Where do the lone pairs go in the "true" resonance structure? Consider the "true" resonance structure of $\ce{CO3^2-}$:

The bottom diagram leaves out the 2 lone pairs, so 4 electrons disappeared in the bottom image as compared to the top one. Are these disappeared lone pairs partial electrons, or shared? To make the bonds in $\ce{CO3^2-}$ correct, doesn't that mean each dotted line is worth double, so each bond is $1.33$ instead of $1.66$? Why don't we draw a $1.33$ electron lone pair if that's the case?

Image source: Resonance, by  Sharon Wei (UCD) and Liza Chu (UCD), Chemistry LibreTexts.

Resonance is a concept resulting from valence bond (VB) theory. Within this theory it is impossible to describe delocalisation. However, every configuration within this framework can be expressed as a Lewis structure. The congruent molecular orbital (MO) theory describes bonding in a delocalised way, therefore it is impossible to express it as a single Lewis structure.

Often we can find Lewis-like structures representing a resonance hybrid. For example from my own questions:

While it is simply impossible to write a valid Lewis-structure (localised electrons) and imply delocalisation, it is often used to denote how a molecule is "special" in a way. We make many assumptions here, and imply that everybody knows that all the electrons are delocalised. Therefore I omitted all electrons, implying the reader knows where to put them.

What you describe as the "true" resonance structure, is an attempt in the same manner to visualise a resonance hybrid (superposition of multiple valid Lewis structures) in a single Lewis-like fashion. The resonance hybrid, like it is used in LibreText, is an insufficient and inaccurate depiction. It is a futile attempt to jumble together VB and MO theory. Even though they are, at their respective theoretical limit equivalent, i.e. congruent theories, at that very limit they become so complex, that a visual depiction is no longer possible.

In this spirit, these depictions of resonance hybrids are overused in the same manner that resonance is often taught wrong and is an overused concept. (Unfortunately the linked LibreTexts article is point in case.) It is important to understand that these simple concepts have their limitations, and to always question the validity.
As you notice yourself in your question, the depiction of the resonance hybrid omits electrons, you have to be cautious to use these simplifications.

On a related, but slightly different note, I especially dislike the article you linked to on LibreTexts. Just to name a few errors:

it omits formal charges in the ozone example;
the delocalisation and resonance structure rules are complete made up, incomplete (1) or even wrong (4, 5);
in the concept of resonance it is inherently wrong to speak of the "correct"Lewis structure;
rules for estimating stability of resonance structures is complete garbage, there is no such thing as a most stable resonance structure;
Lewis dot structure for sulfate uses $\ce{S-O}$ double bonds, which is completely wrong;
it is not generally true, that the structure with fewer formal charges has a higher contribution to the wave function that one that has more charge separation;
etc..


Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution? Why is phenol more reactive towards electrophilic substitution than methoxybenzene? Isn't the lone pair on the methoxy group oxygen more available for participating in resonance into the ring than the lone pair on hydroxyl oxygen due to the +I effect of the methyl group? Or is it because of the steric hindrance of the methoxy group that will favour lesser substitution at the ortho- position of the benzene ring?

Does the hydrogen bonding of phenol have anything to do with this?

First of all we must have to note that the difference in reactivity between anisole (methoxybenzene) and phenol is small. Small enough for the undergraduate texts I own to not mention anisole separately from phenol, just referring to it as having "phenol-like reactivity" (i.e. strong ortho-para directing). We can more or less measure the difference in ortho-para directing strength by looking at the proton NMR spectra of anisole and phenol.

Source.

Source.

As you can see they are almost identical to the point where you can ask yourself if there is any practical difference in reactivity at all. The A protons nicely reflect the ortho-para direction since they haven't shifted from benzene's 7.26 ppm signal and the B and C protons are all upfield as expected.

Also I don't think steric hindrance is a big factor either since aromatic electrophilic substitution goes first trough a $\pi$-intermediate and a single methyl-group is not that bulky.


Why is the expectation value of Hermitian conjugate operators $RR^\dagger$ always real and non-negative? I've been reading through a derivation of the wavefunctions and energy levels for the quantum harmonic oscillator. It defines

$$\hat R^\pm=\frac{1}{\sqrt{2}}[\hat p \pm \mathrm{i}\omega \hat q]$$
in mass-weighted coordinates $q=x\sqrt{\mu},$ such that $\hat p= -\mathrm{i}\hbar(\mathrm d/\mathrm dq)$ and $\hat q$ is the position operator.

It then asserts that that $\langle \psi|\hat R^+\hat R^-|\psi\rangle$ must be real and non-negative, since $\hat R^+$ and $\hat R^-$ are hermitian conjuagtes. I know that $(\hat A\hat B)^\dagger = \hat B^\dagger \hat A^\dagger$ and so $\hat R^+\hat R^-$ is a hermitian operator meaning that the bra-ket must be real, but I can't see why it should be non-negative. Could anyone shed some light on this?

Let $\hat{R}^-|\psi\rangle = |\psi'\rangle$ (a new ket; we don't care what it is). If you take the adjoint of this equation you get

$$\langle\psi'| = \langle\psi|(\hat{R}^-)^\dagger = \langle\psi|\hat{R}^+$$

and hence

$$\langle\psi|\hat{R}^+\hat{R}^-|\psi\rangle = \langle\psi'|\psi'\rangle$$

which by definition of the inner product must be real and non-negative.


Which functional group gets reduced first? The following is a reduction reaction. In the reduction of 2 to 3, keto group undergoes reduction by nucleophillic addition.

Ester could undergo nucleophillic acyl substitution to give an aldo group, further reduced via nucleophillic addition again.

My question is, during such reductions, is keto group reduced first or ester group gets reduced first?

I believe you understand $\ce{LiAlH4}$ is stronger base than $\ce{NaBH4}$. In the reduction of 2 to 1, $\ce{NaBH4}$ failed to reduce ester function at all, but reduce 3-keto function to 3-hydroxy function. That result tells us that isolated carbonyl group is easy to reduce than that in ester (not even to relevant aldehyde). Thus, you can conclude that isolated carbonyl function reduced first before hard to reduce ester function in the presence of stronger reducing agent such as $\ce{LiAlH4}$, which can able to reduce both.


The solubility of haloalkanes in water Haloalkanes aren't very soluble in water because they can't form hydrogen bonds, and the energy required to break hydrogen bonds in water etc is higher than the energy released when new bonds between the haloalkane and water are formed.

But as fluorine forms hydrogen bonds why couldn't a fluoroalkane form hydrogen bonds with water thus become more soluble in water?

I think the answer lies with the fact that alkanes are basically non-polar, even if they have halogen substituents. Water, on the other hand, is polar, so the solvent and solute are incompatible.


Hill system with isotopologue I'd like to use Hill notation (i.e. CH, then alphabetical), but I am not sure how to include isotopologues (i.e. carbon-13) in this notation.

For example, what would be the preferred form of ethylamine (with 13C)?

C13CH7N
CH713CN
12C13CH7N
12CH713CN
Other

I'm leaning toward using #2 but was trying to find the std practice, but I didn't see anything about isotopologues.  Thanks.

I can't point to a definitive authority, but in most of the papers I read, #5 other is used most often.  Specifically for ethylamine:

$\ce{^{13}C\,^{12}CH_7N}%edit$

I've always assumed the reason for this choice is to accentuate the presence of the isotopically unusual atoms.  I suppose a good test for my theory would be formulas with multiple, isotopically distinct atoms of an element such as iron, for which the most abundant isotope is not the lightest.

Hopefully other answers will be along shortly to point to something more authoritative, such as a IUPAC report, guideline, or document.


Does the hydrohelium dication exist? It is well-known that there exists the $\ce{He2^2+}$ dication, and also $\ce{HHe+}$ and $\ce{H2+}$ cations. But I could not find any definite information about the existence or inexistence of the $\ce{HHe^2+}$ cation.

It would be isoelectronic to $\ce{H2+}$, but would have a weaker bonding due to greater helium nucleus charge. It looks unlikely to be stable, but did anyone prove it by calculation or experiment?

I am not sure how to elaborate much beyond saying that yes, there are bound states of $\ce{HHe^{2+}}$. I cite three papers below which give numerical calculations of various states, ref. 1 gives calculations for the lowest 20 states of the system.

It appears to be that the 1s$\sigma$ is not a bound state in this system, but the 2p$\sigma$ state is bound. I would expect there are some other bound states as well.

References:

[1] Winter, T. G., Duncan, M. D., & Lane, N. F. (1977). Exact eigenvalues, electronic wavefunctions and their derivatives with respect to the internuclear separation for the lowest 20 states of the HeH2+ molecule. Journal of Physics B: Atomic and Molecular Physics, 10(2), 285.

[2] Laaksonen, L., Pyykkö, P., & Sundholm, D. (1983). Two‐dimensional fully numerical solutions of molecular Schrödinger equations. I. One‐electron molecules. International Journal of Quantum Chemistry, 23(1), 309-317.

[3] Buehler, E. J., Gooch, E. E., Dial, J. L., & Knudson, S. K. (2000). Semiclassical energies of low-lying states of one-electron diatomics. Chemical Physics, 253(2-3), 219-230.


What are typical runtimes for CASSCF calculations? I'm doing single point energy calculations of FeS using ORCA. I originally used DFT with a variety of functionals, and these calculations all took less than a minute. Now I'm attempting to run CASSCF calculations using 8 active orbitals, 12 active electrons, and the aug-cc-pVTZ basis set. It's been running for 3 days. Is this typical, or should I be attempting to change tolerances and other settings? How long would one anticipate something like this to run? In addition, how can I (generally speaking) evaluate how close a given calculation is to finishing?

There is nothing trivial about MCSCF calculations because it is hard to predict a priori how long a calculation will take. There are well-defined equations for calculating how many determinants

$$
D(n,N,S) = \binom{n}{N/2+S} \binom{n}{N/2-S}
$$

or configuration state functions (CSFs)

$$
D(n,N,S) = \frac{2S+1}{n+1} \binom{n+1}{N/2-S} \binom{n+1}{N/2+S+1}
$$

will be required for your problem; $n$ is the number of active orbitals, $N$ is the number of active electrons, and $S$ the total spin. The calculation cost will scale with this number and can be further reduced with the use of point group symmetry. (Actually, depending on your system and the property of interest, calculations without symmetry can be meaningless, like for excited states and their energies.)

These equations should make it clear that the cost grows very rapidly, considering that for CASSCF, a full CI is being performed within your chosen active space. This is not what makes MCSCF calculations tricky in practice. Here is a plot of the total energy after the first 30 iterations.

This is because the occupation numbers are very close to 0/1/2, and they aren't changing with each iteration. Notice that the occupation numbers have inverted themselves, and repeated at every step is Rot=23,19. Orbital 19 is within the active space, but 23 is external. This is an indication that the orbitals in the active space are not the correct ones. From the paper:

  the 12 or 13 valence electrons of the neutral or anion, respectively, were distributed among 14 valence active orbitals: 3p of sulphur plus 3d, 3d' and 4s orbitals of iron. With this active space we found that all important correlating orbitals are included in the active space.

MCSCF is not a black box; it's one of the few quantum chemical methods left that requires chemical intuition to even get started. This a (12,14) active space; the one in your calculations is (12,8). The procedure I use for running MCSCF calculations is generally:

Converge a single-reference calculation. This could be HF, DFT, MP2, etc. The usual recommendation is to form MP2 natural orbitals, then check the occupations numbers. If the occupation deviates more than 0.02 from an integer value, then it should probably be in the active space. This is in addition to considering the electrons and orbitals you wanted to correlate to begin with, along with antibonding partners. For example, in ethylene this is probably the pi bonding/antibonding pair, leading to a (2,2), and for 3d metal complexes this is whatever coordination (anti)bonds are present, with potentially the double d shell. I've had some luck with using DFT orbitals for metal complexes, where perturbation theory might be a bad idea due to near-degeneracies or cost. The goals here are to confirm your active space and provide the best possible starting orbitals for MCSCF.
Look at your orbitals. Of course, read the Mulliken or Lowdin population analysis, but from personal experience, I've caught many potential mistakes by doing this. It also helps when visualizing how the single-state, single-reference orbitals transform to the natural orbitals (potentially state-average) that result from MCSCF calculations.
Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with. I also tend to look at the reordered orbitals one more time before doing MCSCF to confirm I haven't made any mistakes.
Read in the previous set of (reordered) orbitals into a separate MCSCF calculation and cross your fingers. Monitoring the occupation numbers of your active space is a good guide for seeing how the calculation is converging.
Do whatever post-processing you need for your project. For me, this is looking at final occupation numbers, the orbitals themselves, and then doing MR-CISD or some other multireference correlated calculation.

Doing an auto-occupation procedure without starting from previously-converged orbitals means that 1. you don't have any guarantees about the contents of your active space, 2. you start from orbitals that are close to Hartree-Fock in quality, and 3. the calculation has to do far more work in optimizing both sets of variational parameters (the MO coefficients $\{C\}$ and the CI coefficients $\{c\}$) since the internal and external orbitals (where the CI coefficients are frozen) are still far from convergence using a single-reference method.

To walk through part of this workflow, I ran a PBE0/aug-cc-pVTZ calculation (converges much faster than the SCF before MP2) to look at the Lowdin MO populations:

! pbe0 aug-cc-pvtz cc-pvtz/jk ri rijk tightscf usesym zora

%output
 print[p_orbpopmo_l] 1
 end

* xyz 0 5
Fe      0.000000      0.000000      0.000000
S       1.960000      0.000000      0.000000
*

The MO indices are S 3s: 14, Fe 4s: 20, Fe 3d: 15,16,17,18,19, and Fe 3d': 33,34,36-40. Automatic occupation would have gotten the occupieds right, but the virtuals wrong. I'm not sure which of the 7 virtual MOs that might be the Fe 3d' are the correct ones; there is probably some trial-and-error here, but these should be visualized. Another point is that I've only looked at the ordering for alpha-spin orbitals; for difficult cases such as transition metals, the spatial ordering for beta-spin orbitals can be very different. The reordering for alpha- and beta-spin orbitals might be different. Again, this is speaking from experience.

The final exhibit is the RI-MP2/aug-cc-pVTZ natural orbitals that might enter an MCSCF calculation based on their occupation numbers:

N[ 15]( A2) =   1.97761944
N[ 16]( A1) =   1.97753198
N[ 17]( B2) =   1.97009578
N[ 18]( B1) =   1.97009578
N[ 19]( A1) =   1.03828526
N[ 20]( A1) =   0.99215102
N[ 21]( B1) =   0.98317529
N[ 22]( B2) =   0.98317529
N[ 23]( A2) =   0.02013554
N[ 24]( B2) =   0.01947713
N[ 25]( B1) =   0.01947713
N[ 26]( A1) =   0.01816184

To be thorough, this should be combined with looking at the Lowdin MO populations of these, which might require reading them into another (SCF) calculation and not performing any iterations.

However, it is well-known that converging MCSCF equations is generally difficult. In my limited experience, DALTON, GAMESS, and Molcas have less trouble than ORCA. I haven't used Molpro for MCSCF. Which program you choose dictates what post-MCSCF calculations can be performed. The DIIS algorithm in ORCA, in particular, can stall even when it shouldn't. If you can afford it, the Newton-Raphson algorithm (switchstep nr), where the active space orbital Hessian is calculated at each step, works very well to force convergence once SuperCI is done. ORCA makes it easy to get started with CASSCF, and it has plenty of knobs, but it has a very slow integral engine, doesn't prevent you from doing stupid things like GAMESS, and more advanced features are poorly documented. The workflow described above (namely the plotting orbitals and rotating them) becomes very tedious as well, but that has more to do with MCSCF in general.

Here are some good resources that describe both practical and theoretical aspects of performing MCSCF calculations: 1, 2, 3, 4.


What can I do if I don't get a phase separation between my organic and water phases? Solvent extraction in a separation funnel is a very common method in preparative organic chemistry. But sometimes you don't get a nice phase separation between the organic and the water phase.

What are the possible causes that can prevent the formation of two distinct phases, and how can you force a phase separation when you encounter such situations?

One such cause, where separation does not take place is the nature of the organic solvent. Its better to check the miscibility of the solvent with water. There are several websites, where one can get this info. For example 1 and 2.

Even with hydrophobic solvent, if the problem persists, one can add NaCl and then check the separation.


Counting valency of atoms, in a molecule with python The documentation for the indigo module can be found here

http://ggasoftware.com/opensource/indigo/api#inputoutput

So for instance if I have a molecule object for some SMILES string, e.g. "[C](=[O])", I wish to calculate the valency of each atom, for instance here the desired output would be [atom=C, unbound_electrons=2],[atom=O, valency=0]

If I consider the atom "[C]"
Can anyone explain why is this code printing [atom=C, unbound_electrons=0] not [atom=C, unbound_electrons=4]

from indigo import *
indigo = Indigo()

mol=indigo.loadMolecule("[C]")

print(mol.grossFormula(),"\n")

for atom in mol.iterateAtoms():
        print([atom.symbol(),atom.radicalElectrons()])

EDIT: I could work it out if I could generate a list of the types of bonds on the atom in conjuction with atom.atomicNumber(). E.g. if I could say [C] has a double bond I could take it's atomic number - 2 (second shell) - 2 (double bond)

EDIT#2: This might be useful for visualising what i'm talking about

from indigo_renderer import *
renderer = IndigoRenderer(indigo)
renderer.renderToFile(mol,"mol.png")

EDIT#3: I am not a chemist, so might have got some concepts wrong

I was looking into this, and the results I was getting seemed very anomalous.  Consider this iteration from a single carbon atom at -4 through +4, this is the output I receive (python code at the bottom of the post):

[C+4]
C valence= 4 radicalE= 0 charge= 4 implicitH= 0
[C+3] |^1:0|
C valence= 4 radicalE= 1 charge= 3 implicitH= 0
[C+2] |^3:0|
C valence= 4 radicalE= 2 charge= 2 implicitH= 0
[C+]
C valence= 0 radicalE= 0 charge= 1 implicitH= 0
[C]
C valence= 0 radicalE= 0 charge= 0 implicitH= 0
[C-]
C valence= 0 radicalE= 0 charge= -1 implicitH= 0
[C-2] |^3:0|
C valence= 4 radicalE= 2 charge= -2 implicitH= 0
[C-3] |^1:0|
C valence= 4 radicalE= 1 charge= -3 implicitH= 0
[C-4]
C valence= 4 radicalE= 0 charge= -4 implicitH= 0

I printed them using the canonicalSmiles format.  After I realized something strange was going on (neutral, and +/- 1 charges were 0, but everything else was fine), I dug into SMILES documentation to find out that currently it appears only up to divalent radicals are supported.  Something like a lone carbon atom has technically a tetra-valent radical, so it defaults to nothing, because it cannot be interpreted by SMILES.

I pulled this from ChemAxon which explains what the notation after the atom means. I'm going to assume that |^1:0| is the notation for a mono-valent system.  The 0 indicates the index of the atom, which is always zero for this example.

  Radical numbers
  Atom indexes with:
  - divalent radical center are written after "^2:"
  - divalent singlet radical center are written after "^3:"
  - divalent triplet radical center are written after "^4:"
  - trivalent radical center are written after "^5:"

And from the Indigo website itself it states under the supported extensions

  Radical numbers: monovalent, divalent singlet, and divalent triplet

So, I don't think you've done anything wrong, you've just simply hit a limitation of the SMILES system and the interpreter of the indigo software.  Here is the code I used below:

from indigo import *
indigo = Indigo()

from indigo_renderer import *

renderer = IndigoRenderer(indigo)
mols = []
mols.append(indigo.loadMolecule("[C+4]"))
mols.append(indigo.loadMolecule("[C+3]"))
mols.append(indigo.loadMolecule("[C+2]"))
mols.append(indigo.loadMolecule("[C+1]"))
mols.append(indigo.loadMolecule("[C]"))
mols.append(indigo.loadMolecule("[C-]"))
mols.append(indigo.loadMolecule("[C-2]"))
mols.append(indigo.loadMolecule("[C-3]"))
mols.append(indigo.loadMolecule("[C-4]"))

i = 0
for mol in mols:
    print mol.canonicalSmiles()
    for atom in mol.iterateAtoms():
      print atom.symbol(),"valence=",atom.valence(),"radicalE=",atom.radicalElectrons(),"charge=",atom.charge(),"implicitH=",atom.countImplicitHydrogens()
    filename = 'mol'+str(i)+'.png'
    renderer.renderToFile(mol,filename)
    i += 1


How to calculate the equilibrium constant for an esterification? I have tried to solve it algebraically, using the data I'm provided with, but still, I'm not capable of proving that in this particular case, the equilibrium constant is independent from the volume, maybe because it isn't; however, I don't know how can I possibly know the concentration of each substance in the reaction without the volume, so as to find the equilibrium constant.

  When a flask of constant capacity holds 1.2 mole each of acetic acid and ethanol with a catalyst and the temperature is kept at 25 °C, 0.80 mole of ethylacetate is produced at equilibrium. Calculate the equilibrium constant of the following reaction:
  $$\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}$$
  (Answer option: 0.25, 0.64, 2.0, 4.0, 6.0)

It's given that this whole process is taking place at constant volume, so you can concentrate instead on what the concentrations of the species are (no pun intended) at equilibrium. The information is there, you just need to work through it.

You have been given initial conditions:
$[{\rm CH}_{3}{\rm COOH}] = [{\rm C}_{2}{\rm H}_{5}{\rm OH}] = 1.2\;{\rm mol}; [{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = [{\rm H}_{2}{\rm O}] = 0\;{\rm mol}$

and final conditions, where you are told that $[{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = 0.8\;{\rm mol}$. This means that at equilibrium $[{\rm H}_{2}{\rm O}] = 0.8\;{\rm mol}$.

Your reactant equilibrium concentrations will be $1.2\;{\rm mol} - x$, where $x$ is given to you (I'm leaving that part to you. Hint: It's in the  sentence right above this one). Plug in your equilibrium values and you'll get ${\rm K}_{\rm eq}$ (also up to you).


Can functional groups exist by themselves Just started to study organic compounds and I am curious about functional groups. I want to know if functional groups can exist as molecules themselves.

A functional group on its own essentially has at least one non-bonded electron, thus making it a free radical, this makes them highly reactive molecules, making them react with other substances (including themselves) and can thus only exist in very low concentrations in inert medium/vacuum.

However, there are molecules such as $\ce{NO}$, $\ce{NO2}$ which do have an odd electron in their molecular state, can exist in reasonably large quantities, and can be functional groups in organic chemistry, these are in some sense functional groups that exist as molecules.


How is ethanenitrile protonated? I know that the proton will attack the HOMO, but I'm having difficulty imagining how this will happen.

The structure of ethanenitrile is:

$$\ce{H3C-C#N:}$$

Qualitatively, we can draw some conclusions about the molecular orbitals in the structure. If covalent bonds result in in the involved electrons being lower in energy, then which pair of electrons in the structure has the highest energy?

However, if you are looking at nitriles, you probably already know that both lone pairs and pi bonds are good proton acceptors.

Let's consider what the difference is if the lone pair is the base or if a pi bond is the base:

  Lone Pair

$$\ce{H3C-C#N + H+ -> H3C-C#N+-H}$$

  Pi Bond

Which atom will the proton end up attached to? Which of the two following carbocations is more stable?

$$\ce{H3C-C#N + H+ -> H3C-C+=NH}\ \  or\ \  \ce{H3C-CH=N+}$$

Now... what is the relationship between the two conjugate acids drawn?

$$\ce{H3C-C#N+-H}\ \ and\ \ \ce{H3C-C+=NH}$$

Where is the HOMO?


Identifying a chloride of the unknown element from the masses of chlorine and compound as well as vapor density 1 g of a chloride of an element contains 0.835 g of chlorine. If vapour density of the chloride is 85, find weight of the element and its valency.

I tried to solve it but couldn't get to the answer then I saw its solution and I also could not understand what they want to explain. Can anyone please help me out.
By the way, it's not my homework.

I agree with the starting point of the Matthew's solution, namely the definition of vapor density, but I fail to understand the follow-up math, so I just post my approach.

$$M(\ce{ECl_x}) = D\cdot M(\ce{H2}) = 85\cdot\pu{2 g mol-1} = \pu{170 g mol-1}\tag{1}$$

By definition, molar mass $M(\ce{ECl_x})$ is also

$$M(\ce{ECl_x}) = M(\ce{E}) + x\cdot M(\ce{Cl})$$

$$\implies M(\ce{E}) = M(\ce{ECl_x}) - x\cdot M(\ce{Cl}) = (170 - x\cdot 35.5)~\pu{g mol-1}\label{eqn:2}\tag{2}$$

The unknown variable $x$ is defined by the molar ratio between the elements:

$$x = \frac{n(\ce{Cl})}{n(\ce{E})} = \frac{m(\ce{Cl})\cdot M(\ce{E})}{m(\ce{E})\cdot M(\ce{Cl})} = \frac{\pu{0.835 g}\cdot (170 - x\cdot 35.5)~\pu{g mol-1}}{(1 - 0.835)~\pu{g} \cdot \pu{35.5 g mol-1}} = 24.14 - 5.04\cdot x\label{eqn:3}\tag{3}$$

Solving \eqref{eqn:3}:

$$6.04\cdot x= 24.14 \quad\implies\quad x = 4.0 \tag{4}$$

Using $x$, we can now determine the element from \eqref{eqn:2}:

$$M(\ce{E}) = (170 - 4.0\cdot 35.5)~\pu{g mol-1} = \pu{28.0 g mol-1}\tag{5}$$

which appears to be silicon, valency 4.
The unknown compound is silicon tetrachloride $\ce{SiCl4}$, a fuming volatile liquid at NTP.


pH implications of redox balancing half reaction method I have seen similar questions, but I’m asking a new question as far as I know. Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations? asks about using hydroxide instead of water and that is closest.

I have seen example problems where $\ce{H+}$ is produced or consumed, $\ce{OH-}$ is produced or consumed, and water is used to balance. My thought is that each of these changes affects the $\mathrm{pH}$ or $\mathrm{pOH}$ of each solution.

If the solution is already an acid, there is an abundance of $\ce{H+}$ and it makes sense to me that a reaction may use excess $\ce{H+}$ to make the solution more neutral. One could make the same argument about using up excess $\ce{OH-}$ in a basic solution. What doesn’t make sense to me is adding to the basic or acidic nature. See example below which was specified to be in a basic solution. It appears to be making the solution more basic by producing more $\ce{OH-}$ instead of using it up. Shouldn’t that be less likely to occur? Why not? What am I missing?

$$\ce{6I- (aq) + 4H2O + 2MnO4- (aq) —> 3I2 + 2MnO2 (aq) + 8OH- (aq) }$$

If the reaction happens at neutral or mildly alkalic $\mathrm{pH}$:

$$\ce{6I- + 4 H2O + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 8 OH-}$$

If the reaction happens at mildly aciding $\mathrm{pH}$, it could be as well:

$$\ce{6I- + 8 H+ + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 4 H2O }$$

In any case, either spending  $\ce{H+}$ either production $\ce{OH-}$ increases $\mathrm{pH}$. Both iodide and permanganate carry a negative charge while their spent forms are neutral. This negative charge must be passed to other anions like $\ce{OH-}$ or neutralized by $\ce{H+}$.

If one considers further disproportionation of iodine to iodide and iodate in alkalic solutions, then the total reaction is:

$$\ce{ I-  +  H2O + 2 MnO4- ->  IO3- + 2 MnO2(s) + 2 OH-}$$

In strongly alkalic it could be also
$$\ce{ I-  + 6 OH- + 6 MnO4- ->  IO3- + 6 MnO4^2- + 3 H2O }$$

It may be obvious if the equation is written together with spectator ions:

Mildly acidic:
$$\ce{6KI + 8 HA + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 4 H2O + 8 KA}$$
resp neutral/mildly alkalic
$$\ce{6KI + 4 H2O + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 8 KOH}$$
resp alkalic
$$\ce{ KI  +  H2O + 2 KMnO4 ->  KIO3 + 2 MnO2(s) + 2 KOH}$$
resp. strongly alkalic
$$\ce{ KI  + 6 KOH + 6 KMnO4 ->  KIO3 + 6 K2MnO4 + 3 H2O }$$


How to solve for concentration of two equilibrium reactions with finite equilibrium constants? I have a small problem of equilibrium that I counldn't solve. I want the exact answer. No approximations please.
$$
\begin{array}{ll}
\ce{H2O + H2O <=> H3O+ + OH-} &\quad\left(K_w = 1\cdot 10^{-14}\right) \\
\ce{HNO2 + H2O <=> H3O+ + NO2-} &\quad\left(K_a = 4\cdot10^{-4}\right) \\
\end{array}
$$

Objective: Find the concentration every substance, given that the initial value of $[\ce{HNO2}]_0$ is known. Feel free to insert any numerical value for it, if you wish. I have no idea how to solve for the concentrations, when I have two finite ionization constants. How to solve it?

Of course, in this situation, most people would ignore the contribution of the auto-ionization constant of water (that is, ignore the first reaction), leaving only the second one, which is fairly easy to calculate the concentrations. I do not want that.

The reading of this part is optional. Feel free to skip everything down below if you wish. This is an example that I understood how to solve:
$$
\begin{array}{ll}
\ce{H2O + H2O <=> H3O+ + OH-} &\quad\left(K_w = 1\cdot 10^{-14}\right) \\
\ce{HCl + H2O <=> H3O+ + Cl-} &\quad\left(\mbox{Full Dissociation}\right) \\
\end{array}
$$

Hereby, lets say we had an initial concentration of $[\ce{HCl}]_0$. It will then fully dissociate, thus we shall have: $[\ce{HCl}]_0 = [\ce{H3O+}]_0$. Now we correct considering the water autoionization:

$$
\begin{array}{ll}
[\ce{H3O+}] = [\ce{HCl}]_0 + x \\
[\ce{OH-}] = x \\
\end{array}
$$

We solve for $x$ such that $[\ce{H3O+}][\ce{OH-}] = K_w$. This will give me a quadratic equation, which using bhaskara becomes easy to solve.

Again, most people here would ignore the first reaction. And, in certain circumstances, they could, for instance, find a potentially wrong value of $\text{pH}$. Classical example: Just let $[\ce{HCl}]_0 = 10^{-8}$, and then ignoring water autoionization, it will fully dissociate thus $[\ce{H3O+}]_0 = 10^{-8}$, giving $\mathrm{pH = 8}$. So, we just got a basic $\text{pH}$ after just inserting hydrochloric acid in water!

Completely accurate answer :

The main reactions which take place are
$$
\begin{array}{ll}
\ce{ H2O <=> H+ + OH-} &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
\ce{HNO2 <=> H+ + NO2-} &\quad\left(K_\mathrm{a}= 4\times10^{-4}\right) \\
\end{array}
$$
$\ce{HNO2}$ acid is a weak acid, which does not dissociate completely in water. This means it does not give all its hydrogen ions into the water, the resulting solution contains four species ${\ce{H+}, \ce{OH−},\ce{NO2−}}$, and undissociated acid $\ce{HNO2}$. In order to specify the concentrations of the four species
present in an aqueous solution of $\ce{HNO2}$, we need four independent equations
 between them. These are :

1- Equilibria. We have two equilibrium equations:

The dissociation equilibrium of water
$$\ce{ K_\mathrm{w}=[H+][OH−]}\tag{1}$$
The dissociation equilibrium of acid
$$\mathrm{ K_\mathrm{a}}=\frac{[H^+]_e[NO_2^−]_e}{ [HNO_2]_e}\tag{2}$$
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $\ce[HNO_2]_0$. For the solution of $\ce{HNO_2}$, this equation would be
$$\ce{[HNO2]_0  = [HNO2]_e + [NO2−]_e}\tag{3}$$
3-Charge balance. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$$\ce{[H+]_e = [NO2−]_e + OH−}_e\tag{4}$$
We can use ${\text{Eq}\mathrm{3}}$ to get an expression for $\ce{[HNO2]_e}$, and this can be solved for$\ce{ [NO2−]_e}$; these are substituted into $\text{ Eq}\mathrm{2}$  to yield
$$ K_\mathrm{a}=\frac{\ce{ [H+]_e} {([H^+]_e - [OH^-]_e)} }{[HNO_2]_0 - [H^+]_e + [OH^-]_e}  \tag{5}$$
This equation is cubic in $\ce{[H+]_e}$ when $\ce{[OH−]}$ is expressed as $\frac{K_\mathrm{w}}{\ce{[H+]_e}}$; it becomes as

$$ K_\mathrm{a}=\frac{\ce{[H+]_e} {([H^+]_e – K_\mathrm{w}/{[H^+]_e})} }{[HNO_2]_0 - [H^+]_e + K_\mathrm {w}/{[H^+]_e}}  \tag{6a}$$
Or
$$\
 [\ce{H+}]_e^3
  + K_\mathrm{a} [\ce{H+}]_e^2
  -(K_\mathrm{w} + [\ce{HNO2}]_0K_\mathrm{a})[\ce{H+}]_e
   - K_\mathrm{w} K_\mathrm{a}
  \tag{6b}
$$
 For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.

a- Unless the acid is extremely weak or the solution is very dilute, the concentration of $\ce{OH−}$ can be neglected in comparison to that of $\ce {[H+]_e}$. If we assume that $\ce{[OH−]_e<<[H+]_e}$, then $\text{Eq}\mathrm{5}$ can be simplified to
$$K_\mathrm{a }=\frac{[H^+]_e^2}{[HNO_2]_0 –[H^+]_e}\tag{7}$$
which is a quadratic equation:
$$\ce{[H^+]_e^2 + K_\mathrm{a}[H+]_e – K_\mathrm{a}[HNO_2]_0 = 0}\tag{8}$$
b-If the acid is fairly concentrated (usually more than $\mathrm{10^{−3}}\text{M}$), a further simplification can frequently be achieved by making the assumption that$\ce{ [H^+]_e << [HNO_2]_0}$. This is justified when most of the acid remains in its protonated form$\ce{ [HNO_2]_e}$, so that relatively little $\ce{H+}$ is produced. In this event,$\text{Eq}\mathrm{7}$ reduces to
$$K_\mathrm{a}=\frac{[H^+]_e^2}{[HNO_2]_0}\tag{9}$$
Or
$$\ce{[H^+]}  \approx  \sqrt{K_\mathrm{a}\ce{[HNO_2]_0}}\tag{10}$$
-   Comparing of the$\ce{ [H+]_e }$values  resulted from solution the three
 derived equations when the initial acid concentration equal $\mathrm{0.1}\text{M}$

1-  The completely accurate answer solution using cubic equation $\mathrm{(6b)}$by wolframalpha

$\ce{[H+]_e} = 0.00612772\text{M}$

2- The approximate  answer solution using quadratic equation $\mathrm{(8)}$

$\ce{[H+]_e} = 0.00612771680\text{M}$

3- The further approximate  answer solution using the reduced equation $\mathrm{(10)}$
$\ce{[H+]_e} = 0.00632455532\text{M}$


Gases produced by pyrolysis of cellulose I heated cotton in a sealed container (with a small hole) over a natural gas flame. Some gases and smoke were produced. What would they probably be? I can come up with some guesses based on the composition of cellulose: $\ce{CO2}$, $\ce{CH4}$ or possibly other hydrocarbons, $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, however I do not know which of those they are. Obviously, soot ($\ce{C}$) was also formed, due to the visible smoke particles.

During pyrolysis, organic compounds are thermally decomposed in the absence of oxygen. The pyrolysis products are classified into categories based on their physical state of existence: char (solid), bio-oil (liquid) and non-condensable gases (gas). The relative proportions of these three product fractions significantly vary depending upon the process conditions, as is shown in the table below.

$$ \small
\begin{array}{lcc}
\hline
\text{Pyrolosis Technology} & \text{Residence Time} & \text{Heating Rate} & \text{Temperature} & \text{Char} & \text{Bio-Oil} & \text{Gases} \\
\hline
\text{Conventional} & \text{5-30}\ \mathrm{min} & \text{<50} ^\circ \mathrm{C\ min^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<35%} & \text{<30%} & \text{<40%}\\
\text{Fast Pyrolysis} & \text{<5}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<25%} & \text{<75%} & \text{<20%}\\
\text{Flash Pyrolosis} & \text{<0.1}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{650-900} ^\circ \mathrm{C} & \text{<20%} & \text{<20%} & \text{<70%}^{[1]}\\
\hline
\end{array}
$$

The exact compositions of the products of cellulose pyrolysis at different temperatures can be seen below.

$$ \small
\begin{array}{lcc}
\hline
\text{Products} & \text{Peak Temp,}\ 500 ^\circ \mathrm{C} & \text{Holding Temp,}\ 400 ^\circ \mathrm{C} & \text{Peak Temp,}\ 750 ^\circ \mathrm{C} & \text{Peak Temp,}\ 1000 ^\circ \mathrm{C}\\
\hline
\ce{CO} & \text{0.99%} & \text{0.25%} & \text{15.82%} & \text{22.57%}\\
\ce{CO2} & \text{0.3%} & \text{1.45%} & \text{2.38%} & \text{3.36%}\\
\ce{H2O} & \text{3.55%} & \text{6.49%} & \text{8.72%} & \text{9.22%}\\
\ce{CH4} & \text{0%} & \text{0%} & \text{1.11%} & \text{2.62%}\\
\ce{C2H4} & \text{0%} & \text{0%} & \text{1.05%} & \text{2.18%}\\
\ce{C2H6} & \text{0%} & \text{0%} & \text{0.17%} & \text{0.28%}\\
\ce{C3H6} & \text{0%} & \text{0%} & \text{0.70%} & \text{0.80%}\\
\ce{H2} & \text{0%} & \text{0%} & \text{0.36%} & \text{1.18%}\\
\ce{CH3OH} & \text{0.25%} & \text{0.21%} & \text{1.03%} & \text{0.98%}\\
\ce{CH3CHO} & \text{0.01%} & \text{0.05%} & \text{1.58%} & \text{1.7%}\\
\text{tar} & \text{16.37%} & \text{83.35%} & \text{59.92%} & \text{49.12%}\\
\text{char} & \text{83.63%} & \text{6.17%} & \text{3.32%} & \text{3.91%}\\
\text{other} & \text{0.19%} & \text{0.16%} & \text{2.14%} & \text{1.78%}\\
\text{total} & \text{105.25%} & \text{98.36%} & \text{98.8%} & \text{99.86%}\\
\hline
\end{array}
$$

The holding time for each of these reactions was $30\ \mathrm{s}^{[2]}$. As shown in the table, $\ce{CO}$, $\ce{H2O}$, and $\ce{CO2}$ are the major gaseous products, with $\ce{H2}$ and hydrocarbons being produced in considerably smaller proportion.

$^{[1]}$ Patwardhan, Pushkaraj Ramchandra, "Understanding the product distribution from biomass fast pyrolysis" (2010). Graduate Theses
and Dissertations. Paper 11767.
$^{[2]}$ Hajaligol, M. R.; Howard, J. B.; Longwell, J. P.; Peters, W. A. Product Compositions and Kinetics for Rapid Pyrolysis of Cellulose. Industrial & Engineering Chemistry Process Design and Development Ind. Eng. Chem. Proc. Des. Dev. 1982, 21, 457–465.


Which reaction to use for standard formation enthalpy? What reaction illustrates $\Delta H^\circ_\mathrm f$ for $\ce{CaCO3}$?

Answer: $\ce{Ca(s) + C(s) + 3/2O2(g) -> CaCO3(s)}$.

Ok so why isn't it: $\ce{2Ca(s) + 2C(s) + 3O2(g) -> 2CaCO3}$

Or

$\ce{Ca(s) + C(s) + O3(g) -> CaCO3}$ ?

For example in another similar question:

  Choose the reaction that illustrates $\Delta H^\circ_\mathrm f$ for $\ce{Ca(NO3)2}$.

Answer: $\ce{Ca(s) + N2(g) + 3 O2(g) -> Ca(NO3)2(s)}$

Can someone confirm this assumption of mine:  it's ok to use fractions in these types of equations to blance it out as long as it's in oxygen's coefficient.

As that assumption correct?

The heat of formation refers to the formation of one mole of the compound from its elements in their natural states at $1\mathrm{~atm}$ and $298\mathrm{~K}$.  Your first example forms 2 moles and your second example uses ozone, which is not the "natural state" of oxygen at $298\mathrm{~K}$ and $1\mathrm{~atm}$.

Regarding your question about oxygen, it has nothing to do specifically with oxygen. The important thing is to have 1 mole of the compound on the right hand side of the equation.


Why is density of chloroform greater than water? Hydrocarbons (i.e., molecules containing carbon and hydrogen only) are generally less dense than water.  Yet chloroform, which is a halogenated hydrocarbon, is more dense.
Why is this?  I know water is  more dense than hydrocarbons because its polar nature would cause it to pack more tightly.  But ,why  chloroform, which is less polar than water, has a higher density than water.
It can't be molecular weight, because larger hydrocarbons like hexane and octane have a higher molecular weight than water, yet water is denser.
So is there any theory or concept which will help me find why chloroform is denser than water?

The density of a substance depends on its molecular density and packing fraction (where you need to be consistent in definiting the molar volume in calculating both attributes).
However you are looking for a simpler explanation that gets the general trends right.   Just looking at molecular mass is too simple—larger molecules have greater mass, but take up more space.  So what would be a simple model that works reasonably well?
I suspect the main determinant of density is the average mass of a molecule's atoms, and that relative atomic sizes, as well as packing density, while important, are secondary contributors.
Given this, let's construct our very rough model for predicting relative densities by comparing just the average atomic masses of the molecule's atoms, and see if that, by itself, can explain most of the trends.
Of course, since we are ignoring these other factors, our simple model is imperfect and thus, as expected, will have many exceptions (one of which is included, and bolded, below). Nevertheless, in spite of the very simple nature of this model, it does surprisingly well in matching the general trends (at least for the substances shown here).
In each case I have calculated the average atomic mass of the atoms (u/atom) from $\frac{\text{molecular mass}}{\text{no. of atoms}}$:
Hexane ($\ce{C6H14}$): 86.18/20 = 4.3 u/atom; density = 0.661 g/$\text{cm}^3$
Octane ($\ce{C8H18}$): 114.23/26 = 4.4 u/atom; density = 0.703 g/$\text{cm}^3$
Water ($\ce{H2O}$):  18.015/3 = 6.0 u/atom; density = 0.997 g/$\text{cm}^3$ (at 25$^\circ$C)
Methyl chloride ($\ce{CH3Cl}$):  50.485/5 = 10.1 u/atom; density = 0.911 g/$\textbf{cm}^{\bf{3}}$   (note the exception to the trend; it works well in comparing like compounds—this one vs. the ones below—but fails when comparing very unlike compounds—this one vs. water—when the average atomic masses of the atoms are too close)
Methylene chloride ($\ce{CH2Cl2})$: 84.927/5 = 17.0 u/atom; density = 1.325 g/$\text{cm}^3$
Chloroform ($\ce{CHCl3})$: 119.369/5 = 23.9 u/atom; density = 1.492 g/$\text{cm}^3$
Carbon tetrachloride ($\ce{CCl4}$): 153.8118/5 = 30.8 u/atom; density = 1.594 g/$\text{cm}^3$
Carbon tetrabromide ($\ce{CBr4}$): 331.627/5 = 66.3 u/atom; density = 3.42 g/$\text{cm}^3$ (m.p. = 89$^\circ$C, so it's solid at room temperature)
Carbon tetraiodide ($\ce{CI4}$): 519.628/5 = 103.9 u/atom; density = 6.36 g/$\text{cm}^3$ (m.p. = 168$^\circ$C, so it's solid at room temperature)
Mercury ($\ce{Hg}$): 200.592/1 = 200.6 u/atom; density = 13.534 g/$\text{cm}^3$


Reactions involved in Titanium Anodisation I am writing up a lab report investigating titanium anodisation and producing oxide layers. After some searching, all I can find is the reaction $\ce{Ti + O2 \to TiO2}$, which I am having no trouble understanding.

However, I seem to be confused/mislead by the simplicity of this. The whole unit is about electrochemistry and redox reactions but I cannot find any examples (or seem to produce any either - I have tried generalizing from the rusting of iron in water) of half reactions or anything similar.

I have come about as far as including the redox reactions for the hydrolysis of water and then jumping to the above reaction but am unsure where to go from here.

I was able to locate what appears to be a useful summary of reactions and reduction potentials that are all titanium-specific here.

The reaction $\ce{ Ti + O2 -> TiO2}$ is illustrative of $\ce{Ti}$ being oxidized while oxygen is reduced. I agree with you that it is simplistic-looking and might not appear as traditional an example of redox chemistry as you require.

You might investigate the high-temperature reaction of titanium metal and water (steam), which produces $\ce {TiO2}$ and $\ce{H2}$ gas (titanium is oxidized, hydrogen is reduced), which might be a little clearer. This reaction (and several others) are given on the page I've linked above.


How to draw Fischer projections for carbon chain having more than two carbons? How can we draw Fischer projections for one-five carbon compounds?

For example, say we have pentane-2,3,4-triol ($\ce{CH3-CHOH-CHOH-CHOH-CH3}$; image below). With wedge and dash structure, the alcohol group is below the plane for the 2nd and 4th carbons but above if for the 3rd carbon. How can we draw the Fischer projection of this compound?

I read an extract from the chapter about carbohydrates in Solomon's book (10th edition) but only understood partially how to draw Fischer projections as it is given very briefly. Also, a brief introduction is given in chapter 5 but it is for two-carbon compounds only.

I have tried a lot but cannot understand how to construct it or what are the conventions on which they are made. Apparently, the web doesn't describe Fischer projections in a very detailed way probably because they are generally only used in biochemistry.

The first things to note when drawing Fischer projections are the general rules for drawing them. The first general rule concerns the choices of ‘top’ and ‘bottom’ while the second concerns how each cross is interpreted.

In a Fischer projection, the carbon atom with the highest oxidation state should be closer to the top than the bottom.
In cases where the highest oxidation state is in the centre or where multiple carbons have the same highest oxidation state, the second-highest applies analogously and so forth.

In a Fischer projection, the vertical bonds are always considered to point behind the paper (or screen) plane (hash) while horizontal bonds are considered to point in front (wedge).
This is exemplified in figure 1.

Figure 1: How stereochemical information is encoded in Fischer projections.
You compound presents a problem according to these rules as there is no most oxidised carbon. Rather, all three carbons 2, 3 and 4 are in the oxidation state $\pm 0$ while the two terminal ones are both $\mathrm{-III}$. To ease this limitation, I am going to discuss the Fischer projection with the related molecule shown in figure 2, which has one terminal carbon oxidised to a carboxylic acid breaking the tie.

Figure 2: L-5-deoxyribonic acid; a replacement molecule with the same stereochemistry on the central hydroxy groups.
For this molecule, it is clear that the acid should be put on top. Going down each bond, we need to rotate them until we have the longest chain in the vertical positions and pointing backwards (cf figure 1). I’ll let you use a modelling kit, pen and paper or your 3D viewing to perform the actual transformations; the final product you should end up with is the following in figure 3.

Figure 3: Fischer projection of the molecule in figure 2.
There should be little debate on the carboxylic acid; and whether the carbonyl group or the hydroxy group of it is pointing to the left does not matter; we can rotate around the $\ce{C{1}-C{2}}$ bond. To see the correct orientation of $\ce{C{2}}$ and $\ce{C{4}}$, you can take the skeletal formula from figure 2 and look from above; for $\ce{C{3}}$ you should be looking from below. We find that all hydroxy groups are pointing in the same direction; here: to the left. This means my molecule can be seen as a derivative of ribose; and since the lowest asymmetric carbon’s hydroxy group is pointing to the left, it derives from L-ribose.
Back to your molecule. Not only is the designation ambiguous (see above on the highest oxidation state) but it is also $C_\mathrm{s}$ symmetric, i.e. achiral. I have drawn your molecule along with its plane of symmetry (dashed) in figure 4.

Figure 4: The original molecule in question; 1,5-dideoxyribitol or (2R,3s,4S)-pentane-2,3,4-triol.
We need to arbitrarily choose a ‘top’ carbon here. For reasons of consistency, let’s choose the one on the left that was an acid in the former structure (figure 2). This means, we have a very easy job of drawing the Fischer projection as we already did it. It is shown in figure 5.

Figure 5: Fischer projection of the triol shown in figure 4.
We may be tempted to assign an L-configuration immediately but remember we realised that the molecule should have a plane of symmetry. Well, it does in the Fischer projection, too; the plane is horizontal and encloses the $\ce{HO-C-H}$ bonds of the central carbon atom. We can also turn around our piece of paper or PC screen and realise that we can thereby match it perfectly with its mirror image, shown in figure 5 to the right of the dashed mirror plane. Thereby we have twice shown that your molecule is achiral, that both a hydroxy group pointing to the left and to the right on the bottommost asymmetric carbon is possible and that therefore assigning a D/L stereodescriptor is meaningless.


Why is the relative atomic mass of carbon not exactly 12? Relative atomic masses of atoms of all chemical elements are numbers without units, being the value of proportion compared to  $\frac{1}{12}^\text{th}$ the mass of the carbon atom.

But the relative atomic mass of carbon is never 12! Instead, it is 12.01 or, more accurately,  12.011.

Why is this so?

Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.


What is the difference between Alkanes and Polymers? As far as I understand, Polymers are long molecules made of smaller monomers which are Alkenes. But polymers don't have double bonds, so they are saturated. What is the difference between Alkanes and Polymers structures.

The first one is an Alkane, but if the chain was much longer, it would be structurally the same as the Polymer (Second image).

Some points to note: You said polymers are saturated. This generalization is not true in many cases, e.g. polystyrene, nylon, and much more. Alkanes are always saturated.

Alkanes can have branchings. The ones you have in mind are called normal alkanes, in which no carbon has higher order than two. An alkane can have any kind of branching, as long as it is saturated.

Now, maybe the most important one: polymers are made of so-called monomers. All polymers can be constructed from exactly one monomer (even co-polymers). An alkane which is sort-of tree-like randomly branched, will not have such an unit.

A normal alkane has some similarities with polymers, tho. As others mentioned it, the size is the deciding factor there. A normal alkane with a carbon count of 10000 is for sure a polymer.


HCl synthesis from sulfuric acid and sodium salt I'd like know if this is right. The reaction between $\ce{NaCl}$ and $\ce{H_2SO_4}$ should yield $\ce{HCl}$ and sodium sulfate. This is a double substitution, true? No oxidation took place here. The thing is, when I tried the reaction in the lab, the resulting solution (gas bubbled into water) seemed to be sulfuric acid. Simply enough, the solution didn't react with aluminum. I presume that $\ce{SO_2}$ was formed from the oxidation the sulfur (it a little bit like it).

What am I doing wrong? Shouldn't the substitution take place instead?

Process:

$
\ce{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl}\\
\ce{NaCl + NaHSO_4 \rightarrow Na_2SO_4 + HCl}
$

Adding:

$
\ce{2\space NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2\space HCl }
$

Is this right?

Ideally, that's the reaction. Realistically, there are a few speed bumps.

First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate.

You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a "water-free" combination of the reactants.

Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).

Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.


Collection of gas over water When you collect a gas over water, you have to account for the mixed in water vapor by subtracting the room pressure from water vapor pressure. However, why do you not do something similar for volume, where you subtract the total measured volume from the volume of water?

How can $PV = nRT$ still hold when the pressure you are using is only for the collected gas whereas the volume is for the gas + water vapor volume?

You are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\ce{K}$:  it's  $13.85\,\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is  $$\frac{13.85}{760}\times100=1.83\%$$
As you can see, it's negligible.

P.S. To calculate the vapor pressure of water, I used the first equation in http://en.wikipedia.org/wiki/Vapour_pressure_of_water


Catalyze dehydration of formic acid I know that formic acid will dehydrate upon heating to 100°C (under normal pressure) or adding sulfuric acid. What other catalysts can be used to induce this reaction?

Activated alumina, silica, thoriated silica gel, phosphorus pentoxide-silica gel, aluminum phosphate, titanium dioxide
Catalytic Dehydration of Formic Acid, E. G. Graeber and D. S. Cryder
Industrial & Engineering Chemistry 1935 27 (7), 828-831, DOI: 10.1021/ie50307a019
Mechanism of dehydration decomposition of formic acid over alumina and silica. Adsorption measurements during the catalytic reaction, Kenzo Fukuda, Yuko Noto, Takaharu Onishi and  Kenzi Tamaru, Trans. Faraday Soc., 1967,63, 3072-3080, DOI: 10.1039/TF9676303072

Mixed titanium-vanadium oxide
Sadovskaya, Ekaterina & Chesalov, Yuriy & Goncharov, V. & Sobolev, Vladimir & Andrushkevich, Tamara. (2017). Formic acid decomposition over V-Ti oxide catalyst: Mechanism and kinetics. Molecular Catalysis. 430. 52-64. 10.1016/j.molcata.2016.12.010

Various homogenous transition metal catalysts
Homogeneous Catalytic Dehydrogenation of Formic Acid: Progress Towards a Hydrogen-Based Economy, Gábor Laurenczy, Paul J. Dyson, J. Braz. Chem. Soc. vol.25 no.12 São Paulo Dec. 2014, DOI: 10.5935/0103-5053.20140235

Monoclinic bismuth chromate hydroxide proto-catalyst, $\ce{m-Bi(OH)CrO4}$
Two-step catalytic dehydrogenation of formic acid to CO2 via formaldehyde, Viasus et. al., International Journal of Hydrogen Energy, Volume 44, Issue 3, January 2019, Pages 1534-1543, DOI: 10.1016/j.ijhydene.2018.11.178


Multiday lab courses in classic chemistry This time of the year I usually get some extra income I can use on interesting courses. Usually rescue or sport courses. However, I also have a master degree in chemistry, and would like to spend some days or a week just to do some old fashioned chemistry with a great teacher. But I have never found a "summer school" or something like that.
Does anyone has a any suggestion? I can gladly expand "classical chemistry" in any direction, as metallurgy or microbiology.

Beside asking upfront the research groups, there are summer schools in normal (non-covid 19) circumstances.  For example Oxford's Chemistry Summer School, or ACS' Green Chemistry Summer School.  Given your account name, I speculate you reside in Europe; if so, a suitable directories page not only about core-chemistry may be  www.summerschoolsineurope.eu.  Sadly, because of the current pandemic, many scientific events are cancelled, though (example).  Still, references of courses in the past are retained (example), so you could find help by the former organizers as well.
Maybe you are an outlier by age in comparison to the «typical attendee».  However, the experience you gained after your master thesis, both in terms of chemistry under the hood as well as e.g., project and data management, may be of great benefit to the class providing additional perspectives.  In particular, I think about planning the experiments in the smaller groups, or discussing the results.  Equally, you may become a role model for younger participants in the way you choose to stay continuously engaged in learning.


How can water below a pH of 7 still be alkaline? Alkalinity is a measure of the water's ability to neutralize acidity and is expressed in levels of bicarbonates, carbonates, and hydroxides.
How can water that is slightly acidic, e.g. 5.5-6.5, still boast any alkalinity? Or put another way, is there a maximum level of alkalinity for  water at any given acidic pH?
This is based on my understanding that acidity is, put simply, when there are more H+ than OH- ions. So why don't all carbonates react with the surplus H+ ions at pH ranges 5.5-6.5 until either alkalinity is 0 or the pH has reached a neutral 7?

Classically, alkalinity is determined by $\ce{HCl}$ titration with methyl orange as indicator ( $\mathrm{pH}$ transition range $3.1 - 4.4$ ). IF some $\ce{HCl}$ is spent by lowering $\mathrm{pH}$ from $5.5-6.6$ to $\mathrm{pH}$ of noticing red colour, there is some nonzero alkalinity.
Note that at $\mathrm{pH} \approx 6.3$, the ratio $\dfrac{[\ce{CO2}]}{[\ce{HCO3-}]}$ is 1 : 1. (*)
Generaly, there is no direct relation between $\ce{pH}$ and maximal alkalinity, similarly as there is no direct relation between $\ce{pH}$ and $\mathrm{pH}$ buffer capacity.
But for typical water composition, containing $\ce{CO2/HCO3-}$ $\mathrm{pH}$ buffer, the alkalinity can be estimated from $\ce{pH}$ and expected $\ce{CO2(aq)}$ concentration, which is supposed to be in equilibrium with gaseous $\ce{CO2}$.
$$[\ce{HCO3-(aq)}] = K_{a1,\ce{CO2}}^{*} \cdot \frac {[\ce{CO2(aq)}]}{[\ce{H+(aq)}]}$$

(*) The  Wikipedia page for $\ce{H2CO3}$ has been recently misedited,  confusing $K_\mathrm{a2}$ and $K_\mathrm{a1}^*$(assuming 100% conversion of $\ce{CO2(aq)}$ to $\ce{H2CO3}$.


How to safely clean tantalum jewelry I have some tantalum jewelry that I would like to clean. I've tried just soap and water to no success, so I wanted to try one of the standard at-home cleaning methods, like this one that uses foil, hot water, baking soda, and salt.

I know that tantalum is very non-reactive, so I would expect this to be safe to the metal, but I want to confirm before I accidentally do any damage. Unfortunately because tantalum is such a non-standard metal for jewelry, I haven't been able to get a straight answer from the internet or even the jeweler for whether this will be okay.

Is using the foil and baking soda method safe for cleaning tantalum jewelry? Are there any common cleaning methods that would be safe for most metal jewelry but not for tantalum?

Ultra Met uses tantalum for corrosion resistant coatings and has a list of corrosive substances that won't dissolve tantalum, comprising of hydrochloric, hydrobromic, and hydroiodic acids, nitric acid, sulfuric acid, acetic acid, hydrogen sulfide and even pirahna.

So unless you plan on using aqua regia, hydorfluoric acid, hot chlorine or other hot halogen gas, (none of which are safe for most metal jewelry) your ring should survive—at least the tantalum part will.


Why is the definition of the mole as it is? I have asked my teacher, as she was introducing the concept of mole to us, why that number was chosen, instead of more convenient one. She told me that it came from the definition of the mole, that is the number of atoms in 12g of Carbon 12. When I asked why that definition was chosen, she answered me that my question wasn't really a question, and I understood from her stare that asking why? twice in a row was not appreciated.

However, I certainly hope that this definition wasn't chosen randomly and that there is a reason for it being it.

I gave it some thought and I came to believe that it may be for convenience after all. There are 12 nucleons in a carbon atom and if we say that the electrons mass can be neglected, then we get that a mole is the number of nucleon it takes to get one gram of them. However, can we really neglect the mass of electrons, when the atoms get bigger? And if so, why do the element's mass per mole do very rarely end up being integers?

This makes me think there may be another reason, or some things I do not understand. And if there are no reason, then I guess I might just as well be doing religious studies..!

Why is the definition of the moles as it is?

It is a rather arbitary definition that the mole is the number of atoms in 12g of carbon 12.  This has not always been the definition.  For example, prior to 1960, the definition was based upon oxygen rather than carbon-12.

The first standard was based upon 1 gram of hydrogen.  Later, the standard was changed to 16 grams of oxygen being a mole, for convenience because oxygen formed compounds with many other elements.  Eventually it was realized that elements including oxygen have different isotopes of different mass, and the mass of a particular isotope is a more specific and measurable standard than the mass of natural abundance oxygen.  12 grams of carbon-12 matches the old standard of 16 grams of natural abundance oxygen more closely than choosing 16 grams of oxygen 16.

  can we really neglect the mass of electrons, when the atoms get bigger?

The mass of a proton or neutron is about 1836 times that of an electron. So depending upon the ratio of neutrons to protons, electrons are at most 1/1836 of the mass of a neutral atom.  If accuracy of more than 1 part in 1836 is desired, it is important to consider the electrons, regardless of whether atoms are big or small.

  why do the element's mass per mole do very rarely end up being integers?

No element other than carbon-12 will exactly be an interger.  This is because masses of atoms depend upon number of protons, neutrons, electrons and binding energy.

In other words, considering that carbon-12 has an equal number of protons and neutrons (and electrons), other isotopes with a 1:1 proton/neutron ratio would have essentially integral atomic masses except for binding energy.  For example cadmium 112 is 111.90 instead of exactly 112 because binding energy is slightly more than 0.1 amu, partially offest by a higher number of neutrons than protons, the neutron mass being slightly greater than the combined mass of an electron and proton.

As pointed out by Matt Black, the natural abundance atomic masses further deviate because they are weighted averages of the masses of isotopes of a given element, weighted by their abundance on Earth.


How can I check if this mystery compound is strontium bromide? Trying to identify a mystery compound, here are some of my notes, I'm trying to get a second opinion on what it might be!
Observations
Appears as a white powder, some chunks are formed, sticking to side of tube slightly
Solubility
3 drops of distilled water, all of substance dissolved
Flame Test
Red Flame
Sodium Hydroxide Test
A white ppt was formed when mystery liquid was combined with sodium hydroxide
Concentrated sulfuric acid test
A yellow colored gas formed when 4 drops of sulfuric acid were added with substance in test tube
My thoughts based on these results narrow the cation down to lithium and strontium. After performing a precipitate test again with my two knowns (lithium, strontium)strontium was the only one to form a ppt. So I'm almost 100% sure that my cation is strontium.
I have the anion narrowed down to bromide, nitrate, and nitrite. It has to be one of those three! I'm leaning towards bromide as the anion, just interested on what you guys think from my notes.
Here are the options again, with my choice in bold:
Cation

Lithium
Strontium

Anion

Bromide ?
Nitrate
Nitrite
Iodine

To summarize:
I have a book with 16 possible cations, and 11 possible anions. Based on my results and the information given in the book, I've narrowed it down to 2 cations, lithium, and strontium, and 3 possibly 4 anions. I'm almost certain the cation is strontium, based on the fact that it formed the same white ppt. that the mystery substance did when mixed with sodium hydroxide. Lithium did not. As for the anion, I'm not sure, I'm leaning towards bromide, but was looking for a second opinion. I've been doing some research on how bromide reacts with sulfuric acid, and the results online have been almost identical to what happened with the unknown substance, however in my option book it says bromide and nitrate have very similar reactions.
If you have any more thoughts on how I could identify to anion, please let me know!

Relying on the fact that the compound is almost certainly a strontium compound, I'll look at the reactions with sulfuric acid of the specific anions:

$\ce{SrBr2}$

$$\ce{SrBr2 + H2SO4 ->  2HBr + SrSO4}$$
Both compounds $\ce{HBr}$ and $\ce{SrSO4}$ are white or colourless. BUT! concentrated sulfuric acid oxidizes $\ce{HBr}$ to bromine gas:
$$\ce{2 HBr + H2SO4 → Br2 + SO2 + 2H2O}$$
And bromine gas has a yellow/red colour. So your compound is strontium bromide.

$\ce{Sr(NO2)2}$

This is a yellow compound. As you mentioned, the compound was white, so it isn't strontium nitrite.

$\ce{Sr(NO3)2}$

$$\ce{Sr(NO3)2 + H2SO4 ->   2HNO3 + SrSO4}$$
Both compounds $\ce{HNO3}$ and $\ce{SrSO4}$ are white or colorless. After looking some more stuff up, it turns out that sulfuric acid and nitric acid don't react (at room temperature) to produce any yellow/brown $\ce{N2O4}$ gas.


What are some ways to predict that a double replacement reaction will occur? For instance, say:

$\ce{2KOH + Pb(NO3)2 -> 2KNO3 + Pb(OH)2}$

How would I even go about beginning to predict this?

To determine if a double replacement reaction will occur, you need to look at the products. You correctly balanced the equation and flipped the anions and cations, so let's take a look. A double replacement reaction will occur if one or both of the products is an insoluble precipitate, a liquid, or a gas. To check each of the products you use the solubility rules. You check each rule in order for each of the products. Here is a link to the rules I found online:

http://www.dorettaagostine.com/Solubility%20Rules.pdf

Look at the double replacement reaction chart. The rules are followed in order. Let's look at $\ce{KNO3}$ first. The first rule states that alkali metal salts are soluble. Potassium is an alkali metal, therefore potassium nitrate is soluble. The second product is $\ce{Pb(OH)2}$. Looking through the rules, none of them apply until the 5th rule which states that hydroxides are insoluble. Therefore lead hydroxide is insoluble.

As you can see, 1 of the products is insoluble, therefore the double replacement reaction will occur. I hope that helps! If you have anymore questions or clarifications just ask.

EDIT: How to find the net ionic equation.

First reduce the equation into it's respective ions.
$$\ce{2KOH(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + Pb(OH)2(s)}$$
...becomes...
$$\ce{2K+(aq) + 2OH-(aq) + Pb^2+(aq) + 2NO3-(aq) -> 2K+(aq) + 2NO3-(aq) + Pb(OH)2(s)}$$
However, keep in mind that $\ce{Pb(OH)2}$ does not break down into ions because it is insoluble, as we discovered earlier. Therefore it stays in it's full formula. At this point you do the "cancelling out" by removing duplicate ions on each side of the equation. The cancelled ions are also known as spectator ions because they do not actively participate in the synthesis of the product.
$$\ce{2OH-(aq) + Pb^2+(aq) -> Pb(OH)2(s)}$$
The potassium and nitrate ions are removed because they appear on each side of the equation in the same quantity. These cancel out leaving the net ionic equation show above.

The net ionic equation will show the creation of the insoluble, liquid, or gas precipitate that forms as a result of the reaction. I hope that helped!


Constitutional Isomers or Stereoisomers? With regard to a trigonal bipyramidal molecule, such as $\ce{PIBrClF_2}$, can there be any constituional isomers? I'm thinking no, because constitutional isomers differ in the connectivity of their atoms. The molecule only has single bonds. All are bound to a single central atom. There is no way for the connectivity of the atoms to change versus something such as ethanol, $\ce{C_2H_6O}$. I can think of two constitutional isomers off the bat:

$\ce{H_3CCH_2OH}$ and $\ce{H_3COCH_3}$ in which the connectivity of the atoms differ; in the former, we have a hydroxyl group; in the latter, we are missing the hydroxyl group.

Therefore, $\ce{PIBrClF_2}$ should not have any constitutional isomers but does have stereoisomers, correct?

Correct, no constitutional isomers, but there are stereoisomers (different substituents in the apical and equatorial positions).  Just for the record, what is your fifth substituent?


Why we need the average atomic mass? I know , Average Atomic Mass of an element is - the sum of the Masses of its isotopes,each multiplied by its Relative Abundance percentage (in decimal notation).

But what confusing me is, what is the concept of Average Atomic Mass. In other words,why we want to take average atomic mass of every element on the periodic table,why not Atomic mass ? Why we need it and when to use it?

The average atomic mass is useful because its numerical value is equal to the molar mass of the element. This in turn is useful to know how much of a solid to take when you wish to react it with some known quantity of another reagent, because weighing is typically the easiest way to quantify a substance.


Why oxidation potential of an electrode equals negative reduction potential of the same electrode? My textbook states that:

  $E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$ (of same electrode)

I know that electrons flow from anode half cell (oxidation) to cathode half cell (reduction).

But if $E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$ in each electrode, then potential in each electrode of the galvanic cell should be 0 volts (oxidation cancels reduction).

Hence potential difference in the whole circuit is 0 volts (no electricity is generated).

Do not get me wrong, I am just a high school student and I know my question might sound trivial, I am new to electrochemistry.

There exist two methods for teaching electrochemistry, and of course two schools of teachers. The two methods are equivalent, but they are exactly the opposite of one another. I will take an example to explain it correctly : Let us speak of the Daniell cell Zn/Cu, which uses reduction potentials equal to -$0.76 $ V for Zn and +$0.34$ V for Cu, if the ions concentrations are 1 molar. In this case, the overall voltage is $1.10$ V.
The Daniell cell is based on the two half-reactions :  $$\ce{Zn^{2+} + 2e^- -> Zn}$$  $$\ce{Cu^{2+} +2e^--> Cu}$$In the Zn/Cu cell, Zn is oxidized and acts as emitter of electron, which is in the opposite direction of the above equation, so the equation explaining what is going on in a Daniell cell contains the equation of Zn reversed. This becomes :$$\ce{Zn->  Zn^{2+} + 2e^-}$$  $$\ce{Cu^{2+} + 2e- ->Cu}$$When you want to describe the overall equation that goes in this cell, you add these two half-equations, remove the 2$\ e^-$ and this gives : $$\ce{Zn + Cu^{2+}->  Zn^{2+} + Cu}$$  We are here at the point where the two approaches differ. Strangely enough, one uses an addition, and the second a subtraction. Let me speak of the first method, which is used in your textbook.

First method. To obtain this overall equation, you had to add the two half-equations. To follow the same logic, you may simultaneously add the corresponding reduction potentials. But the reduction potentials of Zn and Cu are respectively $-0.76$V, and + $0.34$ V. And if you add them, you do not get $1.10$ V. So the chemists of this school have decided that when a half- equation is used in the opposite sense, with the electrons at the right-hand side, in an oxidation sense, like Zn here, the sign of the potential of this cell should also be opposite. $\ E_{anode} = - E_{cathode}$. Here : $E(Zn)_{anode} = - E(Zn)_{cathode)}$ = +$0.76$ V. With this assumption, you add $+ 0.76$ V plus $+0.34$ V and you obtain $1.10$ V, which is perfect.

Second method. Here you use the same half-reactions as described previously. But the values of the potential remains the same if the electrode works as a reducing agent or as an oxidant. It is always - $0.76$ V pour Zn and + $0.34$ V for Cu. You never change its sign. But remembering from the electricity course, that a measured voltage is a difference of potential, the voltage of the cell may be obtained by subtracting the redox potential of the cathode minus the potential of the anode, that is $E_{Cu} - E_{Zn}$. The result of this subtraction is: $+0.34$V -($0.76$ V) = $1.10$ V, which is perfect. This remains true even though you have obtained the overall equation with an addition of two half-reactions.

As you see, both approaches are equivalent. Have you understood my development ?


Carbocation stability order? so basically it's between primary benzylic carbocation which is known to be stable (benzlic) and secondary carbocation which is allylic... which is also stablized by resonance, i know that usually benzylic is more stable than allylic but when they are same degree.. here its different...
any help?

The tropylium carbocation (the one on the right) is more stable than the primary benzylic carbocation because it is not only resonance stabilised but also quasi-aromatic (which means that the charge is a part of the aromaticity) which greatly increases the stability of the cation. In contrast, the benzylic carbocation is aromatic but the positive charge is not a part of the aromaticity so it is not stabilised that much.
Hope it helps!


Why do many pharmaceuticals contain fluorine? What exactly does fluorination of some compound do in a biological sense?

I read the following things about what fluorine does for drugs:

Fluorine withdraws electron density and can make an acid stronger.
Fluorine substitution can increase lipophilicity - I don't understand this. Lipids are non-polar; I can't see why adding fluorine would generally decrease the polarity of some molecule ...
Carbon-fluorine bonds are exceptionally stable and resist change.

However, can anyone provide some concrete examples of fluorine influencing the behavior of some drug? I could only find very general points on the internet.

A definitive reference is:
"The Many Roles for Fluorine in Medicinal Chemistry" J. Med. Chem., 2008, 51 (15), 4359–4369.
In reference to the bullet points:

Fluorine obviously affects $\ce{p}{K_{a}}$ and can make an acid stronger but also a base weaker which could be advantageous when optimising a drug candidate (lead) molecule.
I guess here they refer to $\ce{-CF_3}$ which can increase lipophilicity
That's one of the most useful roles of fluorine. A lot of times you may have a very potent and selective molecule, but when you test it in vivo let's say in rats, it has no effect because it gets metabolised i.e. chemically modified to an inactive molecule by the rats liver (and other mechanisms). This can maybe prevented by changing to an $\ce{C-F}$ bond. For example electron rich aromatic sites are prone to metabolic oxidation. You can block this by fluorination of this site, exchanging the proton for a fluorine.

Have a look at the paper on top, you will find a new world of information.


Can rays be considered a type of "element"? First of all, I hardly know any chemistry, so let's get that out of the way.

I have been wondering lately, from a "common sense" point of view (owing to my lack of knowledge), that since air cannot be seen and yet is considered as being made up of elements, then what precludes us from considering things like x-rays and radio waves to be elements? What precludes the electromagnetic spectrum from being considered an element? What about light and gravity, or the four fundamental forces?

If someone could kindly elucidate the differences between these categories of "thing", I would highly appreciate it and it would remove much confusion that I have been experiencing lately.

I like your approach.  "Here's the way I see the categories of 'elements' and 'not elements.'  Why don't they draw the line this way?"  While the line you want to draw to include "rays" alongside the elements that appear in the air isn't the one science draws, it does point out something really important:  We draw the lines between categories because they are useful.

As it turns out, the differences in behaviors between "things" like hydrogen atoms and helium atoms, and "rays" like radio and X-rays is substantial as you dig further into science.  The "elements," for instance, are "massive," meaning they have mass.  Quantum scientists would say they "interact with the Higgs field."  However, you choose to phrase it, these atoms behave very differently than electromagnetic waves (which is the proper term for "rays").  Electromagnetic waves have no mass (other than some secondary effects that you'll learn when you study relativity).  Their propagation through space is well defined by Maxwell's Equations.  They don't form bonds (like most elements do).  They don't take up space (electromagnetic waves don't push each other away, but atoms can only get so close before the repulsive forces drive them apart).

Eventually, you will learn that the lines blur.  Relativity will start to explain why electromagnetic waves indeed do have mass.  Quantum Mechanics will point out that particles have wave-like properties.  However, as you start out, the differences between "massful things" and "electromagnetic waves" is pretty iron clad.  As you continue your studies, you'll become more comfortable with the distinctions drawn, and then we can slowly introduce all those strange line-blurring facets of our reality that makes science really interesting.


How to chemically postpone a chemical reaction This question is about delaying/postponing chemical reactions in general, but I will explain my point based on an example. Most of you will know the famous vinegar and baking soda experiment which allows you to make a volcano or a rocket. What I have noticed when doing this experiment (in particular the rocket experiment) is that if you just put soda in powder form in the bottle and pour in vinegar, a lot of the formed gas escapes before you can put the cork on thus resulting in a less spectacular launch.

In the video I linked above they wrap the soda in paper to allow some time between the addition of vinegar and the start of the reaction. This is a nice trick, but in many cases you don't want any paper inside your 'reaction' vessel. My question is: what are my options to chemically postpone the start of the reaction?

In general the answer to your question is no, you cannot postpone the reaction once the reactants are actually in contact.

Reaction rate usually decreases with temperature.  Thus it is possible, but not certain, that if you freeze a reaction mixture down to liquid helium temperatures the reaction would stop.  This isn't really practical in general.


Do Palladium(II) Chloride and Palladium (III) Chloride follow 18-Electron rule or octet rule? How to determine this? Does $\ce{PdCl2}$ follow 18-Electron rule or octet rule?
How should we determine this?

http://pubchem.ncbi.nlm.nih.gov/compound/24290#section=Top

http://pubchem.ncbi.nlm.nih.gov/compounds/3494742#section=Top

Palladium (II) Chloride follows neither the 18 electron rule nor the octet rule.

Instead, like many Pd(II), Pt(II) and Ni(II) complexes, palldium (II) chloride follows the 16-electron rule.

Pd(II) generally forms square planar complexes, and in palldium (II) chloride each Cl bridges two Pd, with 4 Cl coordinating each Pd in a roughly square planar arrangement.  8 d-electrons plus 2 electrons from each of 4 sigma bonds equals 16 electrons.

I have never heard of "Cl3Pd", which would involve the rare oxidation state Pd(III), but the 16 and 18 electron rules should only be applied to diamagnetic complexes.  Pd(III) complexes are paramagnetic.


Why is trans-1,2-dimethylcyclohexane an enantiomer? Can't the two be superimposed over each other after ring inversion.

No, they can't.

A model from any chemical modeling kit will convince you right away.  Or maybe a cheaper alternative made out of chewing gum and matchsticks will suffice.

Or better yet, you may do it in your head. Place the molecule in a horizontal plane (or rather, as close to a plane as possible, since the thing is not planar). Now place your eyes in the same plane, so as to look at the molecule from the side. Walk around until you are facing the bond that connects two methylated carbons in the ring. Look at that bond. Look also at the two methyl groups to the right and to the left from it. You either have the "left one up, right one down" isomer of your molecule, or vice versa. You might think "what the heck, I'll turn it around." No way. You turn each enantiomer around, and it still ends up looking like itself, and not like the other one.

Another way to think of it is via symmetry elements. For a molecule, not being chiral implies having certain kinds of symmetry. Earlier revisions of your question suggest that you were looking for some symmetry plane. Just where it might be? Think of it. And if you don't find that plane after a while, consider the possibility that it might be simply not there (which is true).

Ring inversion is irrelevant altogether. It can be done, but it will switch all equatorial substituents to axial and vice versa, so you won't be able to recognize the same molecule until you invert it again to the original form.

So it goes.


Why do most drugs (eg: oxycodone) have seemingly arbitrary names? How do some medicines derive their names?

For instance, is the name oxycodone somewhat arbitrary? I am not well-versed in Chemistry (1st semester student) but does the prefix oxy- imply some chemical meaning as well as the -codone such as some sort of functional group or grouping of molecules?

I realize such nomenclature such as di- implies "two".

The reason why the generic name of most drugs have seemingly little semblance to their chemistry is in the interests of practicality; a doctor would find it easier to write sildenafil (generic name) or Viagra (brand name) in a prescription than 5-{2-Ethoxy-5-[(4-methylpiperazin-1-yl)sulfonyl]phenyl}-1-methyl-3-propyl-1$\ce{H}$,6$\ce{H}$,7$\ce{H}$-pyrazolo[4,3-d]pyrimidin-7-one (chemical name)  unless he had a weird fetish for IUPAC nomenclature .
As for how manufacturers (and the relevant regulatory bodies) arrive at those generic and brand names,  msdsmanuals.com has a take on this,

When a drug is under patent protection, the company markets it under its brand name. When the drug is off-patent (no longer protected by patent), the company may market its product under either the generic name or brand name. Other companies that file for approval to market the off-patent drug must use the same generic name but can create their own brand name. As a result, the same generic drug may be sold under either the generic name (for example, ibuprofen) or one of many brand names (such as Advil or Motrin).

Furthermore, they add,

Generic names are usually more complicated and harder to remember than brand names. Many generic names are a shorthand version of the drug’s chemical name, structure, or formula. In contrast, brand names are usually catchy, often related to the drug’s intended use, and relatively easy to remember, so that doctors will prescribe the drug and consumers will look for it by name. Brand names often suggest a characteristic of the drug. For example, Lopressor lowers blood pressure, Glucotrol controls high blood sugar (glucose) levels, and Skelaxin relaxes skeletal muscles. Sometimes, the brand name is simply a shortened version of the drug’s generic name—for example, Minocin for minocycline.

I found equally useful information on popsci.com,

As for commercial names, pharmaceutical companies are reluctant to share their strategies. These are hugely important brands, even after the patents expire--nobody knows what Sildenafil citrate is, but everyone knows its brand name of Viagra. And the odds are you know someone who insists on brand-name Tylenol instead of discount-store acetaminophen.
Though pharmaceutical companies contacted by PopSci wouldn't share their marketing secrets, you can still derive a drug's molecular and popular heritage just by studying its name. Like, for example, Tylenol. Its chemical name is $\ce{N}$-acetyl-p-aminophenol--see the -tyl and -ol in there? Then there's Viagra. It doesn't sound like sildenafil, but it certainly suggests vigor and virility--exactly what Pfizer wants you to think. Xeljanz, which was approved earlier this year after almost 20 years of research, targets a protein called Janus kinase. It works in a different cellular pathway from other arthritis drugs. That's where you get the -jan part of that name, which, to someone with a drug decoder, connotes its uniqueness.

As for oxycodone specifically, it's anyone's guess, since that information can't be found (online) easily enough.

 Note: I've italicized all generic and brand names because, well... they're generic and brand names. Italicizing actual chemical/IUPAC names is against convention, so don't get the wrong idea ;-)


Structural determination of carbohydrates(sucrose) I recently came across a question in which we have to find out the number of moles of $\ce{HIO4}$(Periodic acid) which are required for cleavage of $\ce{C-C}$ bonds in sucrose.

I went to https://en.m.wikibooks.org/wiki/Structural_Biochemistry/Carbohydrates/Carbohydrate_Structure_Elucidation_through_Periodic_Acid_Cleavage . But this article is rather incomplete and doesnot answer the question. In general what are the things to be kept in mind in order to find out the number of moles of periodic acid required for the structural determination of carbohydrates. Any ideas?  Thanks.

First of all, lets be familiar with what per-iodic acid does to organic compounds.

It selects the carbon-carbon bonds that have either an -OH group, a =O(carbonyl) group or if it is the functional group carbon of a carboxylic acid. Both the carbons of the C-C bond must satisfy this condition. Here are some examples of the selection points of per-iodic acid:

Now that per-iodic acid has chosen the center, it cleaves the C-C bond and adds an -OH at each carbon. For every C-C bond, one per-iodic acid molecule is consumed. I am avoiding going throughout the actual mechanism to keep things simple. If any of the carbons has two -OH groups, they will spontaneously dehydrated to give a =O bond instead. For the examples shown above, the final products of the cleavage-oxidation would be:

Now, coming to your main question, the structure of sucrose would be:

By what we have seen, lets mark the C-C bonds that per-iodic acid can cleave:

Since we have 3 centers, so thats why 3 molecules of per-iodic acid are consumed per molecule of sucrose. Now you can even proceed and find out the oxidation products.

I hope this answer helps. Happy carbohydrate breaking!


Synthesis Golf IV: Plaunotol A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This fourth round of golf concerns the synthesis of plaunotol, a peptic ulcer treatment:

InChI=1S/C20H34O2/c1-17(2)8-5-9-18(3)10-6-12-20(16-22)13-7-11-19(4)14-15-21/h8,10,13-14,21-22H,5-7,9,11-12,15-16H2,1-4H3/b18-10+,19-14+,20-13-

In a departure from previous rounds of synthesis-golf, no stereocentres to worry about (hopefully easier for even more people to get involved), though several challenging alkenes for which the geometry must be controlled.

You may start from anything commercially available in the Sigma Aldrich catalogue, but to make this 'simple' challenge a bit harder, you must make each of the C=C bonds (i.e. no buying in the alkenes that form the product, but you may use an alkene if it is not preserved into the product).

16 steps total

Longest linear sequence of 9 steps

Synthesis of Precursor 1:

The hydroformylation of allylamine, Rhodium catalysis (drew the product wrong, it has one more carbon please ignore it for now, I drew the right structure in the one-pot step)
Diazotization of product, occurs immediately before addition to other precursors in one-pot step later on (produces precursor 1)

Synthesis of Precursors 2 & 3

Hydroformylation of methylacrolein, Rhodium catalysis (produces precursor 2)
Protection of aldehyde via hemithioacetal
Protection of ketone through dioxolane
Deprotection of aldehyde by base (produces precursor 3)

Formation of grignard reagent by reaction with magensium of bromoethanal diethyl acetal

One pot reaction, diazo thermolysis resulting in carbene substitution followed by aldol
Weak acid dehydration to form alpha-beta unsaturated ketone
Wittig reaction to establish alkene
Dehydration using Rhenium Catalyst with good selectivity, click here for the paper
Reduction of aldehyde (special reducing agent used just in case free ketone exists)

Deprotection of the ketone (loss of ethane-1,2-diol)
Grignard reaction followed by dehydration to form the ketone of the product
Reduction with lithium aluminum hydride (not shown) to get product, Plaunotol


On nomenclature of certain compounds I have some questions on the nomenclature-
(1) In $H_2O$ we have $H$ written first and $O$ is second but when it comes to Hydroxyl group $OH$ the $O$ comes first, Why?

(2)Why is Propanal called Propion-aldehyde and not Propane- aldehyde?

To answer (2) first, propionaldehyde is the common (Trivial) name and hence is derived and held-onto only due to historical reasons and now obsolete nomenclature reasons. The 3-C aliphatic saturated qroup was historically called Propion- and hence propanoic acid corresponds to propionic acid and propanal corresponds to propionaldehyde, and propanoate(ester) corresponds to propionate in trivial nomenclature. It is a consequence of the history of the original naming of the 3-C aliphatic group at the time of its discovery.
As for (1), it is a consequence of pure convention; just as water is not called hydrogen hydroxide!


Entropy for a closed system Case -1

In a free expansion of an ideal gas in a region or container, we cannot trace a pressure-volume path since the pressure, temperature and volume of the gas fluctuate unpredictably when the gas fills the region (container).

But since entropy is a state property, the difference in entropy must depend only on initial and final state. So we can replace the irreversible free expansion with a reversible process with same initial and final condition.

Consider an insulated container along with a thermal reservoir attached to it whose temperature can be varied by a knob. It's current temperature is T. A piston is also attached to it. Above the piston is kept a small container containing iron balls (to balance the pressure due to atoms below it). The initial state is the same as the initial state of a free expansion.

Removing the iron balls one by one, we increase the volume with an infinitesimally amount and also by keeping the temperature constant. Since it follows an isothermal path, the heat absorbed by gas from reservoir is equal to the work done by the system (positive).

W=Q

After the final condition by the reversible isothermal process is achieved,

Change in entropy = Q/T
And so, Entropy increases

To sum it up:- we can write :-

To find the entropy change for an reversible process occuring in a closed system, we replace that process with any reversible process that connects the same initial and final states.

Case - 2

Since the process is reversible we can proceed in a reversible manner, i.e we can add more and more iron balls to decrease the volume and consequently the energy must be extracted from the gas and be given to the thermal reservoir, i.e the heat Q is negative. Thus entropy must decrease.

But this does not happen since the thermal reservoir's entropy increases and the system consisting of the gas and reservoir, the total entropy change is 0.

Note - By taking the gas as the system, we do not meet the requirement of the entropy's definition of having a Closed System.

To sum up this :-

If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible process. It never decreases.

I have explained here in short about what has been given in my textbook for entropy.

My Question - In case 2, we take the system to be the gas as well as the reservoir and so the total entropy change is 0 while in case 1, we take the system to be only the gas. If I take the system, in case 1, to be gas and reservoir, the entropy should be zero. Since heat is being removed from the reservoir thus Heat Q in the entropy formula for reservoir should be negative. And thus the total change should be zero in case 1 as well.

Why doesn't this happen?

Some people define a "closed system" as one that can exchange neither heat nor work not mass with the broader surroundings.  Others, like us engineers, define a "closed system" as on that can exchange both heat and work with its surroundings, but not mass.  We engineers call a system that can exchange neither heat nor work with the broader surroundings an "isolated system."  Which definition of a closed system are you subscribing to?


Calculate the weight percent of limestone in a rock sample A 1.245 g sample of limestone ($\ce{CaCO3}$) is pulverized and then treated with 30 mL of 0.035 M $\ce{HCl}$. The excess acid requires 11.56 mL of 1.010 M $\ce{NaOH}$ to be neutralized. Calculate the %wt of limestone in the rock.

The first balanced equation would be:

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Then:

$$\ce{HCl + NaOH -> H2O + NaCl}$$

I figured out the moles of $\ce{CaCO3}$ if that's useful for this problem 1.245 g of $\ce{CaCO3}$ / 100.089 g (molar mass)= 0.0124 M

I have this but I don't know the next steps on how to solve the problem.

The question is a bit misleading ;)

Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.

So, there's apparently something else in the rock sample (that doesn't disturb the titration).

However, what you know for sure is:

the stoichiometry of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$
concentration and volume of the $\ce{HCl}$ used in excess to neutralize all the $\ce{CaCO3}$
concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left after reacting all $\ce{CaCO3}$

HINT:

Did you consider to solve the back titration experiment by starting with the second neutralisation step?

   If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...


Solvents and Enzymes I was reading recently about environmental remediation being done, using an esterase to break down / neutralize cocaine that had contaminated waterways in Europe. Are there similar enzymes that can be used to safely break down pharmaceutical drugs, like surplus SSRIs and neuroleptics, that medical staff might otherwise just flush down the toilet thus causing more pollution? A way to neutralize them into safe inert ingredients first would seem beneficial to educate people about.

One of the great properties of enzymes is their specificity. Even broadly specific enzymes such as lipases and proteases are picky compared to, say, a general base catalyst. As a result, if you are interested in degrading a single specific compound (cocaine in your example), it is very likely that you can find or engineer an enzyme to do the job. But if you want to degrade a wide variety of compounds with different structures, you will need a diverse set of enzymes, each specific for one or a few of the compounds.