Substances comprised of large molecules are liquids at room temperature? I am looking for compounds that are liquid at normal temperatures and are physically very large, to be used in an ellipsometer adsorption experiment. I am currently working with decane. The candidates will ideally not be toxic or radioactive (such as $\ce{UF6}$).

You may give porphyrins a try.  These dyes are of interest for molecular electronics, and to render them less crystalline / more glassy, alkyl chains are attached to the central motif.  One recent example are the room temperature liquid porphyrins developed in the group around professor Gryko:

(source, Chem Asian J. 2010, 5, 904-909; doi: 10.1002/asia.200900693.)
Besides standard spectroscopic characterisation, the publication contains the results of DSC:

so compound 4d ($\ce{C176H294N4O12}$) may be a good candidate for an organic molecule, without charges and metal coordination, and suitable for ellipsometry / VASE measurments on surfaces.


Why is phloroglucinol (1,3,5-benzenetriol) not very soluble in water? Wikipedia says that phloroglucinol (1,3,5-benzenetriol) has a solubility of 1 g/100 ml in water.  This is in comparison to:

phenol (hydroxybenzene) 8.3 g/100 mL
catchecol (1,2-dihydroxybenzene) 43 g/100 ml
resourcinol (1,3-dihydroxybenzene) 11 g/100 ml
hydroquinone (1,4-dihydroxybenzene) 5.9 g/100 mL

Why does it have such a low solubility?  I would have thought the OH group would make it more soluble.

Why does it have such a low solubility? I would have thought the OH group would make it more soluble.

Interesting question but there is a logical fallacy. You are connecting the concept of solute's polarity with its water solubility. Note the word polarity is a catch-all term for many interactions. Solubility is the hardest thing to predict or model! It is very hard to predict melting point as well as boiling point. Not sure why general chemistry textbooks only misguide students and they have to unlearn so many things at a later stage.

You made very good observation. Let us look at your list from another angle i.e., chromatography- a special type of chromatography. Lets us say we have a polar stationary phase like silica and use a mobile phase which contains acetonitrile + water at a neutral pH. In this chromatography mode, the most polar compound elutes in the last and the least polar first. If you inject a mixture of phenol, resorcinol and phloroglucinol into a column, what is the elution order?

Phenol, then resorcinol and then phloroglucinol. Lo and behold, this is exactly like your prediction! So there is indeed a trend. However we cannot correlate this single observation with the water solubility.

Let us think of a cyclic sugar, cyclofructan: 21 hydroxyl groups, yet it is insoluble in water! It is insoluble in many organic solvents as well. So it is very hard to correlate the number of individual functional groups and predict water solubility. Exceptions like these will pop up again and again.

However there are softwares, such as Advanced Chemistry Development (ACD) which can predict solubility. No idea what models they use. You can look up their documentation or even contact them.


Catalysist activation energy - the Brønsted-Evans-Polanyi relation I have stumbled upon something called the Brønsted-Evans-Polanyi relation in a study about the design of catalysts (for reactions like those in hydrocracking fuel production.)

The relation shows that the transition state potential energy $E_a$ is linearly related to the CO dissociation energy (over transition metal surfaces):

Reproduction of Fig. 3B from J.K. Nørskov et al, Nature Chem. 2009, doi:10.1038/nchem.121

My question is simply: What exactly is on the axes?

I am not sure if the transition state potential energy $E_a$ is the same as the $E_a$ being described in e.g. the linked Wikipedia article in the first paragraph. Is it a term for the activation energies that belong to the intermediate steps, when the catalyst breaks down the original activation energy jump into more smaller jumps?
The CO dissociation energy is a bit odd for me. Is this the energy for adsorption of CO into the material surface?

According to the Wikipedia article, this relation in the graph should be given by the equation:

$$E_a=E_0+\alpha \Delta H$$

where $\alpha$ seems to be a constant and $\Delta H$ the enthalpy of reaction. This seems to say that my "CO dissociation energy" on the $x$-axis is the enthalpy (or the product of enthalpy with $\alpha$.) I would appreciate if someone could shed some light on the wording "CO dissociation energy".

First, a note on the Bell-Evans-Polanyi (BEP) equation:
The form you've written it in disguises what the constant $\alpha$ tells you. From the wikipedia article, it's said that $\alpha$ is a number between 0 and 1 which tells you how close the transition state is to the reactant state. Thus, this constant actually tells you quite a lot about the shape of potential energy plotted versus the bond coordinate.
To see why this is true, we should look at the derivation given in the article you linked. The BEP equation assumes that the energy of both the reactant and product vary linearly with the bond length of the atom being transferred. This is probably a very bad assumption for finding the actual shape of the well, but the intersection of these two lines seems to give a good approximation for the location of the transition state. Specifically, if one knew the slopes of these lines $m_1$ and $m_2$, we could find $\alpha$ according to $$\alpha=\frac{m_1}{m_1-m_2}.$$
See the wikipedia article you linked for the details on this, but hopefully this motivates the meaning of the constant $\alpha$.
So, what does this have to do with the article you're asking about?
Well, the reaction that is being catalyzed in the figure you have copied is,$$\ce{CO +3H2->CH4 +H2O}$$This seems very unlikely to happen in the gas-phase because the $\ce{CO}$ bond is very strong and we would have to have four molecules collide all at once and magically rearrange their atoms.
Thus, a transition metal surface is used as an adsorption site to get all these atoms close together.

What exactly is on the axes?

Well, now we can see that by the "CO Dissociation Energy" they mean the energy it takes to dissociate the carbon and oxygen atoms once they have been adsorbed onto the metal surface. So, a negative dissociation energy indicates that when $\ce{CO}$ adsorbs, it exists as two separated carbon and oxygen atoms.
Additionally, because they have shown the BEP linear relationship, the slope of that line is $\ce{\alpha}$ and the y-axis is the classical Arrhenius activation energy.
At first, it seems like the best catalyst should be the one which has the smallest CO dissociation energy, but as the article points out,

The reason is that in the limit of weak coupling (Ediss is only a little negative), the BEP relation gives that the barrier for dissociation of the reactants will be high and the rate low. For strong coupling (Ediss very negative) the activation energy of adsorption is small but now the barrier for forming the products will be large.

That is to say, if the reactants adsorb quite strongly to the surface (i.e. Rhenium in the plot) then reaction wil go really slowly because the activation energy (not the on you're seeing on the y-axis here) for forming the intermediates leading to methane will be very high.
On the other hand, if $\ce{CO}$ doesn't adsorb very favorably, the reaction will be slowed down because both $\ce{CO}$ and $\ce{H2}$ need to be adsorbed to react. When they do meet though, because they adsorb weakly, the activation energy to forming the products will be small.
This is why they say,

An optimal interaction strength must exist between these two limits.

The figure below on the right shows this intersection for pure metal surfaces. The closer you are to the peak, the better.

(source: nature.com)
Thus, either Cobalt or Ruthenium would be optimal for his reaction, but as the authors note, these are expensive, so nickel is used in the industrial process. Thus, some kind of alloy which behaves similarly to Ruthenium or Cobalt would be great if it could be found.
Finally, it should be noted that the BEP equation will only work in very specific circumstances where bond stretching is the dominant feature which determines if two things react. Thus, when reactants are adsorbed and thus all they can really do is vibrate, I would expect the BEP equation will do a pretty decent job.
I hope that clarified your questions. Let me know if I missed anything.


A drop of water in a tin of sugar: Which one's the solvent, the sugar or the water? The other day, when we were dealing with the chapter Solutions, our teacher asked us this:

  If I add a drop of water, to a tin full of sugar (without mixing it in), what's the solvent here? The water or the sugar?

Naturally we were taken aback by the 'unorthodox' nature of the question. Seeing that none of us were in a position to answer that any time soon, he gave us the "answer":

  It's the sugar. If you remember the definition we learnt ( solute + solvent = solution; between the solute and the solvent, the solvent is one present in a larger quantity and is in the same phase/state of matter as the solution ) you ought to have realized that, since there's more sugar than water, and at the end of the process the tin's filled with a solid.

Now that may seem like a clever way of testing our command of terminologies and definitions, but I saw a hitch there.

A solution is defined as a homogeneous mixture.

Adding a drop of water to a tin of sugar (without mixing it in) does not result in a homogeneous mixture. So as interesting as the question is, I feel it's flawed on this account... because what we're dealing with isn't technically even a solution.

Now I told this to my teacher, but he (quite deftly) side-stepped my query...a subtle way of indicating that he's not comfortable discussing this.

So I guess my question here boils down to this:

Was my teacher's "answer" correct? Or was the question seriously flawed to begin with?

EDIT-

I wouldn't call this a duplicate. Sure, I mean, both my question and the other one that was linked with this was about identifying the solvent and solute but I feel my question is distinct because:

1) It isn't answered well enough in the other question (that answer was too generalized... I've given a specific instance here)

2) I also want to know if a drop of water added to a tin full of sugar (without mixing it) can be considered a solution.

You could imagine stirring the sugar enough for the water molecules to be uniformly distributed throughout - it would then be homogeneous.

However, even then, to refer to the mixture as a solution of water in sugar is unhelpful, not least because referring to a slightly damp solid as a solution will only confuse.  A definition needs to be useful, as well as being a set of criteria to be met.

Consider adding more water, and stirring till uniformly distributed.  At some point, the mixture will become a thick syrupy liquid.  Add yet more, and it will eventually resemble an ordinary solution.  At what arbitrary point do you say one is a solution of the other?

As a device to get students thinking about the interactions that take place when a solid dissolves in a liquid, or a liquid in another liquid, it's an excellent question, but not one that has either a right or a wrong answer, other than "neither", or indeed "both".

What is definitely wrong is to insist that there's only one right answer to the question.


What is the formula for theoretical buffer capacity for a diprotic buffer system? According to Chembuddy, the formula for theoretical buffer capacity for a monoprotic buffer system is as follows:

  $$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

where $C_\mathrm{buf}$ is the total concentration of buffer and $K_\mathrm{w}$ is the water ionization constant.

However, I have conducted a lab in which I mix $\ce{C4H6O4}$ (tartaric acid; diprotic) with its double salt $\ce{NaKC4H4O4}$ to create a 1.0 M potassium sodium tartrate buffer solution where there are two simultaneous buffer systems. Also, $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ of tartaric acid are quite similar (around 1.6 difference), so I believe that this alternative equation that provides the theoretical buffer capacity for a system with multiple buffers:

  $$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \sum\frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

will not be correct in my case.

Is this right? If so, is there a proper way to theoretically calculate the buffer capacity of the composite system?

The logic is the same used for a monoprotic weak acid, considering the two ionization steps, as indicated by @MaxW.

For a diprotic weak acid we have:

The first ionization

\begin{align*}
\ce{H2A + H2O &<=> H3O+ + HA-} &K_{\ce{a}1}=\frac{\ce{[H3O+][HA-]}}{\ce{[H2A]}}
\end{align*}

and the second ionization

\begin{align*}
\ce{HA- + H2O &<=> H3O+ + A^{2-}} &K_{\ce{a}2}=\frac{\ce{[H3O+][A^{2-}]}}{\ce{[HA-]}}
\end{align*}

To make further calculations easier let's assume that the strong base added is monoprotic:

\begin{align*}
\ce{BOH &-> B+ + HO-}
\end{align*}

Charge balance of the solution is given by the equation

$$\ce{[H3O+]} + \ce{[B+]} = \ce{[OH-]} + \ce{[HA-]} + \ce{2[A^{2-}]}$$

When combined give us formula for the amount of the strong base (for short $\ce{[H3O+]}$ was written as $\ce{[H+]}$)

$$C_{\ce{B}} =
\frac{K_{\ce{w}}}{\ce{[H+]}}-
\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\left(\ce{[H+]}+2K_{\ce{a}2}\right)}
{\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}}$$

when $K_\ce{w}$ is the water ionization equilibrium constant, $C_{\ce{H2A}}$ is the concentration of diprotic weak acid, $K_{\ce{a}1}$ and $K_{\ce{a}2}$ are the acid dissociation constants, and $C_{\ce{B}}$ is the concentration of strong base added.

The buffer capacity of a diprotic weak acid-conjugate bases buffer is defined as the
maximum amount strong base that can be added before a significant change in the pH will occur.

\begin{align*}
\beta &= \frac{dC_{\ce{B}}}{d\ce{pH}} = \frac{dC_{\ce{B}}}{d\ce{[H3O+]}}\frac{d\ce{[H3O+]}}{d\ce{pH}}\\
\end{align*}

Differentiating the equation we should then find that

\begin{align*}
\beta &= \left[-\frac{K_{\ce{w}}}{\ce{[H+]}^2}-1 -
\frac{C_{\ce{H2A}}K_{\ce{a}1} \left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\left(-\ce{[H+]}\ln{10}\right)
\end{align*}

So finally buffer capacity is given by

\begin{align*}
\beta &= 2.303\left[\frac{K_{\ce{w}}}{\ce{[H+]}}+\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\ce{[H+]}\left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\end{align*}

you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.

Look a example based on this question. I hope it be useful.


Calculation of equivalent weight When a solid element is reacted with chlorine, a gaseous chloride of vapor density $68.75$ is formed. If this reaction is performed at constant temperature and pressure, the volume of the system reduces by $\frac{1}{3}$. What is the equivalent weight of the solid element?

I found out the molar mass of the gas $68.75 \times 2$. But don't know how to proceed from here?

I would express equivalent weight $M_\mathrm{eq}(\ce{E})$ of unknown element $\ce{E}$ using the number of equivalents $x$ and corresponding molecular weights $M$:

\begin{align}
M(\ce{ECl_x}) &= x \cdot M_\mathrm{eq}(\ce{E}) + x \cdot M(\ce{Cl}) \tag{1} \\ \to \quad M_\mathrm{eq}(\ce{E}) &= \frac{M(\ce{ECl_x}) - x \cdot M(\ce{Cl})}{x} \tag{1a}
\end{align}

To find $x$, let's apply to the balanced reaction, assuming it is complete and there is no chlorine left:

$$\ce{E (s) + x/2 Cl2 (g) -> ECl_x (g)}$$

The amounts $n$ of gaseous products are bound by stoichiometry:

$$n(\ce{Cl2}) = \frac{x}{2}n(\ce{ECl_x}) \tag{2}$$

At constant temperature and pressure $n_i V_i = \mathrm{const}$. The problem says the volume of the system reduces by $1/3$, which means:

\begin{align}
V(\ce{Cl2}) &= \frac{3}{2}V(\ce{ECl_x}) \tag{3} \\
\to \quad n(\ce{Cl2}) &= \frac{3}{2}n(\ce{ECl_x}) \tag{3a}
\end{align}

Now, comparing/equating (2) and (3a) it's easy to see that $x = 3$:

$$\frac{x}{2} = \frac{3}{2} \quad \to \quad x = 3$$

You already correctly determined molecular weight from vapor density $\rho_\mathrm{v}$:

$$M(\ce{ECl_x}) = \rho_\mathrm{v} \cdot M(\ce{H2}) = 68.75 \cdot \pu{2.016 g mol-1} = \pu{138.60 g mol-1} \tag{4}$$

All that's left now is to do the math with (1a):

$$M_\mathrm{eq}(\ce{E}) = \frac{\pu{138.60 g mol-1} - 3 \cdot \pu{35.45 g mol-1}}{\pu{3 equiv}} = \pu{10.75 g mol-1 equiv-1}$$

I'm not sure this makes any sense from the chemistry prospective as this would suggest that element $\ce{E}$ is sulfur and so the gaseous product is $\ce{SCl3}$. I would've expected the compound to be something like $\ce{BCl3}$, but in this case the vapor density should've been $58.12$.


Covalent atomic radii: oxygen vs nitrogen Many books state that $R_\ce{N} > R_\ce{O}$ which is in accordance with the general trend. However, some books say that $R_\ce{O} > R_\ce{N}$ because of repulsion caused by pairing of electrons.
Which one is correct and should be followed?
Older version of NCERT, cengage, arihant and JD Lee state the latter, while the newer version of NCERT, Wikipedia and many websites say the former. Why was the theory discarded?
Note: van der Waal's radii follow the general trend.

The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time.
$$
\begin{array}{lll}
\hline
\text{Reference} & R_\ce{O} & R_\ce{N} \\
\hline
\text{Pauling, 1939} & 1.40 & 1.5 \\
\text{Bondi, 1964} & 1.52 & 1.55 \\
\text{Zefirov, 1974} & 1.29 & 1.50 \\
\text{Gavezzotti, 1983–1999} & 1.40 & 1.50 \\
\text{Batsanov, 1995} & 1.51 & \\
\text{Wieberg, 1995} & 1.5 & 1.6 \\
\text{Rowland, 1996} & 1.58 & 1.64 \\
\hline
\end{array}
$$
As is evident from the table, $R_\ce{N} > R_\ce{O}.$ An interactive webpage of this reference can be seen here, which has the van der Waals radii of quite a few other elements as well.
Edit:
After OP mentioned the abnormality in Covalent Radii, I checked the radii in Concise Inorganic Chemistry, by J.D. Lee. The Covalent radius of Nitrogen $R_N = \pu{0.77 Å}$, whereas that of Oxygen is the same; $R_O = \pu{0.77 Å}$. Lee mentions that the radii are taken from Tom Moeller's Inorganic Chemistry book (which is from 1952)
OP should also be aware of the method used for calculating covalent radii: The Covalent Radius is defined as half the length of a single bond between two uncharged atoms. According to a more recent reference2 (which is used by Wikipedia), $R_N = \pu{73.4 pm}$ and $R_O = \pu{70.2 pm}$
Even more recent references, such as (3) use a different method of calculating covalent radius:

The covalent radius for nitrogen was obtained from N–N bond
distances in substituted hydrazines (only H or C atoms bonded to
N) with three-coordinated nitrogen atoms, excluding all structures
with an R factor larger than 10% or presenting disorder or
errors. The resulting value, $\pu{0.706(13) Å}$, is the average of 2200
crystallographically independent data and was used throughout
with three decimal figures for calculating other radii from N–
element bond distances to minimize rounding-off errors.
The covalent radius for $\ce{O}$ was derived from a
sample of 10,000 acyclic $\ce{C–O}$ bond distances, in a search limited to
purely organic compounds with two-coordinate oxygen and four-coordinate carbon atoms with R ≤ 5%, and the value obtained was $\pu{0.676(28) Å}$

A comparision of covalent radii has also been provided by the authors:
$$
\begin{array}{lll}
\hline
\text{Author} & R_O\ (Å) & R_N\ (Å) \\
\hline
\text{Cordero et al.} & 0.706(13) & 0.661(19) \\
\text{Alcock} & 0.702 & 0.659 \\
\text{Mingos} & 0.74 & 0.72 \\
\text{Butler and Harrod} & 0.75 & 0.73 \\
\text{Wells} & 0.74 & 0.74 \\
\hline
\end{array}
$$
References

Batsanov, S. S. Van Der Waals Radii of Elements. Inorganic Materials 2001, 37 (9), 871–885. DOI: 10.1023/A:1011625728803. (PDF)
Sanderson, R. T. “Electronegativity and Bond Energy.” Journal of the American Chemical Society, vol. 105, no. 8, Apr. 1983, pp. 2259–61.  doi:10.1021/ja00346a026.
Cordero, Beatriz, et al. “Covalent Radii Revisited.” Dalton Transactions, no. 21, 2008, p. 2832. doi:10.1039/b801115j.


Is ionisation an equilibrium process? I read in a page that ionisation is an irreversible process. But, as equilibrium is always reversible would that mean equilibrium does not involve ionisation process.
And than dissociation is a reversible process therefore only dissociation process is seen in ionic equilibrium.
Now if the above statement is correct would that mean only covalent compounds take part in ionic equilibrium?

An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas.
Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (while temperature is constant) until eventually the boiling point is encountered. At this point the pressure is equal to the vapor pressure (at that temperature) and the gas and liquid can coexist (all of this by the way is simpler to illustrate with a T-p phase diagram). Further compression condenses more gas until all of the substance is liquid. Below the boiling point pressure none of the substance is liquid, and above that pressure none is gas.
Next consider solubilization of a substance in a fixed volume of liquid (with T,p constant). One could traverse the solubility diagram by changing the temperature, but it is for the present discussion simpler to consider changing the concentration. At very low concentrations (small amounts of substance) all of the substance solubilizes. Exactly  none (yes, none) is present as the solid. But as you increase the amount of added substance a concentration is reached at which solid remains. That exact point is the solubility limit of the substance at that temperature. It can be thought of as equivalent to the boiling point pressure (coexistence point) of a gas and liquid at a given temperature. If you increase the amount of added substance at that point, no more substance will solubilize (exactly none).
So when a salt is added below its solubility concentration limit no solid remains, all of the substance is in the liquid, and there is no equilibrium only in the sense that there cannot be solid below the solubility limit.

This is also why the equilibrium equation for solubilization of a substance is written as
$$K_{\mathrm{sp}} = \frac{a(\textrm{soln})}{a(\textrm{s})}= a(\textrm{soln})$$
where a's are activities (for small concentrations, the activity of the solubilized substance is equal to concentration ). We ignore the activity of the solid (it is constant and assumed equal to 1) and above the solubility concentration limit the activity of the solubilized substance is also constant (equal to the solubility product).


Why is the aqueous solution of KMnO4 slightly basic? This is in the context of the various changes in oxidation state the permangananate ion can undergo with the variation of the pH of the medium. Besides the cases of strongly basic and strongly acidic media, the case for neutral media (with reduction to $\ce{MnO2}$) in my notes has a footnote; that this medium is in fact slightly alkaline, due to it being a solution of $\ce{KMnO4}$.
I tried to reason this out. It looks like $$\ce{K+ + MnO4- + H2O -> K+ + OH- + HMnO4}$$ is how it works out.
A quick google search establishes the pKa of HMnO4 to be -2.25. So it makes a pretty good acid in comparison to water.
Thing is, in that case, the product side should have an $\ce{H2O}$ instead of the constituent ions, with the potassium permanganate dissociated. That means the solution's neutral. Which apparently it isn't, going with my notes.
Where did I mess up here?

The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does not work with the latter.
$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$ (in molten KOH solvent; reactants are fused together)
$\ce{2 K2MnO4 + 2 H2O → 2 KMnO4 + 2 KOH + H2}$ (electrolysis in alkaline media; a direct chemical reaction would convert some material back to $\ce{MnO2}$)
Samples are of course purified but must be assumed to contain some leftover alkali from this preparation.


What does it mean that an electron quantum numbers are n1= 4 - l1? I'm trying to solve some exercises from the 2017 IChO, and the problem 6-A1 states that

  in first ionization, an electron with the quantum numbers $n_1 = 4 - l_1$ is removed.

Now, I've never seen this way of expressing quantum numbers, and I don't know what it's supposed to mean, so I would appreciate it if someone could dumb it down to me.

$n_1 = 4 - l_1$ means the same as $n_1 + l_1 = 4$, so the $n$ and $l$ quantum numbers of the first electron added up should equal 4. This could refer to 3p or 4s.

For the second electron removed, the $n$ and $l$ add up to 5. This could refer to 3d, 4p or 5s.

For chromium, the first electron removed is the 4s, followed by the 3d. For copper, this is the case as well, but its mass is higher than that of iron, so it is not the answer.


Need of hyperconjugation? Recently I read about Hyperconjugation. But i have a doubt.

What is the need of hyperconjugation in this structure when there is charge separation resulting in lesser stable structure.

What is the need of hyper conjugation in this structure when there is
  charge separation resulting in lesser stable structure.

I think we'd all agree that the charge separated structure you've drawn is a resonance structure.  Charge has been separated so it may well be higher in energy than the first (neutral) resonance structure.  Nonetheless, to whatever extent the charge separated resonance structure contributes to the description of propene - even if it is just a small amount - it will stabilize propene.

Since there are 3 hydrogens on the methyl group you can draw three identical resonance structures, this will increase the importance of the structure.  Note too the charge separated resonance structures involve an allyl carbanion (resonance stabilized) and a proton; two very stable structures - this will further increase the weight, or importance, of the hyperconjugated structures.

These hyperconjugated resonance structures are really showing us that an $\ce{sp^3}$ carbon can donate electron density to an $\ce{sp^2}$ hybridized carbon.  Electron movement in this direction lowers the energy of the electron because it moves the electron from an orbital that is 25% s-character to an orbital that is 33% s-character.  The more s-character in an orbital, the lower the energy of the orbital (an s orbital is lower in energy than a p orbital).  As we stabilize the electron, we stabilize the molecule.

This hyperconjugation argument comes in handy.  It can be used to explain

why double bonds with more alkyl groups are more stable than double bonds with fewer alkyl groups (ref1, ref2,)
why carbocation stability follows the order: tertiary > secondary > primary (ref1, ref2)
why alkyl groups are ortho-para directing in electrophilic aromatic substitution (ref)


Are glycosides more soluble in dichloromethane than in water? I'm designing an experiment to extract the amygdalin from apricot leaves and kernels. I want to steam the kernels and leaves, grind them, add water and then filter it using a filter paper. Then transfer the solution to a separatory funnel, add dichloromethane (DCM), shake and collect the bottom DCM layer that I'm hoping would contain the amygdalin.

This is the process I used for extracting caffeine from coffee beans and it worked because caffeine is much more soluble in DCM than in water. However, I'm not sure about the solubility of glycosides and whether this process would even work or not which is why I'm making this post.

How soluble are glycosides in DCM, is it enough to make this work? If not, what other solvents do you recommend?

Also, how can I increase the efficiency of this process in general? I've been thinking of using vacuum filtration, but I'm not sure. Tell me what you think.

Anybody with glycoside isolation from bio-matrices would tell you that your plan is not going to work. Amygdalin is classified as a cyanogenic glycoside because it contains nitrile moiety. It is contained in the seeds of bitter almond, apricot, apple, black cherry, peach, plum, etc. Amygdalin was first isolated in 1830 from bitter almond seeds (Prunus dulcis) by French chemists, Pierre-Jean Robiquet and Antoine Boutron-Charlard. Traditionally, the extraction method of the chemical ingredients in the medicinal materials is decoction in boiling water, which was the method French chemists has used to isolate amygdalin. Presently, it has been isolated using several other methods and its structure has been characterized by several groups around the world.

Amygdalin can be dissolved in water and ethanol easily. Solubility of amygdalin in water and ethanol is $\pu{83 g/L}$ and $\pu{1 g/L}$, respectively (Ref.1). Ref.1 described the isolation of amygdalin from plum seeds for the first time, has used ethanol to extract soluble compounds, and finally, used diethyl ether to isolate amygdalin from the complex ethanol extraction mixture. Before extract, the plant material was dried to the moisture content of 6% in a dark place and then ground in a blender to obtain a fine powder with average particle size of $\pu{0.3 mm}$. The ethanol extraction and diethyl ether isolation procedure used in Ref.1 have been described as follows:

  The powdered solid plant material ($\pu{2.0 g}$) was transferred into the Plant material (2 g) was transferred into the $100$-$\pu{mL}$ round bottom flask and treated with exactly defined volume of ethanol. The extractions at higher temperatures were carried out under reflux, while the extractions at lower temperatures were carried out in closed vessels. The system temperature was maintained using a water bath. After extraction, the extract was separated from the solid plant material by filtering process. Plant extracts were evaporated in rotary evaporator under reduced pressure to remove the solvent, and then stored in desiccator until it reached the constant mass. After addition of diethyl ether ($\pu{10 mL}$) to the complex extraction mixture, amygdalin was precipitated, while fatty compounds were dissolved. The dissolved fatty compounds were removed by decanting process. The solid phase was dried at $\pu{30 °C}$ in order to remove the residual solvent. In this way, the sample was prepared for determination of its purity using HPLC analysis and structural characterization by FT-IR, UV, and MS methods.

The effect of extraction time ($\pu{10–120 min}$), ethanol concentration ($20–100\%,\: v/v$), solid-to-liquid ratio ($1: 5\: – 1: 25,\: m/v$) and extraction temperature ($\pu{22–78°C}$) on the content of amygdalin in the dried extract has been studied. As a result, the optimal conditions were achieved: The extraction time of $\pu{120 min}$; $100\% \:(v/v)$ ethanol; solid-to-liquid ratio of $1:25\: (m/v)$; and extraction temperature of $\pu{34.4°C}$. The optimal yield of $\pu{25.30 g}$ of amygdalin content in the extracts per $\pu{100 g}$ of dried extract was reported.

Reference:

Ivan M. Savic, Vesna D. Nikolic, Ivana M. Savic-Gajic, Ljubisa B. Nikolic, Svetlana R. Ibric, Dragoljub G. Gajic, “Optimization of technological procedure for amygdalin isolation from plum seeds (Pruni domesticae semen),” Frontiers in Plant Science 2015, 6, Article 276 (11 pages) (DOI: 10.3389/fpls.2015.00276).


Why are aromatic compounds so stable? I've never actually been told deeper reasons for this. In organic chemistry classes (1 and 2) we're usually told that aromatic compounds are very stable whereas anti-aromatic ones are not, and also, we're taught how to identify them. Why are they, though?

(I'm pretty sure this has something to do with the de-localisation of electron pairs, which enhances stability, but I only have a vague idea. Is that the only reason?)

Let me start by saying that aromatic compounds are not exceedingly stable, they are just much more stable than antiaromatic compounds. However, they can still be hydrogenated exothermically to the corresponding cycloalkanes.

$$\ce{C6H6 + 3 H2 -> C6H12}$$

But why are they more stable than antiaromatic ones? Let’s take a look at the orbital scheme for cyclopentadienyl $\ce{C5H5^{?}}$ (the scheme is the same for radical, anion or cation — at least theoretically). (Taken from this site).

The pictures of the molecular orbitals correspond to their energies. The e means that the two orbitals next to each other are degenerate, i.e. have the same energy. Let’s fill 6 $\unicode[Times]{x3c0}$-electrons ($\ce{C5H5-}$) into this scheme. We get three completely filled orbitals and two rather antibonding unoccupied orbitals. Looking good.

Now let’s extract a single electron to create the $\ce{C5H5^{.}}$ radical. The electron must come from one of the two HOMOs, so we end up with one degenerate orbital completely occupied, the other one only occupied by a single electron (usually drawn with ↑ or $\unicode[Times]{x3b1}$-spin). Now we’ll abstract another electron to arrive at the $4\unicode[Times]{x3c0}$ $\ce{C5H5+}$ cation. Two of the remaining four electrons can be filled into the lower-lying $\ce{a_1}$ orbital, the other two need to be in $\ce{e1}$ and since the latter is degenerate we arrive at a triplet state: Both electrons have $\unicode[Times]{x3b1}$-spin. This is rather unstable.

The triplet state and hence the instability can be avoided by distorting the molecule. If elements of symmetry ($\ce{C5H5^?}$: most importantly the plane of symmetry) are lost, the equivalence of the orbitals here labelled $\ce{e_1}$ is also lost and there is no reason not to pair the electrons into a single orbital. That’s why cyclobutadiene is not a square but a rectangle.

If I remember correctly, the results of simple approximations are actually mathematically equivalent to taking a circular $\unicode[Times]{x3c0}$-system, putting it on one corner and drawing an orbital energy on every carbon — the cyclic Hückel approximation. And since you will always get pairs of degenerate orbitals by this method using perfect shapes, that gives us the Hückel rule of $4n+2$ — two for the lowest orbital (the corner the circle is resting on) and $4n$ for every pair of orbitals that then have the same energy.


Aren't unit cell and the relevant crystal similar? "A crystal is formed by a large number of repetitions of basic pattern of particles in space.
The basic structural unit which when repeated in three spatial directions generates the crystal structure is called unit cell."
Problem:
Assume I take some same perfect geometrical cubes and arrange them in all three dimensional directions, a larger box similar to the smaller ones will be introduced.
From this assumption and the information above it, I deduce that a unit cell should be similar to the relevant crystal. That means, the only difference would be lying between their size.
If this is the case, why consider the geometry of unit cell and not directly the respective crystal? We know in the assumption above; the fundamental cubical boxes would be determined by same geometrical information as would the derived cubical box.
That is; if the lengths of three axes of cube are represented by letters $a,b$ and $c$ and the angles between axes are represented by Greek letters $\alpha$, $\beta$ and $\gamma$ then for a cube,
$$a=b=c$$$$\alpha=\beta=\gamma=\pi^c$$
This information is general. It is for all size of cubes. Then, why to define a unit cell and think it simpler to handle with than the crystal?

The statement from your book is correct, though the devil lies in the details. The point here is that the book is treating a theoretical crystal (i.e. one that is infinitely large). Let me elaborate and then answer your question.

A unit cell is simply a representation of a periodic structure. It is a means of conveying translational symmetry (and perhaps other types of symmetry as well, but let us focus on translational symmetry). If you have for instance a cubic unit cell, than a 3x3x3 grid of these unit cells is also a unit cell of that same lattice. If you extrapolate this idea, then the reasoning you pose in your question makes a lot of sense!

Let us now discuss the details. Assume that we have a theoretical crystal of infinite dimensions. What is then the unit cell? Can we define a unit cell which is also the infinite crystal? Every finite unit cell would obviously not be that crystal. This explains why we even use a unit cell in the first place: By defining the translational symmetry by means of the unit cell, we can define an infinite crystal.

In the field of solid state physics, it is attempted to keep things simple and well-defined and therefore a so-called primitive unit cell is introduced. This unit cell only contains one lattice point and you could see this as the smallest unit cell possible that is still able to convey the translational symmetry of the infinite crystal.

Your last question relates to why we want to use, for instance, the primitive unit cell or some other unit cell rather than the complete crystal? Let me answer this question by giving an example: If you cut a crystal, you will create a surface facet. Such surfaces also have translational symmetry and you can define these using Miller indices. Your textbook will probably treat these as well. Those Miller indices are defined by the reciprocal lattice vectors of your unit cell. You already touched upon on the fact that there exists an infinite set of unit cells that define that infinite crystal, yet the question arises which unit cell to use if we, for instance, want to use these Miller indices. In principle, any unit cell in the set would be valid. To avoid any kind of ambiguity, we tend to use the smallest unit cell possible.

A nice book treating these concepts is "Introduction to Solid State Physics" from Charles Kittel.


E227 Food Additive effectiveness I have been researching on E227 (Calcium Bisulfite), and could find minimal information about how it acts as a preservative and how effective it is. I could only find that it preserves by converting itself to sulfurous acid in an acidic liquid, but could not find out how effective and how exactly it preserves. I would appreciate any help.

Sodium bisulfite is one of several sulfur-based chemicals commonly known as ‘sulfites’. Others include potassium bisulfite, potassium metabisulfite, sodium metabisulfite, sodium sulfite, or sulfur dioxide, and also your calcium bisulfite. They have fungicide properties and stop the natural browning process of produce. Dried apricots stay orange, and golden raisins stay yellow. Sulfites stop the browning and further ripening of fruit by blocking the natural enzyme, polyphenol oxidase. Sulfites are also added to some wines.

Sulfites destroy vitamin B1 (thiamin) and so the FDA does not allow sulfites on food containing vitamin B1 (thiamin). They estimate that 1 in 100 people are sulfite sensitive, so it is illegal to use sulfites on foods sold raw, except potatoes. They get an exception because there is no alternative to keeping french fries from looking brown.

The FDA lists sulfites as generally recognized as safe, but sulfites are linked to neurological, gastrointestinal, and respiratory problems. Trace amounts of sulfites have been known to cause hives, vomiting, wheezing, and anaphylaxis. I know from experience that too many dried apricots can cause intestinal cramping.

Modified, but essentially reporting from  https://traditionalcookingschool.com/preserving/dehydrating-preserving/sulfites-and-food-preservation/


How can molecular mass be calculated from vapour density? An organic compound contains 71.7% carbon, 6.7% hydrogen, 10.4% nitrogen and 11.8% oxygen by mass. Given that its vapour density is 67.5, determine its empirical formula and molecular formula.

From the mass percentages, I have found out that the empirical formula is $\ce{C8H9NO}$. However, to find the molecular formula we usually require the molecular mass.

What does "vapour density" mean, and how can I use it to calculate the molecular mass of the compound?

According to Wikipedia, the vapour density of a molecule is "the density of a vapour in relation to that of hydrogen". The density of a gas, $\rho$, is proportional to its molecular mass, $M$:

$$\rho = \frac{m}{V} \propto \frac{m}{n} = M$$

where $m$ is the mass of the gas, $V$ is the volume, and $n$ the amount of gas. Therefore, if we denote your compound by $\ce{X}$:

$$\begin{align}
67.5 &= \frac{\rho(\ce{X})}{\rho(\ce{H2})} \\
&= \frac{M(\ce{X})}{M(\ce{H2)}} \\
M(\ce{X}) &= 67.5 \cdot M(\ce{H2}) \\
&= 67.5 \cdot \pu{2 g mol-1} \\
&= \pu{135 g mol-1}
\end{align}$$

Finally, you can use this piece of information to find that the molecular formula is $\ce{C8H9NO}$ as well.


What is the state of the oxygen free radical in the following reaction? $$\ce{O2(g) ->[h\nu][\nu\,<\,240~nm]2O^{.}(?)}$$

Oxygen is split by UV light below 240 nm to form oxygen radicals, which then go on to react with oxygen molecules to form ozone.

Firstly, is the oxygen a free radical?

Oxygen has six outer shell electrons, but my teacher mentioned that by the way orbitals work, they don't actually all pair off, rather you end up with two pairs of electrons and two unpaired electrons, hence making oxygen a free radical, though I'm not too sure. But because the free radical is so highly reactive, I'm not sure whether I'm even allowed to assign it a state, because straight after being formed it pretty much immediately reacts.

Going by process of elimination, it can't be aqueous, because it isn't dissolved in water, and I don't think it is solid, because solids are structurally rigid. I am therefore inclined to say that it is a gas, but my main problem is that each reaction only creates two oxygen free radicals, which immediately disappear, so it shouldn't actually show any properties of being a gas.

So what is the state I should assign the oxygen, or does it simply not need one?

According to the reaction $$\ce{O2\,(g) ->[h\nu][\nu\,<\,240~\mathrm{nm}] 2O}$$
photolytic cleavage of oxygen yields two radicals. These will most likely be in the high spin triplet state. And yes these are free radicals according to the definition of the IUPAC Goldbook, as they have two unpaired electrons.

I assume you are talking about state of matter and then of course, you are right. Oxygen radicals can exist neither in aqueous solution, nor liquid, nor solid. Even in gas phase the concentration of oxygen radicals will be very low as they are an incredibly short lived species. They might exist in plasma, too. Keeping that in mind it is probably unnecessary to assign a state to the oxygen radical.


Why is the para carbon in Benzaldehyde NMR more deshielded than the ortho and meta carbons? Assuming this image is correct, why is the Carbon-3 more deshielded than Carbon-2 and Carbon-1? From what I understand, deshielding is the greatest when the Carbon atom is closer to electron withdrawer and it appears Carbon-3 is the farthest.

Resonance effects come into play along with the inductive effects.  I've drawn a few of the resonance structures for benzaldehyde below, notice how resonance effects place some positive charge on carbons 2, 3 and 5 which results in their deshielding.  I might have guessed that the combination of resonance and inductive effects would place C2 below C3, but that's not the case.  At least C5 is at lowest field and C2 and C3 are close and both are below C1. The interplay of inductive and resonance effects is complex and estimating the magnitudes of these effects can get tricky.


How to calculate the concentration H3O+ in a solution with pH=6.99? What is the correct way to calculate the concentration $\ce{H3O+}$ in a solution with $\ce{pH}=6.99$?

Attempt 1.

  pH<7, therefore there are only $\ce{H3O+}$ particles in the solution. $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$

Attempt 2.

  We have $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$ and $[\ce{OH-}] = 10^{-\ce{pOH}} = 10^{-7.01} = 9.77 \cdot 10^{-8}$.   Because of $\ce{H3O+ + OH- -> 2 H2O}$ we are left with $[\ce{H3O+}] = 1.02 \cdot 10^{-7}- 9.77 \cdot 10^{-8} = 4.6 \cdot 10^{-9}$

When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking very large. So I wonder what the correct way is?

If you take a sample of pure water, there will be few hydroxide and hydronium ions. Of course, they can combine to form water and yes they do combine but there will be few water molecules which break/combine to form the ions again. Hence, there exists a dynamic equilibrium between concentration of ions and water molecules.

$\textrm{pH}$ by definition is the negative logarithm of hydronium ion concentration.

$$\textrm{pH} = -\log [\ce{H^+}] = -\log [\ce{H3O^+}]$$

You can obtain the concentration of H+ ions by substituting the value of pH in the following formula,

$$[\ce{H3O^+}] = 10^{\mathrm{-pH}}.$$

Your attempt 2 is flawed because your assumption that all the ions combine to form water molecules is incorrect. There will always be some concentrations of the ions and all of them needn't combine to produce water molecules. Your attempt 1 is correct.

It appears like you are not aware of the concept of equilibrium and self ionization of water, I have picked few good materials which you might(should) want to refer to,

Chemical Equilibrium

Self Ionization of water

The concept of chemical equilibrium is very important and you will come across it frequently in chemistry, so you must learn it. Also, self-ionization of water along with chemical equilibrium are central concepts for learning acids and bases.


Acid-base reaction is screwing up my synthesis I'm gonna get kicked all the way to the Moon, but whatever.

I am doing some nitroglycerin. I know the process, since the reaction itself is pretty basic. I know that it is an electrophilic substitution and that it releases a hell lot of energy. I've handled that with an ice bath and by doing the respective stoichiometry. I add around 5 ml of glycerol every 10 minutes.

All of that works great, until I have to do the wash. I've learned how to do washing and extractions from MIT's OCW. Nitroglycerin is insoluble in water, so I require water to wash the mixture and pull out the acid. <- There's the problem. When I'm doing the wash, the temperature rises and a runway reaction begins.

Any advice, besides the usual: "Just… don't"?

Just do a decantation. Simply discard the acid layer and then do the wash.


How to retrieve Dalton value from m/z value? As I understand, Dalton (Da) is the standard way for representing mass unit. As for $m/z$, which is a specific notation for mass spectrometry, quoting for wikipedia:

  This notation eases data interpretation since it is numerically more related to the unified atomic mass unit

Considering this, could I say that the $m$ in $m/z$ is a $\mathrm{Da}$ unity? If so, given an $m/z$ value, can I retrieve the original mass of the atomic mass by multiplying it by the ions charge, if I have that information?

I ask this because I have a spectrum analysis file with $m/z$ in the $x$ axis and I'm asked to compare masses in $\mathrm{Da}$.

IUPAC is wrong

Martin's answer is useful for its pointer to the IUPAC Gold Book, which is one of the most authoritative sources for chemical nomenclature around.

However, on this particular question, the Gold Book is laughably wrong.

  The abbreviation $m/z$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number.

This sentence is wrong.  The $m$ in $m/z$, as used by nearly every practicing chemist in the world (except IUPAC), is the ion mass, not the mass number.  The mass number is the sum of protons and neutrons in a nucleus.  Therefore it is always an integer.  Thus, according to IUPAC, ions with a charge of $\pm 2$ would always have $m/z$ values that are either integers (if $m$ is even) or half-integers.  Thus, according to their definition, saying the $m/z$ value of $\ce{C7H7^2+}$ is 45.52683903 would be wrong.  However that is exactly what most mass spectrometrists would say it is.

Here is some evidence:

A lipid maps database entry for $\ce{C7H7}$ lists $m/z$ values for a hypothetical $\ce{C7H8+}$ ion that have four decimal places.
I happen to have the latest print edition of the Journal of The American Society for Mass Spectrometry in front of me, and paging through the contents I see things such as:

  ...an intense signal for the $\ce{Cu^2+ - His_3}$ complex at $m/z$ 245.6, instead of the $\ce{[His_3 + 2H]+}$ at $m/z$ 215.1...

  ...MS spectra of standard solutions containing (a) lactose ($m/z$ 365.11, $\ce{[M + Na]+}$; $m/z$ 381.309, $\ce{[M + K]+}$), (b) cytidine ($m/z$ 266.218, $\ce{[M + Na]+}$, ...

These excerpts are not using the (integer) mass number in their calculation of $m/z$, they are using the ion mass, measured in Daltons.  (If they were, the only numbers after the decimal point would be .0 or .5).  That means that $m/z$ values are not really dimensionless, which makes the sentence wrong again.  In fact a unit called the Thomson has been proposed for $m/z$ values in honor of JJ Thomson. However this nomenclature has not really caught on widely.  Nonetheless, $m/z$ values are not dimensionless.

The Gold Book goes on...

  It has long been called the mass-to-charge ratio although $m$ is not the ionic mass nor is $z$ a multiple or the elementary (electronic) charge, $e$.

This sentence is wrong because $m$ is the ionic mass (in Daltons) as explained above.  It is additionally misleading because $z$ is always a multiple of the elementary electronic charge, with the understanding that charge must be measured in multiples of $e$, not in Coulombs.

  The abbreviation $m/e$ is, therefore, not recommended.

Finally something I agree with IUPAC on!

  Thus, for example, for the ion $\ce{C7H7^2+}$, $m/z$ equals 45.5.

Most folks would find that to be an OK approximation to the true $m/z$, which as I said above should be 45.52683903.

For the purposes of your question, Martin was right when he said

  For all intents and purposes the mass number may be seen as the mass of the ion in u or Da...

One thing to keep in mind is that this definition of "mass number" is at odds with IUPAC and nearly every other chemical text, which restricts "mass number" to be integers.

One additional complication

It's a bit unclear in your question if you want to determine the mass of an ion or the mass of the parent (usually neutral) molecule that gave rise to that ion.  If you want to determine the mass of an ion, then the other answer is right and you can just multiply the $m/z$ by $z$, the (absolute value of the) reduced electric charge.

If you want the mass of the parent neutral molecule, you will (a) need more information, and (b) have to do a bit more work. The extra information you need is the ionization mechanism.  In electron ionization mass spectrometry, ionization is usually by electron gain or loss.  So to get the neutral molecule mass, you will have to add back the missing electrons (or substract the gained electrons) from the calculated ion mass to get the molecule mass.  For example, if $\ce{C7H7^2+}$ arose by the loss of two electrons, the $m/z$ would be 45.52683903, the ion mass would be 91.05367806, and the molecule mass would be 91.05477522.

Other types of mass spectrometry involve ionization by gaining or losing protons, not electrons, and so the difference between ion mass and neutral molecule mass is much more significant.


Reaction between 1,2-dicyanoethylene and 2-phenylbuta-1,3-diene in toluene and heat I believe this may be a Diels-Alder reaction, but the benzyl group on the edge does not serve a an electron donating group and more as a stabilizing group.

I am at a loss for what reaction would take place if the Diels-Alder reaction doesn't.

The presence of an electron-donating group on the diene, or an electron-withdrawing group on the dienophile, is not an absolute requirement for the Diels-Alder reaction to occur. There are many Diels-Alder reactions that don't have such features.

In any case, the cyano groups on the dienophile are electron-withdrawing groups. So, it's still a pretty favourable scenario for the Diels-Alder to occur.

Since the dienophile is symmetrical there aren't any regioselectivity issues, and the relative stereochemistry of the two cyano groups in the product will be determined by the configuration of the double bond in the dienophile. A trans or (E) double bond will lead to the trans dinitrile (both enantiomers of it), and a cis or (Z) double bond will give rise to the cis dinitrile.

By the way, according to your title, there's no benzyl group on the diene, only a phenyl group. (Not that it makes much of a difference, though.)


Does rate of Evaporation change with surface area? Two vessels of the same volume, in the
same location and open to the atmosphere. One is of a small diameter, the
other is of a larger diameter.
 The question is will the vessel with the
larger surface area evaporate the liquid
faster than the smaller one?
If not! I need answer to other question ,if i am not wrong.
Why does vapour pressure decreases with decrease in surface area?

Why does vapour pressure decreases with decrease in surface area.

Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}->H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.

That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.

Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.


Electron transfer rates in Marcus Theory Electron transfer rates (k) as derived from Marcus theory assumes i) thermal equilibrium and ii) neglects quantum nature of low-frequency phonon modes. Can anyone explain what is at the heart of these two assumptions? Thanks a lot!

I read this in the context of the electron transfer process in organic semiconductors.

There is a description of Marcus theory here How does the inverted Marcus region explain chemiluminescence? . The quantum nature can be added into the theory as necessary, it leads to a asymmetrical plot of rate constant vs free energy with the inverted region (large -$\Delta G$) decaying away more slowly than the normal region.  The equilibrium is between the solvent reorganisation with reactants and then with products. The solvent plays a controlling role in the process, the proof is that experiment demonstrates that  the rate of electron transfer can be limited by solvent motion. See Harrison et al.  Chemical Physics, Volume 116, Issue 3, 15 September 1987, Pages 429-448, and Irvine et al. ibid.  Volume 104, Issue 2, 1 May 1986, Pages 315-324. The novel aspect of Marcus theory is the role of the  solvent and this was not generally accepted until the inverted region was observed.


If polar non-polar intermolecular forces are stronger, why do polar compounds elute first in GC? I remembering reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.

However, in analytical chemistry classes, we are told that polar molecules elute first in a Gas Chromatograph that has a non-polar column. According to http://www.chem.ucla.edu/~bacher/General/30BL/gc/theory.html

  If the polarity of the stationary phase and compound are similar, the retention time increases because the compound interacts stronger with the stationary phase. As a result, polar compounds have long retention times on polar stationary phases and shorter retention times on non-polar columns using the same temperature.

These two concepts seem to conflict. Is there something else that is going on? Shouldn't polar compounds elute last even if the stationary phase is non-polar.

(Also sidenote: most mobile phases in Gas Chromatography are non-polar: helium, argon, or nitrogen)

I think you're incorrectly remembering something you've been told. People speak of the adage "like-dissolves-like" which really means that polar things interact best with polar things and nonpolar interact best with nonpolar. There are always exceptions and fringe cases, but that adage works pretty well here. Nonpolar molecules interact with other nonpolar molecules pretty much exclusively through dispersion forces. These are the so-called "instantaneous dipoles" you have probably heard of. Polar molecules usually have some way of interacting through dipole-dipole interactions, hydrogen bonding, or simple electrostatics.

Note that the mobile phase, i.e. the carrier gas, does not have a large effect on the retention time in GC, so it really is mostly due to the simplified explanation of intermolecular forces given above.

One might ask why the polar molecules don't also have large dispersion interactions with the stationary phase (they certainly have some), and this is due to two effects. First, polar molecules tend to interact fairly strongly with each other in the gas-phase, and so there is less of an energetic gain for interacting with the stationary phase. This is a problem because interacting with the stationary phase is entropically unfavorable. Second, many polar molecules are smaller than a lot of the nonpolar systems that are studied using GC, and dispersion is correlated with surface area, so the polar molecules tend to have weaker dispersive interactions even ignoring the first point.


Contradictions in acidity and solubility of phenol derivatives o-Nitrophenol is less acidic in comparison to p-nitrophenol due to intramolecular hydrogen bonding. This effect also is the reason for o-nitrophenol being less soluble in water.

At the same time, hydroquinone (para) is less acidic than catechol (ortho).
Of course there is the inductive effect, but a reason mentioned quite often as well is hydrogen bonding. Isn't it contradictional that the reason given for o-nitrophenol being less acidic than p-nitrophenol is the same reason given for catechol being more acidic than hydroquinone?

In the case of o-nitrophenol, the acidic hydrogen is hydrogen-bonded to the nitro group's oxygen atom, making it less acidic. In the case of catechol, one acidic hydrogen is hydrogen-bonded to the adjacent OH group's oxygen atom, but the other, more acidic hydrogen is not hydrogen bonded. This hydrogen is more acidic than that of hydroquinone. The resulting anion is even more strongly hydrogen-bonded than neutral catechol, stabilizing it even further. It is the second acidic hydrogen of catechol that is analogous to the acidic hydrogen of o-nitrophenol.


Why can't sodium salts be produced through a reaction between a metal and an acid? I am writing a lab report for the preparation of sodium sulfate crystals. This question came up: Why can the reaction between a metal and an acid not be used to make sodium salts?

In this case, could one not use sodium metal and sulfuric acid to make sodium sulfate, as in the following:

$$\ce{2Na + H2SO4 -> H2 + Na2SO4}?$$

Normally, reacting a metal and an aqueous acid will produce a salt and hydrogen gas, as in the case of iron:

$$\ce{Fe (s) + 2HCl(aq) -> FeCl2 (aq) + H2 (g)}$$

However, you should not use the same method to produce, for example, sodium chloride:

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

because sodium reacts violently with water.


Amount of substance expressed in yoctomole units I stumbled upon a couple of papers where amounts were expressed in yoctomoles ([ymol], $\pu{10^{-24} mol}$) and I find it somewhat bothersome as 1 ymol would correspond to about 60% of atom/molecule/ion etc., which doesn't make much sense to me:

$$N_\mathrm{A} \times \pu{10^{-24} mol} = \pu{6.022 \times 10^{23} mol-1} \times \pu{10^{-24} mol} \approx \pu{0.6}\,\text{(elementary entities)}$$

Simply put, I'd say that using zeptomoles ([zmol], $\pu{10^{-21} mol}$) is already a stretch as one of the ideas behind a mole concept is to allow to handle large numbers of entities more conveniently, and at this scale there is no reason not to use direct quantities of molecules/atoms/ions.

Should the yoctomoles units really be used, and if so, then how to rationalize fractional elementary entities?

Well, the closer you get to atomic detection limits, the less sense bulk units make, that is correct.

However, coming from lab scale units it is much easier to scale down directly with SI prefixes — the exact reason what they were designed for. Therefore, it makes sense that these papers might say ‘Our predecessors could only detect $\pu{10zmol}$, we have lowered the detection limit by a factor of $100$ to $\pu{0.1zmol}= \pu{100ymol}$.’ And that is basically what they say. I note that both abstracts immediately place a molecule count behind the amount in ymol, so they are well aware of the scale of their unit with respect to molecules.

Once you get to the full molecular scale, you will be dealing with a type of quantisation this way or that; whether you are reporting energy, charge, mass or amount. In none of the other cases is anything special described for atomic scales so why should we make an exception for moles? Just make sure that the value itself corresponds to a whole number of molecules within the boundaries of rounding errors.


Max Ionization Extent How is this formula derived? What does it mean; what is the use of this formula?

I know that at infinite dilution, ionization extent should be very high because of Le Chatlier's principle.

But I'm not seeing how this formula is derived and its significance in actual chemistry. I tried rewriting the formula as:

$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} \cdot 100\%}$

But this isn't elucidating where the formula came from or its significance.

I could also rewrite the formula as this for the hypothetical acid, $\ce{HA}$:

$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} = \frac{\frac{[H_3O^+][A^-]}{[HA]}}{1.0 \cdot 10^{-7}+\frac{[H_3O^+][A^-]}{[HA]}}}$

I think the trouble I'm having is defining maximum ionization extent. For example, I know that in general ionization extent is defined as (where x is the concentration of the derived species - i.e. $\ce{[A^{-}]}$ for the hypothetical acid $\ce{HA}$ and $\ce{M_{i}}$ the initial molarity of the hypothetical acid.

$\mathrm{\frac{x}{M_{i}}}=\text{ionization extent}$

I have a feeling that the $1.0*10^{-7}$ term refers to the concentration of one of the products of the auto-ionization of pure water at 25 degrees Celsius. This makes sense because at infinite dilution the concentration of solute should be negligible (=0).

Upon further thought ...

For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis.
\begin{aligned}\ce{
(1)&&HA + H2O &<=> H3+O + A^{-}\\
(2)&&2H2O &<=> H3+O + {}^{-}OH \\\hline
(3)&&HA + 3H2O &<=> 2H3+O + A^{-} + {}^{-}OH
}\end{aligned}

The degree of dissociation is usually defined as
\begin{aligned}
&&\alpha &= \frac{c(\ce{A^{-}})}{c_0}&\text{with}&&c_0&=c(\ce{A^{-}})+c(\ce{HA})\\
\therefore&&\alpha &= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&\text{with}&&
c(\ce{A^{-}})&=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\
(4)\therefore&&\alpha &= \frac{K_a}{K_a + c(\ce{H3+O})}
\end{aligned}

At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as
\begin{aligned}
&&K_w&=c(\ce{H3+O})c(\ce{{}^{-}OH})\\
(5)\therefore&&c(\ce{H3+O})&=\sqrt{K_w}
\end{aligned}

At $25~^\circ{}C$ you will find your given equation:
\begin{aligned}
\alpha[\%] &= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\%
\end{aligned}


Explain why alumina-containing zeolites possess chemically active surfaces I think zeolites possess acid sites that are catalytically active in many hydrocarbon reactions. The pore system only allows molecules that are small enough to enter, hence it affects the selectivity of reactions by excluding both the participation and formation of molecules that are too large for the pores. Is that right or not ?

Alumina containing zeolites have composition that can be represented as

where M is metal ion with m charge, and so on, respectively.
Meaning you have $[SiO_4]^{4-}$ and $[AlO_4]^{5-}$ tetrahedra which make up negatively charged framework. Positive ions compensate for this negative charge. There is a rule from 1954. called Löwenstein rule, which states that x/y has to be less than 1, which in turn implies there are no Al-O-Al bonds, as two tetrahedra have Al atoms in their tetrahedral positions. This gives really great internal specific area and resistance to high temperature, which makes this type of zeolite a great candidate for catalysis (heterogenous).
The following picture nicely represents the active sites in zeolites - they act as molecular sieves, as only the specific molecules fit in the pores:


What happens to the amino acid when it is in a solution with the pH of its isoelectric point? (zwitterions) I am learning about zwitterions and am confused about the following:

From what I understand, when amino acids are in their solid form, they exist as zwitterions which is why they have such a high melting point and are so soluble in polar solvents.
Then when placed in a solution with pH lower than their isoelectric point, they act as a base and form positive ions, and when placed in a solution with pH above their isoelectric point, they act as acids and form a negatively charged ion and a water molecule.
If the above is correct, what I am confused about is what happens at the isoelectric point. In my textbook it says that the isoelectric point is 'the pH value at which the amino acid exists as a zwitterion', however I thought it also exists as a zwitterion when not in solution. Are they refering just to the case of the amino acid being in solution and ignoring the solid state?
If this is so, I still do not understand how it remains as a zwitterion at the isoelectric point. Why does it not act as both a base and an acid and accept a H+ ion onto the O- atom in the carbonyl group and donate the extra H+ ion from its amine group? Does this have something to do with the fact that then it would no longer have any charges on it so it would not be soluble in the polar solvent?

The accepted definition by IUPAC gold book is: The pH value at which the net electric charge of an elementary entity is zero. pI is a commonly used symbol for this kind-of-quantity. It should be replaced by pH(I) because it is a pH determined under that particular condition.

The solubility of an amino acid is the smallest at its isoelectric point and it easily precipitates.

For question 4 you need to be familiar with dynamic equilibirum. The zwitterion accepts and loses protons all the time, but at the isoelectric point, its concentration is the biggest among all the forms of the amino acid, that is, the species that has a protonated carboxylic group, would easier lose that proton than the zwitterion's carboxylic group would accept a proton.


Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

The answer was only $\ce{AgCN}$ will precipitate, but I don't understand why $\ce{AgI}$ wouldn't precipitate as well since there is more than enough excess $\ce{AgNO3}$ available to precipitate with both $\ce{I-}$ and $\ce{CN-}$?

Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

Assuming that $\ce{Ba(CN)2}$ and $\ce{BaI2}$ dissociate completely.
$\ce{[CN-]_i = [I-]_i =} 2\cdot10^{-10}$ molar
Neglecting any volume change of solution the initial concentration of $\ce{Ag+}$ will be
$\ce{[Ag+]_i} = \dfrac{3.5\cdot10^{-9}\pu{mol}}{0.010\pu{L}} = 3.5\cdot10^{-7}\pu{M}$
Now if both the $\ce{CN-}$ and $\ce{I-}$ are quantitatively removed then the same amount of $\ce{Ag+}$ must be removed.
$\ce{[CN-]_i + [I-]_i =} 4\cdot10^{-10}$ molar
$\ce{[Ag+]_f} = 3.5\cdot10^{-7}\pu{M} - 4\cdot10^{-10}\pu{M} \approx 3.5\cdot10^{-7}\pu{M}$
So the final concentration of $\ce{Ag+}$ is essentially the same as the initial concentration. The concentration of $\ce{Ag+}$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $\ce{CN-}$ is
$\ce{[CN-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{6.0\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{1.7e-10}$
The the final concentration of $\ce{I-}$ is
$\ce{[I-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{8.5\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{2.4e-10}$
Conclusion:
Since $\ce{[CN-]_i > [CN-]_f}$ some $\ce{AgCN}$ will ppt.
Since $\ce{[I-]_i < [I-]_f}$ no $\ce{AgI}$ will ppt.


Change of electron spin in Carbon after Hybridization? During bond formation in Carbon, one of the electrons from the 2s orbital jumps to one the empty p orbitals creating a hybrid orbital as seen in the figure below.

Now clearly one electron from the 2s orbital jumped to the 2pz orbital.
But what happened to the spin?
Before hybridization, one electron had an opposite spin, but after hybridization all electrons have the same spin.
Almost everywhere the same thing is shown.
Did the spin of the promoted electron get reversed?

Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.

The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.

The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2px, 2py and 2pz. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.

Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, but waves, that are spread out across the entire molecule. You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.

For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in this German lecture script, part A, section 4.8)

Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)


Molecules that get charged in Water I am a chemistry super noob and I was watching a MIT video where the professor states that:

"Certain molecules become charged when they come in contact with water in the natural state"

Can someone give me an example of one or more such molecules ?

It depends on the context how he meant it. I could imagine two scenarios:

Certain materials can produce ions when they dissolved in water. A typical example is table salt, NaCl. Table salt is solid form has a lattice that contains Na+ and Cl+ ions, and when you put it in water, the water molecules can help releasing these ions into the solution, where you now can find the ions. Note, that in this case the ions are atomic entities but molecules with charge are also ions, like $\ce{MnO4-}$ or $\ce{SO}_4^{2-}$.
Another scenario when the charged entities are formed as a reaction with water by e.g. giving an electron (oxidation) to water molecules or to other dissolved components e.g. oxygen in water. As you can imagine, it happens generally with very reactive materials (materials that are very sensitive to moisture), but essentially the rusting of iron is a similar reaction.


Dimethylene Cyclohexane to Octalin-1,4-Dione | Hydroboration vs. Ozonolysis Pros Cons With regards to the following synthesis, what are some pros and cons of using hydroboration oxidation followed by oxidation of the alcohols compared to ozonolysis.

More specifically, how do the two compare as to yields of the final product, cost / amount of reagents, speed of reactions, unwanted side reactions?

Here is the synthesis, along with the two ways to prepare the cyclohexadione:

Comparison of the two approaches as far as yield/cost/etc. is void because one of them does not provide the same product as the other. Hydroboration-oxidation of 1,4-dimethylenecyclohexane yields 1,4-bis(hydroxymethyl)cyclohexane with undergoes oxidation to yield 1,4-cyclohexanedicarbaldehyde or 1,4-cyclohexanedicarboxylic acid.

Now, you may be able to take the carbocylic acid, covert to acid halide and then react with $\ce{LiCuMe2}$ to get the ketone, which should undergo Baeyer-Villager oxidation followed by hydrolysis to give the alcohol you want. Or you could covert the diacid to the diamide and subject it to Hofmann degradation to get the amines, which you might be able to convert to the diazo salt with HONO, which may hydrolyze to give the desired diol. Both routes ae significantly more convoluted than ozonolysis, which decreases oveall yield and increases cost.
Now, there are other oxidative cleavage reactions which might be cheaper.


Why does copper(II) carbonate decompose into copper(II) oxide rather than copper(I) oxide when heated? I recently did a lab at school surrounding the decomposition of copper(II) carbonate into copper(II) oxide, and subsequently into pure copper, when sufficient heat is added. What I don't understand is why the copper(II) carbonate can't decompose into copper(I) oxide, rather than copper(II) oxide.

For example, copper(II) carbonate decomposing into copper(II) oxide would be:

$$\ce{CuCO3 ->CuO + CO2}$$

While copper(II) carbonate decomposing into copper(I) oxide would be:

$$\ce{4CuCO3 -> 2Cu2O + 4CO2 + O2}$$

If anyone could help me understand why copper(II) carbonate decomposes into copper(II) oxide rather than copper(I) oxide, I would greatly appreciate it.

Edit: I've been nosing around Wikipedia a bit and discovered that copper(I) oxide reacts with atmospheric moisture to become copper(II) oxide. This would explain why copper (I) oxide is not present; it has degraded into copper(II) oxide. I'm still interested in how copper(I) oxide and water react with respect to reactants and products, so if anyone knows what the equation between the products and reactants is, I would like to know.

It comes to stability of products, in this case interaction of metal ion with ligands (here oxygen). In water solutions $\ce{Cu^2+}$ is usually more stable than $\ce{Cu^1+}$, it's quite different story in non-aqueous solvents like acetonitrile. In solid: $\ce{CuCl2}$ is more stable than $\ce{CuCl}$, but iodide $\ce{CuI}$ is more stable than $\ce{CuI2}$.

The simplest explanation would be hard-soft acid-base theory: $\ce{Cu^2+}$ is a hard acid, and $\ce{Cu+}$ is a soft one.  Oxygen in copper oxide or $\ce{Cl-}$ in chloride are hard bases and $\ce{I-}$ is a soft one. The same to the same: soft acids goes to soft bases, and hard acids likes the hard bases.


Deriving an expression for dU as a function of V and T I'm currently preparing for a physical chemistry exam tomorrow, and was given the following question for practice:

  Derive an expression for $\mathrm{d}U$ as a function of $\mathrm{d}V$ and $\mathrm{d}T$.

I'm confused as to whether they want me to derive the equation below:

$$ \mathrm{d}U=\left(\dfrac{\partial U}{\partial V}\right)_T \mathrm{d}V + \left(\dfrac{\partial U}{\partial T}\right)_V \mathrm{d}T $$

or this

$$ U = q + w $$

given an infinitesimal change

$$ \mathrm{d}U = \mathrm{d}q + \mathrm{d}w $$

where

$$ \mathrm{d}q = C\mathrm{d}T $$

and

$$ \mathrm{d}w = -p\mathrm{d}V $$

hence

$$ \mathrm{d}U = C\mathrm{d}T + -p\mathrm{d}V $$

Essentially, I need help interpreting the direction I should proceed, and advice regarding the correctness of my methods.

I realize now what you're looking for:
Consider $\mathrm{d}S(V,T)$ as

$$\mathrm{d}S(V,T) = \left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T$$ and place into the first law:

$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V = T\left[\left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T\right] - p\mathrm{d}V
$$

Combine like terms of $\mathrm{d}V$:

$$
\mathrm{d}U = T \left(\frac{\partial S}{\partial T}\right)_{V} \mathrm{d}T + \left[T \left(\frac{\partial S}{\partial V}\right)_{T} - p\right]\mathrm{d}V
$$

and you have $\mathrm{d}U$ a function of $V$ and $T$.

It turns out that $\displaystyle T \left(\frac{\partial S}{\partial T}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{V} = \mathrm{C}_V$ and $\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T} = \left(\frac{\partial p}{\partial T}\right)_{V}$ by a second-derivative Maxwell relations of $A$ (below) with respect to $T$ and $V$, and a "prettier" version is:

$$
\mathrm{d}U = \mathrm{C}_V\mathrm{d}T + \left[T \left(\frac{\partial p}{\partial T}\right)_{V} - p\right]\mathrm{d}V
$$

(In the special case of an ideal gas, $pV = nRT$ and the term in brackets is zero - just write $p = nRT/V$ and evaluate.)

To derive $\mathrm{d}A$:
What you're looking for is a Legendre transform. You know

$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V
$$
from the First Law of thermodynamics.

Hence the natural variables for $U$ are $S$ and $V$, not $T$ and $V$ (which is desired). So instead of $U(S,V)$ you want $?(T,V)$ where $?$ happens to be the Helmholtz free energy $A$, sometimes denoted $F$ by physicists.

To do the transform, you take your original function - the exchange variable times its differential form, i.e.

$$ A(T,V) = U(S,V) - S\left(\frac{\partial U}{\partial S}\right)_{V}$$

You know also that $$\left(\frac{\partial U}{\partial S}\right)_{V} = T$$ because

$$\mathrm{d}U(S,V) = \left(\frac{\partial U}{\partial S}\right)_{V}\mathrm{d}S +  \left(\frac{\partial U}{\partial V}\right)_{S}\mathrm{d}V = T\mathrm{d}S - p\mathrm{d}V$$
by total derivative formula for a function, in this case $U$.

So you have $A = U - ST$ or as commonly written, $A = U - TS$ so the differential form of $A$ is $\mathrm{d}A = \mathrm{d}U - T\mathrm{d}S - S\mathrm{d}T = T\mathrm{d}S - p\mathrm{d}V -T\mathrm{d}S - S\mathrm{d}T$, thus $\mathrm{d}A = -p\mathrm{d}V - S\mathrm{d}T.$ That is a function of $V$ and $T$.


What is "super" in superphosphate? This question is inspired from a previous question(marked unclear). I don't know about the context of that question but I was intrigued by a statement in that question:

  Superphosphate is used instead of just phosphate because
  superphosphate is a compound whereas phosphate is an ion.

Now, the question is how the name "superphosphate" describe a compound? Since, it contains the suffix -ate, shouldn't it be considered an ion just like a phosphate?

Just to add some more context, I googled "superphosphate" and it gave results about it being a fertilizer, its various types and its suppliers. How come a fertilizer be named superphosphate? Is it because it is enrich in phosphorus? Or is it just a trademark name? What is the significance of the word "super"?

Searching for more, I came to know that calcium dihydrogenphosphate is also known as calcium superphosphate. Is it the same superphosphate that we are talking about? Is it the same as superoxide? Does IUPAC recommends its usage? To broaden the clarification, what is "super" even supposed to mean? Can we say in general that an acidic anion can be called super-anion?

If there are superoxide and superphosphate, are there any other ion containing the name "super" like supersulfate or supernitrate? The names seem to be too absurd/obsolete to even pronounce. Searching for "supernitrate" gave 2 results: alibaba and super calcium nitrate which is suppose to be nitrogenous ferilizers. Searching for "supersulfate" gave me results of a type of cement(One example here). So, I think that the name "super" isn't bound to ferilizers only.

To clarify my questions:

What is the significance of the name "super" in superphosphate? Is it a real chemical name or a trademark name?
Is the same as superoxide?
Are there any other ions containing the name "super"?

The term superphosphate is really old, even well before the concept of atoms was proposed by Dalton. Therefore it is difficult to rationalize the choice of this terminology.

In the unabridged version of the Oxford English Dictionary, you can see the earliest usage dates back to 1798

  Chemistry. A phosphate containing an excess of phosphoric acid; an
  acid phosphate. Now disused except in superphosphate of lime, calcium
  superphosphate: cf. sense 2.

  1798   Philos. Trans. (Royal Soc.) 88 17   It was..Scheele who discovered, that the urine of healthy persons contains superphosphate,
  or acidulous phosphate, of lime.

Further information from the OED on the usage of "super" in chemical names confirms its use since antiquity. See antique examples

  (b) Denoting the highest proportion of a component, esp. owing to a
  high oxidation state. Now chiefly archaic or hist., except in
  superoxide n.   and in the names of certain substances used in
  industry and commerce, as superphosphate n., supersulphate n. Cf. sub-
  prefix 4b(b).  (i) Categories » [1788   J. St. John tr. L. B. Guyton de Morveau
  et al. Method Chym. Nomencl. 107   New names... Acetite of lead.
  Ancient names... Sugar of lead, Super-acetated lead.] 1811   Jrnl.
  Nat. Philos. June 78   The aqua lithargyri acetati is a saturated
  solution of the proper acetate of lead,..it is an essentially
  different salt from the super-acetate of lead. 1913   Brit. Med. Jrnl.
  4 Oct. 875/1   Dr. Latham..used the superacetate of lead in
  consumption. 1979   A. J. Youngson Sci. Revol. Victorian Med. i. 18
  Acetate or superacetate of lead combined with opium was prescribed for
  haemorrhage of the lungs.


No indicator is suitable when titrating weak acid with weak alkali This was a statement in my book:

A solution of a weak acid cannot be titrated with a weak base using an indicator to find
the end-point because the pH change is too gradual close to the equivalence point.

Could someone please elaborate the bold text.

An indicator works when the solution's pH is in its range. It's not necessarily the equivalence point.  The point is called the end point where it actually changes color.

If the difference in volume is small between the equivalence and end points, then the indicator can be used.  Since a gradual change implies more volume must be added to obtain the same pH change in other acid base titration, the difference between equivalence point and end point(if you use indicator) will be more and therefore indicator will not be suitable for it.


Is a triplet state or singlet state more stable? Is a triplet state or singlet state more stable? How can an atom have a long lifetime in a triplet state even though two electrons have the same spin?

The key issue here that we are comparing excited states. So let's begin by defining Singlet and Triplet Excited States.
Singlet and Triplet Excited States:  A singlet or a triplet can form when one electron is excited to a higher energy level.  In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired).  In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in the illustrative Figure below. Singlet, doublet and triplet is derived using the equation for multiplicity, $2S+1$, where $S$ is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up ($s = +1/2$) or spin down ($s = -1/2$).  If we were to calculated the S for the excited singlet state, the equation would be $2[(+1/2) + (-1/2)]+1 = 2(0)+1 = 1$, therefore making the center orbital in the figure a singlet state.  If the spin multiplicity for the excited triplet state was calculated, we obtain $2[(+1/2) + (+1/2)]+1 = 2(1)+1 =3$, which gives a triplet state as expected.

Illustrative Figure: Spin in the ground and excited states The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. ***For this reason, the lifetime of the triplet state is longer than the singlet state by approximately 104 seconds fold difference.***The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.


Laboratory and commercial preparations of aniline other than reduction of nitroarenes Till now I could only find one laboratory preparation of aniline, i.e. reducing nitrobenzene with $\ce{Sn}$ and concentrated $\ce{HCl}$. I'm confused if Hofmann degradation is only commercial.
I'd like to know if there are other processes to prepare aniline in laboratory.

It is true that until recently, the syntheses of primary anilines mostly rely on the reduction of corresponding nitroarenes. However, recent publications allow the synthesis of primary anilines or their derivatives in laboratory.
One such method is Buchwald–Hartwig amination:

The palladium-catalyzed cross-coupling of amines and aryl (pseudo)halides, now commonly known as the Buchwald–Hartwig amination, was first reported 25 years ago. Since the simultaneous breakthrough reports of Buchwald and Hartwig in 1995, this reaction has transformed the way synthetic chemists think about synthesizing aromatic amines... (Ref.1):

A continuous process for Buchwald–Hartwig amination in different scale synthesis has also been introduced recently (Ref.2). Further, a version of Buchwald–Hartwig amination for the preparation of primary arylamines has been introduced recently as a general method (Ref.3):

The importance of the method is reflected by the development of a synthetic route to manufacture the drug candidate AMG 925 on kilogram scale using Buchwald–Hartwig amination (Ref.4):

Another amine synthetic method not following nitro reduction is Ullmann amine synthesis and its variations (Ref.5 and references therein):

In addition, secondary or tertiary amine can be prepared using Chan-Lam coupling reaction. For example see following reaction (Ref.6):

Other than using aryl halides, there is an another method to convert phenol derivatives to anilines. One such example describes a direct conversion of phenols into primary anilines with hydrazine catalyzed by palladium (Ref.7):

References:

Paola A. Forero-Cortés, Alexander M. Haydl, "The 25th Anniversary of the Buchwald–Hartwig Amination: Development, Applications, and Outlook," Org. Process Res. Dev. 2019, 23(8), 1478–1483 (https://doi.org/10.1021/acs.oprd.9b00161).
Sebastian Falß, Giovanna Tomaiuolo, Antonio Perazzo, Paul Hodgson, Polina Yaseneva, Jacek Zakrzewski, Stefano Guido, Alexei Lapkin, Robert Woodward, Rebecca E. Meadows, "A Continuous Process for Buchwald–Hartwig Amination at Micro-, Lab-, and Mesoscale Using a Novel Reactor Concept," Org. Process Res. Dev. 2016, 20(2), 558–567 (https://doi.org/10.1021/acs.oprd.5b00350).
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New relationship between exponential factors and entropy of a reaction? Consider the reaction $\ce{A <=> B}$. Let $k_f$, $E_A$, and $A$ denote the forwards rate constant, activation energy, and pre exponential factor, respectively. Let $k_r$, $E_A'$, and $A'$ denote the backwards rate constant, activation energy, and pre exponential factor, respectively. I seem to have determined the following relationship:

$\ln{\frac{A}{A'}} = \frac{\Delta S}{R}$

First, is this a correct equation? Second, is this a known equation? I have not seen it in any of the textbooks that I have read. Third, is this a useful equation? I imagine it could be used to determine the ratio between the pre exponential factors from the entropy.

My derivation is below:
The equilibrium constant $K$ is equal the ratio of the forward and backwards rate constants $\frac{k_f}{k_r}$, so $\ln{(K)} = \ln{\left(\frac{k_f}{k_r}\right)}$. Furthermore, the Arrhenius equation expresses forward and backwards rate constants as $k_f = Ae^{-E_a/RT}$ and $k_r = A'e^{-E_a'/RT}$, where $A$ is the pre-exponential factor of the forward reaction and $A'$ is the pre-exponential factor of the reverse reaction. Substituting these expressions yields $\ln{(K)} = \ln{\left(\frac{Ae^{-E_a/RT}}{A'e^{-E_a'/RT}}\right)} = \ln{\frac{A}{A'}}-\frac{(E_a-E_a')}{RT}$.
In Chapter 5 of Atkins' Chemical Principles 7th edition, the difference in activation energies of the forwards and backwards reactions $E_a-E_a'$ is equated to the reaction enthalpy $\Delta H$. Applying this relationship, we obtain: $\ln{(K)} = \ln{\frac{A}{A'}}-\frac{\Delta H}{RT}$.
It is also known, that since $\Delta G = -RT\ln{K} = \Delta H - T\Delta S$, $\ln{(K)} = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$.
Combining these two equations yields: $\frac{\Delta S}{R} = \ln{\frac{A}{A'}}$.

Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation:
$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$


Is 4-methylchlorobenzene an acceptable IUPAC name for this compound? The answer given for this is A and B
My doubt is if toluene is an accepted IUPAC name then why not chlorobenzene?(If that is the reason why the answer is given wrong)

Or else is there any other reason why the option C is wrong?

In the generation of a preferred IUPAC name (PIN), the locant '1' may only be omitted under certain circumstances, as stipulated by Rule P-14.3.4.2 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

  (c) in monosubstituted homogeneous monocyclic rings, e.g. cyclohexanethiol (PIN) or bromobenzene (PIN) ...

Since 1-chloro-4-methylbenzene is not a monosubstituted ring, it does not fall under this category. Therefore, in the preferred IUPAC name, the locant '1' must be cited to avoid any ambiguity.

The omission of the locant '1' when there is no ambiguity is widespread, and is permissible within general nomenclature (refer to Loong's answer here). In this case, the omission of the locant '1' from the name "4-methyl-1-chlorobenzene" is likely to be unambiguous. However, note that the name "4-methyl-1-chlorobenzene" is not in accord with IUPAC nomenclature to begin with. Therefore, the abbreviated name "4-methylchlorobenzene" is probably not acceptable, either.

  P-14.5.1 Simple prefixes (i.e. those describing atoms and unsubstituted substituents) are arranged alphabetically...

Since 'chloro' comes before 'methyl' in alphabetical order, the order of citation of substituents must be chloro followed by methyl; hence the correct name must be of the form x-chloro-y-methylbenzene.