What is an aromatic alcohol? The following question was given in NCERT Exemplar Problems (Class XII):

According to Wikipedia:

  In organic chemistry, the aromatic alcohols or aryl-alcohols are a
  class of chemical compounds containing a hydroxyl group (—OH) bonded
  directly to an aromatic hydrocarbon group, in contrast to the benzyl
  alcohol, where the hydroxyl group is bonded indirectly to an aromatic
  carbon atom.

But, I didn't understood the part: aromatic alcohols or aryl-alcohols (as aromatic alcohols are not always aryl alcohols).

Further, we know that aryl halides are compounds in which halogen group (-X) is directly attached to an aromatic ring. In the same way, aryl alcohols also mean that those compounds in which hydroxyl group (-OH) is directly attached to an aromatic ring.

Hence, according to my understanding, the answer should be (ii), but the answer given is (iii).

Where I am wrong? And, what is an aromatic alcohol?

'Aromatic alcohol' is not a term that appears anywhere in the IUPAC literature - while its meaning is evident, it doesn't make sense in the same way that 'aromatic heterocyclic' does so is probably best avoided.
What is defined by IUPAC is the definition of a phenol, and the definition of an aryl group. The definitions below come from an old IUPAC glossary of  'class names' which, although old, is still referenced in the most recent organic nomenclature guidelines and the most recent IUPAC Gold Book.

  phenols:
  Compounds having one or more hydroxy groups attached to a benzene or other arene ring

  aryl groups: Groups derived from arenes by removal of a hydrogen atom from a ring carbon atom.

  Glossary of Class Names of Organic Compounds and Reactive Intermediates Based on Structure.
  Pure App. Chem. 1995, 67, 1307

To actually answer your question, the answer is (II) - as Jerepierre alluded to, the NCERT books / guides are renowned for being less that accurate, and a quick search around chem.SE will reveal just how many errors there are! In this case, its hard to tell if the error is just due to typesetting (someone typed III instead of II), or it it lies with the ambiguity of the term 'aromatic alcohol' (is it referring to phenols, or is it referring to a compound which contains both an aryl ring and a hydroxyl group).

I think its fair to assume that the question is asking you to identify the phenols, in which case the answer is (II) is because A and D are the only structures in which the hydroxyl group is directly attached to a benzene or other arene ring, which is our working definition of a phenol / aromatic alcohol.


Are any other orthonormalizations schemes besides Lowdin normally used when solving SCF equations? When solving equations of the form:

$$F(C)C = SC\epsilon$$

(e.g. Hartree-Fock-Roothan or Kohn-Sham-Roothan) where $F$ is the fock matrix, $C$ is the coefficient matrix and $S$ the overlap matrix, the usual way to proceed is to first perform a change of basis and orthonormalize the AOs (atomic orbitals) by performing a Lowdin (Symmetric) orthogonalization. I know at least one case (The Psi4 program) where a different orthogonalization scheme is used (Canonical) but I haven't been able to find any information of other alternative orthogonalization schemes implemented in any programs. Are any other schemes used? If so, where and why?

Molcas developer here, currently implementing symmetric and canonical orthonormalization for the RASSCF module in addition to Gram-Schmidt.

As a rule of thumb you should treat those Orthonormalization schemes (ON-schemes) as different tools for different jobs.

If there are $n$ linear independent atomic orbitals (AOs), of which $n$ molecular orbitals (MOs) are to be constructed, then symmetric ON is probably the best, because the squared deviation of the MOs to the AOs is minimized. This means that your MOs are as close to the AOs as possible, while still being orthonormal.
If there is linear dependency, symmetric orthogonalization can detect, but not cure it.

Canonical and Gram-Schmidt orthonormalization have the advantage, that they can be used to construct only $m$ MOs from $n$ AOs ($n > m$) and to cure linear dependence.

If you compare Gram-Schmidt with Canonical ON, Gram-Schmidt has the disadvantage, that it is not as numerically stable. Besides the AOs that are orthonormalized last, can look completely different after the procedure. If you manually constructed your basis with chemical knowledge, this is not what you want. Modified Gram-Schmidt cures the numerical instability, but still has the disadvantage, that MOs might look completely different to the initial AOs.

PS: Just an addition to the comments: Grahm-Schmidt can be written as linear transformation. Although one would rarely do this.


Mechanism for Nagata Alkylation I am looking into the mechanism for the Nagata Alkylation; however, I cannot find anything more specific than the mechanism listed below.

I am only refering to the formation of the first intermediate, not the cyclization of the ortho-quinone methide. I would assume that the phenylboronic acid first forms an ester with the phenol and then the lewis-basic carbonyl attacks the boron. This might then make an electrophillic carbon that can undergo an alkylation reaction. I am not very sure about this proposal of mine, and I was wanting to ask if anyone had any suggestions as to how this mechanism might be occuring.

This reaction of ortho-alkylation of phenols was first indroduced by Nagata and coworkers in 1979 (Ref.1). Accordingly, for example, reaction of a phenol with a large excess of an aldehyde and a  slight excess of benzeneboronic acid in refluxing benzene (or toluene) in the presence of $\pu{0.1 mol}$ equivalent of propanoic acid (or trichloroacetic acid) gave about 97% of the expected product, a 2-phenyl-4 H-1,3,2-benzodioxaborin, as a crystalline product (Ref.1):

Note that the presence of propanoic acid is critical in the reaction (no propanoic acid, no expected product). The authors have suggested that mechanistically, the reaction would have taken place by coordination of an aldehyde molecule to the boron atom of a probable intermediate I (i.e. a substituted  phenyl benzeneboronate), followed by [3,3]-sigmatropic rearrangement (as indicated in II) to give the product, a substituted benzodioxaborin.

References:

Wataru Nagata, Kyo Okada, Tsutomu Aoki, “ortho-Specific $\alpha$-Hydroxyalkylation of Phenols  with Aldehydes. An Efficient Synthesis of Saligenol Derivatives,” Synthesis 1979, (5), 365-368 (DOI: 10.1055/s-1979-28683).


What is the meaning of "equivalent weight" and its formula? What is meant by "equivalent weight" and how does its formula E=M/V support the definition?

Definition of Equivalent Weight (also known as equivalent mass):

The simplest definition of equivalent mass I can think of is the following:

The mass of a compound which will combine with 8g of oxygen or 1g of hydrogen is said to be the equivalent weight of the compound.

Lets take an example,

$\ce{2Ca + O2 -> 2CaO}$

From basic stoichiometry, you know that 2 moles of calcium react with 1 mole of oxygen to produce 2 moles of calcium oxide. We can say the same statement in terms of masses as 80g of calcium reacts with 32g of oxygen to produce 112g of calcium oxide.

Since 80g of Calcium reacts with 32g of oxygen, 20g of calcium should react with 8g of oxygen.

Now by the definition of equivalent weight mentioned earlier, the equivalent weight of calcium is 20g.

$$E = \frac{M}{V}$$

The formula you were talking about gives the equivalent weight of an element (can be extended to be used with ions).

Equivalent weight = (Atomic mass)/(Valency)

This could be intuitive and obvious for some. Something with valency of 1 would combine with just 1 hydrogen atom, or in other words, a mole of compound with valency 1 would react with 1g of hydrogen. If the element has a valency of 2, then it would love to combine with 2 hydrogen atoms, or in other words, a mole of the element will combine with 2g of hydrogen. By definition, equivalent weight is the mass of the substance that would react with 1g of hydrogen. Therefore you must divide the mass by 2 to get equivalent weight. In fact, this formula based definition of equivalent weight of an element is identical with the definition mentioned earlier in this answer.

Previously, we determined that equivalent weight of calcium was 20g using the other definition. We should obtain the same equivalent weight for calcium by using the formula.

Equivalent weight = (40g)/2 = 20g

Definition of equivalents:
Another terminology you might come across is equivalents.

Equivalents is similar to moles. The number of moles is given by (mass of the compound)/(molecular mass). Similarly the number of equivalents is given by (mass of the compound)/(equivalent weight).

For example, 1 equivalent of calcium contains 20g of calcium. 2 equivalents of calcium contain 40g of calcium.

The utility of equivalent weights is amazing. In any reaction, $N$ equivalents of any compound on the reactants side reacts with N equivalents of another reaction to produce $N$ equivalents of the product(s).

If that sounds complicated, this example should make it clear.

20g of calcium reacts with 8g of oxygen to produce 28g of calcium oxide.

An equivalent statement is, 1 equivalent of calcium reacts with 1 equivalent of oxygen to produce 1 equivalent of calcium oxide.


How to determine source of air pollution It should be determined by some metal markers (meaning originating from traffic etc.) but I cannot find any source which would describe it.
Thanks.

The main idea is the following: you will need multiple sampling points and record enough data to obtain a representative dataset. Then when drawing wind roses with high percentiles (let say P95 or above) you will be able to easily point out an unique source of a relevant and exclusive pollutant.

But of course, reality is much complex... And there are some adverse effects that you must address before pointing out the source with a reasonable confidence.

First you better be not so close to the source, in order to measure immission instead of emission. Then environment and local conditions will average out. And wind speed, wind direction and concentrations become meaningful. In this case high percentiles likely originate from your source.

Second, your analyte must be relevant and if possible exclusive from your source. Measuring CO2 is not a good choice when monitoring traffic (it is stable, widespread thus global and not exclusively produced by traffic), NO is a better one because it is local, reactive and strongly related to high compression ratio thermal engine combustion source.

Third, if there are multiple isotropic sources, you will get nothing useful using wind roses. And your problem becomes really tricky.


Difference between Resonance Effect and Mesomeric Effect While studying electron displacement effects in organic chemistry, I read that the Resonance Effect and Mesomeric Effect are the same. Every source I checked used the heading or opening "The Resonance (or Mesomeric) Effect is..."

However, I was able to find the difference between Resonance Structures and Mesomeric Structures is and found out that the two differ in how they depict conjugated pi bonding. The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

So my understanding from this difference between the structures is that the effects must only differ in how they are depicted. But I was unable to find a source that could help confirm this. Can someone please confirm or explain it for me?

The IUPAC gold book (IUPAC are the organisation which 'define' standards for things such as chemical nomenclature) describes them as being essentially synonymous. The following definitions are given (link):

Resonance  In the context of chemistry, the term refers to the
representation of the electronic structure of a molecular entity in
terms of contributing structures. Resonance among contributing
structures means that the wavefunction is represented by 'mixing' the
wavefunctions of the contributing structures. The concept is the basis
of the quantum mechanical valence bond methods. The resulting
stabilization is linked to the quantum mechanical concept of
'resonance energy'. The term resonance is also used to refer to the
delocalization phenomenon itself.
Mesomerism Essentially synonymous with resonance. The term is particularly associated with the picture of π-electrons as less localized in an actual molecule than in a Lewis formula. The term is intended to imply that the correct representation of a structure is intermediate between two or more Lewis formulae.

There does seem to have been some historic difference that has largely been reconciled into a single definition however one use of Mesomerism that still gets commonly used is the definition of groups as having a positive (+M) or negative (-M) effect to describe the effect of a group on a conjugated system.

The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

This statement is not entirely correct.  The 'Mesomeric' structures you talk about are usually refered to as resonance structures/forms. The depiction with the dashed lines and partial bonds is a resonance hybrid.


% coefficient of variation (%CV) for log-linear and log-log regression (calibrations) I have a data set comprising the peak areas of an analyte measured in spiked calibration samples at 'known' and different concentrations levels. For each concentration level I have 5 replicates. My goal is to generate a calibration curve from this data and then, for each concentration level, determine the precision of the 'estimated concentration' expressed as percent coefficient of variation ($\%CV$).
If a linear relationship existed between the measured peak areas and 'known' concentrations, then for each concentration level I would simply calculate $\%CV$ as the standard deviation of the estimated concentration at a given 'known' concentration level and divide by the mean of the same estimated concentrations, before multiplying by 100. For my data set, however, I observe an inadequate linear (i.e. straight line) fit between measured peak areas and 'known' concentrations. Furthermore, there is heteroscedacity of the residuals when fitting a linear model.
To address the above, I've performed a $\log_{10}$ transformation of BOTH peak area and 'known' concentration. An adequate linear fit is observed. I would now like to calculate the precision (coefficient of variation, $\%CV$) of the estimated peak area based on this model.
According to the article cited below, the %CV for log-transformed data would be calculated as:
$$\%CV (\text{estimated concentration}) = 100\%  \cdot \sqrt{10^{\ln(10)\theta^2_{\log}} -1}$$
Where (if I understood correctly): $\theta^2_{\log}$ is the variance of the $\log$-transformed data.
So, I would specifically like to know: is the formula proposed by Canchola, et al. appropriate in the case where BOTH the response (i.e. peak area) and predictor (i.e. 'known' concentration) variable have been transformed?
In my mind, seeing as I would consider the variable of the estimated concentration on the $\log_{10}$-transformed scale, the formula outlined by Canchola, et al. should be fine.
Finally: if I had only $\log_{10}$-transformed the peak areas and then estimated the concentration (i.e. log-linear relationship), would I need to use the Canchola, et al. equation?
Referenced article: Jesse A. Canchola,  Shaowu Tang, Pari Hemyari, Ellen Paxinos, Ed Marins, "Correct use of percent coefficient of variation ($\%CV$) formula for $\log$-transformed data," MOJ Proteomics & Bioinformatics 2017, 6(4), 316-317 (DOI: 10.15406/mojpb.2017.06.00200).

Following the advice of @Buck Thorn and @user1271772, I asked this same question over at the Cross Validated Stack Exchange.
There, I received a very good answer and am providing the link for the interested reader: % coefficient of variation (%CV) for log-linear and log-log regression (calibrations)
To summarise: yes, the formula is appropriate where the predictor (concentration) and response (peak area/other metric) have both been log-transformed.


Are all deliquescent salts highly soluble? $\ce{Na2CO3}$ is deliquescent in nature, and $\ce{NaHCO3}$ is not, for example, so $\ce{Na2CO3}$ should have been more soluble than $\ce{NaHCO3}$ in water ($\ce{NaHCO3}$ I always thought was less soluble since it precipitated easily in the Solvay process). But my book says $\ce{Na2CO3}$ is less soluble than $\ce{NaHCO3}$ in water. Isn't deliquescence though an indicator of high solubility in water?

As already pointed out, dissolution and deliquesence are different phenomenon.

For dissolution to take place, the energy produced by hydration of the salt's constituent ions should compensate for the energy required to break the crystal lattice.

Solubility then depends on temperature, pH and also the amount of salt already present in the solution. (Recall: definitions of solubility products)

Hygroscopic characteristics in general can be attributed to the both absorption and adsorption (at the surface).

Particularly, deliquescence occurs when the vapour pressure of the saturated aqueous solution of a substance is less than the vapour pressure of water in the ambient air.

When water vapour is collected by the pure solid compound, a mixture of the solid and liquid or an aqueous solution of the compound forms until the substance is dissolved and is in equilibrium with its environment; when this happens, the vapour pressure of water over the aqueous solution will equal the partial pressure of water in the atmosphere in contact with it.

A crystalline salt aerosol particle will deliquesce in the atmosphere when the relative humidity surpasses a characteristic value, i.e deliquescence point.


Is it correct to look at plane of symmetry in substituted cyclohexane for determining a meso compound? I read this question and it's answer by @orthocresol. Now, he has written that the compound in question is meso not because there is a plane of symmetry but because of the rapid flipping of the cyclohexane ring, which makes sense to me. In reality the ring is not planar which seems to be in 2D. But, if we don't consider that ring is 3D and that it actually isn't planar, and we assume that it is planar, answer still comes out to be same, that is, for an actual meso compound, we are able to find a plane of symmetry in the 2D drawing and for a non meso compound, there is no plane of symmetry in the 2D drawing.

So, my question is why from that view also answer still comes out to be same? Is it a coincidence?
In this question, @Chakravarthy Kalyan writes "stereochemical analyses such as the preceding one can be done with drawings using planar rings", but with no explanation why. If we consider the compound 1-(chloromethyl)-3,5-dimethylcyclohexane, this is a meso compound with no plane of symmetry in the 2D drawing. So isn't it a counter example to his claim? If that is true, then why at all  people use the method of plane of symmetry for determining meso compounds in cyclic rings?

Operatively:
A plane of symmetry in a 2-D projection of a cyclohexane is a sufficient condition.
Alternatively, and lengthy, you conduct an analysis involving conformers but not limited to the flipping of the ring, that means you must let the chloromethyl rotate, in your example.

In more details:
The answer to your title question is YES providing that the molecule isn't blocked in one conformer.
The latter situation requires low T, or the presence of more than one bulky substituent. At room temperature and within the context of common practice and exercises we can be sure that the condition is met.
I suspect that cases in which the situation can rapidly escalates towards complexity exist, as for heavy substitutions alters the conformational distribution, too.
The second question is basically
"How is that/why a plane of symmetry in a 2-D projection of conformers which themselves have no symmetry but are enantiomers does effectively work as a criterion?".
Well, a laconic but solid answer is that there is no other possible outcome.
There must be a formal mathematical way to express this fact; I am thinking of symmetry and projection operations.
Yet, I can't do better than visualise the situation, but this is again following the reasoning and depicting the sketch within the answer you have linked, although mentally.
It is the flipping between enantiomeric conformers that makes the two simultaneously equalise each other, and this necessarily requires the two stereocentres having opposite absolute configuration and being at opposite sides of a symmetry plane of the molecule's projection, otherwise they would forcedly be different isomers and there won't even be this discussion.
You can also think of the structure as having the axial and equatorial substituents in an averaged  intermediate position, as for the two limits can't be resolved.
Indeed, a 2-D projection is in a way a real representation of the molecule - not brutal alteration is operated. It just says what it's supposed to tell, i.e. how one would see the molecule as watched from above the projection plane.
During the flipping, one would see a rapid on-plane oscillation of the ax and eq substituents in an alternate fashion, and this require a plane of symmetry in order for the two to compensate their effects.
To make this visualisation clearer, I've sketched a rudimental figure showing the motion of ax / eq substituents and their average position as ×

Considering that the motion of the substituent can be taken as a handle for the entire conformational flipping, one can see the relevance of a symmetry plane of the projection. In such a picture there is no reason to expect optical activity, as if the molecule would be flat.
However, while the picture just above is a nice visual, the optical inactivity originates from the simultaneous presence of both enantiomers, and this should be right the reason for deprecating the old term internal raceme in favour of meso compound.


Balancing the reaction of potassium permanganate and sulfuric acid with hydrogen peroxide A problem on redox reactions invites the reader to balance the actual reaction from which the redox simplification arose. After some hunting I found the balanced equation and attempted to re-derive the coefficients using:

$$\ce{a H_2SO_4 + b KMnO_4 + cH_2O_2 -> dK_2SO_4 + eMnSO_4 + fH_2O + gO}$$

We can quickly derive some valid relations by looking at each element:

$$
\begin{array}{cll}\\
\ce{H}&\quad a+c &=& f\\
\ce{O}&\quad 4a+4b+2c&=& 4d+4e+f+g\\
\ce{S}&\quad a&=& d+e\\
\ce{K}&\quad b&=& 2d\\
\ce{Mn}&\quad b&=& e
\end{array}
$$

This can be massaged a little but because there are 5 equations and 7 unknowns it seems that the system is underdetermined.

If we let $a=3,~b=2,~c=5,~d=1,~e=2,~f=8,~g=10$ the equations all work.

My question is:

  Could we arrive at these values without guessing or actually doing an experiment to measure (say) the oxygen evolved?

FWIW, I notice an old paper - J. Am. Chem. Soc. 1889, 11 (6), 94-98 - of which I can only see the first page seems to take a similar approach and gets two possible equations, maybe for the reason above.

It seems to me that this is simply a redox reaction in the presence of acid.

If you write the 2 half reactions, and balance in the presence of acid (sulfuric acid), you end up with the same answer (with 5 O2 instead of 10 O on the product side).

This seems to be more of a mathematical manipulation rather than a chemistry problem, IMHO.


Volume required to dilute solution for a pH change A $\pu{100 mL}$ solution of $\ce{NaOH}$ has a $\mathrm{pH}$ of $13$. What volume of water in $\pu{mL}$ must be added to change the $\mathrm{pH}$ to $11$?

My steps:

Begin by calculating conc. of $[\ce{OH-}]$ for $\mathrm{pH}~13$.
Find the moles by multiplying concentration and volume ($\pu{0.1 L}$)
Determine new concentration of $[\ce{OH-}]$ for $\mathrm{pH}~11$.
Use $\text{amount}/\text{new concentration}$ to find the total volume.
Use total volume ($\pu{100 mL}$) to find volume needed to be added.

The above doesn't produce the right answer, which is $\pu{9.900 mL}$. Can someone please fix my working out, or provide alternative steps to calculate this?

Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:

\begin{align}
c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start}
&= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\
V_\mathrm{end} &= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}
\end{align}

You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):

\begin{align}
\ce{pH}_\mathrm{start} &= 13 \\
c_\mathrm{start}(\ce{OH-}) &= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V_\mathrm{start} &= 0.1~\mathrm{L} \\
\ce{pH}_\mathrm{end} &=11 \\
c_\mathrm{end}(\ce{OH-}) &=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}
\end{align}

Hence

$$
V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}
$$

and therefore

$$
\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.
$$


Applying vacuum at the end of a flash chromatography column I know that flash chromatography uses a positive or pushing pressure to achieve the   goal which is separation. Would the same goal be achieved if you had a vacuum on the other end instead of a pressure at the top, a vacuum on the bottom?

I have used this technique for ion exchange column with efficient and quick results and wonder if this can be achieved with flash chromatography.

All chromatography needs to work is some pressure differential, no matter how it's established. Gravitational chromatography uses hydrostatic pressure from the weight of the solvent, and a typical gravitational column can have a roughly ~20-50 mbar pressure difference between the top and bottom of the column; flash chromatography also has the hydrostatic pressure, but has an additional positive gas pressure at the top from e.g. pumped air, and can typically reach ~100 mbar pressure differences.

The principle of chromatography is to force compounds in a mixture to stay at the interface of a stationary phase (e.g. silica) and a mobile phase. The slight differences in how molecules behave in this boundary are added up over time, creating the observed separation.

Since this is a surface process, having a stationary phase with higher specific surface area and better packing effectively means molecules spend more time undergoing separation at the interface than being dissolved in the bulk of the eluent, where no separation happens. However, a more tightly packed stationary phase also restricts eluent flow. Under a simplified set of assumptions, decreasing the stationary phase particle size by half (and therefore increasing specific surface area by a factor of 4) requires 4 times the pressure differential to produce the same flow rate. So getting better separation in reasonable timeframes requires larger pressure differentials.

This is where vacuum chromatography techniques come in. Vacuum pumps are a staple laboratory instrument, and can easily achieve negative pressures of ~900 mbar with common glassware. Therefore, without too much hassle, it is possible to obtain good flow rates using finer stationary phase particle sizes and get improved separation.

Of course, larger pressure differentials can be achieved with positive pressure at the top of the column. Medium, high and ultra-high pressure chromatography (MPLC, HPLC, UHPLC) make use of stationary phases with very fine particles (from ~10 μm to ~1 μm diameter) which start to look more like silt or clay. The only way to force liquids through these stationary phases is with very large pressure differentials, from ~5 bar to as much as ~1000 bar, but they can achieve incredible resolution of compounds during separation. The reason UHPLC isn't used for every single separation is cost and scalability; it's an engineering challenge to control such enormous pressures safely. Glass is very resilient under compression (which is why glassware can commonly hold a vacuum), but weak under expansion. This means such positive pressures need much more rigid containers, such as thick steel columns. This also limits their feasible sizes and therefore the amount of stationary phase/eluent/compound they can handle.

As an example of a vacuum chromatography technique, I point towards the excellent video by Daniel Pedersen detailing dry column vacuum chromatography (DCVC). There is also a preceding publication which goes into further details. For what it's worth, I've found this vacuum chromatography technique to be far superior to flash chromatography.


Order of activating ability in electrophilic aromatic substitutions From A Guidebook to mechanism in organic chemistry by P. Sykes [1, p. 155]:

The activating effect of $\ce{Y}$ on the nucleus is found to increase, i.e.
the overall rate of substitution increases, in the approximate order:
$$\ce{OCOR} < \ce{NHCOR} < \ce{OR} < \ce{OH} < \ce{NH2} < \ce{NR2}$$
$\ce{NR2}$ is more powerfully activating than $\ce{NH2}$ because of the electron-donating effect of the $\ce{R}$ groups.
It should not be forgotten, however, that in acid solution, e.g. in nitration, these two groups will be converted into $\ce{^\oplus NHR2}$ and $\ce{^\oplus NH3},$ respectively; the nucleus will then be deactivated and substitution will be predominantly m- (cf. $\ce{^\oplus NR3},$ p. 151).
The $\ce{OH}$ group is sufficiently activating to cause phenol to be brominated instantaneously to the 2,4,6-tribromo derivative (the p- and both o-positions all attacked) by bromine water at room temperature.
The groups $\ce{OCOR}$ and $\ce{NHCOR}$ are less powerfully activating than $\ce{OH}$ and $\ce{NH2},$ respectively, because of the reduction in electron-availability on $\ce{O}$ and $\ce{N}$ by delocalisation over the adjacent, electron-withdrawing carbonyl group:

Why is $\ce{OR}$ a less activating group than $\ce{OH},$ but $\ce{NR2}$ is a more activating group than $\ce{NH2}?$
One would expect that that alkyl groups increase the electron density via their inductive effect and thus make the group more activating towards electrophilic aromatic substitutions (EAS) reactions, both in the case of $\ce{OR}$ as well as $\ce{NR2}.$ However, $\ce{OH}$ is a better activating group despite such an effect. Why is this so? Why would the same effect not be applicable for $\ce{NH2}?$
Reference

Sykes, P. A Guidebook to Mechanism in Organic Chemistry, 6th ed.; Longman; Wiley: Harlow, Essex, England; New York: New York, NY, 1986. ISBN 978-0-582-44695-3.

The most likely reason for this is that an $\ce{OH}$ group can act as an acid and exists in an equilibrium with the $\ce{O-}$ anion, which is significantly more activating.
$\ce{NH2},$ on the other hand, is not really that acidic. It does not benefit enough in this manner to be significant.
The loss of the anionic form is making $\ce{OR}$ less activating despite the inductive effect. $\ce{NR2},$ however, does not have as much to lose and only gains in activating ability.


What is charge shift bonding? Can someone explain to me charge shift bonding? I recognize that it is proposed as another domain of bonding - one different from ionic and covalent bonding. I am also told that $\ce{C-F}$ exhibits charge shift bonding. But why—and firstly, what exactly is charge shift bonding? All the literature I have consulted are at too high of a level for me, for example this 1992 article in the Journal of the American Chemical Society.

OK, this was a new one to me too, but I found a few resources which might be useful:

A Computational Organic Chemistry blog post
Michael Scott Long blog post
Nature Chemistry review

The basic idea is this. Most bonds are ionic (i.e., both atoms are charged, but electrostatically attracted) or covalent (i.e., the electron pair is shared).

(This is largely cribbed from the Comp. Org. Chem. blog post. Thanks to Steven Bachrach)

If we were writing a multi-configurational wavefunction to describe the bond, we might have something like:

$$\psi(\mathrm{VB}) = c_1\psi_\mathrm{covalent} + c_2\psi_{\ce{A+B-}} + c_3\psi_{\ce{A-B+}}$$

So normally, we'd look at which is the dominant configuration - maybe it's the covalent one ($c_1$) or one of the ionic ones ($c_2$ or $c_3$). Instead, sometimes the covalent form isn't even stabilizing!

From my perspective, the classic example given for this charge-shift bond is $\ce{F2}$ which of course has many $\pi^*$ anti-bonding electrons, so the classical bond order is 1, but ignores that there's a lot of antibonding character. But you can't really consider $\ce{F2}$ an ionic compound!

I think the big question is the future - can this new type of bonding show general predictability and utility. Clearly most chemists have a general intuition (if sometimes wrong) of covalent and ionic bonding. Besides the authors, I'm not sure if anyone else has an intuition of charge-shift systems.


Why does steam distillation work? I'm aware that there is a similar question here but it didn't really answer my specific questions.
I understand that steam distillation is used to separate an insoluble liquid from an aqueous solution, due to passing steam through the reaction mixture so they both evaporate etc...

I don't however understand what is so significant about this. If we started with a mixture of, let's say, 50% lavender oil and 50% water, wouldn't we also end up with a mixture of lavender oil and water, if both are evaporating below their boiling point? How is that useful - aren't we just ending up with what we started with? Or is it that the distillate will contain a greater proportion of oil? eg 60% to 40% - but then I don't see how that works, as shouldn't water should be more easily evaporated than the oil?

Furthermore, surely the two immiscible liquids form two seperate layers that can be put into a separating funnel and poured into separate containers?

I think that the main purpose (/ historical advantage) of steam distillation was to avoid the exposure of sensitive compounds to sustained heat by either lowering the required distillation temperature, or the distillation duration (at equal temperature), or both, by using steam as a carrier.

The solution that contains the compound to be distillated doesn't necessarily initially contain water itself, though even if it does, there is a significant advantage in that the duration of exposure to high temperatures is as short as the transport time until it is condensed (since the water is boiled separately, and not together with the compound). In traditional distillation the compound has to be exposed to heat for the time necessary to reach the solvant's  evaporation point AND the transport time, which is significantly longer than the transport time alone.


Thermal decomposition of calcium carbonate Thermal decomposition of calcium carbonate.

In general, we heat $\ce{CaCO3}$ to temperature of approximately 825°C it decomposes into calcium oxide and liberates carbon dioxide gas:*

$$\ce{CaCO3} \overset{825\ ^\circ \text{C}}{\ce{ ->}}\ce{CaO + CO2}$$

Is it possible to heat the calcium carbonate at temperatures below 825 degrees Celsius.

Pump a hard vacuum and calcium carbonate will calcine at a lower temperature.  825°C is about one atmosphere of carbon dioxide at equilibrium.  Thermodynamics proposes, kinetics disposes - finely ground goes faster than bulk.


Mono chloro substituted products of naphthalene My chemistry teacher today told that only two mono chloro substituted products are possible for naphthalene. He explained this by saying since the naphthalene molecule is symmetrical  about the axis passing through the C-C bond where the benzene rings are fused, the corresponding positions should be identical & hence only two such products are possible. I personally don't find this convincing & hope to look for a genuine answer. I will be extremely thankful to anyone who helps.

If someone is unable to understand my teacher's explanation, just ignore it. I myself wasn't able to understand it properly so could have mistaken it for something else.

Your chemistry teacher is partially correct by saying that naphthalene molecule is symmetrical about the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (going through the common carbons to both rings). Your chemistry teacher has forgot to mention that naphthalene molecule is also symmetrical about the axis perpendicular to the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (See the top structure of the diagram below). Thus, each quarter of the naphthalene molecule is equal and each quarter contains only two carbon atoms (disregarding common $\ce{C-C}$ bond). I named them as 1 and 2, because they are symmetrical. Therefore, only two different mono-substituted derivatives are possible from naphthalene. If substituent is chloro, then only 1-chloronaphthalene and 2-chloronaphthalene are possible, and so on.


Does increasing concentration of copper sulfate decrease or increase its conductivity? I am a high school student and for my chemistry lab, I did an electrolyis experiment on how the changing concentration of electrolyte (copper sulfate) will change the rate at which the copper is deposited on the cathode.

My hypothesis was that as the concentration is increased, so will the rate of electrolysis. However, my results are completely opposite. When I use 0.1 M solution, loads of copper is deposited on the cathode but as I increase the concentration, the amount of copper deposited becomes progressively less and less. With 1 M solution, there was almost no copper being deposited.

This makes no sense to me. Could somebody explain what is going on? Am I doing something wrong?

I used both copper electrodes. I did not use distilled water, just normal tap water. I used 12 V power supply.

EDIT:

I tried adding a little sulfuric acid (making the pH around 3) to the copper sulfate but it did not make any difference on electrolysis.

I tried to observe the electrolysis very closely today and noticed that there was some sort of oily liquid falling off from the anode. Also, I noticed that initially some flaky brown/black substance fell off from the anode (which may or may not be copper)and soon there was a silvery layer formed on the anode.

Also, I checked using Hanna checker that the overall pH of the solution was around 3 but the pH near the anode read between 9 and 14.

Also, I noticed that when I used a higher concentration, the copper deposition on the cathode was brown and clearly copper. But when I used lower concentration, even though the amount of deposit was a lot more, the substance was probably not pure copper as I had assumed earlier. It was black and flaky. It was certainly different to the deposit I observed when I used higher concentration of copper sulfate.

Does anybody know now what is happening? Is there anything that I can do to fix the problem?

I'm thinking maybe at the higher concentrations your tap water is making a film of $\ce{Cu(OH)2}$ on the surface of your electrode, and that is what's shutting down your electroplating. Your tap water might be mildly alkaline (tap water pH's can range from about $6.5$ to as high as $9.5$).

Try adding a little sulfuric acid to your solutions. If alkaline films are the problem, that will fix it.

It will also increase the conductivity of your solutions tremendously, because hydrogen ions are extremely mobile.

The solubility of copper sulfate decreases with increasing sulfuric acid concentration, so don't add too much acid.


Is Chemical Engineering a good pathway to Biomedical Engineering for graduate school? I have been thinking about my future goals and I have recently decided to go for Biomedical Engineering. sites recommend to start in Chemical Engineering because its a good pathway for biomedical engineering. is this true? please share what you think and from your perspective, thanks

If you want to get into a biomedical engineering program in graduate school, chemical engineering is not a bad option. You would of course probably have better luck if you majored in biomedical as an undergrad as well, since your coursework would more directly prepare you for those programs.

A couple of things to note though:

There is a (big) difference between bio-medical engineering and bio-chemical engineering. Biomedical engineering is more closely related to pre-med and materials science than to biochemistry. That said, bio chem courses probably wouldn't hurt you, but if you have the option, taking real biomedical engineering courses as electives (or as a minor) would be better. For bio-chemical engineering, these days regular chem E programs almost always have at least an elective component in biochem engineering, and often times departments are called "Chemical and Biological Engineering" due to the similarities. In other words, chemical engineering undergrad degrees are very well suited (and possibly over-prepare) you for biochemical engineering graduate work.

The other thing to know is: the real secret to getting into a good graduate program is to get research experience as an undergrad in the field that you want to work in. This demonstrates to the acceptance committee that you meet the two main requirements PIs have for grad students:

You are interested in the subject.
You have the aptitude to actually accomplish something.

A chemical engineering degree demonstrates aptitude on its own, and as @rch mentioned, is a pretty good "fall back" plan in case you change your mind later (although chem E is a tough major if it's just a fall back plan - I would not recommend it unless you really like chemistry and physics and could imagine working in a chemical plant, since that is likely what you will do if you "fall back" on it).

The problem is going to be that lots of other people will be able to demonstrate aptitude, with degrees in biomedical engineering, materials engineering, chemical engineering (lots of them get into biomedical later), pre-med, and other stuff that is just as well suited. In addition to aptitude, a lot of them (if not most of them) will also have research experience through undergrad projects or internships. To be competitive with them, you will want some as well.

The best easiest way to get research experience as an undergrad is to find a professor who is working on something you are interested in, and volunteer. Many times they will have paid positions specifically for undergrads, but if you can put in volunteer hours and are reasonably good at communicating, managing your time, and getting things done, most professors would be happy to put you to work for free. If you can manage to get a publication before you finish your undergrad degree, you will have an excellent shot at getting into the school of your choice, provided your GRE scores and grades are good (another reason to carefully consider chem E - not an easy major).

Another option is internships, although a research internship carries more weight in grad school circles than a regular old "gets the coffee and fills out spreadsheets" internship. They are much harder to get, but usually pay pretty well, so it is definitely worth trying for one.

The main point I am trying to make here is - major in (almost) whatever you want, but if you want to go to grad school in a particular field, get research experience in that field as an undergrad. Go for paid work first, but be prepared to volunteer if you have to, at least for a semester. You can quickly turn a volunteer position into a paid one if you do good work. Work experience also goes a long way toward getting you a job if you don't end up going to grad school. I'd put your odds of getting a good job with a biomedical engineering undergrad degree with experience at higher than getting a job with a chemical engineering degree without experience, especially these days.

Also, just so you don't think I'm trying to talk you out of a Chem E degree, Chem E was my major (BS, MS, and PhD) and I really liked it. If I didn't like chemistry and physics so much though, I probably wouldn't have made it through.


Formation of peracetic acid from acetic acid and hydrogen peroxide and its stability in their presence I've been trying to find out as much as I can about peracetic acid, especially regarding its use as a sanitizer. In the Wikipedia entry it notes that

  Peracetic acid is always sold in solution as a mixture with acetic acid and hydrogen peroxide to maintain its stability.

This is definitely born out by commercial examples: Loeffler Lerasept PAA (a sanitizer I've used before) lists 31% hydrogen peroxide, 17% peracetic acid, 16% acetic acid and 1% phosphonic acid among its ingredients.

But, my understanding was that mixing acetic acid and hydrogen peroxide makes peracetic acid. From the same Wikipedia article:

  It forms upon treatment of acetic acid with hydrogen peroxide.

So my question is: what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid? Does the phosphonic acid inhibit this reaction? Is there something else not listed that does this? Or is it just some bit of chemistry I'm not aware of?

The formation of peracetic acid from acetic acid and hydrogen peroxide is an equilibrium reaction and so in order for the peracetic acid to remain at a constant concentration as desired, acetic acid and hydrogen peroxide must also be present so that the reaction can be at equilibrium. $$\ce{CH3COOH + H2O2 <=> CH3COOOH + H2O}$$

  what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid?

Nothing. The hydrogen peroxide reacts with the acetic acid to form peracetic acid and water but the peracetic acid and water also react to reform hydrogen peroxide and acetic acid. At equilibrium the rates of these reactions are equal and so the concentrations of the species do not change. At room temperature the equilibrium constant is somewhere around 2.5 (see references) which indicates that significant amounts of reactants and products are present, supporting the reasoning above.

The paper Chin J Proc Eng 2008 February, 8 (1), 35–41 has some data on the values of the equilibrium constant and the variation in concentration of peracetic acid formed.

Another value for the equilibrium constant on page vii of Unis, Melod (2010) Peroxide reactions of environmental relevance in aqueous
solution. Doctoral thesis, Northumbria University.


The Reaction Between Nickel and Hydrochloric Acid Having conducted an investigation involving multiple metals, I have found that nickel refuses to react with concentrated hydrochloric acid. I have even left the reaction overnight and still have seen no mass change in the solid piece of nickel. Is the reaction simply extremely slow or does it not react despite being above hydrogen in the reactivity series?

Try 30% and heat it to 80C. All of the chloride should be exhausted after a day or less. You can then evaporate the solution to generate crystals of nickel chloride. Any work with nickel compounds should always be done in a well-ventilated hood.


What salts (or other substances) are available for increasing water's boiling point? Increasing the boiling point of water by adding salts or other compounds would be very useful when working with a water bath while refluxing, because with pure water you cannot reach temperatures over 70°C (at that point too much evaporates from the bath). I know that simple table salt would work, but are there other and better compounds/salts available?

For relatively dilute solutions of non-volatile solutes, the following equation (which assumes ideal behavior) can be used to predict boiling point elevation with reasonable accuracy:

$$\Delta T_b = K_b \cdot b \cdot i$$

Here, $K_b$ is the solvent-specific ebullioscopic constant, $b$ is the molality of the solute, and $i$ is the so-called van 't Hoff factor, which accounts for the extent of solute dissociation. For example, if one were to dissolve $\ce{NaCl}$ in water and assume total dissociation, $i = 2$, since two ions are produced in solution. It follows that, ceteris paribus, a salt that dissociates to yield a greater number of ions will produce a greater boiling point elevation. In practice, the magnitude of boiling point elevation is capped by solubility and deviation from ideal behavior at concentration extremes.

Water also forms negative (i.e., higher-boiling) azeotropes with various substances. Addition of a miscible higher-boiling liquid compound can also effect elevation of the boiling point (assuming the mixture does not form a positive azeotrope), a colligative property similar to the above case for salts. Various polyols (e.g., glycerol, propylene glycol, etc.) are commonly used for such applications.


Product formed under ethoxide/ethanol and heating? What would be the major product of the following reaction sequence?

I believe that the first step will be the formation of a radical at the first carbon of the ethyl group because benzylic radicals are quite stable. However, I'm not really sure what happens after that. The answer key has 'I' as being the correct answer.

You are correct in your prediction for this first reaction:

The second is simply an $\mathrm E2$ β-elimination reaction, which forms product I:


Most activated position on para-terphenyl for EAS Para-terphenyl: it doesn't look pretty with all those math-y numbers, but those are going to come helpful in answering my question!

A question asked me to tell the expected product when this reacts with $\ce{Br2/FeBr3}$. Now, I have done such questions with all types of fancy organic molecules ranging from benzene to picric acid to parantiroanisole etc. I know that I am supposed to find the most activated position1 for EAS. But I have never done more than one phenyl ring at once.

So, in this question, I have drawn the eight resonating structures (without charge separation) of paraterphenyl manually. Here they are:

But, as you can see, every position seems to be activated identically. Each position has their own share of electrons through a pi bond, but it ends up being completely symmetric.

So, how do I find the most activated position in this compound?

1:Clarification: By "most activated position" I mean the position with maximum electron density. For example, in aniline, those are the ortho and para positions.

So, how do I find the most activated position in this compound?

Start by drawing resonance structures of the various possible intermediates (sigma complexes) formed when $\ce{Br^{+}}$ attacks the different ring positions in p-terphenyl. Whichever intermediate has the most resonance structures is likely to be the most stable (lowest energy) intermediate. The lowest energy intermediate will have the smallest activation energy and consequently be the kinetically favored product (barring steric effects).

I've drawn just a few of these resonance structures in the following diagram.

In the top row I've drawn a few of the possible resonance structures for electrophilic attack at the para position on a terminal phenyl ring.  You can draw many more resonance structures delocalizing the charge around the various rings, but notice that it is possible to delocalize the positive charge over all 3 aromatic rings. In the second row I've drawn a few resonance structures showing charge delocalization when electrophilic attack occurs in the center ring.  Importantly, in this case charge can only be delocalized over 2 of the aromatic rings.

This suggests that based on resonance effects, electrophilic attack on a terminal benzene ring would be preferred over attack at the central benzene ring.

Further, we would expect ortho and para attack on the terminal rings to be preferred since the phenyl substituent acts as an activating o-p director in electrophilic aromatic substitution. The rate of attack at the ortho position might also be decreased somewhat due to the steric effect of the adjacent phenyl (actually biphenyl) substituent.

In a paper by Shafig and Taylor (J. Chem. Soc., Perkin Trans. 2, 1978, 0, 1263-1267. DOI: 10.1039/P29780001263), they explore your question by studying the electrophlic protonation of tritiated p-terphenyl.  After correcting the rates for statistical effects (e.g. there are twice as many ortho positions on the terminal rings as there are para positions) they find that:

the para position (your positions $10, 16$) reacts at a relative rate of $273$
the meta position (your positions $9, 11, 15, 17$) reacts at a relative rate of $1.54$
the ortho position (your positions $8, 12, 14, 18$) reacts at a relative rate of $176$, and finally,
the $4$ identical positions on the middle ring (your positions $2, 3, 5, 6$) react with a relative rate of $59.1$

These results support our predictions that

attack on the terminal rings will be preferred over attack on the central ring and
that attack at the ortho and para positions is preferred over attack at the meta position on the terminal rings.


Why is no MS-gas phase NMR coupling possible? With ever stronger magnets coming out every year, LC-NMR coupling becomes more attractive and is now used frequently in the pharmaceutical industry as well. Considering the rapid progress that is made in that direction I wondered why no MS-NMR coupling has been reported yet, especially as gas phase NMR is possible already. So I wondered:
When will it be possible to do MS-NMR coupling? What are the biggest obstacles for achieving MS-NMR coupling?

(The magnets are not really getting stronger. The > 20 T magnets are still too big and expensive to be widely used. First 900 MHz was from 2001, the 1GHz Bruker magnet came out five years ago. Since then seemingly nothing... except this 1020 MHz hybrid prototype by JEOL et al. Looks a bit unwieldy. There is a reason it is a hybrid (inner coils made of Bi-2223, i think): The classic superconducting materials (Nb3Sn, NbTi) have an absolute upper limit of ~25 T. HTSC can go higher, but they are very difficult in manufacturing. )
Fundamental problem is T1 relaxation in the gas phase. It is much too slow. You need hyperpolarisation techniques, that you have to interface with the GC. Difficult. (Impossible with mass spectrometry, because there the sample is in vacuum.)
The amount of analyte would be tinytiny. Likely impossibly small for NMR, even with hyperpolarisation. Certainly too small with mass spectrometry.
Other point: Why? GC is not useful for complex molecules, because the have very little vapour pressure. For less complex molecules, GC-MS is fine. There  are a number of working NMR techniques to separate different analytes. HPCL-NMR, GPC-NMR, DOSY, etc.
I guess nobody has seen a problem that needs GC-NMR to solve it. Or at least he has not been able to convince anyone it would be worth financing it.
(I initially misread the original question to be about a GC-NMR combination. Hence all the GC in my answer.)


Is it possible to compute the colour of *any* molecule? Above are some spectral emission lines for hydrogen, helium and neon.

Using the Schroedinger equation, it's possible to derive the colours that hydrogen will emit when light is fired at it. A little more complicated (although beyond my rudimentary knowledge of quantum mechanics), the same can be done for helium, and so on.

Of course as we progress to molecules, even the relatively small $C_5H_{11}Br$, it's (I'm told) very hard to predict the colours emitted from first principles, and even then incorrect assumptions must be made (like the Born-Oppenheimer approximation), so upon progressing to larger molecules, we find that these simplifications diverge from experiment.

My question is this: with the advent of computers, are we able to (in practice, not theoretically in a few decades) accurately compute the colours given off by a molecule  by numerically solving the laws of quantum mechanics for complicated systems? Much like nonlinear differential equations can now be 'solved' with computers without any simplifying assumptions (like $\sin(x) \approx x$), is it in humanity's capacity today to approximate colours from first principles, so that computing time is the only restraint? Note the difference between an incorrect model of reality (Born-Oppenheimer), and one that converges to the correct one (even if slowly).

It is possible to compute electronic spectra with decent accuracy. Time dependent density functional theory (TDDFT) permits fairly rapid calculation of UV/vis spectra at a quantum mechanical level of theory. Whilst DFT is not truly a first principles method, it can furnish electronic spectra accurate enough to be useful, depending on care taken in selecting appropriate functionals, accounting for solvatochromic effects, etc. Obviously, DFT uses the Born-Oppenheimer approximation, but the beauty of the approximation is that it's actually very close to reality.

See, for instance, this paper by Jacquemin et al., where they calculate $\lambda_{max}$ for a wide range of dyes using TDDFT and get a MAE of 22 nm using CAM-B3LYP (a long-range hybrid functional), reducing that to 10 nm with an empirical calibration curve. The paper only really cares about accurately reproducing the location of $\lambda_{max}$, however this paper by Quartarolo and Russo computes complete spectra with fairly reasonable accuracy. Not perfect, but the general shape of the spectra are reproduced adequately. Part of the divergence between the experimental and computed spectra probably comes down to higher order broadening effects not reflected in the simple Gaussian convolution of the computed data, but I'm not a spectroscopist.

Additional: as for computing the colour of any molecule, well, that does depend heavily on the size of your molecule and any mitigating factors such as symmetry elements. Most QM techniques scale fairly viciously with the number of electrons in your molecule, and whilst you can always reduce the size of your basis set and heap on less and less conservative approximations, accuracy will suffer.


How fast is the interconversion triplet → singlet in organic molecules? When computational studies are performed on transition state geometries, the "thermodynamic" (that is, with no kinetic considerations) energy of the analyzed state is obtained.

I was wondering: if a (multistep) reaction begins in a biradical state, the kinetics of the interconversion from triplet to singlet is so slow that the reaction is likely to proceed in a radical fashion, or if a lower transition state exists for the ionic (singlet) counterpart, then it is likely to proceed in a ionic way, because the rate of interconversion is extremely fast?

I am sorry if the question is dumb, but all I could find was about transition metals, and I don't know if the same applies.

It depends on the organic molecule.

When spin-orbit coupling facilitates intersystem crossing, it can be fast: 100 femtoseconds for nitronapthalene (viz. Chem. Eur. J. 2018, 24, 5379 – 5387).

The Wikipedia article on intersystem crossing cites a timescale for radiative decay (phosphorescence) of triplet to singlet of 10−8 to 10−3 s.


Percentage error. How to find percentage error? The Question

  Calculate the percentage dissociation of $\ce{H2S(g)}$ if $0.1\ \mathrm{mol}$ of $\ce{H2S}$ is kept in a $0.5\ \mathrm L$ flask at $1000\ \mathrm K$. The value for $K_c$ for the reaction $\ce{2H2S(g) <=> 2H2(g) + S2(g)}$ is $1\times10^{-7}$.

MY ATTEMPT
Initially there are $0.1\ \mathrm{mol}$ of $\ce{H2S}$ and nothing for $\ce{H2}$ and $\ce{S2}$

Now at equilibrium let us assume $x$ moles dissociated so

$0.1 - x\ \mathrm{mol}$ of $\ce{H2S}$, $x\ \mathrm{mol}$ of $\ce{H2}$, and $x/2\ \mathrm{mol}$ of $\ce{S2}$

Now I know that degree of dissociation = $\alpha$ = amount dissociated/ initial amount in our case $x/0.1=\alpha$ so $x = 0.1\alpha$. So I substitute.

now we have $0.1-0.1\alpha$ of $\ce{H2S}$, $0.1\alpha$ of $\ce{H2}$, and $0.1\alpha/2$ of $\ce{S2}$

Thus
$$K_c = \frac{\left[0.1\alpha/2V\right](\ce{S2})\left[0.1\alpha/V\right]^2(\ce{H2})}{[0.1-0.1\alpha/V]^2(\ce{H2S})}$$
Substituting value of $V$ and solving I reach here

$$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$

Now my doubt. I assumed $0.1\alpha$ very small in comparison to $0.1$ so I neglected it and assumed the whole $0.1 - 0.1\alpha$ to be $0.1$ only. Now my doubt is that what is the percentage error in my assumption.

On further solving I get percentage of degree of dissociation as $1\,\%$ and that is correct. My method is correct, my only doubt is how to calculate percentage error in my assumption.

Skipping to your question about

  $$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$

I'll point out two things.

(1) Significant figures come into play here. The Kc is only given to one significant figure as is the molarity of the solution. So each of those quantities has a +/- 50% error. Remember that in Chemistry 0.1 m isn't the same as 0.1000 m.

So the molarity of the solution is 0.1 +/- 0.05. If $\alpha < 0.1$ then the assumption that $0.1\alpha$ is negligible is reasonable.

(2) You can solve the equation for alpha exactly by trial and error. Solve as you did for $\alpha$, then use a factor 10 times larger and one 10 times smaller. One number should be too big and the other too small as compared to Kc. Calculate the average of the two values and test that number. That average will be too big or two small. Discard one side of the bracket. So you're creating an ever shrinking bracket about the "true" value. 6-8 iterations should get you very close.

Now with the "true value" and the estimate from the assumption that $0.1\alpha$ was negligible you can calculate the "real" percentage error for $\alpha$.


Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured? Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured?

I understand that temperature can cause either the forward or reverse reaction to be favoured. But then doesn't that mean that the rate of which the reactants or products are being formed differ from each other? So, what does it mean when they define equilibrium as I have said above.

Please keep answers simple and clear.

A chemical reaction can be reversible, and a reversible reaction can be at equilibrium, but they are not the same thing.

Equilibrium is the state where the forward and reverse reactions are the same rate.  This is clearly not true at the start of a reaction because there are no products yet.  As the reaction progresses the forward reaction rate decreases and the reverse reaction rate increases until they are equal.  At that point the concentrations of reactants and products no longer change.

The position of the equilibrium can be far to the left, so that hardly any product is formed before it reaches equilibrium.  This would be because the forward reaction is much slower than the reverse reaction.  If the position of the equilibrium is far to the right, almost all of the reactants have been used up because the forward reaction is much faster than the reverse reaction.

If you do something to change the conditions, like increase the temperature or add more reactants, you break the equilibrium and the reactions are no longer at the same rate.  As above the reaction will progress until the rates are the same again.  Depending on the change, this new equilibrium position may be more to the left or let or more to the right.

It is tempting to think that if you increase temperature, the rate of both reactions will increase the same.  This is not the case.  Although both reactions will increase in rate, because of the usual collision theory, but the same temperature change can have a much bigger effect on the rate of one reaction than the other.


What is the reason for the different solubility of silver halides in ammonia? According to my knowledge, I know that $\ce{AgCl}$ dissolves in dilute ammonia, $\ce{AgBr}$ dissolves in concentrated ammonia and $\ce{AgI}$ does not dissolve even with concentrated ammonia.

What is the reason for this?

Silver chloride is very thermodynamically stable in ammonia solution. The $K$ for the dissolution of silver chloride in ammonia solution is only $\pu{2.9E-3}.$ suggesting that the product — silver ammonia complex ion — is much less thermodynamically stable than the reactant, silver chloride.

$$\ce{AgCl + 2 NH3 <=> [Ag(NH3)2]+ + Cl-}$$

$$
\begin{align}
K &= K_\mathrm{sp}(\ce{AgCl})\times K_\mathrm{f}(\ce{[Ag(NH3)2]+})\\
  & = \pu{1.8E-10}\times\pu{1.6E7}\\
  & = \pu{2.9E-3}
\end{align}
$$

However, silver chloride will dissolve in an excess of concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule $K_\mathrm{sp}$ value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good σ-electron pair donor ligand, combines with the silver ion.

Ions themselves are not precipitates, as all bulk matter must be approximately electrically neutral, and so the silver chloride dissolves when hit with excess, concentrated ammonia.

Also, note that the cyanide ion is also a good complexing agent with respect to silver. The cyanide ion has a surfeit of electrons on the carbon, and carbon is even less able to stabilize negative charge than nitrogen, making the cyanide ion reactive as a complexing agent. Hitting silver chloride and other silver salts with a source of cyanide ion will similarly cause the silver salt to dissolve into ions suspended in solution.


Largest alkane having a given alkane as its base name What is the largest (most carbon atoms) alkane having heptane as its base name?
For example, 2,2,3,3-tetramethylbutane is the largest (most carbon atoms) alkane retaining butane as its base name.

Any alkyl substituent of butane in position 2 or 3 cannot be longer than $\ce{CH3}$ since that would lead to a longer parent chain. And obviously, there cannot be any alkyl substituent at all in the first or the last position of the butane chain. Therefore, the largest structure based on a butane parent chain is 2,2,3,3-tetramethylbutane.

This principle can be expanded to a heptane parent chain. The maximum length for alkyl substituent chains are 0 for position 1 and 7, 1 for position 2 and 6, 2 for position 3 and 5, and 3 for position 4. Therefore, the largest theoretical structure based on a heptane parent chain is 3,3,5,5-tetra-tert-butyl-4,4-bis[3-(tert-butyl)-2,2,4,4-tetramethylpentan-3-yl]-2,2,6,6-tetramethylheptane ($\ce{C53H108}$).


Stoichiometry and POAC Contradiction Lets say we have the reaction

$$\ce{H2 + O2 -> H2O}$$

And we are given 10 moles of $\ce{O2}$ and are asked that
How many moles of hydrogen gas do we need to complete the reaction ?

Stoichiometry gives the answer as 10 moles but by applying Principle Of Atom Conservation,  we get

$$\text{Moles of }\ce{H2}  =2\cdot\text{moles of }\ce{O2}$$

Why is the POAC giving wrong results?

You applied the POAC incorrectly . Even your stoichiometric answer is incorrect.

First Balance the reaction

$$\ce{2H2 + O2    -> 2H2O}$$

We get that one mole of $\ce{O2}$ requires 2 moles of $\ce{H2}$.
Thus 10 moles $\ce{O2}$ need 20 moles of $\ce{H2}$.


Do catalysts shift equilibrium constant towards 1? I want to be able to understand shifts in equilibrium from the maxwell boltzmann distribution. One thing I cannot get my head around is the effect of catalysts on the equilibrium position - supposedly it's none at all.

But consider these diagrams of the M.B distribution of both reactants and products of a reversible reaction.

The top diagram shows the distribution of the products of the reaction which equilibrium favours. It has a bigger concentration of particles. The line dividing the shaded region on these diagrams represents activation energy and the $dE$ represents the shift in activation energy due to the catalyst. We can see that the shaded regions, $A_1=A_2$ because forward and backward rates are the same at eqm. But clearly, the rate of the top reaction increases more as a greater fraction of particles have above activation energy. So adding a catalyst should shift equilibrium to the middle?

Basically, catalysts work better on reactants with higher concentration. So shouldn't the side with the lower concentration be favoured?

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the  shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one).

There is a subtle issue with the way you've presented your drawing. However, we'll come to that slightly later. First, let's establish that the "proportion of molecules with sufficient energy to react" is given by

$$P(\varepsilon) = \exp \left(-\frac{\varepsilon}{kT}\right) \tag{1}$$

Therefore, for a reaction $\ce{X <=> Y}$ with uncatalysed forward activation energy $E_\mathrm{f}$ and uncatalysed backward activation energy $E_\mathrm{b}$, the rates are given by

$$k_\mathrm{f,uncat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f}}{kT}\right) \tag{2} $$

$$k_\mathrm{b,uncat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b}}{kT}\right) \tag{3} $$

The equilibrium constant of this reaction is given by

$$K_\mathrm{uncat} = \frac{k_\mathrm{f,uncat}}{k_\mathrm{b,uncat}} = \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{4}$$

As you have noted, the change in activation energy due to the catalyst is the same. I would be a bit careful with using "$\mathrm{d}E$" as the notation for this, since $\mathrm{d}$ implies an infinitesimal change, and if the change is infinitesimal, your catalyst isn't much of a catalyst. So, I'm going to use $\Delta E$. We then have

$$k_\mathrm{f,cat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f} - \Delta E}{kT}\right) \tag{5} $$

$$k_\mathrm{b,cat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b} - \Delta E}{kT}\right) \tag{6} $$

and the new equilibrium constant is

$$\begin{align}
K_\mathrm{cat} = \frac{k_\mathrm{f,cat}}{k_\mathrm{b,cat}} &= \frac{A_\mathrm{f}\exp[-(E_\mathrm{f} - \Delta E)/kT]}{A_\mathrm{b}\exp[-(E_\mathrm{b} - \Delta E)/kT]}  \tag{7} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \frac{\exp(\Delta E/kT)}{\exp(\Delta E/kT)} \tag{8} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{9}
\end{align}$$

Equations $(9)$ and $(4)$ are the same, so there is no change in the equilibrium constant.

The question then arises as to how eq. $(1)$ is obtained. The simplest way is to invoke a Boltzmann distribution, which almost by definition gives the desired form. However, since you have a Maxwell–Boltzmann curve, I guess I should talk about it a bit more. The fraction of molecules with energy $E_\mathrm{a}$ or greater is simply the shaded area under the curve, i.e. one can obtain it by integrating the curve over the desired range.

$$P(\varepsilon) = \int_{E_\mathrm{a}}^\infty f(\varepsilon)\,\mathrm{d}\varepsilon \tag{10}$$

where the Maxwell–Boltzmann distribution of energies is given by (see Wikipedia)

$$f(\varepsilon) = \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \tag{11}$$

At first glance, we would expect this to be directly proportional to the exponential part of the rate constant, i.e. $\exp(-E_\mathrm{a}/kT)$. Alas, it is not that simple. If you try to work out the integral

$$\int_{E_\mathrm{a}}^{\infty} \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \,\mathrm{d}\varepsilon \tag{12}$$

you don't get anything close to the form of $\exp(-E_\mathrm{a}/kT)$. Instead, you get some "error function" rubbish, and some nasty square roots and exponentials. (You can use WolframAlpha to verify this.)

Why is this so? Well, it turns out that there are other terms that also depend on $\varepsilon$ and therefore need to go inside that integral (they aren't constants and can't be taken out).

The simplest example is that faster molecules tend to collide more often, so even though the right-hand tail of the diagram seems to contribute very little to the "proportion of molecules with sufficient energy", it actually contributes more significantly to the overall rate because these molecules collide more often. In collision theory this is described using the "relative velocity" of the particles $v_\mathrm{rel}$.

There is also another complication, in that the Maxwell–Boltzmann distribution, the direction of the particles is not accounted for. (For more insight please refer to Levine Physical Chemistry 6th ed., p 467.) Therefore, there has to be yet another term that takes into account the direction of movement of the particles. The idea is that a head-on collision between two molecules is more likely to overcome the activation barrier than is a $90^\circ$ collision. The term that compensates for this is the "collision cross-section" $\sigma$.

If you go through the maths (and I don't really intend to type it out here, it's rather long, but I will give some references) then you will find that at the end you will recover the form $\exp(-\varepsilon/kT)$. Once you have arrived at this, it's very straightforward to see that the increases in rate of both the forward and backward reaction cancel each other out.

Now, as for the promised references, Pilling and Seakins's Reaction Kinetics pp 61-2 have a short outline of the proof. Atkins's Physical Chemistry 10th ed. has a slightly longer proof on pp 883-4.


I have a solution of citric acid, tartaric acid, water and glucose. How do I find the concentration of the citric acid present in the solution? If I try to find the concentration of citric acid through back titration or reverse titration, the base I would use, like NaOH, would react with both acids to form sodium citrate and sodium tartrate, which are both white compounds that are colourless in aqueous form. How do I find the concentration of citric acid?

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off.

If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving you an estimate of the concentration. The biggest systematic error is that the apparent $\mathrm{p}K_\mathrm{a}$ is dependent on ionic strength, which you don't know unless you know the concentrations in the first place.


Forming magnetite on cast iron pan I am in the process of seasoning a cast iron pan. (Creating the black non-stick coating typical of such pans.)

This is done by causing thin layers of flaxseed oil to polymerize and bond to the iron.

In the literature there is the recommendation that the polymerized oil bonds better to the pan, if the pan is covered with magnetite rather than blank iron, in part due to the porous nature of the oxide layer.

My question is: is there a DIY method that can either create $\ce{Fe3O4}$ on a polished cast iron pan, or alternatively convert a layer of $\ce{Fe2O3}$ into $\ce{Fe3O4}$?

In the gun manufacturing trade a process called "rust bluing" is used. In it the gun barrel is encouraged to form a thin layer of $\ce{Fe2O3}$ before being immersed in boiling distilled water. This causes the $\ce{Fe2O3}$ layer to convert into $\ce{Fe3O4}$ within minutes. Apparently due to the fact that boiling water lacks dissolved oxygen.

When I attempt to apply this "rust bluing" process to a cast iron pan, the $\ce{Fe2O3}$ layer stays red rather then turning black: Clearly the conversion is not taking place.

I am not clear why this is the case. Is it due to difference in materials, i.e. cast iron vs. gun steel?

Is there any way I can promote creation of/conversion to a magnetite layer on cast iron with DIY-compatible means? The industrial solution of caustic soda solution at 150$^{\circ}$C does not sound like a good idea for home implementation.

I think the two steps that are missing from you approach are "carding", where, after immersion in the boiling water, the steel is rubbed/brushed to knock off the larger rust structures and repetition. The process is repeated several times before the piece builds up a stable blued surface.

Check out this video of Larry Potterfield applying a rust blue surface to a rifle action and barrel: https://www.youtube.com/watch?v=VPD3GWqk-yE


What is the binding mode of the ligand in complexes with quadruple metal-metal bonds? I'm wondering which atomic orbital is left in a complex with quadruple metal-metal bond to bind the steric ligand, since all d orbitals except the x2-y2 are used for the metal-metal bonds?

or is it the x2-y2 atomic orbital that forms the bonding?

and if so: what atomic orbital is used in metal-metal quintuple complexes?

Here is an example complex: https://en.wikipedia.org/wiki/Quintuple_bond

To determine what orbitals do, it is often a good idea to look at the symmetry group as it can provide a rough overview over what can (symmetry-allowed) and cannot (symmetry-forbidden) happen. In the case of a quadruple bond such as $\ce{[Re2Cl4]^2-}$, we can approximate the rhenium(III) centres as being close to $C_\mathrm{4,v}$. A quick glance at the corresponding group theory table (copied below to avoid page loading issues in the large linked page) will show us which irreducible representations we may need to consider.

$$\begin{array}{c|ccccc|cc} \hline
C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$

The final two columns show the relevant orbitals. The s orbital always transforms as $\mathrm{a_1}$, the p orbitals as $\mathrm{a_1+e}$ and the d orbitals as $\mathrm{a_1+b_1+b_2+e}$. The four chlorine atoms will bond using p orbitals (but the orbital type does not matter as we will only be considering σ bonds). They transform as $\mathrm{a_1+b_1+e}$ (calculation not shown). Immediately, we notice that $\mathrm d_{xy}$ does not contribute to bonding to the ligands but only to the $\ce{Re-Re}$ quadruple bond. Furthermore, we notice that $\mathrm d_{x^2-y^2}$ is practically required to bond to the chlorines as there is no other rhenium orbitals of that symmetry.

To determine what is left, it may be easiest to draw out the potential chlorine orbitals. $\mathrm{a_1}$ means that all phases pointing to rhenium are the same. This can bond either with rhenium’s 6s or $\mathrm{5d}_{z^2}$ orbital or a linear combination of both. The $\mathrm p_z$ orbital does not have any contribution in the $xy$ plane and thus cannot participate.

The chlorine’s $\mathrm e$ orbitals correspond to two in which two opposite chlorines have opposite phases and the other two have zero contribution (in one case, the $x$ chlorines have contribution, in the other the $y$ ones). Thus, to bond with these we will use $\mathrm{6p}_x$ and $\mathrm{6p}_y$. Again, this is because the corresponding d orbitals do not have any contribution in this $xy$ plane.

Thus, to sum things up:

$\mathrm{6p}_x,\mathrm{6p}_y$ and $\mathrm{5d}_{x^2-y^2}$ can bond to chlorines
$\mathrm{6p}_z, \mathrm{5d}_{xz}, \mathrm{5d}_{yz}$ and $\mathrm{5d}_{xy}$ can bond to the other rhenium
the $\mathrm{6s}$ and $\mathrm{5d}_{z^2}$ orbitals can bond to either; they probably linear combine with $\mathrm{6p}_z$ to form two bonds and keep an empty orbital.

If the symmetry around a rhenium atom is assumed to be $C_4$ instead of $C_\mathrm{4,v}$, that doesn’t change much of the analysis; only the distinction between $\mathrm d_{x^2-y^2}$ and $\mathrm d_{xy}$ becomes irrelevant.


Reactivity of Alkanes Does the tetrahedral structure of the alkanes contribute to their lower reactivity? I thought that because a tetrahedral structure suggests ${sp^3}$ hybridization, it should contribute to its lower reactivity. Is this right?

Does the tetrahedral structure of the alkanes contribute to their lower reactivity?

Hybridization is at the center of the answer.

Due to poor P-P overlap, it only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about a double bond involving ${sp^2}$ hybridized carbons, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond involving ${sp^3}$ hybridized carbon atoms with highly directional orbitals that result in strong bonds. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane. Alternatively one can view ethene as a two-membered ring (ref. 1), no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring. It is easy to imagine that such a ring system would contain a significant amount of strain (ref. 2).  Either way (and by the way these two views of ethene are equivalent and complementary) - using a poorer overlap argument or a higher strain energy argument, we see why alkenes (and even more so alkynes) would generally be more reactive than alkanes.

From this argument we would expect that if we were to somehow increase the strain in an alkane, then we should increase its reactivity. A good example of such behavior can be found in cyclopropane.  Placing 3 "tetrahedral" carbon atoms into a ring dramatically increases the strain energy in the system (it changes the carbon hybridization too, see ref. 3) and the reactivity increases dramatically.  Indeed, cyclopropane adds bromine much like an alkene (ref. 3)

Since someone above commented that ${sp^3}$ $\ce{C-H}$ bonds are "very stable and typically do not want to react". Let me just point out that $\ce{C-H}$ bond energies vary as follows

\begin{array}{|c|c|c|c|} \hline
\ce{C-H} & \text{Bond Strength}  \\  \text{bond type} & \text{(kcal/mol)} \\ \hline
\ce{sp^3 C-H} & 101 \\ \hline
\ce{sp^2 C-H} & 111 \\ \hline
\ce{sp C-H} & 133 \\ \hline
\end{array}

These bond dissociation energies (ref. 4) don't fit the observed reactivity pattern.  They indicate that an ${sp^3}$ bond is actually easier to break than an ${sp^2}$ or ${sp}$ $\ce{C-H}$ bond


CO2 dry cleaning what is the process From what I understand CO2 drycleaning is essentially heating and pressurizing Co2 until it is at a supercritical state and then using it as a solvent to wash clothing. I just don't understand why clothing that comes out of the resulting process would be cold. Doesn't supercritical carbon dioxide require temperatures of at least ~300 Kelvin to form? Anyway, I would greatly appreciate it if someone with more knowledge on the subject could provide a basic outline as to how CO2 dry-cleaning works. There are very few resources that go into detail on the process.

Thanks!

Like aliphatic cleaning solvents such as hexane, both liquid and supercritical $\ce{CO2}$ are effective solvents for fatty stains. According to Green America, though, "[L]iquid CO2 is pumped into the mix. Clothes are rotated in a cycle that lasts five to 15 minutes at room temperature."

The phase diagram for $\ce{CO2}$ shows that it is a supercritical fluid above ~300 K, just above room temperature, at ~80 kg/cm2, so that it may be either liquid or supercritical during the cleaning. However, at the end of the cycle, "[T]he liquid CO2 is pumped back into the storage tank, to be reused again, if possible. The remaining CO2 is released in the air." That implies a bit of a pressure drop at the end, with concomitant adiabatic cooling.


What is the mechanism for demercuration? The following image is from 'Organic Chemistry Second Edition' by Clayden, Greeves, and Warren.

The final step is the demercuration, but I'm unable to find a definitive mechanism.

The wikipedia page lists a mechanism sourced from the Journal of the American Chemical Society:

Wikipedia Page for Oxymercuration

But I've found videos on YouTube which outline an SN2 displacement of the acetoxymercuri group by hydride. e.g.: Mechanism of Oxymercuration Demercuration (new)

So I would like to know which mechanism is correct.

Apparently, the current assumption implies formation of $\ce{R-Hg-H}$ intermediate, followed by radical chain decomposition This, however, should be taken with a grain of salt, i.e. very skeptically. Proving mechanism is an incredible amount of work, and considering relative unimportance of organomercury chemistry in modern organic synthesis, it is unlikely any considerable resources will be allocated to the task.


Can a reactor be operated at both reflux and above atmospheric pressure? I'm not a chemist by trade so my experience in the lab has probably not been as in-depth as most of you. I've been tasked with investigating a proposed reaction in which an organic mixture (mostly formic acid) is to be heated at reflux for several hours.

The complication being that a pressure of approx. 4 bar is required during the reaction. How would one go about achieving this? The reflux condensors I've browsed through are all open to atmosphere on one end.

Is there some sort of pressure regulator I can attach on the output of the condensor to maintain the pressure at 4 bar?

I'm told that the reaction may produce $\ce{CO2}$ and $\ce{NH3}$ by-products which means pressure is expected to increase and fluctuate as the reaction proceeds. Ideally I'd like to occasionally vent these to drive the reaction equilibrium forward.

The formic acid is quite corrosive to anything not plastic or glass.

You could theoretically boil at the higher temperature but why would that be a requirement? Why would you require reflux specifically if you can actually heat the formic acid to higher temperatures under pressure? Heating should work on its own and degassing to drive the equilibrium should work with a back pressure regulator. However,  most lab glassware will not take that pressure. You normally need some stainless steel pressure equipment.
One way might be to use plug flow reactors rather than bombs.


What is the evolutionary reason for the existence of rings in molecules? Just to give you a bit of context, I don't have any background in Chemistry. I'm actually a Computer Scientist adventuring myself in the world of Chemoinformatics (more specifically, we're developing algorithms to improve drug discovery).

I noticed that many molecules (or at least drug-like molecules) have rings, but I was wondering: from an evolutionary perspective, what does it mean? Why did molecules evolve to have this "feature", do molecules with rings have better properties then similar ones without rings?

I noticed that most molecules (or at least drug-like molecules) have rings, but I was wondering: from an evolutionary perspective, what does it mean?

First note: Delete the ‘most molecules’ bit. The numbers are insane and nigh uncountable, but non-cyclic molecules are similarly common as cyclic ones are.

From an evolutionary persective this means: nothing.

Drugs are designed to inhibit certain enzymes as specifically as possible. The enzyme has a shape that happens to best fulfil it’s purpose according to evolutionary principles. (my draft had ‘design’ here for a second, but it would be the wrong word choice. Nature and evolution do not design.) This will render an active pocket having a certain orientation, size and hydrophobic pattern that drug designers attempt to mimick. If a (poly-) cyclic molecule fits well, it fits well. If a linear molecule fits better it fits better. Think of it as trial and error.


Purpose of ammine in Tollens' reagent In this previous question the mechanism for the reaction of Tollens' reagent was outlined. As I understand it the oxidising agent in Tollens' is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ but the mechanism only showed the silver ion.

What is the purpose of adding ammonia to the silver nitrate to form the diamminesilver(I) complex: why not just use silver nitrate? Is it purely to provide a source of hydroxide ions or is the structure of the complex ion actually important?

The formation of the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ due to addition of ammonia has mainly three functions.

1.
The Tollens test takes place under alkaline aqueous conditions. The complexation of $\ce{Ag+}$ with $\ce{NH3}$ prevents precipitation of brown $\ce{Ag2O}$:

$$\ce{2Ag+ + 2OH- <=> 2 AgOH <=> Ag2O + H2O}$$

2.
The Tollens test shall be selective for aldehydes. $\ce{[Ag(NH3)2]+}$ is a milder oxidising agent than $\ce{Ag+}$.

$$\begin{alignat}{2}
\ce{Ag+ + e- \;&<=> Ag}\quad &&E^\circ = +0.799\ \mathrm{V}\\
\ce{[Ag(NH3)2]+ + e- \;&<=> Ag + 2 NH3}\quad &&E^\circ = +0.373\ \mathrm{V}
\end{alignat}$$

The difference in redox potential can be explained using the stability constant of $\ce{[Ag(NH3)2]+}$:

$$K_\text{B} = \frac{\left[\ce{[Ag(NH3)2]+}\right]}{\left[\ce{Ag+}\right]\left[\ce{NH3}\right]^2}$$

$$\begin{aligned}
E&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Ag+}\right]\\
&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{\left[\ce{[Ag(NH3)2]+}\right]}{K_\text{B}\cdot\left[\ce{NH3}\right]^2} \\
E_{\ce{[Ag(NH3)2]+}}^\circ&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{1}{K_\text{B}} \\
&\approx E_{\ce{Ag+}}^\circ-0.059\ \mathrm{V}\cdot\log K_\text{B}
\end{aligned} $$

3.
The complex formation equilibrium slows down the overall reaction. A slow, controlled reaction is important for creating the desired silver mirror. If $\ce{Ag+}$ is reduced too quickly, colloidal silver metal would appear, which would create a black cloudy liquid.


Calculating reaction enthalpy given other reactions $$\begin{alignat}{2}\ce{X(s) + 1/2O2(g) &-> XO}&&\qquad\Delta H=-895.5\ \mathrm{kJ} \\
\ce{XCO3(s) &-> XO(s) + CO2(g)}&&\qquad\Delta H= +484.3\ \mathrm{kJ}\\
\ce{X(s) + 1/2O2(g) + CO2 &-> XCO3(s)}&&\qquad\Delta H = \;?\ \mathrm{kJ}
\end{alignat}$$

I wish to know how to do it. I have seen some questions where there's 3+ equations and at the end a final equation that I would need to determine the $\Delta H$ for the final equation. I have looked online on how to solve this but the tutorials are very confusing and I can't understand them. Can someone explain how to do this and other similar problems like this?

One of the tutorials talks about making sure they cancel out and flipping the equations? Can someone please explain this?

As enthalpy is a state function, so regardless the multiple stages of a reaction, the total enthalpy change for the reaction is the sum of all changes.This is Hess's law.
So, you can see clearly that the third reaction results from adding  the first reaction to the reverse  of the second reaction, and so is the enthalpy change.


Does the carboxyl group count as a branched chain Do carboxylic acids (or any compound with a carbonyl group) count as a branch chained organic compound due to the C=O? Or does it need a carbon - carbon branch?

In organic chemistry, the terminology of branched vs. unbranched is generally used to refer to the structure of the hydrocarbon skeleton of a molecule. Hence, heteroatomic functional groups are usually not relevant. The structure in your question is therefore unbranched. For an authoritative reference, see the relevant sections of the IUPAC blue book.


How do amines react with flue gases to absorb CO₂? I'm reading this article about carbon capture from the combustion of fossil fuels.  In this article, it says that amines are used as a chemical absorbent to capture the $\ce{CO2}$ from the exhaust of the combustion.

In particular, I have read that N-Methyl diethanolamine (MDEA) is used (a  reference  lists the formula as $\ce{CH3N (C2H4OH)2}$).  I am curious if anyone can explain what happens during the reaction with $\ce{CO2}$ and how it absorbs the $\ce{CO2}$ from the flue gas.

In addition to the reaction in permeakra's answer, some amines may react with $\ce{CO2}$ in absence of water also. This type of reaction is a Lewis acid-base reaction forming a zwitterionic complex: $\ce{R3N + CO2 -> R3NCO2}$.

MDEA adds some stability to this complex. The alcohol groups can stabilize the carboxylate group through hydrogen bonding.


If you placed silver metal into silver nitrate, would a reaction occur? thinking about oxidation and reduction, what happens to the silver metal and the silver ions?

For OP: As I have realized from James Gaidis' answer, you were very minimalistic in describing context of your question, what is very bad practice, as much more burden is put on shoulders of responders via the question interpretation and clarification. As I have just now realized, you may, or may not speak about a solution.

Yes, the reaction will occur, and no, the reaction will not occur.

There is ongoing equilibrium reaction $\ce{Ag <=> Ag+ + e-}$ with zero net result.

Initially, when metallic silver is inserted, very tiny net result of the above reaction happens, until there is achieved the equilibrium potential. It is determined by the initial potential before inserting into the solution and by the effective capacitance of the electrode.

If you charged positively ( e.g. electrostatically ) a piece of metallic silver before inserting into the solution, a very small amount of silver would dissolve. If negatively, a very small amount would deposit itself.

As @MaxW has mentioned in his answer, isotope tracing is frequently used to track the reaction mechanisms of complex processes and also kinetics of systems in equilibrium, that seem macroscopically static.


Compressibility Factor Graph - Which gas attains a deeper minimum? As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?

Is there any temperature dependence of the above?

If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.

I'm looking for more of qualitative comparison, but all sorts of answers are welcome.

There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.

If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,\; P_r=P/P_c,\;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'

$Z$ vs $P_r$ at different reduced temperatures

For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume  and $T_C=8a/(27Rb), \; P_c=a/(27b^2) ,\; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.


Is there a possible effect of mechanical vibration on the stability of organic molecules? Is there a possible theoretical effect of mechanical vibration on the stability of
organic molecules.  E.g. some medicines degrade if they are not stored in a fridge, is there a theoretical foundation that suggests that low frequency vibrations might have the same effect?

What do you mean by "low frequency vibrations"?  The term "extemely low frequency" is used to describe earthquake type frequencies (3-30 Hz, give or take), whereas "low frequency" is used to describe frequencies in the 30kHz - 300 kHz range.  If you mean "low frequency" as defined in the previous sentence, then yes, that frequency range of vibration is known to have chemical effects.  The process involves the formation and collapse of bubbles (cavitation) in the solution.  High temperatures and pressures can be generated during the bubble formation and collapse process.  The typical lab apparatus can homogenize samples, produce catalytic effects by cleaning metal surfaces or increasing phase transfer, disrupt cells, and break some (weak?) chemical bonds.  This reference reports,"Ligand-metal bond cleavage in transition metal complexes to give coordinatively unsaturated species or modified complexes as well as complete strip off of ligands to produce amorphous metals disrupture of the solvent structure altering the solvation of reactants Sonolysis of molecules (homolytic fragmentation to radicals, rupture of polymers, generation of excited states, cell disrupture)."