Nitration of tri-substituted benzene with Acetic acid and HNO3 So I looked it up on the internet and it says that this is to be a nitration reaction. But how is the $\ce{NO2+}$ formed? I simply cant seem to be able to do the first step of the mechanism, that is the generation of the electrophile.

I thought that as $\ce{HNO3}$ would be the stronger acid, it would lose $\ce{H+}$ that would get added to the acetic acid and it would lose water to form a the electrophile that gets added to the ring. But apparently this is not the case. Why? Could someone show the mechanism?

Your reaction goes way back to 1930s when Ingold and coworkers have done pioneering work on  nitration of aromatic compounds using nitric acid in organic solvents in place of sulfuric acid (vide infra). Let's see how this has been developed:

Historical perspectives: The earliest recorded experiments on the kinetics of aromatic nitration has been performed in 1884, when the nitration of benzene and some of its simpler derivatives with various amounts of nitric acid have been examined at various temperatures (Ref.1). No solvent has been used (just nitric acid), and thus, the reaction mixtures were heterogeneous. The first nitration of aromatic substances with nitric acid in sulfuric acid was obtained in 1904, when the velocity constants in the nitration was studied (Ref.1). In their punlication in 1938, Ingold and coworkers have reported the kinetics of nitration using nitric acid in organic solvents (other than sulfuric acid), specifically nitromethane as the solvent (Ref.1). In this studies with nitromethane as the solvent and nitric acid in constant excess, they demonstrated zeroth order kinetics for the nitration of benzene and activated aromatic nucleous such as toluene. They have observed that, when less reactive compounds such as chlorobenzene were used, there were some dependence of the rate on the compound. Hence, they have continued these experiments with considerably less reactive compounds such as 1,4-dichlorobenzene and ethyl benzoate to show that less reactive compounds have established first-order kinetics on nitration.

Originally, Benford and Ingold thought that their results were peculiar to nitromethane, yet they have observed ail the same phenomena with acetic acid as solvent (your case!). Based on these data, Ingold and coworkers proposed the formation of nitronium ion ($\ce{NO2+}$), which may be the rate-determining if it is sufficiently rapidly removed by the aromatic compound used as the substrate. They proposed the actual mechanism of the formation of nitronium ion as follows (Ref.2):

$$\text{In sulfuric acid:} \: \: \ce{HNO3 + H2SO4 <=> H2NO3+ + HSO4-}\\
\text{and} \: \: \ce{H2NO3+ + H2SO4 <=> NO2+ + H3O+ + HSO4-}$$

$$\text{In organic solvents:} \: \: \ce{2 HNO3 <=> H2NO3+ + NO3-}\\
\text{and} \: \: \ce{H2NO3+ <=> NO2+ + H2O}$$

This mechanism is well established now. Based on these fact, I hope I answered your question. In their further work, Ingold and coworkers have proposed the mechanism for nitration of aromatics based on the nitronium ion formation (Ref.3). In that publication, they proposed existence of nitronium ion through rapid equilibrium with concentrated nitric acid, and that mechanism is believed to be the case to date.

References:

Gilbert A. Benford, Christopher K. Ingold, “177. Influence of directing groups on nuclear reactivity in oriented aromatic substitutions. Part V. A dilatometric investigation of the kinetics of aromatic nitration especially in nitromethane solution,” Journal of the Chemical Society 1938, 929-955 (DOI: 10.1039/JR9380000929).
E. D. Hughes, C. K. Ingold, R. I. Reed, “Kinetics of Aromatic Nitration: the Nitronium Ion,” Nature 1946, 158(4013), 448-449 (https://doi.org/10.1038/158448c0).
R. J. Gillespie, E. D. Hughes, C. K. Ingold, D. J. Millen, R. I. Reed, “Kinetics and mechanism of aromatic nitration,” Nature 1949, 163(4146), 599-600 (https://doi.org/10.1038/163599b0).


Nature of bond in oxyacids of phosphorus What is the nature of the bonds between P and O in oxoacids of phosphorous?

Is it a double bond or coordinate bond?

Wikipedia says that it is "double bond" but many times I encountered "coordinate bond".

In general, I'm not sure double bond and coordinate covalent bond are mutually exclusive labels and depending on the specific chemical you're looking at, they might both apply.

My Reasoning:

Coordinate bonds are bonds where both electrons were donated by one atom. Usually, the concept of a double bond is used in the context of valence bond theory and coordinate bond is used in the context of crystal field theory. There is evidence for pi bonding in coordinate complexes (i.e. back bonding) leading to what could be called a double bond in valence bond theory.
If you look at bond length data you can usually determine if there is a double bond. (Or if you have some of the acid and access to an IR, you can do this yourself.) The CRC should contain reference bond lengths.

I 'm not an expert in phosphorous chemistry, but these could be helpful:

Structural calculations for many compounds, including a few acids
Study on structures of phosphates, not acids


What is the magnitude of the current in an electrolyte? if a battery is connected to an electrolytic cell, what is the magnitude of the current in the electrolyte compared to the the magnitude of the current in the wire? This doesn't make sense to me because there are cations flowing in one direction and anions flowing in the other in electrolytes, so is the current twice the magnitude of the current in the wire? does ohm's law still hold? Is current not uniform in the circuit? I hope somebody can help. Thanks!

Edit: Sorry for asking an off-topic question. Thank you so much for the answers and comments!

By Kirchoff's Current Law ("what goes in must come out") the current flowing in total in all the branches of a circuit must be equal at all points. Hence if A amperes are flowing in the wire, as no parallel branches are present we conclude that A amperes must also be flowing in the electrolyte.


How are electronic configurations determined experimentally? Despite the ubiquitous usefulness of electronic configurations (of elements) in explaining their behavious I have not been able to find out exactly how the configurations are determined.

Is there any sure-fire technique to detect them (like x-ray diffraction for crystal structure) or is it based on guesswork, through observation of the various properties and reactions of the elements?

There are lots of techniques you could use. Most useful would be X-ray absorption techniques such as XANES or EXAFS. X-ray emission techniques such as XPF or EDS. You could look at electron emission spectroscopies (ESCA) such as XPS or UPS or Auger techniques and EELS spectroscopy, ELNES and EXELFS.

The important point to remember is that a small perturbation to the chemical environment e.g. by a change in oxidation state or coordination number, etc. will change the signals observed, this can lead to much more than electron configurations.

Each has its own benefit and speciality, depending on the sample and the exact thing you are looking for but that should get you going :)


Where can I find the standard molar entropy for ammonium bicarbonate? I'd like to know the standard molar entropy S0solid for ammonium bicarbonate to use in an example in class, to calculate ΔS0 for the reaction, NH4HCO3(s) → NH3(g) + H2O(g) + CO2(g).

Already checked: Wikipedia, NIST Chemistry WebBook, Google. I don't have access to the CRC handbook.

(Arthur and Chester Miller already found the standard enthalpy of formation for ammonium bicarbonate, in this forum, in 2016; and it's in the German Wikipedia.)

“Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. gives values for the standard molar enthalpy of formation $\Delta_\mathrm fH^\circ$, the standard molar Gibbs energy of formation $\Delta_\mathrm fG^\circ$, and the standard molar entropy $S^\circ$ (all at $T=298.15\ \mathrm K$ and $p=100\ \mathrm{kPa}=1\ \mathrm{bar}$) of crystalline ammonium hydrogen carbonate as follows.
$$\begin{align}
\Delta_\mathrm fH^\circ&=-849.4\ \mathrm{kJ/mol}\\
\Delta_\mathrm fG^\circ&=-665.9\ \mathrm{kJ/mol}\\
S^\circ&=120.9\ \mathrm{J/(mol\ K)}
\end{align}$$


Confused about identifying delocalized electron pairs in Isoniazid I have to identify delocalized electron pairs in Isoniazid (pairs not shown in the image below):

I know the nitrogen in the ring has a localized electron pair, since it already forms a pi bond. I'm confused about the other two nitrogen atoms. When we draw one of the resonance structures, we can see the amide nitrogen has a delocalized electron pair:

And if we push the lone pair from the amine nitrogen, we get another resonance structure (even if it is insignificant, due to all of the formal charges, it still seems to be a valid resonance structure):

Because these two lone pairs engage in resonance, I thought that they were both delocalized. But my textbook (Klein's Organic Chemistry, section 2.12, 2nd Ed.) says only one pair is delocalized:

Therefore, if the electron pairs on both the amine and amide nitrogens in Isoniazid seem to engage in resonance, why is only one pair delocalized?

The reason why the amine lone pair is delocalized is that to make the last conformer you proposed, the nitrogen closest to the oxygen would have to form five bonds. Nitrogen is a second-row element that cannot readily possess 10 electrons from those 5 bonds, even if nitrogen has to possess those 10 electrons for the very short time required for the electrons to switch conformations between different resonance conformers.


How does one find the ground-state term symbol for a configuration that is exactly half-filled? For instance, the ground-state configuration of N atom is a $p^3$ configuration of all parallel spins and one electron in each $2p$ orbital, which has:

Total spin angular momentum, $S = 3 * \frac{1}{2} = \frac{3}{2}$
Spin multiplicity, $2S + 1 = 2(\frac{3}{2}) + 1 = 4$
Total orbital angular momentum, $L = (-1) + (0) + (1) = 0$

and so, $J = {|L - S|, |L - S + 1|, . . . , |L + S - 1|, |L + S|} = {\frac{1}{2}, \frac{3}{2}}$. Thus, I've found that the two candidates are ${}^{4} S_{1/2}$ and ${}^{4} S_{3/2}$.

But my textbook (Pilar, from the '60s!) says that the ground state is ${}^{4} S_{3/2}$. How could I figure something like this out?

Hund's Rules from my text (as far as I can tell---this text is hard for me to read) don't specify how we treat exactly half-filled subshells, only less-than-half-filled, or more-than-half-filled, assuming equal $L_{max}$ and $S_{max}$ already.

(As a side note, my usual McQuarrie textbook is not with me this week, so I cannot simply refer to McQuarrie.)

The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number $s = 1/2$ and magnetic quantum number  $m_s = \pm 1/2$. Orbitals have angular momentum of $s=0, p=1, d=2, f=3$ etc.

The total spin angular momentum is the series of values
$$S = |s_1 + s_2|\cdots|s_1 − s_2|$$
where $s_1 , s_2 , s_3 =+1/2$  are the quantum numbers of each electron. The total angular momentum is
$$ L=|l_1+l_2|\cdots|l_1−l_2|$$
and total angular momentum
$$J=|L+S|\cdots|L−S|$$
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by $1$.

The term symbol has the form $^{2S+1}L_J$. The super-prefix is the spin multiplicity, for spin angular momentum S this is $2S+1$ or in general $2X+1$ for angular momentum X.

When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.

For the spin the S values are
$S = |s_1 + s_2|\cdots|s_1 − s_2| = |1/2+1/2|\cdots|1/2-1/2| = 1, ~ 0 $ .
We call these values $S_1$ and $ S_0$
To calculate the total with the third electron gives
$$S_{s_1}=|S_1+1/2|\cdots|S_1 - 1/2| = 3/2 , ~ 1/2$$
and for the $S_0$ there is one value $S_0=1/2$. Thus for the spin the values are $3/2$ and $1/2$. This means that the multiplicity produces quartet and doublet states, thus so far we have $^4L_J$ and $^2L_J$.

The same method is followed for the orbital angular momentum. The first two p orbitals give
$$ L=|l_1+l_2|\cdots|l_1−l_2|= |1+1|\cdots|1-1| = 2,~1,~0 $$
then for each of these values called $L_1,L_2,L_3$ we combine with the last orbital angular momentum, for example
$$ L_{L_1}=|L_1+l_1|\cdots|L_1−l_2|= |2+1|\cdots|2-1| = 3,~2,~1 $$
and so forth for the other values. The L values are $3,2,1,0$

Putting all these values into a table gives
$$\begin{matrix} L & S &  J &  \text{term symbol}\\
3 & 3/2 & 9/2, 7/2, 5/2, 3/2 & ^4F_{9/2, 7/2, 5/2, 3/2} \\
2 & 3/2 & 7/2, 5/2, 3/2,1/2 & ^4D_{ 7/2, 5/2, 3/2,1/2}\\
1 & 3/2 & 5/2, 3/2,1/2 & ^4P_{5/2, 3/2,1/2}\\
0 & 3/2 & 3/2 & ^4S_{3/2}\\
3 & 1/2 & 7/2, 5/2 & ^2F_{7/2,5/2}\\
2 & 1/2 & 5/2, 3/2 &^2D_{5/2, 3/2}\\
1 & 1/2 & 3/2, 1/2 & ^2P_{3/2, 1/2}\\
0 & 1/2 & 1/2 & ^2S_{1/2}\\
\end{matrix}$$
Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are $^4S$ which is lowest in energy and next $^2D$ then $^2P$


Is osmosis due to osmotic pressure? Okay, I am getting real confused on how water flows from a hypotonic solution to a hypertonic solution through a semipermeable membrane. It is known that hypertonic solutions have higher osmotic pressure than hypotonic ones, so shouldn't it be the opposite?
Is osmosis even related to osmotic pressure or is it just because of difference in concentration?

Osmotic pressure is the additional external pressure that needs to be applied to a solution to compensate its osmotic flow to a second solution, so that there is not net flow between them.

Let's imagine a tube with a piston in each extreme, so we can control the pressure we apply to both sides, with a semipermeable membrane inside it. If the concentration of solutes on both sides is different, there will be an osmotic flow from the hypotonic side to the hypertonic side. To compensate for this osmotic flow, we will need to apply more pressure to the hypertonic side than to the hypotonic side - and in mechanical equilibrium, this additional pressure is equal to the difference in osmotic pressure between the solutions.

Alternatively, if we want to get rid of external pressures, osmotic pressure can be understood not as a pressure exerted on a solution, but by the solution with a certain osmotic pressure on neighbouring solutions with which it can exchange solvent. So, the hypertonic solution exerts more pressure on the hypotonic solution than viceversa, and the result is that solvent moves from the hypotonic solution (which receives more osmotic pressure) to the hypertonic solution (which receives less osmotic pressure) until they become isotonic, just as would happen if it exerted more hydrostatic pressure. This is mathematically equivalent to the original definition.


Why do some gases have lower value of Z for a particular pressure? In the above graph,the minima of the curve for methane is more than that of nitrogen. Also, for a given value of pressure, the value of $Z$ for methane is less than that of nitrogen. They seem to meet at high pressure. Why do some gases have lower value of Z?

The same thing is observed if you compare methane and carbon dioxide. But they don't seem to meet at high pressure. This is because the $Z$ for methane increases at a faster rate. Why does Z increase faster for some gases?

The real gas equation is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$$
I tried expressing $Z$ in terms of $P$ and $T$ by eliminating $V$.

$$V=\frac{ZRT}{P}$$
$$\left(P+\frac{aP^2}{Z^2R^2T^2}\right)\left(\frac{ZRT}{P}-b\right)=RT$$
$$ZRT-Pb+\frac{aP}{ZRT}-\frac{aP^2b}{Z^2R^2T^2}=RT$$
Multiplying by $Z^2$ on both sides,
$$Z^3RT-Z^2(Pb-RT)+\frac{ZaP}{RT}-\frac{aP^2b}{R^2T^2}=0$$
$$f(Z,P)=0$$
(I am assuming constant temperature)

Should I find $\frac{\partial f(Z,P)}{\partial P}$ and make it equal to zero to get information about the minima?

The trend you've noticed here is a very interesting way of studying the interactions of a gas with itself.

You are certainly right about how to find the minimum of these curves, but the important thing here is what that minimum actually means.

$$Z=\frac{PV}{RT}$$By analyzing this, we expect that $Z$ will tell us something useful about the way the volume of a gas changes as pressure changes (holding $T$ constant just gives better information on a graph). Specifically, we see that if pressure increases but volume decreases more, we will have a number less than 1 and if volume decreases less than pressure increases, we will have a number greater than 1. So what does that number tell us?

Well, if pressure increases and the volume gets even smaller than we would expect for a real gas, these gases must be attracted to each other. Then, we can interpret the minimum of this graph as being the pressure (for a given temperature) at which a gas molecule is attracted to itself most strongly. Beyond this point, repulsive interactions become more and more important. Notice that there is also no maximum. Or, there is no point where attractive interactions dominate repulsive interactions again which is what we would expect.

Now, to explain why we see the behaviors we do from certain molecules, we have to imagine what is causing there to be attractive and repulsive interactions at all. This is due to polarizability and the amount of van der waal's forces.

Indeed, if we look up value of the parameter $a$ for the molecules present (I have them in a textbook) we find:$$a_{\ce{H2}}=.24646$$$$a_{N_2}=1.3661$$$$a_{\ce{CH4}}=2.3026$$$$a_{\ce{CO2}}=3.6551$$where $a$ is a measure of how attracted the molecules are to themselves (in the horrible unit of $\frac{dm^6*bar}{mol^2}$).

These values do indeed follow the trend we see in the graph above. That is, larger value for $a$ has a larger attractive deviation initially. $\ce{H2}$ is something of a special case because it is effectivly unpolarizable, so we see a linear increase in the compressibility factor with pressure. This basically means that hydrogen will deviate from the ideal gas precisely the same at all pressures.

My best guess at why the slopes then increase at different rates has to do with the size of the actual molecule. Again, looking up the values for $b$ which represents the excluded volume of a molecule, we have:$$b_{\ce{H2}}=.026665$$$$b_{N_2}=.038577$$$$b_{\ce{CH4}}=.043067$$$$b_{\ce{CO2}}=.042816$$with $b$ in units of $\frac{dm^3}{mol}$.

So, this predicts methane will have a slope increasing more quickly than both nitrogen and carbon dioxide which is what we see and that hydrogen will have the shallowest slope which is again what we see.

Hope that helps.


Describe the function of a salt bridge, What would happen if the half-cell compartments of a galvanic cell were not connected? I know,  The primary function of a salt bridge is to complete the circuit, so that charge can flow from one half-cell to the other. The second function is to balance the mass by allowing the anions to move to the half-cell where addition cations are being produced.But I can't seem to figure out what would happen if the half-cell compartments of a galvanic cell weren't connected?

My electrochemical knowledge is a wee bit rusty, but I believe the salt bridge helps balance the charges and not mass.

What charge you ask? So, let imagine you start up your galvanic cell (below). Electrons from the Zn move to Cu electrode. You start to build up a negative charge on the Cu which would resist/repel further addition of electrons after a period of time, hence production of electricity ceases.

By having a salt bridge, the electrolyte ions can move from one side to the other to mitigate this issue.


Can carbon dioxide be reduced to carbon monoxide and oxygen to produce energy? There are lots of questions about reducing or burning CO2 to carbon and oxygen to solve climate change, but of course that wouldn't work because it takes a lot of energy. But carbon monoxide is more stable than the dioxide, so could CO2 be split into CO and oxygen to create more heat? Of course this is a bad idea since it'd produce a toxic gas but is it at least theoretically possible?
Apologies if I'm misusing chemistry terms or concepts; please inform me in the comments section.

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction:
$$\ce{CO2(g) -> CO(g) + O(g)}$$
This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the row for carbon dioxide that this reaction has an enthalpy change of $\mathrm{+532\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. The proposed reaction is therefore heavily endothermic. That is to say, it must absorb energy. Interestingly, it's true that the extreme strength of the bond in carbon monoxide has a measurable effect, making this process more favourable than expected. However, it is still overall extremely unfavourable, and therefore requires a large input of energy. I stress that this is unavoidable, no matter how fancy your machine - if the end result is the reaction stated above, then you must pay the energy cost somehow.
Part of the problem, though, is that we have monoatomic oxygen as a product, which is a very reactive, high energy species - it doesn't actually exist except in special conditions. A simple adjustment therefore is to have molecular dioxygen, $\ce{O2}$ (the kind in the atmosphere that you breathe). The reaction then becomes:
$$\ce{2 CO2(g) -> 2 CO(g) + O2(g)}$$
So what's the enthalpy change associated with this reaction? Looking up another table, this turns out to be $\mathrm{+283\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. Again, this reaction is endothermic, though much less so than the first one. Regardless, once more this reaction is an energy sink.
If you want more visceral confirmation of this fact, consider the following. It is well known that pure carbon monoxide burns in an oxygen atmosphere. The reaction is self-sustaining and releases considerable heat. If you pay close attention, the reaction in the video is the exact inverse of the second equation. By chemical thermodynamics, if the combustion of $\ce{CO}$ to $\ce{CO2}$ releases heat, then it is necessarily true that cleaving $\ce{CO2}$ to form $\ce{CO}$ and $\ce{O2}$ will consume energy.
As a last point, there are ways to make the production of $\ce{CO}$ from $\ce{CO2}$ feasible, but it requires changing the products. For example, if hydrogen gas is used as a reagent, the following becomes possible:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O(g)}$$
The enthalpy change for this reaction is $\mathrm{+41\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$ , which is still endothermic, but approaching the break-even point. This is not too surprising, as hydrogen gas can behave as a reductant, and the bonds in water molecules are strong, pushing the reaction forwards. Let us make one last tiny modification:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O\color{red}{(l)}}$$
By assuming the water produced is in the liquid state as opposed to a gas, the reaction surrenders a little more energy, and the calculated reaction enthalpy becomes $\mathrm{-3\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. This reaction is very mildly exothermic, which is to say it releases heat (admittedly, so little that it's within margin of error, and slightly different conditions could make the reaction overall endothermic).
If you're not dead-set on having carbon monoxide as a product, then there are further options still. For example, here is the complete reduction of $\ce{CO2}$ to methane ($\ce{CH4}$), a considerably exothermic process with a reaction enthalpy of $\mathrm{-253\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$:
$$\ce{CO2(g) + 4H2(g) -> CH4(g) + 2 H2O(l)}$$
Methane is not an ideal product, as it too is a greenhouse gas and is a low value chemical feedstock due to its abundance and relative lack of useful chemistry. There is much more interest in conversion of $\ce{CO2}$ to compounds such as methanol $\ce{CH3OH}$ and formic acid $\ce{HCOOH}$. These two particular reactions are also exothermic.
There are several issues with using hydrogen reduction of $\ce{CO2}$ as a carbon capture strategy to combat climate change, but perhaps the main one is a real-world factor: most hydrogen we currently produce is derived from fossil fuels, notably from the partial combustion of fossil methane (natural gas) with water at high temperatures, known as steam reforming. Therefore, while alternative sources of hydrogen gas using renewable low-carbon intensity energy are not available, this is a poor strategy to remove anthropogenic $\ce{CO2}$ from the atmosphere.


How does spin flipping of triplet carbenes occur? Below is presented a page from Clayden Organic Chemistry 2ed.

How and why does the spin flipping during the collision with solvent molecules actually occur?

In these reactions you are not observing optical transitions between singlet and triplets, but the interconversion of a triplet to singlet state in competition with rotation of part of the molecule. Only in the singlet can a bond be formed as now each electron has a unique set of quantum numbers. As no photons are involved it is necessary for the triplet and singlet levels to become effectively equal in energy before a transition can occur. Thus there has to some kind of perturbation for this to happen and this has to accommodate one unit of angular momentum change, the difference between singlets and triplets. It can be useful to visualise singlet & triplets with a vector model. This is shown below. The arrows show the spin vectors of the two electrons in a triplet and singlet state.

The triplet state corresponds to the three images on the left and the singlet to that on the right. If you look at the triplet with $m_s=0$ you will see that to convert it to a singlet all that is needed is a re-phasing of the spins from parallel to anti-parallel.  In the diradicals you study, the electrons are rather distant from one another thus their interaction is weak, and the difference in energy between singlet and triplet small. This means that only a small perturbation can cause a crossing between levels. This can, for example, be produced by an external magnetic field which raises and lowers different triplet energy sub-levels and so reach the singlet's energy. A nearby solvent molecule can produce this energy perturbation as it also has electrons and it will be diffusing by rotation and translational motion, i.e. producing a time and spatially varying magnetic field. Similarly the diradical is rotating with whole body motion as are segments of it and in doing so presents different orientation of the electrons to the earth's magnetic field which may also induce changes in energy levels and triplet-singlet crossing. These interactions will also have to involve orbital angular momentum.The time scale for rotation will be sub-nanosecond (in most normal mobile solvents) but only occasionally will a perturbation occur at the same time as the groups are proximal, because segmental rotational diffusion will mainly keep them apart, and the low frequency of perturbations of sufficient magnitude and the geometric requirement just mentioned are likely to make the reactions slow.


How does one prove Avogadro's law from Gay-Lussac's? To understand Gay-Lussac's law, Amedeo Avogadro said

In equal volumes of air under constant temperature and pressure, there are equal number of atoms (or molecules).

Let's take a hydrogen and an oxygen molecule. As the latter is larger in size, when stored in two separate equally-volumed containers, doesn't oxygen increase the pressure inside that container, if the same number of molecules occupied each container? (Does this make Avogadro's statement of constant pressure go wrong?) Again,
$$2mvN\cos\theta = F,$$
where $F$ is the force acting on the walls of the container, $N$ is the number of collisions of the molecules with the walls, $2mv$ is the change of momentum of the molecules, and $\theta$ is the angle of collision of the molecules on the walls.
Therefore, if Avogadro was correct, could anyone give me any reason why $N_\ce{H2} > N_\ce{O2}$ or $v_\ce{H2} > v_\ce{O2}?$
If Avogadro was correct, the above conditions must hold, as $m_\ce{H2} < m_\ce{O2},$ right?
Could you please prove the above phenomena happen when Avogadro's statement above holds?

His name was Amedeo Avogadro, not Amidio nor Amadeo nor Avegadro nor Avagadro. The full name was Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto.
Avogadro's law states that

equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

More exactly, it also includes single atoms, e.g. for a case of noble gases, due the way how IUPAC defines a molecule.
But remember the Avogadro law implies ideal gases, similarly as the Gay-Lussac's law,Boyle's law and Charles' law. For ideal gases, sizes of molecules play no role. Both the size and volume of ideal gas molecules are considered negligible.
Real gases have deviations from this ideal behavior. See  van der Waals equation as the most used real gas state equation.

$$\left(p+a \frac{n^2}{V^2}\right)(V-n b)=nRT$$
$$\left(p+ \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b)=RT$$

where parameters $a$ (addressing cohesive forces ) and $b$ (addresing non zero molecular volume) are specific for the particular gas.
Then the molar amount of a real gas for given $V$, $p$ and $T$ is the solution of the cubic equation for  variable $n$ :
$$pV - n ( pb + RT ) + a \frac{n^2}{V} - ab \frac{n^3}{V^2}  = 0 $$
It clearly shows that the same volumes of different real gases, at the same temperature and pressure, contain ( usually slightly ) different counts of molecules/atoms.


the chemical mechanism for deviations from Raoult's law Mixtures of ethanol and acetone have vapor pressures that are higher than predicted by Raoult's law, while mixtures of acetone and water have lower vapor pressures than predicted by Raoult's law.  Why is this?

That is, why is there a "positive deviation" from Raoult's law for acetone/ethanol solutions but a "negative deviation" for acetone/water?

Isn't an acetone-ethanol hydrogen bond stronger than an acetone-acetone hydrogen bond? If the reason for the positive deviation of the ethanol/acetone solution is that the H bonds there are weaker than ethanol-ethanol H bonds, then there wouldn't that imply that acetone/water solutions would also have a positive deviation, as water-water H bonds must be much stronger than water-acetone, right?

So is there some other explanation?

First some general comments about Raoults law before discussing the particular solutions in question.

Raoult’s law suggests that the partial pressure of each substance above a solution is proportional to its mole fraction x, thus $p=p^ox$ where $p^o$ is the vapour pressure of the pure substance.

Experimentally there are deviations from Raoult’s law and they can be in both the ‘positive’ and ‘negative’ direction. A positive deviation means that p is greater than expected from Raoult’s law and thus a negative deviation has pressure that is less than expected. Alcohol/water mixtures show positive deviations and for example acetone/chloroform negative deviations.

The reason for these deviations is due to the different interaction between the molecules. The change in interaction on adding one liquid to another is due to the difference in interaction energies between them, thus if the energies  between the same type of molecules of type 1 and 2 is $E_{11}$ and $E_{22}$ (which can be positive or negative) we can write the difference in energy as
$$ \Delta E = c(E_{12} - E_{11}/2 - E_{22}/2)$$
where c is a constant to account or the fact that molecule of type 1 and 2 are different.  A positive value of $\Delta E$ indicates that the interaction between types 1 and 2 is less attractive than either molecule to itself and a negative value that types 1 and types 2 will prefer to interact with the other type of molecules than themselves.

All atoms have dispersion interaction (induced dipole-induced-dipole) due to their polarisability, and in addition may have permanent dipoles, (causing dipole-dipole  and induced-dipole energies) and hydrogen bond interactions. Clearly which contribute depends on the particular case being studied.

Starting with the chemical potential and the interaction energy it is possible to calculate the partial pressure as a function of mole fraction and energy $\Delta E$ and this is

$$ \frac{p}{p^o}=x\exp(+(1-x)^2\Delta E^o)$$

where $\Delta E^o = \Delta E/kT$ where k is Boltzmann’s constant and T temperature. This function is shown in the figure for different values of $\Delta E^o$. When this is positive there is a positive deviation from Raoult’s law, a $\Delta E^o = 1$ approximately describes the ethanol/ water data and $\Delta E^o = -0.5$ the acetone/chloroform data. (The maximum $\Delta E^o = 2$ since above this phase separation occurs)

figure: calculated pressure vs molec fraction x. Positive deviation $\Delta E^o > 0$ and negative deviation $\Delta E^o < 0$. Raoult's law $\Delta E^o = 0$

Now for the particular solutions mentioned, ethanol-acetone has a positive deviation and thus $\Delta E^o > 0$. It is not really possible without some (spectroscopic) experimental data to say what interaction causes this; most likely it is slight changes of each type of interaction, but probably mainly changes in dipole-dipole and hydrogen bonding. The data indicates that acetone-ethanol interactions are less favourable than either molecule with itself.

In the acetone - water data the interaction between these two types of molecules is more favourable in such a way as to overcome the sum of water-water and acetone-acetone interaction energies. This is somewhat unusual for water as it is a highly associated (i.e. hydrogen bonded) liquid and a second component tends to break up this association.

We must conclude, however, that the acetone is able to become solvated in such a way as to disrupt the water structure but still be of lower energy than the sum of water-water and acetone-acetone interactions. Again only spectroscopic evidence will indicate what the interactions are; without this we can only speculate but it could be that the main gain comes from removing acetone-acetone interactions and replacing them by water-acetone even though the water-water interaction is favourable, losing some of this interaction must be compensated for by gaining acetone-water ones.


Cyclopropane instability to addition reactions In a question I saw comparing the reactivity (to addition reactions) of cyclopropane, cyclobutane and cyclopentane, it was said that cyclopropane was most prone to addition reactions.

Ring strain does make it prone to reactions, but how can addition reactions make it any more stable?

In that context, how can you compare the three?

You're absolutely right: cyclopropane's maximum ring strain makes it highly prone to addition reactions. Keep the following definition in mind:

  An addition reaction, in organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form a larger one (the adduct).

In your case:

Thus, it breaks down the ring into a straight chain, which makes the low stability of the ring a clear contributing factor to the reaction.

References: Cyclopropane, wikipedia||
Addition reactions, wikipedia


Is it true that heavy water is not blue? I believe I saw this claim somewhere on the internet a long time ago. Specifically, it was claimed that the difference could be observed by filling one long, straight tube with light water and one with heavy water, and looking through both tubes lengthwise (so that light has to travel through the tubes' lengths before reaching the eye), whereupon the light water would appear blue as it does in the oceans, and the heavy water would not. The explanation given was that heavy water has a different vibrational spectrum because of the greater mass of the $^2$H atom, which seemed perfectly plausible.

However, I am no longer able to find a source for this claim, which is strange because if it were true, surely it would not be so difficult to find a source?

Based on your description, I may have found the article you originally saw, or at least one very similar.

Researchers from Dartmouth College published a paper$\mathrm{^1}$ in which they report, among other things, the results of viewing sunlit white paper through two 3 meter lengths of plexiglass; one filled with $\ce{H2O}$ and one with $\ce{D2O}$. Sure enough, because of the lower frequency of the maximum absorption of $\ce{D2O}$ in the red to near IR wavelengths, the blue color that is characteristic of $\ce{H2O}$ is far less pronounced in $\ce{D2O}$. This website is based on the published paper and additionally shows a photograph of the blue colored $\ce{H2O}$ on the left with the far less colored $\ce{D2O}$ on the right:

1) "Why is Water Blue", Charles L. Braun and Sergei N. Smirnov, J. Chem. Edu., 1993, 70(8), 612


Why does the temperature of a tyre rise when it is pumped up rapidly? Why does the temperature of a tyre rise when it is pumped up rapidly?

The way my textbook explains this is that $V\propto T$, so, an increase in volume causes an increase in temperature.

However, I think that this justification is flawed, because air is a real gas instead, and follows (approximately) the van der Waals' equation of state: $\left(P+\frac{a}{V^2}\right)(V-b)=RT$. I'm not not sure how I can neglect the $a/V^2$ term in it, considering the volume in a tyre is not that high.

Moreover, I actually had thought the reverse statement to be true. Pumping air "rapidly" implies the conditions are adiabatic i.e. heat exchange is negligible. So, $TV^{\gamma-1}=\text{constant}$ holds, implying an inverse relation between volume and temperature.

So, why actually does pumping a tyre increase its temperature? And what is the fault in my reasoning above?

Source: KS Verma; Physical Chemistry for JEE (Advanced): Part 1; Exercise 5.1 Question 6

Once the pressure in the tire has risen a little, the volume of the tire doesn't change much any more.  But, you are increasing the number of moles of air inside the tire when you pump air into it.  The new air pumped into the tire compresses the gas that was already in there previously and raises its pressure.  So the new gas is doing work on the previous gas to compress it  adiabatically and approximately reversibly.  This causes the internal energy of the gas and its temperature to increase.


HI/P reduction mechanism I recently answered a question about Breaking Bad's initial methamphetamine production method (i.e. the reduction of (pseudo)ephedrine). The reaction is as follows:

(Source)

It is infamously known as the $\ce{HI/P}$ reduction and was (and still is somewhat) a real plague in the United States until cough medicine containing ephedrine came under tightened control. That for an introduction, now the question: what is actually the reaction mechanism of the $\ce{HI/P}$ reduction?

The first step seems like a simple $\ce{S_{N}2}$ substitution of $\ce{I^{-}}$, catalyzed by the protonation of the alcohol group by hydroiodic acid (making it a far better leaving group). But what is the mechanism after that? How is phosphorus involved? Does the $\ce{HI/P}$ reduction have any legal use?

According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.

(Image source)

The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.


How can I measure the amount of quinine in a home-made tonic water? I'm into mixology, and lately I've been looking into experimenting with home-made tonic water, by extracting quinine (quinine salts) from cinchona barks in an acidic aqueous solution. I've read a bit about the potential dangers of quinine overdose, so I want to make sure I know how much the amount of quinine is in my extract.

I know it is fluorescent under UV; I thought maybe there might be a minimum quinine-to-water ratio for fluorescence. If this is the case, I can dilute the solution gradually until the solution is no longer fluorescent. However my knowledge on this subject is fairly limited.

Is there a home-accessible way of measuring the amount of quinine in an aqueous solution?

Without standard solutions and a fluorimeter, you can do a side-by-side comparison using a commercial tonic water and your solution. I just prevailed in getting a poor photograph of the beautiful blue fluorescence from a commercial tonic water (begins with an S), using 405 nm excitation (BluRay laser diode). The photo is below:

The two cuvettes contain tonic water (undiluted on the right) and diluted about 5 times (on the left). The laser entered from the right. The bubbles must be removed because they cause significant light scatter that gives red artifacts in the camera. By naked eye, it actually looks lots better: no red artifacts.

Removing the dissolved gas, to avoid bubble formation, is often done done by using an ultrasonic bath, sparging or by simply shaking the bottle repeatedly and waiting. Adding NaCl is a no-go because it quenches quinine fluorescence (a nice Stern-Volmer experiment).

Fluorescence is very sensitive, but the standard joke is that only 10% of all substances fluoresce ... and they are the wrong 10%. But this is a good case!

So, you would simply A-B the solutions: make dilutions of the commercial tonic water and see which one most closely matches your home-made tonic water. Cheers!

Note: you can use near-UV LEDs, i.e., 395 or so nm peak emission wavelength. The absorption max for quinine sulfate in dilute sulfuric acid is around 350 nm, and the peak emission wavelength is around 460 nm. But the photo shows that 405 nm works fine: it is not always necessary to be at the optimum excitation wavelength. Driving an LED is trivially simple: just a DC voltage supply, such as a battery or 'wall wart', and a current limiting resistor. Or use a constant current source that can output 20 or so mA.

One more thing: if you dilute the tonic water sufficiently, you will not be able to see the fluorescence with the naked eye. But a fluorimeter can detect it at even lower concentrations. Eventually, with sufficient dilution, you reach the limit of detection. But that is way below whatever you might want in usable tonic water.


Inexact differential-path function From Wikipedia:

  an inexact differential cannot be expressed in terms of its antiderivative for the purpose of integral calculations

What's the mathematical reason that inexact differential cannot be expressed in terms of its antiderivative? Not enough information? Too complicated? $\mathrm{d}U$ is an exact differential. If you integrate $\mathrm{d}U$, what will you get then?

I think you answer comes from the part of the sentence that you left of the quote from Wikipedia:

  ... i.e. it's value can't be inferred by looking just at the initial and final states of the given system.

Inexact differential are the result of integrating nonstate functions (functions which aren't path independent). The fundamental theorem of calculus, which allows the integration of a function over a range by determining the antiderivative at the end points of that range, requires path independence in order to be true. So it won't work for inexact differentials that are path dependent.

As for integrating over $dU$, $U$ is a state function and so you can apply the fundamental theorem and solve the integral using the antiderivatives at the end points.


Balance Repeatability Sample or Population for SD A question for the masses about standard deviation (SD), selection of population vs sample in balance measurements.

Analytical balances are typically checked for repeatability by taking 10 weight measurements and finding the uncertainty by calculating standard deviation.

There are two forms of standard deviation: population and sample. Population divides the sum of differences squared by $n$; where as, sample divides sum of differences squared by $n-1$.

If I take 10 measurements of a standard weight, is this not the entire population? Or does the fact that I will repeat this test in the future make it a sample? I would typically select population in this case; however, most of the balance manufacturers I am reading are quoting $n-1$. What is the correct SD?

Standard Deviation Formulas in Maths is fun is a nice article to review. It explains why we have use two different formulas to find standard deviation for the population and sample. However, I'd like to point out that your explanation of calculation is a bit off. The correct formula to calculate standard deviation is:

$$ \bbox[yellow,5px,border:2px solid red]
{
\sigma = \sqrt{\left(\frac{1}{N} \sum_{i=1}^N (x_i-\mu)^2\right)}\qquad (1)}$$

where $\sigma$ is the standard deviation of the population and $\mu$ is the Mean (the simple average) of the population of numbers:

$$\mu = \frac{1}{N} \sum_{i=1}^N x_i$$

According to the Math is Fun website:

  To calculate the standard deviation of those numbers:

  Work out the Mean (the simple average of the numbers)
  Then for each number: subtract the Mean and square the result
  Then work out the mean of those squared differences.
  Take the square root of that.

The correct formula to calculate standard deviation of a sample is:

$$ \bbox[white,5px,border:2px solid blue]
{
s = \sqrt{\left(\frac{1}{n-1} \sum_{i=1}^n (x_i-\chi)^2\right)}\qquad (2)}$$

where $s$ is the standard deviation of the sample (say you choose $n$ numbers out of total of $N$ numbers population) and $\chi$ is the mean (the simple average) of the chosen numbers, hence: $$\chi = \frac{1}{n} \sum_{i=1}^n x_i$$.

Note: In our old days when I was taking analytical chemistry, we were instructed to use equation (1), if the $N \gt 10$, but to use equation (2) whenever we have to deal with data where $N \lt 10$.


Why does the reduction of Dichromate to Chromium 2 need an acid? The Ecell shows that Zn is enough. Then why, are both of them required?
Also, can I use any acid ? Or only HCl + H2SO4

A big factor is kinetics.  A lump of zinc or even zinc powder sits there undissolved in a neutral solution and does not provide good contact with the chromium species with which you want the zinc to react.  Adding acid creates nascent hydrogen in the solution, and this is what really does the job of reducing the chromium species.  This source, drawn from a related answer by Nilay Ghosh, explores the matter more fully, and I recommend going there.  In addition to the nascent hydrogen, the acid enables keeping the zinc ions in solution as a salt, where otherwise they would form a hydroxide precipitate that interferes with the reaction.

Not every acid is equally good.  To get the most nascent hydrogen you need a strong acid and, even more important, one that does not eat up its hydrogen with other oxidizing functions (nitric acid, you're out!).  Add the fact that cheap is also good, and hydrochloric and sulfuric acids are the usual choices.


How to recrystallize boric acid into flakes? I am putting 100 g of boric acid, 400 g of water and trying to crystallize at low temperature and letting cool slowly, however it recrystallizes as needle like texture. I need soft large crystals like mica. Can someone help out?

The ratio seems about right, though the solution could be a bit more saturated as the solubility of boric acid in $\pu{100 mL}$ of boiling water is $\pu{27.5 g}$, so for $\pu{400 mL}$ of water you can theoretically use $\pu{110 g}$ of boric acid.

From my experience it is crucial that initial nucleation occurs at the surface, then one gets flake-alike crystals. That's said, you need to start to cool down the solution very slowly, preferably using a warm water bath and a closed vessel with large cross-sectional area (wide beaker with a watch glass on top should do). Don't rush the cooling process, especially at the beginning – crystallization should begin on surface, not in the volume, otherwise you get a bunch of crystals of different morphologies.

P. S. When drying the crystals at the end after rinsing them with cold water on a filter make sure not to exceed $\pu{130 °C}$, otherwise boric acid will start to decompose and form metaboric acid $\ce{HBO2}$, then subsequently upon further heating, $\ce{H2B4O7}$, and, finally, $\ce{B2O3}$ [1, pp. 119–120].

Reference

Patnaik, P. Handbook of inorganic chemicals; McGraw-Hill handbooks; New York, 2003, ISBN 978-0-07-049439-8.


How to calculate the solubility of AgCl in a MgSO₄ solution? The solubility product of $\ce{AgCl}$ in water is given as $1.8\cdot10^{-10}$
Determine the solubility of $\ce{AgCl}$ in pure water and in a $0.25~\mathrm{M}$ $\ce{MgSO4}$ solution.

I solve to the $K_\mathrm{{sp}}$ and got $1.34\cdot10^{-10}$
for Part B I was told to use the activity and am not sure if I did it right,
my new $K_\mathrm{{sp}}$ turn out to be $1.123\cdot10^{11}$
could someone help me with part B?

In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$
 the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.

In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:
$\ce{Ag2SO4 -> 2Ag+ + SO4^2-}$ with $K_\mathrm{sp} = 1.4\cdot10^{-5}$

Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't.
$(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.

In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.


At what pKa value do we consider an acid strong? In between the traditional strong mineral acid that dissociate completely in aqueous solution, and obvious weak organic acid like acetic acid, things get murky.

For example, oxalic acid has a $\mathrm{p}K_\mathrm{a} = 1.46$ which is  about 1000 times stronger than acetic acid, but no one seems to qualify it as a "strong" acid. How about trichloroacetic acid, $\mathrm{p}K_\mathrm{a}\approx 0.5$? Well, some people start calling it "strong" acid.

So, what $\mathrm{p}K_\mathrm{a}$ value would qualify an acid as a strong acid?

I would like to know a concrete numbers, not merely “smaller $\mathrm{p}K_\mathrm{a}$ means stronger acid” as found in other questions. Just like a specific body mass of $90\ \mathrm{kg}$ would qualify a boxer to be a heavyweight boxer.

Your question is not as easy answered as you might want it to be. The reasons for that are mainly given in the answer by Zen to What is a "Strong" Acid?
The main problem is, that there hardly can be an absolute acid strength and you will always have to take the reference system into account. Another problem arises, that there is no consistent (or official by IUPAC) definition of the acid constants. Usually, when you are talking about $\mathrm{p}K_\mathrm{a}$ values, then you are most likely already implying aqueous solution.

It is a quite accepted classification of strong acids (in water), to state that they completely dissociate. Hence in the equilibrium is entirely shifted to the right side:
$$\ce{HA (aq) + H2O (aq) <=> A- (aq) + H3O+ (aq)}.$$
The equilibrium constants for this reaction is
\begin{align}
K  &=       \frac{a(\ce{A-})\,a(\ce{H3O+})}{a(\ce{HA})\,a(\ce{H2O})}
    \approx \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})\,c(\ce{H2O})}
    \approx \frac{K_\mathrm{a}}{c(\ce{H2O})}\\
K_\mathrm{a} &=
    \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})}\\
\mathrm{p}K_\mathrm{a} &= -\lg K_\mathrm{a}.
\end{align}

For a strong acid we can see that the concentration of undissociated acid is (nearly) zero. Therefore we write
\begin{align}
\lim_{c(\ce{HA})\to0} \left(K_\mathrm{a}\right) &= \infty&
&\therefore&
\lim_{c(\ce{HA})\to0}\left(\mathrm{p}K_\mathrm{a}\right) &= -\infty.
\end{align}

Strictly speaking, in aqueous solutions the acidity constants cannot be measured, since the hydronium ion is the strongest acid in this medium.

So in priniciple, the value that separates strong from weak acids would be the $\mathrm{p}K_\mathrm{a}$ of the hydronium ion. With that we arrive at a completely new problem, which is quite efficiently discussed here: What is the pKa of the hydronium, or oxonium, ion (H3O+)?

From this question we can take home the suggestion the value of $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = -0.7$ for the purpose of comparing it to other acids. In other words $\mathrm{p}K_\mathrm{a}(\text{strong acid}) < -0.7$ (in water). Taking this literal, however, might lead to many misconceptions, and using a relative approach of “this acid is stronger than this acid” should  be preferred in any case.


Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion? This question has also been asked on physics stack exchange:
Why is the helium atom wavefunction a product of the two 1s wavefunctions?

From [1, p. 116]:

In seeking an approximation to the ground state, we might first work out the solution in the absence of the $1/r_{12}$ term. In the Schrodinger equation thus simplified, we can separate the variables $r_{1}$ and $r_{2}$ to reduce the equation to two independent hydrogen like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be:
$$\psi(r_1, r_2) = \psi_{1s}(r_1)\psi_{1s}(r_2) = e^{−Z(r_1 + r_2)}$$

Why is it only the product and not some linear combination of the two wavefunctions? I heard somewhere that it has something to do with "tensor product". Can someone provide a detailed explanation about this?
Reference:

Blinder, S. M. Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology; Elsevier, 2012,. ISBN 978-0-08-048928-5.

The same mechanics as for the separation of angular and radial parts of the hydrogen electronic wavefunction are in place here: You have a Schrödinger equation that is separable into two (i.e. one for each electron). You solve them and in order to combine them, they must be multiplied.

It so happens in this case that the solution for each equation is known and recognizable: the electronic wavefunction of $\ce{He+}$. The fact that the two wavefunctions are multiplied is not related at all to electronic structure theory per se (no involvement of LCAO etc.), but merely a result of the separability of the Schrödinger equation.


Reductive amination in case of secondary amines When an aldehyde or a ketone undergoes reductive amination when a primary amine is acting,the mechanism follows a pathway through hemiaminal and imine,what happens when a secondary amine is acting on the given carbonyl ? There will be no hydrogen on the amine nitrogen in this case to facilitate loss of water with the hydroxyl group of the hemiaminal.How does the reaction proceed then?

In modern organic synthesis, reductive amination of aldehydes and ketones is considered one of the important and principal ways to make secondary and tertiary amines. It is a powerful and reliable strategy for the formation of $\ce{C–N}$ bonds. One of the advantages of the reductive amination is it can avoid the problem of over-alkylation that is always the problem with direct alkylation of amines with alkyl halides.

Reductive amination can be achieved following either a one-step or a two-step procedure. In both procedures, an amine and a carbonyl compound condense to form an imine (with primary amine) or iminium ion (with secondary amine). This imine or iminium ion is then reduced in situ (one-step reaction) or separated and subsequently reduced (two-step reaction) to form an amine product. Following is the suggested mechanism for reductive amination:

Most common condition is sodium triacetoxyborohydride ($\ce{NaB(OAc)3H}$) in THF or dichloroethane (DCE) under slighly acidic conditions (common acid used is acetic acid)(Ref.1).

Reference:

Ahmed F. Abdel-Magid, Kenneth G. Carson, Bruce D. Harris, Cynthia A. Maryanoff, Rekha D. Shah, “Reductive Amination of Aldehydes and Ketones with Sodium Triacetoxyborohydride. Studies on Direct and Indirect Reductive Amination Procedures,” J. Org. Chem. 1996, 61(11), 3849-3862 (https://doi.org/10.1021/jo960057x).


What does "less hindered hydrogen" mean? In a book they have given a statement,

  In the process of pyrolysis of esters, esters having  beta hydrogen are heated at high temperature and then alkenes are formed along with carboxylic acid.
  If there are two beta hydrogens then beta hydrogen having less hindrance will participate in the reaction.

In the last line, what do they mean by less hindered hydrogen?

Consider these following two structures from the same compound:

The two Newman projections are along the horizontal $\ce{C-C}$ bond which is, of course, freely rotatable. The two structures represent two possible syn-elimination transition states. Note that in the left-hand structure there is a steric clash between the tert-butyl group and the methyl group. The right-hand structure has no such clash. It is therefore the preferred transition state for this reaction.

‘Less hindered hydrogen’ basically means that hydrogen that leads to the more stable transition state.


An issue with Bouger-Lambert Law I'm certain that I misunderstand this law. It states that:

$$I = I_o e^{-kb} $$

and after rearranging:

$$ A = ln{I_o}-lnI $$

where $A$ stands for absorbance.

1) Does is mean that when the intensity of the exiting light is very low than the absorbance is extremely high?

2) But from the graphs in my textbook, I've seen absorbance which was up to 1 or 1,2. Does it mean that most objects give little "resistance" to light?

3) My textbook says that molar absorption coefficient depicts the probability that photon will excite the energy states (not their words exactly :) ). Than why is f(c) = epsilon a constant function? Shouldn't more concentration increase the probability of change in energy states in some for example molecule?

1) Yes that is correct.  Generally, when the intensity of the light transmitted or reflected is low, then the absorbance is high.

2) It is hard to say just what "most" means, but if we are speaking of visible light (remember absorbance is wavelength dependent), there are a LOT of colorless materials that do not absorb in the visible region.

3) Yes, more molecules will increase the probability of light being absorbed by the solution, but as you noted, the absorption coefficient is a "molar" absorption coefficient.  Therefor when using the molar absorption coefficient, the absorbance is adjusted for the number of molecules in the solution.


Non-conventional polymers IUPAC defines a monomer as,

  A molecule which can undergo polymerisation thereby contributing constitutional units to the essential structure of a macromolecule.

Bearing this in mind, I thought about molecules which exhibit hydrogen bonding. Take water for example: one water molecule bonds to another, and that one bonds to another, and another, etc. In the end, you'll have a long chain of water molecules bonded together. This process of water molecules coming together and hydrogen bonding sounds like they 'polymerise' in some way.

To explain my point, consider an infinite amount of water. Here, I'd imagine there to be an infinitely long chain of water molecules bonded together. This sounds like usual polymers I see, where there's an infinitely long chain of repeat units. However, the polymerisation is different for those.

Hence, I have two questions: could a single water molecule be described as a monomer? And, could a long chain of bonded water molecules be described as a polymer?

P.S. This doesn't just have to apply to water. The same could be said about a single alcohol molecule/a long chain of bonded alcohol molecules.

There are actually two ways to answer this question: by lawyering or by trying to capture what a polymer is. I'll try both.
Answer by lawyer
The IUPAC defines the following terms as follows:

monomer: a molecule which can undergo polymerisation thereby contributing constitutional units to the essential structure of a macromolecule
macromolecule: A molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass.
molecule: An electrically neutral entity consisting of more than one atom ($n > 1$). Rigorously, a molecule, in which $n > 1$ must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

Using the data from this paper and a harmonic oscillator approximation, I estimate that the zero-point energy of a harmonic oscillator with two waters and a hydrogen bond is just over 45 kJ/mol.
Since the energy contributed by hydrogen bonding in liquid water is estimated to be between 2kJ/mol and 6kJ/mol, I conclude that the hydrogen bond of water cannot confine a vibrational state.  This means liquid water does not qualify as a single molecule.
Since a long chain of waters connected by hydrogen bonds does not qualify as a molecule, it cannot be a polymer, which means that water cannot be a monomer (at least, not in this sense).

A better answer
The real problem here isn't the technical stuff I wrote above (which, by the way, probably has a few holes large enough to sneak trucks through). It's that words have certain meanings, and we can't capture everything in a simple definition. For the same reason that you'd probably feel a bit weird calling "yellow" a "reddish-green", even though it is literally the result of mixing red and green light, it feels strange to call water a polymer.
Here are some common ideas that spring to mind when I see the word "polymer":

Made of organic monomers bonded together covalently
Structure of covalent bonds (e.g. branching) affects physical properties
Intermolecular forces affects physical properties

These are not strict, but they explain why quartz is not referred to as a polymer, even though it's made of covalently-bonded $\ce{SiO2}$ units: everything is covalently bonded and there aren't any intermolecular interactions to speak of! For another example, depending on the exact context, proteins are sometimes rarely thought of as polymers, even though they're made of repeating amino acids, because their structure is fixed (must be linear chain) and their intermolecular forces don't contribute much.
So the more sensible answer to this question that doesn't rely on lawyering is that liqid water just doesn't have the properties that most people think of when they hear the word "polymer" and so it isn't usually referred to as a polymer.
Supplemental Materials
The code I used to compute the harmonic oscillator energies
from math import pi,sqrt
# Compute the harmonic oscillator energy of two water molecules
# using a two-body QHO approximation
kg_per_amu = 1.66054e-27
watermass = 18.0 * kg_per_amu # Mass of water: close enough for now
reducedmass = watermass**2 / (2 * watermass)

hbar = 1.054571800e-34   # J-s
k = 845.4 # From https://www.princeton.edu/~fhs/paper79/paper79.pdf, N-m

E_zeropt = hbar * pi * sqrt(k / reducedmass) # for n = 0
E_zeropt_kjm = E_zeropt * 6.022e23 / 1000  # Scale from J/molecule to kJ/mol
print(E_zeropt_kjm)


Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane? On the Pauling electronegativity scale, fluorine and chlorine are 3.98 and 3.16, respectively.

Since the dipole moment is dependent on electronegativity, why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?

I guess that the smaller size of the s orbital in the fluorine atom has something to do with this.

Dipole moment is not just about charges, it also has $L$ term. Bond length of $\ce{C-Cl}$ is greater than $\ce{C-F}$ and in this case, that is more dominating factor.

The dipole moment is in order

$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I}$$

You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{Br}$/$\ce{I}$.


1H-NMR and deducing structures here is the data booklet we must refer too:

Here is the exam paper solution:

I dont understand how they conclude this.

Peak areas and the chemical shifts seem to be different for me. For example there is a peak area of 2 for the chemical shift (ppm) of 2.0. How can this be when the group is CH3-COOR... isn't the peak area 3 if there are 3 hydrogens? Can someone please help me clarify reading this table, and deducing structures. How do link the R groups and when they bold H, does that mean the H can be any number?

I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity.

Chemical shift (ppm) and peak area (integral)

These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are in that environment, compared to the number of protons in all the other environments.

That the areas need to be considered as a ratio instead of absolute values is important: if the molecule was symmetric, with an ethyl group either side of a ketone e.g. C₅H₁₀O you would see a peak ratio of 3:2 (two environments), but there would actually be 6 protons in the first environment and 4 in the second. This is a bit like empirical vs molecular formulae.

Some tips for deducing structures

1) Start with 'characteristic' shifts. These are usually the lowest and highest shifts.

Here the 4.1 ppm shift is characteristic of ester protons, i.e. -(C=O)OC H₂R (R can also be a proton too!) where, much like the data sheet above, bold is to tell you which protons are being referred to in a more complicated structure. The shift at 0.9 ppm is characteristic of protons which are attached to a standard saturated alkyl chain. This tells you that those protons aren't very near any oxygen, nitrogen or other electron withdrawing or electron donating groups.

2) Try and explain the other shifts using functional groups that you have identified from your characteristic shifts

The shift at 2.0 ppm is likely to be alkyl chain protons which are next to an electron withdrawing group, which conveniently you have already identified as an ester group by looking at your characteristic shifts! In this case, that means you can put the 0.9 ppm shifted protons at one end of your structure, because there is nowhere else to put them.

3) Check your peak assignment makes sense when you look at your peak areas

The last thing to check is which end of the molecule the 0.9 ppm shifted protons go on. i.e. is the molecule methyl propanoate or ethyl ethanoate? To do this you can look at the peak areas. For this example it is quite easy to check because there are 3 protons at 4.1 ppm, so you know that you have a methyl ester group -(C=O)OC H₃. Our R group from earlier is an H after all! The resulting structure is the one shown in the example: methyl propanoate.


Lewis structure of ammonium ion What would be the lewis structure for ammonium ion?
Is there a coordinate covalent bond as well in the structure?
My confusuion would be dealth with if i get the structure of ammonium cyanate as well.

Coordinate covalent bonds do not exist.  They are all plain old covalent bonds.  We use "coordinate covalent bonds" as a means of counting electrons from atoms that go into bonds; here nitrogen contributes five electrons to four bonds so we figure that the nitrogen must have contributed two electrons to one of the bonds.  But the actual bonds formed between the nitrogen and each hydrogen atom are all alike.  Ammonium ion has the same tetrahedral structure as methane, and for methane we do not need coordinate covalent bonds to count the electrons.


Why is the conductivity of non-metals not equal to zero? My chemistry book says that delocalized electrons are the reason for electrical conductivity in a metal. However it says that nonmetals like sulfur, phosphorus and silicon have "low" conductivity. How can they conduct electricity if they have no delocalized electrons? Silicon has a giant macromolecular structure while sulfur and phosphorus have simple molecular structure. What causes them to conduct even for a little bit?

When we think of molecules we think of their molecular orbitals.  If we place these molecules into an ordered solid like a crystal, then the orbitals on adjacent molecules are close enough to interact and "extended" molecular orbitals are formed.  We call these extended molecular orbitals "bands". Typically these bands are "thicker" (spread over a wider range of energies) than molecular orbitals.  That is to say that because the energy of a molecular orbital in a molecule near an edge will be slightly different than the same molecular orbital in a molecule in the middle of the crystal, there is a spread of energies associated with a band.  Ultimately, because of this "thickening" of bands, all of the bands containing electrons overlap and form what is termed the valence band.  Similarly all of the empty bands overlap and form the conduction band.

(image source)

The valence band and the conduction band are analogous to the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) that exist in individual molecules.  The HOMO-LUMO separation is termed the "band gap" in a solid.

In metals the band gap is essentially zero with the valence and conduction bands overlapping; electrons move readily from the valence band to the conduction band. Electrons in the conduction band then move freely throughout the solid and give metals their characteristic conductivity. In insulators the band gap is very large and very few electrons have the thermal energy needed to be promoted to the  conduction band, hence there is basically no conduction.  Semiconductors lie in between these two extremes and thermal energy can promote some electrons (not as much as in a metal, but far more than in an insulator) to the conduction band, enough to have some conduction.

(image source)


Which biphenyl is optically active? Which biphenyl is optically active?

I know that it can never be 1. I don't think it will be 4 either as I read that it should be ortho-substituted biphenyl.
So, when I look at 2 and 3, I can't make out which one should be the correct answer.
Should it always be two ortho substituents on each phenyl group for it to be optically active?
Also, what if the two groups on the same phenyl are different, will it always be optically active in that case?


Oxidation state of group 2 elements I have came across the following statement : "the oxidation state $+II$ is stable for the elements of the alkalin earth metals" , the same is true for the elements of the first group , i've looked for a possible explanation and heard that it has a relation with the charge densities of the elements , if that's true what is that relation ? otherwise , what could explain the oxidation states for each of the groups in the periodic table ?

The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:

Note that for the alkali metals, each element has one more electron than a noble-gas configuration.  This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:

It is much harder to remove the 2nd electron, since it is contained within a completely filled shell.  (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)

Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.

The same thought process can be applied to the alkaline earth metals.  Here are the first three ionization eneriges for each:

Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove.  Thus, the metals in this group tend to stay in the +2 oxidation state.


Which solvent is good for using SU-8 particle in Optical tweezers? I'm studying Optical tweezers. I've used Polystyrene for samples. Nowadays, I've got SU-8 pillar. I wanna use that for sample, but its density is heavier than water. I can't use that. So I wanna know some liquid for replacing water.

I've done some experiments recently with SU-8 and polistirene and figured it out, that SU-8 solvent melts polistirene. Just wanna say, that this won't work out well.

But to use something instead of water: why not using alcohol (ETOH) or isopropyl. I use the last as much as possible since it's not leaving any stains on stuff I do.


Reversible process in thermodynamics How can we look at reversible processes in an intuitive manner?

How is it possible, that at any moment, it can be reversed by an infinitesimal change?

I sometimes explain it this way: imagine a staircase where each step is $h$ high.  If $h$ is small then it is quite easy to take a single step up or down, i.e. the process is relatively easily reversed. However, as $h$ increases it starts to become increasingly difficult to go up (and hazardous to go down).

Even for small steps there is a difference between going up and down since you expend more energy going up (against gravity) than going down, so it's not truly reversible, but as $h$ decreases the difference becomes minuscule (but never 0).

Similarly, there is no truly reversible thermodynamic process but the more slowly you let a process proceed (and equilibrate) the more it resembles the reversible ideal process


How to find partial pressures of gases only given their percentages? How do I find partial pressures if I am only given their percentages? Where do I begin?

Example: Assume $\ce{O}=21\%$ , $\ce{N}=78\%$ , $\ce{Ar}=1\%$ by volume. How would I calculate the partial pressures if total pressure is $\pu{1 atm}$.

Pressures of gasses are additive. For example, 1 atm X plus 1 atm Y makes 2 atm total pressure. Because all the components of a mixed gas are at the same temperature and pressure, the fraction of each component by volume is also the same fraction of the total pressure.


Finding molecular formula with limited information given I have recently stumbled upon a question I can't seem to solve :

  Chlorine can also be used to synthesise a bromine-containing oxoanion that is mainly used in continuous or batch dyeing processes. When chlorine is passed through an aqueous solution containing potassium hydroxide and potassium bromide, the bromide ions are oxidised to the bromine-containing oxoanion. Careful addition of aqueous lead nitrate to the resulting solution precipitates 4.17g of $PbCl_2$. When this precipitate is filtered off and the resulting filtrate is evaporated, 0.835g of a white crystalline solid is obtained. The white solid has the following composition by mass : $K$, 23.4%; $Br$, 47.8%; $O$, 28.8%. $\mathbb \ Determine\ the\ formula\ of\ the\ white\ crystalline\ solid.$

I attempted by first finding the empirical formula of the solid using by dividing the percentage compositions by their respective elements' mass number, and ended up getting $KBrO_3$ as the empirical formula, which would make my second step to find $n$, where the molecular formula of the white crystalline solid is $K_nBr_nO_{3n}$. However, how can I go about doing this without being given how many moles of the white crystalline solid was precipitated? I can't seem to construct the equations that could potentially lead me to the mole ratio between $PbCl_2$ and $K_nBr_nO_{3n}$. If possible, can anyone tell me how to get a better grip on forming equations in scenarios such as this? I have been struggling in this topic for a while and this is likely to be one of the key reasons why. Sorry for the lengthy question!

Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum.

The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product with the same composition will be obtained. And it will behave exactly the same as the commercial product called potassium bromate with a label saying KBrO3. We prefer the simpler formula.

The numbers given for the weight of PbCl2 and the product KBrO3 do not define a specific reaction, because other reaction products and excess reactants are not specified. And to be realistic, it seems that the filtrate was not evaporated to dryness, because all these other components are not obtained. It seems that this question is posed not just to get you to divide 3 percentages by 3 atomic weights, but to decide what numbers are important to answering the question. The mole ratio between PbCl2 and KBrO3 is unimportant (it's 3) - you got the answer at KBrO3 and should stop there.

Remember this riddle: What has 4 legs and a tail, barks like a dog and concrete?
Answer: A dog. (I just threw in the concrete to make it hard.)

A common test technique of teachers is to throw in extra information to see if you know how to select the correct information to use.


Viewing and capturing real time values of Electronic balance measuring instrument on television set and computer monitor larger screens https://en.wikipedia.org/wiki/Weighing_scale
Is it possible to view and capture real time values of Electronic balance measuring instrument using wired, wireless,CCTV technology on television set or computer monitor larger screens?
Scenario
Chemistry labs - Electronic balance interfaced wired/wireless with the television set or computer monitor i.e.Larger display screen display taking into account.Viewing the mass recording of a object in real time.

Three possibilities come to mind:
(1) Purchase a scale with a digital output e.g, https://www.arlynscales.com/digital-indicator/industrial-scales-with-analog-and-digital-outputs/ , send it to your computer, and output it to a monitor (depending on the scale's output format, it may be possible to send it directly from the scale to a monitor)
(2) Point a USB webcam towards the scale's digital readout, plug that into to your computer, and then display on a monitor.
(3) Same as no. 2, but use a digital camcorder that has a digital output that can be displayed directly on a monitor, without the need for a computer.


Are my protolysis equations right? We had the task to solve several salts in water and measure the pH value of the solution. After that we should "create" the protolysis equation and to tell about Chemical equilibrium[s].

I get some possible result equations, but I'm really unsure about them. The [] brackets were used to make the ionic loads easier visible. solving equations:

$\ce{KNO3 <=> [NO3]- + [K]+}$
$\ce{FeCl3 <=> [Fe]^{3+} + [Cl]-}$
$\ce{NH4Cl <=> [NH4]+ + [Cl]-}$`
$\ce{NaHSO4 <=> [HSO4]- + [Na]+}$

my protolysis equations:

$\ce{[NO3]- + H2O <=> HNO3 + [OH]-}$
$\ce{[Fe(H2O)6]^{3+} + H2O <=> [Fe(H2O)5OH]^{2+} + [H3O]+}$
$\ce{[NH4]+ + H2O <=> NH3 + [H3O]+}$
$\ce{[HSO4]- + H2O <=> [SO4]^{2-} + [H3O]+}$

I would say that all equations run bi-directional, because the result of solving equations can vary dependent on temperature for example, and the protolysis equations should be bi-directional because we're having a solution, which isn't a pure acid.

Additionally, I got the question if I can draw conclusions on the pH-change through those protolysis reactions. I would say yes, but not very well, depending on the sum of all hydrogen-containing ions like $\ce{H3O+}$

yes, they are, all of them are bidirectional.
It is not possible to draw conclusions about the pH value, because of the different abilities to solve in water and because those reaction may cause an increased or a decreased temperature.

(Source: my teacher)


Converting a cis-alkene to trans-alkene I was thinking about simple method to convert a cis-alkene alkene like cis-2-butene to trans-2-butene.
I came up with this method (the method can also be used to convert trans alkene to cis alkene):

First use a peracid like m-CPBA to convert the cis alkene to an epoxide:

Then I used the Wittig reagent $\ce{PPh_3}$ to get a betaine. For the $\ce{- O-}$ to be able to attack $\ce{PPh3}$, they should be on same side. So rotation about $\ce{C-C}$ bond:

Finally, the leaving of $\ce{POPh3}$:

I would like to know if my method is correct and if there are any other methods to bring about the conversion.

Your method looks fine. There is another simple method (with three steps though).

Convert the the cis alkene to a dibromo derivative using bromine;
Use two equivalents of $\ce{NaNH2}$ to get an alkyne (elimination);
Selective reduction to trans alkene by using sodium in liquid ammonia.
However this requires that Hydrogen be present as a substituent on the double bond, on both carbons.


pH Curves for Titrations Why does the pH rapidly increase near the equivalence point during--why does the slope of the graph of a pH curve sharply increase around this area?

I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):

Let's consider the case of  the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:

The pH increases slowly at first because the pH scale is
logarithmic, which means that a pH of 1 will have 10 times the
hydronium ion concentration than a pH of 2. Thus, as the
hydronium ion is initally removed, it takes a lot of base to change
its concentration by a factor of 10, but as more and more
hydronium ion is removed, less base is required to change its
concentration by a factor of 10. Near the equivalence point, a
change of a factor of 10 occurs very quickly, which is why the
graph is extremely steep at this point. As the hydronium ion
concentration becomes very low, it will again take a lot of base to
increase the hydroxide ion concentration by 10 fold to change the
pH significantly.
Let's find the equation of the titration curve in the the region $0<V_b<V_b(eq)$:

The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$

The number of moles of $\ce{H3O+}$ after adding the volume   $V_b$ of the base is $C_aV_a-C_bV_b$

Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
So, the titration curve in the region $0<V_b<V_b(eq)$  $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.


Confusion about the SN1 and SN2 Mechanism My book says, and I understand, that the $\mathrm{S_N1}$ mechanism is much faster for tertiary alkyl halides than primary. And the opposite for $\mathrm{S_N2}.$ This all makes sense. What I don't get is the conclusion it comes to next: hence, we predict that the primary alkyl halides will react primarily via $\mathrm{S_N2},$ and the tertiary via $\mathrm{S_N1}.$

The reason I get confused here is that nothing was said about the comparative rates for a primary alkyl halide via the $\mathrm{S_N1}$ or $\mathrm{S_N2}$ (they just compared primary and tertiary for $\mathrm{S_N1}$ and $\mathrm{S_N2}$). In my mind, which mechanism dominates for primary or tertiary should be dependent on the relative rates for that compound; not rates compared between primary and tertiary.

Is my argument sound? And if so, is there some literature you know of that explains this in such a manner?

Your reasoning is correct. It's the relative rates for a given alkyl halide that matter, and it's difficult to draw conclusions about those, because the relative rates are not fixed, as the $\mathrm{S_N2}$ rate is dependent also on the nucleophile concentration.

We can write the rate of the $\mathrm{S_N1}$ as $k_1[\text{alkyl halide}]$ and the rate of the $\mathrm{S_N2}$ as $k_2[\text{alkyl halide}][\text{nucleophile}]$, so the relative rate at a fixed concentration of the alkyl halide is given by

$$\frac{\mathrm{S_N2}}{\mathrm{S_N1}} = \frac{k_2[\text{nucleophile}]}{k_1}.$$

The identity of the nucleophile will be a factor as well, since that will affect $k_2$.

So all in all, you can't make a general conclusion that one reaction type will always go faster than the other, although it is likely that there are alkyl halides for which one reaction type is faster in a large majority of cases.


Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies? Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?

No, I could not think of any relation between them. Orbital energies are solutions of the electronic Schrödinger equation, and as such, they describe the electronic state of a molecular system. Vibrational frequencies are solutions of the nuclear Schrödinger equation, and as such, they describe the nuclear state of a molecular system. The quantities that describe the electronic state are not related to the quantities that describe the nuclear one simply because they describe two different subsystems: the electronic and the nuclear one.

Note that usually we treat these two subsystems as being independent of each other (the Born-Oppenheimer approximation), but even if you go beyond that, what you'll see is that changes in the nuclear subsystem (say, changes in geometry during vibrations) might affect the electronic state. Consequently, molecular vibrations in principle can alter HOMO/LUMO (as well as any other orbital) energies, but I doubt this interdependence of electronic and nuclear states is the relation you're looking for.


Finding the Oxidation number I have got the homework to balance the following redox reaction.

$\ce{Cr(OH)3- + HO2- -> Cr2O7^{-2} + OH- }$(Basic medium)

To balance this reaction first thing we need to do is to find the oxidation number of each element.

My efforts:

I have found out oxidation number of each element on product side.. But it is bit confusing in reactant side.

Can any help me in finding the oxidation number of elements on reactant side ?

About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: Electronegativity Considerations in Assigning Oxidation States

Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.

   Oxygen, it shall have the oxidation state $-2$

What about the other atoms?

   Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$
Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$

Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.

   Hydrogen gets $+1$ (see above)
 Charge has to remain, i.e. $-1$
 Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$

Now the hydroxyl ion is straightforward.

   Hydrogen gets $+1$ (see above)
 Oxygen gets $-2$

Look at dichromate and apply the above.

   Oxygen gets $-2$, most electronegative
 Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$

In total that will give you:

   $\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -> \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$


Relativistic effect: d-electrons in metallorganic complexes With higher period the d-electrons of the metal are less strong bonded and therefore oxidative addition is easier for $\ce{Ir(I)}$ than for $\ce{Rh(I)}$ and much easier than for $\ce{Co(I)}$.

For metal carbonyl hydrides, electrons with higher period are more stronger bound to the nucleus, so $\mathrm{p}K_\mathrm{a}$ increases in this order: $\ce{HCrCp(CO)3 < HMoCp(CO)3 < HWCp(CO)3}$
This explanation sounds logical!

I also wrote that for $\ce{CO}$-complexes because of the higher effective atomic number for higher period atoms (higher ionization energy) the $\pi$-backbonding is weak and therefore less stable as e.g. for $\ce{Cr(3d)}$.
This explanation sounds logical as well!

But is the written tendency of the oxidative addition not a contradiction to the other tendencies?

First off, be aware that oxidative addition occurs by several mechanisms which are governed by different factors.[1]

As you move down a period, the outer electrons become more easily accessed, as the outer electrons are increasingly screened from the nuclear charge by the inner electrons.[2] This is why both ionization energy and electron affinity decreases down a period.

Classical organometallic chemistry uses metal carbonyl stretching frequencies as a direct measure of π back-bonding and electron density.[3] For a recent discussion, see: Chem. Sci. 2016, 7, 1174. Compare the stretching frequency of the group VI metal hexacarbonyls [4]: $\ce{Cr(CO)6}$: $\pu{2000 cm^-1}$ and $\ce{W(CO)6}$: $\pu{1987 cm^-1}$. Tungsten donates more electron density back into the carbonyl than does $\ce{Cr}$.

The importance of relativistic effects in structure calculations is frequently the subject of reviews.[5]

References:

Crabtree, R. H. The Organometallic Chemistry of the Transition Metals, Ch. 6
Waldron, K. A.; Fehringer, E. M.; Streeb, A. E.; Trosky, J. E.; Pearson, J. J. Screening Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends. J. Chem. Educ. 2001, 78 (5), 635. DOI: 10.1021/ed078p635.
Tolman, C. A. Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis. Chem. Rev. 1977, 77 (3), 313–348. DOI: 10.1021/cr60307a002.
Clark, R.; Crociani, B. Solvent effects on the infrared spectra of chromium, Molybdenum and tungsten hexacarbonyls. Inorganica Chimica Acta 1967, *1undefined 12–16. DOI: 10.1016/S0020-1693(00)93131-1.
Pyykko, P. Relativistic effects in structural chemistry. Chem. Rev. 1988, 88 (3), 563–594. DOI: 10.1021/cr00085a006.