How is the the law of chemical equivalence defined and what are its limitations? I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products. But I'm facing difficulties in understanding double titrations. Moreover, this doesn't work for equations where the n-factor of all the reactants and products don't match.

Let's take an example:

$$\ce{KIO3 + KI + H2SO4 \to I2 + K2SO4 + H2O}$$

Here, the gram equivalence of $\ce{H2O}$ is $2$ which should match with that of $\ce{KIO3}$. But the gram equivalence of $\ce{KIO3}$ is $5$.

Similarly, $\text{g.eq}_{\ce{H2SO4}} \neq \text{g.eq}_{\ce{I2}}$ and so on. When the n-factors don't match, the gram equivalents don't match. Here, we can consider gram equivalent to be $\text{number of moles} \times \text{n-factor}$. Again, this n-factor is what is posing a problem.

  Can someone define the Law of Equivalence and state its limitations because of which I seem to find problems as I've mentioned above?

Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:

$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\
\ce{IO3- -> I2}\tag{reduction} \end{align}$$

In other words, we've just discovered a (reverse disproportionation?) comproportionation reaction. The change in the oxidation state of iodine in the oxidation part is $-1 \to 0 $ ($\Delta_{o.s.} = 1$) and for the reduction half it's $+5 \to 0$ ($\Delta_{o.s.} = 5$). So far so good.

Usually, the questioner would provide you with the number of moles of each species and you're done with number of moles times the n-factor. Nevertheless, let's try to give it a shot. I'm balancing this equation:

$$\ce{1KIO3 + 5KI + 3H2SO4 \to 3I2 + 3K2SO4 + 3H2O}$$

This means for $1$ mole of $\ce{KIO3}$ to react, you'll need $5$ moles of $\ce{KI}$. As this is a proportion, converting it to an absolute value, I'd say you've $1n$ moles of $\ce{KIO3}$ and $5n$ moles of $\ce{KI}$.

So if I try to equate the gram-equivalents, I'll have $1n \times 5 = 5n \times 1$. In other words, $\text{Eq.}_{\ce{KIO3}} = \text{Eq.}_{\ce{KI}}$

  I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products.

Law of equivalence is even more fun, for a balanced chemical reaction:

$$ \ce{aA + bB -> cC + dD}$$

Equivalents of every species will be equal. Mathematically,

Equivalents of A = Equivalents of B = Equivalents of C = Equivalents of D

will hold.

  Why aren't the gram equivalents of $\ce{H2SO4}$ equal to that of water?

You need to strip your equation of the spectating species. For instance $\ce{H2SO4}$ is primarily here because it acts as a medium-provider, it makes the solution acidic. Neither $\ce{SO4^2-}$ nor $\ce{H+}$ underwent any oxidation/reduction so applying the equivalent concept makes no sense in this case.


Nitration vs Nitrolysis I am interested in nitrating an aromatic ester in a solution of dichloromethane. I found the following paper that describes creating a nitric acid solution in dichloromethane using KNO3 and sulfuric acid:

Strazzolini P, et al. "Nitric Acid in Dichloromethane Solution. Facile Preparation from Potassium Nitrate and Sulfuric Acid." Tetrahedron Letters 42 (2001) 1387 - 1389.

This paper describes both the nitration of Benzyl Methyl Ester and the nitrolysis of Benzyl Tertbutyl Ester and Phenyl Tertbutyl Ester using their nitric acid in dichloromethane method.
But I am having trouble understanding how they controlled the reactions to get nitration of one ester and nitrolysis of another ester. The reaction conditions seem similar, but the concentration of nitric acid in the nitrolysis reactions is about twice as high as in the nitration reaction. Could the difference be in the tertbutyl group vs the methyl group on the ester?

Tert-butyl esters are routinely cleaved by anhydrous acid conditions by loss of the tBu+ cation. TFA in DCM is a common system for this. Neat formic acid, HBr/AcOH and PTSA in refluxing benzene or toluene are also used (see Greene - Protective Groups in Organic Synthesis (2nd ed) p246).
Methyl esters usually need aq. acid for cleavage.


Why does this protein seem to have multiple chains? I'm looking at the structure of this protein, which I beleive is supposed to be a single polypeptide chain, since on the PDB page it says:

  Unique protein chains: 1

But using the 3D viewer, there seem to be a few "loose ends" i.e. it looks like the protein is composed of multiple polypeptide chains.

This is was I see  - you can see a separate chain there at the bottom.

Where am I going wrong here?

Some parts a the protein are a bit more flexible, also in the protein crystal. These parts don't give clear electron density after x-ray diffraction, and are removed from the resulting model structure. Often the N- and C-terminal ends are removed, this is something you'll find in many .pdb structures, but sometimes also loops are removed, resulting in two extra loose ends.

The authors were nice to note in the comments section of the pdb file exactly which residues were removed from the model (#465).

There might be tools available to model loops into the structure so you'll have one continuous chain in the .pdb file, but I really don't know which one is best (of course they're all completely unreliable).


Clausius–Clapeyron equation: shape of phase diagrams makes no sense I am trying to model the melting point of a substance at varying pressures (ranging from very small to very very large). All I am trying to do is make an equation that relates melting temperature to pressure, so $T(P)$ is some function. To do this, I am trying to use the Clausius–Clapeyron equation (CC), which states that

$$\frac{\mathrm dP}{\mathrm dT} = \frac{L}{TΔV}.$$

In other words, the slope of the equilibrium line on the phase diagram should decrease as temperature is increased.

However, this is not the case; the curve of the equilibrium line is exponential and the slope $\mathrm dP/\mathrm dT$ increases as $T$ increases. Integrating CC we arrive at a logarithmic function, which again is not what empirical measurements reflect.

As I see it, the empirical results and the equation that is supposed to describe them are mutually exclusive. There is no way to arrive at an exponential curve from a slope that varies with $1/x.$ The CC equation and phase diagrams cannot be both be true at once and it is driving me mad.

Why is this the case? Is the CC equation valid at all because it seems to be totally false? What function do I use to model melting points at different temperatures?

The results that are so stupefying are these:

The shape of the curve is exponential. But, the supposed derivative is $1/T$, in which case the slope of each curve (red and blue here) should flatten as $T$ increases, but it steepens. Also, integrating that supposed derivative gives us $\ln (T)$ which is definitely not the shape of the phase diagram. This discrepancy is true for both the liquid/solid and liquid/gas curves.
I hope this clarifies the question!

This is a summary of the equations to use to calculate phase transitions.

The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion.

The Clausius-Clapeyron equation describes solid-vapour and liquid-vapour changes because the final volume is far greater than the initial one, and is $\displaystyle \frac{dp}{dT}=p\frac{\Delta H}{RT^2}$ where $\Delta H$ the enthalpy change at the liquid–vapour or sublimation transition. Integrating this last equation from pressure $p_1 \to p_2$ and temperature $T_1 \to T_2$ gives $\displaystyle \ln\left(\frac{p_2}{p_1} \right) = -\frac{\Delta_{vap}H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right) $.

The change in the volume during fusion is $\displaystyle \Delta_{fus}V = m\left(\frac{1}{d_l}-\frac{1}{d_s} \right)$ where $m$ is the molar mass and $d_l$ and $d_s$  the densities of the liquid and solid. The pressure variation for the solid to liquid (melting or fusion) change is

$$\displaystyle p_2=p_1+\frac{\Delta_{fus}H}{\Delta_{fus}V}\ln\left(\frac{T_2}{T_1} \right)$$

and for evaporation and sublimation

$$\displaystyle p_2=p_1\exp\left( -\frac{\Delta_{vap}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1} \right) \right)$$

with the appropriate $\Delta H$. This is $\Delta_{vap}H$ for evaporation and $\Delta_{fus}H + \Delta_{vap}H$ for sublimation. Sublimation is treated as two steps merged into one; melting and instantaneous evaporation.

The $p$ vs $T$ plot for benzene is shown in the figure. Notice how this is different to how these are generally shown. This is sometimes due to the fact that log pressure is sometimes plotted but this is not always shown on the figure. Notice also how the solid liquid line is effectively vertical.

The data used is

R = 8.314  #( J/mol/K)
dens_sol = 981.0  #( kg/m^3)
dens_liq = 879.0  #( kg/m^3)
mol_mass = 78.0/1000.0 #( kg/mol)
DH_vap   = 30.8*1000   #( J/mol)
DH_fus   = 10.6*1000   #( J/mol)
p3 = 36.0/760*101325 #( triple point pressure Pa)
T3 = 5.5 + 273.16    #( triple point temperature K)
DV_fus = mol_mass*(1/dens_liq-1/dens_sol)                          # delta volume fusion

Python/numpy  functions for the pressure are:
p_liq_vap= lambda T: p3*np.exp( (DH_vap/R)*(1/T3-1/T))
p_sol_vap= lambda T: p3*np.exp( ((DH_fus + DH_vap)/R ) * (1/T3-1/T) )
p_sol_liq= lambda T: p3 + DH_fus/DV_fus*(np.log(T) - np.log(T3))


Will isopropyl alcohol dissolve the coating/paint of a notebook? The notebook is a Dell Inspiron 6400 and the coating/paint is the silver on the palmrest and back cover.

It says 99,4% on the container of the isopropyl alcohol.

Unfortunately you won't really know until you try it because there doesn't appear to be any info about what they used as a coating. Having said that, isopropanol isn't a superbly strong organic solvent so while it may remove some of the coating, it's not likely to immediately eat through your laptop like how acetone might. Still, follow user137's advice and test it on a very small area before you try anything else, and give it time before you go using it everywhere else. If you don't see any damage and you decide to go through with whatever it is you want to do (I assume cleaning it), just make sure you wipe it off with a damp cloth (water) followed by a dry cloth as soon as you get the results you want to minimize any possible damage.


Oxidation of sulfur dioxide to trioxide and its decomposition under same condition at higher temperature I found that $\ce{SO2 + 1/2 O2 -> SO3}$ when catalysed by platinum and heated to $\pu{700 ^\circ C}$. On the other hand I found that $\ce{SO3 -> SO2 + 1/2 O2}$ with same catalysator and temperature about $\pu{800 ^\circ C}$. How depends these two cases in real reaction? What controls which reaction will happen?
I thought that one reaction is exothermic and other endotermic, and duration of reaction and temperature controls which reaction happens. But facts I found are different. Sorry for maybe too simple question. I would be pleased with some tips which chapters of chemistry to study to understand my problem more deeply.

There is an equilibrium between $\ce{SO3}$ and $\ce{SO2 + 1/2 O2}$, at least at high temperatures and with an appropriate catalyst.  We could write the equilibrium as
$$\ce{SO2(g) + 1/2 O2(g) <=> SO3(g)}$$
Just from writing this reaction this way we can glean several things:

The reaction combines 1.5 gas molecules into one gas molecule, a net reduction in the number of molecules.  Since fewer molecules (but the same number of atoms) means there are fewer translational degrees of freedom, the entropy will decrease the more $\ce{SO3}$ "product" is formed.  Higher temperatures will thus disfavor the right-hand $\ce{SO3}$ side of the equilibrium and favor the left-hand $\ce{SO2 + 1/2 O2}$ side.

Total pressure will also affect the equilibrium but in the opposite way.  Higher pressure (crudely speaking) is a force that squeezes things together, so at high pressures the right-hand side will be favored.

The other principle factor that influences the equilibrium would be the partial pressure of $\ce{SO2}$, of $\ce{O2}$, and of $\ce{SO3}$.  So for example  if you had 99.9 mol % oxygen gas and only a trace of $\ce{SO2}$, that would favor the conversion of nearly all sulfur to the LH sulfur trioxide side.

The exact balance will depend on the precise enthalpy change of the reaction, the partial pressure of each species, the temperature, and the total pressure.  You could modify the system to favor one side or the other, and could model this accurately and quantitatively with appropriate numerical data.


Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus? My chemistry book explains that even though electrons in the $\mathrm{2p}$ orbital are closer to the nucleus on average, electrons from the $\mathrm{2s}$ orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g., $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{3d}$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?
Although AcidFlask already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.
I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states (which are in turn lower in energy than $\mathrm{3d}$ states).
$\ell$-degeneracy of the hydrogen atoms energy eigenvalues
The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the Kepler problem known from classical mechanics.
This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant.
Now, it is known from physics that symmetries of a system lead to conserved quantities (Noether Theorem).
For example from the rotational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the magnetic quantum number $m$, which are all energetically equivalent as long as the system maintains its rotational symmetry.
So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom.
Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry.
The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.
(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)
This dynamical symmetry conserves the Laplace-Runge-Lenz vector $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found here.)
Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms?
As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g., by applying a magnetic field, the $\ell$ degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form.
This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.
(Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)
Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?
Two effects are important here:

The centrifugal force puts an "energy penalty" onto states with higher angular momentum.${}^{1}$ So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.

The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
\begin{equation}
{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad
\end{equation}
The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.

Core repulsion (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.

Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.

The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance
between centrifugal force and core repulsion, that both act against the nuclear Coulomb
attraction.
In the real atom the balance between centrifugal force and core repulsion is broken,
The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.
Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states which are in turn lower in energy than $\mathrm{3d}$ states.
Of course, one has to be careful when using results of the hydrogen atom to describe effects in multielectron atoms as AcidFlask mentioned. But since only a qualitative description is needed this might be justifiable.
I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.


Is zero dipole moment possible in a square pyramidal complex? Say, the complex is [ML5] and is repelled in magnetic field. Given, its dipole moment is zero. Can its shape be predicted as trigonal pyramidal by the given information or square pyramidal structure may also fullfill them?

A square pyridimal structure can have a dipole moment of zero, e.g. a squarepyramidal cluster of atoms of the same element would have no dipole moment. However for most cases where the bonds are polarized, the dipole moments will not cancel as they would for other symmetries (e.g. tetrahedron). So basically, most square pyramidal complexes with polarized bonds will have a net dipole.

If you have a complex in mind just calculate the dipole computationally or look at the ir spectrum.


What is the purpose of subtracting the blank when constructing an analytical calibration curve? When building a calibration curve for quantification of some analyte using any spectroscopic technique, what is the purpose of subtracting the intensity of the blank from the intensity of each point of the curve?
I can build two calibration curves using the same measurements, one plotting the total intensity recorded at each point (F) and the other plotting the total intensity recorded minus the intensity when the concentration of the analyte is zero (F-F0), both against analyte concentration.
When making a linear regression for each curve, the slope will be the same and the difference between the intercepts will be the intensity of the blank (F0). If I try to quantify an analyte in a given sample with a different matrix than the one used in the standard (different solvent, for instance) maybe the "blank of the sample" so to speak would be different than the blank of the standards (F0). In that case, the result would be equally wrong, using either of the curves, because I would either be comparing the response of the sample-F0 to analytic curve based on F-F0 or the total response of the sample to analytic curve based on F.
It would make sense to me if I could, when analyzing an unknown sample, measure an exact blank of the sample, and use this measurement to compare with the analytical curve built using F-F0. Then I would be comparing two intensities associated only with the analyte, but that is't possible.
Am I missing something? Does subtracting the blank helps in any way when the objective is to quantify the analyte in samples?
I've checked two different undergraduate level analytical chemistry books and found no answer. They just say it is done this way.

Good to see interesting analytical chemistry questions after a long time.
I can give a qualitative example. Suppose you wish to collect the spectrum of a star. What time would be the best to carry out the measurement? You would agree that night time would be the best, because constant unwanted sunlight background is missing at the time of measurements. Subtraction of the blank serves a similar purpose, you wish to remove the background signal from your analytical signal so that you can assess the magnitude of your analytical signal. The blank also helps to serve to establish the minimum concentration that can be detected reliably. This is why blank subtraction is carried out
One of the commentators on your question has written a whole book on this subject.
You did not mention the analytical instrument you are using, it would have helped to discuss more. Anyway, calibration curve is another story. In this case, you certainly can make two calibration curves, and you are very correct, their slopes are the same. The sample concentration will turn out to be the same from both.
With blank correction, your equation should have a form y= mx with negligible intercept.
Without blank correction, your equation has the form y=mx+c, where c is the intercept, the average blank reading.
I may ask another thought provoking question which is certainly not discussed anywhere.
For example, what if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear.
a) What will happen to the magnitude of slope?
b) Will the calibration curve sensitivity increase? "Apparently, yes!"
c) If the sensitivity increases by this simple mathematical manipulation, why shouldn't we do it?
Coming to the second part of the question:

If I try to quantify an analyte in a given sample with a different
matrix than the one used in the standard (different solvent, for
instance) maybe the "blank of the sample" so to speak would be
different than the blank of the standards (F0).

In this case, you should not/ perhaps cannot use a simple calibration curve. One of the key assumptions of a calibration curve is that the matrix of the sample is the same as the standard. If you violate this rule, then we cannot apply this technique.
Consequently, you will have to use a technique called Standard Addition Method. It produces very accurate results, when the sample matrix is significantly different from the standards.


Space groups: Herman-Maugin notation, diagrams and Wyckoff positions I am currently taking an Introduction to Crystallography course. After studying and understanding the punctual symmetry groups (2mm, 4/m 2/m 2/m, 32,...), the flat groups and the diverse elements of symmetry that exist: helical axes, sliding planes, symmetry planes, among others.

I have doubts when building the diagram of a certain spatial group. For example, in an exercise, I am asked to construct the diagram of the spatial group Pmab.

Pmab implies the existence of:
 an ordinary plane of symmetry perpendicular to the crystallographic axis a,
 a sliding plane of type a perpendicular to the crystallographic axis b,
a sliding plane of type b perpendicular to the crystallographic axis c.

Also, Pmab is an abbreviated notation, because, in reality, there are also two helical binary axes parallel to the axes "a" and "b", and a binary axis parallel to the axis "c".

On the other hand, Pmab belongs to the orthorhombic (rhombic) system and derives from the point symmetry group 2/m 2/m 2/m 2/m.

This is all the information I get from the Herman-Maugin notation. If there is something wrong that I have understood, I hope you will tell me.

Once here I don't know how to start building the diagram, which would look like this:

As far as Wyckoff positions are concerned, I am often asked for them, but I am not quite sure what they are.

Well, knowing crystallography implies being able to deduce the picture above solely from the group's name $\rm Pmab$, and that's how it is done.

First you write down the symmetry elements mentioned explicitly in the group's name (that is, $\rm m,\;a$, and $\rm b$ in our case; in other examples there could have been axes among them, or even worse) and orient those accordingly, which you already know how to do. Then you combine these elements in all possible ways to find out what else is hidden beneath.

Essentially, any symmetry element is but a linear transformation that takes any vector $\bf\vec x$ to another vector $\bf A\vec x+\vec b$, where $\bf A$ is the $3\times3$ matrix of our transformation (rotation or reflection or otherwise), and $\bf\vec b$ is the shift, that is, a half-translation along the axis for a screw axis, or a half-translation in whatever direction for a glide plane, or $\bf\vec0$ in all other cases. The typical matrices are:
$$
\begin{array}{cc}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} &
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}&
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}\\
\text{Reflection $\perp$ X} & \text{Rotation about X}& \text{Center}
\end{array}
$$

Given two symmetry elements, you simply combine the two transformations by doing some matrix multiplication, which I sincerely recommend, if only to know how it feels. Alternatively, you may rely upon the wisdom of the predecessors, which says:

Two perpendicular planes give a twofold axis parallel to both.
A plane and a parallel twofold axis give another plane, perpendicular to the plane and parallel to the axis.
A plane and a perpendicular twofold axis give a center.
Two perpendicular twofold axes give a third axis, perpendicular to both.
All of the above remains true if you change

every "plane" to "plane or a glide plane" and
every "axis" to "axis or a screw axis".

When some of the combined elements have shift in them, the following happens:

If the shift is parallel to the resulting element, it becomes a part of the said element (that is, the result is going to be a screw axis or a glide plane).
If the shift is perpendicular to the resulting element, it moves the said element by half the value of the shift (that is, by $1\over4$ of a translation).

If you end up with a center of symmetry someplace other than at the origin, you relocate your origin there.

Now try to apply the rules 1-7 and see where this gets you.


Significant Figures Interpretation Please correct me if I’m wrong, but I’ve read that the reason why we use significant figures is to avoid making the result of a calculation more accurate than the starting values prior to the calculation. For example, if we were seeking the ratio of two weights, lets call them A and B, using a cheap scale and an expensive scale we could get the following varied results:

Cheap Scale
$$A = 34 \space oz$$
$$B = 23 \space oz$$

Expensive Scale
$$A =  34.0000 \space oz$$
$$B = 23.0000 \space oz$$

Computing the ratio $\frac{A}{B}$ using a calculator we get the following value:
$$\frac{34}{23} = 1.47826$$
Now this is a correct ratio depending on the scale used (ie the cheap scale could not have produced such an accurate measurement given the starting values)
This leads to the following question:

  Why are the zeros behind a decimal point ignored if they are just place holders?

For example, the value 0.000023 oz only has two significant figures, but isn’t that value something only a very fine tuned scale could measure (keeping in line with the previous weight example). So why are these zeros ignored?

This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.

What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz.

The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.

To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.

  but isn’t that value something only a very fine tuned scale could measure?

Not necessarily. The cost of an instrument is much more closely related to the relative precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.


Degassing HI from Reaction (Via Ultrasound Maybe?) I'm just going to put the questions first and the explanation after in case y'all don't have the patience.
I want to know if it's reasonable to think that :

Using an ultrasonic bath on something like a rotovap would prevent bumping, with the cavitation effects serving as basically the ultimate boiling chip.

Whether ultrasound could be used to degas an organic solution while a reaction is occurring, in order to definitely confirm that a gas is not supersaturating the reaction mixture.

Does it make any sense whatsoever that hydrogen iodide could stay in solution in toluene?

I'm considering strapping an ultrasonic transducer onto a metal pan we use as our oil bath. Is there a potential reason that this wouldn't be effective for 1) and 2)?
Context: working on a mild cannabinoid conversion (at a Canadian university, definitely not sketchy) via the following method:
https://pubs.acs.org/doi/10.1021/acs.jnatprod.7b00946
Long story short, we're investigating why side products are being produced and halving the yields when any larger amounts are used in the reaction. I'm seeing brown residue (immiscible with the reaction solvent of Toluene) collect and return down the condenser into the mixture, formed via this reaction:
4 HI + O2 → 2 H2O + 2 I2
Of which the HI and H2O then form concentrated hydroiodic acid droplets which return to the mixture.  I think this means that purging with nitrogen and otherwise degassing the solution is necessary. I think HI should basically boil out as it is formed due to very low boiling point and negligible solubility in toluene, so just doing it under nitrogen flow should be fine. Supervisor thinks some is sticking around in solution (which is why larger volumes at identical reagent concentrations also lower yields, as the smaller volumes means higher surface:volume and takes less time for the entire volume to boil/recondense/cycle compared to larger volumes), so basically what I'm trying to do is go so ham on degassing this solution that we can diagnose whether this is accurate.

In short, to answer your question "Why side products are being produced",  you appear to have ignored all possible radical activity!
Per this source: Free radical formation from sonolysis of water in the presence of different gases, for example, to quote:

$\ce{H2O + Sonolysis -> .H + .OH}$

Here, replacing H2O with HI likely results similarly in:
$\ce{HI + Sonolysis -> .H + .I}$
Some possible further reactions:
$\ce{.H + .H -> H2 (g)}$
$\ce{.I + .I -> I2 (s)}$
And, in the presence of an organic (RH):
$\ce{.H + RH -> .R + H2 (g)}$
$\ce{.OH + RH -> .R + H2O (l)}$
$\ce{.R + .I -> RI }$
$\ce{.R + .R -> R2 (g)}$
Also, with any dissolved oxygen (or even nitrogen, see cited reference), expect even more, as for example:
$\ce{.H + O2 -> .OOH}$
$\ce{.HO2 + .HO2 -> H2O2 + O2}$
$\ce{.R + .OOH -> ROOH}$
....
As such, expect a range of products, as would occur, by the way, if you performed photolysis in lieu of sonolysis. However, there are related reactions, like in the presence of ammonia, where the final photolysis products can display a high degree of product selectivity.


Are all enolised molecules called enolates? From what I've gathered in my lectures, enolates are enolised cabonyl compounds. However e.g. a nitrile with an H on the alpha C can undergo enolisation. Is the resulting molecule called an enolate or an enolate equivalent, or something else?

There are two questions in here.

First enols/enolates. It is correct that all enolates are enolised carbonyl compounds but not all enolised carbonyl compounds are enolates. Take for example pentan-2,4-dione. While in protic solvents this will indeed have two $\ce{C=O}$ groups, if dissolved in unpolar solvents it will enolise to allow for a cyclic intramolecular hydrogen bond for stabilisation (see scheme).

Obviously, the resulting compound is an enol, as the functional group $\ce{R2C=CR-OH}$ (a hydroxy group bound directly to a vinylic double bound) is present. However, it is not an anion so it is not an enolate.

Most carbonyl compounds prefer the keto-configuration but if there are electronic factors that favour enolisation (e.g. enolisation results in a sufficiently large conjugated system) then an enol will be preferred.

Enols can be easily deprotonated at the oxygen so the generation of an enolate is straightforward. However, since ketone and enols are tautomers of another with the key limiting step being hydrogen transfer ([1,3]-suprafacial hydrogen migration is forbidden), once the proton is removed the difference becomes purely electronic and thus deprotonated ketones are routinely referred to as enolates whether or not the cation seems more attached to the oxygen or the carbon. (Reaction mechanisms differ; sometimes only one attack (often carbon) results in a productive mechanism so it seems as if the other had not reacted or be not present but this is often a hindsight fallacy.)

The term enolate is not limited strictly to ketones. For one of the more synthetically relevant examples, see the Evans enolate used in Evans alkylation or Evans aldol reactions or the Myers enolate used in Myers alkylation reactions; both are amide enolates.

However, these compounds are still characterised by the presence of a $\ce{(C-)C=O}$ bond which upon enolisation becomes a $\ce{(C=)C-OH}$ functionality. In the case of a nitrile or an imine, the oxygen is not present so even though a similar reaction may take place it is not correct to call the resulting product enol. For imines, the compound is known as an enamine and happens to be the standard outcome if a secondary amine is reacted with a carbonyl group.

For nitriles, I am not aware of a similar shorthand and would probably simply call it a nitrile anion (as does Wikipedia). (It is worth pointing out that nitrile anions are somewhat different from enamides/enolates as they are essentially an aza-ketene, not an aza-allyl anion.)


Why is water an Arrhenius acid? Is water an Arrhenius acid? We define Arrhenius acidity with respect to water, but what will water itself be? It can donate $\ce{H+}$ as well as $\ce{OH-}$.

We define Arrhenius acidity with respect to water

That definition is new to me, whatever it is. The definition I know is:

An Arrhenius acid is a substance that dissociates in water to form $\ce{H+}$ ions.
An Arrhenius base is a substance that dissociates in water to form $\ce{OH-}$ ions.

The key concept to answer the question is the self-ionisation of water given in equation $(1)$.
$$\ce{H2O <<=> H+ + OH-}\tag{1}$$
This ionisation happens in water, and the ion product as written in equation $(2)$ is $10^{-14}$.
$$K_\mathrm{w} = [\ce{H+}] [\ce{OH-}] = 10^{-14}\tag{2}$$
Thus, water qualifies as a substance that dissociates in water to form $\ce{H+}$ ions. It also qualifies as a substance that dissociates in water to form $\ce{OH-}$ ions. It is both an Arrhenius acid and an Arrhenius base and thus the only Arrhenius amphoteric compound.


Feasibility of phenyl migration I am learning the fundamentals of organic chemistry,
here is the reaction which got me stumbling me for hours,
in the below reaction i could see that the amine group would get converted into diazonium salt, but i am clueless how to proceed further..

The compund A(solution) is given below, it seems like a phenyl group has migrated to the adjacent carbon, i have never seen something like this and it is also not clear for me why a carbonyl group is formed in the middle. Overall I dont understand the reaction...it would be helpful if someone points out what did i miss out in reading that made me unfit to solve this problem. Thank you!

This is similar to pinacol-pinacolone rearrangement. Only difference is in pinacol-pinacolone rearrangement, carbocation forms by dehydration while here, it forms through diazonium reaction. Following is the tentative mechanism of the rearrangement:

Since reaction progress through carbocation intermediate, the final product is a racemic mixture. $\ce{CH3-C+-CH3}$ (in pinacol) or $\ce{CH3-C+-H}$ (in given compound) can rotate about middle horizontal bond before alkyl or aryl group migration.


IUPAC naming of 1,2-disubstituted cyclohexane derivative I'm facing the challenge of naming this molecule:

We have to name it as an alkyl-substituted cycloalkane, since the ring has more carbons than the longest substituent (isopropyl). My questions:

When numbering carbons on the ring, we can start counting from either the methyl group (1-methyl...), or the isopropyl group (1-isopropyl...). Which one should I start with? Or do I need to start numbering with the substituent which name comes first, alphabetically?
I set the carbon number 1 on the methyl substituent, giving me the following name: 1-methyl-2-isopropylcyclohexane. However, my book says the correct name is: 1-isopropyl-2-methylcyclohexane. So the solution managed to sort the name both numerically and alphabetically. Is this what I need to do? Is the name I gave incorrect?
Which prefixes count towards alphabetically-naming a compound? Which prefixes don't? All I know is that the numeric prefixes like di- or tri- don't count. Are there any others which don't count either?

Simple prefixes (simple substituent groups such as methyl and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘methyl’ is considered to begin with ‘m’; ‘isopropyl’ is considered to begin with ‘i’.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Thus, the correct alphanumerical order in your example corresponds to the name x-isopropyl-y-methylcyclohexane.

Furthermore,

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g; however, Rule f does not permit a decision to be reached between ‘1-isopropyl-2-methylcyclohexane’ and ‘2-isopropyl-1-methylcyclohexane’ since both names correspond to the locant set ‘1,2’.

According to Rule g, your example is named as 1-isopropyl-2-methylcyclohexane rather than 2-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.

Note, however, that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.

Therefore, the preferred IUPAC name for your example is actually 1-methyl-2-(propan-2-yl)cyclohexane because ‘methyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).


Water-water vapour equilibrium above the boiling point Consider $\ce{H2O_{(g)} -> H2O_{(l)}}$ at $\pu{105^{\circ} C}$. Under these conditions it is false that:
(1) $T\Delta S_r$ is negative.
(2) Relative humidity is 100%.
(3) $|\Delta H_r|>|T\Delta S_r|$
(4) $P_{\ce{H2O_{(g)}}}>\pu{1 atm}$
(5) $|{\Delta G_f}_{\ce{H2O_{(g)}}}|=|{\Delta G_f}_{\ce{H2O_{(l)}}}|$

Been trying to understand this question. Here is my reasoning:

True, because $T$ is positive $\pu{(105 + 273)K}$ while $ΔS$ should obviously be negative because the final entropy is lower than the initial entropy. We are after all going from water's gaseous state to its liquid state. There are fewer degrees of freedom in water's liquid state.

If there is an equilibrium between water vapor and water liquid, then by definition relative humidity is 100%. The air is saturated with water vapor to the point where it can't even hold the water vapor and water vapor exists in a dynamic equilibrium with liquid water. From Wikipedia:

Saturated air is really just air which contains water vapor in equilibrium with a liquid water source.

False, because $ΔG$, not $ΔG^{\circ}$. $ΔG$ for a system at equilibrium is always $0$. $ΔG^{\circ}$, not necessarily $0$. So we have $0 = ΔG = ΔH - TΔS$. This implies $ΔH = TΔS$.

True. For liquid water to boil, vapor pressure must equal atmospheric pressure. Similarly for water to condense, atmospheric pressure must equal vapor pressure. Also, vapor pressure is solely dependent on temperature of the pure liquid. The below statement is from UC Davis' ChemWiki. I would think that adding solute would change the vapor pressure (and yes it does, and it's discussed further down the page of the site).

Vapor pressures are dependent only on temperature and nothing else.
Taking into consideration that water boils at $\pu{100^{\circ}C}$, we may conclude that at $\pu{105^{\circ}C}$, the vapor pressure of liquid water is greater than $\pu{1 atm}$. And therefore, for this equilibrium to exist, both the vapor pressure of liquid water and the atmospheric pressure of gaseous water must be equal and thus greater than $\pu{1 atm}$.

This is true. This would make $ΔG = 0$ since $ΔG = G_{\text{(products)}} - G_{\text{(reactants)}}$.

I would like to know if my reasoning is correct.

All the arguments are perfect except the fourth.

The statement is true, but the premise that the vapour pressure of water must equal the atmospheric pressure is incorrect. A phase equilibrium between the gaseous and the liquid phase can be established at a particular temperature without the liquid boiling. Boiling is a special case of this equilibrium when the atmospheric pressure is entirely constituted by water vapour in the vicinity of the liquid surface and hence the vapour pressure can no longer rise at that pressure. In normal cases, as is possible in this case also, vapour pressure of water is just one component of the atmospheric pressure and the liquid need not be boiling.

A more accurate sentence would be $P_{atm}\geq P_{vap}$. This would still mean that both vapour pressure and the atmospheric are higher than 1 atm but not necessarily equal.


How to identify which of two given electrodes with corresponding solutions is reducing more? For the sake of the question, let’s assume that you are provided with two labelled rods of the pure metals $\ce{A}$ and $\ce{B}$ and two unlabelled bottles containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{A^m+}$ and  the other bottle containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{B^n+}$.

I want to find out, which metal out of the given two is more reducing, and identify each solution that is given in the unlabelled bottles. What are the procedures that I should follow?

If I immerse $\ce{A}$ in a solution of  $\ce{A^m+}$, I guess  there will be no change. Because if I do so there would be an equilibrium prevail with the ions in the solution and the electrode, and I think this is true for $\ce{B}$ also.

But I don't know how to identify the one which gets reduced more, yet I know that the anode gets reduced. So I am confused about that.

I feel that the only clue I have is immersing $\ce{A}$ in  $\ce{B^n+}$ and vice versa for $\ce{B}$ also. But  really I don't know what  would happen if I do so.

The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that $\ce{A}$ is the more reducing metal. We can consider what happens in each case:

If we immerse $\ce{A}$ in a solution of $\ce{A^m+}$ nothing will happen.
Similarly, if we immerse $\ce{B}$ in a solution of $\ce{B^n+}$ nothing will happen.
If we immerse $\ce{B}$ in a solution of $\ce{A^m+}$ nothing will happen because the electrode potential for $\ce{A|A^m+}$ is more negative (more negative means its a stronger reducing agent) than for $\ce{B|B^n+}$ so the reaction potential is negative and the reaction is not feasible.
However, if we immerse $\ce{A}$ in a solution of $\ce{B^n+}$, $\ce{B}$ will precipitate out on the surface of $\ce{A}$. This is because $\ce{A|A^m+}$ has a more negative electrode potential than $\ce{B|B^n+}$ and so the reaction potential will be positive. $\ce{A}$ will reduce $\ce{B^n+}$ ions to form $\ce{B}$ metal and $\ce{A}$ will be oxidised to form $\ce{A^m+}$:
$$\ce{A (s) + $\frac{m}{n}$B^n+ (aq) -> A^m+ (aq) + $\frac{m}{n}$B (s)}$$

In short, the most reducing metal is the one which causes precipitation of the other metal out of its solution.


Are difluorine and hydrogen chloride considered isoelectronic? Why are $\ce{F2}$ and $\ce{HCl}$ not considered isoelectronic? Both have 18 electrons.

$\begin{align}
\ce {H}\!:&\  \ce{1 e-}\\
\ce{Cl}\!:&\  \ce{17 e-}\\
\ce{F}\!:&\  \ce{9 e-}\\
\end{align}$

$\ce{F2} \implies 2 \times \ce{9 e- = 18\ e-}$

$\ce{HCl \implies 1 e- + 17 e- = 18 e-}$

If we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have the same number of electrons, they are actually isoelectronic. Please see "Ultraviolet Spectroscopy And Uv Lasers"
 edited by Prabhakar Misra, Mark A. Dubinskii, page 35 (available in google books)

But if we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have a similar electron configuration and the same structure, they are not.


Why does water condense on grass and iron grills in cold weather? In my book it is said that in cold weather fog appears. This fog is nothing but water droplets. After a hot day if the night is cold then the moister in air starts to condense. What I do not understand is why does water condenses heavily on specific places like grass and iron objects only? I first thought that since iron is good conductor of heat so it becomes cold at night early which then captures nearby moisture more effectively than usual objects. But then grass is a poor conductor of heat. Another thought is that most of the objects absorb water. Those which do not absorb are metals and grass.

Please explain.

While grass IS a poor conductor of heat, it is super thin and will cool off really quickly because of of that alone. The increased surface area also means a greater area is exposed to the night air during the time when the moist air drops to the ground and will thus collect more dew than maybe a stone or a tree trunk.


Treatment of copper (0) with sodium hydroxide I recently made self-assembled monolayers (SAMs) of alkanethiols on a copper substrate. After scrubbing off the $\ce{Cu2O}$ top layer and rubbing the surface with ether and ethanol, I let my copper plates stand in a 5% (w/w) $\ce{NaOH}$ solution for 5 minutes, which I then throughly rinsed off with deionized water.

I'm wondering if I could skip this step. I'm not sure what it is needed for. My knowledge of chemistry tells me that $\ce{Cu(0)}$ should not react with sodium hydroxide. I'm also positive that any grease or other contaminants were washed off in the previous steps.

What do you think?

$\ce{Cu(0)}$ will not be attacked by $\ce{NaOH}$, and will not decompose in this aqueous environment at room temperature. So, this is not to create some reaction with the copper. Two uses I can think of:

$\ce{NaOH}$ increases the solubility of $\ce{Cu2O}$, so maybe this is a way of getting rid of the last remaining copper oxide.
To bind the thiols with copper, you'd better be in alkaline conditions rather than acidic conditions, right? If so, then maybe this is the way of making sure you don't have any acidic groups or molecules remaining adsorbed on the copper surface, to improve bonding at the next step.


What is an alternate way of preparing a sodium hydrogen phosphate/sodium dihydrogen phosphate buffer through titrating an acid with a strong base? So my friend and I are stumped. This is in the context of achieving the sodium hydrogen phosphate/sodium
dihydrogen phosphate buffer by titrating the acid with a strong base.
My friend says the answer would be H3PO4 + Na3PO4. In my case, I'd think that sodium hydrogen phosphate would be the acid (but I genuinely am not sure) that we'd titrate in that scenario and all we'd need then is a strong base (although I honestly don't know how to figure out what strong base in particular you must use.)
What's correct? Or what would be the correct way?

This picture says more than a thousand words:

You want a mixture of $\ce{HPO4^2-}$ and $\ce{H2PO4^-}$. You can start with any species (in a solution with a certain pH) and add another species whose pH is "on the other side" of the desired pH. If the pH of your starting solution is lower than the desired pH, you can also titrate with NaOH.
What you actually do (of the six options) is determined by solubility of the solution and by availability. If you titrate two phosphate solutions of the same concentration, there will be no dilution effect. If you titrate with NaOH, you would used fairly concentrated NaOH to avoid dilution, but would have to add water to a defined volume (or measure the volume after reaching your desired pH) to figure out the concentration of your buffer.
You should not titrate with HCl because then you end up with extra sodium choride in your buffer.


What is this piece of glassware called? Here's an image:

The left side is designed to be fitted into a beaker, and it captures gas evolved during a reaction.  The right side can be filled with a liquid to then react with that gas.  We use it to capture ammonia from a decomposition reaction, with an indicator in the right half as part of an identification test.  I'm sure it must have a name, it's a commercial piece of glassware, but the box we keep it in in the lab doesn't identify it.

The picture shows parts of an arsine generator.

The apparatus can be used to determine trace amounts of arsenic by the silver diethyldithiocarbamate photometric method.

A typical method description can be found here (this is also the source of the drawing):

https://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/c211.pdf

or

http://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/GC_pdf_USP38/c211.pdf


Aldol of 7-oxooctanal I want to do intramolecular aldol condensation of 7-oxooctanal:

I am talking about the major product. Clearly, a 6-membered ring would be a major product, but there are 2 possibilities: taking hydrogen from $\ce{C^2}$ or $\ce{C^6}$. Which one would be preferred?

If in doubt, draw:

Deprotonation at C-6 under thermodynamic control results in the more stable enolate (1c) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction.

As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene.

However, if that is your target molecule, addition of acetylene to cyclohexanone, followed by an acid-catalyzed rearrangement of the propargylic alcohol would be an alternative route.


Calculating stoichiometry of a reaction between xenon and fluorine Xenon and fluorine were mixed in a 1:1 ratio by volume and sealed in a vessel attached to a barometer. After some time, crystals were found to have formed on the walls of the vessel and the barometer showed that the pressure had dropped to 70% of the original pressure. Analysis of the gases remaining indicated 4 parts xenon to 3 parts fluorine. Construct a balanced equation for the reaction. Assume constant temperature throughout.

Equal volumes of gas at same temperature and pressure have the same number of molecules, thus I deduce that the mole ratio of the 2 gases initially is 1:1. Since $pV = nRT$ then the number of moles of the two gases combined would be 70% of the total number of moles of the 2 gases at the start of the experiment.

I am unsure of how to proceed from here.

Xenon and fluorine were mixed in (1) a 1:1 ratio by volume and sealed in a vessel attached to a barometer. After some time, crystals were found to have formed on the walls of the vessel and the barometer showed that (2) the pressure had dropped to 70% of the original pressure. Analysis of the gases remaining indicated (3) 4 parts xenon to 3 parts fluorine. Construct a balanced equation for the reaction. Assume constant temperature throughout.

You have properly analyzed (1) and (2). It remains to analyze (3). We started with ten parts of gas, 5 parts xenon and 5 parts fluorine. From (2) and (3), we see that 4 parts xenon and 3 parts fluorine are left in the gas phase. Hence 1 part xenon ($\ce {Xe}$) and 2 parts fluorine ($\ce {F2}$) must have reacted to form crystalline $\ce{XeF4}$.

The balanced reaction is left as a simple exercise.

The choice of starting with ten parts of gas (as opposed to fifty parts, or a hundred parts) is arbitrary. Since we're given only ratios, we can choose an arbitrary number of parts of gas and work with that throughout. Choosing ten makes the math simple.


Removing denatured Crystalbond? I have some water-soluble crystalbond that is attached to silicon and, because of some overheating (to about 150 Celcius) during a final processing step, the crystalbond seems to have charred (white/grayish colour, see photo at the bottom) and is no longer dissolvable in water. Since this silicon is important and cannot be replaced, I really need to find a way to chemically remove this overheated crystalbond residue without using abrasives that could scratch of the thin SiO$_2$ coating on silicon.
Looking at the MSDS (see here), crystalbond is composed of the following

97% Nonylphenol Polyethylene Glycol Ether(127087-87-0) EC-No.
500-315-8
3% Polyethylene Glycol(25322-68-3)EC-No. 500-038-2

My question is: does anyone have any ideas for what crystalbond decomposes into around 150 C and how one could remove this overheated crystalbond residue?
Things I have tried (unsuccessfully)

Hot DI water (in a cup, and in a ultrasonic bath)
Acetone
2-propanol/ethanol
"Piranha" etch (asked a lab friend to do this, too scary for me!!)
Photoresist remover (lab friend also tried this)
Hexane
Dunking in LN2
Putting on a hot plate at 225C for about 24 hours
Burning under a 1100C torch for about 30 seconds

Any bright ideas, especially from organic chemists? All I can say is: damn this stuff doesn't want to come off.
Some examples of the residue below.
Lots of crystalbond residue on this Silicon wafer

Another silicon wafer with crystalbond residue. Less than the one above, but still significant.

Just a quick answer without looking too much into it.
You have been using a lot of polar solvents without any luck. The reason is that the molecule no longer has the groups that make it water-soluble because of the thermal decomposition. Only hydrocarbons chain remain. You need to use an organic solvent (toluene, benzene, hexane). Since you probably don't have access to a lab, I would try with gasoline.
Let me know if it works.
Cheers.


Calculate the number of Carbon atoms in .0400g C7H5NO3S I understand how to get from grams to moles and from moles to molecules with Avagadro's number. That gives us 1.31x10^20 molecules of C7H5NO3S. Now we take that and multiply by 7 to get the individual Carbon atoms. This is where I get confused because my calculator says 9.17x10^20 but my book says 9.20x10^21. Please help explain!

.0400g is a really non-chemical way to write that number.

I expect the text meant 0.400g, which is 10 times higher than the number actually printed, and this would explain why the book's answer is 10 times higher than the actual answer: they are working from the number they didn't print.


Why can't primary alcohols react with sodium bromide via an SN2 mechanism? I was told that a primary alcohol reacts with $\ce{HBr}$ for example, but not with $\ce{NaBr}$ through the $\mathrm{S_N2}$ mechanism in the same reaction environment. Why?

This reaction can occur:

$$\ce{R-OH + HBr -> R-OH2+ + Br- -> R-Br + H2O}$$

But this reaction does not occur:
$$\ce{R-OH + NaBr -> R-OHNa+ + Br- -> R-Br + NaOH}$$

For $\mathrm{S_N2}$ to occur, the $\ce{-OH}$ has to be "helped along" to leave, by converting it to a good leaving group (good leaving groups = weak bases). If you protonate it with a strong acid, $\ce{Br-}$ can attack, and a molecule of water will be kicked out as the leaving group.

So why doesn't it work with $\ce{NaBr}$? A proton is a strong lewis acid, meaning $\ce{HBr}$ can protonate a small fraction of the $\ce{-OH}$ groups. $\ce{Na+}$ is not, on the other hand; it is too electropositive (just as $\ce{Na}$ tends to give up an electron and become $\ce{Na+}$). It simply cannot hold on to the electrons from a formally neutral oxygen-atom at all. Hence, it's like you are still trying to get $\ce{HO-}$ to leave.

  Extra - when oxygen has a formal negative charge

If you go on and study carbonyl chemistry you will find places where $\ce{Li+}$ (which has a better size to interact with $\ce{O-}$ than $\ce{Na+}$) can stabilise the so-called tetrahedral intermediate, and help keep it from collapsing.

This is useful to stop the reduction of, for example, esters at the aldehyde, instead of continuing the reduction to the primary alcohol.

To get the lithium-ion to dissociate, you just add water. As DHMO explained, as soon as you have water, the positively charged group I metal ions are simply solvated so well by the water, that they are virtually inert.


Wetting agent for H2O2 solution I need to increase the wetting of an $\ce{H2O2}$ solution on ceramic surfaces. Does anyone know of a common chemical I can use for this?

The properties it would ideally have are:

Increases wetting of $\ce{H2O2}$ - water mixture on ceramics (particularly glazed tiles and tile grout).
Does not react or reacts very slowly with $\ce{H2O2}$ (and water).
Non-toxic / low toxicity. Food safe would be ideal.
Reaction products are non-toxic / low toxicity.
Reaction products are all either soluble in water or evaporate under atmospheric conditions.
Commonly available and reasonably cheap.
Non-foaming or anti-foaming.

Thank you so much to anyone who has a suggestion.

You might consider chlorhexidine gluconate. It is a cationic surfactant used as a skin cleanser and disinfectant prior to surgery, and is an ingredient in certain mouthwash products, so it must have low toxicity. Drug stores sell scented and unscented solutions for hand washing -- choose the unscented.


Shortcut for calculating pH of mixtures My teacher gave me a shortcut to calculate the resultant pH if 2 or 3 solutions are mixed, which are relevant only if the difference between the pH of individual solutions is 1.

If there are two solutions, the pH of the mixture $\approx$ (the mean pH − 0.24).
If there are three solutions, the pH of the mixture $\approx$ (the mean pH + 0.56).

What is the logic behind this approximation?

Case 1: Mixing equal volumes of two solutions with a pH difference of 1.
Initially we have:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404
$$
Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.
Before mixing:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
c&\frac{x}{100} & -\log x + 2 & \frac{xV}{100}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682
$$
(It seems you got the sign wrong on your last equation.)
Caveats:
Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below $\mathrm{pH}$ of 0 and above $\mathrm{pH}$ of 14.
You also get a slight deviation if you mix solutions that have $\mathrm{pH}$ too close to neutral, because the auto-dissociation of water affects the equilibrium values of $[\ce{H+}]$ and $[\ce{OH-}]$.
For example, if we mix two equal amounts of liquid, one being pure water($\mathrm{pH} = 7$) and the other being aqueous $\ce{HCl}$ solution with $[\ce{HCl}] = \pu{10^{-6}M}$. Considering the auto-dissociation constant for water as $K_\mathrm{w} = 10^{-14}$, the $\mathrm{pH}$ for the last solution is actually approximately $5.9957$ because water auto-dissociation already affects it slightly.
Neglecting that, we mix the liquids creating a solution with $[\ce{HCl}] = \pu{5\times 10^{-7} M}$. Directly using $[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$ yields a $\mathrm{pH}$ of $6.3010$.
Using the trick, we get the pH for $[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$ as $6.2596$. However, the real value for $[\ce{H+}]$, taking into account the shifting of the water auto-dissociation equilibrium, is $[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$, such that the real $\mathrm{pH}$ is $6.28461$. A small deviation surely, but it's present.


What is the Crystal System of MoS2? I'm confused about the crystal system of MoS2.

In the case of the monolayer MoS2, it has 3-fold rotation symmetry when viewed from the top of its unit cell (c-axis). Even in the case of bulk (AB stacking) MoS2, it has only 3-fold rotation symmetry from the same direction. So, I expected that it must be the trigonal system. However, 'Materials Projects' doesn't seem the trigonal system.

Here are some database from 'Material Projects'

List of MoS2

https://materialsproject.org/#search/materials/{%22reduced_cell_formula%22%3A%22MoS2%22}

Bulk MoS2

https://materialsproject.org/materials/mp-2815/

Bilayer MoS2

https://materialsproject.org/materials/mp-1025874/

and so on.

Some of them are the trigonal system, and the others are the hexagonal system. I studied the minimum symmetries determine the crystal system, or that is a misconception. But now, it might be wrong by the web data.

Which one is correct and if both are correct, how can I check the 6-fold rotation symmetry for the hexagonal system?

(appended--------------------------------------------------------------------)

This is the figure for the HCP structure. Its symmetry along the c-axis can be called 3/m or bar(6).

This view is the bulk MoS2 along c-axis. In the unit cell, it has two layers of MoS2. However, it has 6/mmm point group. How does this happend?...
https://materialsproject.org/materials/mp-1018809/#

Naturally occurring molybdenum disulfide can occur in either of two crystal systems, hexagonal or "trigonal"; the latter is also known as rhombohedral (see here).

In both cases the coordination geometry around molybdenum has only threefold rotational symmetry, but in the hexagonal form the crystal will add a center of symmetry to become hexagonal.  Such a conversion is seen in other cases such as hexagonal close-packed metals like magnesium and zinc, which actually have only threefold symmetry about any axis perpendicular to the basal plane and passing through an atom, yet the Crystal's are hexagonal.  In general, when you have a unit cell with $D_{3h}$ symmetry -- a mirror plane perpendicular to the threefold exis -- you will get hexagonal crystals in both molybdenum disulfide and hexagonal close-packed metals.

The rhombohedral crystal, with a different arrangement of mirror planes (the point group symmetry is $D_{3d}$ instead of $D_{3h}$, and $D_{3d}$ has no mirror plane perpendicular to the threefold axis), has the required center of symmetry without going from threefold to sixfold rotational symmetry.


Name of "the" benzodiazepine Wikipedia has a thorough article on the benzodiazepines. Of course, it is written primarly  from the pharmacological point of view; but there is also the "Chemistry" section, which includes the IUPAC name. The parent hydride for the compound class is named 1H-benzo[e][1,4]diazepine. I do not understand why it is benzo[e][1,4]diazepine and not benzo[f][1,4]diazepine: shouldn't the bond alphabetic designation sequence go around the original diazepine ring, starting with the 1,2-bond (a), then continuing to 2,3 (b), 3,4 (c), 4,5 (d), 5,6 (e) and, finally, 6,7 (f), to which side the benzene ring is attached? Could please someone clarify the matter to me?

Let's look at the „[1,4]diazepine“ part more closely. (Standalone, it's correctly named 1,4-diazepine.) The „-epine“ in the Hantszch-Widman system stands for unsaturated seven-membered ring. More precisely, one with the maximum number of noncumulative double bonds. (IUPAC uses that term instead of „conjugated“ probably to avoid confusion with physical electronic conjugation, which might not be the case, as we're at the level of topology, not even geometry.)

The exact structure is specified with indicated hydrogen. There are four 1,4-diazepine structures, 1H-1,4-diazepine, 2H-1,4-diazepine, 5H-1,4-diazepine, and 6H-1,4-diazepine (not considering their stability/existence in tautomeric pair/multiple):

In the world of abstract organic chemistry nomenclature, 1,4-diazepine without indicated hydrogen can be represented as following structure (dotted bonds for unknown bond order, any arrangement of noncumulative dbl. bonds assumed; no explicit hydrogens):

When fusing with benzene, 1,4-diazepine part has to be numbered as follows (not the other way), to get lowest numbers/letters when starting with heteroatom:

… obtaining benzo[e][1,4]diazepine (new numbering of the resulting structure marked, partially; coincidentally we get the same set of numbers for nitrogens):

Indicate the hydrogen to get our 1H-benzo[e][1,4]diazepine:

(But note that fused ring nomenclature is quite long and complicated, generally the final numbering could also depend on the indicated hydrogen position.)

Finally, note that the new nomenclature has simplified the names of benzo-fused rings, the current preferred name is 1H-1,4-benzodiazepine.


Why is HCl a stronger acid than Acetic Acid? Drawing the conjugate bases,  Oxygen is more electronegative than Chlorine so negative charge on Oxygen is more stable than Chlorine,  Thus My argument that Acetic Acid is more acidic.

Also the O¯ ion is in resonance with another Oxygen so that should also add to the acidic nature.

Where am I going wrong?

A simpler question, would be why is water a weaker acid than $\ce{HCl}$?  If we were to look at that issue, we might clear some of the mess up.

Strongly reactive compounds produce weakly reactive products.

In the case of acid base theory, chloride ion is a much weaker base than hydroxide.  Why is this so?  The hydroxide is much smaller, and what is referred to as 'harder'.  The guts of the issue is that in order for a base to deprotonate anything it has to donate two electrons to a bonding region near a proton (and in doing such render the orbital manifest, and the bond existent).  In oxygen based anions, those two electrons are more likely to be in a confined region, and hence more likely to donate to the putative $\ce{HO-H}$ bond that is forming, while $\ce{Cl^-}$, being larger has a harder time getting two electrons into the bonding space required to form a $\ce{Cl-H}$ bond.  So hydroxide is a stronger base.  This means that when donating a proton, it is likely to just yank it back again from whatever it just gave it up to, as compared to $\ce{Cl^-}$.  $\ce{HCl}$ gives up the proton and is more likely to not recapture it.

Groovy?  Back to acetate.  The resonance in acetate creates some diffusion of the negative charge and makes it harder to form said $\ce{AcO-H}$ bond, and as such it is a weaker base than hydroxide.  However, the smaller oxygen atoms are still better and forming bonds than $\ce{Cl^-}$ and as such it is a stronger base than chloride, and therefore its conjugate acid is less acidic.

Strong acids: weak conjugate bases.
Strong bases: weak conjugate acids.


Production of hydrogen using vinegar and metal for experiments If
 I cannot, or do not want to use electrolysis of water to produce
hydrogen, is it a viable alternative to react vinegar (ethanoic acid)
and iron (or steel, or copper) to form iron(II) or copper acetate? (Assume steel wool or copper wires.)

If I tried to fill a balloon with the produced gas (assuming no leaks
apart from balloon), would the hydrogen leak out of the balloon faster
than it would be produced?

Acetic acid will probably react to slow. If you want to produce hydrogen in the litre scale I would suggest to use diluted hydrochloric acid and zinc flakes or iron. But be aware that it can be difficult to control the reaction. In a laboratory you could use a Kipp's apparatus for this.


Computational Chemistry Modelling Redox Potentials How could I accurately determine the Fe3+/Fe2+ redox potential in an aqueous solution? What would be a good computational model(s) for this?

Here is a non-electrolysis scenario that could be of interest. Add deoxygenated water to FeCl2 in a closed vessel filled with oxygen or air. Record the oxygen consumption.

Chemistry:

$\ce{4 Fe(2+)  + 2H2O + O2 (g) -> 4 Fe(3+) + 4 OH- }$

The wet ferrous chloride should be presented in a fashion to maximize oxygen exposure from air.

[EDIT] I would also place the system (till the reaction is completed) under pressure to limit adjustments owing to the speed of oxygen assimilation.

To be clear, as apparently required by a comment, the purpose of this experiment, per my understanding of the Question, is to serve as a frame for real world validation of a sophisticated theoretical/computational based model.


Grignard vs organolithium reagent opening cyclohexene oxide Could someone please explain the reason for the two different products formed? One with RLi and the other with RMgBr?

$\ce{RLi}$ is much more reactive than $\ce{RMgBr}$ and attacks through the SN2 process, thus inversion of configuration occurs. But in case of $\ce{RMgBr}$, $\ce{MgBr-}$ coordinates with epoxide O atom and forms carbocation on the adjacent carbon. Then a $\ce{C-C}$ bond migrates and the above product is formed.


Getting to use the lab! Today I finally mustered the courage to go and talk to the director of the labs to politely ask him if I could use the labs out of class hours. Long story short, our lab sessions "are bad" because we are given a protocol, and like babies we just need to follow it. No brainer, and I am against that. So I got permission to use the lab for experiments that don't pose a mayor risk, as well as a slight restriction of the reagents.

I'm not sure of which reagents we have in the lab, but there aren't that many there. So, now to the point. I was hoping that you could tell me which experiments I should try. I came up with the idea when we worked with ferric ammonia ferrocyanide $[Fe(CN)_6][Fe(NH_4)_2]$ and I had to balance the chemical equation of the reaction that we used. However, I had no idea of what the byproducts where, because I had no way to test for their presence! My idea is to bring an hypothesis and then, starting pretty much from scratch (except reference tables or such) prove or disprove it.

Could you share a nice example of something I could test in the lab? If not, where can I find stuff that I can work with, without getting the answer right out of where I got the info. Did I get to express my idea correctly? More or less?

Thanks for the help!

P.s.: I kindly asked a professor to supervise my work, so I would get help regarding methodology.

http://books.google.com/books/about/Crystal_Growth_in_Gels.html?id=pQ-JuNXEL6wC
  The classic text.  Set them up, come back in a few weeks.  (Potentially any gel - polyacrylamide, agarose, agar, gelatin...)  Meanwhile,

The Prussian Blue prep works across the transition metals, with different colors.  Alas, it is not crystalline (but maybe in gel...).  Vanadium oxidation states are a spectrum of colors.  Vanadyl, V=O, is especially fine (but not as a food supplement).

Cobalt can be preparatively more fun than iron.  Co(II) is substitution labile - put anything on it you want, in real time.  Co(III) is substitution inert.  Anything you put on Co(II) then oxidize to Co(III) stays forever (absent a trace of reducing agent to reopen the door).   There is vast and colorful Co coordination chemistry.

http://onlinelibrary.wiley.com/bookseries/10.1002/SERIES2146
  Everything in there works to spec.  Fine some interesting things to do (and scale down to grams).

Cobalt sepulcrate and especially sarcophogate (Alan Sargeson)!   Tris(ethylenediamine)cobalt(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 6 186 (1960)) or made entirely as one hand.  Net charge is 3+.  Tris(oxalato)cobaltate(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 8 204 (1966)) or made entirely as one hand.  Net charge is 3-.  Tris(ethylenediamine)cobalt(III) tris(oxalato)cobaltate(III) as a salt with same or opposite hands could be interesting.  The crystal will certainly be tightly bound.


How potassium ferrocyanide is coloured? The colour of a complex compound is due to unpaired electrons. As per crystal field theory, $\ce{K4[Fe(CN)6]}$ has no unpaired electrons so it has to be colourless. But then why is it coloured?

There is no direct connection between having unpaired electrons and having a colour. In fact, many colourful compounds have no unpaired electrons and yet display a bright colour: potassium permanganate, potassium dichromate, azobenzene, sudan red, phenolphthalein, iodine and many, many more. Arguably, there are more coloured compounds known in the singlet state (which means no unpaired spins in the vast majority of cases) than in the dublet, triplet or any other state.

So what causes the colour? It is the possibility of electron transitions from a lower to a higher orbital. If that transition is equivalent to a photon with a visible wavelength, we observe a colour. Examples:

Most phenyl groups absorb light at c. $\pu{250 nm}$ — to short a wavelength to be visible; they are ‘ultravioletly coloured’.
The electronic transitions of most transition metal complexes (except $\mathrm{d^0}$, $\mathrm{d^5}$ and $\mathrm{d^{10}}$ complexes) are well within the visible range of $400$ to $\pu{700 nm}$.
In rare cases, extensive electronic systems may even allow the absorption maximum to be shifted outside of the visible range into the infrared range. Chlorophyll comes rather close to this, absorbing red light (the longest visible wavelengths).

For potassium hexacyanidoferrate(II), the colour we observe is yellow so the absorption must be the complementary blue/violet, indicating a relatively large energy difference.


Which is the strongest base relative to ammonia? Among these bases,

  ammonia
  methylamine
  ethylamine
  dimethylamine
  trimethylamine
  quinuclidine

  1) Which is the strongest base relative to ammonia? Why?

  2) Which conjugate acid would be the strongest and which will be the weakest? Why?

I'm going to assume I need to look at the relative basicity (which is in kcal/mol)

ammonia: 0
methylamine: 10.1
ethylamine: 13.2
dimethylamine: 16.6
trimethylamine: 21.1
quinuclidine: 28.9

Looking at these values I would want to guess that means quinuclidine is the strongest base; and that ammonia is the strongest conjugate acid and quinuclidine is the weakest conjugate acid.

Is this correct?

The more I think about it I think I should be looking at proton affinity because the largest proton affinity is going to be the strongest base, which is still quinuclidine. but I'm still unsure about the conjugate acid.

Let's have a look at some of the definitions in the gold book and see how they relate to your problem.

  proton affinity
  The negative of the enthalpy change in the gas phase reaction (real or hypothetical) between a proton (more appropriately hydron) and the chemical species concerned, usually an electrically neutral species to give the conjugate acid of that species. Proton affinity is often, but unofficially, abbreviated as PA.

  gas-phase basicity
  The negative of the Gibbs energy, $\Delta G_\mathrm{r}^\circ$ change associated with the reaction:
  $$\ce{B + H+ -> B+-H}$$
  in the gas phase. Also called absolute or intrinsic basicity.

I was unable to find an exact definition for relative basicity, so I am going to assume based on the following reasoning. First we define the basicity for ammonia, $Bf(\ce{NH3})$, and an arbitrary amine $Bf(\ce{NR3})$.
\begin{align}
\ce{NH3 + H+ &~-> NH4+}  -\Delta G_\mathrm{r}^\circ(\ce{NH4+}) &
Bf(\ce{NH3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NH4+})\tag{1} \\
\ce{NR3 + H+ &~-> NR3H+}  -\Delta G_\mathrm{r}^\circ(\ce{NR3H+}) &
Bf(\ce{NR3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NR3H+})\tag{2} \\
\end{align}

We can follow up and define the relative basicity of an arbitrary amine  to ammonia $Bf_\mathrm{rel}(\ce{NR3})$ as the difference of the independent basicities.
$$Bf_\mathrm{rel}(\ce{NR3}) = Bf(\ce{NR3}) - Bf(\ce{NH3})\tag{3}$$

With (3) we would actually describe the following reaction:
$$\require{cancel}
\begin{align}
\ce{NR3 + \cancel{H^+} + NH4+ &~-> NR3H+ + \cancel{H^+} + NH3}\\
\ce{NR3 + NH4+ &~-> NR3H+ + NH3}\\
\end{align}$$

So the relative basicity describes in this scenario how likely it is that the arbitrary amine is abstracting a proton from the conjugate base of ammonia, i.e. ammonium. The higher the relative basicity, the stronger the base.
The conjugate acid $\ce{NR3H+}$ is the strongest if its parent compound has the smallest basicity, meaning most likely to loose the proton. The weakest conjugate acid is obviously the strongest base.

But what have we actually compared? In principle we have discussed with this comparison how likely it is for the amines to bind a proton and we have chosen to do that in comparison with ammonia. We could chose to do the comparison to water and with a couple of transformations and simplifications we would arrive at a $\mathrm{p}K_\mathrm{a}$ which we would then compare. To make things easier, we could also directly compare the values of $\Delta_\mathrm{r}^\circ$ in (1) and (2); the lower the value, the higher the basicity.

If we now choose to compare the enthalpies instead of the Gibbs energies, then we would be ultimately comparing (gas-phase) proton affinities. We would be still looking at the same reactions, just a different characteristic property. We would still be observing the same trend.

TL;DR It does not matter if you look at proton affinities or relative basicity, you are still comparing the same effect with different numbers, that are all linked.


Why some reaction have no transition state? All stationary point are local minimum, if any parameters changed the energy should be increase. There is no way for a stationary point that energy go straight down to reach another.

I was told that radical reaction have no transition state, but why? What kind of reaction also do not have transition state?
Is there any way to confirm there is no transition state using quantum chemical calculation method?

There is nothing special in having an activation barrier between reactants and products although, obviously, very many reactions do. But some reactions, such as electron transfer, are observed experimentally to have no barrier which happens when the product potential crosses that of the reactant at its minimum (i.e. between the 'normal' and 'inverted' regions).  In fact  electron transfer is an interesting case as it is possible to control the barrier height by changing the redox potential. Some atom diatom reactions such as H+F2 also have very very small barriers.  Whether you want to call the crossing point a transition state when it has a zero barrier is a moot point. The behaviour is more like water going over a waterfall, so limited by the transport of the wavepacket along the reaction path, rather than sometimes being reflected at the barrier and sometimes crossing.


Is exposure to amyloid beta proteins dangerous? Recently, I began working in a lab in which our main area of study is Aβ peptides. My concern is whether or not exposure to high levels of Aβ proteins is hazardous.

I suspect that the peptides cannot pass through one's dermis, but this is just speculation. Also, if one were to accidentally ingest Aβ, it would likely be broken down into its monomers by your digestive system before it could cause any harm.

I've tried to find resources covering my concerns but to no avail.

Amyloid proteins form insoluble aggregates and they cannot pass through the dermis. The aggregates can be (and frequently are, in the intracellular context) resistant to digestion but they would not be absorbed from the gut. Moreover, these peptides are not "toxic" per se. It is their property of forming aggregates and plaques that gives rise to the pathology observed in the associated neurodegenerative diseases. It is important to note that neural tissue does not regenerate like many other tissues in the body and that is one reason why neural tissue is vulnerable.

So unless you inject some of these proteins inside your cranium, it is unlikely that you will suffer from any problems beacause of Aβ exposure.


Which of the free access molecular modelling program apply the B3LYP/G-31G(d,p) level of theory? Please, let me know which an open free molecular modelling program will apply
the B3LYP/G-31G(d,p) density functional level of theory? Thanks.
Helen

There are several free packages, see here.

Like orthocresol mentioned, you might want to be tutored by someone of experience at first. Quantum calculations are very easy to start (too easy, some might say), but harder to actually interpret correctly. The actual method can also greatly influence the results and needs to be chosen carefully (for example, B3LYP has fallen out of favor after some weird behaviours have been observed (Why the Standard B3LYP/6-31G* Model Chemistry Should Not Be Used in DFT Calculations of Molecular Thermochemistry: Understanding and Correcting the Problem); consider M06 or M062X as good alternatives)


Major product of the nitration of benz[cd]indol-2(1H)-one Decide major product for given reaction of electrophilic aromatic substitution:

Is the major product a because of steric hindrance, or is it b because of hydrogen bonding?

According to this patent here, N-ethyl 1,8-naphtholactam nitrates in the 4 position i.e. structure a. It seems entirely likely the N-unsubstituted material will do likewise.


Why does hand sanitizer heat up? While washing my hands with soap I added some hand sanitizer just because it was right there on the counter and couldn't resist mixing them... Ha!  However, when I added the hand sanitizer it heated up for a moment and then cooled back down.  Why does this happen?

Hand sanitizers are generally ~50% - 70% ethanol or isopropanol. When these alcohols mix with water, they may release heat. Starting at room temperature, mixing ~100% ethanol with an equal volume of water will raise the temperature of the mixture ~8 0C.


Azeotropes and separation by distillation? I have just recently learnt the theory of fractional and normal distillations and the basics of maximum and minimum boiling azeotropes.

Most books say that the components of the mixture forming an azeotrope cannot be separated by distillation because on heating, the residue (in case of negative or maximum boiling azeotropes) and the distillate (in case of positive or minimum boiling azeotropes) is closer in composition to the azeotropic composition and always approaches it.

This being understood, we at least separate one component from the mixture as the residue (for positive azeotropes) and the distillate(for negative azeotropes) is richer in one component than the other since the remaining part approaches the azeotropic concentration thereby increasing the concentration of one of the components depending on which side of the azeotropic "point" are we on the graph of boiling temperature and composition.

If this is true, for any mixture, we can at least theoretically obtain one purified component (provided we have enough quantities of the mixture)??
Reference azeotrope basic

You seem to have the right idea.

To elaborate, take the specific example of the positive azeotrope of (roughly) 96% ethanol and 4% water. If you begin with a mixture containing less than 96% ethanol, distillation will result in a distillate more abundant in ethanol and nearer to the azeotropic composition. That implies the residue will necessarily be proportionally lower in ethanol. If you continuously repeat the distillation procedure using the remaining pot residue, it will approach 100% water. On the other hand, if you begin with a concentration above 96% ethanol, the resulting distillate will actually be lower in ethanol (and therefore nearer to the azeotropic point), while the residue will actually become more concentrated in ethanol. Hence, in theory, with many repeated distillations you can approach purity in the residue when dealing with positive azeotropes, and the ultimate composition of the residue will depend on which side of the curve you're on to start with.

When dealing with negative azeotropes, the situation is analogous but the particulars are basically reversed. Take the example of hydrofluoric acid and water, which form an azeotrope at approximately 37% HF. If you initially begin distilling a mixture comprising less than 37% HF, you'll obtain a distillate more dilute in HF, while the residue will contain more HF and be nearer to the azeotrope. If you were to collect the distillate and repeat the distillation, you would again end up with an even more dilute distillate, with the concentration of the distillate approaching 100% pure water. If you begin distilling a mixture containing more than 37% HF, then the residue will end up being more dilute as it approaches the azeotrope, while the distillate will actually be richer in HF. If you repeat this process many times over, you can eventually obtain a distillate approaching 100% pure HF (in theory).

So, in summary, you're correct to suggest that one of the components can be isolated with something approaching complete purity from an azeotropic mixture, assuming a sufficient volume to enable an adequate number of simple distillation cycles or theoretical plates in fractional distillation. Of course, the problem is that you really have no choice in determining which of the two components you can isolate, as that depends entirely on the starting composition of the mixture. In practice, you can often separate azeotropeic mixtures by modifying the distillation process in various ways.


Can single molecules of C and O2 react in isolation, and if so how will momentum be conserved? I am trying to figure out how is it possible to reconcile the reality of exothermic reactions, which means that kinetic energy is transferred to (heats) the surrounding matter, with the principle of conservation of momentum, in particular when the product is a single molecule.
I have been exploring the question but found no sufficiently explanatory answer to the question of how the bond energy that is released may create movement (heat). The nearest to an answer that I can imagine is that the reactants enter some transitory combined state of vibration and that further contact with some other molecule results in the separation of each of them in opposite directions with equal but opposite momenta increments thus preserving the overall momentum, while at the same time reducing the vibration and stabilizing the product molecule.
If that were the correct answer (at least in some cases), I would like to know a bit more about the details of the process, which I suppose entail some description about how binding energy is transferred to that vibration.
In any case, it would be interesting to know whether is it actually possible for a single carbon atom and oxygen molecule to react (if they collide with the necessary energy) and produce carbon dioxyde or not, be it for the reason above or another one, since if they are in isolation, the transfer of vibration energy cannot be realized and thus the reaction could not be completed, and then I suppose that would eventually end with the spontaneous separation of the components.

$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this:
In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved?
That's a legitimate concern all right. After all, we are taught that this reaction does happen instantly, once given a chance, and that's in fact true. Also, we know that it releases a lot of heat. Now, heat is nothing but the motion of molecules; how does energy convert to the motion of one molecule as a result of one single reaction act?
It doesn't. The conservation of momentum forbids that, just as you reasoned. Chemical reactions are collective phenomena. Nobody cares about a single molecule.
Now what really happens to a single molecule which has just formed as a result of the mentioned reaction? That's really simple: the molecule is vibrating wildly, ready to break apart. Is has just enough energy to do so. It will do so half of the times, or maybe more often. But that doesn't matter. What matters is that sometimes the vibrating molecule will hit another molecule and sent both of them flying away in opposite directions, thus releasing a part of its energy and becoming more or less stable.
An emission of a photon is also an option, but that's another story.
So it goes.


How do you list all possible reactions and intermediates of a reaction? Ins this video https://youtu.be/txk-VO1hzBY?t=2391 the speaker claims that $$\ce{H2 + O2 -> H2O}$$ involves 23 reactions and 11 species. How do you list all of that ?

You've discovered a secret. In undergraduate chemistry classes, the field of chemical kinetics is treated with a level of detail very far from reality. The truth is that virtually any reaction you can devise is vastly more complicated than we would like to think. This is especially true for gas phase reactions like the one you write down because it is possible for very unstable intermediates to exist for long enough to collide with lots of different species or undergo an internal conversion (when the molecule is large enough to do so) which introduces minute components that could nonetheless make a big difference in the kinetics.

Some people actually attempt to study these reactions without just ignoring the small sideways paths that lead to weird intermediates or weird products. Generally they do this by using what is called a Master Equation. I would encourage anyone to read about these and a very nice overview can be found here: Miller, J. A., & Klippenstein, S. J. (2006). Master equation methods in gas phase chemical kinetics. The Journal of Physical Chemistry A, 110(36), 10528-10544.

What you'll find when reading this is that they care a lot about what the potential energy surface looks like everywhere. This is because one formulation of the Master Equation is as follows:$$\frac{dn_i}{dt}=\sum_j(p_{ij}\cdot n_j(t)-p_{ji}\cdot n_i(t))$$This basically says that we are looking at the change in probability of finding the system in a state $n_i$ as a function of time and we are doing this by describing the probability per unit time $p_{ij}$ of moving from state $j$ to state $i$ as well as the process which goes the other way. Note that $n_i$ is essentially the same as a number density for a large number of molecules, but strictly this is a probability.

You can imagine then that in order to use this equation (which isn't very usable in its current form), one would need a detailed picture of what possible intermediates can form immediately from the state the system is in right now.

Thus, the answer to how you write all these down is that you look very carefully at the state of any system and ask yourself, what intermediates could I possibly imagine forming? Many of these intermediates are very unstable at low temperatures, but these kinds of models are frequently used when studying combustion chemistry or flames, so the number of intermediates can becomes very large and the number of reactions those intermediates can undergo (the number of states $j$ you can jump to from state $i$) is even larger. To operationalize this, you would basically write out all known intermediates, from experiment or theory, and imagine what reasonable combinations you could make of those that eventually lead to your product.

In the video you link, some unusual intermediates are listed, such as $\ce{HO2}$, and it is said that this is a simplified model. Basically what that means is that they had even more possible combinations of intermediates, but for computational simplicity or good experimental reasons, decided that many combinations of those intermediates contributed negligibly to the reaction of interest.

One final note is that it's not really as simple as combining everything you can imagine because you have to consider where a given intermediate came from. For instance if a single molecule is formed by means of a bimolecular reaction, the angular momentum must be conserved. This can lead to situation where certain states you could imagine hopping to are not actually possible because it would violate conservation of angular momentum. Indeed, one could imagine this leading to some reaction mechanisms which are very different from what intuition would dictate. The paper I linked above notes that when angular momentum becomes an important factor, the intermediate usually undergoes an internal conversion to basically store that angular momentum before reacting with other things... So, when you're writing out all of these intermediates don't forget that $\ce{HO2}$ might very well exist as $\ce{HOO}$ or $\ce{OHO}$ (though that seems unlikely I only meant it as an illustration).

I hope this addressed the spirit of what you were really asking.