Why are only sigma bonds used in determining the geometry and bond angle in VSEPR theory? Why aren't pi bonds used? I have read that in VSEPR theory, multiple bonds are considered or treated as single bonds when predicting the geometry of a molecule? I've read in Yahoo! Answers that it is because only sigma bonds are used in determining their shape. What's the reasoning behind this?

VSEPR is used to predict highly idealized geometries corresponding to Platonic solids or combinations thereof (e.g., the trigonal bipyramidal geometry being essentially the result of fusing two faces of a tetrahedron). The basic rationale behind its use, and the reason for its success in many situations, is the recognition that electrons, being like-charged, repel one another electrostatically. Hence, as a first approximation, it's reasonable to assume that the most stable spatial configuration of any molecule would have the atoms spaced as uniformly and as distant from one another as possible. VSEPR achieves this by placing the central atom of a simple molecule (or some molecular sub-structure) at the center of a geometrically perfect polyhedron having as many vertices as non-central atoms and lone pairs, and then placing the non-central atoms and lone pairs at those vertices.

The reason that $\sigma$ bonds (and lone pairs) determine the geometry is that they form the basic skeleton of the molecule. Recall that $\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented along the imaginary axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the two nuclei. $\pi$ bonds, on the other hand, are essentially orthogonal to the $\sigma$ bond skeleton, and are substantially weaker. Moreover, $\pi$ bonds do not exist in isolation, meaning any $\pi$ bond between two given atoms is always formed secondarily to the $\sigma$ bond between said atoms. As such, $\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by $\sigma$ bonding, although in practice, because they do actually introduce additional electron density and require closer orbital overlap, bond multiplicity affects bond length and bond angles (as does, for that matter, the size of the atoms involved). The magnitude of those deviations from the idealized VSEPR geometry could, in principle, be quite large, although most often those deviations are slight enough that VSEPR remains useful as a first approximation.


What happens when a solution of sodium chlorite is acidified? I'm told that when sodium chlorite $\ce{(NaClO2)}$ is mixed with aqueous acid, chlorine dioxide ($\ce{ClO2}$) is produced and remains mostly dissolved in water.

I would like to know how the strength of the acid (concentration of acid and/or the type of acid) affects this reaction. I would also like to know what happens (afterwards) to the $\ce{ClO2}$ over time. For example, if undisturbed, does the $\ce{ClO2}$ enter the air, convert into another compound, or something else?

When acid is added to sodium chlorite in solution (note: $\ce{ClO2-}$ is the chlorite ion), chlorous acid ($\ce{HClO2}$) is transiently formed which then goes on to decompose. However, you need hydrochloric acid, not just any acid, because chloride ion ($\ce{Cl-}$) itself acts as a catalyst for the decomposition. What happens is that chlorous acid disproportionates; that is, some of it gets oxidized while the rest gets reduced.

Specifically, out of five $\ce{Cl}$ atoms in chlorous acid (oxidation state +3), one will end up as chloride in the oxidation state $-1$, and four will end up as $\ce{ClO2}$ in the oxidation state +4:

$$\ce{5HClO2 -> 4ClO2 + Cl- + H+ + 2H2O}$$

Pretty much any concentration will do. Calculating the proportions is easy if you know a bit of stoichiometry: you need 5 moles of chlorite for 4 moles of hydrochloric acid. The reaction proceeds to completion, and you get 4 moles of $\ce{ClO2}$.

The $\ce{ClO2}$ remains in solution and is a good bleaching agent; see Wikipedia. It can undergo further disproportionation but if the solutions are kept in the dark, this process is slow.


Is it really appropriate to measure entropy change using enthalpy? One statement by Peter Atkins in his book Elements of Physical Chemistry confused me quite a bit:

If a chemical reaction or phase transition takes place at constant pressure , we can identify $Q$ with the change in enthalpy of the system, and obtain
For a process at constant pressure: $$\Delta S = \frac{\Delta H}{T}.$$

Is it really correct to measure the entropy change using change in enthalpy?
I'm telling so because enthalpy change only measures how much heat energy enters or gets expelled from the system. It doesn't measure how much energy gets expelled out of the system when the system does $PV$ work.
As Peter Atkins points out:

If $10~\text{kJ}$of energy is supplied as the heat to the system that is free to change its volume at constant pressure, then the enthalpy of the system increases by $10~\text{kJ}\; ,$ regardless of how much energy enters or leaves by doing work and we write $\Delta H = +10~\text{kJ}\; .$

I know it is the heat energy that is concerned in the definition of entropy change. But, suppose the system is heated which means it is in more disorder; however if it does work, then the energy the system got as heat would decrease which would therefore decrease the chaos & disorder of the system. If we use only the enthalpy change, wouldn't we exclude that work which decreases the heat energy gained by the system?
I'm rather confused on the use of enthalpy for defining the entropy change. Could anyone please help me explain this how enthalpy change actually measures the entropy change?

Yes.  So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B.  If it is a phase change, you start out with liquid  at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B).  This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor.  During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor.  Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$.  From the first law, the change in internal energy is equal to the heat added minus the work done:  $(U_v-U_l)=Q-P(V_v-V_l)$.  Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T).  Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$.  This completes what I wanted to say about the phase change situation.  I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.


Purifying/Recrystallising N-Iodosuccinimide I'm planning on an Iodination reaction of a 2-aminothiazole derivative. Specifically, 2-aminothiazole, with two Boc groups on the amine.

What practical advice do you have for recrystallising NIS? Literature recommends dissolving in hot dioxane and then adding $\ce{CCl4}$ to precipitate out the NIS.
If I do not have access to $\ce{CCl4}$ would $\ce{CHCl3}$ suffice?
Do NIS and NBS (I've previously used NBS for this reaction in $\ce{CHCl3}$) have similar solubilities in $\ce{CHCl3}$?

Purification of Laboratory Chemicals (a generally accepted reference book for methods of purification of common chemicals) appears to agree with your procedure:

  N-Iodosuccinimide [516-12-1] M 225.0, m 200-201 °C. Crystallise it from dioxane/CCl4. It iodinates arenes in triflic acid. [Olah et. al. J. Org. Chem. 1993, 58, 3194, Beilstein 21/9 V 544.]

As you point out, carbon tetrachloride is difficult if not impossible to get hold of now, so the methodology will not work.

The most obvious thing to try would be the method for the purification of NBS, however NIS is not tolerant to recrystallisation from water and from previous attempts doesn't achieve anything.

Depending on what you're doing with the NIS, I've often had success in simply dissolving in solvent and washing with sodium thiosulfate to remove any liberated iodine. Drying and evaporation then gives cleaner, though probably not pure NIS.

Also note that NIS should be stored in the dark at reduced temperatures to prevent decomposition (which is noted by the appearance of an orange/brown decoloration).


Why is boron trifluoride a lewis acid? $\ce{BF3}$ does back bonding with fluorine and still accepts a pair of electron and is considered lewis acid why?

A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.


How does diluting an acid change pH? By the Henderson–Hasselbalch equation, log (salt)/(acid) term is a ratio of the concentrations. So, how can dilution change the pH of the acid as the concentrations of both salt and acid would change with dilution and the term would remain the same?

Consider a strong acid such as $\ce{HCl}$ which — within experimental error — can be considered fully dissociated in aquaeous solution. This can be expressed as:

$$c(\ce{H3O+}) = c_0(\ce{HCl})$$

Assume you have a concentration $c = 1~\mathrm{\frac{mol}{l}}$. pH is defined as:

$$\mathrm{pH} = - \lg [\ce{H3O+}]$$

Thus our solution has a pH of:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 1 = - 0 = 0$$

(Remember that to calculate pH we need dimensionless concentration. In this post, $c$ shall always contain a dimension while square brackets shall not.)

Take $1~\mathrm{l}$ of this solution and add $1~\mathrm{l}$ of water. Our new solution has a concentration of $c = \frac{1~\mathrm{mol}}{2~\mathrm{l}} = 0.5~\mathrm{\frac{mol}{l}}$. Our pH is:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 0.5 = - (-0.301) = 0.301$$

The solution’s pH value obviously changed with concentration change.

Now, let’s do the same thing with a weak acid such as acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$). But first, I need to do a bit of maths. Remember that we cannot use the Henderson–Hasselbalch equation, because it assumes a buffered system. A simple acid is unbuffered. Instead, I will start at the definition of the acid constant:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{OAc-}]}{[\ce{HOAc}]}$$

And since chemistry dictates that $[\ce{H3O+}] = [\ce{A-}]$.

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]}$$

Because $c_0(\ce{HOAc}) = c(\ce{HOAc}) + c(\ce{H3O+})$, we can transform to:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}$$

Which finally gives us an equation with only one unknown. Solving this:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}\\
K_\mathrm{a} \left ( [\ce{HOAc}]_0 - [\ce{H3O+}] \right ) = [\ce{H3O+}]^2\\
[\ce{H3O+}]^2 + K_\mathrm{a} [\ce{H3O+}] - K_\mathrm{a} [\ce{HOAc}]_0 = 0$$

Since the determinant is positive, we can immediately write:

$$[\ce{H3O+}]_{1/2} = \frac{- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0}}{2}$$

Now plug the values $c_{0,1}(\ce{HOAc}) = 1~\mathrm{\frac{mol}{l}}$ and $c_{0,2}(\ce{HOAc}) = 0.5~\mathrm{\frac{mol}{l}}$ as above. You will need to add the value of the square root or you will get a negative concentration. This is what you should be getting (assuming OpenOffice isn’t buggy):

$$\begin{array}{ccc}
\hline
\text{expression} & c_0 = 1~\mathrm{\frac{mol}{l}} & c_0 = 0.5~\mathrm{\frac{mol}{l}}\\
\hline
K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0 & 6.95 \cdot 10^{-5} & 3.48 \cdot 10^{-5}\\
\sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.34 \cdot 10^{-3} & 5.90 \cdot 10^{-3}\\
- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.32 \cdot 10^{-3} & 5.88 \cdot 10^{-3}\\
[\ce{H3O+}] & 4.16 \cdot 10^{-3} & 2.94 \cdot 10^{-3}\\
-\lg [\ce{H3O+}] & 2.38 & 2.53\\
\hline
\end{array}$$

Obviously, once again, the pH went up when diluting the solution.

The Henderson–Hasselbalch equation, as I skimmed past earlier, is meant for buffer systems. The concentrations $[\ce{A-}]$ and $[\ce{HA}]$ are those concentrations you added to the buffer system. Since we did not add any acetate ions (other than acetic acid), the logarithmic argument would become zero causing an undefined error. We cannot use Henderson–Hasselbalch here.

However, if you have a solution containing $1~\mathrm{\frac{mol}{l}}$ acetic acid and $1~\mathrm{\frac{mol}{l}}$ sodium acetate, the logarithm’s argument will be $1$, therefore the logarithm will disappear and the pH value will be $4.76$ — exactly the $\mathrm{p}K_\mathrm{a}$ value of acetic acid. This buffer system is concentration independent. Diluted acids to whom the conjugate base has not been added as a salt are not!


What is the entropy of mixing a gas into the whole atmosphere? Suppose I release 1 mole of pure argon into the atmosphere. What is the entropy of mixing? The equations I have found seem to only apply to the case of a fixed volume. Should I use an estimate of the volume of the whole atmosphere? Some sort of infinite approximation?

Performing an actual calculation could be quite complicated, even if we use the ideal gas approximation, because the atmosphere has enormous variations in both temperature and density.

To simplify the problem, let's think of the atmosphere as an ideal gas of uniform temperature and density.   Then we need to determine how the entropy of the argon released into the atmosphere increases as a result of this release.

Since we're doing an ideal gas approximation, the argon doesn't "know" about the other atmospheric gases – the only thing that affects its entropy is the increase in volume.  But, when doing an entropy calculation, we do need to account for the fact that the atmosphere already contains 0.93% Ar.   Effectively, the presence of the existing argon removes a fraction of the configurational possibilities, since if we replace argon with argon, we don't change the entropy.   Hence the entropy increase is calculated from the effective configurational volume that remains for the argon relased from the box.

Specifically, for the purpose of calculating entropy, the effective final volume for the argon released from the box is:

$$V_\mathrm f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}\approx V_\text {atmosphere} - V_\text {Ar in atmosphere} = (1- x_\text {Ar})V_\text {atmosphere},$$
where $x_\text {Ar}$ is the mole fraction of argon in the atmosphere, which is 0.93% = 0.0093.

Proceeding, we start with the following general expression for a pure substance (again, the ideal gas approximation allows us to ignore the non-argon molecules):

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{\alpha}{\kappa},$$
where $\alpha$ is the coefficient of thermal expansion, and $\kappa$ is the isothermal compressibility.

For an ideal gas, $\alpha = \frac{1}{T}$ and $\kappa = \frac{1}{p}$, hence:

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{p}{T}$$

If we additionaly assume the expansion is isothermal (i.e., T of the box is the same as that of the atmosphere--which we've already assumed is of uniform T) we obtain, for an ideal gas:

$$\mathrm dS = \frac{p}{T}\,\mathrm dV = \frac{n R}{V}\,\mathrm dV \Rightarrow \Delta S = n R \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V} = nR \ln\frac{V_\mathrm f}{V_\mathrm i}=nR \ln\frac{0.9907 \,V_\text {atmosphere}}{V_\mathrm i}$$

where $V_\mathrm i$ is the initial volume available to the argon, and $V_\mathrm f$ is the final volume available to the argon.

As you can see, releasing a mole of argon into the atmosphere will result in a significant entropy increase, once the argon is fully dispersed.

Note that, in the case where the atmosphere is 100% Ar:
$$V_f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}=V_{\text{Ar in box}} \Rightarrow \frac{V_f}{V_i}=1\Rightarrow \Delta S = 0, \text{ as expected.}$$

We can also calculate the corresponding free energy change using:

$$\Delta G = \Delta H - T\,\Delta S$$

With the ideal gas approximation, there is no enthalpy of mixing, hence:

$$\Delta G = -T\,\Delta S = -nRT \ln\frac{V_\mathrm f}{V_\mathrm i}.$$

If we assume STP ($273.15\ \mathrm K$, $1\ \mathrm{bar}$), and take the volume of the atmosphere to be the area of the earth's surface times, say, 50 kilometers*, then we have:

$$\Delta S = (1 \ \mathrm{mol})\ R\ \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = 0.403\ \frac{\mathrm{kJ}}{\mathrm K}$$

$$\Delta G = - (1 \ \mathrm{mol})\ R\ (273.15\ \mathrm K) \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = -110\ \mathrm{kJ}$$

*This is all arbitrary, since we're unrealistically assuming constant pressure and temperature; but I needed to pick a number for the calculation, so I chose the top of the stratosphere ($50\ \mathrm{km}$).  $94\ \%$ of the atmosphere's gases are found below this altitude, so about $94\ \%$ of the argon would end up there as well.

Then again, because of the logarithm, the choice of atmospheric volume doesn't significantly affect the calculation, as you can see if you compare my answer with Buck Thorn's. Indeed, even if we allow the argon to expand to fill the observable universe ($4 \times 10^{83} L$), this only increases $\Delta S$ and $\Delta G$ by a factor of four! [$\frac{83}{22}\approx 4$.]

Calculation done in Mathematica (in Mathematica, ln is represented by "Log"):


How to make copper plating solution? I intend to make the following composition of copper plating solution:
\begin{array}{c:c}
\rm material   & \rm concentration \\\hline
\rm Copper sulfate & \pu{200 \frac{g}{L}}  \\
\rm sulfuric~acid  & \pu{100 \frac{g}{L}}  \\
\ce{Cl-}       & \pu{50 ppm}   \\
\end{array}

Do I use pure $\ce{H2SO4}$ ($95$–$98~\%$) and pure $\ce{HCl}$ ($37~\%$)?

Please check if I am correct with calculation.

first, copper sulfate: $$\pu{200 g} \times 5 = \pu{1000g}$$
then ($98~\%$) $\ce{H2SO4}$: $$\pu{100 \frac{g}{L}} \times \frac{\pu{5 L}}{\rm 1.84 g.ml^{-1}~ (\rho~of~H2SO4)} = \pu{271 \frac{ml}{L}}$$
then use ($37~\%$) $\ce{HCl}$: $$\pu{0.05 g/L} (= \pu{50ppm}) \times \pu{5L} \times \pu{1.19 g/mL}\ (\rho~ \text{of}\ \ce{HCl}) = \pu{0.21 ml/L}$$

so $$\pu{5000ml} - \pu{1000ml} - \pu{271ml} - \pu{0.21ml} = \pu{3728.79 ml}$$ — fill up this much water at final

I don't know what you are plating (nor do I care...) but here are some general observations.

(1) Plating is messy and exacting analytical measurements are overkill. Your calculation of 3728.79 ml water for instance is wild. I'm not sure what you intend to use for a container, but just diluting to 5 l +/- 100 ml would be more than enough precision.

(2) It is unclear if 200 g/l for copper sulfate is anhydrous copper sulfate or the pentahydrate. You're near the saturation limit for dissolving anhydrous copper sulfate at room temperatures.

(3) Your sulfuric acid calculation has the right number but wrong units. You need 271 ml per 5 liters.

SAFETY: You will require eye protection and appropriate clothing.  When diluting sulfuric acid always add the sulfuric acid to the water. The reaction is very exothermic. (If you try to add water to the acid it will boil violently and splatter sulfuric acid which is more than extremely dangerous.)

You should also not work alone, and you should have some sort of plan as to how you would handle an accident. What are you going to do if you get acid in your eye, or what happens if you drop the bottle of sulfuric acid? It turns out that concentrated sulfuric acid is very slippery...

(4) Your hydrochloric acid calculation is wrong. You have the right density, but HCl acid is only 37% HCl, the rest is water. Also HCl is only 97.3 % Cl, that pesky H is the rest. Your calculation is also again for 5 liters not 1 so the units that you show are wrong. I'm not sure about copper sulfate, but it wouldn't surprise me if there were some chloride in the sulfate.  That little hydrochloric acid isn't going to change the acidity of the solution. I'd be tempted to use copper (II) chloride as the source of chloride. It stores a lot easier, and would be easily measured. (Hydrochloric acid releases HCl fumes unless absolutely sealed - like in a glass ampule.)


What kind of compound absorption the 365nm UV light on TLC What kind of compound absorption the 365nm UV light?

compounds that has benzene ring or conjugated pi bond, make a fluorescence on 254 nm UV on TLC.
what kind of moiety(functional group) make fluorescence on 365nm UV on TLC

Your question is almost impossible to answer - it simply is too broad.

Excitation spectra of fluorescent organic compounds typically aren't sharp lines but consist of one (or more) bands. Even if the maximum of the excitation spectrum isn't located at $\lambda = \textrm{365 nm}$, monochromatic irradiation at that wavelength will excite a lot of different compounds (and even compound classes) which then deactivate by fluorescence.
Your light source does not emit monochromatic radiation! Hand lamps used to inspect TLCs typically have coated low-pressure mercury lamps. While the maximum of the radiation density might be centred around 360 nm, the full with at half maximum is - roughly estimated - about 30 nm or more. This means that there's even more compounds that might get excited by the emission of the hand lamp.

In summary, there's no exhaustive answer to your question, but just to name some compounds that are potential candidates:

Diphenylacetylene (1)
1,6-Diphenylhexatriene (DPH, 2)
4',6-diamidino-2-phenylindole (DAPI, 3)
6,8-difluoro-7-hydroxy-4-methylcoumarin (DIFMU, 4)


What is the difference between the reactivity of Lead subacetate and Lead II acetate undergoing a hydrothermal process? I am attempting to adapt a hydrothermal method for synthesizing $\ce{PbS}$ nanocrystals to my school that contains a mixture of cationic and anionic surfactants sodium dodecyl sulfate (.05 mmol, SDS) and cetyltrimethylammonium bromide (.05 mmol, CTAB) and ethylenediaminetetraacetic acid (.2 mmol, EDTA) as a chelating agent that diminishes the reactivity of lead ions in solution. The lead source within the method is Lead II Acetate and the sulfur source is an organosulfur compound with the formula $\ce{SC(NH2)2}$.

The company that we bought our chemicals from sent us Lead subacetate and while I wait for them to send us the right one back I am wondering if using the appropriate ratio of moles of Lead subacetate which has 3 moles of lead per 1 mole of lead subacetate to the documented lead II acetate which contains 1 mole of lead would lead to lead sulfide $\ce{PbS}$ nanocrystals.

The structure of subacetate contains two crystalline basic acetates containing hydroxide ions in addition to acetate ions. Does the subacetate create a conjugate acid that lowers the pH?

Compared to lead(II) acetate, lead subacetate contains additionally $\ce{2 Pb(OH)2}$. The hydroxide can either be neutralized by protons from EDTA or take part in the hydrolysis of thiourea:

$$\ce{SC(NH2)2 + OH- \rightleftharpoons OC(NH2)2 + HS-}$$

$$\ce{HS- + OH- \rightleftharpoons S^{2-} + H2O}$$

The pH of the solution should be higher unless all hydroxide is neutralized or consumed by the hydrolysis reaction.


How do the toxicities of arsenate and arsenite compare? So I was browsing the detoxification metabolistic mechanisms for E. coli, and came across an arsenic detoxification mechanism that converted arsenate to arsenite using a glutaredoxin. So my question is, how does the toxicity of arsenate compare to arsenite? And what is the use of this pathway in E. coli?

I see now.

According to the same website with the reaction mechanism,

  Although the arsenite formed is more toxic than arsenate , it can be extruded from some bacteria by EC 3.6.3.16 , arsenite-transporting ATPase; in other organisms, arsenite can be methylated by EC 2.1.1.137 , arsenite methyltransferase, in a pathway that produces non-toxic organoarsenical compounds. cf EC 1.20.4.4, arsenate reductase (thioredoxin) .


Weight of sodium acetate required to get to certain pH How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$

Like solving a puzzle, I found from the asked $\mathrm{pH}$ the concentration of hydrogen cation and hydroxide anion, and wrote to myself that I need to find the final concentration of $\ce{CH3COONa}$ to seek from it the amount and then the weight.

But my problem was in all the middle steps. I understand the the concentration of hydroxide anions that I found is after the equilibrium, but I miss the point of how to use them to find the concentration on the acetic acid by given the $K_\mathrm{a}$. I'm guessing that I need to get the $K_\mathrm{b}$ from the $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$ equation. But why exactly and how then, I need to find the concentration of $\ce{CH3COONa}$?

Maybe I lack the knowledge and thus the imagination for it.

Support highly appreciated!

It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:

$$\ce{CH3COO- + H2O <=>[$K_\mathrm{h}$] CH3COOH + OH-}$$

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\
&= \frac{K_\mathrm{w}}{K_\mathrm{a}}\label{eq:1}\tag{1}
\end{align}
$$

At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}\label{eq:2}\tag{2}$$

Knowing that

$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}}\label{eq:3}\tag{3}$$

one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as

$$
\require{cancel}
c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}\tag{4}
$$

Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:

$$m = cVM = \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} = \frac{\pu{2.50 L}\cdot\pu{82.03 g mol-1}\cdot 1.75\cdot 10^{-5} \cdot 10^{-14}}{\pu{10^{-2\cdot 9} mol-1 L}} = \pu{35.89 g}\tag{5}$$


pH of an acetate buffer after dilution 100 mL of a 0.1 mM buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?

The pKa value for acetic acid is given as 4.75.

The answer provided in my textbook is pH = 5.53. The closest that I can get to that is pH = 5.507. I know that's an almost insignificant difference but I'm just confused how they got to their answer, so if anyone can give me some suggestions that would be awesome. I'll outline my approach below:

Using the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] in the 100ml:
$$10^{5-4.75} = \frac{[A^-]}{[HA]}$$

$$[A^-] = \left(\frac{10^{0.25}}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 6.4{\times}10^{-5}M$$

$$[HA] = \left(\frac{1}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 3.6{\times}10^{-5}M$$

The addition of 900ml of water to 100ml of buffer is a 10x dilution:

$$[A^-] = 6.4{\times}10^{-5} \times 0.1 = 6.4{\times}10^{-6}M$$

$$[HA] = 3.6{\times}10^{-5} \times 0.1 = 3.6{\times}10^{-6}M$$

$$\ce{[H+]} = 10^{-5} \times 0.1 = 10^{-6}M$$

ICE table to determine equilibrium concentrations after dilution:

$$
\begin{array}{c|c@{hello}c@{}c@{}c@{}c@{}c@{}c}
   &   [CH_3 COOH] & + &[H_2 O] & {}\rightleftharpoons{} & [H_3 O^+] & + &[CH_3 COO^-] \\
\hline
I & 3.6 \times 10^{-6} && - && 10^{-6} && 6.4 \times 10^{-6} \\
C & -x && - && +x && +x \\
E & 3.6 \times 10^{-6} -x && - && 10^{-6} +x && 6.4 \times 10^{-6} +x \\
\hline
\end{array}
$$

$$K_a = \frac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$
Then substituting in Ka and the equilibrium values from the ICE table: $$(10^{-4.75}) = \frac{(10^{-6}+x)(6.4 \times 10^{-6} + x)}{(3.6 \times 10^{-6} -x)}$$

$$x^2 + 2.518 \times 10^{-5}x -5.762 \times 10^{-11} = 0$$

$$x = 2.111 \times 10^{-6}$$ $$ [H^+] = 10^{-6} + 2.111 \times 10^{-6} = 3.111 \times 10^{-6}$$

$$ pH = -log_{10}(3.111 \times 10^{-6}) = 5.507 $$

Depending on your first calculations before dilution:
\begin{align}
[\ce{CH3COO-}]&=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\
[\ce{Na+}]_\text{(After dilution)}&=\pu{6.4\times{10^{-6} M}}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-5} M}}~\text {(After dilution)}
\end{align}
So,you have the the following $5$ independent equations in order to specify the unknown concentrations of $5$ species as present in an aqueous solution consisting of $\pu{1\times{10^{-5} M}}$  buffer solution made from acetic acid and sodium acetate :

\begin{align}
6.4 \times 10^{−6} &= [\ce{Na+}] \tag{1}\\
K_\mathrm{a} &= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}
     = 1 \times 10^{−4.75} \tag{2}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{3}\\
1\times{10^{-5}}
    &=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\
[\ce{H+}]+[\ce{Na+}]
    &= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5}
\end{align}


Stability of Allenes and Alkynes, In Allenes, the $\ce{C=C=C}$ structure makes sure that the $p$-orbital of central carbon overlaps with 2 other carbons and the $p$-orbital of central carbon spreads out over the 3 carbons.

In Alkynes, the $\ce{C#C-C}$ structure makes sure that the energy released during formation of alkynes.
I also read that more the number of non-delocalized $\pi$-electrons on an atoms, the lesser its stability, though I do not know why.

When, it comes to comparing Allenes and their isomeric Alkynes,it has always been stated that  "Allenes are more stable than Alkynes".

One reason I found was that $\pi$-electrons in Allenes are less strongly bound than those in Alkynes.

Why are $\pi$-electrons less strongly bound?  If the reason in incorrect, what is the actual reason?
Why is it that as the number of non-delocalized $\pi$-electrons increases on a carbon, the smaller its stability?

The shape of the two bonds (see the first page of this .PDF) shows why pi bond electrons are less tightly bound. Sigma bonds are formed pretty much along the line connecting the two atomic nuclei, whereas the pi bonds are formed above and below that same line. Naturally, then, the distribution of the latter less concentrated (having more space to cover) than that of the former. Loosely stated, the shape of the pi bond distribution is more roundabout whereas the sigma bond directly touches the two nuclei which makes the electrons in it less strongly bound.
When it can, delocalization of electrons provides greater stability for the molecule because the electron density is distributed well over the delocalized regions, rather than, say, confining it in a small high-energy region surrounded by low-energy regions.

Think about resonant structures. Benzene (C6H6) is a good example. So the inverse of that is: less delocalization, less stability. Hope this helps.


Naming of 2-methylpropane I can't understand why 2-methylpropane has/needs the 2 in front of it?

It appears to me that if you were to add the methyl group at the first or last carbon on the chain then you get plain old butane and so defining 2-methylpropane is unnecessary when it could just be methylpropane.

Have I missed something or is this just a convention? Apologies if it's a really simple question; it's my first ever time studying organic chemistry.

You are correct. It is unnecessary to include the 2 in order to provide an unambiguous name. This compound is also often called isobutane which is an older name that is still in common usage.


Formation of thermodynamic product from 1,4-addition Question:

The temperature is high enough for formation of the thermodynamic product. Now, there are two possibilities:

Starting with the top double bond

2

Or, starting from the bottom double bond,

The answer given specifies that the second product is formed. Why is this, why isn't the first product formed?

Both of your products have a tetrasubstituted double bond.  This makes me think that they have similar stabilities, therefore there is no thermodynamically preferred product.

In both the top and bottom lines of your drawing you show the formation of allyl carbocations.  You use an arrow like this "$\ce{->}$" to connect the second structure from the left to the third structure from the left. The allyl carbocation is a resonance stabilized carbocation.  The second and third structures are really resonance structures and should be connected with a resonance arrow "$\ce{<->}$".  Both resonance structures taken together describe the allyl carbocation.

If you compare the allyl carbocation on the top line to the one on the bottom line, you'll notice that the one on the bottom has the positive charge delocalized onto 2 tertiary carbons while the allyl carbocation on the top line has the positive charge delocalized onto 1 tertiary and 1 secondary carbon. Therefore, the bottom allyl carbocation is more stable and the product derived from it will be preferred.


Are different enclosing mark types in chemical names needed to differentiate between substances? What is the significance of enclosing marks in chemical names in determining the uniqueness of a substance (organic and inorganic)?

Can a simple change in the type of enclosing mark used indicate that we are dealing with a different substance or does it just mean that the names were written slightly differently (or incorrectly)? If I decided that I was always going to use parentheses and never use brackets when naming chemicals would it stop me from being able to differentiate between two substances based on their chemical name or would it just be non-standard?

This article makes me think that switching from a parentheses to a bracket has a different meaning but that a brace could be used in place of brackets just because it makes the name look nicer.

To help illustrate what I mean, the following chemical names differ only by the use of curly braces instead of brackets. Are these two different substances or are they the same substance named slightly differently?

  Poly(oxy-1,2-ethanediyl),.alpha.-[4-[(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo]phenyl]-.ome
  Poly(oxy-1,2-ethanediyl),.alpha.-{4-{(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo}phenyl}-.ome

Update: The Naming and Indexing of Chemical Substances for Chemical Abstracts talks a lot about how punctuation is used to distinguish between different types of substances. It talks a lot about brackets and parentheses in these discussions but almost ignores curly braces. The publication backs up my theory that the differentiation between parentheses and brackets can be important but that curly braces are not. If I am correct then the two chemical names that I listed above are two different names for the same substance. If I can get verification on this or even opinions from chemists, that would be helpful as I am far from a chemist.

There is no real difference between the types of bracket - indeed, in online searching in the CA Registry file on STN all brackets are replaced by round () brackets. However, there are conventions with regards to square brackets in display in some names and these should be kept.


What is the major product of the reaction of 2-methylbuta-1,3-diene with HBr? What is the product of the reaction of 2-methylbuta-1,3-diene (isoprene) with HBr?

What I understand is that carbon (2) will get a positive charge after protonation.
But after this, which should be the major product?

3-bromo-3-methylbut-1-ene – because in this case the carbocation would be stable because of being tertiary, but the double bond would have less number of hyperconjugated structures and will be less stable

1-bromo-3-methylbut-2-ene –  as the carbocation would be less stable because it would be primary but the double bond would be much more stable because of 8 $\alpha$ hydrogen available

The regioselectivity of the overall reaction depends on the temperature.  The carbocation intermediate can pass through the two different transition states  that lead to the 1,2- and 1,4- products, respectively. If we choose a temperature low enough, then the product distribution will reflect the difference in energy between the two activation energies Ea (1,2) and Ea (1,4). So long as the reaction is not reversible, the product with the lower energy transition state will dominate which is decided by the stability of the intermediate, so in that case 3-bromo-3-methylbut-1-ene (product-1) will be more stable product.
At higher temperatures, the reaction has the potential to be reversible, this means that both the 1,2- and 1,4- products can ionize This sets up an equilibrium. The product ratio will now reflect the relative stabilities of the 1,2- and 1,4- products, not the transition states leading to their formation. In the case of 2-methylbuta-1,3-diene, the product 1-bromo-3-methylbut-2-ene (product-2) will be more stable and dominant at higher temperatures.


What exactly does the "per mole" unit of heat of atomization mean? I am facing a confusion in the definition of the standard enthalpy of atomization. The confusion arises because of the "per mole" in its units ($\pu{kJ mol^-1}$).

My question: When I say "per mole", do I mean "per mole" of the product or per mole of the reactant?

When I mean "per mole" of the product, my representative reaction for $\Delta H_\mathrm{a}^\circ$ is:

$$\ce{1/4 P4(s) -> P (g)}$$

And when I mean "per mole" of the reactant, my representative reaction for $\Delta H_\mathrm{a}^\circ$ is:

$$\ce{P4 (s) -> 4P (g)}$$

Using either of them leads to different calculations involving Hess's law or other chemical reactions, hence I feel there is a need to resolve this confusion.

To add to this confusion, NCERT Chemistry (Part 1 page 171) writes:

  It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

but then gives a representative reaction as:

$$\ce{CH4 -> C(g) + 4H(g)}$$

which involves breaking 4 moles of bonds instead.

If someone could write the correct definition of the standard enthalpy of atomization, with more than one reputed reference to back their claim, it'd definitely resolve this confusion.

According to Wikipedia,

  Standard enthalpy of atomization is the enthalpy change when $1$ mole of gaseous atoms is formed from its element in its defined physical state under standard conditions $(\pu{298.15K}, \pu{1 bar})$.

$\Delta H_{\text{at}}^{\circ}$ focuses on the production of $1$ mole of atoms from the element. Hence it's defined for the reaction:

$\ce{\dfrac{1}{8}S8 ->S}$

but not for the reaction:

$\ce{S8 -> 8S}$

So whenever you see the term "per mole", it is, by definition, implicitly referring to the product and not to the reactant.


Understanding what is happening when something is dissovled Let's start with a saturated solution of NaCl in water. At this point, any further addition of NaCl at the same T and P would not dissolve. However, if I then add another fully soluble salt (say AgNO3) it will also dissolve until it reaches equilibrium. If I keep adding different soluble salts (assume I don't encounter things like the common ion effect), would I ever overwhelm the solvent such that it could not dissolve anything else?

My intuition is that eventually, there would be no more solvent moleclues available for solvation and thus nothing would dissolve. Is this even valid to think about?

Your question seems to be exclusively about salts dissolving to form ions in water. The answer is fairly trivial: a saturated solution of a salt in water is not water.

At (thermodynamic) equilibrium, the solubility depends on the net free energy of the system. When the interactions in solution no longer energetically justify charge separation of the ionic solid into aqueous ions, then you've got equilibrium, right?

So, consider a solution as being (in a very crude and simplified way) Solvent-M1 interactions, Solvent-X1 interactions, solvent-solvent interactions, and M1(aq)-X1(aq) interactions as well as M1(aq)-M1(aq) and X1(aq)-X1(aq) interactions. Which of those will change when you add M2X2 ions to it? Well, at low [M2X2] none of those will change significantly, but you'll add the interactions between M2 and X2 and ALL of the species present. (as already explained, silver nitrate being added to NaCl is a very poor choice, AgCl2 is quite insoluble)

At high enough concentrations you'll begin to significantly change the energetics. In other words, adding M2X2 to a solution of M1X1 changes the interactions of all species present, so you would expect that the mixed salt solution will have different equilibrium concentrations for both M1X1 and M2X2.

If you were to carefully choose highly soluble salts, I think what would be surprising is how well their solubility was predicted by just using each independent Ksp, rather than the (expected) deviations from those predictions. Note that nothing I've said here should be taken to mean that adding a different salt to a saturated solution will decrease the solute concentration of the first salt. (Although it is the most likely scenario, just as the common ion effect would suggest. In some cases solubility will increase!) The reality is that highly concentrated solutions are far more idiosyncratic and generalizations (especially "laws") are much harder to come by.

While there probably is justification to claim that water is dissolved in sodium nitrate (at equilibrium), I've never found such foolishness useful. And I've been around for a long while. (ultra) Pure water does not freeze at 0°C, negative absolute temperatures (negative Kelvin) are attainable, and a bunch of other stuff which while true, doesn't really aid understanding (for the novice) probably should be ignored until one wants to deep dive into a subject.


Separating sugar from soda I want to do a lab with my students in which they will try to prove how much sugar is in a can of coke. I know that distillation is the go-to method for separating out the syrupy sugar mix, but does anyone know of a method that goes further to separate out the pure granulated sugar and not just a syrup?

The simplest approach would be to degas a sample and measure its specific gravity with a vintner's hydrometer. It will have scales of specific gravity and Brix (sugar content) on it.The Brix (Bx) reading is the percent of sucrose, w/w, in a solution with the same density/specific gravity as the one you are testing. There are, of course, other things in coke besides sucrose (fructose, coca extract, phosphoric acid) but they should be in small enough concentration that the sugars will swamp everything else and all sugars behave amazingly the same with respect to density of their solutions).

Based on the reported sugar content of Coke it contains 108 grams of sugar per liter. Thus its specific gravity should be about 1.041 and the corresponding Bx reading about 10.4. The density of water at 20 °C is 0.998203 so the density corresponding to 1.041 SG is 1.039 and 10.4% of 1039 grams (weight of a liter of 10.4 Bx sucrose solution) is 108 grams.

If you don't have a hydrometer with the Bx (or Plato) scale or if you determine density by weighing a known volume (convert to specific gravity) you will need to be able to determine the sucrose content of a solution of given specific gravity. There is lots of opportunity here to go into the history of wine and beer making, which you male students will love (though I understand that in modern schools discussion of things related to alcohol may not be politically correct) but in any case determining the w/w sucrose content of solutions of given SG is a well studied problem. There are various tables but the easiest way to get w/w sucrose from specific gravity is a polynomial promulgated by the ASBC (American Society of Brewing Chemists). It is

°P = ((135.997*S -630.272)*S + 1111.14)*S -616.868

I believe that separating out the pure sugars at anything close to 100% recovery would be difficult (barring HPLC or something of that sort). Sugars seem to be assayed by indirect methods such as the gravimetric ones I proposed or measurement of refractive index.


Representing conformers using SMILES and Graph representations I was wondering if either of SMILES or graph representations (in their vanilla form) can distinguish between two conformers of the same molecule.

Can this be enhanced by adding additional metadata on Graph nodes?
what info are necessary then?
Can this be enhanced for SMILES representations?

For the SMILES strings as defined by Daylight, you can expect these representations tell about the molecular constitution plus information about ($E$/$Z$)- and CIP's $(R)$/$(S)$-configuration. But nothing about the conformation like the boat / twist form of cyclohexane which generally is dynamic, or a faithful preservation of information like s-cis to describe the conformation of butadiene favorable for the Diels-Alder cycloaddition reaction.
Further, I speculate SMILES strings are not a faithful representation of locked conformations as in the axial $P/M$ configuration of 2,2',6,6'-substituted 1,1'-biphenyls. However, perhaps someone else may clarify this point.


Can rate constant for forward reaction be equal to rate constant for backward reaction? For a particular reaction, can we have
$K_c = 1$?

My confusion arised when I was studying the effect of temperature on $K_c$ . I see that the temperature must affect forward and backward rate constants, differently for a reaction so that it changes $K_c$ (which is ratio of the two rate constants). So I thought for a reaction with $K_c$ equal to one, both the rates must be equal, so both will get equally affected by temperature, and thus no change in $K_c$ , or does it have something to do with the Arrhenius equation?

I know that there is a lot of lacking in my intuition as I have just began studying Le Chatelier's principle and all the stuff. So a detailed explanation would be much welcome.

When forward and backward reaction have the same rate, $K_c$ is equal to $1$ in the case of a reaction where the total number of reactants is equal to the total number of products in the chemical equation describing the change. If the total of these stoichiometric coefficients is not the same before and after the horizontal arrow, $K_c$ may be equal to $1$, but must have a dimension, like mol/L or L/mol, or any combination of these units.

As the forward and backward rections have different activation energies, their rates change differently with increasing the temperature. So the equilibrium constant $K_c$ can be equal $1$ at only one temperature.


Free database of polymers with Hansen parameters I am developing a product that needs a component to expand when in contact with crude oil.

Found the wikipedia article about Hanses solubility parameters and thought it would be a good idea to run the parameters of crude oil against a list of polymers to find maybe the five closest polymers.

Where can I find a database of solvents and polymers online?

It should be noted that I know very little about chemistry, but enough about computer programming to be able to run the test against any database format.

I'm afraid that due to copyright claims the only database available is HSPiP, plus there is the most recent handbook Hansen, C. M. Hansen solubility parameters: a user’s handbook, 2nd ed.; CRC Press: Boca Raton, 2007. The book costs ten times less than a database and contains tables for solvents on pp. 345 -- 485, and for polymers on pp. 487 -- 500.


Difference between Lindlar and Rosenmund catalysts Is there any difference between the Lindlar and Rosenmund catalysts? I've checked around, and it seems the same compounds are used to make both. Is there a difference in their reactivities or are they essentially the same?

This is a rather interesting question because these names actually refer to classes of reactions (specific to certain reagents and products), and aren't constrained by specific proportions of substances or even the identity of these substances.

$\hspace{4cm}$

A Rosenmund catalyst is used to reduce acyl chlorides to their corresponding aldehydes, and is typically composed of palladium supported on $\ce{BaSO4}$, and poisoned by sodium acetate, N,N-dimethylaniline, thiourea, thiophene, dibenzothiophene, ethyldiisopropyl amine, or, most commonly,
quinoline.

A Lindlar catalyst is used to reduce alkynes to cis alkenes, and is typically composed of palladium supported on $\ce{BaCO3}$, $\ce{CaCO3}$, or carbon, and poisoned by lead acetate and quinoline.

The broad list of poisons which cause nearly identical catalytic activity of the supported palladium in both Lindlar and Rosenmund catalysts tells that the poisons are not involved in the catalytic process itself, but rather are affecting the surface of the catalyst. Interestingly, Lindlar-active palladium shows no activity for the conversion of benzoyl chloride to benzaldehyde under various conditions. This demonstrates that despite similar catalyst preparation procedures, the two reactions are catalyzed by different active sites.$^{[1]}$

$[1]$ Mcewen, A. B.; Guttieri, M. J.; Maier, W. F.; Laine, R. M.; Shvo, Y. The Journal of Organic Chemistry 1983, 48 (23), 4436–4438.


Bisulfate vs Hydrogen sulfate What's the difference between bisulfate and "something" hydrogen sulfate?

Has there historically been a preference for one over the other?

Nope, no rule, and again there is no difference, like your question about the names for iron oxidation states.  The "bi" form is the older form, and the "hydrogen" form is now preferred, again because it more explicitly describes the form of the compound.

EDIT: Some additional detail was requested.  When I say "preferred", I mean the name sanctioned by the IUPAC (International Union of Pure & Applied Chemistry), also known as the systematic name, so called because the rules to form the name are "systematic"; i.e. consistent and clear. Older names are called common names, which were in used before the IUPAC.  So "bisulfate" is a common anion name, while "hydrogen sulfate" is systematic.


Why is the entropy of solidifying solution higher than that of a pure solvent? According to a source,

  Solutions have a greater entropy than pure liquids as the particles in a solution are more separated and solvent molecules separate the solute particles. If the number of moles of gas increases in a chemical reaction (as shown by the stoichiometry of the reaction) then the entropy also increases.

My questions are as follows:

1) How do we know the particles in a solution are more separated ?

2) How do the solvent molecules actually separate the solute particles?

3) Also, we know that the entropy of solid is less than that of a liquid, so shouldn't the addition of solute particle (which are obviously solid) actually lead to lesser entropy than pure liquid?

1) How do we know the particles in a solution are more separated?

When a solution forms, intermolecular bonds in the solid solute are broken and replaced by interactions with the solvent. The solute is then free to diffuse in the solvent. If the solute molecules still interacted in a highly favorable way they would tend to bond and precipitate out. This is exactly what happens when a solute has a low solubility.

  2) How do the solvent molecules actually separate the solute particles?

What drives the dissolution process is an overall favorable combination of the change in the energy of intermolecular interactions and motional freedom. This is described by the Gibbs free energy. The solvent provides an environment for the solute to move (diffuse), which increases the solute's entropy, and the solvent forms weak interactions with the solute that make up (perhaps only partially) for the lost interactions when in the solid. In general, there is an enthalpic component (bond-breaking and making) and an entropic component (changes in solute configurational freedom) to the free energy. If the energy of the interactions of a solute with itself in the solid is identical to the interaction energy between solute and solvent in the solution, then the change in free energy is described entirely by the entropic contribution.

  3) Also, we know that the entropy of solid is less than that of a liquid, so shouldn't the addition of solute particle (which are obviously solid) actually lead to lesser entropy than pure liquid?

The solid particle is no longer "solid" when dissolved, it is in a new state. The entropy in solution reflects the new state of the solute upon dissolution. The solute does not retain a memory of what it was before. At low concentrations, the solute is diffusing with few other solute molecules nearby. The new entropy is not an average over the entropy of the starting states (solid and solvent).

(That is unless you are referring to a dispersion of solid particles, but that is called a colloidal solution and is another matter.)


Explain why these ions are more strongly hydrated in an aqueous solution? I've already answered this question. All I want to know is if my reasoning is correct.
Case 1:

$\ce{Rb+}$
$\ce{Na+} \leftarrow$ This ion is more strongly favored

Case 2:

$\ce{Mg^{2+}} \leftarrow$ This ion is more strongly favored
$\ce{Na+}$

Reasoning:
For the first case, my guess would have to be because the electrons are much closer to the center of the nucleus (the atomic radius of the atom is very small) and makes it much easier to interact the positive dipole end of the water molecule (the hydrogen molecules).
For the second case, the reason why the magnesium ion is more favored is because even though both ions have roughly the same molar mass, one is more positively charged than the other and thus is more likely to attract the negative dipole end of the water molecule (the oxygen molecule).
Is my reasoning justified or is there something in particular that I'm not paying attention to?

I believe your reasoning is correct (more or less). Solvation is influenced by charge to size ratios. Higher the charge to size ratio, the easier it is for the species to be solvated by water molecules.

Apply this idea to both your cases. (note: in the second case, I don't get your argument about molar mass. The idea is much simpler, magnesium's atomic radius is smaller than sodium's and it has a higher positive charge. What can we conclude from this?

Also, reconsider the first case: a positive ion would not be attracted to the positive end of a dipole.


Order of Lewis acid character among sulfur hexafluoride, sulfur tetrafluoride and tetrachloromethane Problem

Compare the order of Lewis acid character of $\ce{SF6},$ $\ce{SF4}$ and $\ce{CCl4}.$

My attempt
Oxidation state of $\ce{S}$ in $\ce{SF6}$ is $+6,$ so it is very electrophilic.
$\ce{S}$ has a lone pair on it in $\ce{SF4},$ but can expand its octet if attacked by a donor.
$\ce{C}$ cannot accept lone pair from donor.
So, the order should be $\ce{SF6} > \ce{SF4} > \ce{CCl4},$ but that is a wrong answer.

S is in its highest oxidation state (+6) it has no electrophilicity hence not a Lewis acid.
Also, CCl4 can't have more than 8 electrons in its outermost orbit and it can not also increase it's coordination number also so it can't behave as a Lewis acid.
SF4 can accept electrons as S isn't in its highest oxidation state.
Hence order of Lewis acid character is : SF4 > CCl4 = SF6


An example to illustrate the difference between an inversion operation and a twofold rotation Reading in Atkin's Inorganic Chemistry book, I cam accross:

  Although an inversion and a twofold rotation may sometimes achieve the
  same effect, that is not the case in general and the two operations
  must be distinguished.

He does not give an example of an occasion where such cases are not the same.

I will appreciate if someone can give an example of when inversion and twofold operation won't give the same result.

If the body is chiral, a ${C_2}$-axis maintains chirality.  An inversion (${S_2}$, ${i}$, the product of ${C_2}$ and ${σ}$h) inverts chirality.

https://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/CHEM_3111_6170/Symmetry_and_group_theory.pdf
Frame 5
http://alpha.chem.umb.edu/chemistry/ch612/documents/MolecularGeometryandPointGroups_002.pdf
More technical
http://www.b-u.ac.in/sde_book/msc_phychem.pdf
The math


Organic reaction of dry ice and Grignard reagent Which of the following product is formed in the reaction $\ce{CH3MgBr}$ in DryIce/acid?
  A) $\ce{CH3COOH}$
  B) $\ce{CH4}$
  C) $\ce{CH3OH}$
  D) $\ce{CH3CH2CHO}$

My answer is A, since dry ice is $\ce{CO2}$ and it has one carbon atom and reactant has one carbon, so in all the product should have two carbon atoms, which is option A.

Am I correct?

You are correct. Grignard reagents $\ce{RMgX}$ (where $\ce{R}$ stands for some hydrocarbon group and $\ce{X}$ is a halogen, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) react with carbon dioxide to produce carboxylic acids after acidic workup:

$$\ce{RMgX ->[1)\ \ce{ CO2}][2)\ \ce{ H3O+}] RCO2H}$$

$$\ce{RMgX + CO2 -> RCO2- MgX+}$$
$$\ce{RCO2- + H3O+ -> RCO2H + H2O}$$

  In general your reasoning when approaching this problem is a good one. If you are reacting two 1-carbon reagents, then you probably have a 2-carbon product.


Diluting methanol in water: is it exothermic? Can someone explain in simple terms why does temperature increase when mixing water with methanol? I do not have a strong background in chemistry, but I usually have to mix both liquids (sometimes also water + ethanol) (as part of my work in a biology lab) and have always wandered about the reason behind the evident increase of temperature upon mixing. I guess that the break/formation of H-bonds is involved but I haven't found a clear ("easy-to-understand") explanation in online searches.

When the methanol and water are separate, they both exhibit hydrogen bonding with themselves. When the water and methanol are mixed together, some of the existing hydrogen bonding (water-water or methanol-methanol) is disturbed and now there is hydrogen bonding between water and methanol. This new hydrogen bonding pair (water-methanol) releases energy when it forms as it is more favorable than maintaining only water-water and methanol-methanol hydrogen bonding; the temperature increase you observe is the result of the released energy warming the solution.


Methods for dissolving Scotch/Kaptop/etc. Tape for Electron Microscopy? The famous method for the Nobel prize in studying Graphene is the so-called "scotch tape" method. Here, one takes a weakly-bonded van der Waals-type material and peel a layer off. From that layer, another piece of tape is taped onto the original piece of tape with the layer to split it into two pieces. This is repeated until nanometer thick layers are produced.

At this point, you have a single monolayer on a piece of tape, and need to transfer it onto a grid to look at it with a TEM (transmission electron microscope). However, the monolayer is much too thin to peel off the tape, so the tape must be completely dissolved.

The question I have is: what are the best options for dissolving tape?

Obviously there are many, many variables at play here. But here are some relevant constraints:

Completely dissolves tape, leaving little sticky residue
Solvent is relatively non-reactive to inorganics, metals
Safe and inexpensive

For example, I found a question on Research Gate that suggests using an alkaline solution to completely dissolve Kapton. I wonder if there are other methods for scotch tape

Since this method for TEM preparation is so common there must be some great literature resources on this topic, so resource recommendations would be great!

I don't fully understand the mechanics of "stickyness" so a description from the atomic perspective would be useful, especially in understanding what good solvents should do.

Generally you use Acetone. The following is taken from
Enoki, Toshiaki, and Tsuneya Andō. 2018. Physics and Chemistry of Graphene : Graphene to Nanographene p. 94

The Scotch tape method is very common in the fabrication of
graphene devices, which are mainly used in research into the
fundamental properties of graphene. It can produce high-quality
graphene with lateral sizes ≤ 0.1 mm; however, controlling the size
and position of the graphene is almost impossible. The following is
the typical procedure of the Scotch tape method:

(i) Place a few flakes of graphite (≈ 1 mm) on the adhesive side
of a plastic sticky tape with tweezers (Fig. 3.1(a)). Scotch tape
(3M) and Nitto tape are commonly used. Natural graphite,
Kish graphite, or HOPG (highly oriented pyrolytic graphite) is
usually used for the starting material.

(ii) Fold the adhesive side to sandwich the graphite flakes and
press the tape firmly (Fig. 3.1(b)).

(iii) Peel the tape apart slowly, so that the graphite flakes are
cleaved and attached on both sides of the tape (Fig. 3.1(c)).

(iv) Repeat the second and the third steps with slightly shifting
the fold line, so that the graphite flakes do not overlap (Fig.
3.1(d)).

(v) Stop repeating the cleavage when graphite flakes spread over
the sticky tape (Fig. 3.1(e)).

(vi) Prepare a silicon substrate with a silicon dioxide layer on
the top surface (Fig. 3.1(f)). Place address markers on the
surface in advance using photolithography so that one can
easily locate the position of a graphene flake in an image.
These markers are indispensable for further microfabrication
on graphene. An example of the address markers is shown in
Fig. 3.2

(vii) Attach the adhesive side of the tape with the graphite flakes to
a silicon substrate, and gently rub the surface to remove any
air between the substrate and the tape (Fig. 3.1(g)).

(viii) Slowly peel the tape off the substrate (for example, more than
2 min for a 1 cm substrate) (Fig. 3.1(h)). Not only graphite
flakes but also some adhesive remains on the substrate. The
latter can be removed by submerging the substrate in acetone
for a few seconds.

It is also reported that in the scotch tape method, not only graphene flakes but also a large amount of adhesive is attached to the substrate. You get rid of it immersing the substrate in Acetone. Other methods are reported in the reference too.

About adhesion
There are lots of aspect about adhesion. About scotch tape the most interesting answer you can find here "https://www.scientificamerican.com/article/what-exactly-is-the-physi/" is the following

"The simplest answer that I can give to the question is that pressure-sensitive adhesives (which are polymers) are 'tacky' or 'sticky' because they are essentially very high viscosity liquids that also have some elastic characteristics--in technical terms, they are 'viscoelastic.' This property means that they exhibit some of the characteristics of liquids, and so they will 'wet' a surface to which they are pressed. But then, because of their elasticity, they will resist separation when stressed. Thus, 'stickiness' is strictly a physical (viscoelastic) phenomenon, not a chemical one."


What is the difference between reversible and irreversible work done in adiabatic process? The work done in an adiabatic reversible process is given by:
$$W = R\frac{\Delta T}{1-\gamma}$$

The work done in an irreversible process is given by:

$$W = P_{ext}\Delta V$$

By the 1st law of thermodynamics, the change in internal energy must be equal to the work done. The change in internal energy is given by:

$$\Delta U = nC_v\Delta T$$

where $C_v$ is the molar heat capacity at constant volume.

Can't this equation be applied to both the processes? Why are there different formulae for the two types of adiabatic processes?

Your problem lies in your analysis of the irreversible expansion.

In the case of both a reversible expansion and an irreversible expansion, the gas pressure at the interface with the piston matches the externally applied pressure during the deformation. However, in the irreversible expansion, the gas within the cylinder is not uniform with respect to pressure or with respect to temperature.

So the interface pressure (which determines the amount of work done) does not match the average pressure of the gas within the cylinder. In addition, the pressure at the interface depends not only on the gas volume but also on the rate at which the gas volume is changing. So for an irreversible process, we have very little control over what is happening inside the cylinder.

The only thing we can control is the pressure at the interface $p_{ext}$ where the work is being done. We do this by whatever means necessary, including feedback control systems. In the typical scenario, we drop the pressure at the interface to a constant value, less than the initial pressure in the cylinder, and then hold it at that value until the gas volume has increased to the point where the gas in the cylinder is again at equilibrium. At this point, the gas pressure throughout the cylinder is described by the ideal gas law again.

If we follow this game plan for the irreversible process, even if the final pressure matches the value be obtain in the reversible process, the final temperatures and the final volumes will not match.

If you would like me to provide a detailed analysis of the irreversible case, illustrating in detail what transpires, I will be glad to do so. But right now, I'll give you a chance to ask some questions.

Irreversible and Reversible Expansions

Initial conditions:  $T_i, P_i, V_i$

Irreversible expansion:

$$\Delta U=-\int_{V_i}^{V_f}{P_{ext}dV}$$
In the irreversible expansion we are considering, $P_{ext}$ is controlled to be constant at the final pressure $P_f$ with $P_f<P_i$. The expansion is allowed to continue until the gas re-equilibrates at the final pressure and volume. Therefore, we have:
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)\tag{1}$$

As noted in my discussion above, the ideal gas law cannot be applied to the intermediate states during the irreversible expansion, but it can be applied to the two equilibrium end states. Therefore, $$V_f=\frac{nRT_f}{P_f} \tag{2}$$
and
$$V_i=\frac{nRT_i}{P_i}\tag{3}$$If we combine equation $(1)$, $(2)$ and $(3)$, we obtain:$$C_v(T_f-T_i)=-R\left(T_f-\frac{P_f}{P_i}T_i\right)\tag{4}$$

We can solve this equation for $T_f/T_i$ and a function of $P_f/P_i$ to obtain:
$$\frac{T_f}{T_i}=\frac{1+(\gamma-1)(P_f/P_i)}{\gamma}\tag{5}$$
Note that this equation differs from the relationship one obtains if the expansion is carried out reversibly.

Reversible expansion:
For a reversible expansion, we have $$dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$$ or $$C_vd\ln T=-Rd\ln V=-R(d\ln T-d\ln P)$$This integrates to:
$$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{(\gamma-1)}{\gamma}}\tag{6}$$
Note from equations 5 and 6, that if the temperature ratios are the same in both cases (equal work), the pressure ratios are not, and if the pressure ratios are the same in both cases, then the temperature ratios are not (unequal work).


Copper substituting silver in Tollen's reagent I am an artist attempting to use Tollen's Reagent in a number of my projects.

Is there any reason I could not substitute Copper Nitrate for the Silver Nitrate in the reaction? Would the copper adhere to the glass?

For the particular reaction where you get a silver mirror with an aldehyde, I doubt it'll work for copper, due to the reduction potentials. I'll take acetaldehyde $\ce{<->}$ acetate as an example here, but I believe that the reduction potentials of most aldehydes will be similar. The nitrate is a spectator ion here, so we need not consider its presence.

$$E^o_{\ce{Ag^+,Ag}}=+0.80$$
$$E^o_{\ce{Cu^{2+},Cu}}=+0.34$$

$$E^o_{\ce{CH3COO^{-},CH3CHO}}=+0.580$$

Our reaction involves the metal ($\ce{Ag}$ or $\ce{Cu}$ being reduced, so it must have a higher reduction potential than the acetate/acetaldehyde pair. Of the two metals, only silver has a higher SRP, thus the reaction involving silver nitrate is spontaneous--whereas the copper nitrate reaction is  infeasible.

Of course, you could find some substrate other than an aldehyde which reacts with copper nitrate to achieve the effect you want--though I'll have to look up some stuff to figure this out. I think the Tollen's reaction is a single step everyday redox reaction, which means that we just need to look for a stronger reducing agent which doesn't harm the glass.

Actually, it looks like the standard process for plating nonconductors with metals is to give them a conducting layer by dipping them in something, and then using electroplating. Not sure about this, though.


Effect of pressure on solubility of salts The volume of certain saturated solution is greater than the sum of the volumes of the water and salt from which it is made. On increasing pressure solubility of this salt
(a) Increases
(b) Decreases
(c) Remains unaffected
(d) Can't be predicted

The solution says

Density of product < Density of reactant [mass of reactant = mass of product]
So on increasing pressure equilibrium shift toward higher density solid or liquid.

According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?
Is my interpretation of the Principle wrong or is the solution of the answer wrong?

The answer is "(b) Decreases."  The reason is that the saturated solution occupies more volume than the starting reagents.  At higher pressures, creating this extra volume requires more $P dV$ work be done by the dissolving reagents.

  According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?

I find it easier to think in terms of volumes, not densities.  High pressure should favor low volumes.  Imagine a balloon filled with saturated solution.  If we squeeze the balloon by raising the pressure on the outside of it, the balloon will "want to" shrink.  In the example given, the balloon can shrink only by saturated solution converting back into solid salt and liquid water, so that is what would be thermodynamically favored.

Update: Based on discussion in the comments, I wanted to add a point about compressibility and thermodynamic stability.  The degree to which a system opposes a change in pressure is called the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic systems.


What is the difference between a 500W high pressure mercury lamp and a 400W high pressure mercury lamp I know that the Wattage of a UV lamp is only one factor in the UV output of a light, and therefore cannot be used as a measure of how powerful a light is. Also it doesnt have any effect on the produced wavelength. Is it like this that a lamp with higher wattage can cause higher temperature on substrates that are being irradiated ?

I would say all other things being equal, i. e. Spectral distribution,  quantum efficiency,  that more wattage = more photons per second. And yes, more heat.


How can I estimate reaction rates in water with computational chemistry? I want to estimate rates of substitution reaction of esters with nucleophiles in water.
How can I estimate them with a computational chemistry software (preferably GAMESS(US))?
Or is it practically impossible?
I know that this is possible in gas phase, but I don't know how to do it in solution.

Finding solvent  effect in computational chemistry is still a crucial part and subject to deep research effort. Now there are two types of models that are being used in computational chemistry.
One is explicit solvation model: Here you will use explicit solvent molecule with your reactant or products and then using first principles you will find the equilibrium configuration but the problem is, it is very expensive and the configuration space is so much high that it is practically impossible. For that you have to use molecular dynamics or monte carlo techniques. In our group we use 1000 to 5000 explicit water molecule but we can't do it for large reaction.
Another method is implicit solvation model: Here electrostatic potential is applied to account the solvent effect but this procedure is so expensive that now that potential is replaced by the solvent induced reaction field energy.
In our group we are using a hybrid approach that involves communication of several software at different level of theory. In my quantum chemistry class i used spartan and qchem and they have some implicit solvation but i didn't use GAMESS.
From GAMESS manual i found that they have solvation module. They are using QM/MM approach. That's actually a very good approach. Please see page 9 of this file:  http://www.sdsc.edu/~kimb/gamesstut.pdf
In this following file at page 3 & 4 you will also find the list of keywords that are needed to start a solvation model run
http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf


Why is charcoal useful as a fuel compared to wood? Don't we lose energy by making it? Charcoal is produced by partially burning wood. However, as the wood is burned, some of its energy is lost. Why is it that charcoal is still preferred as a fuel instead of wood in many cases?

Doesn't charcoal contain less energy than wood? Is the main reason that wood contains a lot of water which would form unwanted vapor?

As I understand it, charcoal --mostly carbon, plus some amount of inorganic ash-- burns cleaner - and hotter - than wood, and has more energy content per unit of weight than wood. The process of making charcoal involves heating the wood in the absence of air or with limited contact with air - driving off water and low molecular weight oils and odor components that produce a lot of the smoke from burning wood, and converting cellulose (a major component of wood) into carbon plus water.


Difference between isopropyl and isobutyl Knowing the meaning of ''iso-'' prefix, i could not understand why it is used in some cases. For instance

n-propyl and isopropyl

are named according to the carbon atom which has a lack of hydrogen. The same goes for

isobutyl

where branches define the common name as well. In the end my question is: What is the logic of using iso- prefix (especially in those cases), and how can i know when to use it?

p.s- if you know the answer to my question, it is mostly appreciated to state your source of answer!

The IUPAC Organic Chemistry Nomenclature does not explicitly define the "iso-" prefix meaning, as it is not used in the systematic naming. However, it is mentioned on several places when listing acceptable semi-systematic names. In the section A-2 (1979 Blue book, some also in 1993 Recommendations, section R-9.1, table 19a) those are (not mentioning neo- compounds)

isobutane (CH3)2CH−CH3
isopentane (CH3)2CH−CH2−CH3
isohexane (CH3)2CH−CH2−CH2−CH3

And in the section A-2.25 (some also in R-9.1, table 19b), where it lists acceptable "isoalkyl group" names:

isopropyl (CH3)2CH−
isobutyl (CH3)2CH−CH2−
isopentyl (CH3)2CH−CH2−CH2−
isohexyl (CH3)2CH−CH2−CH2−CH2−

Some related isomeric alkyls mentioned in this section are

sec-butyl CH3−CH2−CH(CH3)−
tert-butyl CH3−CH(CH3)2−
tert-pentyl CH3−CH2−CH(CH3)2−

There are also some other, unsaturated "isoalkyl" group names mentioned:

isopropenyl CH2=C(CH3)−
isopropylidene (CH3)2C=

From the above we can induce that "iso-" prefix, when related to alkanes or alkyl groups, it means the simplest (n-)alkane/alkyl isomer branched at the end of the carbon chain.

Isopropyl happens to be secondary (sec) alkyl (there's lack of carbon atoms to create primary alkyl isomeric to the n-propyl), all higher isoalkyls are primary (no branching at the alkyl "attachment").


Do you say "reduce something to polystyrene"? this question may not fit in this forum because it's about English, but biochemistry knowledge is required, so please allow me to make this thread.

When you measure the molecular weight of a polymer, you may use polystyrene standards and say "polystyrene equivalent Mw of --- is..." or "the molecular weight relative to polystyrene (standards) is..."
I have seen these usages in various articles and documents. The other day, however, I ran into an expression, "the Mw was reduced to polystyrene." I understand that it means the same thing, but my question is, do native speakers really say like this?
I googled "reduced to polystyrene" and got many hits, but it seemed many of them were written by Japanese. And when I did the same thing on Google Books, there was only one hit and the "reduced" meant "converted/changed."
So my question is, do you say "reduce to polystyrene" in this particular situation, meaning you use polystyrene standards?

Thank you very much!

Just write what you actually did: You used the calibration curve of your SEC column for polystyrene in solvent xx. Those phrases you cite are meant to sound impressive, but are actually just scientifically wrong, bad English, and therefore stupid.

Nothing gets reduced here like a reduced temperature or a chemical reduction. Also there is no "PS equivalent Mw", because it depends strongly on the solvent. "Relative" is as wrong, because that implies that there is simply a factor between the two scales, while in fact there are two calibration curves, one of which is unknown.


How do you Benzylate (add a benzene group to) cyclic compounds? How do you add a benzene group to cyclic compounds like piperidine, piperazine and phenyl derivatives?
Piperidine

Piperazine

Brenton, the term "benzene group" is a bit misleading; it seems that you'd like to add a benzyl group, i.e. $\ce{C6H5-CH2}$.

Typically, these reaction are carried out using freshly distilled benzyl halides. In the case of the diamines, monoalkylation is usually achieved using the bis-hydrochloride of the amine and a stochiometric amount of base, thus keeping one nitrogen atom protected.

Note however, that the Australian regulations, kindly mentioned by Richard Terrett, are likewise true in the EC: 1-benzylpiperazine, an infamous psychoactive party drug, is a controlled substance!


Why does ionization violate the stable or lowest energy rule in atoms? Electrons that fill orbitals always do so in such a way that the resulting structure has the lowest energy state possible, though there are anomalies like chromium and so on.

But ionization no matter which element tends to violate this rule. When electrons are ionized they are from the outermost (valence) shell. But this doesn't always produce the lowest energy state possible like the case with 3d and 4s.

To clarify, all I am asking is why in that specific case, is the electron taken out of the 4s orbital first instead of the 3d orbital, since the 3d orbital is a higher energy level?

Textbooks won't bother to dig into details. Most of them are satisfied with saying that such thing exists. They're right I guess. I'll do a chemguide approach to the answer since I find it most understandable.

We draw this diagram, and then assume that it works for all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.
If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, you are adding extra protons to the nucleus, and extra electrons around the nucleus.
The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.

Going through the Aufbau principle, when you pass Ca (Z = 20) you will face Sc, the first element in which a d orbital is going to be filled. But, you must make a really important change in your train of thoughts.
As chemguide explained, with change in the number of protons and electrons due to the change of Z, the way their orbitals' energies are defined changes. Till Ca, the behavior insisted that 4s has a lower energy state than 3d and thus, fills sooner. But, starting from Sc, the pattern changes: In Scandium, 3d has a lower energy than 4s. The behavior changed.
This change in behavior results in the current confusion of yours. This much answers the problem you have, and you can visit the page at chemguide to learn more. If you want more info, I can freely add. Hope I've helped!


NGP mechanism vs the simple carbocation mechanism Question:

Predict the product of: ($\ce{C^*}$ is a $\ce{C^14}$ carbon)

My attempt:
I simply assumed the super-leaving group $\ce{-OTs^-}$ would leave, forming a primary ethylbenzene carbocation. That would rearrange via a hydride shift to form a secondary carbocation on $\ce{C^*}$ atom. The nucleophilic $\ce{-OH}$ group of the $\ce{-COOH}$ group will then attack and lose a proton to form 1-phenylethylethanoate (with the ester linkage at the $\ce{C^*}$ carbon):

However, my book has instead taken a different route. In the very first step, after $\ce{OTs-}$ leaves, they have taken the NGP mechanism and formed this transition state:

and said that now the $\ce{OAc-}$ will attack either of the two carbons (in that trigonal ring). Hence, according to their logic, both types of 1-phenylethylethanoate will be formed - one with the ester linkage at the $\ce{C^*}$ carbon, and the other at the $\ce{C}$ carbon.
My question:
I have confirmed from Clayden (Chapter 37 Rearrangements page 976) that the product ratio is indeed 50:50. So:

Why is the NGP mechanism being favored here over the other mechanism I described, even though the NGP mechanism involves a loss of aromaticity?
Are there other cases where the NGP mechanism dominates over the simple carbocation mechanisms, or is this the only such case?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q16 in  Alkyl Halides (Substitution)

From Clayden (Chapter 37 Rearrangements page 976):

  Intramolecular reactions (including participation of a neighbouring group) that give three-,
  five-, or six-membered rings are usually faster than intermolecular reactions.

It also gives the exact same example as your question:

Notice that both products are given to be exactly 50% each.

NGP is very often faster than bimolecular reactions as for the latter to take place two molecules have to collide first.

1:1 ratio can be attributed to the fact that in NGP the transition state involves loss of aromaticity BUT NGP supports the ionization of the starting material.

There are a lot of examples  this one reaction isn't a black swan. Interested readers may see chapter 36: Participation, rearrangement and fragmentation in Clayden to find more details and examples.


Retinol — an oligomer of isoprene? The following image is taken from Wikibooks

How does isoprene polymerise to form double bonds between two monomers and in each monomer? As far as I have learned is that upon polymerisation two monomer join by a single bond and form a double bond within each monomer which can be explained by the following image:

Biomolecules that consist of ‘isoprene oligomers’ — more typically labelled terpenes do not actually polymerise in a way that you may be used to from plastic synthesis. When generating plastics, a double bond of a monomer unit (e.g. propene) is somehow made more reactive (e.g. by the addition of a nucleophile to generate an anion or by the addition of an electrophile to give a cation) and this reactive species then continues to attack monomer units giving a saturated chain of polypropylene.

Nature chooses a different path. It is not actually isoprene ((3E)-2-methylbut-1,3-diene) that is oligomerised — the resulting reaction would be far too hard for nature to control efficiently. Instead, the actually employed monomers are isopentenyl pyrophosphate (IPP) and dimethylallyl pyrophosphate (DMAPP).

Figure 1: The terpene biomonomers isopentenyl pyrophosphate (IPP; left or top) and dimethylallyl pyrophosphate (DMAPP; right or bottom).

As you can see, these only have one double bond but a diphosphate group attached to C4 that can be eliminated to give another. The biosynthetic pathway to couple these into oligomers begins by the removal of diphosphate from DMAPP resulting in a delocalised 3,3-dimethylallyl cation. This is captured by IPP’s terminal double bond to give a tertiary carbocation. In a final step, one of the two α-hydrogens in direction of former IPP’s diphosphate is eliminated to give geranyl pyrophosphate (figure 2).

Figure 2: Geranyl pyrophosphate; the dimerisation product of IPP and DMAPP.

Another addition of IPP gives farnesyl pyrophosphate which is then fed into multiple different biosynthesis pathways.

Thus, terpenes can be considered formal oligomers of isoprene but one should never think that they are generated by actual condensation of isoprene units.


Given volume of reagents find the chemical formula of nitrogen oxide $$\ce{NH3 + O2 -> N_xO_y + H2O}$$

I know that there is $\pu{1 m^3}$ of both the nitrogen oxide and the ammonia and the volume of oxygen on the left side is $\pu{1.25 m^3}$.

I think that I should use the ideal gas equation. It is also obvious that there is only one atom of nitrogen in the oxide - I guess that there is also only one oxygen.

Could anyone give me any hints?

As a hint, you could write balanced equations for oxidation to N2O, NO, NO2 and N2O5.  (Did I forget any nitrogen oxide?)  Then you can compare moles (or volumes, since we are talking gases) of starting materials and products.


Methanol skin exposure serious or not? I have just minutes ago been splashed with industrial grade methanol. It seemed like a small amount (1-3mL) on my skin. I already washed the affected area thoroughly with soap and water. Online I've reviewed SDS's for methanol and they all seem to say that immediate medical attention is needed, and that the person exposed can be asymptomatic for a while. Does a small splash like that warrant medical evaluation?

"[A] factory worker who spilled a gallon of methanol down his trouser leg, was dizzy on the following day, took a short nap, and woke up total blind" Biochemical aspects of Methanol Poisoning, Biochemical Pharmacology, vol. 11, page 405.

So spilling methanol on your skin is serious, but the amount in your case is relatively small.

The article also explains that ingestion of as little as "a teaspoon" (5 mL) has caused permanent blindness.

In any case, the internet is no place to get medical advice.  If you are concerned you should call your local poison control center.


Periodate oxidation of sugars When Methyl D-glucopyranoside is treated with $\ce{HIO4}$, what is the amount of $\ce{HIO4}$ (in moles) consumed per mole of the sugar?

I have studied that $\ce{HIO4}$ oxidises only those vicinal diols which are cis to each other. But here only one pair is cis, so shouldn't the answer be 1?

The vicinal-glycols can be cleaved to corresponding aldehydes and/or ketones in high yield by the action of periodic acid ($\ce{HIO4}$) or lead tetraacetate ($\ce{Pb(OAc)4}$). This oxidative cleavage of a carbon-carbon single bond provides a two-step involving cyclic intermediate reaction mechanism with high-yield. A generally accepted equation for these oxidations (A) is shown below:

The reaction using $\ce{HIO4}$ is called Malaprade Reaction (Ref.1), while that involving $\ce{Pb(OAc)4}$ is called Criegee Reaction. Malaprade Reaction is described as:

  The oxidation of adjacent diols with periodic acid or its salt in aqueous solution is generally known as the Malaprade reaction. Several solvents have been used to increase the solubility of organic substrates and the reaction proceeds faster under acidic conditions. This reaction has been further extended to the cleavage of $\alpha$‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, $\alpha$‐amino alcohols, $\alpha$‐amino acids, and polyhydroxy alcohols, and successfully applied for structural analysis.

Both cis- and trans-glycols (e.g., cis- and trans-cyclohexane-1,2-diols) undergo the oxidation with $\ce{HIO4}$ or its salts. As a rule, cis-glycols react more rapidly than trans-glycols and there is evidence for the presence of heterocyclic intermediates as shown in the diagram (B). For example, the publication by Buist, et al. (ref.2) revealed that cis-cyclohexane-1,2-diol cleaves about 20-times faster than trans-cyclohexane-1,2-diol. However, the equilibrium between the diol and the cyclic periodate intermediate (faster first step) slightly favors the trans-isomer (the ratio of cis/trans equilibrium constants is $400:1000$; Ref.2) while the rates of cleavage of cis-cyclohexane-1,2-diol to products is much faster compared to its trans-isomer (the ratio of cis/trans rate constants is $3300:165$; Ref.2). Thus, theoretically your believe of only cis-vicinal-diols would get oxidized with $\ce{HIO4}$ is completely wrong.

Therefore, theoretically, each available $\ce{-C(OH)-C(OH) -}$ function would get oxidized ($\alpha$-methyl-D-glucopyranoside has 4 such functions), costing one $\ce{HIO4}$ molecule per function (Ref.3). However, rate of oxidation is depend on conditions, steric effects, $\mathrm{pH}$, etc. For example, the oxidation of $\alpha$-methyl-D-glucopyranoside by a solution of $\ce{HIO4}$ in pyridine has cleaved only the $\ce{C}$3$-\ce{C}$4 bond while the reaction of 6-O-trityl-$\alpha$-methyl-D-glucopyranoside is less selective under same condition and is directed to a greater degree to the $\ce{C}$2$-\ce{C}$3 bond (Ref.4).

References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).
Glen Dryhurst, In Periodate Oxidation of Diol and Other Functional Groups: Analytical and Structural Applications; First Edition, Pergamon Press: Oxford, United Kingdom, 1970. ISBN-10: 1483128385.
R. G. Krylova, S. N. Ryadovskaya, L. I. Kostelian, A. I. Usov, “Periodate oxidation of α-methyl-D-glucopyranoside, 6-O-trityl-α-methyl-D-glucopyranoside and 6-O-tritylcellulose in pyridine,” Bulletin of the Academy of Sciences of the USSR, Division of chemical science 1972, 21, 2005-2009 (https://doi.org/10.1007/BF00854627).


Why is LDL more easily oxidized/damaged than HDL? Why is LDL more easily oxidized than HDL is? By LDL and HDL, I mean the lipoproteins that carry cholesterol.

Oxidation of LDL ultimately contributes to heart disease.

Just a brief background about lipoproteins:

Lipids and proteins associate noncovalently to form lipoproteins, which function in the blood plasma as transport vehicles for triacylglycerols and cholesterol. Simple, unesterified fatty acids are merely bound to serum albumin and other proteins in blood plasma, but phospholipids, triacylglycerols, cholesterol, and cholesterol esters are all transported in the form of lipoproteins.

In general, lipoproteins are classified according to their densities. The densities are related to the relative amounts of lipid and protein in the complexes.  Thus, there are:

high-density lipoproteins (HDLs),
low-density lipoproteins (LDLs),
Intermediate density lipoproteins (IDLs),
very-low-density lipoproteins (VLDLs), and also
chylomicrons.

Chylomicrons have the lowest protein-to-lipid ratio and thus are the lowest-density lipoproteins. They are also the largest.

There are a couple of things to take into consideration:

How they function

LDL appears to be the major circulatory complex for cholesterol and cholesterol esters. At various target sites, particularly in the capillaries of muscle and adipose cells, these particles are degraded by lipoprotein lipase, which hydrolyses triacylglycerols. Lipase action causes progressive loss of triacylglycerol (and apoprotein) and makes the lipoproteins smaller. This process gradually converts VLDL particles to IDL and then LDL particles, which are either returned to the liver for reprocessing or redirected to adipose tissues.  This results in even more increase in LDL than other lipoproteins.

Environment

Another explanation is based on environment conditions -the LDL binds to specific LDL receptors, which cluster in domains of the plasma membrane which subsequently form coated vesicles which pinch off from the plasma membrane and form endosomes. In the low pH environment of the endosome, the LDL particles dissociate from their receptors. The endosomes then fuse with lysosomes, and the LDLs are degraded by lysosomal acid lipases into amino acids, fatty acids, and cholesterol.

Newly formed HDL contains virtually no cholesterol ester (however, over time, cholesterol esters are accumulated through the action of lecithin:cholesterol acyltransferase).

Another associated protein, cholesterol ester transfer protein, transfers some of these esters to VLDL and LDL. Alternatively, HDLs function to return cholesterol and cholesterol esters to the liver. This latter process apparently explains the correlation between high HDL levels and reduced risk of cardiovascular disease. High LDL levels, on the other hand, are correlated with an increased risk of coronary artery and cardiovascular disease.

Reference

Biochemistry (Grisham)

Biochemistry (Campbell and Farrel)


Enthalpy of Zinc and Sulfur Reaction I was working on my homework which was to write a description of what would occur if zinc and sulfur reacted and what steps would a scientist have to make in order to make them react. When I researched I bit, I saw that it would be a exothermic reaction, and I wanted to show this by using enthalpy. I haven't been taught this however, and as a result, I'm confused. I know that enthalpy equals the heat transfer and there is some sort relationship with bond energy, but I don't know how to apply this to the reaction.

Zinc Powder + Sulfur Powder = Zinc Sulfide + Energy

$\ce{Zn_{(s)} + S_{(s)} -> ZnS_{(s)}}$ + $205.98 \text{ kJ}$

Because $205>0$ this is an exothermic reaction. But where did this $205.98\text{ kJ}$ come from?

The free energies of formation of the elements in their standard states is zero, by definition.  You can look up the (variously reported) standard enthalpy of formation of the product, −204.6 kJ/mol (exothermic!).  It is roughly the binding energy of the crystal lattice less the ionization energies of the inputs.  So, by the numbers, do you obtain spalerite or wurtzite?  ZnS has crystal polymorphs and polytopes with differently populated unit cells

$\ce{Zn_2S_2}$ Wurtzite-2H
$\ce{Zn_4S_4}$ Wurtzite-4H and Sphalerite
$\ce{Zn_6S_6}$ Wurtzite-6H
$\ce{Zn_15S_15}$ Wurtzite-15R


How were urolithins discovered and named? Urolithin is a product of the degradation of ellagic acid by gut flora and has significant biological activity. (Wikipedia)
I am curious as to how it was discovered but cannot find the story. Although I can imagine how it might have been discovered through the study of fecal flora, I cannot imagine why it was named with reference to kidney stones.

The identification of urolithins as products of human metabolism of ellagic acid was reported by Cerda et al. in the European Journal of Nutrition in 2004. The full reference is
Cerda, B., Espin, J.C., Parra, S., Martinez, P. and Tomas-Barberan, F.A. (2004) The potent in vitro antioxidant ellagitannins from pomegranate juice are metabolised into bioavailable but poor antioxidant hydroxy-6H-dibenzopyran-6-one derivatives by the colonic microflora of healthy humans. Eur J Nut 43:205-220.
However, these researchers did not name the compounds as urolithins. Rather, they note that the compounds

have also been found as constituents of ‘clover stone’, a type
of renal calculus found in sheep which can also ingest large
amounts of ellagitannins[26]. In fact, the metabolite d
(3,8-dihydroxy-6H-dibenzo[b,d]pyran-6-one) has been previously named
as ‘urolithin  B’ [22, 26].

That explanation seems sufficient to understand the source of the name, but if you are interested in the specifics of the original discovery, reference 26 is to the Dictionary of Natural Products published in 1992 by Chapman & Hall, which is available only by subscription. Perhaps someone with access to that Dictionary can look up the entry on urolithin and provide more info.
UPDATE: Since this answer has received a decent amount of attention, I dug a bit deeper to find the references for the first isolation and naming of the compounds (thereby actually answering the question).
In 1963 at the 430th meeting of the Biochemical Society in Oxford, England, Nottle and Pope first reported isolating two compounds that they named "urolithin A" and "urolithin B". Their report was published among the proceedings of the meeting in the Biochemical Journal, which are available for free from the journal here. The full reference is
Nottle, M.C. and Pope, G.S. Two Constituents of 'CloverStone', a Type of Urinary Calculus Found in Sheep in Proceedings of The Biochemical Society. Biochem J (1963) 89 (1): 1P–67P.
The Nottle and Pope report is on p. 67P. In that report, however, the structures of the urolithins were not described, as they had not yet been determined. In a subsequent paper, Pope reported the structures of the urolithins, noting that they were actually previously known compounds (specifically urolithin A is identical to pigment I of castoreum, isolated from beaver scent glands), but despite the compounds already being known, Pope and others continued to use the new name, which seems to have stuck. That report is also freely available from the Biochemical Journal here. The full reference is
Pope, G.S. (1964) Isolation of Two Benzocoumarins from 'Clover Stone', a Type of Renal Calculus Found in Sheep. Biochem J 93:474.