Why is hexane called skellysolve-B https://pubchem.ncbi.nlm.nih.gov/compound/hexane
Another name for hexane is skellysolve-B, is that just to be silly?

It's probably considerably notorious historical trade name. In many older publications, e.g. "Skellysolve B (petroleum ether)"[1,2] appears for solvent used.

It was a product or trade name of Skelly Oil Company[3]
 founded in 1919.

It's not a pure hexane, rather narrow-boiling hexane oil fraction.[3]

I could find analysis of similar product Skellysolve C, which is C7 heptane fraction:[3]

24.9% heptane
22.9%

2,3-dimethylpentane
2-methylhexane
3-methylhexane

19.3%

trans-1,2-dimethylcyclopentane
trans-1,3-dimethylcyclopentane

15.0% methylcyclohexane
5.5% cyclohexane
5.2% heavy nonaromatics
2.7% toluene
2.5% hexane
…

Other ones:[4]

  Skellysolve A is essentially n-pentane, b 28-30°/atm,
  Skellysolve B is essentially n-hexane, b 60-68°/atm,
  Skellysolve C is essentially n-heptane, b 90-100°/atm,
  Skellysolve D is mixed heptanes, b 75-115°/atm,
  Skellysolve E is mixed octanes, b 100-140°/atm,
  Skellysolve F is petroleum ether, b 30-60°/atm,
  Skellysolve G is petroleum ether, b 40-75°/atm,
  Skellysolve H is hexanes and heptanes, b 69-96°/atm,
  Skellysolve L is essentially octanes, b 95-127°/atm.

Unfortunately, Chemical Abstracts Service, mixes names for different substances, including this tradename: “n-hexane”, “hexane (all isomers)”, “Skellysolve B” under single CAS RN 110-54-3.[5]

References:

Newman, H.; Zilversmit, D. Quantitative Aspects of Cholesterol Flux in Rabbit Atheromatous Lesions. Journal of Biological Chemistry 1962, 237 (7), 2078–2084.
Holick, M. F.; DeLUCA, H. F. A New Chromatographic System for Vitamin D3, and Its Metabolites: Resolution of a New Vitamin D3, Metabolite. J. Lipid Res. 1971, 12 (4), 460–465.
Griswold, J.; Andres, D.; Berg, C. F. V.; Kasch, J. E. Pure Hydrocarbons from Petroleum - Separation of Straight-Run Fractions by Distex Process. Ind. Eng. Chem. 1946, 38 (1), 65–70.
Armarego, W. L. Purification of Laboratory Chemicals; Butterworth-Heinemann, 2017.
CommonChemistry.org – CAS Registry Number: 110-54-3


Way to create acidic/HCl gel? Is there a way to turn azeotropic hydrochloric acid into a gel?

I am trying to remove barnacles from the bottom of a boat and normal hydrochloric acid works, but it is tedious because in liquid form it just runs off, so I have to keep applying it over and over, kind of wearing the barnacle away.

If I could turn the HCl into a gel then it could just be applied for a few minutes and remove the whole barnacle.

It turns out somebody already answered the question in a different post:

  Xanthan Gum can handle hydrochloric / muriatic acid (HCl) for a while,
  depending on how concentrated the acid is. Here is how I made an HCl
  gel to clean an old toilet that was heavily coated with lime scale...

  I first dissolved a level teaspoon of xanthan gum into 1/4 cup of
  rapidly stirring water. After a few hours of letting it sit, with
  occasional stirring, I had a very thick, viscous goo. To make the HCl
  gel, I mixed one part of this goo with 1.5 parts of 31% hydrocholoric
  acid, by volume. This gave a gel of a suitable viscosity to get the
  lime dissolved off the toilet. I don't know how long this gel would
  last before the HCl broke down the gum, but my gel lasted several
  hours, which was enough to get the toilet clean.


What happens to the bleach after mopping? At home, I clean the tiled floors by mopping with highly diluted bleach.  After mopping, I let the floor dry.  What happens to the acid in the bleach?

Will it crystallize on the floor, stay there invisibly in solid form, and get solved again next time the floor gets wet?  Will it instead evaporate, or turn to fine dust that floats in the air and leaves through the window?  Will it decompose in some sort of chemical reaction?

Sodium hypochlorite decomposes by a couple of different pathways:

$\ce{NaOCl\leftrightharpoons Na^+ + OCl^-\\
2OCl^- \leftrightharpoons O_2 + 2 Cl^-}$

In that case, the oxygen escapes (if it didn't react with "things to bleach", which is of course how bleach works in the first place), and you are left with ordinary salt.

The other pathway is

$\ce{3 OCl^- -> ClO_3 + 2 Cl^-}$

In that case you get some sodium chlorate, and again some ordinary NaCl salt.

Both NaCl and NaClO3 are crystalline solids that may be blown around like fine dust; some NaOCl may also remain behind. Note that according to wikipedia most commercial bleach solutions contain a small amount of sodium hydroxide to slow down the decomposition along the pathways mentioned above.

Straight dope link


Kinetics of air evolving from water on heating Is there a simple relationship describing the kinetics of air evolving from water on heating?

For example, if I let water at room temperature and pressure saturate with air, it should have some concentration. Then, if I heat that water up to 90°C, the solubility of air will be much lower by Henry's law. The evolution of air is, presumably, not instantaneous. Is it fast or slow compared to a reasonable heating rate? Does it depend on just how far the system is from equilibrium or on other conditions like the amount of air-water surface boundary?

Is there a simple relationship describing the kinetics of air evolving from water on heating?

There is no simple relationship. The problem is that you can't heat the water absolutely uniformly. Thus all the convection currents in the water would create a mathematically chaotic relationship. There is also the problem of the water being supersaturated - especially if you heat from the bottom.

  Is it fast or slow compared to a reasonable heating rate?

This would seem to be sort of a judgement call, but I'd say that it would be faster to dissolve a gas in a liquid (say via bubbling) than heating to expel it. The bubbles mechanically stir the solution rather than just equilibration by diffusion from the surface. Also think of liquid nitrogen. It boils so fast when spilled on the floor that the drops skitter around. The escaping gas insulates the liquid drop limiting the heat transfer to the liquid. So there is a limit as to how fast you can heat a solution.

  Does it depend on just how far the system is from equilibrium

Yes. The further from equilibrium, the less likely a solution could be super-saturated.

  Conditions like the amount of air-water surface boundary

The time to reach equilibrium surely depends on the surface area to volume. Think of a cylinder of water. A cylinder which has water one inch deep would certainly come to equilibrium faster than cylinder of the same diameter which has water a mile deep.


Can I use a digital multimeter as substitute for a zero resistance ammeter? I am testing the galvanic corrosion between two metals, following ASTM G71. In this method, a zero resistance ammeter was the material presented to determine the galvanic current measurements. Unfortunately, we don't have access with it. So can I use a digital multimeter instead?

Yes, using Ohm's law: I=V/R.

The only true "zero" resistance ammeter are superconductive. In practical use, the voltage drop across a small resistance is measured. If your DVM can measure down to a millivolt (mV), then a 1 mV drop across a 0.001 ohm resistor shows a flow of 1 ampere current. You'll need to consult the guide (~$45 online) for the maximum permissible voltage drop, and use an appropriate resistance.

Recently, I used the voltage drop across a piece of #16 copper wire to measure currents in the 1 to 5 A range... a switch in the circuit had greater resistance!


On the angle subtended between the angular momentum vector and the applied field for a single spin According to Dr James Keeler of Cambridge University in one of his lectures at Queensland University, the angle theta can be any angle, but I struggle to see how this is true. There are only a certain number of I values that nuclei take and as illustrated in the figure above, each magnetic quantum number corresponds to a fixed theta. Is my logic flawed?

You are correct that a given value of $m_I$ and $I$ does correspond to one single value of the angle $\theta$. For the most commonly used nuclei in NMR spectroscopy, the "allowed" values are $I = 1/2$ and $m_I = \pm 1/2$. The two states are commonly labelled $\alpha$ and $\beta$; these are the eigenstates of the angular momentum projection operator $\hat{I}_z$, such that

$$\begin{align}
\hat{I}_z|\alpha\rangle &= \frac{1}{2}|\alpha\rangle \\
\hat{I}_z|\beta\rangle &= -\frac{1}{2}|\beta\rangle
\end{align}$$

(treating $\hbar \equiv 1$). The problem is that you are assuming that the wavefunction (of a spin) can only be in an eigenstate. This is not true. It is perfectly valid, and indeed necessary (see below), to have states that are superpositions of eigenstates. In general, a spin can have any state

$$|\psi\rangle = c_\alpha |\alpha\rangle + c_\beta |\beta\rangle$$

where $|c_\alpha|^2 + |c_\beta|^2 = 1$ (normalisation), and in general the coefficients are time-dependent (see below). This is no longer an eigenstate of $\hat{I}_z$, and consequently is no longer well-described by one value of $m_I$, or for that matter, $\theta$. But that is not intrinsically a problem. The only circumstance in which a wavefunction must be described by an eigenstate is immediately following an observation, where the wavefunction $|\psi\rangle$ collapses to the eigenstate $|i\rangle$ with probability $|c_i|^2$. Otherwise, superposition states are entirely consistent with quantum mechanics.

A more generalised example is to think of the time-independent Schrödinger equation:

$$\hat{H}|\psi\rangle = E|\psi\rangle$$

which is an eigenvalue equation for the Hamiltonian, $\hat{H}$. It is true that only eigenstates of the Hamiltonian are admissible solutions to this equation. But that does not mean that the only physically permissible states for the system are eigenstates of the Hamiltonian. The time-independent SE only governs systems that are not evolving in time, or stationary states. We need something else to describe systems that do evolve with time, which is the time-dependent Schrödinger equation:

$$\hat{H}|\Psi\rangle = i\hbar\frac{\partial |\Psi\rangle}{\partial t}$$

The most general solution to the TDSE is, indeed, a superposition of stationary states (the Internet has lots of more thorough explanations, or read Griffiths' Introduction to Quantum Mechanics):

$$|\Psi\rangle = \sum_i c_i \exp\left(-\frac{iE_it}{\hbar}\right) |\psi_i\rangle $$

where the $c_i$'s are time-independent. Indeed, it is necessary for us to have superposition states, because this allows the coefficients of each eigenstate to evolve in a time-dependent fashion (because of the exponential term, which evolves at different frequencies for different eigenstates). If a system is prepared in an eigenstate of the Hamiltonian, then it can never evolve into another eigenstate (the exponential evolves, but if there is only one eigenstate, it is no more than a simple phase factor).

This might not seem immediately relevant for NMR, but recall that the (free) Hamiltonian in NMR is simply $-\gamma B_0 \hat{I}_z$. So, the eigenstates of $\hat{I}_z$, $|\alpha\rangle$ and $|\beta\rangle$, are also the eigenstates of the Hamiltonian. If we never had superposition eigenstates, then the spins would simply sit there forever and do nothing.*

* As long as the Hamiltonian doesn't change. But if you change the Hamiltonian, say from $\hat{H}_0$ to $\hat{H}_1$, then the eigenstates change, and so your system (which was originally an eigenstate of $\hat{H}_0$ is no longer an eigenstate of $\hat{H}_1$). Indeed, it would be in a superposition of eigenstates of $\hat{H}_1$, and would evolve in time according to the TDSE, so it doesn't invalidate any of the preceding discussion.


Proposed mechanism for the catalyzed oxidation of CO by O3: Are NO and NO2 catalysts or intermediates? The steps below represent a proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$:

  $$\begin{align}
&\text{Step 1: } & \ce{NO2(g) + CO(g) &-> NO(g) + CO2(g)} \\
&\text{Step 2: } & \ce{NO(g) + O3(g) &-> NO2(g) + O2(g)} \\
\end{align}$$

  Which of the following statements is true based on this mechanism?

  The overall products are $\ce{NO2}$ and $\ce{O2}$.
  The overall products are $\ce{NO}$ and $\ce{CO2}$.
  The overall products are $\ce{NO}$ and $\ce{O2}$.
  Either $\ce{NO2}$ or $\ce{NO}$ can be viewed as a catalyst. (correct answer)
  Both $\ce{NO2}$ and $\ce{NO}$ are intermediates.

Based on the given proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$, the answer for above question suggests that either $\ce{NO}$ or $\ce{NO2}$ can be considered as a catalyst.

Thus, I'm very confused as to why the answer is 4. I would think it is 5) because $\ce{NO}$  is produced first and than consumed later on. I understand $\ce{NO2}$  does not have that same pattern, however I thought that it cannot be a catalyst because $\ce{O3}$  is already a catalyst.

My take on this problem is as follows. Suppose a flask contains an equimolar mixture of $\ce{CO}$ and $\ce{O_3}$. The net reaction between the two gases will be

$$\ce{CO (g) + O3 (g) -> CO2 (g) +  O2 (g)} \tag{1}$$

But it needs a catalyst. If $\ce{NO_2}$ gas is introduced, then the reaction proceeds via the two reactions given in the OP's question:

$$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{2}$$
and $$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{3}$$

So the nitrogen oxides cycle around. On the other hand, if $\ce{NO}$ gas is introduced, then the reaction proceeds via 'flipped' equations:

$$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{4}$$
and $$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{5}$$

and again the nitrogen oxides cycle around. So it does not really matter whether $\ce{NO_2}$ gas or $\ce{NO}$ gas is introduced first: the two nitrogen oxides are the overall catalytst and either could be considered to be an intermediate. Basically, this is a distinction without a difference.


How to calculate the probability of finding an electron in a box between 0.25L and 0.75L? If the wavefunction of the electron in a box of length $L$ is $$\psi{(x)}=\sqrt{\frac{2}{L}}\cdot\sin{\frac{n\pi x}{L}}$$
What would be the probability of finding the electron between $0.25L$ and $0.75L$?.

From postulate 1 of quantum mechanics I know that $$1=\int_{0}^{L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}.$$

Trying out a similar reasoning leads me to think that the required probability is the integral $$ \int_{0.25L}^{0.75L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}$$ which gives the answer as $0.5$. But the book gives the answer as $0.82$.

What is the problem with my method and can anyone explain me why the answer is $0.82$?

You have the reasoning correct, so I suspect you made a mistake in your integral calculation. The probably of finding the particle between $a$ and $b$, for the $n$-th state, is indeed given by:

$$P_n(a,b) = \int_a^b \left|\psi_n(x)\right|^2\, \mathrm{d}x $$

but if I do the calculation for the ground state ($n = 1$), I find the same results as your textbook:

$$ P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818 $$


What should the Rehm-Weller equation look like? The Rehm-Weller equation, which defines the energy change for photoinduced electron transfer according to Lakowicz "Principles of Fluorescence Spectroscopy" [1, p. 337] should look like:
$$\Delta G=E^\mathrm{red}(D^+/D)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \frac{e^2}{\epsilon d} \tag{1}$$
So you can see the first two components are reduction potentials of donor $D^+$ and acceptor $A$. But sometimes you can find another definition:
$$\Delta G=E^\mathrm{ox}(D/D^+)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \dfrac{e^2}{\epsilon d} \tag{2}$$
Which one is correct? I'm totally confused.
References

Lakowicz, J. R. Principles of Fluorescence Spectroscopy, 3rd ed.; Springer: New York, 2006. ISBN 978-0-387-31278-1.

Both expressions are identical. It's just juggling of the signs, since

$$E^{\mathrm{red}}(D^{\cdot+}/D) = - E^{\mathrm{ox}}(D/D^{\cdot+})$$

Relevant primary references would be:

Rehm, D.; Weller, A. H. Isr. J. Chem. 1970, 8, 259-271.
Rehm, D.; Weller, A. H. Ber. Bunsen-Ges. Phys. Chem. 1969, 73,
834-839.
Weller, A. H. Z. Phys. Chem. NF 1982, 133, 93-98.

A normative glossary of terms in photochemistry is available from the IUPAC.


Comparison of COSY and TOCSY for detection of small couplings between protons I routinely run COSY spectra to help with assignment of my proton NMR.

The experiment is fast to run and generally useful, however some of the intermediates I've recently been making have had sufficiently small coupling constants that the COSY hasn't been terribly useful in certain regions(a 'standard' COSY isn't giving the resolution needed).

Typically, I get around this by running a set of HMBC and edited-HSQC, which allows me to trace the protons along the molecule, in a similar way as I'd be able to do with the COSY. The issue with this is that with the quantity of material I have, the HMBC takes significant amounts of time to run, making this method unhelpful for day-to-day work.

A colleague suggested using 2D-TOCSY to see which protons are in the same spin systems (not the same information as in the COSY, but would give me what I need).

Looking through the ever useful Claridge, it seems that the mixing pulse from the COSY gets replaced with a 'spin lock' sequence. What is less clear to me from this though, is whether TOCSY is also dependant on the magnitude of the coupling (i.e. if the J values are minimal, will the TOCSY suffer from the same issues as the COSY), or whether the spin systems are detected indirectly/through some other property.

Standard 2D COSY spectrum

Standard 2D TOCSY spectrum

The TOCSY transfer also depends on the size of the J coupling, you don't get around that.

Generally speaking it is found that cross peaks due to direct coupling build up to their maximum intensity in a time which is of the order of $1/(2J)$. Peaks which arise from transfer through two successive couplings take longer to build up, and if three successive couplings are involved, even longer mixing times are needed
Understanding NMR spectroscopy, James Keeler, page 230

The TOCSY has the advantage of creating in-phase signals instead of the anti-phase signals of the COSY, which increases sensitivity. But that doesn't mean that it is generally more sensitive.

Whether or not the overall intensity of the COSY or TOCSY cross peaks is greater depends intimately on the choice of the experimental parameters and on the molecule being studied. However, the general experience is that for small to medium-sized molecules, COSY is the preferred experiment for detecting small couplings.
Understanding NMR spectroscopy, James Keeler, page 229

I'm not sure if you can get a general answer for your case, you might have to directly compare the experiments for the kind of molecule you want to look at. You could also try longer mixing times for the TOCSY and see if they help with your small couplings (which they should, as long as relaxation doesn't eat up that advantage).


how to mathematically predict reaction results I've been trying to computationally achieve the following (I have no lab, I'm 14 years old, and I want to make a Minecraft mod to teach real Chemistry):

I want to mathematically predict the range of temperature and pressure (if the reaction is between gases) or temperature and concentration (if the reaction is in a solution) in which a given chemical reaction can happen, and the rate at which it happens, with the variation of those two arguments.

EDIT 1:

Regarding the first part of @sixtytrees answer I suppose I must calculate $K$.

I assume $\Delta G$ would be:

$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - 137.3 - 0\\[6pt]
&= -299.8\ \mathrm{\frac{kJ}{mol}}\end{align}$$

Now to solve for $K$:
(I believe $R$ must be $0.0083144598\ \mathrm{\frac{kJ}{mol\ K}}$, because $\Delta G$ is in $\mathrm{\frac{kJ}{mol}}$
Also, I think we have to take $25\ \mathrm{^\circ C}$ as the temperature which is $298\ \mathrm K$)

$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-299.8}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-299.8}{-2.4777090204}\\[6pt]
&= 120.99887336713996006405304847878\\[6pt]
K &= \mathrm e^{120.99887336713996006405304847878}\end{align}$$

And now I realize I must have made some mistake because this number will be too big.
☹

EDIT 2:

I just woke up and I noticed I miscalculated $\Delta G$, because I added them together when I should be subtracting. Let's see how it goes now.

$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - (-137.3 - 0)\\[6pt]
&= -25.2\ \mathrm{\frac{kJ}{mol}}\end{align}$$

Now $K$ will be:

$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-25.2}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-25.2}{-2.4777090204}\\[6pt]
&= 10.170685820053125395644218884808\\[6pt]
K &= \mathrm e^{10.170685820053125395644218884808}\\[6pt]
&= 26126\end{align}$$

Still not as small as 10 that @sixtytrees predicted.

$$26126 = \frac{[\ce{CH3OH}]^1}{[\ce{H}]^2[\ce{CO}]^1}$$

What now?

EDIT 3:

I've been thinking, and decided Chemical Equilibrium is not something I need to bother with, because as the reaction progresses, it is assumed that the final products are drained out as they are formed, and additional pressure is added to account for the reduction of total molarity. This way I don't need to distinguish between reversible reactions, irreversible reactions, and virtually irreversible reactions (when there are trace amounts of the reactants in the final products).

This only works if the reactants are gases, I don't know how to sustain the rate of the reaction if I take away the products from a solution.

The original question remains, now that equilibrium is irrelevant, is it possible to predict the temperature and pressure/concentration range of the reactants in which the reaction can occur?

I downloaded Orca, by the way. No idea how to use it.

EDIT 4:

Ok, so let´s say this is all happening in a sealed container with 1L of volume, so

$\begin{align}
26126 &= \frac{[CH_3OH]}{2^2 \cdot 1^1}\\
[CH_3OH] &= 26126 \cdot 4 = 104,504
\end{align}$

What does this mean? It can't mean I end up with more than one hundred moles in 1 liter. It makes no sense, I don't have enough reactants to get so much product. So, I am clearly doing it all wrong, can someone explain it to me?

There are two parts to quantification of reactions:

(1) How far it goes? (Thermodynamics, equilibrium). This all is dictated by ground state energy.

(2) How fast it goes? (Kinetics, reaction rates). This is dictated by  transition state energy vs ground state energy.

Equilibrium is much easier to understand. Let [Y] be concentration of reagent Y. You say that for 2A + B = C+3D +4E you have:

K = [C]$^1$[D]$^3$[E]$^4$/([A]$^2$* [B]$^1$)

Note how powers are equal to coefficients in the equation.

$\Delta$G= -RT*ln(K)

$\Delta$G is Gibbs energy (measured and tabulated for all of inorganic chemicals from school textbook).
R is a constant (Universal gas constant)
T is temperature.

How to use it? Well, imagine that you have

2*H2+ CO <-> MeOH

Lets say that K=10 under that conditions. I am sure I am off. Correcr way to find it is to use $\Delta$G= -RT*ln(K). You know T, R is constant. $\Delta$G is known for MeOH, H2 and CO. Calculate the difference to find $\Delta$G. You will need to find $\Delta$G for methanol vapor or take it for liquid MeOH and add $\Delta$G of evaporation.

You start with 1M of CO ( atm at 273C) and 1.5M of H2 (67.2 atm at 273C). Total pressure 112 atm. (Mol to pressure conversion using ideal gas law). What will be the equilibrium amount of MeOH if ?  You say that

Let x be the amount of product at equilibrium. H2 = 1.5 M - 2*x and CO = 1-x.

x/(1.5-2*x)$^2$*(1-x)=10=K

Now you have a cubic equation to solve (1.5-2x)$^2$*(1-x)*10 - x = 0;

With more reagents you have algebraic equation of higher power, but there are methods to solve those numerically and the idea is simple enough to explain.

This gets a bit more complicated when you have multiple reactions going. But This is the basis. When system gets complex you start ignoring some small members to simplify it. (1.0001)$^5$ ~ 1.0005 (reasonably accurate). YOu do the same in chemistry.

Kinetics is trickier. For A+ 2B -> C the first approximation is that

rate = k*[A]*[B]$^2$.

Also "k often increases by a factor of 2-4 per 10C of increase in temperature".

Now to treat it accurately you need to solve a system of equations. each equation is either

r$_i$ = $\delta$[A$_i$]/$\delta$t

or

r$_i$ = ($\delta$[A$_i$]*$\delta$[B$_i$])/$\delta$t

Rate if i$^{th}$ reaction = change of concentration of i$^{th}$ reagent per time period as the period goes to zero).

You end up with a system of ordinary differential equations. But don't get tricked by the word "ordinary". Most of them don't even have solutions in "elementary functions" (functions that you study in school or university as a chemist.

To understand what is a differential equation you would need to know what is an integral. We study differential equations in second year of university and that probably the most difficult subject in university.


Is it possible for Cytosine and Hypoxantine to make three hydrogen bonds? I think the O from Hypoxanthine can bond with one of the Hs in NH2 from Cytosine and the H that is connected to N in Hypoxanthine with the N in Cytosine but is it possible also for the H connected to C in Hypoxanthine to bond with the O in Cytosine?

[Hypoxantine]

[Cytosine]

I have almost the same question between Xanthine and Adenine.(please note that I'm not talking about Xanthine and Adenine inside RNA, I know they can't bind at all in RNA because of their sizes)

I strongly believe that it won't be possible for these compounds to form three hydrogen bonds.

Revisiting the basic definition of a hydrogen bond:

  A hydrogen bond is an electrostatic attraction between two polar
  groups that occurs when a hydrogen (H) atom covalently bound to a
  highly electronegative atom such as nitrogen (N), oxygen (O), or
  fluorine (F) experiences the electrostatic field of another highly
  electronegative atom nearby.

Such electronegative atoms as oxygen  and nitrogen with free lone pairs are potential hydrogen bond acceptors.

Hydrogen atoms attached to very electronegative atoms like O and N have strong partial positive charge and these are potential hydrogen bond donors

These two crucial requirements for hydrogen bond formation, have thus been summarised:

  In order for a hydrogen bond to occur there must be both a hydrogen
  donor and an acceptor present. The donor in a hydrogen bond is the
  atom to which the hydrogen atom participating in the hydrogen bond is
  covalently bonded, and is usually a strongly electronegative atom such
  as N,O, or F. The hydrogen acceptor is the neighboring electronegative
  ion or molecule, and must posses a lone electron pair in order to form
  a hydrogen bond

Here is an impression of how these two would bond:

Note the "X" illustrating the missing requirement(s) to form a 3rd H-bond

missing potential donor group (electronegative atom) just C present.

Acknowledgements

Hydrogen Bonding
Voet & Voet Biochemistry: Genetic code

.


What elements in Tomato and Spinach make them cook quickly when cooked together? I have started experimenting on cooking these days and I have noticed that when I add spinach leaves and tomato, they cook very quickly. The spinach melts very fast and the tomato melts fast also - after adding a spoon of table salt. My question is what is the reason behind it. Is it something to do with oxidation?

The table salt is probably irrelevant to why the veges cook down quickly, although it does cause veges to sweat, it may leach the moisture out of the tomato and spinach faster than if you didn't add it but I think heat would be the main variable. Also do you stir the food more when you add the salt than if you didn't?

Most organisms have a ratio of about 《80% water, spinach and tomatoe may have a high water content and also contain oils and other volatile components that are released when heated so when you heat / cook them the water and volatile components evaporate and the solid mass(other 20% of initial weight) is left behind.

They are soft veges too so the material can be sweated easier than say a carrot and will lose its structure and water content at a faster rate.

Some oxidation could be occurring along with other chemical reactions between components such as decarboxylation, hydrolysis, hydration, dehydration and many more depending on what compounds are present in the process, the kind of environment (eg. aqueous) and the techniques used (eg. frying, baking).


Industrial production of alkali metals My professor says that $\ce{Li}$ and $\ce{Na}$ are produced by electrolysis, while $\ce{K, Rb}$, and $\ce{Cs}$ are made via the reaction $\ce{Na + MCl -> NaCl + M}$, where $\ce{M = K, Cs, Rb}$.

I don't understand how the reaction $\ce{Na + MCl -> NaCl + M}$ can take place, given the reduction potentials of the alkali metals ($-3.04, -2.71, -2.93, -2.98,$ and $-3.03$ V for $\ce{Li, Na, K, Rb, Cs},$ respectively). From the data, we see that sodium is the weakest reducing agent and the least electrochemically active among these alkali metals. Why then, would it prefer to displace a more electrochemically active species?

One aspect to consider is that the boiling points of alkali metals go down as they get heavier.
From The Encyclopedia Britannica:
Lithium:  bp = 1342°C
Sodium:  883°C
Potassium:  759°C
Rubidium:  688°C
Cesium:  671°C
Francium:  677°C (probably this one is merely predicted)
Thus the heavier alkali metals could be volatile enough to effectively escape as vapors while sodium and the chloride salts remain largely liquid.  Thereby the equilibrium that would otherwise limit the displacement of the heavier metal is pulled forward in favor of that metal being lost to the gas phase.  The vapor would later be condensed to get the metal as a solid.


(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule The Re2Cl8 2- ion is composed of two square planar ReCl4 units joined by four Re- Re bonds (quadruple bonding) and has D4h symmetry.

(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule.

This is the entire question. However, I only need help genereating the irreducible representations. There are four sets of irreps that need to be generated that are based on the dz^2 orbital, dxy orbital, dyz orbital, and dxz orbital. I have only been able to determine the irrep for the dz^2 orbital. Here is what I did:

I transformed the dz^2 orbital under every symmetry operation from the D4h point group. (E, 2C4, C2, 2C2', 2C2'', i, 2S4, sigma h, 2 sigma v, and 2 sigma d. )
I got a reducible representation of E=2, 2C4=2, C2=2, 2C2'=0, 2C2''=0, i=0, 2S4=0, sigma h=0, 2 sigma v=2, 2 sigma d=2
Lastly, I used tabular reduction to get that dz^2 = A1g +A2u

For the other three d orbitals I can not seem to get the correct irreps.

I'll try to walk through how it works for $d_{xy}$. You will need to consider $d_{yz}$ and $d_{xz}$ as a pair later when you attempt those.

Obviously $\chi(E)=2$. Now for the $C_4$ rotation, just think about rotating one of the orbitals in the $xy$ plane (the same thing will happen to the other orbital, so we will then double the character). When we do this, the orbital remains in place, but it's positive and negative lobes invert. So it will contribute -1 and for both orbitals $\chi(C_4)=-2$. For $C_2$, the rotation again leaves the orbital in place and leaves the lobes in each location with the same sign so $\chi(C_2)=2$.

$\chi(C_2')=\chi(C_2'')=\chi(i)=\chi(S_4)=\chi(\sigma_h)=0$ since all of these operations involve a rotation/reflection around a parallel line/plane so they will move the orbitals.

$\sigma_v$ runs along the $\ce{Re-Cl}$ bonds. Along and on either side of this plane, the sign of the lobes is the same and so $\chi(\sigma_v)=2$. $\sigma_d$ runs between the $\ce{Re-Cl}$ bonds and so this plane inverts the sign and $\chi(\sigma_d)=-2$.

Plugging these into a calculator I get a $b_{1g}$ and a $b_{2u}$ irrep. I will leave it to you to figure out which of these is the bonding combination and which is antibonding.


Beckmann rearrangement v/s dehydration of Oximes In the above question, A is an oxime, which is then reacted with phosphorus pentoxide. Now, phosphorus pentoxide is both a dehydrating reagent and a reagent used for Beckmenn rearrangement, but which reaction will form the major product mostly, ie, which reaction will dominate? My teacher told to consider beckmenn rearrangement with Phosphorus pentoxide, but the answer of this question is A, that is, dehydration is favored . What I think from my experience is that aldoximes will favor dehydration, and ketoximes beckmenn rearrangement( dehydration obviously cannot  occur in them), but I am not sure. Please tell the conditions under which any reaction will dominate over other. Also what  would be the answer if phosphorus pentoxide is replaced by phosphorus pentachloride?

Do you think that the mechanism of "Beckmann" and "dehydration" are different?
Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments)
The mechanism for Beckmann rearrangement is as follows,

(Source: Wikipedia, Beckmann rearrangement)
Note: Just replace $\ce{R1}$ with $\ce{H}$ and then, in the intermediate (upper one) after second step, the arrow will be from $\ce{N-R1}$ bond to $\ce{N}$. That's dehydration!
Also, if you replace $\ce{P2O5}$ by $\ce{PCl5}$ or $\ce{H2SO4}$, the answer would be same for all.


Minimizing energy of a molecule I am minimizing energy of butane on the CHARMM Force Field and I am confused on how to go about getting a uniform unit for each different PE  as in the potential energy for bond, angle, torsion, VDW, ES of butane. For instance if you compute the angle PE you will be using angles vs bond length PE which will be in Ångstrom. I am confused on how to set everything within the same units or parameter.

Also I am programming this in python using numerical methods and will not be using a simulator.

First I only wanted to comment, but then it got too long... But I wanted to pass on this info... this isn't a real answer...

Every simulator I know uses atomic units. I would check out the NAMD and CHARMM manual for their units, I imagine they use nanometer for distance and ultimately kJ/mol for energy. You will need a conversion for the electrostatic since you always input +/- charge, where charge is usually between 0 and 1, but units of charge are actually Coulomb. The conversion factor for this is ~138.9

This is from my own code... qq_convert is the electrostatic conversion I mentioned above. I use nm for all distances. I try to follow the same procedure as Gromacs.

const na = 6.02214129e23      # mol^-1
const R = 8.3144598e-3        # kJ mol^-1 K^-1
const kb = R / na             # kJ K^-1
const e = 1.6021765653e-19     # C
ϵ₀ = 8.854187817e-12 # C²/(J ⋅ m)
ϵ₀ *= 1e-9  # C²/(J ⋅ nm)
ϵ₀ *= 1000   # C²/(kJ ⋅ nm)
const qq_convert = e^2 / ϵ₀ * na / 4.0 / pi # 138.935458  # kJ mol^-1 nm e^-2

after each electrostatic interaction I multiply the energy by qq_convert

For instance the real part contribution of the EWALD summation is

@inline function ERealSingle(diff::MVector{3,Float64},qqCut²::Float64, coord1::SVector{3,Float64},
        coord2::SVector{3,Float64},q1::Float64, q2::Float64, kappa::Float64,
        overflow²::Float64, L::SVector{3,Float64})

    @inbounds for i=1:3
        diff[i] = vector1D(coord1[i],coord2[i],L[i]) # mirror image separation
    end

    rij_sq = diff[1]*diff[1] + diff[2]*diff[2] + diff[3]*diff[3]
    rij = sqrt(rij_sq)

    if rij_sq < qqCut²
        return q1 * q2 * erfc( kappa * rij ) / rij
    else
        return 0.0
    end
end

This is for a single charge interacting with another charge... I loop through all charges etc and then I multiply the summed energy by qq_convert.

I highly recommend checking your code against these SPC/E water benchmarks offered by NIST... really saved my bacon having a test set to duplicate... NIST Test Set. This set allows you to test your Lennard-Jones, all parts of the EWALD Summation and also tail corrections.

Lastly, I recommend using the Julia programming language... it is much faster than Python... Python can almost keep up if you use the Numba JIT decorators, but Julia has other perks.

Edit:

It looks to me like charmm uses Angstrom as you have said, so that is the only units of length you should be using. It would be dangerous for them to deviate.

In my experience with Gromacs, the parameters of each potential energy function will have units such that everything cancels out leaving you with $\ce{kJ/mol}$ (or $\ce{kcal/mol}$ in CHARMM I assume). In Gromacs all I need to do is make sure all units of length are nm. I imagine in CHARMM it is the same, but for Angstrom. Look at the angle parameter units, they do have units and it will work out that the energy is $\ce{kcal/mol}$ or else I say leave CHARMM.


Is cyclohexol the same as cyclohexanol? I recently came across different names for the following compound:

Is the IUPAC name for this cyclohexol or cyclohexanol? Is the use of cyclohexol as seen in a google patent and in this Chinese thesis accurate?

Note that I can only comment on the English translations!

Cyclohexanol is correct, cyclohexol is just wrong. This is not the only error in the abstract of the thesis.

  Cyclohexone is the most [...]

isn't correct either. The name is cyclohexanone!


Why do self-replicating molecules dissociate? I am new to chemistry (have only done a GCSE) so I apologize for my lack of knowledge. But I am confused about why self-replicating molecules dissociate.

It is my understanding that a self-complimentary molecule can gather the parts needed through bonding to create another copy of itself on its recognition surface. Once the new molecule is constructed I am unsure why it would then depart from the original one during this dissociation phase. Why would all of these bonds suddenly break apart? What exact mechanisms are involved in this process?

Well, allow me to answer since I worked on self-replication during my master’s thesis.

The thing about all these associationen is that there is always an equilibrium to be considered. That is also true for chemical reactions: Technically, every reaction is at an equilibrium of some kind.

$$\ce{A + B <=> C + D}$$

or in the special case of associations:

$$\ce{A + B + AB <=> A-B-AB}$$

For nomal bond-creating and bond-breaking reactions, the energy involved in the whole process is relatively large — breaking a carbon-carbon bond requires approximately $400\,\mathrm{\frac{kJ}{mol}}$! However, if we are merely talking about association as in complexation, the energies involved (i.e. gained or lost) are rather small. This es especially also true for the activation energy, meaning that forming and dissociation of such complexes are rather fast and at equilibrium the whole time. It is important to realise that the bonds formed for the self-replicating complex are all intermolecular and only the bond between the fragments is a covalent, intramolecular bond.

Let’s take a look at an exemplary self-replicating complex:[1]

On the right half of this picture we have the two fragments, on the left we see the template. All solid lines are actual (intramolecular) bonds while all dashed lines are intermolecular associations — in this case hydrogen bonds. The intermolecular dashed interactions build up and break down rapidly with low energy barriers. The new intramolecular bond is formed between carbon and nitrogen on the aromatic rings (not formed on the right, formed on the left) and represents the energy liberated in the reaction. This one is only formed, not broken.

To form the bond, the two fragments need to linger on the template long enough. That is generally no problem, despite rapid association and dissociation.

In fact, form too stable of a complex (i.e. one that won’t dissociate quickly) and you have effectively eliminated the complex’ ability to self-replicate.

[1] A. Terford, G. von Kiedrowski, Angew. Chem. 1992, 104, 626. DOI: 10.1002/ange.19921040523
In English: Angew. Chem. Int. Ed. 1992, 31, 654. DOI: 10.1002/anie.199206541


Chemical Test for tin How can I distinguish whether a metal that I have is tin ($\ce{Sn}$)? I am specifically looking for a chemical reaction that will happen only with tin and not with other metals such as $\ce{Al,~Cu,~Fe,~\ldots}$. Do such reactions exist? What are they?

Beside determining the volumic density (tin: 7.31 g/ccm, aluminum: 2.7 g/ccm, etc. pp) and as asked here on chemistry SE, there is a quick pre-test.  In German it is known as "Leuchtprobe", literally "illumination probe".  Be reminded to protect yourself properly (safety googles, fume hood, gloves).

In a first instance, you have to reduce the tin compound.  For this, a little of your solid under investigation is finely mixed together with some grains of zinc (Zn). In a porcelain dish, to this mixuture is added 5 mL of 20 m% aqueous HCl.  Await this reaction ceased.  Zn will produce hydrogen, and because it is freshly created (statu nascendi), it is of superiour reducing power compared to bottled hydrogen gas, and the solution is cooled to room temperature.

Then dip a cool test tube, partially filled with cold water, into this solution.  While still wetted by the solution, rapidly  place the test tube into the upper part of the non-luminous flame of a Bunsen burner.  Blue fluorescence is an almost specific evidence for tin (caused by $\ce{SnCl2}$, and stannanes), already at low concentrations.

Presence of arsenic may hamper this fluorescence.  Niob-compounds yield some similar fluorescence, too.  Beside using tin-free chemicals and equipment to be used -- and prior to the test of your unknown -- do at least once the test in known absence of tin, and once with a sample with tin to learn how to discern the two instances.


Reduction at the cathode I'm learning electrolysis and we learned that the positive ions go to the cathode which is where reduction takes place.
However, I got confused because when you learn about ionic bonding, the positive ions are the ones that lose electrons. And we learned that reduction is the GAIN of electrons. These two things clash. Can somebody please explain this to me?

Positively charged ions wouldn't usually lose more electrons, they've already lost some.

The positively charge ions you have are called cations. These go to the cathode, which is negatively charged, as opposites attract. So, positive ions go to an area of negative charge and gain electrons, i.e. reduction takes place.

Now how can we negate any confusion between cathode and anode (it's very easy to get mixed up)?

Let's examine the word cathode a little more closely. Essentially there are two parts, 'cath' and 'ode'. The key part here is the ending of the word, which is 'ode'. The Greek root of this means, according to the Oxford English Dictionary, way. So as long as you can remember that 'cath' usually refers to a positive and 'an' to a negative, you can put the two together. Cathode meaning 'positive-way', as in positive things go this way.


Synthesis of Copper(I) Iodide Copper(I) iodide is usually made by the reduction of copper(II) salts. An example of using potassium iodide and copper(I) sulphate is shown below.

$$\ce{2CuSO4 + 4KI + 2Na2S2O3 -> 2CuI + 2K2SO4 + Na2S4O6 +2NaI}$$

I did another method of synthesising $\ce{CuI}$ by preparing it directly from its elements, with $\ce{NaI}$ as the solvent for solid $\ce{I2}$:
$$\ce{2Cu + I2 -> 2CuI}$$

May I know why the 2nd method (direct synthesis) is better than the 1st in terms of green chemistry?

According to the United States Environmental Protection Agency, there are 12 principles of green chemistry. The second method of synthesis is more green than the first method of synthesis as it adheres to the first, second and fourth principles while the first method does not. We can easily observe why this is so.

Firstly, the second synthesis method prevents wastage of any reactants, all the reactants go into the making of the product, copper (I) iodide while for the first method, there are other products generated along with the desired product. In other words, there is a wastage of reactants if the first synthesis method is used.

Secondly, the second synthesis method maximises atom economy (i.e. waste few or no atoms). This is similar to the first principle. Atoms are wasted in the first synthesis method in making the other products.

Thirdly, the second synthesis method produces what is desired and does not produce any other chemicals that may be hazardous. According to this and this, sodium tetrathionate and sodium thiosulfate are irritants to the human body.

Hope I have answered your question.


Calculating collision diameter with viscosity, but not having the mass given I've got a homework question i'm stumped on.

  The viscosity of the ethylene gas was measured at $99.8\times10^{-7}\ \mathrm{N\ s\ m^{-2}}$ at $288\ \mathrm K$. Calculate the collision diameter in $\overset{\circ}{\mathrm{A}}$ for this gas molecule.

I started with

$$\eta=\frac{m\overline v}{3\sqrt2\pi d^2} $$

then i said

$$\overline v=\sqrt{\frac{8RT}{\pi M}}$$

Molar mass $M=28.05\ \mathrm{g\ mol^{-1}}$

So my equation looks like

$$\eta=\frac{m}{3\sqrt2\pi d^2}\sqrt{\frac{8RT}{\pi M}} $$

but $m$ is still a variable. What should I do from here? Is there another substitution I can do?

So apparently my confusion about solving this problem, came from a lack of understanding what the $m$ variable was supposed to represent. I had thought that it was the total mass of the system, but it is actually the mass of one molecule.

Answering my own question here (just received the answer to the problem.)

the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:

$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$

The solution came out looking like

$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$

$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$

$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$

$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$


Is glycinate ligand a monoanion or a zwitterion? Glycinate as far as I can guess from the structure is a monoanion with negative charge on oxygen atom (though that negative charge is resonating). But when I used this logic in my exam it costed me 4 marks. I don't understand why.
Is it because it is derived from glycine in strongly basic medium which is not a monoanion but a zwitterion? Still, it doesn't seem to be making sense.

,Assuming glycinate is in its anionic form ($\ce{H2N-CH2-CO2^-}$), one thing to consider is it may not act just as a simple anionic ligand like chloride or hydroxide.  You can count off four atoms from the nitrogen to either carboxyl-ate  oxygen and those atoms both have available electron pairs.  So you add the metal atom and you get a five-membered ring similar to that formed with ethylenediamine.  Thus, glycinate anion may act as a chelating agent.


If Gibbs energy of activation is positive, how is the change to the transition state possible? The Gibbs energy increases from the reactants to the transition state, but according to the Second Law of Thermodynamics, entropy of the universe can only increase. Why is the transition state possible then?

...according to the Second Law of Thermodynamics, entropy of the universe can only increase.

Well, there was a nice similar question in bio.SE about this. A system is not universe. The overall entropy of the universe increases, but that isn't true for only a particular thermodynamic system.

Also, no one said the transition state is stable!

  The transition state of a chemical reaction is a particular configuration along the reaction coordinate. It is defined as the state corresponding to the highest potential energy along this reaction coordinate. Wikipedia

(High potential energy = lesser stability)

This chemwiki.ucdavis page is also good to help reshape your understanding of second law of thermodynamics.


How to tell the difference between lysine and the N-terminal amino acid in DNP labeling? The N-terminal amino acid of a protein can be identified by reacting the N-terminal amine with FDNB (1-fluoro-2,4-dinitrobenzene), or Dansyl chloride, or Dabsyl chloride, then digesting the protein with acid.
This question points out that the epsilon amino group of lysine can also react with the labeling reagents, creating a side-chain labeled lysine. Dinitrophenyl (DNP) derivatives of amino acids (specifically epsilon-DNP-lysine)
But if we're just looking for the labeled amino acid, how do we tell the difference between the N-terminal amino acid and lysine?
I suppose that if we ran the amino acid mixture on HPLC or TLC that the epsilon labeled lysine would have different retention times, and we could learn to ignore those peaks once we knew what lysine looked like.

Actually DNP-labeling of amino groups was first introduced in 1923 when 2,4-dinitrochlorobenzene (DNCB) was used for the identification of the terminal groups of a partial hydrolysate of silk fibroin (Ref.1). Then, Sanger has introduced 1-fluoro-2,4-dinitrobenzene (DNFB) in 1945, which bears his name, the Sanger’s reagent. DNFB was first used to detect free amino acids of insulin (Ref.2). In a nutshell, DNFB undergoes nucleophilic aromatic substitution with the $\ce{N}$-terminal amino group of a peptide or protein. DNP-group is stable to acid hydrolysis. After the DNP-labeled peptide or protein is hydrolyzed, the individual amino acids can be separated and only the DNP-labeled $\ce{N}$-terminal amino acid can be detected by a colorimetric detection at specific wavelength. Hence, DNFB can be used in protein sequencing to determine $\ce{N}$-terminal amino acids.
Not only the $\ce{N}$-terminal amino acid of the protein is labeled by Sanger’s reagent, but also the backbone amino acids with an active group to DNFB such as $\epsilon$-$\ce{NH2}$ in lysine, $\ce{SH}$ in cysteine, $\ce{OH}$ in tyrosine, and $\ce{NH}$ in histidine would be labeled as well. The labeled protein is then hydrolyzed (DNP derivatives are stable to acid hydrolysis), and extracted with ether. If lysine, cysteine, tyrosine, histidine are terminal amino acids, they should be di-DNP-labeled amino acids. If any of them are in the backbone (within $\alpha$-helix), they should have free amino group after hydrolysis. These mono-DNP-labelled amino acids with free amine group are soluble in acid phase and won't come to ether layer when extracted (Ref.2).
The concentrated ether layer is loaded on silica gel chromatography column and the elution profile is matched with the standard profile obtained from individual DNP derivative of all the amino acids. The reagents produce yellow colored derivatives which can be easily detected by absorbance (Ref.2). This method can be applied to any protein (best works with less than 50-70 amino acids). For larger proteins, before DNFB treatment, the protein should be hydrolyzed to small fragments by enzymes such as trypsin or pepsin, or by chemicals like cyanogen bromide.
Note: Extraction and separation methods are detaied in Ref.2, hence I attached the PDF file. For instance, in insulin (MW: ~$\pu{12000 Da}$), there are four terminals (two glycines and two phenylalanines), and two $\epsilon$-DNP-lysines.
References:

E. Abderhalden, W. Stix, Hoppe-Seyl. Z. 1923, 129, 143 (Reference 2 of Ref.2).
F. Sanger, "The free amino groups of insulin," The Biochemical journal 1945, 39(5), 507–515 (doi:10.1042/j0390507)(PDF).


Alkylation of conjugated nitriles - regioselectivity Why does the enolate of the nitrile attack the alkyl chloride from the bond shown in the picture and not from the far end pi bond?

I found this article that sheds some light on deprotonation mechanism - it differs significantly from presented in the question and although it's for acetonitrile not crotonitrile it still provides some useful information.

Lithium amide exists as $\ce{(LiNH2)2}$ dimers which tend to coordinate to nitrile groups creating Li-N bonds. In acetonitrile deprotonation occurs intramolecularly with opening of $\ce{Li2N2}$ rings, not pericyclicly, but for crotonic nitrile analogous mechanism isn't possible what suggests intermolecular reaction.

  On the basis of our MP2(full)/6-31+G*//MP2(full)/6-31+G* calculations (Scheme 2 and Figure 1), the adduct, which is formed from CH3CN (7) and the dimeric base 8, is stabilized by about 12 kcal/mol relative to the separated reactants. It is likely that in solution this coordination proceeds by replacement of a solvent molecule from the dimeric lithium amide. The existence of a complex comparable to 9 and the dimeric character of solvated lithium amides 6 are confirmed by NMR spectroscopy and structural studies. To allow hydrogen transfer from the nitrile to the amide moiety, the central four-membered ring of 9 must open.  The resulting intermediate 10, an “open dimer”, is also supported by various investigations of related reaction pathways. 7 The cleavage of an Li-N bond requires about 11 kcal/mol, while the transition state for proton transfer, 11, lies only just above slightly more than 1 kcal/mol higher in energy than 10. This low barrier is indicated in a lenghtened C-H bond of the open dimer (1.094 Å, compared with 1.082 Å in acetonitrile). After the formation of another “open dimer”, 12, which is stabilized by 14 kcal/mol relative to the transition state, and extrusion of ammonia, the final intermediate, 14, is a mixed (dianionic) aggregate of LiNH2 and LiCH2CN, comparable to Boche’s QUADAC.

Another article provides some info about anions of nitriles:

  The  exceptional nucleophilicity stems primarily from the powerful inductive
  stabilization  of  metallated  nitriles  rather  than  resonance
  stabilization as is the case with enolates.

and influence of nitrile group on carboanion:

  Essentially, the  powerful  electron  withdrawal  of  the  CN  group stabilizes the adjacent negative charge in a manner akin to an ylide. Consistent with this inductive stabilization is the range of geometries for the metallated carbon which spans a continuum varying from planar in 12 to pyramidal in the cyclopropanecarbonitrile 13.

Mainly inductive stabilisation suggests that in metallated anion of crotonitrile negative charge is concentrated on alpha carbon and thus it's more nucleophilic then gamma carbon and in low temperature, under kinetic control, attack via alpha carbon should be favoured.


Questions Regarding Osmosis From  Costanzo's Physiology:

Osmosis of water is not diffusion of water: Osmosis occurs because of a pressure difference, whereas diffusion occurs because of a concentration (or activity) difference of water.

Also from Costanzo, describing two solutions 1, and 2 , seprated by a semipermeable membrane, and 1 having a higher concentration of solute:

The solute in Solution 1 produces an osmotic pressure and causes, by the interaction of solutes with pores in the membrane a reduction in hydrostatic pressure of Solution 1. The resulting hydrostatic pressure difference across the membrane then causes water to flow from Solution 2 into Solution 1.

My question is:
How does the solute produce an osomtic pressure, and what type of interactions with the membrane pores causes the reduction in hydrostatic pressure that ultimately drive water movement?

Osmotic pressure

is consequence of non zero net water diffusion,
which is consequence of non equal water activities on both sides of semipermeable membrane,
which is consequence of the fact dissolved solutes decrease activity of water.

Osmotic pressure of a free solution is formally an external pressure needed to be acting on this solution to keep diffusion equilibrium with pure solvent on the other side of a semipermeable membrane.
Applying external pressure then leads to the well known reverse osmosing, widely used in production of deionized water ( replacing expensive distillation ) from sweet water, or in desalination of sea water.
Osmosis continues until fully counteracted by extenal, e.g. atmospheric pressure. E.g. if there is osmotic pressure $\pu{1 atm}$, osmosis leads up to $\pu{10 m}$ high solution column ( eventually corrected by solution density ). Similarly, you would need to use external pressure $\pu{p > 1 atm}$ to start reverse osmosis.
If a book says the primary reason is hydrostatic pressure difference, it is wrong. We can have 2 solutions of the same density, but different osmotic pressures. Combining them on the opposite membrane sides would lead to disbalancing of solution levels, even if they would be initially the same and even if the hydrostatic pressure difference would be zero.


Reaction of alcohols with bromine In a reaction between $\ce{R-OH}$ and $\ce{Br2}$, what will the product be?
Initially I thought it would be $\ce{R-Br}$, but that doesn't seem right.

I'm oscillating between the product being either $\ce{R-OBr}$ or no reaction occurring at all. I've searched a bit on the web too, but couldn't get an answer.

What is the product?

No reaction occurs with the reagents you have stated at room temperature. Maybe with different conditions (eg. basic conditions) the haloform reaction that Ivan commented would occur.


How did Noe et al. derive $k_{AB}$ in their PNAS 2009 paper? This question is regarding Constructing the equilibrium ensemble of folding pathways from short off-equilibrium simulations which uses Markov State Model to find the equilibrium protein folding pathways from off-equilibrium simulations.
I find the notation a bit hard to follow. How does Frank Noe and others derived equation 1?

In the supplementary information they don't really explain and aren’t careful with the math notations:

I do see how a rate equals $k_{AB} = P(A^C|A)/P(B^C \cap A^C)$, but I don’t see how $\sum_{i=1}^m \pi_i q_i^{-}$ gives $P(B^C \cap A^C)$ which is the probability to be outside of $A$ and not in $B$.

Let us start by defining all the terms in Eq. 1 carefully, as the OP does not do so. I have always found it easier to interpret such models in the context of particles flowing from state to state, and that is the picture that I will try to present here.

$T_{ij}$ measures the probability of going from state $i$ to state $j$ within a time $\tau$.
$\pi_i$ is the equilibrium probability of being in state $i$.
$q_i^+$ is the probability that, starting at state $i$, a particle will reach the set of states $B$ before the set of states $A$.
$q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$. (At equilibrium, all trajectories are reversible, so we can alternately think of $q_i^- = 1 - q_i^+$ as the probability that, starting at state $i$, a particle will reach $A$ before $B$, but that is irrelevant for this problem.)
$k_{AB}$ is the normalized rate of particle flow from $A$ to $B$. In other words, given particles that start out in $A$, $k_{AB}$ is the fraction of particles per unit time that end up in $B$.

Now we simply need to interpret the numerator and denominator of Eq. 1. To think about these quantities in the particle picture, let us multiply top and bottom by the number of particles $N$ evolving in the system, defining $N_i = N\pi_i$, so that $$k_{AB} = \frac{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+}{\tau\sum_{i\in\text{all states}}N_iq_i^-}.$$ In the numerator $$\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+,$$ the quantity $N_iT_{ij}$ is the number of particles that leave $A$ (going from $i$ in $A$ to $j$ not in $A$) in a time $\tau$. A fraction $q_j^+$ of these particles will end up in $B$. Now, in steady state, this must also be the number of particles that are arriving at $B$ from $A$, or otherwise we would have an accumulation of particles in the system.

In the denominator $$\tau\sum_{i\in\text{all states}}N_iq_i^-,$$ the quantity $\sum_{i\in\text{all states}}N_iq_i^-$ is the number of particles in the system that started out in $A$, which serves to normalize the number of particles in the numerator, and $\tau$ is the time interval in which this all occurs.

Putting all this together, then, $$k_{AB} = \underbrace{\frac{1}{\sum_{i\in\text{all states}}N_iq_i^-}}_\text{normalized}\underbrace{\frac{1}{\tau}}_\text{rate of}\underbrace{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}}_\text{particle flow from A}\underbrace{q_j^+}_\text{to B.},$$ as desired.


Bandgap of a semiconductor for photocatalytic water splitting I'm reading a review article on photocatalytic water splitting (using semiconductors) and I came across the following:

"To achieve photocatalytic water splitting using a single photocatalyst, the bandgap of the semiconductor must straddle the reduction and oxidation potentials of water, which are +0 and +1.23 V vs the normal hydrogen electrode (NHE) at pH =0".

I have a few questions about this statement:

What exactly does it mean when you say that the bandgap has to "straddle" the reduction and oxidation potentials? Does it mean that you need an electron/hole with energy at least equal to 1.23 eV for oxidation/reduction to take place. So, the bandgap has to be greater than 1.23 eV in order to ensure that. Is my understanding accurate?
When using a single photocatalyst, I assume the oxygen and hydrogen evolution take place on the same material. If you had separate oxygen evolution and hydrogen evolution catalysts, what would the condition on the bandgaps of these semiconductors be?

Thanks! Sorry if these questions are too basic, I have little training in chemistry.

The original purported advantage of semiconductor|liquid junctions proposed by those in the field of photoelectrochemistry arose from the hypothesis that charge carriers would have a shorter distance to travel than in solid-state photovoltaic systems, thereby allowing lower quality materials to be used in an integrated solar energy system and still achieve high conversion efficiencies. In theory, one would also be able to avoid forming metallurgical junctions, say those p-n junctions made by using different dopants and creating a well-defined interface to separate charge carriers. Instead of forming these solid-state structures, you can define the electric field by selecting a suitable redox couple in solution. The specific energetics of the semiconductor and redox couple will then define the maximum power obtainable from the given system, so this is why the valence band must be below the energy of the OER and the conduction band above the energy level corresponding to the HER. By natural induction, this means that the bandgap must meet or exceed 1.23 eV, but this is only a thermodynamic constraint.

In reality your system’s bandgap must actually exceed 2.0 eV, this is due to the kinetic constraints based in part on the OER (and to a lesser extent the HER ) and losses encountered within the light absorber. Of course, for a single light absorber there is a tradeoff between the percentage of utilized solar flux and the bandgap.

This explains why many efforts are focused on using dual light absorbers, with one dedicated to the anode and another for the cathode. In this tandem structure they can efficiently utilize more of the solar spectrum. If you were to stack these light absorbers and use a 1.2 eV and 1.8 eV materials (with each one carrying out one of the two half-reactions required to split water), you could in theory get twice the efficiency compared to a single absorber with a bandgap of 2.2 eV.

Now are these semiconductor|liquid junctions going to beat a dedicated photovoltaic solar panel/electrolyzer any time soon? Maybe, but probably not. Is water splitting really worth doing? Maybe, but probably not. Is it a beautiful idea? Definitely.


Is it possible for this Iron-Carbon Compound to exist? Can this compound be created?

Where the orange is Iron and the black is Carbon.

Short answer: No

Long answer: Assuming you had hands tiny enough to manipulate individual atoms, you could make this molecule. The amount of time it would stay together would be as close to 0 as it gets. To start iron can't accommodate that amount of electrons, and the formal charges of this molecule are extreme to say the least.


Do enzymes that digest D-glucose react with L-glucose? D-glucose most common in nature and L-glucose is synthesized in the lab.
I know that humans can't use L-glucose in their aerobic pathways because it doesn't match the active site of the enzyme ...but I'd like to confirm and understand:

Why can't our cells use L-glucose?

It is like the problem of the key and the lock. If you put the good key in a given lock, it will go all right.It will open the lock. But if you make up a new key which is the image of the original key, it will not open the lock.


Calculating the change in pH upon diluting an acid What volume of water is added to $\pu{400 ml}$ of $\ce{HCL}$ solution so that the pH of the solution increases by 1 unit

This is how far i went but the answer says $\pu{1000 ml }$ where am I wrong is there any simplet alternative  to this question?

As Ivan said in the comments, $\ce{HCl}$ is a strong acid. It dissociates as follows,
$$\ce{HCl -> H+ + Cl-}$$
Now, let us assume the intial pH to be x. Since, $$\mathrm{pH} = \ce{-log[H+]}$$ $\ce{[H+]_i}$ = $\mathrm{10^{-x}}$, where subscript i means initial. Likewise, since final pH= x+1, $\ce{[H+]_f}$ = $\mathrm{10^{-(x+1)}}$ where f means final. Since $\mathrm{V_i = 400 mL}$, since total moles remain the same,
$$\mathrm{M_iV_i = M_fV_f}$$
we get $\mathrm{V_f = 4000 mL}$.
Hence amount of water to be added is $\mathrm{4000-400 = 3600 mL}$.


Hydrogen gas' reducing property I just went though a reaction wherein hydrogen gas is liberated and hence restricts the product formed to ferrous and not ferric salt. I just want to know how H2 gas gets its reducing property? It has a moderate bond strength also if I am not wrong.

Iron on reaction with HCl forms Fe(II) salts and not Fe(III) salts because even if Fe(+3) is formed, as we can see in electrochemical series Fe(+2) will be favoured ie. H2 which gets can reduce Fe(+3) to Fe(+2). So thermodynamically Fe(+2) salt is favoured. If you want to make Fe(+3) salt then you can react Fe and Cl2 directly.

  How H2 gets its reducing property?

H2 gas will react with FeCl3 to form FeCl2 and HCl. I don't know the exact mechanism but it can be explained well on the basis thermodynamic data available.


Why doesn't the oxidation of cellulose oxidise all the alcohol groups, and how does it cause discolouration? In cellulose, there are several alcohol groups - the two on the ring being secondary alcohols, and the one on the branch being a primary alcohol. I would expect all of these to be oxidised in the presence of any oxidising agent, such as heating under reflux with acidified potassium dichromate - this would give an aldehyde, and then a carboxyl group on the branch, whilst the two on the ring would be oxidised to ketones. However, as you can see in the diagram below (ignoring the sulfonation step which is irrelevant to the question), this is not the case with cellulose.

Why is the branched alcohol not also oxidised? Is it to do with it being a primary alcohol?

Further to this, the oxidation of these groups in paper causes discolouration. Is this because the double bonded oxygen atom then joins the delocalised system, increasing the size of the chromophore and hence bringing the frequency at which the material absorbs into the visible region? Or is there a different reason?

Well Noah,
Primary alcohols, in general can be oxidised more easily, if we are using suitable reagents like acidified dichromate/permanganate etc.

By looking at the reaction it looks like there has been some reagent used like lead teraacetate (LTA) or NaIO4, which attack syn, vicinal diols/(alpha)-hydroxy aldehydes/glyoxal to cleave the C-C bond and produce the oxidized product:

It has been mentioned elsewhere, that cellulose can be oxidised by NaIO4 into some carbonyl derivatives:

This reaction has been responsible for yellowing [at times referred to as discolouration] of paper:

These carbonyl groups arranged in some definite fashion give rise to yellow colour, sometimes called chromophore, which are any group responsible for proving colour to an otherwise colourless compound.


Example of chemical reaction with a given form for the kinetic equations I’m an applied mathematician looking for examples from various applications fields (right now, chemistry, obviously) to illustrate the following reaction-diffusion system
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\partial v}{\partial t} - d_v \Delta v = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4}
\end{cases}
$$

Above, all constants $d_u$, $d_v$, $k_i$, $p_i$, $q_i$ are positive, and $p_i,q_i\geq 1$. $k_2-k_3$ and $k_7-k_8$ might be of any sign. The system need not be self-contained — if you need to add a third equation in order to make these two exist, feel free to do so.

If the diffusion part is bothering you, then I’m also fine with
$$
\begin{cases}
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4}
\end{cases}
$$

And if the $p_i$ and $q_i$ are bothering you, then one example where they are all equal to $1$ is
$$
\begin{cases}
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u v - k_5 u^2 \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u v - k_{10} v^2
\end{cases}
$$

One more precision: if $k_2-k_3\leq 0$, then $k_5=0$ is allowed, and similarly, if $k_7-k_8\leq 0$, then $k_{10}=0$ is allowed. Therefore the following system would be a satisfying answer:
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_2 u - \tilde{k}_3 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_2 u - \tilde{k}_1 v - \tilde{k}_4 uv
\end{cases}
$$
(where the constants $\tilde{k}_i$ differ a priori from the previous constants $k_i$).

I looked for actual examples of first-order reversible reactions. I found $\ce{Cu+ <-> Cu^{2+}}$ there but I do not know if having an irreversible reaction “$\ce{Cu^+ + Cu^{2+} ->}$ some product” in this configuration is possible.

Can you think of any “real” chemical system that would satisfy such kinetic equations?

A chemist friend found an answer for the simplest case:
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_{-1} u - \tilde{k}_2 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_{-1} u - \tilde{k}_1 v - \tilde{k}_2 uv
\end{cases}
$$

$u$ is the concentration of ethenol;
$v$ is the concentration of ethanal;
the tautomerization of ethenol and ethanal gives the first-order part of the reaction;
the aldol reaction of ethanal with ethenol, giving the 3-hydroxybutanal, gives the second order part.

This is definitely a satisfying answer.

Nevertheless I’m still interested in other answers, especially with autocatalysis of either $u$ or $v$ (that would decrease the first-order consumption of it) or with autoreactions (that would give the term $-k_i u^2$ or $-k_i v^2$). As a matter of fact, the French Wikipedia claims that an aldol reaction of ethanal with itself in presence of a base can also produce 3-hydroxybutanal, so that the same example would also give a term $-k v^2$? But maybe these two aldol reactions correspond in fact to the same elementary mechanism?


How to calculate the activation energy of diffusion of carbon in iron? At room temperature, $298~\mathrm{K}$, the diffusivity of carbon in iron is
  $9.06\cdot 10^{-26}\frac{m^2}{s}$.
  At $573~\mathrm{K}$ the diffusivity is $1.66\cdot 10^{-15}\frac{m^2}{s}$. What is the activation energy, $Q$?

I am wondering if my solution is correct:

$D=D_0\exp\left\{\frac{-Q}{RT}\right\}$ and $D_0$ is constant (is this correct), thus

$\displaystyle \frac{D_{298}}{\exp\left\{-Q/(RT_{298})\right\}}=\frac{D_{573}}{\exp\left\{-Q/(RT_{573})\right\}} \implies Q\approx 122~\mathrm{\frac{kJ}{mol}}$

I would be very greatful if someone could check my math here as well, if the logic is correct, since this was a test problem.

Maybe I'm wrong but my calculation has given another result:
\begin{align}
D_{298} &= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\
D_{573} &= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\
\end{align}

Thus, going through the $\ln$ form, you get
\begin{align}
\ln\{D_{298}\} &= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{573}\} &= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{298}\} - \ln\{D_{573}\} &= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right)
\end{align}

And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.


Comparing Methods of Determination of Dissolved Oxygen Content in Water Sample I am currently doing an IB Chemistry Individual Investigation on dissolved oxygen in water samples by using Winkler method. However, my teacher suggested that I have to use oxygen probe as a standard and comparing the efficiency of dissolved oxygen content determination between Winkler method and other methods.

My problem is I don't know what method I have to compare with the Winkler method that I had done. Are there any suggestions or improvements on my investigation to let it have more exploration aspect?

Let's abstract your task to determine dissolved oxygen (DO) to something more general:

You have to tie an observable to a concentration.
It would be nice if the method would work over a broad range of concentrations.
The method should be specific, or, alt least, very selective.
If possible, this relation should be linear.
It would be nice if the method could be packed into a sensor that works on its own (permanently or in cycles) and does not require a lot of personal interaction (= autonomous probe)

What properties of molecular oxygen dissolved in water can be measured?

In water, $\ce{O2}$ can be reduced electrochemically: $\ce{O2 + 4e- + 2 H2O -> 4 OH-}$→ Have a look for galvanic and polarographic oxygen sensors
Molecular oxygen is known to quench the luminescence of electronically excited states of dyes. The observed luminescence intensity (and lifetime) is reciprocally proportional to the the concentration of the quencher(Stern-Volmer equation)
→ Have a look for luminescence-based oxygen sensors


Why is the reactivity of primary alkyl halides with nucleophiles (SN2 mechanism) greater than secondary and tertiary alkyl halides? For the reaction between alkyl halide and a nucleophile, following the SN2 mechanism, the reactivity of alkyl halides is in the order: primary halide > secondary halide > tertiary halide.

If you consider the reactivity of alcohols with halogen acids to form alkyl halides, the reactivity of alcohols is in the order: tertiary alcohol > secondary alcohol > primary alcohol.

Why is there a difference in order of reactivity in both the cases?

The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an SN2 mechanism. A $\ce{C-Nu}$ bond forms and a $\ce{C-X}$ bond breaks at the same time:

$$\ce{CH3CH2Br + NaOH ->[\text{acetone}] Na}\bigg[\ce{CH3CH2(Br)(OH)}\bigg]^{^\ddagger_{-}}\ce{->CH3CH2OH +NaBr}$$

All else equal, the rates of these reactions are controlled by sterics on the alkyl halide, which control how well the nucleophile can react the $\ce{C-X}\ \sigma^*$ orbital. Thus:

$$\text{Methyl}>1^\circ>2^\circ>>3^\circ$$

The reactions of alcohols with hydrohalic acids follow a different mechanism (mostly). These reactions happen in a protic medium ($\ce{HX}$ is protic) with a poor leaving group ($\ce{OH}$), at least initially. These are conditions that favor the step-wise SN1 mechanism, where $\ce{C-X}$ bond breaking occurs before $\ce{C-Nu}$ bond forming. A carbocation is formed as a key intermediate.

$$\ce{(CH3)3COH + HBr -> [(CH3)3COH2+]Br- -> [(CH3)3C+]Br- + H2O -> (CH3)3CBr + H2O}$$

According to the Hammond postulate, the stability of this carbocation influences the rate of reaction. More stable carbocation intermediates suggest lower energy transition states for the formation of those carbocations, which implies lower activation energies and faster rates. Thus:

$$3^\circ>2^\circ>>1^\circ,\text{Methyl}$$


Struvite Crystals: How does it form? I'm in general chemistry, and I was wondering how ionic species crystallize and form the shapes they do, specifically in regard to the Struvite precipitate $\ce{MgNH_4PO_4. 6H_2O}$.
Deconstructing the ionic species, we get $\ce{NH_4^+}$, which does not have a dipole moment (I think), $\ce{PO_4^{3-}}$, which also does not have a dipole moment, $\ce{H_2O}$, which has its dipole moment pointing towards the oxygen, and $\ce{Mg^{2+}}$, which is just an ion floating in solution.
Drawing the lewis structures of these molecules and writing the lewis structures, I'm not really sure how the crystal would form. But I do have a guess.
I feel as if the $+1$ formal charge on the nitrogen in $\ce{NH_4^+}$ will be attracted to one of the $-1$ formal charges on the oxygens in $\ce{PO_4^{3-}}$, while the magnesium ion will take in some of the free electrons on the other two oxygens. The $\ce{H_2O}$ will then form hydrogen bonds with the net dipole moment generated?
But this all sounds so hand-wavy to me, and I know it's probably wrong. Furthermore, I can't really visualize what this would look like. I honestly don't have a good understanding of how things work with crystallization. This is, after all, my first semester of chemistry.
If someone could give me tips and hints on how a struvite unit arranges its ions via electrostatic interactions to make something stable, I would appreciate it. Resources and pages in textbooks would be appreciated too!

As Ed V points out, the conditions of formation of struvite have many variables and the struvite crystals produced are variable. On the one hand, this could be a pleasant exercise of playing with atomic models, but on the other, it may be unproductive unless you were to use computer graphing to model various configurations.
Perhaps you could start with a simpler, but similar compound and develop a crystallization mechanism from that. Ammonium phosphate seems simpler, but isn't, because the third proton of H$_3$PO$_4$ is so weak that pH is still a significant variable, and you are also dealing with diammonium phosphate and double salts (Ref 1).
Another possibility is plain magnesium phosphate, but it appears in several forms and with several degrees of hydration, so it is not a simple crystal formation. (Ref 2) The anhydrous form is too unrelated to struvite. (Ref 3)
Stretching your imagination further, replacing the ammonium with potassium gives an interesting material that has applications in cementing. One article gives a simple picture of the "components" of KMgPO$_4$.6H$_2$O:

The octahedron of Mg(H$_2$O)$_6$$^+$$^2$, the tetrahedron of PO$_4$$^3$$^-$, and the sphere of K$^+$ are objects that could be played with to make a model crystal. The sphere could also suggest how the NH$_4$$^+$ ion (just assume it is almost spherical) fits into struvite. (Ref 4)
Struvite was the basis for a rapid repair patch for roadways some years ago (Ref 5), and has current applications in immobilizing radioactive waste. (Ref 6)
Ref 1. https://en.wikipedia.org/wiki/Ammonium_phosphate
Ref 2. https://en.wikipedia.org/wiki/Magnesium_phosphate
Ref 3. https://byjus.com/chemistry/magnesium-phosphate/
Ref 4. https://pubs.acs.org/doi/10.1021/acs.jpcc.7b01396#
Ref 5. US Patent 3,960,580 J.M. Gaidis et al.
Ref 6. https://www.sciencedirect.com/topics/engineering/magnesium-potassium-phosphate


Trying to understand the contradiction in the data of Proust/Dalton/Berzelius vs Gay-Lussac I am trying to understand some of the historical context behind the discovery of atoms (pre-Avogadro).  In particular, I am getting stuck in reading the following documents:

http://physics.unl.edu/~klee/phys261/readings/class09.pdf
http://physics.unl.edu/~klee/phys261/readings/class11.pdf

My understanding is that "atoms" (as they understood them at the time) were discovered to react in whole-number ratios by Proust/Dalton/Berzelius (who measured by mass) and by Gay-Lussac (who measured by volume).  While this was highly suggestive of some sort of "atom" concept, there was a discrepancy between the two sets of measurements.  But I'm having trouble following the specific discrepancy pointed out by the above PDFs.

The first PDF gives the following mass data:

                         Mass of Nitrogen    Mass of Oxygen
Nitrous Oxide (N₂O)             100 g             58 g
Nitric Oxide (NO)               100 g            127 g
Nitrogen Dioxide (NO₂)          100 g            239 g

The text doesn't explicitly mention how this data was interpreted by Proust/Dalton/Berzelius.  Looking at the data, I don't see that much can really be said:

? N + 1 O = ? Nitrous Oxide (N₂O)
? N + 2 O = ? Nitric Oxide (NO)
? N + 4 O = ? Nitrogen Dioxide (NO₂)

The question marks are there since they didn't know what was "in" 100 g of nitrogen, which would mean that they didn't really know what was in the product, either, right?

So that's my first question:  Am I correct in thinking that the above is about as far as Proust/Dalton/Berzelius could go with that data?  Or were they pretty sure they knew how to fill in the question marks?

Now, continuing on to the second PDF, when Gay-Lussac measured by volume, he had:

                        Volume Nitrogen  Volume Oxygen  Volume Product
Nitrous Oxide (N₂O)             100           49.5             100
Nitric Oxide (NO)               100          108.9             200
Nitrogen Dioxide (NO₂)          100          204.7             200

Now, the document goes on to say that the above data might suggest the following formulas:

2 N + 1 O = Nitrous Oxide (N₂O)
1 N + 1 O = Nitric Oxide (NO)
1 N + 2 O = Nitrogen Dioxide (NO₂)

The text then says that the equation for nitrous oxide represents a contradiction.

But I don't see how that contradiction follows from the data.  As the text suggests, if we say that a volume of "50" represents one "atom" (so that volume of "100" nitrogen represents two atoms), wouldn't we say:

2 N + 1 O = 2 Nitrous Oxide (N₂O)
2 N + 2 O = 4 Nitric Oxide (NO)
2 N + 4 O = 4 Nitrogen Dioxide (NO₂)

Wouldn't that mean really it's the other two equations that don't make sense based on the number of atoms not working out?  I feel like I'm missing a basic idea in the text.

(FYI, I am not in school and so I'm not a student, nor is this homework.)

In regards to your first question, I would think that Dalton would have a sense of the molecules' formulas based on the information at hand. Not only did Dalton have the mass ratios for these compounds, but he had also developed a measure of the relative mass of the atoms, as shown in your first document. Given this information, it would not have been difficult to compute that there were $\sim2$ atoms of nitrogen for every one of oxygen in nitrous oxide, nitric oxide was one to one, etc.

In regards to your second question, I think you are misinterpreting the contradiction. The authors proposal for the reactions was that if the compound formed was $\ce{N2O}$, it should only form one volume, since it should only form as much volume as there was volume of $\ce{O}$ originally. This is because there can only be as many molecules formed as there were atoms of the limiting reactant. The two volumes of $\ce{N2O}$ that are actually produced runs counter to this, as it would seem to require having $2$ volumes of $\ce{O}$ to start with.

For all the reactions, the products and reactants were related by whole number ratios, but not the same ratios suggested by their predicted molecular formulas.


What is the behavior of a Zwitterion during ion exchange chromatography? We're interested in purifying a protein that has an azide moiety using ion exchange chromatography. Due to its unique structure, the moiety is a Zwitterion with a novel chemical behavior. My question is how will the molecule behave when I run it through an ion exchange column? Will it be considered positively charged, negatively charged, or neutral or is it much more complicated?

I would expect it to depend on the pH of the mobile phase. The charge of your stationary phase and whether you want the azide to elute will determine what charge you want on the azide. Given that, you should set the pH of the mobile phase either acidic or basic enough to ensure that the azide is fully charged in the direction that you want.

From Wikipedia:

  Proteins have numerous functional groups that can have both positive and negative charges. Ion exchange chromatography separates proteins according to their net charge, which is dependent on the composition of the mobile phase. By adjusting the pH or the ionic concentration of the mobile phase, various protein molecules can be separated. For example, if a protein has a net positive charge at pH 7, then it will bind to a column of  negatively-charged beads, whereas a negatively charged protein would not. By changing the pH so that the net charge on the protein is negative, it too will be eluted.

If changing the pH is problematic, apparently increasing the ionic strength can help shield the azide's charge from the column's charge:

  Elution by changing the ionic strength of the mobile phase is a more subtle effect - it works as ions from the mobile phase will interact with the immobilized ions in preference over those on the stationary phase. This "shields" the stationary phase from the protein, (and vice versa) and allows the protein to elute.


NMR of trigonal bipyramidal molecules I'm slightly confused about NMR.

I seem to remember at school that if a molecule is symmetrical in $\ce{^1H}$ NMR, that only one peak would be shown e.g. propane's terminal $\ce{-CH3}$.

Consider the two molecules shown below.

My professor remarked that both these isomers would show a doublet and triplet in their proton-decoupled $\ce{^31P}$ NMR and I was struggling to work out why.

Consider the molecule on the left as molecule A and the right as molecule B.

My train of logic was that the axial $\ce{Ph3P}$ group on molecule A would be split by each $\ce{P}$ atom that is equatorial, so using the expression $2nI + 1$ where $I = \frac{1}{2}$ so our singlet would be split into a triplet.

Likewise, the equatorial $\ce{Ph3P}$ groups would both only see one axial $\ce{Ph3P}$ group, hence only a singlet would be observed.

I used the same logic on molecule B.

Is this the correct reasoning for the doublet and triplets observed in both molecules?

“Likewise the equatorial Ph3P groups would both only see one axial Ph3P group, hence only a singlet would be observed” - I think you’ve just made a little mistake. The equatorial groups can see one phosphorus other than themselves, so (2nI + 1) = 2, not 1!


Is there a database on chemical reactions, similar to NIST, but far more complete? The only thing I'm interested in is to be able to search all reactions involving that compound in the database. Equilibrium constant, reaction rate and energy of activation would all be very welcome, but in case these data are absent it's no big deal.

Doesn't need to be an online database either, I can download a software if it contains such data.

The best known databases are Beilstein and Chemical Abstracts with around 22 millions reactions each since the end of the 19th century but they are commercial.

Organic Syntheses has made available its database for free and it covers 6000 reactions. One of the greatest advantages is that each step is described in depth and each reaction has been thoroughly tested prior to publication.

Have a look here for a list of databases.


How do I list these elements according to their radius? $\mathrm{Z=33,\ Z=37,\ Z=9,\ Z=31}$

Since we shouldn't look at the periodic table, I found their groups and periods, what should I do next?

We have these elements:(just for ease, I'm naming them, you can just avoid it)

  $$
\begin{array}\\
Z=33&\text{Arsenic}&\ce{As}\\
Z=37&\text{Rubidium}&\ce{Rb}\\
Z=9&\text{Fluorine}&\ce{F}\\
Z=31&\text{Galium}&\ce{Ga}
\end{array}
$$

Now, let's see the periodic table (the useful region, thanks to Martin) :
$$
\begin{array}{ll|cccccccccccccccccc}
&\text{Gr.} &1 &2 &13  &15  &17 \\
\text{Per.}&\\\hline
2&&. & .& .& .& \color{red}{\ce{\tiny^9F}}\\
4&& .& .&  \color{red}{\large\ce{^{31}Ga}}& \color{red}{\small\ce{^{33}As}}& .\\
\hline
5&& \color{red}{\huge\ce{^{37}Rb}}\\
\end{array}\\
$$

Now consider this:

While going down or towards left, size increases. Equivalent, is to say the size is larger towards bottom left corner

It wouldn't be now difficult to conclude that size is as follows:
$$37>31>33>9$$


Subtracting wave functions? In molecular orbital theory bonding of two atoms can be seen as addition or subtraction of the wave functions of these atoms. Addition seems pretty logical and straight-forward - atoms overlap their electron clouds and probability to find an electron in it increases. Subtraction is the thing I don't get. How can the probability to find an electron be negative? Am I missing something?

I think you are confusing the wavefunction $\Psi$ with the probability density function $\Psi^2$. The former is a state function describing the behavior of an electron. The latter describes the "probability of finding an electron".

Wavefunctions are usually complex functions, but even the simplest sine wave functions can be negative in the algebraic sense: the value of $\Psi=f(x)<0$. These regions where $Psi=f(x)$ has different algebraic signs are the regions corresponding to the orbital lobes of different phase, like in a p-orbital. However, $\Psi^2$ is always $\Psi^2 =f^2(x)\ge 0$. Information about the "location" of the electron is gained, information about its phase is lost.

Take as an example the two wave functions representing two atomic orbitals $AO_1$ and $AO_2$ on the graph below. $\Psi_{AO_1}=\cos{(\frac{\pi x}{2})}$. $\Psi_{AO_2}=\cos{(\frac{3\pi x}{2})}$. $\Psi_{AO_2}$ has regions where $\Psi_{AO_2}<0$ when $\frac{1}{3}<|x|<1$.

Now let's look at the probability density functions $\Psi^2_{AO_1}=\cos^2 {(\frac{\pi x}{2})}$ and $\Psi^2_{AO_2}=\cos^2 {(\frac{3\pi x}{2})}$. Notice that both functions are always $\ge0$.

Okay... now, when atomic orbitals combine to form molecular orbitals, one of those conservation laws (angular momentum) gets involved, and the number of orbitals has to remain constant. Thus, molecular orbitals are linear combinations of atomic orbitals. For example, if two atomic orbitals combine, two molecular orbitals are produced.

Let's go back to our two sample sine wavefunctions above and pretend they are atomic orbitals overlapping. One of the molecular orbitals could then be: $\Psi_{MO_1}=\frac{1}{2}\Psi_{AO_1}+\frac{1}{2}\Psi_{AO_2}$. This is the orbital representing constructive overlap between these two wavefunctions. To get the other orbital, we can't just add again, or we would get the same orbital. If we subtract, we get a different orbital - the one representing destructive overlap. The graphs of these two "molecular orbitals" and their probability density functions are below.

Again note the difference between the MO wavefunctions and their density functions. If these were atomic nuclei at $x=\frac{1}{3}$ and $x=-\frac{1}{3}$, then $\Psi_{MO_1}$ would be bonding and $\Psi_{MO2}$ would be antibonding (or perhaps nonbonding).


Will SnO2 or SnO form from this reaction? I have two possible chemical equations for a reaction between $\ce{SnCl2}$ and $\ce{H2O}$. The first one is:

$$\ce{2SnCl2 + 2H2O +O2 -> 2SnO2 + 4HCl}$$

where the reaction draws on oxygen from the air. The second possibility is:

$$\ce{SnCl2 + H2O -> SnO + 2HCl}$$

I know that in the first reaction, tin is oxidized and oxygen is reduced, but I can't figure out if that reaction is favorable, or if it is possible that it's using oxygen from the air. The second equation is the only other possible reaction that I can find, but it doesn't form the product that we want.

The actual reaction of tin (II) chloride with water in the presence of air is given here.

$$\ce{6SnCl2 + O2 + 2H2O → 2SnCl4 + 4Sn(OH)Cl}$$

  Tin(II) chloride react with oxygen and water to produce tin(IV) chloride and hydroxotin(II) chloride. Interaction with oxygen and air humidity. This reaction takes place slowly.

As Mithoron noted, the reaction takes place very, very slowly and thus get hydrolysed partially to form the basic salt, $\ce{Sn(OH)Cl}$ instead of forming the hydroxide or the oxide.

The basic salt will also form when tin (II) chloride reacts with water in the absence of air, thanks to Gannex.

$$\ce{SnCl2 + H2O ⇄ Sn(OH)Cl + HCl}$$

  Tin(II) chloride react with water to produce hydroxotin(II) chloride
  and hydrogen chloride. This reaction is proceeds by diluting the
  solution tin(II) chloride. (source)

As noted earlier, the reaction is very, very slow and maintains an equilibrium between the reactants and the products.

However when tin (II) chloride is present in a concentrated, acidified solution, it will form a complex.(source)

$$\ce{SnCl2 + H2O → [Sn(H2O)Cl2]}$$

  Tin(II) chloride react with water to produce dichloridoaquatin.
  Tin(II) chloride - concentrated solution. pH < 7.


Why do enthalpy changes of combustion and formation not take into account oxygen? I've seen many enthalpy changes in my studies and I don't understand why the oxygens on either side of reactions are not calculated (especially if the unreacted oxygens don't balance). For example:

$$\ce{2C(s, graphite) + 3H2(g) + 1/2O2(g) ->  C2H5OH(l)}$$

By convention elements in their most stable polymorph at standard conditions have at standard conditions (298.15 K) a standard enthalpies of formation $\Delta{}_fH^\circ$ of zero (one reference).  This is why thermodynamics discerns C(graphite) from C(diamond), and C(fullerene), for example.  Hence on first view it looks like zeros in your example (graphite, Hydrogen, Oxygen) got forgotten; but they are not.


Do Sulfuric Acid and Calcium Hydroxide Completely Neutralize One Another? I just completed a unit test at school in which there was a simple fill in the blanks question which asked:

  Would 1 mol of sulfuric acid and 1 mol calcium hydroxide in water form an acidic, basic, or neutral solution?

The answer I wrote was neutral.

My reasoning was that sulfuric acid ($\ce{H2SO4}$) has two protons it can donate, and calcium hydroxide ($\ce{Ca(OH)2}$) has two protons it can accept. However, after the test I realized that H2SO4 dissociates into HSO4- which is a weak acid and won't dissociate completely.

After some research I have consistently found that the neutralization equation is written as: $\ce{Ca(OH)2 + H2SO4 -> CaSO4 + 2H2O}$

Do sulfuric acid and calcium hydroxide completely neutralize  one another? Or was I wrong to say that the resulting solution is neutral.

Could someone please explain why I am right or wrong?

The first proton transfer is simple enough and you have no doubts.

$$\ce{H2SO4 + Ca^2+ + 2 OH- -> HSO4- + Ca^2+ + OH- + H2O}\tag{1}$$

Now, you wonder whether hydrogensulphate is a strong enough acid to combine with the remaining hydroxide. Well, hydrogensulphate may be a weak acid but hydroxide is a strong base. So even the second step will go to completion:

$$\ce{HSO4- + Ca^2+ + OH- + H2O -> SO4^2- + Ca^2+ + 2 H2O}\tag{2}$$

Frankly, it does not end there. Calcium and sulphate form a salt with a very low solubility so this salt will precipitate out of solution.

$$\ce{SO4^2- + Ca^2+ + 2 H2O -> CaSO4 v + 2 H2O}\tag{3}$$

So the only thing you are left with in reasonable amounts is water and a precipitate. This, of course, is a neutral solution.

However, you don’t really always have to go that far. Check if there are strong acids and bases (don’t forget hydroxides), cancel them away and then check what remains. If you had done that, you would have noticed that there are as many hydroxide ions as acidic protons, overall giving an approximately neutral solution.


Which of the following compounds will form significant amounts of meta product during mono-nitration reaction? Hi, I tried to answer this question based wholly on the concepts of inductive effects. But haven't gotten anywhere with it.
I would appreciate any help regarding these questions:

What condition makes a compound to give meta product(instead of ortho or para)?
I know there is some flaw with my approach, so what is the correct way to do it?
Can this condition be applied to other compounds as well, or is it just for aromatic compounds?

The answer to the question is (1) Aniline.

This occurs because under standard strongly acidic nitrating conditions (typically cH2SO4/HNO3) the aniline NH2 is protonated removing the ability of nitrogen lone pair to donate any electrons. Consequently the electron-withdrawing polar effect makes it m-directing. Some p-nitroaniline is formed because there is a small amount of the aniline freebase in equilibrium with the anilinium, but that is much more reactive than the anilinium.

Further reading here

Also discussed in this question  Why does nitration of aromatic amines give meta directed nitro compounds?