How was cholestyramine discovered? How were the properties of bile acid sequestrants like cholestyramine discovered? The earliest reference to the chemical I could find was the following article which does not discuss the history of the drug:

Glueck CJ, Fallat R, Tsang R. Pediatric familial type II hyperlipoproteinemia: therapy with diet and cholestyramine resin. Pediatrics. 1973; 52:669-79. [IDIS 37416] [PubMed 4355362]

Does this kind of drug exist in nature or is it purely synthetic?

It is definitely synthetic, a trimethyl-ammonium derivative of polystyrene.

The first to use it as a human medication seems to be Stanley S. Bergen, Jr. et al.

See Effect of an Ion Exchange Resin on Serum Cholesterol in Man Proceedings of the Society for Experimental Biology and Medicine, Volume 102, 1959, Pages 676–679:

  It is well known that bile acids are a major
  end product of cholesterol metabolism and
  that cholesterol is probably the only precursor
  of bile acids([references]1-3 ) . Rate of oxidation of cholesterol
  to bile acids is believed to be regulated
  by need for bile acids for digestive purposes,
  cholesterol normally being oxidized at
  a rate sufficient to replace the bile acid that
  escapes reabsorption and is lost in the feces
  ([references] 4 - 5). If an agent could be found that would
  sequester bile acids and thereby promote their
  excretion in the feces, 2 consequences of clinical
  interest might be anticipated: (1) increase
  in rate of oxidative degradation of cholesterol.
  and (2) decrease in serum cholesterol level.
  Tennent ([reference] 6) described lowering of serum cholesterol
  levels in animals fed the chloride salt
  of a basic anion exchange resin. This material
  (MK-135) has marked affinity for bile
  acids in vitro and is believed to exert its serum
  cholesterol-lowering effect by virtue of this
  property( 6 ) . The present report summarizes
  results obtained when MK 135 was administered
  to 26 patients, many with elevated serum
  cholesterol levels.

Reference 6 is Tennent, D. M., Siegel, H., Zanetti, M. E., Kuron, C. W., Ott, W. H., Wolf, F. J., Circulation, 1959, v20, 996.

Even though Bergen's paper only uses the term "MK 135", according to the editor's note on the current online version, this paper is the first use of cholestyramine as a medication in humans.


The meaning of different parts in the naming of glucose There are two main types of glucose, $\alpha$-D-(+)-glucose and $\beta$-D-(+)-glucose.
Can anyone please explain what is $\alpha$,D or (+)?

So (maybe by permutation and combination), how many types of glucose are there?

Glucose can exist in the open form or cyclic form.

In the open form, there are only two types, D-glucose (illustrated in Fischer projection above) and L-glucose which is the mirror image of the above.

Experimentally, D-glucose rotates plane polarize light clockwise, hence the symbol "+", and L-glucose rotates plane polarize light counterclockwise, hence the minus symbol.  For more information on the meaning of + versus - see: http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation

In the cyclic form, there is an additional chiral center, so there is alpha (where 3 OH groups are on the same side of the ring) vs beta (where 2 OH groups are on each side of the ring) as ilustrated in the drawing of the question.

Each of alpha and beta have a D and an L form, so there are 4 types in the cyclic form.

So in summary, there are only two types in the open form, but 4 types in the cyclic form.


Can I replace potassium nitrate with sodium nitrate in a smoke bomb? A very famous experiment goes like this:

$\ce{KNO3}$, molten wax and sugar are mixed and later burnt; the resulting mixture produces quite a lot of smoke.

After hearing about the experiment, I threw in the idea, "What would happen if $\ce{NaNO3}$, or other alkaline compounds were used instead?". However, the teacher wants us to be sure that the building doesn't explode. So here I am. Has this experiment been attempted before? What would happen if it was used?

Here is a link with the smoke bomb experiment.

In pyrotechnics, $\ce{NaNO3}$ is normally avoided because it is very hygroscopic.

With dried $\ce{NaNO3}$, the effect should be identical.

In any case, the usual safety rules apply!


Are all cubic point groups encountered? My symmetry lecture notes state that there are seven high-symmetry 3D point groups, which have more than one rotation axis of order greater than 2: $T$, $T_d$, $T_h$, $O$, $O_h$, $I$, $I_h$. I sometimes see them called the “cubic point groups”.

My point is: most molecules (or molecular fragments) I have encoutered in these groups are always $T_d$ (tetrahedral) or $O_h$ (octahedral). What are examples of molecules in the other groups?

I refer you to the second link in my answer for How does one recognized Td/Oh symmetry in molecules?, which (according to the abstract) describes some high symmetry species of unusual point groups $T$, $O$ and $I$. Additionally, the molecule [6.6]chiralane has $T$ symmetry.

Buckminsterfullerene is a well known $I_{h}$ symmetric molecule.

Don't know of any $T_h$ molecules but I'm confident they can be devised.


Negative potential of the oxidation of iron(II) to iron(III) The oxidation potential which I found on the internet the following reaction

$$\ce{Fe^{2+} -> Fe^{3+} + e^-}$$

is $\pu{-0.77 V}$. But how can it be negative? Negative oxidation potential means oxidation tendency is very less or we can say negative oxidation potential means $\ce{Fe^2+}$ has very less tendency to lose electrons. But it doesn't make sense, because $\ce{Fe^3+}$ is in $\mathrm{3d^5}$ configuration, which is half-filled $\mathrm{d}$ orbital, which is very stable than $\ce{Fe^2+}$. So, $\ce{Fe^2+}$ should get readily oxidize to $\ce{Fe^3+}$. Then why do we have negative oxidation potential for this reaction?

As IanB2016 mentioned in his answer, the redox potential of a half-cell reaction is always in reference to another half cell.  The usual "reference" half cell in electrochemistry and thermodynamics is the standard hydrogen electrode (SHE).  Some key features of the SHE:

the half cell reaction is $\ce{2 H+(aq) + 2e- <=> H2}$, i.e. the reductant is hydrogen gas and the oxidant is aqueous protons.
the pressure of hydrogen gas for a standard hydrogen electrode is 1 atmosphere.
the concentration of acid in the aqueous electrolyte is 1 molar, which corresponds to a pH of 0.

The oxidation potential of a species is related to its strength as a reducing agent.  Things that have a negative oxidation potential are less reducing than hydrogen gas, and things that have a positive oxidation potential are more reducing than hydrogen gas.  Lithium metal, for example, has an oxidation potential of more than three volts, i.e. $\gt+3\;\mathrm{V}$.  It is an extremely powerful reducing agent, far more powerful than hydrogen gas.

As you note, $\ce{Fe^{2+}}$ has an oxidation potential of $-0.77\;\mathrm{V}$, meaning it is a less powerful reducing agent than hydrogen gas.  This makes sense in several ways:

aqueous solutions of iron(II) salts do not spontaneously reduce aqueous protons to hydrogen gas.  If you mix iron(II) sulfate with water, for example, hydrogen bubbles do not spontaneously form.
Hydrogen gas releases more energy when burned in oxygen than iron(II) salts do when they react with oxygen.  Hydrogen reacts explosively, whereas iron(II) salts do not spontaneously explode or catch fire when they are oxidized by oxygen.

Finally, note that the oxidation potential of water (via the reaction $\ce{O2(g) + 4 H+ + 4 e− <=> 2H2O}$ at pH 0 at 1 atm of pressure) is -1.229 volts.  Water is a very weak reducing agent, which means that oxygen is a strong oxidizing agent, stronger than iron(III).  This means:

iron(III) salts do not spontaneously oxidize water to produce oxygen.
conversely, oxygen does spontaneously oxidize iron(II) to iron(III).  Solutiosn of iron(II) salts are thermodynamically unstable under aerobic (oxygen-rich) conditions.


What makes an X-ray machine's white photographic film go black? I read on a   bbc gcse bitesize webpage "In older X-ray machines, white photographic film" [edit- I now know that bbc webpage is wrong, and it's not a white film. I'll leave the rest of the post after the next sentence, as is, but note that from the next paragraph, the rest of the post is based on the misunderstanding from that bbc link that there is white film  (And I guess it's talking of any pre digital when it says 'older' - and anyhow prior to film they used photographic plates - glass rather than plastic/film ).]

I understand that X-ray images are black and white. You have the X-rays themselves, the white film, and an object/person in-between.

The white parts of the image are bones and/or things the X-rays couldn't get through, the white film remains white, and the black part is where the X-rays did pass through, and it blackens the white film.

I've also heard that they have a cassette/casing, and in that is fluorescent material or coating and film.  The X-rays hit the fluorescent material and (I guess by absorption and reflection, or absorption and emission) converts it into regular light, and there is a phosphorescent material/coating there, too, so the light remains for a while, and the light, I guess, blackens the film.

I understand they don't work like that anymore and now you can have other types of X-ray detection mechanisms. But I want to understand at least how the traditional ones are black and white.

I understand that X-ray machines used to use a white photographic film.

What is the white photographic film made from that distinguishes it from a black photographic film? Can I buy white photographic film?

I have seen old-fashioned camera film before.

In black and white photography, was it a black photographic film that went white when exposed to light?

X-ray film is very similar to standard black and white negative film. It's basically a transparent plastic base coated with an emulsion of silver halide particles (a salt of $\ce{Ag+}$ with a halide, usually $\ce{Cl-}$ or $\ce{Br-}$). As shown below, when light hits one of these particles, it reduces a small amount of the halide to elemental silver (1). When this is exposed to a photographic developer (essentially a mild reducing agent), more elemental silver is produced around the initial sites (2).$$\ce{Ag^+ +e- -> Ag}$$

Finally, a fixing agent is added which solubilizes the unexposed silver halide so it can be removed from the emulsion, preventing further light exposure from fogging the film.

Thus, the regions that are exposed to light appear dark when developed as the tiny silver particles ($\ce{Ag}$) block light passing through the film and the unexposed regions are transparent, taking on the colour of the plastic base (often blue or grey). X-ray films are typically viewed on an illuminated white background, making the unexposed areas appear white (or off-white, coloured by the plastic base).

Normal B&W film is actually slightly sensitive to x-rays but, the emulsion of dedicated x-ray film contains other components to facilitate the silver reduction from x-ray exposure.

Source


How to explain Born repulsion between ions in gas phase? In simple terms, the collision of two atoms $\ce{A}$ and $\ce{B}$ will result in ions $\ce{A^+}$ and $\ce{B^-}$ if $$I_a(\ce{A})+E_a(\ce{A})<I_a(\ce{B})+E_a(\ce{B})$$ where $I_a$ and $E_a$ are the ionisation energies and electron affinities, respectively. Let us view $\ce{A^+B^-}$ as a molecule of sorts, having ions of rigid, non-polarisable spheres. Say $r$ is the sum of the radii of such spheres. The potential energy $E_p$ of such a system would then be $$E_p=I_a(\ce{A})-E_a(\ce{B})-\frac{e^2}{r}+\frac{b}{r^n}.$$

It is intuitive that

the term $I_a(\ce{A})-E_a(\ce{B})$ characterises the energy required to form isolated ions $\ce{A^+}$ and $\ce{B^-}$ in the gas phase;
$-e^2/r$ takes into account culonic attraction between two ions.

It is not intuitive for me why ${b}/{r^n}$ is  how it is.

  The final term, introduced by Max Born, encompasses the repulsion generated by shells of electrons at either ion.

Of course, such effects cannot be ignored. But I am interested in why is it $b/r^n$. What is the justification? Feel free to provide the rigorous approach, even though it will probably$^{[1]}$ be above my abilities of comprehension.

Browsing on Wikipedia, the information might be given in the book Advanced Inorganic Chemistry by F. Albert Cotton, Geoffrey Wilkinson, Carlos A. Murillo, and Manfred Bochmann.

The pertinent section “1-6. Energetics of Ionic Crystals” starts at page 18$^{[2]}$, however I cannot find the derivation. There is only that the value of $n$ relies on a compressibility measurement, given as fractional change in volume per unit change in pressure, or

$$\frac{\Delta V}{V\Delta P}.$$

Also, this is seems somewhat contradictory to the earlier assumption of rigid spheres as written in another book$^{[3]}$. So, if anything, the linked book confused me further.

$^{[1]}$ Most certainly.

$^{[2]}$ Page 28 in the PDF.

$^{[3]}$ U. Palm, V. Past. Physical Chemistry. (1974) [To my knowledge, not available in English.]

The term $+b/r^n$ is a power law repulsion term between the ions, where $b$ is the bare ion radius (twice the van der Waals radius)  and $r$ the separation.

The hard sphere model considers the ions as 'billiard balls' which means that $n=\infty$ since $+b/r^n$ is effectively zero when $r \gt b$ and infinite when $r \lt b$.

Two other types of potentials are used (a) the power law potential when $n$ is some integer value, usually between 9 and 16  and (b) the exponential potential $+a_0exp(-r/a_1)$ where $a_0$ and $a_1$ are adjustable parameters with $a_1 $ of the order of 0.02 nm. The justification  for using these potentials is weak (the power law particularly so) and they are used as they represent a better approximation than the hard sphere potential and are mathematically convenient. The exponential form is presumably justified since the combination of all orbitals decay exponentially with distance.


Addition of Methyl Grignard and use of HClO4 in this question Okay so the first step is the use of Grignard. I know that Grignard does a 1,2 addition(generally) and the attack happens directly on the carbonyl,and I think HClO4 basically does protonation.Not sure about the role of NaOH. Also the final answer given is 3,5 dimethylphenol. The final product is devoid of the oxygen which is present in the ring. How should I approach this question?

Consider the resonance structures of the pyranone. MeMgCl attacks the structure shown second. The perchloric acid workup hydrolyses the enol ether to give the diketone shown bottom right. NaOH promotes the internal aldol cyclisation to give 3,5-dimethylphenol.


How to explain the concept of mol and reactions involving mol quantities From time to time I do little pro bono lessons for second grade in math, physics, and chemistry. Recently I came across this question.

  In a container is $2~\mathrm{mol}$ of $\ce{SO3}$ (sulfur trioxide). This compound under certain conditions breaks down into $\ce{SO2}$ (sulfur dioxide) and $\ce{O2}$ (oxygen). At the same those two resulting compounds react together back to form $\ce{SO3}$. In equilibrium we have $1.52~\mathrm{mol}$ of $\ce{SO3}$.
  How much of remaining resulting compounds do we get?

The answer: It is obvious that $0.48~\mathrm{mol}$ of $\ce{SO3}$ broke down into $0.48~\mathrm{mol}$  of  $\ce{SO2}$ and $0.24~\mathrm{mol}$ of $\ce{O2}$.

The problem: A few of students cannot grasp how could we start with $2~\mathrm{mol}$ of some substance and end up with $2.24~\mathrm{mol}$ of other substances. I have tried to explain to them that $\mathrm{mol}$ is not a quantity as they are used to as in weight or volume, but that a $\mathrm{mol}$ is a number of molecules. I have tried to simplify an example using just 2 molecules of $\ce{SO3}$ and showed that the break down to 2 molecules $\ce{SO2}$ and 1 molecule of $\ce{O2}$. That was understandable. But as soon as we got back to moles: "How can we get more than we started with?"

What is the best way to explain this? Any experience? To me, it seems so obvious.

The fundamental misunderstandings here are related to two concepts:

Conservation (atoms are conserved, molecules are not)
A mole is just a number (like the chemist's dozen)

I would start with conservation, then move on to moles.

Explaining why atoms are conserved, but not moles

For second-graders, hands-on activities are probably best, so molecular models or something similar would be a good approach. I like your example using two molecules - if you can get some models, you could try having groups of students start with two (or more - 4+ would make it easier to see) molecules of $\ce{SO3}$. Have them count the number of molecules and the number of each atom. Write the results down where everyone can see them. Then, have them do the "reaction" - let them take apart the molecules and put them back together to make$\ce{O2}$ and $\ce{SO2}$. Again, have them count the molecules and the atoms. Show them that the number of atoms stays the same (for each element and in total), but the number of molecules can change.

Once your students are comfortable with this idea, introduce the concept of moles as a way of counting very large numbers. This is the more difficult part, and in my experience is the "make or break" concept for most chemistry students.

A mole is just a very large number

Goal: The number of atoms in macroscopic samples is huge

You can motivate the discussion by calculating the number of molecules/atoms in very small samples. In my classes (college level) I use the number of molecules in one mL of hydrogen gas at STP, and show that if you counted one molecule per second, it would take longer than the age of the universe (a lot longer) to count them all. Sometimes I use the number of silicon atoms in a bag of sand and show them that it is larger (much larger) than the number of stars in the universe. Other examples would work as well - the idea is to get them thinking about the largest quantity that they can possibly imagine and show them that the number of atoms in even a small amount of matter is much larger than that.

Goal: A mole is just a number

Now you want to link this to moles. I start with the concept of a dozen. I ask things like "how many donuts are in a dozen?" "Ok, how many cookies?" "How many cats?" "How many people" until someone points out that it's always 12. Then I ask them why we say a dozen instead of 12. If they don't figure it out quickly on their own, I tell them that for bakers and their customers, it is a convenient way to count stuff. Then I tell them chemists also have a convenient way to count atoms and molecules, which is good because there are a lot of them! The number they use is called a mole, and similarly to how a dozen = 12 of anything, a mole = 6.022 x 10^23 of anything.

Goal: A mole is a HUGE number

Depending on how good the students are with scientific notation, you might want to actually write out that number of zeroes - it's an impressive demonstration! When there is time, I also like to demonstrate the scale by comparing it to different things as the number of zeroes grows. For example, I say (or have them say) "thousand," "million," "billion," "trillion," for every three zeroes. At around 11 zeroes I start comparing it to unimaginably large things. As examples:

The number of stars in the Milky Way galaxy is about 100 billion ($1 \times 10^{11}$).
The number of sheets of paper that would fit in a stack between Earth and the moon is about 4 trillion ($4 \times 10^{12}$).
The number that would fit between Earth and Mars (at their average distance) is about $2 \times 10^{15}$.
Age of the universe in seconds is $4 \times 10^{17}$.
There are $10^{22}$ stars in the universe.

I think that this is effective because it connects the concept of moles being huge and the number of atoms being huge to a bunch of astonishing facts - people tend to remember weird little surprising facts, and because of the way memories form, the moles stuff gets stored along with it, leading to an artificially created "intuitive" understanding of the size scale of moles and their relationship to counting atoms.

Now you need to tie it all together.

Goal: Moles are used to count atoms and molecules

I connect the concept of a mole being a very large "convenience" number used for counting to the stoichiometry and conservation concepts with a simple exercise. I write out a chemical equation, which in your case would be:

$$
\ce{2SO3 <=> 2SO2 + O2}
$$

Underneath, I write the number of molecules of each, starting with the smallest. So, I would write...

$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
\end{array}
$$

... and say "If we have two molecules of $\ce{SO3}$, we will end up with two molecules of $\ce{SO2}$ and one of $\ce{O2}$." Then I ask them: "What if I start with 10 molecules $\ce{SO3}$?" and let them figure it out, writing on the board as I go. Next, I ask

"What if I start with 12?"
"What about a dozen?"
"How about 100?"
"A billion?"
"$6.022 \times 10^{23}$?"
"A mole?" - at this point, most of them will have the idea
"10 moles?"

At the end, the board looks like this:

$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
10\space \rm{molecules} &  &  10\space \rm{molecules} & 5 \space \rm{molecules} \\ \hline
12\space \rm{molecules} &  &  12\space \rm{molecules} & 6 \space \rm{molecules} \\ \hline
1\space \rm{dozen} &  &  1\space \rm{dozen} & 0.5 \space \rm{dozen} \\ \hline
100 \space \rm{molecules} &  &  100 \space \rm{molecules} & 50 \space \rm{molecules} \\ \hline
1 \times 10^9 \space \rm{molecules} &  &  1 \times 10^9\space \rm{molecules} & 0.5 \times 10^9 \space \rm{molecules} \\ \hline
6.022 \times 10^{23} \space \rm{molecules} &  &  6.022 \times 10^{23}\space \rm{molecules} & 3.011 \times 10^{23} \space \rm{molecules} \\ \hline
1 \space \rm{mole} &  &  1 \space \rm{mole} & 0.5 \space \rm{mole} \\ \hline
10 \space \rm{mole} &  &  10 \space \rm{mole} & 5 \space \rm{mole} \\ \hline
\end{array}
$$

At this point most everyone will "get it" - and the advantage of this method is that they sort of figure it out for themselves along the way, which means they will remember it more clearly. If anyone still struggles with the conservation part, you can have them calculate the number of each atom and each molecule for all or some of the examples above, in the same manner as at the beginning of the lesson.


Why is zinc deposited on copper when boiling an aqueous zinc sulfate solution? I did an experiment where I dissolved 30 grams of zinc sulfate in water and boiled it with strips of solid zinc metal and strips of copper metal and a layer of zinc was deposited on the copper.

I know that if you place solid copper in zinc sulfate aqueous without the solid zinc nothing would happen because zinc is more reactive than copper.  Why does the addition of zinc sulfate cause the zinc to coat the copper?  What is happening?

Zinc, being in contact with copper, is easily oxidized, because copper drives electrons out of it.

Metallic copper becomes electron-rich and is therefore capable to yield electrons to $\ce{Zn^2+}$ (stemming from $\ce{ZnSO4}$), which reduces to $\ce{Zn^0}$. The latter is finally observed to adsorb at the copper strips.

In other words, copper metal works as an auxiliary electrode whose function is to transport electrons, without being $-$ by no means $-$ transformed during the experiment.

In conclusion, the overall reaction uniquely involves zinc and its ions.

This is similar to what is observed when copper, in contact with zinc, is dipped into an acidic solutions. Hydrogen bubbles are seen even at the copper surface, without the latter being oxidized.


Why is this β hydrogen in trans position to the leaving group? Problem:

One of the most widely used methods for the formation of $\ce{C=C}$ is β - elimination reaction of the following type:

The spatial arrangement for a bimolecular elimination (E2) of the type shown above requires a trans geometry of the β - hydrogen and the leaving group (X). In most of the cases, the most substituted olefin is the favored product. However, in the case of ammonium and sulphonate salts, the least substituted olefin is formed as a major product
Identify the major product in the following reaction:

My query:
As you can see in the diagram they've drawn, the $\ce{NR3+}$ is in the plane of the paper, while $\ce{H}$ is NOT in the plane of the paper. Thus, I would not consider them as trans, since they are not exactly opposite to each other, but are in fact, inclined to each other at certain angle. In the current diagram, I would only consider (1) the two H atoms on a wedge and a dash position, (2) the left phenyl and right $\ce{NR3+}$, (3) the right phenyl and the methyl - as trans to each other, as then they are exactly opposite.
As is obvious in the answer key, they've considered the given H atom and the $\ce{NR3+}$ as trans and have eliminated them to yield:

Thus I wish to know exactly where my reasoning is wrong? Thanks!

Complete problem 2 and its answer key for reference

Your assessment of which groups are trans to each other is correct but you have not taken into account that the molecule can adopt other conformers.
The classic confomation for E2 elimination is when the hydrogen and the leaving group are trans. This is because it maximises the orbital overlap between the $\ce{C-X \sigma ^*}$ and the $\ce{C-H \sigma}$ orbitals which are key to the reaction.

As shown in the question, the molecule will not undergo elimination without rotating into the appropriate conformation where the hydrogen and the leaving group are trans. This leads to the product shown.
Other points to note:

However, in the case of ammonium and sulphonate salts, the least substituted olefin is formed as a major product

This seems to be a complete red herring because there is no option for variably substituted products in this case (or in any other case of E2 elimination that I can think of?).
E2 elimination is also possible when the hydrogen and leaving group are syn, although it is usually less favourable. However, in this case the positive charge on the leaving group and the negative charge on the base might promote this pathway, leading to a different product with the two phenyl groups cis.


Confusion regarding calculating the pH of a salt of weak acid and weak base I need to calculate the $\mathrm{pH}$ of a salt of weak acid and weak base in an aqueous solution.

Method I used for calculation

For a salt of weak acid and weak base (assuming complete dissociation of salt) both cation and anion part of salt undergo hydrolysis. When cation part undergo hydrolysis, since they are the conjugate base of weak acid, they grab $\ce{H+}$ from water and releases $\ce{OH-}$ ions. Similarly, when anion part undergo hydrolysis they release $\ce{H3O+}$ ions.

This $\ce{H+}$ and $\ce{OH-}$ may react to give water. If we have same concentrations of these ions the solution will have a $\mathrm{pH}~7.$ If there is any excess of any of these ions after reaction, then the $\mathrm{pH}$ of the solution may be greater than or less than $7.$ But my answer didn't match up with what is given in my text.

Textbook answer

Formula given in my textbook is

$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right)$$

where $K_\mathrm{b}$ is dissociation constant of base, $K_\mathrm{a}$ is dissociation constant of acid and $K_\mathrm{w}$ is ionic product of water.

When I try to derive this formula I need to make some assumptions like degree of hydrolysis of both anionic and cationic part of salt is equal. And the $\ce{H+}$ in the solution is contributed by the dissociation of weak acid how can this be possible. The dissociation of the weak acid formed from salt hydrolysis is been suppressed by already available concentration of its conjugate base (anion part of salt).

Another approach, I believe, is this:

Definitions
$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right) \\ 2\,\mathrm{pH}=\log(\frac{K_b}{K_a\,K_w})$$

It suggests equations' sum, like this:

$$\ce{HA <=> A- + H+} \tag{$\ce{K_a=\frac{[A-][H+]}{[HA]}}$}$$
$$\ce{BOH <=> B+ + HO-} \tag{$\ce{K_b=\frac{[HO-][B+]}{[BOH]}}$}$$
$$\ce{A- + B+ + H2O <=> AH + BOH}\tag{$\ce{K=\frac{[AH]\,[BOH]}{[A-][B+]}=\frac{K_w}{K_a\,K_b}}$}$$
Finding $\ce{[H+]}$ in relation to $\ce{K}$

If salt's concentration is $\ce{C_s}$, and the mass balance:

$$\ce{C_s = [AH] + [A-] = [BOH] + [B+] } \tag{Mass balance}$$

Dividing by $\ce{C_s}$ we define $\alpha$ and $\beta$ such that
$$\ce{\frac{C_s}{C_s} = \alpha{} + (1-\alpha) = \beta{} + (1-\beta)} \tag{Mass balance}$$

And if we admit that $\alpha$, $\beta <<1$ which also implies $\alpha\approx\beta$ (this is possibly a weakness of the answer) then $$\ce{\alpha^2\approx\frac{K_w}{K_a\,K_b}}$$

And $\ce{K_a=\frac{(1-\alpha)[H+]}{\alpha}\approx\frac{[H+]}{\alpha}}$

$$(\frac{[H^+]}{K_a})^{2}\approx \frac{K_w}{K_a\,K_b}$$
$$[H^+]^{2}\approx \frac{K_w\,K_a}{K_b}$$

And we get the equation.


What is inhibiting polystyrene's ability to dissolve in traditional solvents like ethanol or acetone? My friend and I have been struggling with this problem for a couple days. Because polystyrene has a similar structure to many plastics, we assumed that it would dissolve in acetone. However, after running several tests, we found that polystyrene would not dissolve in a measurable quantity even if we let it sit in acetone for a significant amount of time.

Here is the plastic we are attempting to dissolve:

https://imgur.com/rdawCIP

And here is a plastic more easily dissolved:

https://imgur.com/hAVMUsM

We understand that most molecular-bonded plastics dissolve via van der Waals dispersion forces, so why isn't polystyrene dissolving? My theory is that the benzene ring is interfering with the dissolving process -- that only half the compound of polystyrene is interacting with the acetone and thus it does not dissolve like a polyethylene (which obviously has twice as much surface area to dissolve).

Polystyrene is soluble in acetone, MEK and other ketones, but solubility and the speed with which it dissolves depends, in part, on the molecular mass of the polymer.

If the solvent diffuses very slowly into the polystyrene, only the surface will appear to be affected. Applying a bit of energy with heat or mechanical crushing of the plastic greatly speeds the action of the solvent.

For more technical information, see Solubility Parameters: Theory and Application.


Favorskii rearrangement Aren't alpha halo carbonyl compounds extremely fast and reactive via $S_N2$ (the fastest i guess)? If this is so, why isn't the major product in the Favorskii rearrangement an alpha hydroxy carbonyl compound?

And if the alpha carbon(to which the halide is attached) is tertiary, there should be a high chance of an elimination product via $E_2$ (leading to the formation of alpha beta unsaturated carbonyl), right?

This is an old question, but in case anyone comes down this road...

  Aren't alpha halo carbonyl compounds extremely fast and reactive via $\ce{S_{N}2}$
  (the fastest i guess)?

I don't think so. In an $\ce{S_{N}2}$ reaction a nucleophile is attacking a positively polarized carbon atom.  The more positive charge on the carbon, the more the nucleophile is attracted.  Look at structure I below, it shows one of the resonance structures for a carbonyl that leads to a polarized carbonyl with positive charge on the carbonyl carbon.  Now to whatever degree the adjacent C-Cl bond is polarized with positive charge on carbon, we wind up with an unfavorable electrostatic situation of adjacent positive charges.  Said differently, the presence of a carbonyl group suppresses polarization of the adjacent C-Cl bond, which will make the C-Cl carbon less prone to nucleophilic attack.

  And if the alpha carbon(to which the halide is attached) is tertiary,
  there should be a high chance of an elimination product via E 2  (leading to the
  formation of alpha beta unsaturated carbonyl), right?

No, you want to remove hydrogen D in the diagram below.  Hydrogen D is much less acidic than hyrdrogen A because removal of the latter results in a resonance stabilized carbanion.  The difference in acidity between hydrogens A and D is so large that effectively only hydrogen A will be removed by a nucleophile.

These two arguments effectively eliminate pathways C and D from consideration.  Pathway B is simply an equilibrium that goes nowhere, so it has no real effect on the outcome of the reaction. This leaves pathway A, the Favorskii reaction, as the only viable alternative.  Pathway A, besides being the only game in town, is actually favored by two factors, 1) as noted above, hydrogen A is acidic and it's removal produces a resonance stabilized carbanion, and 2) completion of the Favorskii reaction is strongly favored by entropic factors since the carbanion and the C-Cl carbon are contained in the same molecule and held relatively close together.


How does dilution affect result of titration of jam? We currently do some pH adjustments of jams (target pH of 3.2) using citric acid, however, when there is a large difference between the starting pH and the target pH this can be quite a time consuming process (add to vessel, mix, take sample, repeat).

It's been a long time since I've done chemistry, but I believe that because the pKa of the citric acid is close to the target pH and the jams are a complex mix of buffering agents that it isn't easy to come up with a simple formula for the amount to add (in fact, I observe, even at a similar starting pH, the amount of citric acid required varies a  lot).

My idea was to titrate citric acid into a sample of jam that has been diluted with distilled water for easier mixing:

Jam @ 50% dilution
Citric Acid dissolved at a ratio of 1:9 in distilled water

My thought is that if I keep taking samples and titrating the citric acid with the diluted Jam that I will get a nice correlation between the titration amount and the amount of citric acid required to add to the batch.

My question is:

Will this produce a nice correlation?
Is it possible to do this theoretically?

Thanks,

You can do this at any dilution of both titrate and titrant, as long as you "keep the books" up to date when calculating the results you should be fine. I would probably only dilute the titrate though, unless you have good reasons. (If the dilution of jam is to make it less viscous then that is a good reason for doing it)

Whenever I needed to perform titration something of unknown pH and contents (and I had a large supply of the solutions) I would do a low resolution "sweep" with high concentrates and low tech (old school burettes are usually the fastest, valve fully open and stirring at... vigorous). I'd do two or three of those, just to pinpoint the area of interest. (Sometimes there are several, but then you just do the below for all of them). Assuming one specific point of interest was at 100% addition of titrate, I would add 80% of the titrant to a sample of titrate, then do another titration at 10x dilution of titrant. Three fast samples. Repeat the above, and now at 100x dilution i'd put it in the autotitrator and get some graphs.


How does chloroform react with cyclopentadiene in base? Is my reaction mechanism feasible?

The worry I had is that the normal way to make dichlorocarbene from chloroform is to treat it with a strong base, this will convert the cyclopentadiene into the cyclopentadienyl anion. This anion is an aromatic species which is unlikely to be able to react with the carbene.

But in one paper (Grignon-Dubois, M.; Dunogues, J.; Ahra, M. - Recueil des Travaux Chimiques des Pays-Bas, 1988, vol. 107, # 3, p. 216 - 225) it has been reported that chloroform, base and cyclopentadiene will react together to form the first intermidate.

I think that the reaction step showing the formation of the cyclopropane ring is perfectly reasonable. We need to consider how we would form the carbene. I know that it is possible to form carbene by a photochemical reaction on ketene which forms the carbene and carbon monoxide. I know from personal experience that chloroketene reacts in a [2+2] reaction with cyclopentadiene to form a cyclobutanone. So that route might not be a good idea.


Why do BH3 and NaBH4 have different selectivities? I've always known borane ($\ce{BH3}$), as a reagent for alkene hydroboration. Recently in class we talked about its use in the reduction of carboxylic acids to alcohols as well.

Now obviously $\ce{BH3}$ is an electrophile and I don't think it reacts by donating hydrides like sodium borohydride ($\ce{NaBH4}$) does - in any case it's not charged like $\ce{BH4-}$. Also $\ce{BH3}$ should be a pretty weak hydride donor, considering the covalent nature of the bonds and lack of any charge. It can reduce carboxylic acids, however, which $\ce{NaBH4}$ can't do! Where's the difference and what's the reason for this contradiction?

And what prevents borane from reducing acid chlorides (my instructor told me this)? This is something $\ce{NaBH4}$ is successful at.

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon:

$\hspace{2.4cm}$

$\ce{NaBH4}$ is incapable of reducing carboxylic acids only because the acidic hydrogen is more reactive towards the hydride ion than is the carbonyl carbon. It is, in fact, capable of reducing carboxylate anions to primary alcohols, though the reaction is very slow. Electron-rich carbonyls favor reactions with electrophilic reducing agents. To help in understanding what makes a carbonyl electron-deficient or rich, let us consider the contributing resonance structures of amides, esters, and acid chlorides:

$\hspace{4.7cm}$

Nitrogen atoms are much better than bearing a positive charge than are oxygen than is chlorine. In this way, you can think of nitrogen donating its lone pair into the carbonyl the most, making it electron-rich, and the chloride donating its lone pairs the least, making it electron-deficient.

The $\ce{BH3}$ reduction mechanism is quite a bit longer and has some interesting intermediates, but here it is:

$\hspace{2cm}$

Note that this mechanism does in fact include hydride transfers (it has to). Though you were right in thinking that $\ce{BH3}$ wouldn't immediately donate a hydride, coordinating to a lone pair weakens the $\ce{B-H}$ bonds and makes hydride transfers possible. $\ce{BH3}$ and $\ce{NaBH4}$ really aren't as different as you might first think.

Though $\ce{BH3}$ is more reactive towards electron-rich carbonyls, it is also capable of reducing electron-deficient carbonyls like acid chlorides and esters through similar pathways. This should show you that there isn't any contradiction here, and that just because two compounds have different reactivities does not mean that they cannot perform the same type of reactions—they just won't perform it in the same way.


Second order reaction half life equation? For the reaction:
$$2A\to p$$
The rate, $v$ may be given as:
$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$
Correct?

Integrating this gives:
$$-\frac12\int_{[A]_0}^{[A]}\frac{d[A]}{[A]^2}=k\int_0^tdt$$
$$\frac1{2[A]}-\frac1{2[A]_0}=kt$$
Rearranging...
$$\frac1{[A]}=\frac1{[A]_0}+2kt$$
Now substituting $t=t_{1/2}$ and $[A]=\frac{[A]_0}2$...
$$\frac2{[A]_0}=\frac1{[A]_0}+2kt_{1/2}$$
Minus $1/[A]_0$ from each side...
$$\frac1{[A]_0}=2kt_{1/2}$$
This then gives:
$$t_{1/2}=\frac1{2k[A]_0}$$

However, I have seen in textbooks that this should be written:

$$t_{1/2}=\frac1{k[A]_0}$$

Have I gone wrong somewhere? If so where?

I need to know which equation is correct because when figuring out the rate constant, would the gradient be equal to $\frac{1}{2k}$ or $\frac{1}{k}$?

What you did isn't wrong, but instead of:

$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$

some write:

$$v=-\frac{d[A]}{dt}=k[A]^2$$

which is also true.

Twice a constant is still as constant (a different constant of course).

See http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html


Hydrolysis of 4-chloro-3,3-dimethylbut-1-ene forms which products much faster than primary alkyl halides? Question

My Attempt

First a primary carbocation is formed and then methyl shift occurs to form a more stable secondary carbocation hence making (C) the major product, consequently I also arrived at answer (D) via hydride shift, and also option (A) would anyway be a possible product.

But the answer given is (B). Where am I going wrong?

Apparently, (B) is wrong for obvious reasons.

The product formed fastest will be kinetically determined. Shifting of bonds takes time so the fastest formed product will be the the one where minimum migrations occur hence (A).
As @orthocresol stated below, a stable compound similar to cyclopropylmethyl carbocation is forming here. "The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual". Hence product (A) will be favored thermodynamically too (refer).


Gaussian and Surface Reactions - Doable? Independent from my general kinetics question here - Estimating Surface Reactions - Liquid, Turbulent Flow - I am only interested in responses regarding Gaussian in this question.

I would like to know if it is feasible to predict or model the influence of a surface (metal) on a reaction between two hydrocarbons (aromatic and aldehyde for example) in Gaussian.
If yes, how would I go about it? (Building a small metal grid and working with it is very computationally intensive - it is the only approach I can think off...)

(I have access to Gaussian at my university - however I do not have access to Linda, so no shared nodes.)

You did go to Gaussian's Documentation on the subject, right?

In principle, it should work.  In practice, you should consider using massive firepower, and the plane-wave codes (NWChem, abinit, VASP, etc. etc.)

Also, I question whether any of these codes properly model bond breaking on a surface, i.e.  metal-catalyzed reactions.  Big simulations don't guarantee correct wave functions.

I should point out that surface reactions with plane waves are also difficult. At least they used to be.  There are/were issues with Ewald summation having to be done in 3D, and so the surface normal dimension used to have to be HUGE to avoid image effects.

This is a tough business.


Why is ΔS the same regardless of whether a process is irreversible or reversible? Why do we take the entropy change of the system to be the same in both irreversible and reversible processes? What is the idea behind this?

It would be nice if anyone could give me an example with some data.

In an irreversible process, we can't use the integral of dQ/T to calculate the change in entropy between the initial and final states of the system, because there is entropy being generated within the system (as a result of mechanical energy dissipation to internal energy) which the integral does not account for.  However, since entropy is a function of state, we can focus on the exact same initial and final states of the system, and devise a path between these two states that does not involve dissipation of mechanical energy, and calculate the integral of dQ/T for this alternate path.  Such a path is a reversible path, and the integral along this (or any other) reversible path gives us the change in entropy.


Why are peroxides unstable but disulfide bridges considered stable? Why are esters stable but thiolesters are unstable? I can not understand why a peroxide $\ce{R-O-O-R}$ is considered reactive and unstable.

Going down one row on the periodic table, a disulfide bridge ($\ce{R-S-S-R}$) is apparently super stable and super important to proteins 3d structure.

At the same time, a thioester is considered to be as unstable/reactive as ATP and the ester is the stable one?

What am I conceptually missing here to explain this contradiction?

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$)

$$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} &
     \text{Bond} & \text{Enthalpy}  \\ \hline
\ce{C-C} & 350 & \ce{Si-Si} & 226\\
\ce{N-N} & 163 & \ce{P-P} & 201\\
\ce{O-O} & 146 & \ce{S-S} & 226\\
\ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\
\end{array}$$

We observe a decrease in bond energy from carbon to nitrogen and from silicon to phosphorus. It is clear and probably have same reason: appearance of lone electron pair of the atom. There is a repulsion between lone pairs, that is in partially compensated by stronger bonds formed by smaller atoms (the atomic radii falls down to the end of the row). When moving from second to third row, this repulsion is reduced thanks to larger size of atoms of the elements of the third row.

The general decrease in energy from carbon to silicon is thanks to increased size of the orbitals of valence level, making them more diffuse with less efficient overlap and longer bond length.

The stability of esters vs thioesters is ruled by different reasons. Sulfur is not as electronegative as oxygen, but when you consider an anion, sulfur anions have larger radius and are less prone to grabbing first positive charge they find. So sulfur anions are more stable intermediates (kinetically), and thus polar dissociation of thioester bonds is easier.


Do modern dispersion-corrected DFT methods give more accurate molecular geometries? I was discussing with a colleague the use of modern dispersion-corrected density functionals. I take it almost as a given that the methods generally produce "better" (for some definition) geometries, in addition to more accurate predictions of thermodynamic and other properties.

Naturally, since the dispersion correction involves intermolecular interactions (e.g., van der Waals complexes), they should do better than "regular" functionals. I'm curious about intramolecular geometries.

Are there any benchmark comparisons on bond lengths, angles, etc. comparing different computational methods (DFT and otherwise) to crystal structure or other geometries?

In other words, how accurate are different computational methods at getting geometric properties of individual molecules?

I am afraid I do not know of a paper specifically discussing the effects of DFT dispersion corrections on intramolecular properties. However, it should be noted that for short-range descriptors (such as bond lengths and bending angles) the dispersion corrections will not influence those as they correspond to distances below the typical distance at which the corrections take place. The various schemes differ in how they work, but typically the corrections (in Grimme's schemes) are damped below a distance of ~6 Å. It is also my experience that such local structural characteristics are not affected by dispersion corrections.

However, you are right that dispersion corrections will influence intramolecular structure at larger scale, and in particular for “soft” or “flexible” long molecules with many conformations. There, it is worth noting that the so-called “dispersion” corrections actually correspond to the lack of accounting in local DFT functionals of middle-to-long range correlation effects. Thus, it is actually expected that the corrections — or nonlocal functionals including dispersion terms — actually perform better for those too.

Unless, like Grimme's 2006 two-body dispersion correction scheme (often called “D2”), the dispersion corrections happen to be often too strong and make intramolecular distance somewhat too small (instead of much too large)!

Edit: Now that I come to think of it, one extreme example of intramolecular system that has been studied in detail w.r.t. dispersion corrections is that of helicenes. It was shown (for example there and there) that dispersion corrections improve the accuracy of computational structures and energies compared to experimental (or high-level computational) data.


Reaction control on molecular level Imagine I have a bottle with a random quantity of molecular hydrogen, and can apply exactly enough energy to break only one molecule of hydrogen. Is the reaction going to occur or the energy would be shared between other molecules in the bottle?

The questioner asks that exactly enough energy is added to a molecule of hydrogen just to dissociate it and will energy be shared. The answer is both yes and no. It is better to take a step back and consider what happens when a photon is absorbed. In the case of hydrogen, dissociation must occur from an electronically excited state as it is a homo-nuclear diatomic and does not have a permanent or varying dipole. This means that the ground state cannot absorb a photon's energy from v=0,1,2 etc to any other vibrational level. (I'm assuming absorption into continuum just above v=$\infty$ is zero.) So excitation has to be with a UV photon. If the photon is just at dissociation, the H atoms will have effectively zero kinetic energy and so not depart from one another. In fact at this point life becomes complicated and interesting. The potential energy between two atoms increases very slowly with increasing separation as one reaches dissociation. In fact it is possible to form Rydberg atoms which may be a micron in diameter when excited to within fractions of wavenumbers of dissociation. This size is huge, about ten thousand times larger than the minimum H$_2$ atom separation, bigger than a protein and approaching the size of a bacterium! Clearly the two atoms hardly influence one another and consequently a small perturbation, a minute magnetic field for example, and certainly a nearby molecule may cause the Rydberg molecule either to break apart or recombine depending on whether some energy is added or subtracted in the interaction. If there is not enough energy then the H$_2$ will remain in a highly excited vibrational level until it suffers a collision with another molecule and some vibrational energy is transferred to this molecule. Alternatively the electronically excited H$_2$ could fluoresce and any vibrational energy left in the ground state shared on collision with another H$_2$.  If there is far too much energy than needed to dissociate, then the H atoms ballistically fly apart in opposite directions and transfer some of their energy at each collision with other H$_2$ molecules. Possibly reaction also occurs, but H$_3$ is almost certainly unstable.


Dilution with 3 Different Percentages I'm trying to solve this problem.

  How many grams of a solution at $25\%$ do you have to add to $100\ \mathrm{g}$ of a solution at $40\%$ to obtain a solution at $30\%$.

My Attempt

It's a bit more than a simple diluition. So far I've calculated the mass of the solute in $100\ \mathrm{g}$ at $40\%$: $(40\times 100\ \mathrm{g})/100 = 40\ \mathrm{g}$

Then the mass of the solution at $30\%$ so: $(40\ \mathrm{g}/30)\times 100 = 133\ \mathrm{g}$

$133\ \mathrm{g} - 100\ \mathrm{g} = 33\ \mathrm{g}$ (of solvent you need to add)

Now, how does that $25\%$ fit in all this? Can anybody point me to a useful formula or an example? Thanks!

OK, for what it's worth:
Let $x_f$ be the final solution mass and $x$ be the mass of the 25% solution.
The dissolved mass can be defined by
$$0.3x_f~=~40+0.25x$$

The solution mass is:
$$x_f=100+x$$

Substitute $x_f$ into the first equation and solve. You should get the 25% solution mass to be added and then solve the easy solution mass equation to check your answer.


Racemic mixture of meso compound My solution manual implied that we can have a racemic mixture of a meso compound.

Is this possible?

I was able to rotate the molecule to make it symmetrical. Sure, the bromination in the nucleophilic solvent gives you an anti product, but from my understanding, if you can rotate the molecule so that it is symmetrical, the molecule can be meso.

I'm thinking no because meso compounds by definition have no enantniomers and a racemic mixture has both enantiomers.

You are correct and you enumerated the reasons why.  A racemic mixture is a mixture of d and l enantiomers in equal proportion.  A meso compound has no d or l isomers, it is achiral.


How to read these graphs and what are they called? How does one interpret/read the following graph (found here) as well as on wikipedia or

What are these called?

This depiction is a ribbon diagram of a protein structure. It shows the protein backbone (alpha carbons and the peptide bonds) smoothed out to a continuous line. In certain areas where the protein adopts alpha-helices or beta-sheets, these are indicated by the elaborate spiral in your picture (alpha-helix) or a wide, flat arrow (beta-sheet; not present in your structure). These depictions are supposed to give you an idea of what the protein looks like.

I would assume the two arrows in the bottom to be two of the coordinate axes of the crystal since ribbon diagrams are typically obtained from crystal structures. However, I cannot know without additional information (e.g. in a footnote of a picture).


Net conductivity of mixtures If we have a solution, of conductivity k1, and another solution of same volume having conductivity k2, then what will be the conductivity if I mix the 2?
I think that net conductance should be the sum of individual conductance of the 2 solutions in the final beaker (where volume is doubled). But then won't the individual conductance of the 2 solutions in double the volume be different from that in the initial volume?

Also, can we say that conductivity is the conductance of unit volume of solution?

This concept of electrolytic solution conductivity is complex and dependent on a variety of factors aside from mere concentration of electrolytes. There are separate laws used to describe the relationship between solution conductance and concentration dependent on the dissociative strength of electrolytes and the extent of overall solution saturation.

In your simple case, just recognize that the net conductance of the final solution may not work out to be the mere sum of conductances in the original solutions

From Wikipedia:

  The specific conductance of a solution containing one electrolyte
  depends on the concentration of the electrolyte. Therefore, it is
  convenient to divide the specific conductance by concentration. This
  quotient, termed molar conductivity, is denoted by Λm
  $$\Lambda _{m}={\frac  {\kappa }{c}}$$

Hence, since conductivity is proportional to concentration, and concentrations aren't additive, you cannot expect conductivities to be additive.


Why is carbon monoxide a monodentate ligand Carbon monoxide has lone pairs on both the carbon and the oxygen atoms. However, I have never seen any complex where the oxygen atom or both the oxygen and the carbon atoms are the donors.

I know that the much lower electronegativity of carbon makes it a much stronger donor. However, this difference in electronegativity exists in the cyanide ion too, albeit a little less. However, cyanide ion is ambidentate.

My question basically is :
Why is carbon monoxide not an ambidentate ligand? Further, why is it not bidentate?

It is possible for carbon monoxide to coordinate through oxygen, as reported here.  However, there is indeed a strong tendency for carbon monoxide to coordinate through carbon, essentially due to the electronegativity difference noted by the OP.  Cyanide is more likely to coordinate through N than carbon monoxide through O, simply because the electronegativity difference is not so large.  Even cyanide requires special circumstances to get it to coordinate through nitrogen, often involving the carbon being "preoccupied" with another atom -- just like carbon monoxide in the referenced article.


Why does enantiopure sec-butyl alcohol retain its optical activity over aqueous base but forms a racemic mixture over dilute sulphuric acid? Why does enantiopure sec-butyl alcohol retain its optical activity over aqueous base but forms a racemic mixture over dilute sulphuric acid ?

My reasoning is that, sec-butyl alcohol get dehydrated into a alkene because of sulphuric acid and then get hydrolysed to form alcohol again. Thus a equilibrium is formed between alkene and alcohol and between $R-$ and $S-$ sec butyl alcohol.
Dehydration cannot occur with a base, so the alcohol remains optically active.

I feel that conversion of alcohol to alkene and then back to alcohol seems ridiculous and impossible. What do you think, is this reasoning correct or just plain non-sense ?

To be specific, I have some doubt whether alkene will convert back to alcohol or not in dil sulphuric acidic ?

Although sulphuric acid is a notable dehydrating agent, it is concentrated rather than dilute sulphuric acid that would be typically used for such a purpose.

Here, it is due to the improvement of $\ce{H2O}$ as a leaving group compared to $\ce{OH-}$, due to the lack of any charge as a leaving group.

Under acidic conditions, the sec-butyl alcohol will be protonated in a small amount creating a much better leaving group.  The protonation also makes the alcoholic carbon centre more electrophilic and vulnerable to $S_N2$ attack by other water molecules in solution, leading to inversion of the chiral centre.

This in not possible under neutral conditions as the unprotonated alcohol would have $\ce{OH-}$ as a much poorer leaving group, preventing elimination or substitution reactions.

Under acidic conditions, with neutral water as a nucleophile $S_N2$ substition will be the preferred path.

A secondary alcohol on a small, un-branched carbon chain represents a relatively accessible spot for nucleophilic attack.
$E_1$ eliminiation or $S_N1$ substituion would require the formation of a secondary carbocation with relatively poor stabilisation compared to a tertiary carbocation.
$E_2$ elimination would require the removal of a hydrogen of low acidity from the carbon chain by a very weak base, water.

This allows a equilibrium converting molecules back and forth between the $R$ and $S$ forms of the alcohol to be set up and leads to racemisation in solution.


Since PbCl2 is insoluble in water, how is it that lead leached from pipes due to Cl- reaction can become ingested? I have reviewed a discussion concerning the salt content in the Flint River being directly connected to lead being leached from civic water pipes in Flint, Michigan. Wikipedia has a site dedicated to $\ce{PbCl2}$ and line 1 states that this salt, once formed is insoluble in water, unless that water is heated. Is there a secondary process that results in the lead returning to a form, which can be ingested — after it has been leeched from the wall of a pipe?

Lead(II) chloride is not the sitting-at-the-bottom-like-a-brick type of insoluble. Rather it is the well-ma~ybe-I’ll-dissolve-a-little type of insoluble. So enough lead chloride will be dissolved — as Ivan mentioned, enough to poison somebody. Its solubility is given as $10.8~\mathrm{g/l}$ on Wikipedia. There are much less soluble salts like mercury(I) chloride ($2~\mathrm{g/l}$), barium sulfate ($2.4 \cdot 10^{-3}~\mathrm{g/l}$) or silver(I) iodide ($3 \cdot 10^{-6}~\mathrm{g/l}$). Note that for mercury(II) sulfide a solubility is not even given, it is merely termed insoluble (and is the sitting-at-the-bottom-like-a-brick type of an insoluble compound).

All values were taken from Wikipedia and are given at $20~\mathrm{^\circ C}$.


Polarity of amino acids If I understand correctly, solubility of organic compounds depends mainly on the presence of polar groups plus putting in the factor of the number of carbon atoms in the molecule.

So putting that in mind I do not understand why Glycine, an amino acid with two polar groups, the amine and the carboxyl groups, is considered non-polar and insoluble in water .

According to Wikipedia and my experience, glycine is soluble in water and is a polar molecule. The reference you saw, uses polar and non-polar to mean if there is another group other than the carboxylic acid and the amine group.

  Solubility in water   24.99 g/100 mL (25 °C)


Base catalyzed rearrangement of an alkyl substituted cyclopent-2-enone I have been asked to provide a mechanism for the following transformation:

I don't see any viable way to make this happen. All I can see that is possible is the formation of an acetal or a Michael addition. Neither of which seem useful. Can you help?

I believe that the mechanism for rearrangement lies in a retro-aldol condensation (technically a Michael attack as a first step) followed by a new aldol condensation; both catalysed by the excess base which can both catalyse the mechanism and be the Michael attacker in step 1. My version of the mechanism is shown here:

One should be able to support the mechanism by labelling either of the ketone carbons with $\ce{^{13}C}$. (Oxygen labelling would not work since in the hydroxide conditions the oxygen will exchange rapidly via a hydrate intermediate.)

As I forgot to say and as Orthocresol correctly noted: Every step is to be seen as being in equilibrium and the overall equilibrium is shifted towards the thermodynamically most stable product: the tetrasubstituted double bond.


Degree of association, van't Hoff factor and dissociation constant $\ce{CH3COOH->CH3COO- + H+}$
  \begin{array}{c|c c c}
  \mathbf{Initial} &  \mathrm{1~mol} & \mathrm{0~mol} & \mathrm{0 ~mol} \\\hline
  \mathbf{Final} &   \mathrm{1-\alpha ~mol}&\mathrm{\alpha ~mol}&\mathrm{\alpha~ mol}
 \end{array}
  Where $\alpha$ is the degree of dissociation.

Therefore, $i = 1-\alpha+\alpha+\alpha =1+\alpha$, and $\alpha=(i-1)$.
This is what is written in my book. And I understand this.

  $\ce{CH3COOH->CH3COO- + H+}$
  \begin{array}{c|c c c}
  \mathbf{Initial} &  \mathrm{c~ mol}&\mathrm{0~mol}&\mathrm{0~ mol}\\\hline
  \mathbf{Final} &   \mathrm{c(1-\alpha)~ mol}&\mathrm{c\alpha~ mol}&\mathrm{c\alpha~ mol}
\end{array}
  Where $\alpha$ is the degree of dissociation.

Therefore,
$$i = \frac{c\alpha+c\alpha+c(1-\alpha)}{c} =1+\alpha$$
 (same as in case 1), and $\alpha=(i-1)$.

$$K_a=\frac{c^2\alpha^2}{c(1-\alpha)} = \frac{c\alpha^2}{(1-\alpha)}$$
This is also written in my book, and I understand this, too.

But, my question is why can we not use the former method to calculate the dissociation constant, $K_a$, which will give me:
$$K_a =\frac{ \alpha^2}{(1-\alpha)}$$
as opposed to what I got in the second (and the correct) case, which was:
$$K_a = \frac{c\alpha^2}{(1-\alpha)}$$
I can see the difference of a $c$ in the two equations, but can someone explain why do we get the difference?

There is absolutely no difference in your two equations but you forgot the units which makes it seem like there is a difference.

Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$
since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?

   $$k_a=\frac{\alpha^2}{1-\alpha}~\ce{mol~L^{-1}}$$

How do I know that the c is the concentration?

Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$
and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.


Trying to understand the necessity of orbital approximation or the independent electron model I'm reading the approximations used for solving the Schroedinger's equation for molecules; the first is the the Born-Oppenheimer approximation:

  Freeze the molecule with a fixed internuclear separation $R;$ then carry
  out calculations to obtain the total energy, $U,$ and wavefunctions for that $R$ value.$^1$

Then another approximation is used which is known as the orbital approximation to get rid of the inter-electronic repulsion as said by the same source.

  Consider each electron to move in some sort of "average potential" which incorporates the
  interactions with the two nuclei and an "averaged interaction" with the other electron.

  The electronic Hamiltonian can then be separated into two parts:
  $$\hat{H}_e = \hat{H}_1 + \hat{H}_2 \\ \text{where}:\\ \hat{H}_1 \;\text{is dependent only upon the properties of electron (1) and upon R .}\\ \hat{H}_2 \;\text{is dependent only upon the properties of electron (2) and upon R .}$$

I couldn't understand for what purpose this approximation is meant for. Also, I couldn't able to conceive what average potential actually means.

I then googled the term finally to come before a pdf file where it is written that

  $\color{red}{\bullet}$ Electrons repel each other according to Coulomb's law,
  with the repulsion energy $r^{-1} _{ij} .$

  $\color{red}{\bullet}$ Hartree-Fock replaces this instantaneous electron-electron
  repulsion with the repulsion of each electron with an
  average electron charge cloud.

Prior to that it is written:

  $\color{red}{\bullet}$ It assumes that each electron interacts with an average
  charge distribution due to the other electrons. This is
  wrong!

I've spent a hard time understanding this approximation.

Could anyone please explain why this approximation is taken & how it actually solved that problem? The source cites the reason of inter-electronic-repulsion as the problem; how does this term in the Hamiltonian cause problem & how is it solved by the approximation? Also, what is meant by average potential?

An intuitive explanation would be appreciated.

$^1$ http://www.chem.qmul.ac.uk/software/download/mo/3.pdf

Setting the stage

Once the Born-Oppenheimer approximation is introduced, we have (first) to solve the so-called electronic Schrödinger equation that could be written as follows,
\begin{equation}
    \newcommand{\el}{_{\mathrm{e}}}
    \newcommand{\elel}{_{\mathrm{ee}}}
    \newcommand{\elnuc}{_{\mathrm{en}}}
    \newcommand{\nucnuc}{_{\mathrm{nn}}}
    \newcommand{\op}[1]{\hat{#1}}
    \op{H}\el \psi\el(\vec{r}\el) = E\el \psi\el(\vec{r}\el) \, . \tag{1}
\end{equation}
Here $\op{H}\el = \op{T}\el + \op{V}\elel + \op{V}\elnuc$ is the electronic Hamiltonian, $E\el = U - V\nucnuc$ is the electronic energy, and $\psi\el(\vec{r}\el)$ is the electronic wave function. The electronic Schrödinger equation, which is a partial differential equation, is fairly complicated. One of the most frequently used methods of simplification of such equations is technique known as the separation of variables.

Separation of variables technique

The separation of variables is aimed to broke a partial differential equation of many variables into a set of simpler partial differential equations of fewer variables. Note though, that the separation of variables is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.

The most common way of separation of variables for a partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is by writing its solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
\begin{equation}
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) = \prod_{i=1}^{n} f_i(x_i) \, , \tag{2}
\end{equation}
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed, one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into (2) obtain the solution of the starting equation.

Separation of particles coordinates

It is fairly well known that an attempt to separate the coordinates of individual particles in the time-independent Schrödinger equation
\begin{equation}
    \op{H} \psi(\vec{r}) = E \psi(\vec{r}) \, ,
\end{equation}
by making a substitution $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$
will be successful only if the potential energy function can be expressed as the sum of the separate potentials for each individual particle
\begin{equation}
    \op{V}(\vec{r}) = \sum_{i=1}^{n} \op{v}(\vec{r}_{i}) \, ,
\end{equation}
which implies that the particles do not interact with each other. For such a system of non-interacting particles the time-independent Schrödinger equation, partial differential equation of a function of $n$ variables, breaks up into a system of $n$ independent partial differential equations of functions of $1$ variable only,
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\op{H} = \sum\limits_{i=1}^{n} \op{h}_{i}$ and $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$. All this is relatively easy to establish, and besides, this derivation is done in many textbooks in great details, so I leave at as an educational exercise to OP.

Separation of electrons coordinates

Electrons, on the contrary, interact with each other, so that a simplification of the electronic Schrödinger equation by means of separation of coordinates of individual electrons won't work. Mathematically, it is the interaction potential $\op{V}\elel$ that will prevent the separation of the electronic coordinates,
\begin{equation}
    \psi\el(\vec{r}\el) \neq \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}

But what if we can approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? Obviously, such an approximate potential should have the form of the sum of one-electron potentials each of which depends only on the coordinates of $i$-th electron as we have already seen. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence the name of the approximation, the mean field approximation. The corresponding approximate potential can be formally written as follows
\begin{equation}
    \op{V}_{\mathrm{MF}}
    =
    \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
The electronic Hamiltonian $\op{H}\el$ is then approximated with the sum of effective one-electron Hamiltonian operators $\op{h}(i)$
\begin{equation}
    \op{H}_{\mathrm{MF}}
    =
    \sum_{i=1}^{n} \op{h}(\vec{r}_{i}) \, ,
\end{equation}
where
\begin{equation}
    \op{h}(\vec{r}_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{α=1}^{ν} \frac{Z_{α}}{ r_{α i} }
    + \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
And in contrast to eigenfunctions $\psi\el(\vec{r}\el)$ of the true electronic Hamiltonian $\op{H}\el$, eigenfunctions $\psi_{\mathrm{MF}}(\vec{r}\el)$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ defined by the corresponding eigenvalue equation
\begin{equation}
    \op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el)
    =
    E_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) \, ,
\end{equation}
can be written as products of one-electron wave functions
\begin{equation}
    \psi_{\mathrm{MF}}(\vec{r}\el) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equation
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
\end{equation}

The main problem with the mean field approach described above is that so far we have just assumed that $\op{V}_{\mathrm{MF}}$ exist in one form or another, but we have no idea of how it actually might look like.
There is a well-known workaround, but its a different story.

Note that I ignored spin completely in the discussion above which is, of course, fundamentally wrong, but spin won't make a big difference with respect to the key concept discussed, the mean-field approximation.


What is this glassware for a distillation apparatus called? I'm messing around with building a solar distillation apparatus, and one of the parts I think I need looks kind of like the following, except I have no idea what it is called (and thus can't look it up to see if that site even sells them).

So, can anyone tell me what this thing is commonly called?

The "A" on the glassware gives it away as an Aldrich glassware component;
and Ron is correct calling it a Kugelrohr bulb. It is actually a single bulb ball tube.

This specific item can be found in the catalogue at Aldrich's website


Tips for maximizing yield, purity and crystal size during recrystallization I have already asked a similar question on titrations before. In a few months, I have a test on chemical synthesis. It is marked on the percentage yield, purity and crystal size of the product. I do not have a lot experience in doing syntheses and I am looking for any tips that will help maximize the yield, purity and crystal size of my synthesis.
Most of the synthesis that we do, the products are obtained by recrystallization. The solid is dissolved in a solvent at a high temperature and then it is placed in a cool bath. Then the product will crystallize on the bottom of the beaker.
Reading through an analytical textbook, it states a few ways to improve yield and purity but it doesn't actually state how you would do it practically:

allow the solution cool down as slowly as possible. This will maximize crystal size.

maintain a low relative supersaturation of the solution, this will increase crystal size and improve purity

describes the process or digestion or Ostwald ripening where larger crystals grow at the expense of smaller crystals

Does anyone have any suggestions on how I would actually do these 3 above tips? Also does anyone have any other tips?

When I grew crystals of aluminium potassium sulphate I filled a jar with boiling water and placed it in a cooking pot that was also full of water. Doing this helps to maintain a constant temperature. I set the cooker to a low heat and began dissolving the alum until the solution was saturated. At this point I removed the jar containing the solution and suspended a piece of string for a crystal to grow on. This would be my seed crystal. I left the jar on a window sill to cool and waited.

A few days later a crystal had formed so I lifted out the string and set it aside. Then I reheated the solution to dissolve any crystals that had formed on the jar as they would compete with the suspended seed crystal, reducing its size. It is also a good idea to filter the solution at this point. Once I had finished this step I placed the string with the seed crystal back into the jar and placed it on the window sill again. After a week I moved the jar to a refrigerator (around 3 degrees C)  to further cool the solution. This left me with a pretty big, well shaped crystal.

I didn't do this but you could try evaporating off the water after the previous step to grow the crystal further.


Are there any tricks I can use to pull SO2F2 out of the air? A massive apartment complex near where I live is being fumigated with Vikane gas (sulphuryl flouride, $\ce{SO2F2}$).  The process will take a long time (weeks) because there are a lot of buildings, and their tents appear to have rips in them, clearly visible: several neighbors have complained of symptoms of exposure to the gas, and I'm worried about it in my home.  Obviously we should try to get them to stop / etc., but that's a matter of law and politics not meant for a chemistry site.

In the meantime, what I would like to know is if there is any safe way I can use to pull the gas out of the air if it should waft its way into my home.  My understanding is that activated charcoal will not work for this molecule.  I've heard that either strong bases (usually photo-developing chemicals are mentioned) or heavy water vapor reacts with the gas; but I'm not sure what that reaction is, or if it would result in something better or worse than the gas itself.  I've tried Googling but with just a high school chemistry exposure some decades ago, I can't make much headway.  Please note that I do not have access to any chemicals or machines I can't readily buy in a nearby store.  My goal is to turn it into some liquid or solid, or bind it to such, so that I can't breath it in.

Sulphuryl Fluoride is chemically fairly stable, considering it's a pesticide - toxic effects are caused by chemical reactions which means they can't be inert.

However, according to the Vikane documentation, it's not UV stable. UV lighting is fairly widely available, but TBH a sunny day will do far more.


Reversible and irreversible processes of thermodynamics My text book states the following:

(Edit:It should be p1/p2, sorry.)

My understanding of a reversible process is that it us a process that is executed so slowly that no energy would be dissipated at all and it is possible to bring it back to the original state again due to this.
(The classic jar and piston example)
An irreversible process is every other process.

I understood the derivation for work for reversible processes but for the irreversible process I am unable to relate to the formula.
Take the example of the jar again.
If we suddenly compress the gas, we are changing the external pressure, right? What would we substitute for p ext ?
Also, how did this formula come and how does it accommodate the extra work we had to do that gets dissipated to get to the same state?

Please correct my understating if it
Is wrong.

For both reversible and irreversible processes, the pressure of the gas at the interface with the surroundings exactly matches the external pressure imposed by the surroundings $P_{ext}$, and the work done on the surroundings is always equal to $\int{P_{ext}}dV$.  (The equation you wrote for an irreversible process applies only if $P_{ext}$ is held constant).  However, for an irreversible process, the pressure of the gas at the interface is not given by the ideal gas law, because (a) there are viscous (dissipative) stresses that contribute to the pressure at the interface and (b) the pressure and temperature of the gas is not uniform within the cylinder.  So we need to specify how $P_{ext}$ varies during the process (say, as a function of the volume V) manually in order to analyze an irreversible process.

For more details on this, see the following link:  Calculate Work Done for Reversible and Irreversible Adiabatic process


What is the most stable structure amongst the keto-enol tautomers of pentane-2,4-dione? I was assigned the following question and I cannot decide which is the correct answer.

  Which is the most stable structure?
  (a) pentane-2,4-dione
  (b) 4-hydroxypent-4-en-2-one
  (c) (Z)-4-hydroxypent-3-en-2-one
  (d) (Z)-penta-1,3-diene-2,4-diol

I am confused between options (c) and (d).
Both seem to be stable due to resonance, but I think (d) should have more resonance as the oxygen's lone pair in the $\ce{-OH}$ bond will also participate in resonance making it more stable.

But the answer is given as (c).

Could you please explain why (c) is more stable than (d).

The compound in option (c) is highly stable because it forms a 6 membered ring by tautomerism and that ring is highly stable because all bonds in that ring have developed partial double character. Whereas in option (d) there is ring formation but only two bonds of that ring have partial double character. This makes (c) more stable than (d).


How could you predict degenerate pi molecular orbitals just knowing what symmetry point group a molecule belongs to? If you were to apply the Huckel method to some molecules which belonged to, say, $C_{2v}, D_{2h}, $ and $ D_{4h}$, which ones would you expect to have degeneracies?

I'm not sure if this is mixing up math jargon with chemistry jargon here, but I know the Mulliken symbol E is doubly degenerate and T is triply degenerate. So would the $D_{4h}$ molecule have degeneracies because it has the irreducible representations $E_g$ and $E_u$?

Which orbital symmetries you can get in a molecule always depends on the molecule’s point group. Only for those point groups which have degenerate irreducible representations can you get degenerate orbitals; so out of your examples only the $D_{4h}$ molecule could. This is entirely independent of Hückel’s method to determine energy levels.

But just because your molecule has the irreducible representations, doesn’t mean that a $\pi$ orbital has to have that degeneracy. Instead, it again depends on what you get when following symmetry rules.

You can calculate whether you will get $\pi$ degeneracies by drawing your molecule of interest, including all the p atomic orbitals and determining the molecule’s point group. Once you have that, for each symmetry operation, check how many p orbitals would be transformed onto themselves by said operation (and note whether with or without a phase change). Add that up. Once you’ve done that, you basically have the representation of the p-orbitals of the molecule that you can reduce to irreducible representations, finding out how many discrete and degenerate energy levels you have (albeit not including energy values).


Enzyme Kinetics - Given Km find substrate concentration at a certain velocity From what I understand about the Michaelis Menten Model

Km defines the amount of substrate required to reach half-saturation.
1/2 Vmax corresponds to Km on the x axis
Generic formula is v = Kcat [ES] [S] / (Km + [S])

Given Km and the desired velocity in terms of reaction efficiency, can one just plug in Km into the equation and get a relational expression with [S] on one side?

Or do you manipulate the 1/2 Vmax expression?

You can solve for [S]. There are two common ways to write the equation, like you have it:
$$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$

or after dividing both numerator and denominator by [S]:

$$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$

In this form, it is easier to see that [S] occurs once and you can solve for it.

Get [S] out of the denominator of the "big" fraction and isolate the "little" fraction:

$$\frac{K_\text{m}}{[S]} + 1 =  \frac{k_\text{cat} [E_0]}{v_0}$$

$$\frac{K_\text{m}}{[S]} = \frac{k_\text{cat} [E_0]}{v_0} - 1$$

Finally, take the reciprocal and multiply by $K_\text{m}$:

$${[S]} = \frac{K_\text{m}}{\frac{k_\text{cat} [E_0]}{v_0} - 1}$$


Diamagnetism in sulfur Sulfur is a diamagnetic element, meaning that it has no unpaired electrons. But its electronic configuration for valence shell is just like oxygen because they are in the same group i.e ns2, np4.

My question is that why is sulfur diamagnetic? If there are 4 electrons in p-orbital of its valence shell then shouldn't it have 2 unpaired electrons?

You seem to be confusing the elements with the pure simple substances of the elements. In atomic terms, both the oxygen atom and the sulfur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the substances oxygen ($\ce{O2}$) and sulfur (generally intended as $\ce{S8}$) are not the same as the atoms.

In the simple picture of Lewis structures, both substances have all atoms forming two bonds, completing the octet rule and therefore without unpaired electrons. This would imply that both are diamagnetic. However, simple valence bond theory is inadequate to explain the bonding in oxygen. Simple molecular orbital theory suggests that there are in fact two unpaired electrons in the $\ce{O2}$ molecule. This gives rise to paramagnetism in the oxygen molecule, also.


Stoichiometry question I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.

In 50.3 gr of a compound whose formula is $Na_2X_4O_7$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.

a) What is the number of Na atoms in the sample?

I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $0.5*N_A$.

b) What is the molar mass of X?

I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.

I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:

$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$

Moreover, in a certain process 0.84 moles of $XH_{3(g)}$ were formed.

1) How many grams of gaseous oxygen were formed in the process?

I first balanced the reaction thus:

$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$

I figured that if 0.84 moles of $XH_3$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.

2) How many grams of $Na_2X_4O_{7(s)}$ reacted in the process?

I figured that 0.84/4 moles of $Na_2X_4O_{7(s)}$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.

3) What are the volumes of the oxygen and the $XH_{3(g)}$ formed in the process at room conditions?

I figured $V_{O_2}=n_{O_2}*0.082*298$ and $V_{XH_3}=n_{XH_3}*0.082*298$, where $n_{O_2}=1.26 moles$ and $n_{XH_3}=0.84 moles$

4) What is the concentration of $OH^-$ ions in the solution obtained (500ml)?

I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $OH^-$ were formed.

I'd appreciate any comments on my attempt above.

One thing I see:  when you work out the MW (or AW) of X, you assume 4 moles of X based on the chemical formula.  But, properly, you have twice as many moles of X as Na ($\ce{Na_2X_4}$), and you found just 0.5 mole of Na.  So instead of 4 moles of X you have just one mole of X making the atomic weight if X 10.8 g/mol not 2.7.  Hmmm, isn't there an element in the second row of the Periodic Table with that atomic weight?


Does the Hofmann Rearrangement extend to Secondary Amides? I have read where it is explicitly stated that the Hofmann Rearrangement is a reaction whereby primary amides are converted into primary amines with one less carbon. I have tried to find examples where a secondary amide is used, but haven't been able to find it addressed anywhere. I have been able to find examples where a N-hydroxylamide is used, but that's still a primary amide.

What happens when a N-methylamide or an N-methylhydroxylamide is subjected to Hofmann Rearrangement conditions?

I think it's not possible, because if you go through the reaction mechanism, you will see that in the 1st and 3rd step two protons are needed to be abstracted by the base to proceed the reaction. So if you take a secondary amide, i.e., one of the H being substituted by any alkyl/acyl/aryl group, the base will not be able to abstract that group and hence the reaction will not progress to any fruitful product.

Pic reference: http://www.name-reaction.com/hofmann-rearrangement


The differences between integrals evaluations of contracted an uncontracted basis functions I wanna to evaluate molecular integrals (overlap, kinetic and etc) using Gaussian basis functions. To do so, there are two approaches: contracted and uncontracted basis functions in which the former means the primitive Gaussian functions ($G_1$ and $G_2$) are summed to form contracted ones and finally by summing the contracted functions  ($\phi_1$ and $\phi_2$) using appropriate coefficients, the final molecular orbital is formed, for two contracted basis functions and three primitives we have:

$$
\psi=c_1\phi_1+c_2\phi_2
$$
$$
\phi_1=d_{11}G_{11}+d_{12}G_{12}+d_{13}G_{13}
$$
$$
\phi_2=d_{21}G_{21}+d_{22}G_{22}+d_{23}G_{23}
$$
$$
G_{11}=e^{-\alpha_{11}r^2}
$$
where the exponents are given as a 2x3 matrix (generally exponents matrix in contraction basis functions approach is a [number of electrons]x[number of primitive Gaussians] matrix and for uncontracted functions approache is a [number of electrons]x[number of basis functions] matrix).
In this way the final overlap matrix for example is

$$
\begin{bmatrix}
c_1c_1\phi_1\phi_1 && c_1c_2\phi_1\phi_2\\
c_2c_1\phi_2\phi_1 && c_2c_2\phi_2\phi_2\\
\end{bmatrix}
$$

For simulating total wave function using two uncontracted basis functions, we have

$$
\psi=c_1\phi_1+c_2\phi_2
$$
where it's not necessary to use any summation to obtain $\phi_1$ and $\phi_2$, instead we can construct them directly by knowing Gaussian exponents:

$$
\phi_1=e^{-\alpha_{11}r^2}
$$

Again the overlap matrix has a form like above matrix.

Are the mentioned explanations true? How should we apply these differences in a Mathematica code?

A lot of stuff is going on in this question. So for clarity let's make this answer very explicit. We can try to consider the two cases of using an uncontracted basis and a contracted basis to clarify the differences. As a case let's look at STO-$2$G for $\mathrm{H_2}$. In the STO-$2$G basis set $\mathrm{H_2}$ would normally be described by two basis functions each consisting of two primitives.

Uncontracted case

If we use the STO-$2$G basis set in the case of using uncontracted Gaussian basis functions, each hydrogen will be described by two basis functions. Our overlap matrix will therefore be of the form:

$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_3^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_4^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$

In this notation we have that:

$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
N_iN_j\int_{-\infty}^\infty P_i\exp\left(-\alpha_ir_i^2\right)P_j\exp\left(-\alpha_jr_j^2\right) d\tau$$

With $P$ being a function of the angular momentum, in cartesian coordinates:

$$P=x^ly^mz^n$$

It can be seen that $P=1$ for s-functions. As we can see we have 4 basis functions in total (two on each hydrogen). You wrote something about, you would expect that in the case of uncontracted that we would have ""[number of electrons]x[number of basis functions] matrix"". As we can see this is not the case. Right now our matrix is in the atomic basis. When we solve the Hartree-Fock equations and find the coefficients that will take us to a molecular basis, we will see that the orbitals that are occupied in the atomic orbital basis, will have a contribution to occupied orbitals in the molecular orbital basis, and are therefore still important.

Contracted case

Lets now take a look at how the STO-$2$G basis set will be used in the case of contracted basis functions. Now our overlap matrix will take the following form:

$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_2}\left|\phi_2^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$

At first glance the overlap matrix look just like the one in the uncontracted case, but smaller. What has changed is how the elements are defined:

$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
\sum_I\sum_J c_Ic_JN_IN_J\int_{-\infty}^\infty P_I\exp\left(-\alpha_Ir_I^2\right)P_J\exp\left(-\alpha_Jr_J^2\right) d\tau$$

We can see that we now sum over all the primitives included in the contraction. To make it more clear, let's write out how the second element in the overlap matrix looks like. Now remember that each of the basis functions consists of two primitives:

$$
\begin{eqnarray}
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right>
&=& c_{11}c_{21}N_{11}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{11}c_{22}N_{11}N_{22}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau \\
&+&c_{12}c_{21}N_{12}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{12}c_{22}N_{12}N_{22}\int_{-\infty}^\infty\exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau
\end{eqnarray}
$$

In this notation the first number refers to the basis function and the second refers to the primitive. As we can see each element of the overlap matrix in the contracted basis is made from four integrals. Even though our overlap matrix is four times as small compared to that of the uncontracted basis, we still have to evaluate the same number of primitive integrals. In this case we have investigated here, where we used STO-$2$G, we have no unoccupied orbitals. In general, this will not be the case, most basis sets (also when used in the contracted basis) will have unoccupied orbitals.

Other comments

In your question I can see that you mention molecular orbitals. Just to be clear. When we do the integrals we are in the atomic orbital basis. The molecular orbital basis is the one we find when we solve the Hartree-Fock equations. I.e. the molecular orbitals have nothing to do with uncontracted vs contrancted basis functions, though we will get more molecular orbitals if we use uncontracted basis functions, because we will get the same number of molecular orbitals as we had atomic orbitals.

Regarding how to implement this in Mathematica I have no idea. Though as my understanding of how Mathematica is ought to be used, I would recommend that you would implement your integral code in a language like Python. For inspiration you can take a look at this guide, which implements all the integrals with contracted basis functions.


What happens inside salt bridges? I don't understand how a salt bridge work exactly: what happens inside it and how it affect the cell's lifetime.
I have already read:
Salt Bridges/Porous Disks
Salinic bridges in voltaic cells
Difficulties understanding how salt bridge works
What I understand:
The salt bridge's goal is to maintain electro-neutrality in each of the half-cells so that the voltaic cell keeps working and it helps avoiding the mixing of the half-cells' ions in solutions.
When the salt bridge is being used in the cell, its ions (anions and cations which constitute the salt bridge's electrolyte) are moving towards the half-cell that attracts them.
What I don't understand:

Do the salt bridge's ions leave the salt bridge to go all the way within the hall-cells' solutions or do they stay at the border ? Is it the case in all bridges ?
Do the half-cells' ions move inside the salt bridge or do they stay in the salt bridge at the border solution/salt-bridge ? Is it the case in all bridges ? And is it in the same proportions as the salt bridge's ions ?
How do the ions move exactly in the salt-bridge ? It seems like the propagation in, let's say agar-agar, is much slower than the displacement of ions in the liquid solutions in the half-cells, is it not a problem for the whole cell ? Is it like a longitudinal wave, ions pushing others ? If so, are they two speeds of propagation (the wave speed and the ions speed) ?
If ions are moving inside it, we can consider that there is a current in the salt bridge, correct ?
Are they families of salt bridges depending on how they work ?
Finally, can the salt bridge be exhausted before the half-cells (whether the solutions or the electrodes) ? It seems the salt bridge is quite small in basic cells examples compared to the sheer volumes of the solutions and sizes of the electrode, thus containing much fewer ions.

Ions leave the salt bridge and move further, without accumulating at the bridge end. For all salt bridges where there is no other reactions involved, so all simple salts in solution or molten salts.
Ions move all the way through. There is some concentration difference in salt bridge and main half cell volume. It is true for simple salt solutions and molten salts. Same ions move through half cell and salt bridge.
The speed of a wave is on the order of 100 000 km/s. The speed of ions moving is on the order of 0.001 mm/s. You can estimate the ion speed from calculating amount of ions participating in a reaction per second and dimensions of the bridge and ion density in the solution or molten salt.
Yes. But it is important to remember that there are two cancelling each other currents in the salt bridge. Thus magnetic field is close to zero.
Most important is a number of ion types. If there are just two types, such as in solutions and molten salts, then this intuition applies. If more ion types are present, and especially if ions can convert between types, things will get more complex. Another option to consider is a fuel cells where only one type of ions passes through, and electrons use outside route to complete the loop. So, cases with one negative and one positive ion types are considered here. All other collections of ions will behave very differently.
No. Ions are constantly being replenished by the bigger volume of the both half cells.

Electroneutrality strictly speaking is not conserved when the bigger volume than an atom is considered and precise measurments are available. There is a gradient of potential due to different concentration of ions. Concentration of ions is the reason for the effect you are interested in. Important parts to consider are electrodes, where ions react, thus dissapearing from what we calculate, and narrow areas, where ions' speed increase, such as in salt bridge.