A "function of proportionality" in a rate law I am currently studying rate laws and the determination of the order of a reaction. So first-order reactions remind me of linear functions $f(x)=kx$ while second-order reactions remind me of quadratic functions like $g(x)=cx^2$ in high-school algebra. Since we are dealing with reactants here, the input of the functions would be the concentrations of our substrates $[\ce{S_1}]$ and $[\ce{S_2}]$. I am wondering if there are any fancier functions that arise in "real-life" chemistry?

If by "fancier" you mean "uglier", then absolutely yes. Real chemistry is messy, much more than we can fully handle. The simple rate laws we learn in undergraduate physical chemistry are just approximations.

I've written elsewhere that, in general, a full description of chemical kinetics for any overall reaction requires studying an entire tree of simultaneous chemical reactions. There can be dozens, even hundreds of intermediate species involved, all mutually linked in complex ways, and with more than one final product ("by-products"). The difficulty in performing this full description has lead to the whole field of chemical reaction network theory. I hope you like matrix algebra.

To make chemical kinetics tractable in most cases, two major approximations made are:

Forget about the whole reaction network, and look at only the overall kinetically fastest path from reagents to products
Forget about all the steps and intermediates in the fastest path, and describe only the bottleneck - the slowest step

Even with these two crushing approximations, it's possible to describe many useful reactions with a fair level of accuracy. This is partly luck, and partly unavoidable (due to simplicity naturally arising in randomness). This level is what most undergraduate and even graduate work sticks to. Even when it's a bad fit, we'll sometimes stubbornly fit a system into a particular rate law, do some variational analysis, and hope the error isn't too great.

A nice example of a reaction whose kinetics is slightly more complex than the standard "nice" rate laws is the radical reaction between hydrogen and bromine.

$$\ce{H2 + Br2 -> 2HBr}$$

According to experimental data, it's possible to obtain the following rate law (which may still be approximate!):

$$\mathrm{r_{HBr}=\frac{k_1[H_2][Br_2]^{3/2}}{[HBr]+k_2[Br_2]}}$$

The deduction is nicely described in this site. As you can see, depending on the reaction parameters, the denominator can potentially be simplified down to a single term, with some degree of approximation, and a simple-ish rate law could be used. But in the general case, it's messier. Because of these reaction networks, even the concept of a "reaction order" does not exist in general. An example of a natural but extremely difficult reaction to describe kinetically in full would be the combustion of a moderately sized alkane, such as octane ($\ce{C8H18}$).

When taking full chemical networks into account, there appear cases with pretty unique kinetics. For example, if you haven't heard of oscillating reactions, they're a topic of considerable experimental and theoretical study. With a particular set of coupled reactions, it's possible for chemical systems to display the hallmarks of mathematical chaos. Even if individual steps can be described by simple kinetics, the global process cannot. There's a nice discussion of oscillating kinetics in this Chem.SE question (and possibly others).

I'm sure there are other ways to get strange rate equations. Some reactions just don't follow kinetic theory in its simplest form. This can lead to weird conclusions, such as negative activation energies. Furthermore, rate laws almost always assume macroscopic amounts of matter. When a tiny amount of reagents are involved, the discrete nature of matter calls for statistical corrections to chemical kinetics, because an integer number of molecules can't follow continuous analytical curves.


In an O-H bond, will polarity increase/decrease if some of the electron density on the oxygen atom is reduced? Like when an e- withdrawing group (sp2 carbon) is attached to oxygen of the OH bond, it attracts the electrons. So will the polarity of the OH bond increase or decrease?

The polarity on the oxygen atom will decrease. This is because of the sp2 hybridised carbon atom which is highly electronegative. This will result in a decreased polarity in the OH bond and hence it will disassociate more easily in solvents such as water. One example of this effect is the increased acidity of phenol compared to most alcohols. This is because the OH is bonded to sp2 hybridised carbon in phenol and hence will disassociate in water easily when compared to normal alcohols.


Solubility of white phosphorus Why is white phosphorus soluble in $\ce{CS2}$ but red phosphorus not? Is it something in relation with the structures of both? Please explain.

White phosphorus has P4 molecules packed into a crystal,these dissolve readily in Carbon Disulfide.

Whereas Red phosphorus is polymeric in nature. It is a derivative of white phosphorus where one P-P bond is broken and an addtional bond is formed with neighboring tetrahedron molecule resulting in a chain like structure .

Like all long chain polymers this will lower its ability to dissolve and its solubility in solvents


Are all complexes with a polydentate ligand examples of chelation? I'm getting a little confused about the definition of chelation and its implications (I should probably point out I'm only an A2 Chemistry student). The IUPAC defines chelation as the following:

  The formation or presence of two or more separate coordinate bonds between a polydentate (multiple bonded) ligand and a single central atom

Does this mean that any ligand that is not monodentate (e.g. even ethanedioate) is an example of chelation? I thought chelation was only for complexes with a large coordination number (e.g. $\ce{EDTA^4-}$), it would seem odd to call a single metal ion with 1 ethanedioate ligand chelated?

I think you should not focus on the ligand to define chelation. You should instead focus on the complex that is formed. In fact, the very same multidentate ligand may participate in binding in more than one way:

by forming more than one bond with the same central atom, therefore forming a chelated complex;
by forming only one bond with that central ion, even though it could form two; this may be the case when a vast excess of this ligand is present with respect to the metal ion;
by bridging between two adjacent central ions.

Only the first of these possibility is referred to as chelation. In other words, forming a chelated complex doesn't depend solely on the fact that the ligand is multidentate. Being multidentate is a necessary, but not sufficient condition for a ligand to form chelated complexes.


What turns clear liquid soap blue? Liquid soap that sits in metal pump handles turns blue after a week or so, as in this photo comparing the soap in its initial clear state and then after pumping from the dispenser.

The pumps are presumably brass with either nickel or chrome electroplating on the exterior.  Can't say for sure whether or to what degree the interior where the soap makes contact is plated.  They have been in use for years.

The ingredients of this particular soap are:

water
sodium laureth sulfate
cocamidopropyl betaine
decyl glucoside
sodium chloride
"fragrance"
DMDM hydantoin
polyquaternium-7
tetrasodium EDTA
citric acid
sodium sulfate
PEG-120 methyl glucose dioleate
PEG-7 glyceryl cocoate
poloxamer 124
D&C red #33
FD&C blue #1

The color of the precipitate is strongly reminiscent of copper(II) hydroxide. I hypothesize:

Chloride corrosion of copper from exposed brass (ref 1 || ref 2)

a. Per ref 2 above, sulfate may also participate
Alkaline precipitation of $\ce{Cu(OH)2}$ in the neutrally-buffered soap
Insufficient $\ce{Na_4EDTA}$ to complex the amount of $\ce{Cu}$ corroded from the brass

It's also possible the blue material is $\ce{Ni(II)}$ complexed in such a way that its color is changed from the typical green-blue of $\ce{Ni(OH)2}$.


Open resource for spectroscopic raw data Is anyone aware of reliable resources (ideally public) which allow the user to download the raw spectroscopic data such as infrared spectrum of small organic molecules, or rotational spectrum of HCl/DCl mixtures or NMR spectra of molecules? I was looking at the NIST database but apparently there is no option for downloading the spectrum as an Excel, csv or txt file. Thanks.

NIST Spectral Database

As a quick fix: To access the IR spectral data of $\ce{HCl}$, I opted for a seach by (Hill) formula  on NIST's web page, which yielded three entries here.  For $\ce{HCl}$ in particular (here) I opted for the first one here, indeed offering the spectrum as image file or as JCAMP-DX.

As in the comment by @EdV, this JCAMP-DX an ASCII file indeed.  Note JDXview by Norbert Haider (Department of Drug Synthesis, University of Vienna, Austria) is a very handy program (only 560 kb, installation free executable) for a quick display of this file by NIST. Via File -> Save, the export of the data in a two-column, tabulator separated .txt file is just a click away.  The program works perfectly well in Windows, and with wine libraries equally in Linux, too.  (The author equally distributes the underlying Delphi / Pascal code.)

Note 1: Because of the tags physical chemistry, and spectroscopy, you may be interested in a python module like jcamp to access either IR, or NMR spectra in the JCAMP-DX formats (there are multiple).

Note 2: Sometimes, all you have indeed is an image (a scanned) spectrum.  If you pay attention and proceed in sections (e.g., $4000 - 2000$, then $2000 - 800\,\pu{cm^{-1}}$) to account for the break on the abscissa in IR spectroscopy, a digitizer like WebPlotDigtizer may be worth to use.  If you want, the program works via a web browser, installation free, with «normal image files» like .png, and since some time with images embedded in a .pdf (think about publications), too.


How to compute the NMR shift in Gaussian09 for only one specific atom in a bigger unit cell? I am trying to compute the NMR shift for a specific atom in the middle of the surrounding crystal unit cell (1 surrounding anion sphere and 1 cation sphere) in Gaussian09. I only need the chemical shift for the middle atom.

How can I only compute the shift for this atom, and leave out the rest? And would it even save any computation time if I did?

Short answer, you can't.

It might save a bit of time, but probably not much. The calculation would still need to go through the normal SCF on the entire system to calculate the NMR shifts.

In principal, in evaluating the property calculation, you could save a small amount of time by only computing the one atom, but for most systems, I would think the overall SCF calculation would be vastly greater than computing a few properties.


Do metals form covalent bonds besides ionic and cordinate bond? My chemistry textbook says that metals form ionic or cordinate bonds whereas non metals form covalent bonds. But in another textbook I read that Lithium, Beryllium, Aluminium, Chromium, Manganese etc, which are metals, form covalent compounds like LiCl, BeH2, AlCl3, H2CrO4, H2MnO4 respectively

Now my question is whether metals form covalent bonds or not? And if yes then how to predict which metallic compound contains ionic and which contains covalent bonds?

The first thing that should be said is that there's no difference between a coordinate bond (dative bond) and an ordinary covalent bond. Yes, the electrons "come from different places"; but the molecule doesn't actually know this, nor does it care. Once a covalent bond is formed, it is a covalent bond, regardless of where the electrons "come from".

Some of the examples you have given can equally be described as coordinate bonds or covalent bonds. For example, $\ce{MnO4-}$ could be viewed as four $\ce{O^2-}$ ligands forming coordinate bonds to a central $\ce{Mn^7+}$. Or, you could also just view them as plain old Mn=O covalent bonds. It doesn't actually matter.

Having established that there is no real difference between coordinate bonds and covalent bonds, the only real question is how can you tell a covalent bond apart from an ionic bond. The answer is mostly to do with polarisability. If you imagine an ionic bond between a strongly polarising cation ($\ce{Al^3+}$) and a polarisable anion ($\ce{Cl-}$), the cation will pull over some of the anion's electron density: this is exactly what is meant by "polarising". This changes the nature of bonding from an ionic model (where the bonding electrons reside entirely on one atom) to a covalent model (where the bonding electrons are somewhere in between the two atoms). You'll notice that all the cations you've listed: $\ce{Li+}$, $\ce{Be^2+}$, $\ce{Al^3+}$, (formally) $\ce{Cr^6+}$, and $\ce{Mn^7+}$, are very highly polarising. The first three have a very small size, and the last two have exceptionally large positive charges.

That said, covalent bonds between two transition metals are reasonably common, too: a simple example is $\ce{Mn2(CO)10}$ (Wikipedia), which is essentially $\ce{(OC)5Mn-Mn(CO)5}$ with a single bond between the two.

The bond order between two metals can go up to slightly ridiculous numbers. Cr–Cr and Re–Re quadruple bonds are well-known in $\ce{Cr2(OAc)4(H2O)2}$ and $\ce{[Re2Cl8]2-}$ respectively. Even quintuple and sextuple bonds between metals have been described (although some of these accounts are not without controversy).

How can you predict this, though? It's not trivial at all, and in truth you probably don't need to care about these exotic examples until you have enough experience to understand the bonding.


I heard aluminium doesn't produce fumes when burnt but the opposite happened, how? One day I took aluminium foil(sanitary brand,  don't know if you've  heard it)  then I casually burned it over  a gas lighter then it burned and gave off Grey fumes and the foil became a dull Grey where it had burnt.  I mistakenly may have inhaled a bit of fumes.  I had a slight pain in my right nostril the whole day. How is that?  Is that dangerous?

It doesn't. Consumer aluminium foil often is a thin film on thicker polymer film, it is much more durable this way. The most common polymer for films is likely polyethylene, and when heated it produces nasty mix of pyrolysis and oxidation products. While not extremely toxic, they are definitely irritating and unhealthy.


Reaction of 1,3-dimethyl-2,5-dintrobenzene with NaOMe/MeOH I have looked through articles and closest I feel is in this question there will be a nucleophilic aromatic substitution. Now, a lot of questions are available where there is a halogen and OMe group substitutes the halogen.
My take on this is maybe due to SIR effect (steric inhibition of resonance) The NO2 group present in between the two methyl groups would be substituted by OMe? Will this be the approach?

You are correct that the nitro group can be displaced by methoxide, this has been reported examples here.
My instinct is that the less hindered nitro group is the one that will be displaced.


PV=nRT approximation on other planets? In Chemistry class yesterday, I learned that real gases with low atomic masses behave like an ideal gas at high temperatures and low pressure. Since on earth at sea level the pressure is close to $1\ \mathrm{atm}$ (which is a low pressure) and temperature is close to $288.15\ \mathrm K$ (high temperature), I can use the equation $PV=nRT$ to approximate properties of the gas. (I got these values from Wikipedia, under Standard Sea Level).

$$\left(\text{pressure}\right)\cdot \left(\text{volume}\right)=\left(\text{number of moles}\right)\left(R\right)\left(\text{temperature}\right)$$

In physics we learned that pressure is the force per unit area. On earth the atmospheric pressure is earths gravitational pull on the air. The gravitational pull of any object on earth is $G_\text{pull}=\left(\text{mass}\right)\cdot \left(9.8\ \frac{\mathrm m}{\mathrm s^2}\right)$. Because the gravitational pull is $G_\text {pull}=\left(\text{mass}\right)\left(\text{acceleration due to gravity}\right)$.

Since different planets have different masses the acceleration due to gravity on these planets are different. Below I listed the values:

$G_\text{Mercury}=3.59\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Venus}=8.87\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Mars}=3.37\:\frac{\mathrm m}{\mathrm s^2}$

For planets like Mercury, Venus, Mars the approximation of $PV=nRT$ still can be used since they would have even lower pressures than Earth. But what about planets like Jupiter and Neptune

$G_\text{Jupiter}=25.95\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Neptune}=14.07\:\frac{\mathrm m}{\mathrm s^2}$

Jupiter and Neptune have more pressure at ground level than that of Earth. So can $PV=nRT$ still be used? If yes, how well? If not, is there a different formula that can be used to find the volume and number of moles on these other planets?

Any input is appreciated. I am just being curious and didn't find anything on Google.

The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet.

The main factor is the mass of gas each planet's atmosphere contains.

Mercury has almost no atmosphere.  The total mass of all gas in Mercury's atmosphere is only 10000 kg! The pressure is less than $10^{-14}$ bar. The approximation will be very accurate.

The mass of Venus's atmosphere is $4.8 \times 10^{24}$kg and pressure is 92 bar, so PV = nRT will be much worse of an approximation than on earth.

Mars has a small atmosphere, less than Earth more than Mercury.

Jupiter, Saturn, Uranus and Neptune are gas giants with no distinct surface.


Propagation of the uncertainty in measurements The inside dimensions of a box that is cubic is 12.7 cm on each edge with an uncertainty of 0.3 cm.

What is the volume of the box?
What do you estimate to be the uncertainty in the calculated volume?

The first part is simple enough. To obtain volume of the box: $(\pu{12.7 cm})^3 = \pu{2.05E3 cm^3}$
Now for the second part, "estimate the uncertainty in the calculated volume?" – I would imagine that because the uncertainty is $\rm\pm0.3\ cm$ on each side the uncertainty in the volume of the box would be $\rm\pm(0.3)^3 = \pm0.027\ cm^3$? The textbook gives $\rm\pm0.60\times10^3\ cm^3$. I am (excuse the pun) uncertain of how they have reached this answer.

The volume $V$ can be calculated from the length $x$, width $y$, and height $z$.

$$\begin{align}
V&=x\cdot y\cdot z\tag1\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times 12.7\ \mathrm{cm}\tag2\\[6pt]
&=2048.383\ \mathrm{cm^3}\tag3
\end{align}$$

If we assume that the uncertainties of length $x$, width $y$, and height $z$ are not correlated (which is not necessarily true when the same ruler was used to measure length, width, and height), i.e. their correlation coefficient is $\rho=0$, the relative uncertainty of the volume may be estimated as

$$\begin{align}
\left(\frac{u(V)}{V}\right)^2&=\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2\tag4\\[6pt]
\frac{u(V)}{V}&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\tag5
\end{align}$$

Thus, the absolute uncertainty of the volume is

$$\begin{align}
u(V)&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot V\tag6\\[6pt]
&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot x\cdot y\cdot z\tag7\\[6pt]
&=\sqrt{\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag8\\[6pt]
&=83.809\ \mathrm{cm^3}\tag9
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(V)&=\sqrt{\left(\frac{\partial V}{\partial x}\right)^2\cdot u^2(x)+\left(\frac{\partial V}{\partial y}\right)^2\cdot u^2(y)+\left(\frac{\partial V}{\partial z}\right)^2\cdot u^2(z)}\tag{10}\\[6pt]
&=\sqrt{\left(y\cdot z\right)^2\cdot u^2(x)+\left(x\cdot z\right)^2\cdot u^2(y)+\left(x\cdot y\right)^2\cdot u^2(z)}\tag{11}\\[6pt]
&=\sqrt{\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2\\+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2}\tag{12}\\[6pt]
&= 83.809\ \mathrm{cm^3}\tag{13}
\end{align}$$

However, if we have to assume that the uncertainties of length $x$, width $y$, and height $z$ are correlated, i.e. their correlation coefficient is $\rho=1$, then the equation for the estimate of the relative uncertainty of the volume simplifies to

$$\frac{u(V)}{V}=\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\tag{14}$$

and the absolute uncertainty of the volume is

$$\begin{align}
u(V)&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot V\tag{15}\\[6pt]
&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot x\cdot y\cdot z\tag{16}\\[6pt]
&=\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag{17}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{18}
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(V)&=\frac{\partial V}{\partial x}\cdot u(x)+\frac{\partial V}{\partial y}\cdot u(y)+\frac{\partial V}{\partial z}\cdot u(z)\tag{19}\\[6pt]
&=y\cdot z\cdot u(x)+x\cdot z\cdot u(y)+ x\cdot y\cdot u(z)\tag{20}\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\\
&\hphantom{={}}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\tag{21}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{22}
\end{align}$$

The numerical values of an estimate $y$ and its standard uncertainty $u(y)$ should not be given with an excessive number of digits. It usually suffices to quote $u(y)$ to at most two significant digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.
Thus, the result for the volume $V$ according to $\text{(3)}$ and its standard uncertainty $u(V)$ according to $\text{(9)}$ or $\text{(13)}$ may be expressed as

$$V=\left(2048\pm84\right)\ \mathrm{cm^3}\tag{23}$$

or

$$V=2048(84)\ \mathrm{cm^3}\tag{24}$$

Using the uncertainty according to $\text{(18)}$ or $\text{(22)}$, the result may be expressed as

$$V=\left(2.05\pm0.15\right)\times10^3\ \mathrm{cm^3}\tag{25}$$

or

$$V=2.05(15)\times10^3\ \mathrm{cm^3}\tag{26}$$

(According to various international standards, the ± format should be avoided.)


Keeping one component concentration constant and varying another in a solution This is a simple question but it has been so long since I have done any chemistry. Basically I would like to keep the concentration of A constant and the concentration of B variable and vice versa.

Assuming we want to make a 50 ml solution with constant concentration of A and variable concentration of B.From my knowledge I would use C1V1=C2V2 in order to dilute A to its fixed concentration lets say from 4% to 3%. But how do i go about keeping it constant and B concentration variable for a final 50 ml solution.

EDIT:
Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?

2nd EDIT:
Apologies, this is my first time using this site and was told to edit my question and add my comment. I didn't put values as I believed it was clear enough but I guess not.

@ y chung that is what I have carried out in order to create the solutions, so I'm guessing I did it correctly then.

@ bon thanks for the comment. I will tell you exactly what Im trying to accomplish. I have two solutions one is A with a concentration of 9% and one is B with a concentration of 7%. I want to create 5 solutions each 50 ml with a constant 5% concentration of A and varying concentration of B from 1-5 %.

What I have done is as follows:
I reduced my A concentration to the required 5% by C1V1=C2V2 for a 50 ml solution getting 28 ml. I put 28 ml in each volumetric flask of 50 ml capacity.

I then calculated how much volume of B with the 7% concentration I would need to get it to say the 15 concentration based on my remaining volume of 22 (50-28). For the 1 % concentration this yielded a volume of B at 7% of 3.14 ml. I added that to my flask contain the 28 ml of A and made up the solution to 50ml.

The question is, is this correct? from what I gathered from y chung ,it is.
Thanks

Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.


Melting points and boiling points of primary alcohols do not follow the same trend If one considers boiling points (in °C) of primary alcohols, one finds the following:

methanol: 65
ethanol: 79
1-propanol: 97
1-butanol: 117
1-pentanol: 138

This trend is due to Van der Waals forces increasing with molecular weight.

Now if one focuses on melting points, I would expect the same trend, but these are the experimental values (in °C):

methanol: −98
ethanol: −117
1-propanol: −127
1-butanol: −90
1-pentanol: −79

the lowest value being the melting point of 1-propanol.

The general idea stating that with all else being equal, more massive molecules have higher boiling and melting points, does not fully apply for this set of alcohols.

So my question: why melting and boiling points of primary alcohols do not follow the same trend?

Melting points are a bit trickier to compare than boiling points, especially when you're looking at the lightest examples of a group of molecules. This graph for the unbranched alkanes exemplifies nicely how the trends differ in smoothness.

As you increase the weight of molecules, the observed trend depends on how you increase the weight. For example, if you start with methanol and increase the weight by adding more $\ce{-HCOH -}$ units (methanol, ethylene glycol, glycerol, and so on) you would find a very clear increase in both melting and boiling points. In some sense this is a more legitimate trend to analyse, because the relative amounts of different types of intermolecular interactions stays approximately constant (in particular, all molecules have one hydroxyl per carbon atom, and hence they all have about the same amount of hydrogen bonding per atom in the molecule). In your sequence of alcohols, the strong hydrogen bonding allowed by the hydroxyl group becomes "diluted" as the molecule gains an ever larger alkyl chain, which can only sustain much weaker van der Waals interactions. In the limit of alcohols with very large alkyl chains, the alcohols' melting and boiling points do not differ tremendously from their parent saturated hydrocarbons. Compare the melting and boiling points of 1-hexadecanol (49°C and 344°C) and hexadecane (18°C and 287°C).

Also, the temperature at which a substance freezes is dependent not only on the strength of intermolecular interactions, but how the solid packs. Even substances with strong interactions may only freeze at very low temperatures if they pack very poorly when solidified. Extreme examples to which this effect contributes are ionic liquids. They tend to have a very low vapour pressure at reasonable temperatures (as you would expect of many salts), but some melt even below 0°C. The freezing of such bulky entities would entail lots of ordering and therefore a very large decrease in entropy, so it is disfavoured (or equivalently, their melting is highly entropically favoured, so the free energy of the liquid phase becomes lower than the free energy of the solid phase even at relatively low temperatures). The exact geometry of the molecules in the solid is also important, as some favourable interactions may be suppressed in the solid state due to steric hindrance which may be less present in the more freely-moving molecules of a liquid. Furthermore, molecules with a high amount of symmetry tend to have high melting points, because freezing is entropically favoured due to molecules more readily fitting in the solid lattice.

Yet another effect to consider is how large a surface area each molecule has available for intermolecular interactions. This is often brought up to explain differences between melting and boiling points of branched and unbranched organic compounds; when comparing structural isomers, branched compounds have a lower available area for interaction than unbranched compounds, so the latter tend to have higher melting and boiling points (branching may sometimes create a combination of opposing packing and surface area effects, though).

So why does your sequence of alcohols behave the way it does? The fact that the trend in boiling point is monotonic while the melting point is not suggests that even though there is a relative shift in importance of types of intermolecular interaction as the alkyl chain increase (from hydrogen bonding to van der Waals interactions), it is probably not the source of the melting point drop. Hence it likely has to do with geometric factors in the solid. The alcohol's alkyl chain probably disrupts the hydrogen bonding network of the solid as it grows from methyl to propyl in a very severe fashion, by making the solid pack less well or by partially hindering the number or strength of hydrogen bonds in the geometry of the solid. Perhaps looking at the crystalline structures of the solid alcohols may provide further insight.

Edit: I wrote a wrong answer initially. I've updated it, but now I realize I'm still not quite sure it's correct. I originally meant to write just a comment, and as I kept writing more I forgot that my very first thought was actually incorrect, so the whole text wasn't built as well as it should be. Anyone is free to pick apart or repurpose what I wrote!

Edit 2: User Uncle Al posted an interesting list of compounds and their melting/boiling points, showing the importance of molecular symmetry and solid packing.


Reaction of amines with diethyl oxalate (Hofmann amine separation method) Why does only one molecule of secondary amine react with diethyl oxalate and two molecules of primary amine react with diethyl oxalate?

I feel secondary amine also should have reacted similar to primary amine.

From JEE-Main 2015:

  Reaction with diethyl oxalate

  1°, 2°, and 3° amines can be distinguished by their reactions with diethyl
  oxalate. Primary (1°) amines react with diethyl oxalate forming N,N-oxamide, which is a solid.

  Secondary (2°) amines react with diethyl oxalate forming oxamic ester,
  which is a liquid.

  Tertiary (3°) amines do not react with diethyl oxalate.

When a 1° amine attacks to a carboxyl group, The structure below (left) is generated which lowers the energy level of the intermediate such that the second amine can easily react with the other carboxyl group.
It is clear that 2° amines (right) don't have such proton for hydrogen bonding with the oxygen. Why the second 2° amine doesn't attack to the other carboxyl at all is maybe due to the decreased partial positive charge of the amide (less electronegativity of nitrogen) toward the carboxyl.
Finally, 3° amines don't have even a leaving proton to substitute with the ethoxide.


How to measure band gap of powder sample with using diffuse reflectance method We have UV SHİMADZU 2550 and its diffuse reflectance attachment. I want to measure band gap of single crystal samples. What should I do? Which formulas should I use to calculate band gap? I don't have much sample...

It's not too hard to determine band gap.  When the energy of a wavelength of light is higher than the energy of the band gap the light will be absorbed, any wavelengths that are lower in energy will reflect or transmit. Simply make the spectrum, there should be a cliff-looking cut-off in transmittance, determine the wavelength and corresponding energy and that is your band gap.


What is the correct regular expression for InChI? When I look for a regular expression for InChI, I succeed on EBI's MIRIAM Registery website:

^InChI\=1S?\/[A-Za-z0-9]+(\+[0-9]+)?(\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/]+)*$

And indeed it works for some examples (using Python's re module):

import re

pattern = '^InChI\=1S?\/[A-Za-z0-9]+(\+[0-9]+)?(\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/]+)*$'

s = 'InChI=1S/C2H6O/c1-2-3/h3H,2H2,1H3'

re.findall(pattern, s)

will print

[('', '/c1-2-3/h3H,2H2,1H3')]

However, if I take other examples, as e.g. this one

s2 = 'InChI=1S/C6H12O6/c7-1-2-3(8)4(9)5(10)6(11)12-2/h2-11H,1H2/t2-,3-,4+,5+,6?/m1/s1'

it fails

re.findall(pattern, s2)

will return

[]

What is the correct and general regular expression for InChI?

EDIT

If I see this correctly, in the pattern from above, question marks are not taken into account which are used to mark both “unknown” and “undefined” chiral centres. If one adjusts the regex accordingly, it at least works for the example above; whether it then holds for all InChI is unclear, of course.

pattern = '^InChI\=1S?\/[A-Za-z0-9]+(\+[0-9]+)?(\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/\?]+)*$'

s2 = 'InChI=1S/C6H12O6/c7-1-2-3(8)4(9)5(10)6(11)12-2/h2-11H,1H2/t2-,3-,4+,5+,6?/m1/s1'

re.findall(pattern, s2)

will return

[('', '/c7-1-2-3(8)4(9)5(10)6(11)12-2/h2-11H,1H2/t2-,3-,4+,5+,6?/m1/s1')]

As long as you are not looking for something very specific, and you are not performing a sanity check of what you are matching against, the provided regular expression will match more often wrong than falsely not matching. It will even match more nonsense InChI than total number of existing ones. It is therefore not much better than the trivial catch them all:

^InChI\=1S?\/[^\s]+(\s|$)

The space (and some few others) is a character that will not be part of a correct InChI. The above will therefore match all correct InChI strings (and many more), and certainly more correct ones than the one given in the question (or in your answer).

The principal problem with the whole approach is basically given in the technical FAQ in section 5. Depending on what we are looking at, not all the layers will be present. There is not even a guarantee that a sum formula is present, as represented by the proton: InChI=1S/p+1. This one (and its relatives) is basically responsible why regex is doomed from the start, and that there is no correct regex in the first place.

If we wanted to implement only a tiny fraction of sanity control to reduce the number of mismatches, we will soon see that the whole string becomes about a page long.
Let's examine the given regex a little more to actually understand what it and why it's attempt is futile.

^InChI\=1S?\/[A-Za-z0-9\.]+(\+[0-9]+)?(\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/\?\;\.]+)*$

Beginning with

^

matches the start of the string, which is not really necessary

 InChI\=1S?

matches the prefix, while leaving it open if standard or not

           \/[A-Za-z0-9\.]+(\+[0-9]+)?

matches a sum formula (even the silly ones) or anything that doesn't have a main layer and goes straight to isotopes or charges, etc.. Better would here be to make the sum formula optional, i.e. (\/[A-Za-z0-9\.]+)?

                                      (\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/\?\;\.]+)*$

this is pretty much a fancy way of saying match any kind of layer, as often as you want,( or not,) in what order is just convenient, with whatever content.
However, it does not match *, which it should: InChI=1S/2C2H5.Zn/c2*1-2;/h2*1H2,2H3; as given in the FAQ.

A much better approach would be to match the layers separately, to have a little control, because they have a predefined order. For example start with the connectivity 1.1 layer (\/c[0-9\-\*\;\(\)]+)?, move to the connectivity 1.2 or hydrogen layer (\/h[H0-9,\(\)\;]+)?. I am not too sure whether I caught all special characters that might be inserted. Then match on with charge (\/q(\+|\-)+[1-9]+[0-9]*)? and the proton balance \/p, the various stereochemistry layers, fixed layers, etc. There are plenty. I probably got the charge already halfway wrong, as I am not sure whether it needs ; or not.

Coming up with a correct and not too greedy regex for each layer is already a nice task, and if you can do something useful with it, good. For everything else, there is a reason that a software exists to create and parse those strings. For every other purpose the greedy match-it-all from the start should work perfectly fine. Now, InChIKeys are something very different ...


Volume of products - stoichiometry $\rm 20\,cm^3$ of ethyne, $\ce{C2H2}$, is reacted with $\rm 500 \,cm^3$ of oxygen.
  The equation for the reaction is:

  $$\ce{2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)}$$

  What is the total volume of gas remaining at the end of the reaction?

  (all volumes are measured at room temperature and pressure)

My effort: I think the answer should be $510\,\mathrm{cm^3}$ because the volume of the leftover oxygen is $450\,\mathrm{cm^3}$ and the volume of the products is $60\,\mathrm{cm^3}$ but the answer is $490\,\mathrm{cm^3}$. Can someone please explain why?

Here's a breakdown: First, you identify that ethyne is the limiting reagent and would be consumed completely. Second, you correctly calculated the volume of oxygen left that is $450\ \mathrm{cm^3}$. $20\ \mathrm{cm^3}$ of ethyne will completely react to produce $40\ \mathrm{cm^3}$ of carbon dioxide.

Thus, the total gas volume when the reaction reaches completion is $490\ \mathrm{cm^3}$.


Possible Synthesis of MVK from 2-Butanone? My professor wanted me to set up a lab to synthesize Methyl Vinyl Ketone (AKA Butenone) from some MEK we have lying around (2-Butanone).

Now, I know there are multiple ways to do this, but the tricky part is finding out a safe way to go about this as MVK has a very low flashpoint.

I figured we could do an acid catalyzed alpha halogenation with Bromine and some Acetic Acid, and then follow up with an Elimation using a bulky base such as tert-butoxide.

The only problem is I have no idea how long, how much, or how cold/warm these materials need to be for the reaction to occur.  Ideally the entire reaction would take place in an ice water bath due to the danger, but at the same time I don't want to take five hours synthesizing one compound.

Let me know what you think, and feel free to recommend a different (but safe) route of synthesis, as our lab has plenty of chemicals to work with.  By the way, we only need to create about 10-20ml of the stuff, but more is better.

As pointed out in the comments, there is no way to make it safe, as methyl vinyl ketone itself is a very nasty compound.

I am also not an expert on organic synthesis, but I found some procedures, that might help you in adopting a synthetic route. The one you suggested might work, but be aware, that you are dealing with a methyl ketone, which can undergo the haloform reaction quite easily.

For example, in [1] they describe the bromination with bromine and absolute methanol at the methyl group for 3-methyl-2-butanone at a nice moderate temperature range 0-10 °C. The yield is 95%. But this is not what you want. Adding an acid as catalyst might work in your favour though, as after protonation the partial positive charge is better stabilised at the higher substituted carbon.

In [2] they use N-bromosuccinimide and aqueous hydrogen bromide to introduce the halogen in α position. They convert hexanoic acid with thionylchlorid to the acid chloride and subsequently to 2-bromohaxanoylchloride. The reaction takes place in the also quite nasty carbon tetrachloride at about 85 °C. This might also work for your system, it's a bit warmer than you expected though.

In [3] they use sulfuryl chloride in carbon tetrachloride to convert pantane-3-one to 2-chloropentane-3-one at about 45 °C with about 80% yield. How selective this will work with an asymmetric ketone is not clear.

Once you have obtained the 3-halogeno-butan-2-one, things will not get easier. I found a patent [4], where they convert 3-chlorobutan-2-one via pyrolysis with sodium acetate at 500-550 °C.

I could not find a reference for an elimination, but that should also work as long as you use a bulky base.

Wikipedia [5] has a different route for the synthesis, which might be a bit more economical and easier.

Well, good luck!

Org. Synth. 1976, 55, 24.
Org. Synth. 1976, 55, 27.
Org. Synth. 2002, 79, 251.
J. R. Long: Preparation of methyl vinyl ketone. US 2256149 A, 16. 09. 1941.
http://en.wikipedia.org/wiki/Methyl_vinyl_ketone#Production


Trouble to predict the effect of pressure and temperature for a reaction in equilibrium I have trouble to predict the effect of decreasing pressure and increasing temperature on the reaction $\ce{N2 + 2O2 <=> 2NO2}$. Increasing temperature caused this reaction to shift in the forward direction and decreasing pressure brought no change. I know increasing temperature favour endothermic reaction and decreasing temperature favour exothermic reaction. I also know on increasing pressure the system will shift to the direction where less moles are present. Please tell me how to use this information to predict the effect for increasing temperature and decreasing pressure.

When I posted this question I had no idea how changing each factor bring change in equilibrium. Then I bombarded my brain with everything I know about equilibrium and I found a fairly easy way to understand this.

Change in concentration

Imagine two vessels A and B connected to each other through a thin hollow pipe through the middle of the vessel. Let vessel A be reactant and vessel B be product. The vessels are filled with water a little above the pipe. If I add water to vessel A  water flow from A to B. Thus increasing concentrated of A caused the water in vessel A to flow to B; which in other words is the forward reaction.

Change in temperature

Think of a reaction which is in equilibrium.

1)Decreasing the temperature of reaction in equilibrium.

We know that decreasing the temperature favour exothermic reaction
 and exothermic reaction are favoured by low temperature. Exothermic mean giving out heat which remove the deficiency of temperature to bring it back to normal in equilibrium.

2) Increasing the the temperature of reaction in equilibrium

We also know that increasing the temperature favour endothermic reaction or endothermic reaction are favoured by high temperature. Endothermic means to absorb heat energy which causes the excess heat energy ( from high temperature) to bring it back to normal in equilibrium

Effect of pressure

Changing pressure play a significant role in gaseous reaction which are accompanied by change in moles. On increasing pressure of the system equilibrium will shift to the direction where number of moles decrease or where pressure decrease (because pressure is directly proportional to the number of moles). If pressure on the system is increased equilibrium shift in the direction in which decrease in number of moles take place.

This is the second time I answered on stack exchange so please let me know my mistake if any. If I had known how to create the diagrams I usually find in other answers I would have added them.


Why is the inversion barrier larger in PH3 than it is in NH3? The inversion barrier in $\ce{NH3}$ is approximately $5~\mathrm{kcal~mol^{-1}}$ and that of $\ce{PH3}$ is $35~\mathrm{kcal~mol^{-1}}$. This has well-known stereochemical consequences in that amines are not chiral whereas phosphines can be. How can the larger inversion barrier be explained?

Ammonia is the classic system for $\ce{sp^3}$ hybridisation save methane. The lone pair (and each of the $\unicode[Times]{x3C3}$-bonds) has almost $25~\%$ s-character which corresponds nicely to $\ce{sp^3}$. However, the whole system can also swing around, changing its hybridisation to $\ce{sp^2}$ and back; a process during which the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.

Elements outside of the second period show a much smaller tendency to involve the s-orbital in bonding. The bonding orbitals only have an s-character of approximately $16~\%$. This also means that the phosphorous lone pair has a much higher s-character of approximately $50~\%$, while the bonding $\unicode[Times]{x3C3}$-orbitals have a larger p-character. (This also explains the much smaller $\ce{H-P-H}$ bonding angle of approximately $90^\circ$.) An orbital of high s-character has a long way to go to turn into a p-type orbital, and the three largely p-type $\unicode[Times]{x3C3}$-orbitals have an equally long way to go to give $\ce{sp^2}$-type orbitals so the interconversion and thereby the inversion of $\ce{PH3}$ is strongly hindered.


Do We change sign of electrode potential when reversing an equation? I have a cell equation as:

$$\ce{2Cr^0 + 3Cd^2+ -> 2Cr^3+ + 3Cd^0}$$

I am provided the standard reduction potentials as:

$$\begin{align}\ce{Cd^2+ + 2e- &-> Cd} &&    E= \pu{-0.4V}\\
\ce{Cr^3+ + 3e- &-> Cr} &&      E= \pu{-0.74V}\end{align}$$

Now do I use chromium’s reduction potential as $\pu{-0.74V}$ or do I flip the equation and use $\pu{+0.74V}$ (as it is getting oxidised and I have the equation for reduction)?

This is using the formula $E(\text{cell}) = E(\text{cathode}) -E(\text{anode})$, all reduction potentials.

The reduction potentials are a special case of standard reaction enthalpy, because you can transform $\Delta E$ into $\Delta H$ just by calculating with a few constants.

Standard reaction enthalpy and all other thermodynamic values related to reactions change sign when the direction of reaction is reversed.


Can ammonia be detected by a 1D-1H-NOESY experiment? I'm not detecting ammonia in my samples while using a NOESY pulse program for a 1H-NMR experiment.

Is it possible for ammonia to be detected by a 1D-1H-NOESY experiment?

In the 1H,1H-NOESY experiment you are using the nuclear Overhauser effect to transfer magnetization from one proton to another proton in close proximity (< 5 Å).  This only works within a single molecule, unless you have a reasonably stable complex you can't observe signals in the NOESY between protons of different molecules in your solution.

Ammonia has three protons which are at least chemically equivalent, their signals will appear at the same chemical shift. In a 1D NOESY you irradiate at the frequency of a specific proton and then observe the transferred magnetization at other frequencies. You can't observe magnetization transfer between protons at the same frequency that way, as your irradiation will destroy any signal at this specific frequency.

You can ignore the explanation above, as the NOESY experiment you are using is different. In metabolomics the NOESY 1D is used for water suppression, this is a different experiment than the 1D NOESY I explained above.

The common variant of this I found also uses presaturation to suppress water. This presaturation will also affect all signals that exchange quickly with the solvent, and the protons of ammonia are exchanging with water. So I would expect the ammonia signal to be far weaker or even not visible in such an experiment.

If you want to observe ammonia, you have to use a water suppression method that doesn't affect exchanging protons.


Effects of Changing Avogadro's Constant The Avogadro project suggests that we redefine the Avogadro constant to be equal to our best known estimate, $N_\mathrm{A} = 6.02214179 \times 10^{23}$, and redefine the kilogram based on the Avogadro constant.
What happens if we define the Avogadro constant to be exactly $6 \times 10^{23}$ and reassign atomic weights correspondingly?
Would any unforeseen complications arise from here?
Example:
Under current definitions, $N_\mathrm{A}$ atoms of carbon-12 weigh 12 g. If we redefine the constant to have the value $N_\mathrm{A}^\prime = 6.0\times 10^{23}$, then $N_\mathrm{A}^\prime$ atoms of carbon-12 would weigh $\pu{11.955879 g}$, which we would have to assign as the new atomic mass of carbon-12.

As of May 20th 2019, the Avogadro constant will be set to $6.02214076 \times 10^{23}\ \rm{mol}^{–1}$, i.e. the Avogadro constant times one mole will be an integer, but not the one the OP asked about. In preparing for the switch of definition, the BIPM looked carefully at two ways of determining the constant, and then set it to the best estimate from experimental values, truncated to nine significant figures.

This way, there will be minimal practical changes of units even though the definitions have changed. Also changed is what is defined vs. what is measured. For example, the molar mass of 12-C is now based on measurement, and could change (certainly in terms of how many significant figures are known) as measurement techniques improve.

Setting the Avogadro constant to a dimensionless number as the OP asked about would make lots of relationships between quantities and constant dimensionally incorrect, such as $$ R = k_\mathrm{B} N_\mathrm{A}$$

with $k_\mathrm{B}$ the Boltzmann constant, R the universal gas constant and $N_\mathrm{A}$ the Avogadro constant.

  Would any unforeseen complications arise from here?

Setting Avogadro's constant to $6 \times 10^{23}\ \rm{mol}^{–1}$ (I added the proper units) would have lots of complications, but they would not be unforeseen, as @Zhe discusses in the comments.


Are we technically reducing electrodes/wires for the instant that an electron passes through them? In lecture I was taught that, in a galvanic cell at standard conditions, platinum is often used as an inert electrode when the species being oxidized or reduced exists only in solution, because platinum is "inert," which means it is very stable at its oxidation state and resists reduction or oxidation. This concept confuses me, because "Reduction" means to lose charge, a.k.a. gain an electron(s). When the electrode has an electron passing through it in order to be transferred to another material, isn't it being reduced, albeit for a very miniscule amount of time?
For example, consider a galvanic cell with platinum electrodes and an iron wire. If we were to examine the cell at an instant and examine the platinum electrodes/wire, we would see a charge of -1 on some of the platinum atoms and iron wire, correct?
What I'd like to imagine is that, while platinum is capable of having a negative charge because of the room for electrons in its d-block, the species being reduced in the half-cell will always have a higher affinity for accepting the electron due to its higher standard reduction potential. This still leaves me confused on why platinum is willing to accept the electron and thus making platinum a medium for electron transfer in the first place, though. This all applies to the iron wire (or any other element that may be used for electron transfer) as well, but I would like to tackle the case of platinum overall.

Reduction and oxidation are defined as gaining and donating electrons from the valence shell of an atom respectively.
Whereas in the case of electrons flowing in the electrodes and connecting wires the electrons are passing from the interatomic voids and not being released by the wires and the electrode.
(There will be some electrons which will interact with the conductor's atoms but what will happen is that they will get deflected. There will be a slim chance that a few electrons will replace an electron from the conductor's electron cloud but that doesn't change the situation as both the electrons are same in a classical sense. )
If your claim was true than the anode would have got some extra negative charge and the cathode extra positive which would have driven up the voltage higher than that expected from the cell alone.
In a similar way when we connect a wire it doesn't magically change the measured voltage.
Even when we perfom the same experiment with different inert electrodes and different kinds of wires the measured reading doesn't change much. (There are some changes in current flowing through the circuit due to the difference in resistances of different materials)


Cadmium Selenide, Cadmium Sulfide, Cinnabar and Lead Oxide toxicity I have a question about the toxicity of the pigments Cadmium Selenide (Cadmium Red), Cadmium Sulfide (Cadmium Yellow), Mercury Sulfide (Cinnabar or Vermillion) and Lead Oxide (Lead White).

Not long ago the EU banned lead-white pigments from use, even for art. It is now generally impossible to get hold of lead-white oil paint on the internet. However, the cadmium pigments were spared a ban recently. And it is quite easy to find Cinnabar and Vermillion oil paint online even of European production. For example, the dutch company Royal Talens supplies Vermillion paint online, without even warning that it contains mercury: https://www.royaltalens.com/en/catalog/rembrandt-oil-colour/oil-colour-paint-vermillion-311-40ml-tube/

I am not a chemist but my limited knowledge tells me that mercury is a lot more poisonous than lead. How come Cinnabar wasn't banned and what is the relative toxicity of the four pigments I listed, given that they are suspended in oil?

Thank you very much.

I found an answer to my question. Typically oil paint manufacturers will write what pigments are used in their paints. The pigments are classified according to the 'color index' or C.I. These pigments may actually have different chemical composition, but nevertheless the color index can be used to search the pigment database here: http://www.artiscreation.com/red.html

For example, PR106 stands for genuine Vermillion, which is true Cinnabar and has a toxicity rating of C according to the table linked above, where A is non-toxic and D is extremely toxic.

Most colors that are marketed as 'cinnabar' are actually composed of various different pigments, most often not actually containing mercury.


Term used to describe scalar coupling over a single bond? I know that coupling on NMR over two bonds is referred to as geminal coupling and over three bonds is referred to as vicinal coupling. However is there a term for coupling over one bond (such as a carbon-13 bonded to a fluorine?

I've never heard a special term but you can describe it as $^1 J$ or $^1 J_{CF}$ coupling using the systematic nomenclature.


Why does NOCl dissociate as +NO and Cl-? $\ce{N}$ is more electronegative than $\ce{Cl}$ (albeit by a very small difference) but it should ideally dissociate as $\ce{Cl+}$ and $\ce{-NO}$ (similar to the dissociation of $\ce{Cl}$ in $\ce{HOCl}$). Moreover, due to the stronger bonding between $\ce{N}$ and $\ce{O, N}$ should also be pulling electrons even more tightly to itself due to the very electronegative $\ce{ O}$  by its side.

I was doing a problem on the electrophilic addition of $\ce{NOCl}$ on alkene and ended by getting the wrong product (according to the book) because I took $\ce{Cl+}$ and $\ce{-NO}$ in the mechanism at the very beginning.
Could someone explain?

I've attached pictures of what I have done and what the book has done too. Please correct me if I am wrong.

What I did:

What the book did:

Firstly, the correct comparison to make to determine which is more likely of $\ce{NO+}+\ce{Cl-}$ and $\ce{NO-}+\ce{Cl+}$ is the ionization potential vs. the electron affinity of the the two species. The electronegativities are correlated with these other metrics, but it is important to emphasize that those are really the important things.

Relatedly, nitrosonium $\ce{NO+}$ is stabilized by resonance and so this is a considerably stabilized cation (i.e. IP not too high). On the other hand, the IP of $\ce{Cl}$ is huge while the EA is small; it would much rather get an electron than lose one. The hyponitrite anion $\ce{NO-}$ actually usually dimerizes for reasons outside the bounds of the question. So $\ce{NO+}+\ce{Cl-}$ is way lower in energy.

In general, a red-flag should go up when you see halogens with positive charges. They do happen, but very rarely, especially at the level of chemistry in an intro organic course.


Belousov-Zhabotinsky reaction The Belousov-Zhabotinsky reaction is a very fascinating one. The Chemistry behind the reaction concerns thermodynamics and equilibrium. I want to perform the experiment, but I'm not able to find detailed instructions about the reaction. Like, proportions of the chemicals, explanation of the reaction, duration of the reaction, sequence of mixing the reagents, etc.

The Belousov-Zhabotinsky reaction is a video I found on YouTube.

But that does not give any details at all. Can anyone give me a detailed reaction procedure?

That is indeed a fascinating reaction!  The following is a part answer, providing links to some resources that may be of use.

The following resources may be of assistance:

The Belousov-Zhabotinsky page from the Kansas Academy for Mathematics and Science Education provides a bit about the background, chemical processes and most relevant to your request, an experimental procedure.

Mary Newton of the Georgia Regent University has provided a reasonably detailed procedure on the webpage Procedure for B-Z reaction.

Slightly different, is the information provided in the Belousov-Zhabotinsky Reaction (Leigh) page from the Flathead Valley Community College takes a different approach, with a focus on the mathematical derivations relevant to the experiment.


Does chromium oxide form on Stainless-steel in 'deoxygenated' water? Will a passive layer form on stainless-steel in pure (0.055 μS/cm) water with an oxygen concentration < 50 ppb. If so, why is this type of water generally called 'deoxygenated'? If a passive layer (chromium oxide) does not form with such low oxygen concentration why is this, since there IS still oxygen in the water.

From a practical standpoint, once the stainless steel item is made the passive layer immediately begins to form. Passivation with acids is a process used to accelerate this process and have parts ready to use in a shorter period of time. So if you have any high quality SS parts, they will come to you with the passive layer already developed.

Theoretically speaking, if a completely naked section of stainless was put into this "pure water", there would still be some oxide formation because the specification of <50 ppb still allows for some oxygen to be present. However, this should not have any practical implications. I suspect the name "deoxygenated water" is used to indicate there was a conscious process used to reach such a low amount of free oxygen in the water but not to claim o% free oxygen content.


What are the properties of various solutions at pH <2 that may cause injuries? For example, why do some solutions not cause any injuries at pH 2 while HCl does?

Injury is not solely related to how strongly acidic a solution is, as there are several mechanisms by which organic matter is damaged from exposure to acids, not just protonation. It is also highly dependent on the type and length of exposure. Aqueous $\ce{HF}$ is likely considerably corrosive after skin contact even at pH 2, mostly because of $\ce{F^-}$ which is a rather reactive ion with a strong tendency to abstract $\ce{Ca^{2+}}$ (in fact, the fact that  $\ce{HF}$ is a weak acid makes it even more dangerous because the undissociated $\ce{HF}$ is much less polar and the molecules can cross the skin barrier more easily, allowing direct access to muscle and bone which can be severely damaged). Aqueous $\ce{HCN}$ also forms weak acids, but again $\ce{CN^-}$ ions are extremely toxic as they block cell respiration. The risk is compounded by the fact that $\ce{HCN}$ is a very volatile substance and can be released as a gas from concentrated or acidic solutions.

With regards to $\ce{HCl}$, you need to be a bit more specific. What kind of injury? $\ce{HCl}$ itself is a relatively safe acid in dilute solution, even though it's a strong acid, as it does not have strongly oxidising/reducing/dehydrating/toxic properties and is only volatile at rather high concentrations. Remember that $\ce{HCl}$ is even used as an integral part of our digestive system, being the species that allows stomach acids to reach pH 1-2.


Can i use deionized water instead of distilled water My question is really simple : can i use deionized water instead of distilled in experience ?
Because i have a air dehumidifier at home and it produce a lot of deionized water
I know that deionized water ($\ce{diH2O}$) is :

  Water that has had almost all of its mineral ions removed, such as cations like sodium, calcium, iron, and copper, and anions such as chloride and sulfate.

And distilled water ($\ce{dH2O}$) is :

  Water boiled and then condensed into a clean container, leaving solid contaminants behind

So basically the difference between $\ce{diH2O}$ and $\ce{dH2O}$ is that $\ce{dH2O}$ keep his ions mineral right ?
Can i use deionized water anyway ? Or not ?

Water from a dehumidifier is similar to distilled, in that it contains little in the way of dissolved materials.   It condensed on (typically) anodized aluminum,
and dripped through polymer collection trays into a bucket.

How clean is the bucket?  There may be residue from dust/lint etc.

It's water.   You can moisten plants, fish can swim in it, and it's pure enough
to drink or wash your windows (in fact, it's softer than most tap water, it
makes a good final rinse for windows).  I'd stick to deionized or distilled
for lab use, though, because those didn't come from a dirty bucket.


Dissolution of SO2 in water Majority of texts (at high school level) say that when $\ce{SO2}$ is dissolved in water sulphurous acid $\ce{H2SO3}$ is formed. However my text book states that it forms clathrate $\ce{SO2.6H2O}$ when dissolved in water, and can significantly exist as $\ce{H+ + HSO3-}$.
I found similar finding on Wikipedia:

Sulfur dioxide is fairly soluble in water, and by both IR and Raman spectroscopy; the hypothetical sulfurous acid, $\ce{H2SO3}$, is not present to any extent. However, such solutions do show spectra of the hydrogen sulfite ion, $\ce{HSO3−}$, by reaction with water, and it is in fact the actual reducing agent present:
$$\ce{SO2 + H2O ⇌ HSO3− + H+}$$

However there's no mention of clathrate on the whole page. Do what's the actual product on dissolution of $\ce{SO2}$ in water?

Thanks for bringing up this topic, and I would have appreciated it a few years earlier, however!
The important topic I am referring to is the apparent exclusive gas-phase formation of the molecule H2SO3, as correctly noted in Wikipedia on H2SO3, to quote:

There is no evidence that sulfurous acid exists in solution, but the molecule has been detected in the gas phase.[1] The conjugate bases of this elusive acid are, however, common anions, bisulfite (or hydrogen sulfite) and sulfite. Sulfurous acid is an intermediate species in the formation of acid rain from sulfur dioxide.[2]

The implication for acid rain formation has previously been noted, for example, in an MIT article, with cited  Reactions (1) to (3) below:

$\ce{SO2 + .OH + M → .HOSO2 +M (1) }$
$\ce{ .HOSO2 +O2 → SO3 + .HO2 (2) }$
$\ce{SO3 + H2O → H2SO4 (3) }$

However, in this recent 2019 work: A New Mechanism of Acid Rain Generation from HOSO at the Air–Water Interface, some important chemistry:

The photochemistry of SO₂ at the air–water interface of water droplets leads to the formation of HOSO radicals. Using first-principles simulations, we show that HOSO displays an unforeseen strong acidity (pKₐ = −1) comparable with that of nitric acid and is fully dissociated at the air–water interface. Accordingly, this radical might play an important role in acid rain formation.

where the net photolysis of gaseous sulfurous acid (in addition to SO2) likely proceeds as follows:
$\ce{H2SO3 (g) + hv -> .OH (g) + .HOSO (g) }$
Supporting source: See Page S6,Table S2, Eq (1), Eq (2), Eq (5) and Eq (12) in this available supplement. Also, related results for the photolysis of nitric acid, to quote:

Here we present both field and laboratory results to demonstrate that HNO3 deposited on ground and vegetation surfaces may undergo effective photolysis to form HONO and NOx, 1–2 orders of magnitude faster than in the gas phase and aqueous phase. With this enhanced rate, HNO3 photolysis on surfaces may significantly impact the chemistry of the overlying atmospheric boundary layer in remote low‐NOx regions via the emission of HONO as a radical precursor and the recycling of HNO3 deposited on ground surfaces back to NOx.

Note, there is thus an implied similar possible acceleration in the rate of photolysis of gas-phase SO2 after becoming H2SO3 at the air–water interface.
The implications of the above chemistry is that in addition to the cited Reaction (1) above (which is a sink for the removal of the hydroxyl radical, that otherwise could be involved in an ozone depletion cycle), the UV photo-induced decomposition of also gaseous H2SO3 likely leads to more problematic radicals cited in the acid rain formation and even ozone depletion.


Polyethylene biodegradation, how? Worm guts have bacteria that apparently can degrade polyethylene:

The only agents that I have heard of that can degrade a fully-saturated hydrocarbon at room temperature are fluorine (and related nasties), other halogens (with UV light), magic acid (which is extremely reactive with almost anything). None of these seem biochemically feasible.

How is the polyethylene being attacked? Are the ends of the molecules more reactive double bonds that can be be broken off one carbon at a time like beta oxidation of fats?

How is the polyethylene being attacked?

Polyethylene is attacked by microbes that contain oxygenase or oxidase enzymes.  These enzymes can attack C-H bonds even at moderate temperatures.

  The only agents that I have heard of that can degrade a fully-saturated hydrocarbon at room temperature are fluorine (and related nasties), other halogens (with UV light), magic acid (which is extremely reactive with almost anything).

One prominent example of a hydrocarbon degrading enzyme is methane monooxygenase, which is capable of catalyzing methane oxidation to methanol even at temperatures as low as 0 °C.  The C-H bonds of polyethylene are about as reactive as methane's.  (Actually, they're slightly more reactive.)  Thus enzymes can do the job.  They don't need elemental fluorine or magic acid.

However, your point about halogens reacting with hydrocarbons is very interesting.  Several research groups have reported that adding halide salts to aqueous (and aerobic) suspensions of bacteria accelerates the rate of oxidation.

It's possible that this finding can be explained by the involvement of haloperoxidase enzymes in the degradation.  However, the mechanism of microbial polyethylene degradation is a very new, and very active, area of ongoing research.


How is indigo carmine used in Lowenthal-Permanganate titration as a redox indicator? In a Lowenthal-Permanganate titration, cider tannins are quantified by titrating the cider with indigo carmine indicator and sulfuric acid against potassium permanganate. Tannins are antioxidants, so that means they give up electrons to reduce the potassium permanganate. The endpoint of this titration is when the blue analyte turns a yellow colour. According to Wikipedia, indigo carmine turns yellow when reduced.
What I don't quite understand is why the indigo carmine didn't reduce when it was in the presence of the antioxidative tannins before the potassium permanganate was added and why it turned yellow only after potassium permanganate was added.

I did some brief looking but given that the carmine is also pH indicator that remains blue at below 11.4 pH it would appear as though your sulfuric acid is preventing the reduction from occurring in your indicator. Under acidic conditions KMnO4 is a strong oxidizer where carmine is a weak oxidizer under these conditions. However I am not sure how the carmine gets reduced to yellow upon equilibrium. Perhaps the pH changes. It appears as though permanganates acid conjugate is strong so its a weak base. Just throwing out ideas there though.
http://chem.lapeer.org/Chem1Docs/IndigoCarmine.html
https://www.sciencedirect.com/topics/earth-and-planetary-sciences/permanganate


Why are geminal diols unstable? The most stable of:

(a) $\ce{(CH3)2C(OH)2}$

(b) $\ce{CH3 C Cl (OH)2}$

(c) $\ce{CH3CN}$

is $\ce{CH3CN}$, because the other two are geminal diols. How can we tell that geminal diols are unstable?

Formation Of Gemdiol

Carbonyl compounds react with water to give gem diols. This reaction is catalysed by acid. The reaction is reversible reaction.

Gem diols are highly unstable compounds hence equilibrium favours the backward direction. The extent to which an aldehyde or ketone is hydrated depends on the stability of gem diol.

Stability of gem diols depend on the following factors:

(i) Steric hindrance by +I group around α-carbon decreases the stability of gem diols. +I group decreases stability of gem diol and hence decreases extent of hydration.

(ii) Stability of gem diols mainly depends on the presence of –I group on α-carbon. More is the –I power of the group more will be stability of gem diols.

(iii) Intramolecular hydrogen bonding increases stability of gem diols. –I groups present on carbon having gem diol group increases strength of hydrogen bond.

More is the strength of hydrogen bond more will be the stability of gem diol.

Overall stability : (CH3)2C(OH)2 < CH3CCl(OH)2 < CH3CN


What are high-energy electrons? I read that (in cellular respiration) the transported electrons in NADH have a higher energy than those in FADH2.
I can't find a (simple or otherwise) explanation of what a "high-energy" electron is.
I came across a post on Reddit which asks the same question:

I am not a chemist but I am trying to understand what my biochemistry book
means by "high energy electrons" that are obtained from glycolysis and
the citric acid cycle. I am assuming that this "high energy" is not a
property of the electrons themselves, but it is not clear to me in
what way they are "high energy". Do they mean that when these
electrons are transferred to the electron carriers they are part of
molecular structures in which they sit in high energy positions?

(There were no satisfactory answers to the question).
Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

transported electrons in $\ce{NADH}$ have a higher energy than those in $\ce{FADH2}.$

This is jargon describing the redox potential of the electron carrier $\ce{NADH}/\ce{NAD+}$ vs the electron carrier $\ce{FADH2}/\ce{FAD}.$ So if you compare the two reduction half reactions
$$
\begin{align}
\ce{NAD+ + H+ + 2e- &-> NADH} &\quad E^{\circ '}_\mathrm{red} &= \pu{-0.320 V}\tag{1}\\
\ce{FAD + 2H+ + 2e- &-> FADH2} &\quad E^{\circ '}_\mathrm{red} &= \pu{0.031 V}\tag{2}
\end{align}
$$
$\ce{NADH}$ is the stronger reducing agent.

Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

Another way of saying this is that the reaction of $\ce{NADH}$ with dioxygen is more exergonic (the equilibrium lies further on the side of the products, more free energy is available from it) than the reaction of $\ce{FADH2}$ with dioxygen. For completeness, the reduction half reaction for dioxygen/water is:
$$\ce{1/2 O2 + 2 e- + 2H+ -> H2O} \quad E^{\circ '}_\mathrm{red} = \pu{0.816 V}\tag{3}$$

Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?

It has to do with how tightly the electrons are bound in one and the other electron carrier. The OP is correct to place the term "high-energy" in quotes when referring to electrons; electrons are indistinguishable, so their energy is with respect to the bound state that they are in.


Different solder composition affects the connection strength? I was wondering if the tin/lead/other metals bond with the metal parts that are low-temperature soldered. I know that the flux in the solder is used to remove the oxide layer, leaving a clean metal surface ready to form metallic bonds (I think).

So for example, should I use a tin/silver/copper eutectic to bond a silver surface and copper wire? Will that make a difference or can I use tin/lead with the same result? For example, do we get a different connection strength? In every case the solder material is mostly made of tin, which is why I was wondering what's the real impact of the other components.

Good question!

The short answer: if you are soldering bulk silver to copper, 63% Sn/ 37% Pb eutectic would probably be OK, but for silver plating, you might be able to make a successful join using tin/lead, if heat is quickly applied and removed, but might need to use a silver-bearing solder for thin silver plate or if the base metal under the silver is not easily soldered.

BTW, "cold soldered" is perhaps better expressed as low-temperature soldering, though both indium and gold can be joined at room temperature.

A longer explanation: reasons to use a specific solder composition.

To some extent, solder dissolves the surface of the metals to be united. This can be problematic with plated parts, e.g. gold plated over nickel, where the solder can completely remove the plating, making it difficult to adhere to the base metal. For that purpose, solders containing the plating metal are made, but they have other liabilities: expense and brittleness. Similarly, copper-containing solders are used for copper-plated electronic parts.
Not only alloys form between soldered parts, but also intermetallic compounds form, such as $\ce{Ag3Sn}$, $\ce{Cu6Sn5}$ and $\ce{AuSn4}$, some of which are brittle. Changing solder composition may not be practical to prevent some intermetallic compounds from weakening a joint, so barrier coatings and specific temperature cycling and storage may be used to retard their formation.
Pure tin solder, as well as zinc, has the property of forming whiskers, which not only short-circuit electronics, but was rumored to have caught Napoleon's troops with their pants down.

For these and other reasons, soldering can be considered as much a craft as a science. You may have to make some tests of the solder joints if they are critical, as in aerospace or medical applications.


How do atomic radius and nuclear charge influence ionisation energies? This is actually a multiple choice question.

Why is the first ionisation energy of neon higher than that of fluorine?

A: Fluorine is more electronegative than neon
B: Neon has a complete octet, but fluorine does not
C: The atomic radius of fluorine is less than that of neon
D: The nuclear charge in neon is greater than that in fluorine

I understand why A is incorrect and why B is correct. But I am confused about C and D. I did try to search online but my questions just increased.
Firstly, why is the atomic radius of fluorine less than that of neon? Both fluorine and neon are in the same period which means that they have the same amount of shells. Since the nuclear charge of neon is more, shouldn't the nucleus attract the electrons in the outer shells closer? Which would result in neon having a smaller atomic radius than fluorine.
Secondly, according to a source, the largest factor is that the 2p sub shell in neon is full whereas in the fluorine it is not. They said that a full sub shell means that the atom is more stable which is the way neon has a higher first I.E. but since the 2p sub shell is full, wouldn't there be mutual repulsion?
I'm sorry if this is a lot, I'm trying to get a head start for my next education level and I cannot find any textbooks for that level's chemistry. I've been reading a summarized guide book and trying to do assessments whilst trying to find guides online as well.

When we are considering atomic radius:
The atomic radius of noble gases are considered the highest in a period since we only consider their Van der Waals radius (distance between two similar atoms in their closest approach). Although technically they should have lower atomic radius due to increased $Z_\text{eff}$.

As for ionisation energy:
The ionisation energy of noble gases are way higher than the other elements of the same period due to stable configuration


Will I get bromine by electrolysis of the brine from the Dead Sea? I know that electrolysis of a saturated solution of sodium chloride with an inert anode (eg., graphite) is a handy way to obtain pure chlorine. I have a sample of saturated salt solution from the Dead Sea which is reported to have up to 5% bromine concentration, the domination anion being $\ce{Cl}^{-}$, of course.

Will I get bromine on the anode by electrolysis of Dead Sea water?
In general, if I have a mixture of chlorides and bromides, what determines the ratio of electrolytically produced pure $\ce{Cl}_2$ and $\ce{Br}_2$?

The short answer is probably a bit, depending on the potential you apply.

There are at least a few factors involved:

How easily are $\ce{Cl-}$ and $\ce{Br-}$ oxidized?

$$\begin{align*}\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} +2e-}\; &E°_{\mathrm{ox}}=-1.36\ V\\\ce{2Br^{-}_{(aq)} -> Br_{2(aq)} +2e-}\; &E°_{\mathrm{ox}}=-1.09\ V\end{align*}$$
At standard activities, the two ions are pretty similar in oxidation potentials, so there is probably a range of potentials where both reactions occur to a similar degree or it might be possible to achieve some partial selectivity for bromine formation. The Nernst equation can be used to calculate equilibrium concentrations for a given potential, but so long as the potential applied is well over the standard oxidation potential, the exact values don't really matter.
What difference do the concentrations make?

In most cases, what determines how much of either reaction occurs is mass transport. For a redox reaction to occur, the reacting molecule must be very close to the electrode surface (<10 nm). For a planar electrode at a single potential in a standing solution, mass transport occurs by diffusion and the current that flows should be directly proportional to concentration in the bulk. In principle, the Cl to Br ratio should be the same as their activities in solution, though in some situations, other factors like the speed of the electron transfer can be limiting, so this may not be the case, especially since there is so much chloride compared to bromide.
What about the electrodes?

As electrolysis is a surface phenomenon, the electrode and its interactions with the reactants can be important. Assuming you pick an electrode which doesn't itself react with the chloride or bromide, the exact material can affect the behaviour of the cell. Even though the Nernst equation may predict a certain oxidation potential, interactions of the reactants and the electrode may result in varying amounts of additional potential (overpotential) needed to cause the reaction.
What happens to the bromine?

A certain amount of bromine can dissolve in water, but depending on the pH, it can rapidly disproportionate back to bromate and hypobromate: $\ce{Br2 + 2OH- → Br- + OBr- + H2O}$ Also, the chlorine gas can react with the bromide to make bromine and bromine is also known to react with metals (use carbon electrodes?). In short, you can probably create an electrolytic cell that produces some bromine, but it's difficult to say in what ratio with chlorine and what its exact fate would be.


Finding pH of a buffer What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, $\ce{CH3NH2}$,
with 25.00 mL of 0.10 M methylammonium chloride, $\ce{CH3NH3Cl}$? Assume that the volume of
the solutions are additive and that Kb = 3.70 × 10-4 for methylamine.

1) First, I multiplied 0.025 L with 0.10 M and received 0.0025 moles. I then divided 0.0025 moles with 0.050 L (The total volume of the solution) and received 0.05 M, the molar concentration of the methylamine and its conjugate acid in this mixture.

2) I then created an ICE Chart (my reactant was the methylamine so that I would not need to find the Ka) and I solved for "x" (which would give me the concentration of $\ce{OH}$ ions at equilibrium). First, I solved for "x" but I assumed that is was insignificant. In the end I received 10.57 as the pH, which was the wrong answer. Then when I solved for "x" using the quadratic formula (assuming that "x" was significant), I got 0.0003646 as my "x".

3) When I took the negative log of 0.0003646, I got a pOH of 3.46. I subtracted this number by 14, and I still received a pH of 10.57

4) I even attempted to use the Henderson Hasselbalch equation and still received 10.57 because the ratio of the base and acid is 1, meaning that the log is 0.

I don't understand what I am doing incorrect. The answer is supposed to be 10.87, can someone please assist me?

Your approach is correct and the result you obtain is right. Checked here.

As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).


dSsur, dS and dStotal relation By definition,

  dS=dqrev/T.

Based on my understanding, the subscript rev means that the dq is heat exchanged in a reversible pathway from a state i to another state f, regardless of the actual pathway by which the process proceed. I am always curious why ΔS is defined only for reversible process, it would be good if someone can explain it, nonetheless it is not our main concern here-let's just accept the definition first.

So, similarly,

  dSsur=dqrev,sur/Tsur......(1)

From what I have read, V of surrounding can be regarded as constant, so

  dqrev,sur=dUsur

and since dUsur is a state function,

  dUsur=dqsur

and therefore dqrev,sur=dqsur implying

  dSsur=dqsur/Tsur......(2)

(a) If V is constant, q=ΔU, so q is a state function at constant V?

(b) Let's stop at equation (1). Suppose we would not like to simplify it further, and since the equation (1) is an intermediate result it must be true, we have

  dS=dqrev/T and

  dSsur=dqrev,sur/Tsur

I suppose dS without any subscript is dS of system. By First Law,

dqrev= -dqrev,sur

and then if T=Tsur and the chemical reaction has not reached equilibrium (is this scenario possible?),

  dS= -dSsur

By summing up all entropy changes for system and surrounding, we have
ΔStotal=0 for non-equilibrium state? But as I know ΔStotal=0 if and only if surrounding is in equilibrium with system. What's my misconception here?

(c) How can dqsur,irrev=dqsur,rev? Does not make sense for me physically, without looking at the equation with dUsur

In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir for which $\Delta S$ is always equal to $Q_{surr}/T_{surr}$, irrespective of whether the process that the combined system and surroundings experiences is reversible.  As such, the surroundings is assumed to have an infinite heat capacity and thermal conductivity, and to not change significantly in temperature (its temperature is always $T_{surr}$), no matter how much heat it receives or rejects.  So it always presents the temperature $T_{surr}$ to the system at its interface with the system.  And, all the entropy generated in any process is confined to the system.

In evaluating the entropy change for system and surroundings, the system and the surroundings must first be separated from one another, and then each must be separately subjected to a reversible path.  These separate reversible paths do not have to match up in any way whatsoever.  For the surroundings, the final state is that which was obtained with the irreversible path when $-Q_{irrev}$ was transferred from the system. So the change in entropy for the surroundings is $-Q_{irrev}/T_{surr}$.  And the change in entropy for the combination of system and surroundings is then given by:
$$\Delta S=\Delta S_{sys}-\frac{Q_{irrev}}{T_{surr}}=\left[\int{\frac{dQ_{rev}}{T}}\right]-\frac{Q_{irrev}}{T_{surr}}$$


Synthesis of 2-Phenylcyclopentanone from cyclopentanol My best try was to use aldol condensation after the reduction of the substrate(cyclopentanol) but that gives me a product with a methylphenyl substituent(2-Phenylcyclopentanone), nor a phenyl only.

Step 1 dehydrate cyclopentanol to cyclopentene with conc Sulfuric acid

Step 2 epoxidise cyclopentene with m- Chloroperbenzoic acid

Step 3 open the epoxide with phenyl grignard to give 2-phenyl cyclopentanol

Step 4 oxidise to required product


What are the products of alkene oxymercuration with diols? I was wondering what would be the product of the reaction of alkene with a diol (such as glycol) in alkoxymerucuration/demercuration, i.e. $\ce{(CF3CO2)2Hg/NaBH4}$?
I tried to look online but I couldn't land on a paper or website that goes through such reaction. I also tried Scifinder, I marked both compounds as reactants, but zero results stemmed from the reactants!
Any input?

I'm thinking if these possibilities (I have more possibilities in my mind, but for example I selected three) are structurally accurate and energetically stable to be produced in such reaction and which is favored?

To classify them more ... to single and di addition, here is what I'm thinking:

There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE.
To start, the alkene is the (E)-propenylbenzene 1 and the diol is butan-2,3-diol 3, devoid of stereochemistry. Formation of mercurinium ion 4 with mercuric trifluoroacetate (2) is drawn as unsymmetrical because in the transition state for addition the carbon bearing the greater positive charge is the site of SN2, C-O bond formation. Mercurial 5 is a racemate, ignoring any stereochemistry of 3, which is what is formed after the first reaction. The second reaction, reduction with NaBH4, proceeds via radical 6 to form alcohol 7. A minor product may be bis-ether 9 arising from alkene 1 and mercurial 5. For mercurial 5 to form 1,4-dioxane 8 requires an oxidation of the carbon bound to mercury and a reduction of Hg+2 to Hgo as shown by the red arrows in structure 5. How do we know that this process does not occur? Because if it did, NaBH4 would not be required! Consider the oxymercuration of alkene 1 in the presence of water. If what you are suggesting were true, then 1 would form a diol rather than the observed mono-ol prior to the participation of NaBH4.


Density and concentration of hydroiodic acid I weighted my hydroiodic acid sample: 25 ml weight 34.1 g, so the density is about 1.364 g/ml.

I want to know its concentration based on density. Does anyone have the corresponding density chart?

In 1981, Nishikata et al. published Viscosities of Aqueous Hydrochloric Acid Solutions, and Densities
and Viscosities of Aqueous Hydroiodic Acid Solutions in the Journal of Chemical Engineering Data.

Table III from this paper shows what you are after.

I don't know what temperature you used to make your density measurement, but assuming it was 20 °C, then your $\ce{HI}$ looks to have a concentration of about 39 weight percent.


HCl/diethyl ether + Nitrogen? Just wondering if anyone can tell me why bubbling nitrogen through a solution of $\ce{HCl}$/diethyl ether catalyst would remove said mix from my solution.

My current thought is that the $\ce{N2}$ will form ammonia and deprotonate the $\ce{HCl}$. But what of the Diethyl ether? I can't think of how it would be removed.

My only final explanation is that it is the $\ce{HCl}$ which is the important species to remove and the diethyl ether is just solvent ... but I would really like a second opinion!

HCl is a gas at room temperature and diethyl ether is a low boiler.  Both will have significant vapor pressures and, therefore, exist in the vapor phase above your  solution.  The HCl and diethyl ether in solution will continuously maintain an equilibrium with the vapor phase. As you bubble nitrogen through the system it will continuously sweep the vapor out, removing HCl and diethyl ether.  But since the HCl and diethyl ether in solution continuously maintain an equilibrium in the vapor phase, they will be continuously swept out until all of the HCl and diethyl ether in solution have been removed.


Theory behind Experiment: Extraction and Identification of DNA A few days ago, in my chemistry lab class, I performed an experiment aimed at extracting and identifying DNA. The class is mostly practical oriented, so I thought of asking my theoretical doubts on ChemSE.

First of all, let me explain the experiment:

We used onion extract, sodium dodecyl sulfate (SDS) solution, 4% $\ce{NaCl}$ solution, papain extract, ethanol, TE buffer solution, and diphenylamine reagent. I shall ask my questions while explaining the procedure followed-

I took $\pu{10ml}$ of the onion extract in a boiling tube, and added $\pu{1.5 ml}$ of SDS solution to it, following which the mixture was allowed to stand in ice for nearly $\pu{10 minutes}$.
What does onion extract + SDS do? What's the chemical reaction if any? Why is low temperature (ice bath) important?
I added 5-6 drops of papain extract to the mixture, and stirred gently. A quick Google search says Papain, also known as papaya proteinase I, is a cysteine protease enzyme present in papaya. I wonder how it 'reacts' with the mixture prepared in step 1?
Next, I slowly poured nearly $\pu{25ml}$ of ice cold ethanol down the wall of the test-tube, so that ethanol formed a separate layer on top of the mixture. Note that this won't happen if ethanol is poured in rapidly, as that disturbs the mixture at the bottom of the tube. Why so, though? Difference in densities should make sure separate layers are formed, even if so after a long time? What's wrong here?
After allowing the tube to stand straight in a test-tube holder for ten minutes, I noticed some stringy white substance collect in the alcohol layer. We were told that that substance was DNA! How did this happen?
Using a hooked glass rod, I spooled out whatever DNA I could from the top of the test-tube, very carefully. This was used to carry out the following confirmation tests:

Diphenylamine Test:

Some crude DNA was collected in a test-tube, to which $\pu{2ml}$ of 4% sodium chloride solution was added. Next, $\pu{2ml}$ of diphenylamine reagent was added and the mixture was placed in a boiling water bath for an hour. The colour of the mixture changed to pale blue. How does this work? What are the reactions involved? Why the colour change?

UV Absorption:

DNA was dissolved in $\pu{2-3ml}$ of TE buffer solution (why?), and ratio of absorbances was determined at $\pu{260nm}$ and $\pu{280nm}$. Furthermore, our instructor told us that this ratio should be less than 2 (I guess?), for a confirmation. I wonder how that number was determined. (Also, some of my classmates got the ratio slightly greater than 2, due to some error. What error could they possibly have made?)

I'm trying to make sense of what I did practically, with the help of theoretical inputs, and would appreciate any help with the same. No doubt, the experiment was really fun to do!

1) Sodium dodecyl sulfate (SDS) is a solid anionic detergent. It can solubilize proteins and lipids that frame the cell membranes by degrading the cell (the proteins from the cell membrane get damaged and cell gets broken) and nuclear membranes that protect the DNA. This will help the cell membranes to separate and expose the chromosomes that contain the DNA. SDS can also release the DNA from histones (a group of basic proteins found in chromatin) and other DNA binding proteins by denaturing them. Breaking the nuclear membrane exposes the DNA to reactive and possibly dangerous chemicals in the cell. To slow down these chemical reactions, which might damage or break up the DNA, you must use cold ingredients and add cold alcohol. Also, it is also known fact that the colder the alcohol is, the more DNA it will extract.

2) Once the DNA has been released, adding papain extract helps untangle and unfold the DNA from the other parts of the cell. The enzymes in papain break down protein attached to (or tangled with) DNA molecules, making them free from proteins.

3) DNA is not soluble in absolute alcohol but soluble in water. If you pour alcohol rapidly it will mix with water and you may not have DNA precipitation. Therefore, you should slowly pour ethanol (or isopropanol) down the inside of the test tube with a Pasteur pipette, so alcohol layer build up on the extract.

4) DNA is soluble in water, but not in salty water. All the other cell contents are also soluble in water and salt water. This difference in solubility allows you to separate DNA from the rest of the cell material. When the DNA leaves the salty water mixture, it comes into contact with the alcohol but cannot dissolve (DNA is not soluble in alcohol either). When DNA comes in contact with alcohol, it uncoils and precipitates. All the other cell materials stay in the salty water solution. Note that an uncoiled piece of DNA can be $\pu{2.8 inch}$ ($\pu{7.2 cm}$) long, and visible when few of them tangled each other.

5) Diphenylamine will react with deoxyribose sugar in an appropreate condition to give blue-colored complex. This is te principle underlying estimation of DNA using diphenylamine. This is called Dische Diphenylamine Test For DNA:

One of the acid degraded products, $\omega$-hydroxylevulinylaldehyde, under acidic conditions, reacts with diphenylamine to produce a blue-colored complex that absorbs at $\pu{595 nm}$. The intensity of the blue color of this complex is proportional to the concentration of DNA.

About using UV-vis-Spectroscopy to determine the purity: The fact is that nucleic acids have an absorption maximum at $\pu{260 nm}$ UV wavelength. However, this absorbance suffers from interference from othe molecules such as single nucleotides, proteins, and single-stranded nucleic acids (as well as RNAs). Thus:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= (A_{260} - A_{320}) \times 50\times$ dilution factor where $50$ is DNA extinction coefficient, $A_{260}$ and $A_{320}$ are absorbance of DNA at $\pu{260 nm}$ and $\pu{320 nm}$, respectively. In general, since the DNA has limited absorbance at $\pu{320 nm}$, this is simplified as:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= A_{260}  \times 50\times$ dilution factor

Yet, the DNA purity is determined with the help of software. This is done by measuring the absorbance of the DNA against protein:

$$\text{DNA Purity} = \frac{(A_{260} - A_{320})}{(A_{280} - A_{320})}\approx \frac{A_{260}}{A_{280}}$$

If $\frac{A_{260}}{A_{280}}$ is in the range of $1.7-2.0$, the content of the solution is considered to be pure DNA. In other words,, constructing an absorbance ratio between these two absorbance wavelengths can provide an estimate of sample purity. As a general rule any preparations with a $\frac{A_{260}}{A_{280}}$ ratio greater than approximately $1.7$ is called pure.

Sources:

http://www.biotech.iastate.edu/publications/lab_protocols/DNA_Extraction_Onion.html

https://openlab.citytech.cuny.edu/bio-oer/chemistry/biologically-important-macromolecules/nucleic-acids/nucleic-acids-dna-extraction-and-disches-diphenylamine-test-activity/

https://www.news-medical.net/whitepaper/20140203/DNA-Analysis-with-UV-Visible-Microspectroscopy.aspx

https://www.bmglabtech.com/fileadmin/06_Support/Download_Documents/Application_Notes/AN263.pdf