Edward F. Degering and Fred W. Upson, J. Biol. Chem. 1931, 94, 423-431. DOI: n/a; URL: http://www.jbc.org/content/94/2/423.short (pdf) (no https available)

Eat it.$\mathstrut$

We prefer to not use MathJax in the title field, see here for details.

@Martin-マーチン huh. Complicated issue, I guess--thanks!

If you care for some more on the issue: https://chemistry.meta.stackexchange.com/q/2774/4945 It's totally okay not to care ;)

@Martin-マーチン I see that much thought has been given to this. Looks like I'm on your side, then--it seems that unless this issue is handled in this sort of minimalist way, the slug likely will come back to bite at some point.

I'd suggest you read into the Fenton process. Using an iron catalyst with hydrogen peroxide will break down (potentially) any organic matter into $\ce{CO2}$ and $\ce{H2O}$ within minutes or hours. The only drawback from your requirements is the pH, since I think this is only down in acidic solution, but you could do a little research to find out more.

Looked at it--very interesting, thanks! From the first look, it does appear like the Fenton process is a rather aggressive reaction, involving both hydrogen peroxide and two radicals. Perhaps I can learn more about it.

Converting glucose into water and carbon dioxide through a process other than burning involves cellular respiration. Cellular respiration is a metabolic pathway that cells use to extract energy from glucose and convert it into water and carbon dioxide. The process occurs in multiple stages, primarily glycolysis, the citric acid cycle (Krebs cycle), and the electron transport chain.

Your answer does not seem to contribute anything significant to a question answered almost six years ago. This looks like an AI produced answer. Actually, I have obtained a very similar response to yours from an AI chatbox.

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text:
Why does NCl3 use sp³ hybrid orbitals?
I have a question regarding the hybridization of $\ce{NCl3}$.
From what I've understood, hybridization occurs when electrons belonging to different orbitals mix.
For example an electron belonging to the orbital 's' moves to one of the 'p' orbitals creating a certain number of hybrid orbitals. This occurs when more individually occupied orbitals are needed in order to complete the bonds and form the molecule.
Now, considering the fact that in $\ce{NCl3}$, nitrogen already has three individually occupied 'p' orbitals, it doesn't need to hybridise these orbitals in order to create three bonds with chlorine.
So why does $\ce{NCl3}$ utilise sp3 hybridization?

Let us suppose that your assumption is correct and nitrogen has already $\ce{2p^3}$ configuration. Let's further assume that it doesn't undergo hybridisation. Instead, the pure atomic orbitals undergo overlap with 3 $\ce{Cl}$ atoms and the filled $\ce{2s}$ shell electron pair acts as a lone pair and thus $\ce{NCl3}$ is formed. This consideration creates problems.
If pure atomic orbitals of nitrogen, i.e. $\ce{2p_x}$, $\ce{2p_y}$ and $\ce{2p_z}$ undergo σ-bond formation with the chlorine atoms, then the bonds should be $90^\circ$ to each other, as $\ce{p_x}$, $\ce{p_y}$ and $\ce{p_z}$ orbitals are mutually perpendicular. But $\ce{NCl3}$ has 3 σ-bonds with bond angles between $\ce{Cl-N-Cl}$ close to $107.1^\circ$, which is close to the tetrahedral bond angle of $109.5^\circ$. Therefore the pure orbital undergoing bond formation doesn't support this observation.
If we consider $\ce{sp^3}$ hybrid orbitals of $\ce{NCl3}$, this tetrahedral arrangement of three σ-bonds and a lone pair (disregarding  the lone pair, it's pyramidal) is nicely explained. That's why, the hybridisation is $\ce{sp^3}$ which supports every observed chemical property of $\ce{NCl3}$.

Although the above-written answer is true, one can always ask Why doesn't $\ce{NCl3}$ form 3 $\sigma$ bonds out of the pure p orbitals? What is the reason that $\ce{NCl3}$ adopts such tetrahedral geometry?

@ApoorvPotnis I would not say that the above is correct. It assumes hybridisation prior to bond formation, which is dead wrong. Hybridisation is a mathematical concept, it only helps us understand certain aspects of the wave function. You are absolutely correct in asking for the reason of the mixing of s and p orbitals to allow more flexible bond angles, because this is essentially the reason why a description with sp³ orbitals is possible in the first place. This has partially be answered here.

@Martin-マーチン Yes, my mistake. I did not read the last paragraph properly.


text:
Covalent bond in NO3
Can anyone explain how the bonding is formed in NO3 because I can't understand how N can satisfy electron hunger of 3 oxygen atom

Resonance. Count the number of electrons on the oxygens and nitrogen in each of the resonance structures.

Please, clarify: are you talking about $\ce{NO3-}$ anion or $\ce{NO3}$ neutral radical .

@permeakra While I agree to clear up ambiguity here, for the electronic structure and the "electron hunger" it shouldn't matter that much.

Nitrogen has 5 valence atoms:
2 electrons are donated to an oxygen atom with 6 valence electrons.
2 electrons are shared with an oxygen atom to form a sigma and a pi bond.
1 electron is shared with a negatively charged oxygen atom forming a sigma bond.
Since the double bond can be anywhere, there are three possible Lewis structures. You can symbolize this by using one single drawing to show that the double bond is delocalized (that's called a resonance structure).


text:
What's the chemical reaction involved when adding Caesium to water?
According to the Notting Hill and Ealing High School, the reaction of caesium $\ce{Cs}$ with water $\ce{H2O}$ gives the following equation,
$$\ce{2Cs (s) + 2H2O (l) → 2CsOH (aq) + H2 (g)},$$
producing caesium hydroxide $\ce{CsOH}$ and hydrogen $\ce{H2}$, thus it's not hard to understand why is the reaction so violent.
But I don't really understand the reaction. Is this a single displacement reaction?
If so, why does the water break so fast into $\ce{H2}$ and $\ce{OH}$ ?
Is this related to caesium's weak electronegativity? ($0.79$ on the Pauling scale)

$\ce{Cs}$ donates a single electron to a hydrogen of water and forms $\ce{Cs+,}$ $\ce{H^\bullet}$ and $\ce{OH-}$. And then, two hydrogen atoms combine to form $\ce{H2}$ gas.

So this is a redox reaction? The Cs is reducted by the hydrogen?

The oxidation number of $\ce{Cs}$ is increased from $0$ to $+1$. So, I think $\ce{Cs}$ is oxidised. s-block metals generally function as reducing agents.


text:
How to evenly mix NaCl and Lactose
Q:
You have 10 g NaCl (s) and 90 g Lactose (s) (both easily soluble in water.) You want to achieve a 100 g mixture of these two compounds in solid form where they are completely evenly mixed. (A 1 mg sample of this mixture will always have 0.1 mg NaCl and 0.9 mg Lactose no matter from where in the mixture the sample is taken.)
Attempted A:
Boil a litre of distilled water in a clean container. In a sterile container put the pulverized NaCl and Lactose. Wait until boiled water cools to at least 50 C because high temperature increases risk of hydrolyzing lactose. While stirring with a sterile glass stir rod add the hot water step-wise until just enough water has been added to completely dissolve all the powder. Leave the solution on a warm plate (temperature shouldn't exceed 40-50 Celsius) and use a fan or similar device for airflow to speed up evaporation. Once most of water has evaporated put in a dessicator chamber to finish drying.
What I'm not sure about:
Would this actually create a completely even mixture because lactose and NaCl have slightly different solubilities so as the amount of water drops wouldn't one start precipitating earlier? If this is true how could you prevent that and are there any other issues?

Well you could freeze dry it rather than evaporate of the water

The final solid mass would be a powder of sort, right? If you grind it / mix it will be homogeneous irregardless of the selectivity of their precipitation in the prior solution.


text:
What changes does zinc chloride undergo when ‘fusing’ on an atomic level? How does this influence reactivity?
I have been using a reaction with an organozincate starting material. This zincate is freshly prepared from zinc chloride and a Grignard reagent. The experimental procedure ‘handed down’ from more senior members of the lab include fusing the zinc chloride before use. ‘Fusing’ means:

heating under high vacuum with a heat gun until the salt melted
cooling the liquid down rapidly with the cold air stream of another heat gun to receive a glass-like solid

This is then dissolved in THF and the Grignard added.
In a recent group seminar I spoke about the trouble I had in achieving consistency with that reaction. Part of the resulting discussion revolved around the method of fusing and what I had been doing wrong. The PI noted how important it is for the fusing to be performed correctly as it influences the reactivity.
What actually happens to the zinc chloride during this process? My little understanding of inorganic solid-state chemistry leads me to believe that the initial crystals break down but the resolidified glass should essentially have the same structure. I am also at a loss how to explain any differences in reactivity following this method.
While it is not the most easy task to search for such references, the concept of fused zinc chloride does exist in the literature; the following two references use it in their experimental sections:

P. Dotta, A. Magistrato, U. Rothlisberger, P. S. Pregosin, A. Albinati, Organometallics 2002, 21, 3033–3041. DOI: 10.1021/om020314q. Synthesis of the ligand 4 on page 3040.
P. G. Gassman, J. G. Schaffhausen, J. Org. Chem. 1978, 43, 3214–3223. DOI: 10.1021/jo00410a024. Synthesis of compound 28 on page 3220.

Zinc Chloride is highly hygroscopic so what you are doing by melting it under high vac is driving off the water it has absorbed. When it resolidifies I doubt its structure will be greatly different (except for the missing water)

If there's enough water around to formally form the hydrate of zinc chloride that gives Zn(OH)Cl on heating and could possibly account for the variability you're observing

@Waylander One thing to note is that our zinc chloride is stored and filled in a glove box. Not sure how much water there can still be remaining.

But if you use it without first fusing it, what happens?

@Waylander Haven’t tried that; only wrong fusing (i.e. letting it cool down slowly rather than rapidly which was identified as one of the possible points of concern.)

@Jan do a KF of your zinc chloride solution in THF, it will tell you how much water is present.  The commercial stuff has < 0.05% water, see http://www.albemarle-lithium.com/sap/pdf/T408518.PDF

The difference between rapid and slow cooling does seem to suggest some difference in crystal form

One purpose is to eliminate water as @Waylander has already pointed out.
Another is to have a clean crystal surface. Aging is a common process of crystals. The surface is transformed slowly. The crystalline structure changes since it is not the same to be surrounded with other atoms (molecules, ions) of the same composition than with nothing or a gas.
Finally, another purpose is to have a crystal size as small as possible. The smaller the crystals the bigger the surface area, so more is available and the reaction proceeds faster.
 The small crystals are obtained in the rapid cooling. The faster, the smaller.
The zinc chloride solidifies in one lump, but there are still crystals.
They only have grown together.
If it were a real glass, the reactivity would be the highest since there would be no packing.
This can only be achieved with extreme rapid cooling.
When cooling is fast but still takes some time, crystals grow.
When the cooling is slower, the crystals grow bigger.
The smaller they are, the more surface the lump has.
Melting must be done for the other reasons, but when you cool it must be done fast because, if not, you will get a less reactive mass.
It is not a comparison between molten and non molten product, but between fast or slow cooling of the molten salt.


text:
Are bi-cycles chiral molecules?
I was studying how chirality influences the diastereoselectivity of reactions when I saw 1,4-dihydro-1,4-epoxynaphthalene.

I determine chirality by looking for C with 4 different groups. I see 2 C that accomplish this requirement (sp3 C bond to the O) but both are in a rigid structure, so I can't tell if they're (or not) chiral at all. If they are, can I use the same rules to determine R or S?

"Is the mirror image superimposable on itself" is sometimes a more useful question to ask, rather than "is there a C with four different groups".

Yes you use the same rules to determine R or S. So what if the thing is rigid? It could have been rigid the other way just as well.

There are bi-cycles that are chiral, the one shown in the question is not. This can be shown, as pointed out in the comments, by superimposing the molecule on its mirror image, which is possible. Another fast way of ruling out chirality is finding the mirror plane within the molecule (in this case. perpendicular to the both cycles).


text:
What is the Italian ‘Acido Psammico’ (Psammic acid?)
I am reading an Italian detective novel — Lo stato delle anime by Giorgio Todde — in which someone is poisoned with ‘acido psammico’.

Marini trova nello stomaco della vittima un ostia con all’interno dell’acido psammico (usato per conciare le pelli), l’acido ha ucciso la donna avvelenandola.

Which I think translates into English as:

In the stomach of the victim Marini finds a communion wafer, inside of
  which is ‘psammic acid’ (used to tan hides): the acid had killed the
  woman by poisoning.

I have spent quite a time with dictionary and Google searches using the Italian, French (acide psammique) and English (psammitic acid comes up but is not defined) terms without success. The term ‘psammico’ seems to come up in a geological context, but infrequently and without clarifying things.
I should add that the novel is set in rural Sardinia in the 1890s and the implication is that shepherds would have access to it.
Does anyone know what the modern name and chemical composition of this is?
Addendum: Further Information
I am now reading further in the book (slowly, very slowly) and on p. 101 the protagonist (Marini) is in the capital city (Cagliari) and enters a chemist’s  (pharmacist’s) shop to ask for ‘acido psammico’. He receives the following reply:

La sabbiolina mortale?

In which ‘sabbiolina’ (not in my dictionaries) appears to be a diminutive related to ‘sabbia’ — sand. (‘mortale’ = ‘deadly, fatal’, of course.)
Consistent with this, further down the page we have:

La sabbia velenosa…

“The poisonous sand…”

@RaoulKessels — Yes, although I'm not quite sure whether it is arsenic or what. However thanks for your contribution which I found very interesting.

Did you get to the end? If it was As related please accept my A :)

@Alchimista — I did get to the end, and it doesn't add anything else except the social class of the people using the term. I'll give you the points by accepting your answer, but before that I will write my own, explaining why I think the author invented the term so that he could ascribe whatever properties he wanted to it.

It was more for curiosity than for points :) do not worry

@Alchimista — There were two subsequent murders in the novel, not involving poison, and the only additional fact relating to the poison was where it was purchased (a pharmacist's at Cagliari), by whom (femme fatale), for what ostensible purpose (tanning hides) and at whose request (that would be giving the plot away). Although the author has pretensions to be writing more than a ‘giallo’ (detective novel) with an interest in the backwardness of rural Sardinia, psychology and the historical embalmer, Efisio Marini, I found the plot rather a cheat. I much prefer Gianrico Carofiglio.

(Edit: this answer turned indeed to be the right one)
Italian Psammico - from the greek Psámmos for Sand - reads as "sandy". It can thus refer to a soil or the organisms living in or about sandy soils and waters, or to something derived from sands.
We as chemists know that sand is mostly silicates, so it seems difficult to think of a poison.
However if we take "sand" as to mean "from the soil, mineral, inorganic" and combine this with the fact that the most historical and literary mineral poison is Arsenic, than I would suggest that the character has been killed by As2O3 or some other compound containing this element.
Arsenic and its compounds can (were) indeed use to tan skins and leathers, to preserve wood and  timbers, and even in mummification processes.
In particular Wikipedia mentions cupper arsenochromate as a (modernly banned) ingredient in non-tannin based tanning.
At least in XIX century Italy, the word "acid" could have referred to non formally and/or non chemically acid compounds. Given the negative connotation of the word among those workers with no chemistry background, acid could have referred to any dangerous substance.  Alternatively, it could have referred as it does today, to accepted traditional nomenclature or dehydrated forms, like in the case of tannic acid or in that of As2O3, respectively.
My answer remains speculative but I think has quite reasonable bases. The novel author in this case did not use any license, just found a more cryptic and archaic but "original" name for arsenic as poison.

Very good!!! +1 We will have to read the book to see if there is some other hint. Unfortunately, for finding which acid it is, my Italian is not even basic ;-)

@Raoul Kessels. Couldn't be As2O3? Perhaps we shall look for what Agatha Christie used?  Can't Google now but later. ..

I think both are possible since the acido psammico is unknown. I just wanted to say that now I am curious from where the murderer has obtained the poison. Maybe in the book there is some hint if it came from organic or inorganic sources.

@Raoul Kessels. Correct!  When OP will finish it :))

@RaoulKessels — Alas, we are reading the book at an Italian class, 20 pages per week (with questions to discuss at the class). So it will probably be a month before I can tell you more. Sorry about that, as it's not the way SE works. I'm thinking of writing to science departments in Sardinia to see if anyone there knows. I'll update if I hear anything.

Thanks for this. One odd thing was when I Googled for "acido psammico" the Italian Wikipedia page for Arsenic came up. However the phrase did not occur anywhere in the text. Perhaps it was in the HTML header. I'll check.

@David. Type just psammico Acido psammico bring nowhere. Might be to arcaic and just in scanned materials.

@David. Ah see what you mean!  Let us check.

Can't find it scanning the source of the page. I am begining to think that it is an obsolete term for arsenic — I don't think it is likely to be Saridinian dialect with the Greek root — or an invented term so that the author could attribute the particular toxicolgical properties that fitted his plot. btw the sand reference fits with the village being between the sea and the mountains.

@David . I really think has nothing to do with Sardinian dialects. On mobile I cannot see the source of the page. Ok when you will find something more let us know.

I agree it has nothing to do with Sardinian dialect (I said I didn't think it likely and the characters are all speaking standard Italian, and my Sardinian teacher assumed it was standard Italian). I have changed 'skins' to 'hides'. Technically I think that leather is the product of tanning skins or hides (or hide if you prefer to use this alternative plural).

@David. Thanks for the english info.  I suspended that "la concia delle pelli'" is something more specific than "tanning of skins". ;

It is certainly related to sand! See my addendum. But relating that to a specific chemical? And I think that’s about all the info I will get, unless anyone would be helped by my describing the pathological effects as revealed by the autopsy…

@Alchimista Congrats!!! You were right

Actually, the meaning is:
used to tan leather.

The acid used traditionally in tanneries is tannic acid, from there its name.
Tannic acid can be found in several plants one of which is the Sicilian Sumac (Rhus coriaria). This plant is also found in Sardegna (as per Flora Italiana), where the novel is placed.
The Italian name for this shrub is Sommacco siciliano.
The description of this plant is:
piccolo albero tipico di luoghi sassosi e rupestri, con fiori giallastri in
pannocchie, le cui foglie e corteccia, ricche di tannino, si adoperano nella
concia delle pelli (fam. Anacardiacee)

[the last part says: the leaves and bark, rich in tannin, are used in the tanning of leather.]
From sommacco to psammicco there is not too much distance.
But I agree that it is speculative.
Tannic acid is toxic, but the oral LD50 in rats is 2.26±0.083 g/kg body weight, so too much to hide in an ostia.
However, that can be a writers license.
Tannic acid is a complex polyphenol with structure:

Many Thanks. Seems very probable. Will check a few things myself before accepting. Have an Italian class on Friday and the teacher is Sardinian, but no chemist.

it is indeed a very interisting question ! I need to research a bit more since my parents are from Sardinia and I'm  a chemist. In sardinia the most common plan used for the tanning of the leather was Pistacia Lentiscus (that is very abundant) I need to research a bit more to be sure ;)

checked ! I think it's correct it is indeed a very interesting topic (at least for me)

Plus 1 and bravo! for the sommaco psammico catch. However I am on a more poisonous trace as in my answer below.

The term ‘acido psammico’ is from the Greek and literally means an acid derived from sand, as @Alchimista explains in his answer. He argues that it could be a term for arsenic, and, although I do not think there is strong evidence for this, I think it fair to accept his answer, especially as my own answer is speculative. However, for the record, here it is.

I believe that ‘acido psammico’ is not a genuine Italian chemical (or psuedo-chemical) expression, nor a Sardinian dialect term, but a name invented by the author so that he could ascribe particular properties to the poison that suited his story.

My reasons for thinking this is the term is used and recognized by a pharmacist, a doctor, a midwife, a priest and a professor of medicine. One would therefore expect it to have been in common use at the time and some record of it would have been found by an internet search or in the heavy-weight Italian dictionaries I have consulted. It is further understood by the daughter of a shepherd (although she, exceptionally had a formal education) in relation to tanning of hides, again suggesting wide general use not supported by any sources I have consulted.
The author of the book — Giorgio Todde — is still alive and there is an interview with him and a presentation by him on YouTube, so it may be possible to contact him and ask him directly. I will try to follow that avenue, with the help of colleagues with better written Italian, and update this answer if I have any successs.


text:
Hamiltonian 2nd positional derivative analogous to acceleration?
In Quantum Mechanics, we learn that the Hamiltonian operator for an electron confined to a 1-D space is:

We learn in QM that many operators have analogous interpretations familiar to us from classical mechanics. My question is then: is there a classical mechanics interpretation for the second positional derivative? Is its physical significance at all related to the second positional derivative in classical mechanics (i.e. acceleration)?

It's related to kinetic energy
$$T = \frac{1}{2} mv^2 $$
which can be rewritten in terms of momentum $p$
$$ T = \frac{p^2}{2m} $$
Replacing $p$ with its quantum mechanical operator
$$ p = -i\hbar\frac{\partial}{\partial x} $$
gives
$$\hat{T} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} $$
I'm not sure why you bring up acceleration. Acceleration $a$ is the second time derivative of position, $$a = \frac{d^2 x}{d t^2}$$

Ah, I had a brain burp. I mixed up the second time derivative and the second position derivative.

If you really want a classical physics comparison, a second spatial derivative comes up in physical problems in the context of diffusion systems. The 1-D heat equation for example
$$\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}$$
This equation is fairly similar to the Shrodinger equation at a zero potential. You could even argue that the potential term is analogous to adding a source term to the equation, which would physically mean some form of heat generation within the system.
The only important difference is that $\alpha$ is a real value in the heat equation, and thus the solution to it with proper boundary conditions leads to a vanishing time derivative at infinity: in fact, the equilibrium state. The wavefunction however does not travel to equilibrium but oscillates forever. This attribute can only be achieved by having an imaginary coefficient of the second-derivative term: remember $e^{-t}$ vanishes at infinity, but $e^{-it}$ corresponds to a sine wave according to Euler's formula. And that is what you get from Schrodinger's equation.


text:
Why do these reactions not occur?
The following reactions do not occur:
$$\ce{HCl(aq)}\ +\ \ce{Pb(NO3)4(aq)} \to$$
$$\ce{HCl(aq)}\ +\ \ce{Cu(NO3)2(aq)} \to$$
Is this because in both cases both reactants are entirely dissolved as their components ion?
$\ce{NO3}$ is a conjugate base of a strong acid  so it is a very weak base so it doesn't participate in reactions, right?

Check the subscripts on the $\ce{Pb}$ compound, $\ce{Pb^{2+}}$ seems more likely. It depends on concentrations, but $\ce{PbCl2}$ should precipitate.

@TAR86 I checked and it is 4

When put in water a mixture such as $\ce{HCl}$ and $\ce{Pb(NO3)4}$ goes to complete ionic dissociation. Lead nitrate is soluble in water and hydrochloric acid is a very strong acid. Nitric acid is also strong, which means those dissolved nitrate ions are not reversing back to the molecular form.
Their ions will be fairly stable in solution and will only combine in case a more stable compound is formed. You could come up with reactions in which some compounds will go out of solution (like $\ce{Cl2}$), such as $\ce{ 2HCl + HNO3 \to HNO2 + Cl2 + H2O} $. Lead could also act as an oxidizing agent. But I would argue that if you consider the change in free energy or the net redox potential of those reactions, you'd find that they are quite unfavourable. If you want to calculate it itself, you could start from a reduction potential table like this one.

$\ce{Pb}(\rm IV)$ can definitely oxidize $\ce{Cl-}$ to $\ce{Cl2}$.

OK! I definitely did not look up at the redox potential of each possible reaction. But the overall advice as to how to approach the problem is still valid.

I agree, but there are more to consider besides redox. For example, as written in the comment, $\ce{PbCl2}$ precipitates.

Isn't the lead in $\ce{PbCl2}$ reduced with respect to $\ce{Pb(NO3)4}$?

Yes. It is, and it actually helps the redox reaction to go in the forward direction.


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Xylan vs. Teflon for non-stick food pans?
I am looking at non-stick pans and some are Xylan® others, and most, are Teflon™.

What are the chemical differences?
What is better for cooking?
What is better for health safety?

I do know that Teflon can be consumed without a problem, as it is never degraded, but at very high temperatures the fumes are dangerous.

Welcome to Chemistry! Take the tour to get familiar with this site.

Xylan and Teflon are basically the same thing but are trademarks belonging to different companies for their specific products.

Both Xylan and Teflon are chemically polytetrafluoroethylene. But a polymer just by name can mean a lot of different things if you consider the structure/arrangement, additives and impurities. These are highly process-dependent and, guess what, we do not know their chemistry for sure because they are industrially protected. For actual matters, both Xylan and Teflon are unsafe above 260 ⁰C (500 ⁰F), when they start to break down to toxic fluorine polymer fumes.


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Ozonolysis Reaction
Why does the starting compound not need an -OH group? I thought ozonolysis only produced carbonyls?

The product of ozonolysis depends on how you work up the intermediate ozonide. You are clearly familiar with the reductive workup (with dimethyl sulfide or PPh3) that gives two carbonyl compounds, aldehyde or ketone. But if you work up under oxidative conditions, as is the case here, you get carboxylic acids from mono-substituted double bonds and ketones from di-substituted.

Thank you! What do you mean by oxidative conditions?

Conditions that oxidise the ozonide, such as hydrogen peroxide as used here, chromate, permanganate or other oxidising agents. Any aldehyde formed would be oxidised to carboxylic acid under these conditions. Google 'Criegee Mechanism' for further information