Why does the nature and oxidation state of a central metal affect field split? I know the following factors affect the value of field splitting in a coordination compound:

No. of ligands surrounding the central atom.
Strength of the ligand (as per spectrochemical series)
Nature of the metal ion
Oxidation State/charge on central atom

I have understood why the first two are contributing factors, but the 3rd and 4th points do not ring a bell.

For the 4th point, i.e., ox. state of central atom, my guess was that if there is a lower oxidation state present, then less number of electrons will be present in the orbitals of the central atom, and hence due to lesser repulsion between ligands, value of field splitting will decrease.

Am I correct in this?

As for the 3rd point, i.e., the nature of the metal ion, I am completely at a loss. What does nature specify here, and how does it affect the value of field splitting?

Let me start this answer — as usually for an answer I give about coordination complexes — by showing the molecular orbital scheme of a typical octahedral complex ignoring $\ce{L\bond{->}M}$ π interactions.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

It is easy to see where field splitting comes from: two of the ligand group orbitals transform as $\mathrm{e_g}$ as do two of the metal’s $\mathrm{d}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). These interact in a bonding/antibonding manner to give $\mathrm{e_g}$ (considered a ligand-centred orbital due to the electronegativity difference) and $\mathrm{e_g^*}$ (a metal-centred orbital).

It is important to note that this scheme depends strongly on the relative positions of the orbitals. The closer the energies of the ligand and metal group orbitals, the larget the split will be.

If you consider different aquacomplexes, you can always be sure that the ligand orbitals start off from the same energy (they are the same water, after all). However, the metal orbitals will be differently positioned. The higher the first ionisation enthalpy (typically corresponding to the removal of an electron from $\mathrm{4s}$ — here $\mathrm{a_{1g}}$) the lower these orbitals will be in this scheme. Loosely, the ionisation enthalpy depends on the number of protons in the nucleus, so we might expect nickel to have lower-lying orbitals than titanium. Whatever the actual difference is (there are many more effects), no two different metals will ever have the same orbital energies and therefore any two different metals will always have different field splits corresponding to a different colour. You know this: hexaaquachromium(III) is green while hexaaquairon(III) is yellow-orange.

The oxidation state influences the orbital scheme in a similar way. Each oxidation corresponds to the removal of an electron meaning less electronic repulsion meaning lower-lying orbitals (in slightly larger contexts, this corresponds to the contraction of cations). Thus, if you change the oxidation state of a metal, e.g. from $\mathrm{+II}$ to $\mathrm{+III}$, the metal orbitals will be lowered in energy and the field split will most likely increase since there is better overlap with the ligand orbitals. Therefore, iron(II) and iron(III) have different colours in solution.


Why is proton transfer so fast? Why is proton transfer always kinetically favored? In other words, why are Brønsted acid-base reactions so quick?

Is it because protons are generally unhindered, sterically? This seems plausible; consider an alcohol. The proton isn't blocked by any bulky groups. It's on the end of a oxygen atom.

What else could contribute to the kinetic favorability of a proton transfer? Reasoning backwards - given that protons transfers are very fast - this implies that the activation energy of a proton transfer reaction is small. And as outlined above - the activation energy is small in part because most of the time, protons are unhindered. What else might contribute to the small activation energy of a proton transfer/kinetic favorability of a proton transfer?

Due to their low mass, protons can tunnel through barriers, they don't always have to go over the barrier in an activated process.  Proton tunneling becomes particularly important at low temperatures where the available thermal energy may be inadequate for an activated process (e.g. over the barrier) to occur.  A nice, brief and not overly-mathematical introduction to tunneling can be found here.

dashed line - thermally activated process

solid line - tunneling process

A second reason why proton transfer is so facile relates to hydrogen bonding.  If hydrogen bonding is present, then the proton only needs to elongate one bond (the initial bond) while simultaneously shortening the hydrogen bond in order to effect transfer from one molecule to the next.  This energetically compensating process reduces the activation energy required for proton transfer.  In solvents with hydrogen bonded clusters or chains (water would be one such example), this process can be repeated within the cluster thus allowing the proton to cover large distances (from one side of the cluster to the other) on a very short timescale.  In other words, only minor mass movement is required; by simply shortening some bonds a bit, and slightly elongating others, a proton at the left side of a group of molecules (top row in the following figure) can "reappear" on the right side of the chain (bottom row) extremely rapidly. This process is referred to as the Grotthuss mechanism after its discoverer.  Some of the finer points of the mechanism still constitute an active areas of research.


Where can I find the infrared band positions for phosphorus compounds? I am trying to interpret an IR spectrum where I have $\ce{(NPCl2)3}$

I have $\ce{P-N}$, $\ce{P=N}$ and $\ce{P-Cl}$ bonds and it is cyclic.

Where can I find or does anyone know what the general ranges for these stretches are?

I found a website that told me $\ce{P=N}$ is around $1170 \, \mathrm{cm}^{-1}$.

A good resource for IR and Raman bands is "Infrared and Raman Characteristic Group Frequencies: Tables and Charts" by George Socrates. There you can find a chart containing the positions and intensities of bands observed in the infrared spectra of phosphorus compounds (explanation: "w", "m", "s"  specify the observed intensity of the bands means whereby "w" stands for "weak", "m" for "medium", and "s" for "strong"; additionally the prefix "v" can be given meaning "very", e.g. "vs" means "very strong"):

More specifically you find the following numbers for the different bands you are looking for:

There is also another comment on $\ce{P-Cl}$ bands:

  Compounds containing the $\ce{P-Cl}$ bond have a medium-to-strong absorption
  at $605 - 435$ $\mathrm{cm}^{-1}$ due to the $\ce{P-Cl}$ stretching vibration.
  The position of the band due to the $\ce{P-X}$ ($\ce{X}$=$\ce{F}$ or $\ce{Cl}$) stretching vibration is affected by the oxidation state of the phosphorus atom. 'In the presence of
  more than one halogen atom directly attached to the phosphorus atom, two
  peaks are observed due to the asymmetric and symmetric stretching vibrations
  respectively. Difluorides of the type $\ce{-P(O)F2}$ absorb at $930-895$ $\mathrm{cm}^{-1}$ and $890-870$ $\mathrm{cm}^{-1}$


How was the composition of DNA studied before the knowledge of its structure? In a large polymer molecule (such as a DNA chain) which is made from multiple component monomers containing many atoms, how was it determined that the chain was made up from the four groups of specific combinations of monomers of elements (in this case, adenine, thymine, guanine and cytosine)?
In other words how were they able to determine that the DNA contained the four specific bases before they knew the structure of the DNA molecule?
In general, doesn't breaking up a large molecule into its constituent atoms also break up the monomers?

In the early experiments performed by Erwin Chargaff and co-workers to determine the composition of DNA, leading to the deduction of the Chargaff's rules, the total DNA was extracted, chemically digested and the products were analysed using paper chromatography. The DNA was not broken down to the level of atoms which is certainly impossible using the standard chemical treatment procedures.

This is an excerpt from one of the earliest papers by Chargaff [1]:

  Two procedures were compared. In one, which will be referred to here
  as Procedure 1, the nitrogenous components were liberated and
  determined by the methods described previously from this laboratory
  (5, 8, 12, 13); i.e., the purines were set free by N H2S04 (1 hour,
  100o) and chromatographed in aqueous butanol-diethylene glycol
  (NH3 atmosphere); the pyrimidines were liberated with
  concentrated formic acid (2 hours, 175o) after removal of the purines
  as the hydrochlorides by treatment with methanolic HCl, and
  chromatographed in aqueous butanol.

  The second method, Procedure 2, is considerably simpler and permits the
  use of even smaller quantities of DNA (2 to 6 mg. for a complete
  analysis) than was possible heretofore; its use was briefly mentioned
  in a recent paper (14). It is based on the simultaneous liberation of
  all purines and pyrimidines by the hydrolysis of the DNA with
  concentrated formic acid and on their separation, on the same
  chromatogram, by means of ammoniacal butanol. When it is desired to
  establish complete balances, i.e. to express the molar distribution of
  individual nitrogenous constituents not only in terms of P present in
  the hydrolysate, but also in relation to the total N and P of the
  nucleic acid, about 6 mg. of DNA are used in the analysis. The sample
  (dried in a high vacuum at 60o for 3 hours) was heated in a sealed
  small bomb tube (about 220 × 6.5 mm.) with 0.8 cc. of concentrated
  formic acid (98 to 100 per cent) at 175o for 2 hours. The hydrolysate
  was transferred quantitatively to a 1 cc. volumetric flask and
  adjusted to volume by means of small portions of distilled water which
  served to rinse the hydrolysis vessel. Aliquots, measured with an
  accurate microburette in the usual manner (12), were employed for the
  calorimetric P determination (0.04 cc. of hydrolysate) and for the
  chromatographic separation (0.01 cc.). When very small quantities
  of DNA were to be analyzed, 2 mg. of the substance were heated with
  0.2 cc. of concentrated formic acid and the hydrolysate was, without volume adjustment, analyzed for P and for individual purines and
  pyrimidines. In this case, all computations were based on the ratio of
  moles of nitrogenous constituent to moles of phosphorus present in the
  hydrolysate (13, 14). The chromatographic and spectroscopic techniques
  were, in general, similar to those described previously (8, 12). The
  paper sheets (16 × 46 cm.), carrying five lanes 3 cm. wide, were left
  for 20 minutes in an ammonia atmosphere in order to neutralize the
  acid. The solvent used for the chromatographic separation consisted of
  a mixture of 6 volumes of n-butanol and 1 volume of 0.6 N aqueous
  ammonia. Two beakers, one containing the solvent mixture, the other
  0.6 N ammonia, were placed along- side each paper sheet at the bottom of the jar used for chromatography; the paper sheets were suspended,
  the solvent trough being left empty, and allowed to equilibrate with
  the solvent atmosphere for 1 hour in the closed vessel. Then the
  trough was filled with the solvent mixture and the separation
  permitted to proceed for 18 to 20 hours, at which time the solvent
  front had passed over the lower edge of the paper. After the papers
  had dried in air, the position of the separated components was marked
  under a Mineralight lamp or a similar ultraviolet lamp.4 The
  components, in the order of increasing distance from the starting
  point, were aligned as follows (the distance of thymine arbitrarily
  taken as 100) : guanine (26), cytosine (47), adenine (71), thymine.
  The quantitative determinations were carried out as described before
  (12).

 Chargaff, Erwin, et al. "The composition of the desoxyribonucleic acid of salmon sperm." Journal of Biological Chemistry 192.1 (1951): 223-230.


Why do carboxylic acid derivatives show a more upfield shift than carbonyls in 13C-NMR? The prototypical 13C NMR shifts of carbonyl carbons is in the range 190–200 ppm, however, carboxylic acid derivatives have a 13C NMR shift in the range 160–180 ppm instead. What structural factors cause this shift upfield? Shouldn't the carbon in the carboxylic acid be more deshielded than the analogous carbonyl, since there are two oxygens instead of one?

Shouldn't the carbon in the carboxylic acid be more deshielded than the analogous carbonyl, since there are two oxygens instead of one?

Well, no. I love this misconception and I'm always having fun when doing oral exams about reduction reactions and grignards, which is basically the same problem.

If you look at reactivity of acids/esters and ketones/aldehydes against nucleophilic attack (grignard reagents or hydride reagents) you will see that ketones and aldehydes are much more reactive towards nucleophiles than acids/esters are. So why are ketones and aldeyhdes more electron deficient on the carbonyl carbon?

Yes, in case of acid/ester you got two oxygens both with electron withdrawing -I properties, but the additional oxygen also got lone pairs so a +M effect is possible. This actually pushes more electron density towards the carbon than the -I is pulling away leaving the carbonyl carbon in acids and esters overall more electron rich than in aldehydes and ketones.


Why does DMSO help facilitate halohydrin reactions? It would make sense, due to bond energy, that water molecules would have a hard time competing with $\ce{Br}$ in a bromonium ion intermediate.

However, I was reading that by doing the reaction in DMSO, it encourages the protonation of the water molecules. How? What is the rational behind this?

Alkenes predominantly gives relevant 1,2-dihalo product when reacted with a halogen (e.g., $\ce{Br2}$) in a organic solvent such as chloroform or dichliromethane. If the reaction is performed in water, the product is the corresponding 1,2-halohydrin due to restriction applied by polar water molecules caging $\ce{Br-}$ ion. Organic molecules are sparingly soluble in water as solvent. Thus, the halohydrin reaction is often done in a mixture of organic solvent and water. For example, bromohydrin can be achieved by using N-bromosuccinimide as the electrophilic bromine source in $\ce{DMSO/H2O}$.

Use of $\ce{DMSO/H2O}$ as a solvent has few advantages:

The bromine source, N-bromosuccinimide (NBS) is readily soluble in this solvent system [Ref. 1]).
$\ce{DMSO}$ is also participated in reaction mechanism. Reference 1 has detailed labeling procedure using $\ce{DMSO^{18}}$ to show that isotopic $\ce{^{18}OH}$ was incorporated in $\ce{OH}$ group of bromohydrin [Ref. 1].
Evedence also suggest the participation of water in the medium, probably due to increasd basic nature of water in $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}$ of water in $\ce{DMSO}$ is $31.4$ at $\pu{25 ^\circ C}$ [Ref. 2]).

Suggested mechanism for involvement of $\ce{DMSO}$ as follows:

References:

Bromohydrin Formation in Dimethyl Sulfoxide: D. R. Dalton, V. P. Dutta, D. C. Jones , J. Am. Chem. Soc., 1968, 90(20), 5498–5501 (https://pubs.acs.org/doi/abs/10.1021/ja01022a030).
Acidities of water and simple alcohols in dimethyl sulfoxide solution: W. N. Olmstead, Z. Margolin, F. G. Bordwell, J. Org. Chem., 1980, 45(16), 3295–3299 (https://pubs.acs.org/doi/abs/10.1021/jo01304a032).


Why don't we produce water from combustion? My friend who is a biology major proposed that we can produce water by burning any hydrocarbure or wood or ... to  obtain water.

$\color{red}{f(x,y,z,\Lambda)}\ce{C}_x\ce{H}_y\Lambda_z+\color{red}{g(x,y,z,\Lambda)}\ce{O_2->}\color{red}{x}\ce{CO_2}+\color{red}{\frac{y}{2}}\ce{H_2O}+\color{red}{\text{other substances}+\text{heat}}$

I know this may be a stupid question but why isn't this method used to obtain pure water? Is it maybe because it releases a lot of heat and a lot of carbon dioxyde or is it because the fuel is more expensive than water?

The answer is in the formula.  The 'other substances', which are either volatile matter or ash byproducts, are formed along with the water.  The water is not pure because some of volatile substances can condense with the water which will require further treatment for removal.  Removal of the volatile matter and the disposal of the rest of the 'other substances' will cost too much money for this solution to be economically feasible.


Protection of Alcohol I carried out an experiment and it goes like this:

2-bromoethanol + dihydropyran (DHP) ---> 2-(2-bromoethoxy)oxane

Acidic ion exchange resin was added alongside as a source of H+ ions (catalyst).

If, instead of using acidic ion exchange resin, a conventional acid such as p-toluenesulfonic acid was used as a catalyst. How can it be removed from the product?

I thought of using a weak base, but would there be any side products formed?

I have run this exact reaction with pTSA in CH2Cl2 many times. As @NotEvans says above, wash with saturated  sodium hydrogen carbonate solution, dry over anhydrous sodium sulfate, filter and concentrate under reduced pressure


Is tropone aromatic? Looking at resonance structure 1a, it doesn't seem that tropone is aromatic. However, the resonance structure 1b with $\ce{C+-O-}$ satisfies Hückel's rule for aromaticity. Is it correct to say that tropone is then aromatic?

Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon.  If you look at the resonance structures in the drawing below you'll see that structure B

depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons.  Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is considered aromatic, or stabilized, if it contains 4n+2 pi-electrons and anti-aromatic if it contains 4n pi-electrons.  On this basis, molecules like the cyclopropenyl cation (4n+2, n=0), benzene (n=1), cycloheptatriene (tropylium) cation (n=1), cycooctatetraene dianion (n=2) would be considered aromatic systems.  Correspondingly, the cyclopropenyl anion (4n, n=1), cyclobutadiene (n=1), cyclopentadienyl cation (n=1), cyclooctatetraene (if it were a planar molecule, n=2) are considered destabilized or anti-aromatic molecules.


Hybridization of oxygen in Nitrate ion and the location/bond of nitrogen's lone pair I have tried to apply the rules and basics that I learnt so far. But I am confused about the hybridization of oxygen atoms which are making the single bond in nitrate ion.

Following are the steps I used. The average bond order of 4/3 makes the molecule unstable.

Based on the steps, the single bonds are made of $\ce{sp^3}$ of oxygen and $\ce{sp^2}$ of nitrogen, the bonds are sigma bonds. However that doesn't seem like the case. If I look at $\ce{HNO3}$ dissociation equation in the water, lone electron pair of nitrogen is entirely shared with (co-ordinative covalent bond) one of the oxygens making a single bond.

$\ce{ HNO3(aq) -> H+(aq) + NO3^{-}(aq) }$

But considering $\ce{NO3}$ ion individually, without thinking about nitric acid, I can't find the long pair to be in the same type of co-ordinative bond. Isn't there an unbonding sp2 electron with nitrogen (probably one from the lone pair)?

The other issue, why isn't oxygen hybridizing to be $\ce{sp^3}$ here? What am I doing wrong here?

The actual structure of the nitrate ion is symmetrical. It forms a trigonal planar molecule.

This means that the bonds between oxygen and nitrogen are all the same. The resonance structures are a simple way to explain that structure using two-electron bonds. In reality the "extra" electron is delocalised across all three bonds (this will involve the nitrogen lone pair and nitrogen will be sp2 with no free lone pair).

The average bond order will be, as you calculate 4/3 but this doesn't, as you claim, make the molecule unstable.


Naming of [Pt(py)4][PtCl4] Why is $\ce{[Pt(py)4][PtCl4]}$ called tetrakis(pyridine)platinum(II) tetrachloridoplatinate(II) and not tetrapyridineplatinum(II) tetrachloridoplatinate(II)?

I thought you use tetrakis for ligands that contain numerical prefixes. I don't understand how pyridine contain numerical prefixes.
I saw this in this site.
The same applies for copper(II) glycinate in IUPAC naming.

Is this because of the property of the ligand?

I know this seems to be a silly question, but I can't understand why this is.

Py and Cl both ligands are monodentate . Therefore, the coordination number of Pt in both the cases are 4 and Pt in general shows +2 oxidation state for 4 coordination number. Again py is neutral ligand whereas Cl is anionic (-1). So, $\ce{[Pt(py)4]^2+}$ is positively charged and $\ce{[PtCl4]^2-}$ is negatively charged. In IUPAC nomenclature, for a coordination complex, the name of cation is written first and of anion at end (cation or anion may be simple or any coordination entity). Therefore the second name, that you told can't be accepted.

Now , the prefixes bis, tris, tetrakis are used-

When the name of the ligand is lengthy or complicated (polysyllabic or multiplicative prefixes are already present in the name of the ligand)
In cases of organic ligand.
In all other cases where using bi, tri, tetra, ... creates ambiguity because in that case a single name may indicate more than one compound .

In case of py first it is an organic ligand and despite that if you use bi/tri/tetra that may indicate a different ligand (for example bipyridine and pyridine are two different ligands) . So, the prefix tetrakis will be used here.


Will there be change in electronegativity difference in C-O and C=O? Is there any change in EN difference in C-O and C=O. If yes, why?
Does EN difference change if its bond is changed to single, double, or triple?

I think that you didn't understand the answer to this question.

A dipole is created when two equal (only in magnitude) and opposite charges are separated by some distance. The dipole moment depends on the magnitude of the charges and the distance between them. Electronegativity is just a concept that helps us to get an idea of the magnitude of the charges. So, in order to find the dipole moment, you should focus on the magnitude of the two charges rather than the electronegativity difference.

Look at the hybridization of the atoms in the single and double bond. That will tell you that both the atoms should be more electronegative after forming a double bond, but won't help you comment on the electronegativity difference.


Can I damage a biotinylated antibody in solution by vortexing or mixing too vigorously? I am running an immunoassay using biotinylated antibodies and have noticed a decreasing trend from duplicate assay runs (completed within minutes of each other) from the same vial.
In between each run:

I switch to a new low protein binding pipette tip.
I vortex the biotinylated antibody in solution in a low protein binding vial.
Then I prewet the pipette tip and deliver the reagent.

Note that degradation seems unlikely.
Any ideas on what may cause this?
Could it be vortexing or shearing during pipette tip mixing?

It is good practice to avoid vortex mixing any protein solutions, especially vigorously.  Vigorous vortex mixing can produce foaming and proteins can denature (unfold and lose activity) when they are present at an air/liquid interface (as in a foam).  This is because the hydrophobic regions of the protein will want to be in the air region and the hydrophilic regions will want to be in solution, and so unfolding may occur.
In practice, antibodies are usually quite stable as proteins go and you can often vortex them without much negative effect to your experiment. That said, if you routinely vortex proteins in the lab and come across a less stable one you will suffer degradation.  Since you will not always know in advance which proteins will be OK to vortex, it is best practice to mix gently using a pipetman (again avoid introducing air that creates a foam) or inversion, or other method. If you must vortex, then be gentle with short pulses, avoiding foaming.
I have purchased some antibodies where the datasheet specifically says not to vortex mix.


Is there a relationship between acidity of the water solution and oxidation/reduction reactions? What I want to know is if there is a relationship between the acidity of the water solution and redox reactions and if there is a relationship what favors the oxidation and what favors the reduction? Or does it depends on the reaction?

Fenton chemistry based on iron is definitely more reactive at lower $\mathrm{pH}$. This is due, in part, to speciation issues and associated solubility. Source: 'Effect of $\mathrm{pH}$ on Fenton and Fenton-like oxidation', by
Jung YS, et al., at https://www.ncbi.nlm.nih.gov/pubmed/19278159.

However, the use of chelates (like EDTA) can expand the $\mathrm{pH}$ range of reactivity (see full paper at https://www.jstage.jst.go.jp/article/jwet/11/1/11_21/_pdf ).

There are other transition metals, like Copper, which appear to be active in a much wider $\mathrm{pH}$ range where the chemistry is referred to as Fenton-type. There is an old french patent on water purification (from microbes,...) based on copper ions ($\ce{CuSO4}$ in small amounts) and natural waters containing dissolved oxygen. Interestingly, the addition of EDTA is still recommended, but now as a means to TERMINATE the redox reaction at near-neutral $\mathrm{pH}$.

There is a similar property for Cobalt ions, however, its toxicity has generally precluded it from commercial applications (like remediation of polluted waters through advanced oxidation processes referred to as AOP). Here is a reference on non-iron fenton processes at https://www.sciencedirect.com/science/article/abs/pii/S0304389414003239 and a more general review here, click on pdf download feature at https://www.researchgate.net/publication/322020986_A_Review_on_Advanced_Oxidation_Processes_for_Effective_Water_Treatment/link/5a4b0cd5458515f6b05b5221/download).


What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation? We know that the Gibbs free energy is related to the equilibrium constant by the following equation:
$$\Delta_\mathrm{r}G^\circ=-RT\ln K$$
We also know the Van't Hoff equation:
$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$

My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?

The first equation actually contains the definition of the standard equilibrium constant:
$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$
With this definition the equilibrium constant is dimensionless.

Under standard conditions the van't Hoff equation is
$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$
and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.
$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =
  \frac{{\ - \Delta H^\circ }}{R}
  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$

Now the ordinary equilibrium constant may be defined in various forms:
$$K_x = \prod_B x_B^{\nu_B}.$$

Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction,
$$\ce{\nu_{A}A + \nu_{B}B -> \nu_{C}C + \nu_{D}D},$$
this resolves in
$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$

The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = \pu{1 mol/L}$.
At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
See also a very detailed answer of Philipp.

The partial pressure is connected to the activity via
$$a(\ce{A}) =
\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
\begin{align}
\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &=1, &
p(\ce{A}) &= y_{\ce{A}}\cdot{}p.
\end{align}

Of course concentrations and partial pressures are connected via the ideal gas
\begin{aligned}
pV\ &=nRT\\
p\ &\propto \text{const.} \cdot c,
\end{aligned}
and therefore it is valid to write:
$$K_c\propto \text{const.} \cdot K_p.$$
It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.

While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.


Why does water change phase so easily? I did some undergraduate level Chemistry as part of my degree. It left me with the distinct impression that the polarity of water is responsible for its unusual phase properties: it's the only everyday substance that can exist in all three states under easily replicable conditions.

My daughter is learning about states of matter at school, and I thought this might be a fun "extra" for her homework, so I sat down to explain it to her. At which point I realised that I couldn't explain it myself. Polarity explains why water solidifies relatively easily, and why ice floats/expands - but I couldn't see/remember why it might have anything to do with transition into a gas.

So I taught her about phase diagrams instead, as a nice visual thing to take to class.

But it left me with the question: does water's polarity help explain its unusual phase properties? Or is it something else? Or are those phase properties perhaps not as unusual as I seem to remember?

does water's polarity help explain its unusual phase properties?

Not just its polarity, but its ability to form hydrogen bonds, together with the size and shape of its molecules.

Hydrogen bonds give water an unusually high boiling point for a molecule its size; many molecules of similar size that can't hydrogen-bond are gases at room temperature. They also give water a relatively high heat capacity and high enthalpies of fusion and vaporization. These latter two properties mean that ice has to absorb more heat to melt and liquid water has to absorb more heat to boil than would otherwise be necessary without hydrogen bonding.

Water's shape lets it form 4 tetrahedrally arranged hydrogen bonds. This lets it hydrogen-bond into networks of chair-shaped hexagonal rings in ice. The voids in the center of the rings make ice less dense than water. This is unusual; the solid form of a compound is most often denser than the liquid form.


Synthesis of sodium methoxide For synthesis of sodium methoxide, can we use sodium hydroxide in methanol and slow evaporation after dissolving sodium hydroxide?

No, all that does is give you back your NaOH. Its not a strong enough base to completely deprotonate methanol.
I use sodium metal in methanol, cooled in ice/water bath and stirred under nitrogen for safety.


Writing a electrochemical cell representation in correct way The answers to following question seem wrong to me. The books has plainly told correct answer is (3), but it seems incorrect to me. Please help.

The chemical reaction $\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$ taking place in a galvanic cell (under standard condition) is represented by the notation
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M KCl(aq) | AgCl(s) | Ag(s)}\tag{1}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq) | 1 M Ag+(aq) | Ag(s)}\tag{2}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} ||  1 M HCl(aq) | AgCl(s) | Ag(s)}\tag{3}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq)  | Ag(s) | AgCl(s)}\tag{4}$$

Photo of problem: https://i.stack.imgur.com/Oflyg.jpg

The chemical reaction is:

$$\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$$

If we split this into half reactions, we get an oxidation half reaction at the anode of:

$$\ce{H2(g) -> 2 H+(aq) + 2 e-}$$

And a reduction half reaction at the cathode of:

$$\ce{2AgCl(s) + 2e- -> 2 Ag(s) + 2 Cl-(aq)}$$

If you want all the species at standard state, you need a partial pressure of $\pu{1 bar}$ of hydrogen gas, a $\mathrm{pH}$ of about 0, and a chloride concentration of $\pu{1 mol/L}$.

In cells where the reagents can't react directly, you don't need a salt bridge. In the reaction we are looking at, $\ce{AgCl}$ is confined to the cathode because it sticks to it, and hydrogen gas is confined to the anode because that's where we are releasing it, and it has low solubility in water.

The cell notation starts with the conductor of the anode half reaction, and ends with the conductor of the cathode half reaction. Different phases are separated by vertical lines. If there is more than one species in a phase, they are separated by commas.

$$\ce{Pt(s) | H2(g, \pu{1 bar}) |  H+(aq, \pu{1 mol/L}) , Cl- (aq, \pu{1 mol/L}) | AgCl(s) | Ag(s)}$$

  The books has plainly told 3. to be correct.

The answer is (5) none of the above. Using the format of the answer key, the best answer would be:

$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq) | AgCl(s) | Ag(s)}$$

You get this answer from answer 3 by changing the salt-bridge boundary ("||") to a phase boundary ("|"). Having a salt bridge between the hydrogen gas and the aqueous hydrogen ions would mean that you are generating hydrogen ions at the electrode where they are not in contact with $\pu{1 M}$ hydrogen ions.


How do transmembrane protein alpha helixes bind to the membrane? I would post this in biology but as the question is biochemical in nature I decided this would be a more appropriate platform.

What I'm asking is how exactly does a transmembrane protein bind to the cell membrane? I realize that the transmembrane alpha helix is about 20 amino acids long with mostly hydrophobic amino acids as the inside of the membrane is hydrophobic.

What I don't realize is how exactly this alpha helix binds to the membrane and sticks to it? I've tried searching on the subject but I can't seem to find the answer. I found out that the alpha helix satisfies the hydrogen bonds internally leaving no polar R groups for the membrane as most groups are hydrophobic, but that only explains how it doesn't bind to the membrane.

Well, it binds via hydrophobic interacions - non-polar R groups are attracted to hydrophobic interior of the membrane via dispersion forces, and also don't disrupt the structure of water as if they would outside of membrane. It's the same thing which holds molecules of the membrane together.


Is a time domain spectrum obtainable from a frequency domain spectrum? I know for a fact that a frequency domain spectrum can be obtained from a time domain spectrum using a Fourier transform  - but can you do the reverse?

Also what are the advantages and disadvantages of the frequency and time domain spectra?

Can you do the reverse?

Yes, mathematically the Fourier transform can be reversed. If we define the spectrum $S(\omega)$ as the Fourier transform of a time-domain signal $f(t)$,

$$S(\omega) = \mathcal{F}[f(t)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\mathrm{e}^{-\mathrm{i}\omega t}\,\mathrm{d}t$$

then the inverse Fourier transform is simply given by

$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}S(\omega)\mathrm{e}^{\mathrm{i}\omega t}\,\mathrm{d}\omega$$

Wikipedia has a page on this which is rather technical, but any decent book on Fourier transforms should cover this, so a mathematics book targeted at physicists/chemists may be easier to digest.

  What are the advantages and disadvantages of the frequency and time domain spectra?

The time domain signal (in NMR, "free induction decay") is for the most part, not particularly useful. It is the Fourier transform which makes it valuable for chemists, because the frequencies that appear in the spectrum are related to transitions between different energy states:

$$\Delta E = \hbar\omega$$

In general, probing these energy levels is the main point of spectroscopy, regardless of what range of energies you are using.


Why is NAD+ regioselectively reduced at C-4 to form NADH? We could draw a resonance form with the pi electrons shifting, placing a positive charge on C-4, to which a $\ce{H+}$ along with 2 electrons are added. Thus we get an NADH molecule as shown in the picture.

If this is how the reaction happens, then why doesn't the hydrogen get added to C-2 (or C-6 which is equivalent)? There's also a resonance form where C-2 has a positive charge and could accept a hydride ion. Why doesn't this happen?

Living organisms utilize protein based enzymes to catalyze otherwise unfavorable reactions under very mild conditions. They achieve this by bringing the substrates together in a very specific manner. Dehydrogenases are a class of oxidoreductases that oxidize substrates using NAD$^+$. Here is the active site of lactate dehydrogenase:

As you can see, carbon-4 of the nicotinamide ring is oriented towards the substrate. The orientation of the molecules is why carbon-4 is selectively reduced.


What results from the reaction of 4-phenylbutanoyl chloride in presence of AlCl3? Please note that this is 4-phenylbutanoyl chloride:

My guess was that another molecule of 4-phenylbutanoyl chloride would react with the phenyl, thus forming a structure like the one below:

But this molecule seems somewhat awkward, and I am unable to name it. Is my guess correct? If yes, what would be its name?

My guess was that an other molecule of 4-phenylbutanoyl chloride would react with the phenyl [...]

Your guess is not that bad and it seems that you have identified the type of reaction.

   Apparently, it is a special type of electrophilic aromatic substutution, known as the Friedel-Crafts acylation. In your guess, one molecule of the starting material would serve as the acylating agent (through the acyl chloride), while another is acylated at the phenyl.

What if you only had one molecule of the staring material? Can you imagine a similar reaction?

   The alkyl chain of $\ce{CH2}$ that links the acyl chloride unit to the phenyl is flexible! There's no reason why the Friedel-Crafts acylation could not happen as an intramolecular reaction, where acylating agent and the attacked phenyl group are part of the same molecule.

What would the the outcome?

   The intramolecular reaction would furnish tetralone.


Electrophilic Substitution with Ceric Ammonium Nitrate and Lithium Bromide as reagents The 2020 IChO Preparatory Problem List (source here) had the following reaction as part of the solution to the first question:

(CAN = Ceric Ammonium Nitrate).
What is the mechanism for this 'Electrophilic Substitution' of the ring? I have not encountered such a reaction before. As far as I'm aware, CAN is an oxidizing agent, whereas Lithium Bromide would dissociate to give $\ce{Br-}$, which is not an electrophile. Aromatic Nucleophilic Substitution (ArSN) is out of the question here, as the ring is highly activated by the methoxy group.
The only probable reaction I can think of is that CAN successively oxidizes $\ce{Br-}$ to $\ce{Br2}$, then to $\ce{Br+}$, but this seems unlikely, and I could not find such a reaction mentioned anywhere in literature. A mechanism for this reaction, preferably along with a reference, would be much appreciated.

First of all, the linked paper clearly points out the following:

Hint: In second step (A $\rightarrow$ B), combination of lithium bromide and cerium(IV) ammonium nitrate (CAN) is used as a brominating reagent

From a problem solving perspective, this step is trivial for someone who has read the problem well (not me, unfortunately :)

Ceric Ammonium Nitrate is a very strong oxidizing agent and has the ability of oxidizing benzene compounds first to Nitrobenzene, then to quinone or quinoid derivatives1. Assuming this line of reasoning works, the quinoid derivative could then undergo nucleophilic addition with $\ce{LiBr}$, followed by workup with acid to give the p-subsituted benzene derivative. This could be a probable reaction mechanism, if the substitutent was $\ce{-OH}$ instead of $\ce{-OMe}$ and if an acid workup step was involved. Also, this mechanism can give the unwanted product 4-bromo-3-isopropylphenol, as shown below:

However, this reaction mechanism is very speculative, and it also won't hold good here, because the substituent is $\ce{-OMe}$. Some further research on the links provided in the comments2 yielded the following results:

Alkyl aromatics were brominated in the side chain using sodium bromide and CAN. Asakura et al. reported bromination of C-5 of uracil nucleosides using CAN and lithium bromide in acetonitrile.

This is also the reaction present in (3), which seems to have been lifted without a citation to Asakura et al.

Finally, Asakura et al. 4 provides a suitable mechanism for this reaction, which I correctly assumed arose from the oxidation of $\ce{Br-}$ to $\ce{Br2}$, and then to $\ce{Br+}$

References:

Nikishin, G. I., et al. “Reaction of Aromatic Hydrocarbons with Cerium Ammonium Nitrate.” Bulletin of the Academy of Sciences of the USSR Division of Chemical Science, vol. 24, no. 6, June 1975, pp. 1328–29. doi:10.1007/BF00922077.
Nair, Vijay, and Ani Deepthi. “Cerium(IV) Ammonium NitrateA Versatile Single-Electron Oxidant.” Chemical Reviews, vol. 107, no. 5, May 2007, pp. 1862–91. doi:10.1021/cr068408n.
Cerium (IV) Ammonium Nitrate mediated oxidative transformations of diaryl methylenecyclopropanes. Shodhganga link
Asakura, Junichi, and Morris J. Robins. “Cerium(IV)-Mediated Halogenation at C-5 of Uracil Derivatives.” The Journal of Organic Chemistry, vol. 55, no. 16, Aug. 1990, pp. 4928–33. doi:10.1021/jo00303a033.


Understanding the synthesis of twistane What are the products P and Q?

This was a question in my test series, but I don't have any experience with bridge head compound regarding the cis and trans transformation. I need help understanding this process.

The first step is a conversion of the nitrile into a carboxylic acid, in the literature Martin Olbrich, Peter Mayer, and Dirk Trauner (J. Org. Chem., 2015, 80, 2042-2055) used KOH in ethylene glycol at 155 oC for a couple of hours. The next reaction is attack of iodine on the alkene. The cyclic cationic intermidate will be attacked by the carboxylic acid.

I am not sure what SBH is, but it is some reducing agent which functions in the same way as hydrogen with platinium and a base. This combination of reagents removes the iodine by converting the alkyl halide into a sp3 CH. This gives us P.

Another option I would consider is to use nickel chemistry, I would put some hydrated nickel(II) chloride in a flask with ethanol, I would then add a solution of sodium borohydride in water. Do this with care as it emitts a lot of hydrogen gas. It will change the nickel solution from a nice green clear solution to an ugly black slurry. This is not a problem. Add the organic material to this. This black slurry acts like raney nickel. I have used it a few times to remove things like thioether groups from aromatics. For example if you react 2-acetyl pyridine with a base and carbon disulfide and methyl iodide you can make a new compound. If this is then reacted with 2-acetyl pyridine and base, before being boiled up with ammonium acetate in acetic acid you can form terpy with a SMe group on the 4 position of the central pyridine ring. The way to convert this SMe version of terpy to terpy is to boil it with the black slurry made from nickel and sodium borohydride. The problem is that it is hard to do oxidative addition of alkyl halides, many reagents do not work well.

If we consider the Monsanto process for acetic acid then the rate determining step is the reaction of the methyl iodide with the rhodium, if we move to the more modern BP version which uses iridium then the oxidative addition is much more easy. I think that the platinum catylst will work a lot lot better than either the nickel method I mentioned or palladium on carbon.

I worry that sodium borohydride with no modification will be unable to deal with the alkyl iodide in the moelcule.

Another option would be to use tributyl tin hydride and AIBN, this is a selective radical based reducing agent which will convert alkyl iodides into alkyl hydrides (alkanes) without touching alcohols, esters or other carboxylic acid based groups. This could be a method of putting a deuterium label into the product if we very much wanted it.

Keep in mind that if you are ever asked for an example of a free radical chain reaction other than halogenation of an alkane that you can always offer reduction of an alkyl halide by tributyl tin hydride or a free radical polymerisation reaction.

Now a lot of people hate tributyl tin chemistry, there are several disadvantages of it. Firstly it is often hard to separate the tributyl tin side products from the product. Secondly organo tin chemistry is rather toxic, the worst effects are on marine life such as oysters. Tributyl tin compounds can cause female marine organisms to turn into an intersex creature which then dies. But in our case it is not so bad. What we can do is to use a tributyl tin reducing agent and the use the acid base properties of the carboxylic acid to separate it from the neutral tributyl tin iodide.

If shaking a solution of the product in benzene with sodium hydroxide was to fail, I would suggest making a solution of the carboxylic acid in methanol and then passing this through a pad of a weak base ion exchange resin (amino groups on a polymer such as polystyrene). I would then elute the column with more methanol and then with formic acid dissolved in methanol.

The next thing is that we use aqueous caustic potash to break open the lactone ring. The lactone is then oxidised with permanganate. I would have used a chromium oxidant as they tend to be a little more gentle.

Next we react the ketone with ethylene glycol using acid and some dehydration method. One good method is to use toluene and a dean-stark head. Another one which I like is to use triethyl ortho formate with a little acid. This triethyl orthoformate method works under more gentle conditions to remove the water. We might get some ester formation but in the ideal world I would want to do the next step with an ester not a carboxylic acid.

We reduce the carboxylic acid using the cyclic acetal to protect the ketone from the lithium aluminium hydride. Then using acid in water we deprotect the ketone. We then convert the alcohol into a leaving group with tosyl chloride. We then form an enolate with the LDA which then will attack the tosylate to form the ketone.

It is important to keep in mind that under some conditons that enolates can be attacked on the oxygen by hard electrophiles such as tosylates. But I suspect in this case that the tosylate will be unable to reach the oxygen. It would be best to run this reaction at a high dilution. One method would be to slowly add the ketone-tosylate to a large flask containing a large amount of solvent and some LDA. Another method would be to convert the tosylate to the alkyl iodide by reacting the tosylate in acetone with sodium iodide first.

It may be a good idea to read the paper by H. W. Whitlock in Journal of the American Chemical Society 1962, 84 (17), 3412-3413. This is the original synthesis of the compound.


Where can I find reliable data for melting points of organic compounds? I'm interested in the melting/freezing points of organic compounds, but the online literature has data that's all over the map. I can understand that especially for isomers of alkanes, pure samples can be difficult to obtain; but as basic a property as melting point ought to be well-established and easy to reference.

For example, the m.p. listed for 3-methylpentane varies from -118 °C to -163 °C depending on source. That's an enormous spread; in fact, those two numbers reoccur frequently enough that I suspect that sources are quoting older sources possibly back to the early 20th century. And that's far from the only example I could cite.

Are there any reliable sources that have experimentally confirmed the values?

You are absolutely right about suspecting the reliability of sources, based on their value spread. I think, I found what's going on with those internet (I assumed) sources. Yet, I know everybody agree with me that CRC handbook of chemistry and physics is The Bible for Physical Constants of Organic Compounds. Here is what I found on 2005 Internet Version of CRC handbook of chemistry and physics:

3-Methylpentane (CAS #: 96-14-0): physical form: liquid; Melting Point: $\pu{-162.90 °C}$; Boiling Point: $\pu{63.27 °C}$; and Density: $\pu{0.6598 g/cm^3}$ (pp. 3-398 of Ref.1)

Perfluoro-3-methylpentane (CAS #: 865-71-4): physical form: liquid; Melting Point: $\pu{-115 °C}$; Boiling Point: $\pu{58.4 °C}$; and Density: not listed (pp. 3-458 of Ref.1)

I think some reference sources may have mistaken with values of perfluoro-3-methylpentane as those of 3-methylpentane.

Nonetheless, evem PubChem listed two other values than $\pu{-162.90 °C}$ for melting point of 3-methylpentane (Ref.2): $\pu{-162.90 °C}$ from Lide, D.R., G.W.A. Milne (eds.). Handbook of Data on Organic Compounds. Volume I; 3rd ed.; CRC Press, Inc.: Boca Raton, FL. 1994., p. V4: 3865 (most trusted value since it is PEER REVIEWED - https://toxnet.nlm.nih.gov/cgi-bin/sis/search2/r?dbs+hsdb:@term+@rn+@rel+96-14-0); $\pu{-118 °C}$ from http://www.ilo.org/dyn/icsc/showcard.display?p_version=2&p_card_id=1263 (International Chemical Safety Cards (ICSC) are data sheets intended to provide essential safety and health information on chemicals in a clear and concise way); and third value of $\pu{-245 to -148 °F}$, basically, $\pu{-153.9 to -100 °C}$, from  OSHA Occupational Chemical DB (http://www.osha.gov/chemicaldata/chemResult.html?RecNo=764). I can only speculate that they all listed wrong values because they got hits for some other substituted 3-methylpentanes with search keyword of 3-methylpentane. For example, CRC Handbook of Chemistry and Physics listed $\pu{118.27 °C}$ for boiling point of 3-ethyl-3-methylpentane (CAS #: 1067-08-9).

Yet, only CRC Handbook of Chemistry and Physics consistently listed $\pu{-162.90 °C}$ as the melting point of 3-methylpentane since 1994. Thus, I'd take that value at any time.

References:

"Section 3: Physical Constants of Organic Compounds", in CRC Handbook of Chemistry and Physics, Internet Version 2005, David R. Lide, ed., http://www.hbcpnetbase.com, CRC Press, Boca Raton, FL, 2005 (pp. 3-398).
https://pubchem.ncbi.nlm.nih.gov/compound/3-methylpentane#section=Melting-Point


Stereospecific in SN2 reaction (S)-1-bromo-1-fluoroethane reacts with sodium methoxide to give pure (S)-1-fluoro-1-methoxyethane.

This is an SN2 reaction so inversion should take place , hence the product should be R but it is not so . Why?

You're right in that the SN2 Reaction Mechanism will invert the stereocenter. However, the naming convention is based on relative priority and the mechanism by which the product is formed has little to do in how it is named. This appears to be where your confusion lies since you're associating the mechanism and the named configuration when they are not related at all

In the case of an SN2 reaction:

If the attacking nucleophile had the same relative priority as the leaving group (in this case anything with an atomic number > F, since Br is the heaviest among H, F, and C), then you're right, S would turn into R.

However, in the case you quote the nucleophile is MeO-, and O has an atomic number lower than F. Following the CIP rules, we would have to assign the absolute configuration of 'S' even though the stereocenter has been inverted relative to the original compound.


Mixing solvents and other compounds - safety issues I would like to know if it is safe to mix organic solvents together or organic solvents with some compounds listed below.

In a mixture with random ratios (greater than zero - all chemicals present) we have:

Ethanol
Methyl ethyl ketone (as denaturant for ethanol)
Bitrex (aka Denatonium - as denaturant for ethanol)
Isopropanol
n-butanol (as denaturant for isopropanol)
n-propanol (as denaturant for isopropanol)
Ethyl acetate
Acetic acid (in case that ethyl acetate hydrolyzes)
Water (from ethanol and isopropanol)
Acetone
Methanol
Naphtha (Hydrogenated)

In general, we have alcohols, ketones, denatonium, ester, organic acid, water, and a liquid hydrocarbon mixture (Naphtha).Ratios can be taken as equal for all chemicals except those used for denaturation.(I think ratios for denaturants  wouldn't matter in case they would react.)

Could some of these chemicals react together and produce any unstable or hazardous compound?

What conditions should be avoided? (Light,heat,etc.)

Any reaction (slow) that could happen with atmospheric oxygen? (opening and closing a bottle containing all these chemicals)

I don't know too much about organic chemistry, just asking to be sure this is safe.I want to test something that possibly involves mixing all the above chemicals and I don't know if they could probably react together.

Note: Question unrelated to waste disposal. I'm experimenting with a DIY electronics cleaner.

First, I would suggest you read the material safety data sheets (MSDS) for each of those products to be sure of any incompatibility that may arise. They are available throughout the web or on distributor web sites like Sigma or Fisher.

For your list of waste disposal (as it seems to be) most of your products are alcohols or water so they can be mixed without issues.
The ethyl acetate will react with water to create ethanol and acetic acid and acids may be an issue if you have great quantities. So I would suggest to react and neutralise with $\ce{NaOH}$ prior disposal.
Same thing for acetic acid: you should neutralise prior disposal.

You can also request an evaluation of price from your chemical waste service contractor, which will give you the price of each disposal category with the incompatibility management included.


How many d-electrons are in the outer shell of metals like Cadmium (this is counting the 4d electrons)? First I know similar questions have been asked about counting d-electrons.

However, my question is about the electrons in the 4d orbital. For example if I have CdS. The cadmium has a charge of +2. So does this affect the number of d-electrons?

Would it be 10 d-electrons as the 2 extra have been removed from the 5s orbital?

Also is the electron configuration of Cd in CdS:

[Kr] 5s2 4d10

or is it:

[Kr] 4d10

Without doing some sort of at least half-serious calculations, you cannot really answer the question. For all we know, a cadmium(II) ion could be $[\ce{Kr}]\,\mathrm{4d^{10}}$, $[\ce{Kr}]\,\mathrm{4d^9\,5s^1}$ or $[\ce{Kr}]\,\mathrm{4d^8\,5s^2}$.

However, given the analogies to other transition metals and the general structure of practically all transition metal ions, the first configuration, $[\ce{Kr}]\,\mathrm{4d^{10}}$ is most likely. I can’t recall a case where a transition metal ion had any s electrons in the valence shell before the d electrons were completely populated; i.e. even if you have two s electrons in the neutral atom and ionise only once you would end up with only d electrons in the ion.


Is peroxynitrous acid an isomer of nitric acid? Structure of nitric acid:

Structure of peroxynitrous acid:

Nitric acid and peroxynitrous acid have the same structural formula, $\ce{HNO3}$ and in both the compounds, O.N. of nitrogen is +5. So, is peroxynitrous acid an isomer of nitric acid?

It seems that peroxynitrous acid is a very, very unstable compound and I don't think it has been synthesised in pure, stable form. This video shows the preparation of peroxynitrous acid but it readily converts into nitric acid and thus the peroxynitrous acid yield is very poor.

Keeping aside its stability and considering only their structure and structural formula, is peroxynitrous acid an isomer of nitric acid?

Keeping aside its stability and considering only their structure and structural formula, is peroxynitrous acid an isomer of nitric acid?

Yes.

It seems that peroxynitrous acid is a very, very unstable compound and I hardly doubt that it has been synthesised in pure, stable form. This video shows the preparation of peroxynitrous acid but it readily converts into nitric acid and so the peroxynitrous acid yield is very poor.

Indeed, it is impossible to isolate peroxynitrous acid $\left(\ce{ONOOH}\right)$ in aqueous solution, due to the rapid decomposition to nitric acid. The yield is not just poor, it is zero.
However, the conjugate base of peroxynitrous acid, peroxynitrite $\left(\ce{ONOO^-}\right)$, happens to be extremely stable. (Apropos the question, peroxynitrite is an isomer of nitrate.) So, as long as a synthetic approach is devised that keeps the product strictly in its anionic/alkaline form, one can actually produce peroxynitrite solutions and salts and work with them in experiments.  In fact, I synthesized sodium peroxynitrite solutions myself as part of my Ph.D. research.  You can read about the procedure in Section 4.2.4 of my thesis (free for download).  A short literature review of various $\ce{ONOO-}$ synthesis methods is also included there.
Briefly, I loaded aqueous solutions of (1) $\ce{NaNO2}$, (2) $\ce{H2O2}$ plus $\ce{HCl}$, and (3) $\ce{NaOH}$ into syringes fitted onto a syringe pump.  I then connected the syringes to a tubing network that first mixed the acidified $\ce{H2O2}$ with the $\ce{NaNO2}$, forming $\ce{ONOOH}$, and then quickly mixed the product solution with the $\ce{NaOH}$.  The alkali "caught" the $\ce{ONOOH}$ while a significant concentration still remained, providing a bright yellow solution of alkaline peroxynitrite.  After decomposing residual $\ce{H2O2}$ by treatment with $\ce{MnO2}$, these solutions held their concentrations for a couple of years when stored at $-80\,^\circ\mathrm C$.


Is there a word for a compound that has both ionic and covalent bonds? For example, calcium carbide (CaC$_2$) has covalent C‒C bonds and ionic Ca$^{2+}$‒ C$_2^{2-}$ bonds.

There may be a strong case to call this a "salt", like other ionic compounds.  Very often, the anion (negative ion) of a salt is an ion comprised of different atoms with covalent bonds.  For example, $\ce{ SO_4^{2-} }$ is the sulfate anion, and the bonds between $\ce O$ and $\ce S$ are covalent to some degree.  So $\ce{ Na_2SO_4 }$ is an ionic compound under this definition.  Thus, any compound with some component that is strongly ionic can be called an ionic compound; i.e. a salt, regardless of the bonding structure of the components.


1H NMR of cyclohexanecarbaldehyde Can someone please explain how to assign the 1H nmr peaks of substituted cyclohexane group? I am more familiar with interpreting substituted benzene.
I get that the signal near 10ppm refers to the aldehyde group. The rest of the signal refers to the hydrogens on the cyclohexane group. I also understand that the assignment of Hb (see the picture below).

However, I don't know how I would compare the relative chemical shift between the remaining hydrogens. Since there are multiple peaks, the "Hd"s are chemically distinct. That is also observed when we use do the NMR analysis with higher resolution (See the second spectrum below).
Indeed, I have 1H NMR results with slightly better resolution. For "Hc" and "Hd"s, we can indeed resolve 4 peaks corresponding to signals at ~1.9, ~1.8, ~1.7 ~1.3ppm. The integration for these peaks are 2:2:2:4.

How would you assign the chemical shift for those 4 peaks?

the data is coming from http://www.hanhonggroup.com/nmr/nmr_en/B21091.html

Here is spectrum with a better resolution.

The picture is retrieved from https://spectrabase.com/spectrum/LEvG58hCD1J?a=SPECTRUM_LEvG58hCD1J

How would you assign the chemical shift for those 4 peaks?

Since there is overlap between some of the resonances I would run a COSY experiment to resolve the peaks and find correlations, and make assignments on that basis. A number of things can complicate assignment. First, the CHO group can be in axial or equatorial positions, leading to two overlapping subspectra under slow exchange. In addition you have different shifts for geminal Hs, in equatorial/axial positions, separated by ~0.5 ppm (equatorial further downfield), with resolution into two peaks depending on the exchange time (e.g. temperature). The equatorial/axial H are coupled with a relatively strong J-coupling (~12 Hz). The chemically equivalent hydrogens are not magnetically equivalent, but this is not very important since long distance couplings are small.


Crystal Field Explanation of Colors of Coordination Compounds Crystal Field Theory explains colors of Coordination compounds as follows : A d-orbital splits into multiple orbitals, the process being called crystal field splitting. When white light falls on the compound, an electron makes a transition into a higher state thus absorbing a particular wavelength of light. The rest of the light is reflected. This reflected light is responsible for the color of the compound.

Problem: Suppose, all electrons have now transit into a higher state. If more light falls on the compound, there should be no more transitions as all the electrons are already in the highest state. This means the compounds should lose color after some time in the light.

Question: How is the light absorbed again and again? If we suppose that the electron falls back to lower state, that would emit the same wavelength it absorbed. Even in this case, how do we explain color?

P.S. Also, if they can absorb only a particular wavelength(given quantisation), that should not bring about a color change. There must be a range of wavelengths(however narrow) being absorbed. What is the reason for that?

Your assumption that the electron emits a photon when transitioning back to the ground state is not entirely correct.

The excited electron can also transition back by transferring the excitation energy to heat energy. The compound then warms up ever so slightly and the electron is buzzing away happily in the lower $d$ orbitals.

As to your P.S.-ed question, I'll quote Wikipedia:

  The main factors that cause broadening of the spectral line into an absorption band of a molecular solid are the distributions of vibrational and rotational energies of the molecules in the sample (and also those of their excited states). [...] In gas phase spectroscopy, the fine structure afforded by these factors can be discerned, but in solution-state spectroscopy, the differences in molecular micro environments further broaden the structure to give smooth bands.


What is the name of "cis/trans" isomerism when all four groups are different? How would you call a pair of isomers on the picture below?

They look a lot like regular cis/trans isomers, except all four groups are different. I can tell they are spacial isomers (stereoisomers) but is there a more specific name for this kind of isomerism?

When all four substituents are different cis-trans type of isomerism cannot be applied.  Moreover the Cis trans system is often ambiguous because configurational descriptions have not been defined according to any general and clear set of rules. For these types,  there is system called E-Z system of nomenclature and is based on the sequence rules of Cahn, Ingold  and Prelog (CIP) rule, originally developed for naming optical isomers on R-S system


Finding structure of a compound given its reactions Question:

  An organic compound A (molecular formula $\ce{C6H12O}$) does not change the colour of acidic dichromate solution. Compound A on treatment with $\ce{H2SO4}$ produces alkene which on oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ which gives positive iodoform test. Find out structure of A.

So I know about ozonolysis, but I do not know when acidic dichromate solution changes colour. From the question an alkene is produced so I guess it may be an alcohol, but if someone an explain about it and iodoform test, then I will be able to give another try on this problem.

I have taken the liberty of identifying your unknown compound.  If you'd like to work through the problem yourself, then carefully scroll down to the part of the answer that deals with the iodoform test.  Good luck!

What we know

Just looking at the molecular formula we see that there are too many hydrogens for the 6 carbons to be in a benzene ring.
does not change the colour of acidic dichromate solution - acidic dichromate readily oxidizes primary and secondary alcohols to aldehydes and ketones respectively; tertiary alcohols are not oxidized. The aldehyde produced from a primary alcohol can be further oxidized to a carboxylic acid, this results in a solution color change from orange to green. But there was no color change, therefore there is no primary $\ce{-OH}$ or aldehyde group in compound A.
Compound A on treatment with H2SO4 produces alkene - this is a common reaction that causes deydration (elimination of water) and converts an alcohol into an olefin.  Therefore compound A is (as you suggested) an alcohol (but not a primary alcohol).
oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ - ozonolysis will cleave the double bond in the compound produced upon dehydration.  The key fact here is that the resulting product still contains 6 carbon atoms.  If the double bond involved carbons outside of a ring we would wind up with an ozonolysis product containing fewer than 6 carbons.

(image source)

Only if the double bond is contained within a ring will we retain all 6 carbon atoms.  Therefore our double bond must be in a ring.

positive iodoform test - This is our final clue, the iodoform test (or more generally the haloform reaction) is a test that detects the presence of methyl ketones and generates iodoform ($\ce{HCI3}$) in the process.

Therefore we know that there is a methyl ketone in our ozonolysis product.

Combining all of this information (non-aromatic, ring, non-primary alcohol, methyl group) leads to the following structure

(mouse over to see the answer)

which is called

   1-methylcyclopentanol.

Note: some methylated-cyclobutanols might also be considered, but when these dehydrate and produce an alkene, the double bond is likely to be placed external to the ring (or [advanced topic] external to a cyclopropane ring formed upon rearrangement) and we would not wind up with 6 carbons in our ozonolysis product.


Why is the Vitamin B complex, a "complex"? I often come across the term "Vitamin B Complex" in my biology classes and innumerable times on the back of multivitamin packets, but what does the term "complex" here, even mean?

I'm still in high-school, and the only times we ever come across the term "Complex" in our Chemistry classes is when we're dealing with Coordination Compounds, but I don't think that's what our textbook is getting at when it says "Vitamin B Complex". I compared the structures of a few B-vitamins, but they don't appear to have anything to do with the coordination Chemistry kind of "Complexes".

My teacher tells me, that the term "Complex" here is supposed to indicate the existence of a wide range of B-Vitamins (Like $\ce{B1, B2, B3,}$ etc.). But I didn't find that 'explanation' satisfying, since there are all sorts of other vitamin categories (such as D and K) that have quite a few subtypes ( $\ce{D1, D2, D3, K1, K1, K3,}$ etc) but I've never heard of a Vitamin D Complex or a Vitamin K Complex for that matter.

So my question stands:-

What does the term "Complex" mean, with regard to the Vitamin B Complex? Do  Vitamin D Complexes and Vitamin K Complexes exist as well?

It is essentially exactly what your teacher says: the term vitamin B complex is used because the structures assigned a name ‘vitamin Bx’ don’t really have anything in common although they are all somewhat important.

As to why nobody ever uses vitamin K complex or vitamin D complex: well, at least in my country those vitamins are very under-represented when compared to ABCE. ABCE vitamins are those often seen as the most important, the most valuable or similar.[citation needed] They are often advertised as being included in supermarket products. The advertising industry probably doesn’t want to ‘establish’ yet another thing they could add but would be rather cost-intensive.


How to show that when the equilibrium constant is larger than one, some real-life equilibria are still reactant favoured? My instructor assigned us with this question:

  When real examples of chemical equilibrium are used, the conclusions about reactant- or product-favoured versus the value of $K$ must be modified (e.g. real equilibria can have $K > 1$ but still be reactant-favoured). Discuss this idea, using examples of real chemical equilibria and assumed values of equilibrium concentration.

I was taught that when $K > 1$, it will be product favoured, assuming it’s already at equilibrium ($Q = K$).

How would I go about answering this question? Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

The terms reactant-favored and product-favored
As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.
First possible definition
One straight-forward definition would be to call reactant-favored reactions those for which $K$ is smaller than one, and product-favored those for which $K$ is larger than one. You would have to specify the reaction (with phase information), the temperature and the standard state you are using. With this definition, you would be correct to say that such a reaction, if initiated at standard state, would go in the net forward direction to reach equilibrium if $K$ is greater one (and therefor, according to this definition, it is product-favored). Of course, if you take this definition, the question does not make any sense.
Second possible definition You could make a definition based on whether there is "more" reactant or product at equilibrium. This is problematic for multiple reasons. First, you could have a lot of one product, little of another product, and something in between for the reactant. So is there more reactant or more product? Second, if you use amounts of substance (moles) as the criteria for "more" and have a heterogeneous reaction, the relative amounts depend on the volumes (compare oxygen in equilibrium between hemoglobin and the dioxygen in a test tube vs in the entire atmosphere).
Third possible definition If you look at the concentrations to define "more", it still is confusing under some circumstances, even for a homogeneous reaction (everything in the same phase). If the sum of coefficients of reactants is not equal to that of products, diluting the system will disturb an equilibrium, so reactant: product ratios will depend on the dilution of the system.
Answering the question

How would I go about answering this question?

You would start by asking for the definition of reactant-favored and product-favored.

Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

Once the terms are clearly defined, it would be sufficient to look at equilibrium states (and you need to know the definition of the standard state, otherwise the value of $K$ has no meaning).


Does almond extract made from bitter almonds contain any cyanide? NOTE: This question is not seeking medical advice. I am just trying to understand the chemistry behind why bitter almonds are poisonous but bitter almond extract (presumably) isn't.

A friend made pancakes and decided to add flavor by adding almond extract, the ingredients of which were "water, organic alcohol, glycerin, natural bitter almond oil". She added about a teaspoon.

Later she was feeling very nauseated and had a stomach ache. After a couple hours, the nausea went away followed by the stomach ache. Other people in the house ate the same pancakes and she was the only one who got sick. (And there similarly wasn't evidence that it could be food poisoning from the night before since she ate the same as others.)

(She does have an allergy to sunflower seeds and certain nuts, but she has no problem eating regular sweet almonds.)

I read that eating a small number (like 10) of bitter almonds can be fatal because of the cyanide. Does almond extract made from bitter almonds contain any cyanide, and could this be related to her getting sick? I tried to Google it and found lots of info about bitter almonds being dangerous, but nothing that mentioned problems eating a lot of almond extract.

Crude extracts from bitter almonds or apricot seeds do contain amygdalin, which is hydrolyzed in the human body and releases cyanide.

Food-grade bitter almond extract does not contain amygdalin, because the crude extract is typically treated with a with a mixture of calcium hydroxide and $\ce{FeSO4}$. In this process, the amygdalin is decomposed and the cyanide is extracted in the form of $\ce{Ca2[Fe(CN)6]}$.

The typical flavour of the resulting product stems from benzaldehyde.


How do you identify reducing / non-reducing sugar by looking at structure? Identifying reducing / non-reducing sugar been confusing me for a while now , I know that reducing sugar contain aldehyde or ketone group . It's easy to identify them in monosaccharides but this becomes confusing in case of disaccharides!

I fail to see carbonyl group here, How can I identify reducing / non-reducing sugar by looking at structure in picture? can anyone help me out?

Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a hemiacetal.

Once you realize that a hemiacetal can equilibrate with a carbonyl (e.g. it is a carbonyl in disguise), identification of reducing sugars becomes easier.

Here is an example from Wikipedia, it is Maltose, the same as your third compound

See the aldehyde in the open isomer drawn on the right; see how it can close to form a hemiacetal on the left.


How can we synthesize para-nitrophenol During nitration of phenol, both para- and ortho-nitrophenols will be formed.

Is there any way in which we can synthesize para-nitrophenol  only?

This can be done by first reating the phenol with $\ce{HNO2}$ Which gives us this

Then reacting with dil $\ce{HNO3}$ we get 90% yield of para nitro phenol

This happens because $\ce{N=O}$ has a very strong $- I$  effect thus it concentrates the electrons more towards the para position so on reacting with dil $\ce{HNO3}$ we get max yield of para

P.S - Another way of getting only para is using a protecting agent but  i cant recall the reaction now. I will post it later


Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron? We all know that $\ce{CuSO4.5H2O}$ is blue and it turns to $\ce{CuSO4}$ on heating which is white. I also learnt about coordination complexes and d–d transitions and how transition metal compounds are colored. Recently I came to know that $\ce{CuSO4.5H2O}$ is also a coordination compound and in its structure four water molecules are coordinated with $\ce{Cu^2+}$ and one is hydrogen-bonded to $\ce{SO4^2-}.$ So, I thought lets try to apply the theory here and see what causes the blue color of copper sulfate.

$\ce{CuSO4}$ has $\ce{Cu^2+}$ ions and I wrote down the electronic configuration as $\mathrm{(3d)^9(4s)^0}.$ Since no d–d transition is possible, it should be white.

$\ce{CuSO4.5H2O}$ is actually $\ce{[Cu(H2O)4]SO4.H2O}$ and the complex ion is $\ce{[Cu(H2O)4]^2+}$. $\ce{H2O}$ is a weak field ligand, therefore the electronic configuration is $\mathrm{(3d)^9(4s)^2(4p)^6}.$ But still no d–d transition is possible, so it should also be white!? What went wrong in my estimation?

Although the question is a bit old, I think it still hasn't been answered yet. And by looking at the given answers it seems like the discussion went into a different direction at some point.

So let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data.

As you can see, it changes from a $\ce{[Cu(H2O)4[SO4]2]^2-}$ to a let's call it $\ce{[Cu[SO4]6]^10-}$. So all effects that are typical for $\ce{Cu(II)}$ should remain, it is still an octahedron, there is still Jahn-Teller present, we only got the change from four $\ce{H2O}$ ligands to four $\ce{[SO4]^2-}$ ligands.

Those two are similar in their ligand field splitting, where the latter one is slightly weaker, so we expect a red-shift of the absorption energy. And that is also what Nagase et al. investigated on (Ref.1):

  As shown partly in Fig. 1, the $\mathrm{d–d}$ band maxima of the hydrated and anhydrous copper sulfates were observed at $\pu{13200 cm-1}$ ($\ce{CuSO4.5H2O}$), [...] and $\pu{12500 cm-1}$ ($\ce{CuSO4}$), respectively, in their reflectance spectra. This is in agreement with the expectation from the spectrochemical series, and indicates that, in a series of complexes of comparable tetragonalities, the substitution of a coordinated water molecule with a sulfate oxygen weakens the apparent ligand field strenght about $\pu{150-200 cm-1}$. In addition to this shift of the absorption maxima, the following spectral changes were observed on dehydration: (i) the width of the $\mathrm{d–d}$ band becomes gradually narrower with the advance of dehydration; (ii) a broad and very weak absorption appears near $\ce{20000 cm^{-1}}$ in the spectrum of $\ce{CuSO4.H2O}$, and, with somewhat increased intensity, in that of $\ce{CuSO4}$; and (iii) a strong absorption (probably charge-transfer band) appears in the ultraviolet, and shifts remarkably to the lower frequency side with the advance of dehydration, indicating that the $\ce{[SO4]^2-}$ ions are getting more and more strongly bonded, as more and more sulfate oxygens are getting coordinated in the $x-y$ plane of the complex and shared among many $\ce{Cu^2+}$ ions.
  [...]
  Both the slight red-shift and the narrowing of the $\mathrm{d–d}$ band contribute to the disappearance of the blue color [...]

Or, as it is summarized in the book by K. Sone and Y. Fukuda (Ref.2):

  Therefore, the disappearance of the blue color of $\ce{CuSO4.5H2O}$ caused by dehydration is only an apparent phenomenon, which is brought about by a relatively small shift of the $\mathrm{d–d}$ band to the red end of the visible spectrum (and partially into the near infrared) and perhaps also by the spontaneous pulverization of the dehydrated product; if we had eyes which could register radiation from near IR to near UV, we should certainly feel that $\ce{CuSO4}$ is much more darkly colored than $\ce{CuSO4.5H2O}$!

  Since the ligand field strengths of $\ce{H2O}$ and $\ce{[SO4]^2-}$ are seemingly only slightly different from each other, these changes will only bring about a small shift in the $\mathrm{d–d}$ band towards the red. On the other hand, the $\ce{[SO4]^2-}$ ions, which are weakly held in the axial positions of $\ce{CuSO4.5H2O}$, are gradually brought into the equatorial coordination sphere, so that their CT interaction with the $\ce{Cu^2+}$ ion will be strongly enhanced. The observed spectral changes thus come about. Somewhat similar spectral changes are also observed when the double sulfates $\ce{M2Cu[SO4]2.nH2O}$ (where $\ce{M = Na^+, K^+, Rb^+, Cs^+}$ and $\ce{[NH4]^+}$) are heated.

References:

K. Nagase, H. Yokobayashi, K. Sone, “Spectrophotometric and thermal analytical studies on the dehydration of copper(II) sulfate and its double salts,” Thermochimica Acta 1978, 23(2), 283-291 (https://doi.org/10.1016/0040-6031(78)85070-9).
Kozo Sone, Yutaka Fukuda, In Inorganic Thermochromism (Inorganic Chemistry Concepts, Volume 10); Christian K. Jørgensen, Michael F. Lappert, Stephen J. Lippard, John L. Margrave, Kurt Niedenzu, Heinrich Noth, Robert W. Parry, Hideo Yamatera, Eds.; Springer-Verlag: Berlin, Germany, 1987 (ISBN 978-3-642-51019-9).


Difference between Soap and Detergent In most articles and texts that I've read, soaps and detergents seem to be talked about collectively. I know that they are both surfactants and that soaps are usually natural, in comparison to detergent which is artificial. However, what are the differences in terms of chemical structure and properties (between soap and detergent)?
NOTE: not the same as Is there an important difference between soap and detergent?, as I'm asking about chemical structure and properties rather than the effects in everyday life.

There is no restriction that soap be natural or unnatural. Soap are salts and detergents are not necessarily salts. The latter is a very broad term. Soap has a narrow meaning.
IUPAC clearly defines soap as

"A salt of a fatty acid, saturated or unsaturated, containing at least
eight carbon atoms or a mixture of such salts."

Just note that not all soaps are used for cleaning nor there is any condition that they should be soluble in water. Insoluble soaps of metallic salts serve as lubricants.
Similarly,
IUPAC defines a detergent as

"A surfactant (or a mixture containing one or more surfactants) having
cleaning properties in dilute solutions (soaps are surfactants and
detergents)"

Just like banana is a fruit but not all fruits are bananas, in the same way, all water soluble soaps are detergents and surfactants. Not all detergents are soaps.
Visit the IUPAC Gold Book and search the entries detergents and soap.


What is the minimum operating temperature of chemical hand warmers? I have been looking at chemical hand warmers such as Hot Hands, Grabber and HotSnapZ. Unfortunately, I cannot find a minimum operating temperature for these types of devices. Logic tells us that they won't work at -273°C and will probably work at 0°C, but I'd like to have an idea of where the cutoff lies between them.

From Wikipedia here and here, it looks like they all contain some amount of saltwater. The lowest freezing temperature I can find for saltwater at saturation (26%) is -21°C. Is it reasonable to assume that a frozen hand warmer will not work? Would the other components in the hand warmer lower the freezing temperature even more?

TL;DR
What is the minimum temperature at which a chemical hand warmer can still provide heat?

They are often made of the crystallic $\ce{CH3COONa.3H2O}$, that dissolves in its own crystal water, when heated up.

As the cooled down liquid state is supercooled solution, it is hard to predict, at what temperature it would spontaneously solidify.

It will provide heat in any case. But as it warms up itself typically by  $\pu{40-50^{\circ}C}$, it would barely warm itself to room temperature, if it was previously cooled down to e.g.  $\pu{-60^{\circ}C}$.

The air activated heaters, based on oxidation of fine metal powders (thanks @MaxW ), may have problem with too slow initial reaction when too cold. So they may need to be kept at pocket temperature for some time before activation.


Using non-food grade NaOH for pH control I would like to use non-food grade $\ce{NaOH}$ for $\ce{pH}$ water treatment. Any problem with this mixing with water that contains iron?

The $\ce{NaOH}$ will be pumped into a well. As a $25\%$ solution. Then sent through an ozone filter and a water softener.  Eventually, it will be used for pet and human consumption.

Iron hydroxide is insoluble, so it will precipitate out. If there are other metals that aren't  alkali metals or alkaline earth metals heavier than calcium, they will also precipitate out. Most of these reactions will release some heat, but if the concentration of the metal ions is low, that shouldn't matter too much.

If you plan to drink the water, using non-food grade $\ce{NaOH}$ might cause a problem, because you won't know what else is in there.

If you add more details, we can probably tell if there are any other things to be concerned about.


Question about a line from Rosenfeld's paper on optical rotation There are a lot of papers on optical rotation which cite Rosenfeld's (German) 1928 paper "Quantum mechanical theory of natural optical rotation..." [Quantenmechanische Theorie der naturlichen optischen
Aktivitat von Flussigkeiten und Gasen, Zeitschrift für Physik, Volume 52, Issue 3-4, pp. 161-174] as background. Here is a paid link--

Apologies that I cannot link to the paper. I have a paper copy and can't find a paywall-free version.

The symbols below are: $\lambda$--wavelength; $c$ is the speed of light; $\nu$ is frequency; $n$ is the index of refraction; $\rho$ is a quantity that does not change between equations (42) and (43).

The last two equations in that paper (42), (43), are a conclusion about an angle of rotation $\theta:$

$$\Delta n = \frac {2\pi ~\rho}{\lambda~ n} \tag{42}$$

and

$$\theta = \frac {2\pi \nu~\cdot \Delta n}{c~\cdot ~ 2} = \frac{2 \pi^2\rho}{\lambda^2} \tag{43}$$

My question is pretty dumb. If we insert the expression for $\Delta n$ into the equality in (43) we get:

$$\frac{2\pi\nu}{c}\cdot \frac{2\pi \rho}{2\lambda~ n} = \frac{2\pi^2\rho}{\lambda^2}\to \frac{\nu}{c~n}=\frac{1}{\lambda}\to\lambda =\frac{cn}{\nu} $$

We know already--and Rosenfeld mentions this a few lines up and at eq. (21)--that

$$\lambda = \frac{c}{\nu~n}$$

which forces $n$ to be $1$. In general $n \neq 1$.

Can someone tell me what, if anything, has gone awry here?

Edit: After looking at several versions of Rosenfeld's equation I see $\theta$ is proportional to the trace of the rotation matrix...the constant of prop. can take various forms and is not the focus of this approach.

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome.

We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index.

In equation (42) of Rosenfeld's paper we have (using $\lambda n =\lambda_o$):

$$\Delta n =  \frac{2\pi\rho}{\lambda n}= \frac{2\pi\rho}{\lambda_o}.\hspace 74 mm
 ~~~(42)$$

Then in equation (43) above we get $\lambda_o^2$ in the denominator:

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{\Delta n}{2}=\frac{2\pi\nu}{c}\cdot\frac{\pi\rho}{\lambda_o}=\frac{2\pi^2\nu\rho}{c\lambda_o}=\frac{2\pi^2\rho}{\lambda_o^2}\hspace 30 mm (43) $$

and finally substitution of (42) into (43) gives

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{2\pi\rho}{2\lambda n}= \frac{2\pi^2\rho}{\lambda_o\lambda n}=\frac{2\pi^2\rho}{\lambda_o^2}. $$


Reason for vibration of molecules Why do molecules of a body exhibit vibrational motion? Is it due to interaction between various molecules of the body, or interatomic interactions between atoms in the same molecule? If it is because of any of them, how can these interaction cause the molecules to vibrate?

Here is a picture and an equation that may add to the comment LordStryker made.  Molecules vibrate because they have energy. First the equation, as $T \to 0$ (absolute zero), the first term will vanish, but the molecule will still have an energy given by $h\nu/2$.  So even at absolute zero a molecule will have some vibrational energy, this is called the "zero point energy".  The parabola (representing the hydrogen well, for example, that two hydrogen atoms will "fall into" when they bond and become a stable hydrogen molecule) gives a pictorial representation of why this must be - energy levels are quantized.  Even the lowest level is above the bottom of the parabola, so even in the lowest energy level at absolute zero, the atom or molecule will still have a non-zero energy which will cause it to vibrate.

$\displaystyle\varepsilon = \frac{h\nu}{\mathrm e^{h\nu/kT} - 1} + \frac{h\nu}{2}$


Why does N₂ react with O₂ to Form NO at high temperatures? This also raises questions that I have about the Haber Process which produces ammonia ($\ce{NH3}$) from molecular nitrogen ($\ce{N2}$) and hydrogen ($\ce{H2}$).

I have heard multiple times that bond between diatomic nitrogen is one of the strongest bonds in nature due to the fact that it is a triple covalent bond that fills the valence shells of both atoms.

I understand that at high temperatures it is possible to break this bond, but I don't understand why the resulting Nitrogen atoms wouldn't simply return to their previous bonds as the temperature cooled.

For example, I read that lightning can result in this reaction:  $\ce{N2 + O2 -> 2NO}$

Why would the atoms not return to their original bonds since they would be more stable in that manner? Is bonding indiscriminate at high energy levels? Completely random and dependent on luck?

$\Delta G = \Delta H - T \Delta S$

In the case of the $\ce{N2 + O2 -> 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature.

If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to decompose to nitrogen and oxygen.

However, that NO is unstable at room temperature tells us nothing about the rate of the decomposition reaction.  In fact there was an interesting 40 year study showing very little decomposition of NO sealed in glass tubes over that time period.  The authors' calculations show that without a catalyst the timescale of decompsition could be $10^{29}$ years!


Significant figures used to imply stated error? The question reads, " express each of the following quantities using significant figures to imply the stated error".

a)$2.3 \pm 0.001$

b)$1.989 \pm 0.0003$

I think the first one is 2.300 because the zero in the thousands place implies an error of $\pm 0.001$ but how do I use sig figs to imply an error of $\pm 0.0003$?

The use of significant figures always implies some error, that is not explicitly stated. It is therefore irrelevant how big this error is. I merely states that the last digit is rounded.

In the first case the quantity is given as $2.3\pm0.001$, which gives you the interval $(2.301;2.999)$, which is expressed as $2.300$ in significant figures.

The same applies to the second quantity, which is given as $1.989\pm0.0003$, which in turn gives you the interval $(1.9893;1.9887)$, which is expressed as $1.9890$.


Mechanism of arene side chain oxidation by permanganate When treated with hot, concentrated acidic $\ce{KMnO4}$, arenes are oxidised to the corresponding carboxylic acids. For example, toluene is oxidised to benzoic acid.

I've tried to examine how this happens, using the mechanism of oxidation of double bonds via cyclic intermediate as a reference, but I can't manage to cook up a satisfactory one.

In an older book, I have read that there is no (known) mechanism for many organic oxidation reactions. I'm inclined to think that this may have changed.

So, is there a mechanism for this? If so, what is it?

If not, what are the hurdles in finding this mechanism? For example, what problems are there with other proposed mechanisms (if they exist)?

Some general information on side-chain oxidation in alkylbenzenes is available at Chemguide:

  An alkylbenzene is simply a benzene ring with an alkyl group attached
  to it. Methylbenzene is the simplest alkylbenzene.

  Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by
  an alkaline solution of potassium manganate(VII) (potassium
  permanganate).

  Methylbenzene is heated under reflux with a solution of potassium
  manganate(VII) made alkaline with sodium carbonate. The purple colour
  of the potassium manganate(VII) is eventually replaced by a dark brown
  precipitate of manganese(IV) oxide.

  The mixture is finally acidified with dilute sulfuric acid.

  Overall, the methylbenzene is oxidised to benzoic acid.

  Interestingly, any alkyl group is oxidised back to a -COOH group on
  the ring under these conditions. So, for example, propylbenzene is
  also oxidised to benzoic acid.

Regarding the mechanism, a Ph.D. student at the University of British Columbia did his doctorate on the mechanisms of permanganate oxidation of various organic substrates.1 Quoting from the abstract:

  It was found that the most vigorous oxidant was the permanganyl ion ($\ce{MnO3+}$), with some contributing oxidation by both permanganic acid ($\ce{HMnO4}$) and permanganate ion ($\ce{MnO4-}$) in the case of easily oxidized compounds such as alcohols, aldehydes, or enols.

The oxidation of toluene to benzoic acid was one of the reactions investigated, and a proposed reaction mechanism (on pp 137–8) was as follows. In the slow step, the active oxidant $\ce{MnO3+}$ abstracts a benzylic hydrogen from the organic substrate.

  $$\begin{align}
\ce{2H+ + MnO4- &<=> MnO3+ + H2O} & &\text{(fast)} \\
\ce{MnO3+ + PhCR2H &-> [PhCR2^. + HMnO3+] & &\text{(slow)}} \\
\ce{[PhCR2^. + HMnO3+] &-> PhCR2OH + Mn^V} & &\text{(fast)} \\
\ce{PhCR2OH + Mn^{VII} &-> aldehyde or ketone} & &\text{(fast)} \\
\ce{aldehyde + Mn^{VII} &-> benzoic acid} & &\text{(fast)} \\
\ce{ketone + Mn^{VII} &-> benzoic acid} & &\text{(slow)} \\
\ce{5 Mn^V &-> 2Mn^{II} + 3Mn^{VII}} & &\text{(fast)}
\end{align}$$

The abstraction of a benzylic hydrogen atom is consistent with the fact that arenes with no benzylic hydrogens, such as tert-butylbenzene, do not get oxidised.

Reference

Spitzer, U. A. The Mechanism of Permanganate Oxidation of Alkanes, Arenes and Related Compounds. Ph.D. Thesis, The University of British Columbia, November 1972. DOI: 10.14288/1.0060242.


Stability of CI4 and PbI4 I have read in my textbook that $\ce{CI4}$ is more stable than $\ce{PbI4}$ due to inert pair effect. But shouldn't we also consider the effect of steric repulsions? In $\ce{CI4}$, four large iodine atoms surround the small carbon atom, which should make it less stable than $\ce{PbI4}$.

Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms.  Even so, carbon tetraiodide is definitely established and in use as an iodinating agent.  Storage below normal room temperature is recommended.

Carbon tetraiodide is also very red.  Ordinarily saturated carbon compounds are white or colorless because the electronic transitions involving single bonds to carbon require ultraviolet radiation to go, but the relatively weak bonds in carbon tetraiodide enable transitions in the visible light region.  Iodoform is more lightly colored for a similar reason.

Lead tetraiodide, meanwhile, does not seem to be definitely known to exist.


Why doesn't water evaporate in a balloon filled with water? I know for sure that if you fill a balloon with water (without leaving any air inside) and you start heating it, the balloon won't expand until the temperature reaches 100 °C.
I think it is because while the temperature is below 100 °C, the vapor pressure of the liquid is <760 mmHg so any expansion of the balloon is prevented by the atmospheric pressure.
However, let's suppose to put some water in a water bottle and close it. Initially, there was no water vapor in the bottle, there was only air. But, after a certain amount of time, water will evaporate until the pressure of the water inside the bottle reaches the vapor pressure of the water at that given temperature.
So, my question is: why can water evaporate if there is some air over it but it cannot without air, although in both cases it has to overcome the atmospheric pressure? What is the difference between a closed container with no air inside (only liquid) and a closed-one with some air inside?

The difference is the empty space in the latter case and lack of it in the former. By "empty" I mean "filled with air", but that's irrelevant; it could have been truly empty with the same result. Air pressure is also irrelevant; it could have been 10 atm, and that still wouldn't prevent water from evaporating.

See, when several gases occupy the same volume, they do not even know about each other. They do not care if another gas enters the same volume and starts to share it with them. But they care a great deal about the volume itself, and try to increase it with all their might; that's what we call pressure.

So it goes.


How does neutralization of free radicals by $\beta$-carotene work? In my understanding, by bonding to the free radical species, say by one of the double bonds, a new free radical on the $\beta$-carotene develops.

$\beta$-Carotene is a highly delocalised $\pi$ system.
According to George Britton there are numerous mechanisms to achieve this.

Oxidation
Oxidising radicals can remove an electron from the carotene molecule.
$$\ce{CAR + .R+ -> .CAR+ + :\!R}$$
Reduction
Analogous it can accept an additional electron.
$$\ce{CAR + .R- -> .CAR- + R}$$
Hydrogen Abstraction
It is possible to abstract a hydrogen atom from a saturated carbon in allylic position.
$$\ce{H-CAR + .R -> .CAR + HR}$$
Addition
You are correct in assuming, that a free radical may form a bond with that system. This is most common for hydroxyl radicals.
$$\ce{CAR + .OH -> .CAR-OH}$$

You are also correct in assuming, that another radical will always be created. However, the single electron will be very delocalised (resonance stabilised) throughout the whole system and not nearly as reactive as the free radical before.

The remaining molecule can then of course attract more free radicals to form other singlet species, which the body can digest or excrete.


Reimer-Tiemann reaction on indole According to me net reaction should be at $\ce{C3}$ when Reimer-Tiemann reaction is done on indole (see image), since in the resonance structure, negative charge is on this carbon.

But, the answer given is on $\ce{C2}$, and the solution given is

  In pyrolle, the most electrophilic site is $\ce{C2}$. Think in a similar way.

But, I can't either find fallacy in my reasoning, not do I convince with the solution.

TL;DR Where will dichlorocarbene attack on indole (am not interested in ring expansion).

The paper by Order and Lindwall [1] agrees with you that the R-T reaction gives indole-3-carboxaldehyde.

  Since its isolation in 1903 by Hopkins and Cole (1), 3-indole aldehyde has been investigated very little. It was first prepared by Ellinger (2) from indole through the use of the Reimer-Tiemann reaction. This method was improved on later by Boyd and Robson (3).

References

Order, R. B. V.; Lindwall, H. G. 3-Indole Aldehyde and Certain of Its Condensation Products. J. Org. Chem. 1945, 10 (2), 128–133. DOI: 10.1021/jo01178a006.