Crystallisation and crystal surface It is my understanding that if you put table salt (as an example of polar compound) in water, the Na+ and Cl- ions will freely move through the solution. They will make the solution conductive and will independently travel through it to carry electric current.

Now the thing is, if you dry the solution, how come every Na+ finds it's Cl- to recombine into NaCl? How come there are no lonely Na+ and Cl- pairs who couldn't find their mate in time, before the solution dried completely? Can someone give me a better understanding of how do polar solutions work?

The crystal structure of NaCl forms at the growing surface of the crystals, where there is plenty of time for the ions to sort themselves into a regular lattice. Further, each ionic species is very evenly distributed also in the solution, because of the repellent electrostatic forces between the members of each. Whenever there is a local excess of positive or negative charges, it will level itself out very quickly.

If you have something larger than sodium and chloride ions and try to crystallise very fast, you will indeed get an increasing amount of errors in the structure. Freeze-drying also tends to leave behind rather unordered structures. In both cases, this is, of course, more likely with noncharged or even nonpolar substances.


Chemical reactions involved in staining I'm a geology student completing a project on chemical staining and was after some chemistry help on some of the reactions taking place. I am using various different stains to identify carbonate and feldspar minerals.

One of the stains, in particular, is alizarin red S (1,2-dihydroxyanthraquinone) and is used to stain calcite ($\ce{CaCO3}$) red and Fe-calcite/Fe-dolomite blue. The stain powder is mixed with a weak (2%) HCl solution to react with the carbonate and produce the stain. However, I am just unsure about what reactions are taking place between the stain solution and the mineral that produce the precipitate on the surface of the grains.

I'm not finding literature on this reaction - probably because it has been mainly used by geologists without a solid chemistry background. Any help would be much appreciated.

Alizarin red S can stain calcium deposits through interaction of its sulfonic acid and hydroxy groups with calcium ions.

I found a reference for you: http://jhc.sagepub.com/content/17/2/110.full.pdf+html

Interestingly, however, this research seems to suggest using a solution of pH $\approx$ 9, stating that solutions of acetic acid (pH 5) did not work that well.


When expressing a quantity, do the non-significant figures determine whether you round up the last significant figure? I had thought that the non significant figures determined whether the least significant figure was to be rounded up. For example, given the number $9.1145$, if the number is to be expressed as three significant figures, then, I had thought that, it would be expressed as $9.12$.

However, I noticed that in one of my chemistry professor's calculations in which the quantities were expressed as significant to three figures, she expressed $8.125 \times 10^2$ as $813$. In another calculation she expressed $9.1145 \times 10^{-3}$ as $0.00911$. However, it seems to me that $0.00912$ is the correct expression of $9.1145 \times 10^{-3}$, as the $5$ requires rounding up the $4$, which requires rounding up the least significant $1$ to $2$.

Is there a rule for rounding that my professor is following that I don't know?

(Note: this isn't a homework question, but the homework tag seemed to be the most appropriate one.)

The international standard ISO 80000 Quantities and units – Part 1: General includes the normative Annex B Rounding of numbers, which stipulates:

Rounding in more than one stage (…) may lead to errors. Therefore, the rounding shall always be carried out in only one step.
Example: 12.254 should be rounded to 12.3 and not first to 12.25 and then to 12.2.

(Note that, in the given example, the ISO standard follows the round half to even rule in which the even multiple is selected as the rounded number. This is the default rounding mode for floating-point computation. Anyway, the rule “rounding shall always be carried out in only one step” is independent of the used tie-breaking rule.)


Do acids and bases have the same coefficients if they neutralize one another? The questions states: Ammonia has the formula $\ce{NH3}$. Household ammonia is a dilute aqueous solution of $\ce{NH3}$. Aqueous ammonia is a base that can neutralize acids.
If 18.2 mL of 0.800 M $\ce{HCl}$ solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?
Express your answer with the appropriate units.
I used the values I was given; converted the $\ce{HCl}$ to moles; divided the moles of $\ce{HCl}$ by the liters of ammonia solution. However, I do not understand why this is correct.
$\ce{HCl}$ is an acid and $\ce{NH3}$ is a base. How does finding the molar mass of $\ce{HCl}$ help one calculate the molarity of $\ce{NH3}$?

Look at the chemical equation. One mole of $\ce{HCl}$ will neutralize one mole of $\ce{NH3}$. $\ce{HCl}$ is an acid so it loses a hydrogen ion and ammonia is a base so it gains a hydrogen ion.

$\ce{HCl + NH3 -> NH4Cl}$

So, you found the number of moles $\ce{HCl}$ added, and this equals the number of moles in the 5.00 mL sample of $\ce{NH3}$. Divide that by the volume and you get concentration, which is moles per liter. You don't need to find the molar mass of $\ce{HCl}$ and you didn't find it as described in your calculations above.


Bonding in the phosphate ion I'm looking for an explanation of the bonding in the phosphate (PO43−) ion:

(Image courtesy of Wikipedia)

Phosphorus (15P) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:

1s2
2s2 2p6
3s2 3p3

This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the Phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).

Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?

Finally, why are the oxygens shown as carrying a negative charge?

First of all, let me state the obvious: Phosphorus is awesome.
After we got that out of the way we can focus on why.

There are many different modifications of phosphorus in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -> P_{red} -> P_{violet} -> P_{black}}.$$

Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.

Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.

But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples:
$$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$
While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals.

In case of phosphane $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.

In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.

Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.

Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.

A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method.
\begin{array}{rlrr}\hline
       &             &\mathrm{\%P (hyb)}  &\mathrm{\%O (hyb)}\\\hline
3\times&\ce{Bd(O-P)} & 24 (\ce{sp^3}) & 76 (\mathrm{sp^{2.3}})\\
       &\ce{Lp(O)}   &                & 100 (\mathrm{sp^{0.4}})\\
2\times&\ce{Lp(O)}   &                & 100 (\ce{p})\\\hline
\end{array}

This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.

The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.)

The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)


How do we obtain the π molecular orbitals for allene via Hückel theory? For a conjugated system, for instance 1,3-butadiene, we can try to obtain the π molecular orbital energies and wavefunctions via Hückel theory. In our given molecule, there are 4 interacting Pz orbitals. By creating a 4x4 Hückel matrix and solving for the energy eigenvalues, we can arrive at 4 possible linear combinations (of p-orbitals) to get 4 molecular orbitals.

However, how can we use Hückel theory to deal with a molecule like 1,2-propadiene (allene)?

My attempt: I thought that since this molecule has essentially 2 separate π-systems:

we may be able to apply Hückel theory to each one individually. Then, we can take all the possible combinations of the individual energies to arrive at the possible total energies. But I am not sure if this is correct. What would be the correct way of approaching this problem?

For p-orbitals $i$ and $j$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero:
$$\langle i | H | j \rangle = 0$$
If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form:
$$\mathbf{Hc} = E\mathbf{c},$$
or in explicit form, using the fact that the matrix elements of $\mathbf{H}$ are $\mathbf{H}_{ij} = \langle i | H | j \rangle$,
$$\begin{pmatrix}
\alpha & \beta & 0 & 0 \\
\beta & \alpha & 0 & 0 \\
0 & 0 & \alpha & \beta \\
0 & 0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}
=
E
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}$$
This matrix has block-diagonal form, so solving for the eigenvalues of the entire matrix simply reduces to solving for the eigenvalues of each individual $2\times 2$ block (lots of discussion can be found on Google). For a physical interpretation, this means that both π systems have no interaction with each other, so you can indeed solve them individually.
Each block gives rise to one MO with energy $\alpha + \beta$ ("bonding") and one with energy $\alpha - \beta$ ("antibonding"), so all in all you will have two degenerate MOs with energy $\alpha + \beta$ and two degenerate MOs with energy $\alpha - \beta$.

* If you don't ignore the overlap matrix $\mathbf{S}$, whose matrix elements are defined by $\mathbf{S}_{ij} = \langle i | j \rangle$, this becomes a "generalised eigenvalue equation"
$$\mathbf{Hc} = E\mathbf{Sc}$$
However, because the overlap matrix also has the same block-diagonal form, ignoring the overlap matrix doesn't change the fact that the problem is separable into two $2\times 2$ blocks. It merely changes the exact eigenvalues and eigenstates by a small amount.


Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode? We all know that pure water can't conduct electricity. But during electrolysis, if add a small amount of $\ce{HCl}$ acid inside, water could be decomposed to hydrogen gas and oxygen gas, as it is reduced at the cathode and oxidised at the anode. So why can't pure water just conduct electricity by electrolysis?

Indeed, water can conduct electricity by electrolysis. But this is hampered by the fact that water has very high resistance. From Electrolysis of water:

  Electrolysis of pure water requires excess energy in the form of
  overpotential to overcome various activation barriers. Without the
  excess energy the electrolysis of pure water occurs very slowly or not
  at all. This is in part due to the limited self-ionization of water.
  Pure water has an electrical conductivity about one millionth that of
  seawater.

And from Self-ionization of water we can see how small it is:

  Water molecules dissociate into equal amounts of $\ce{H3O+}$ and $\ce{OH-}$, so their concentrations
  are equal to 1.00×10−7 mol∙dm-3 at 25 °C


How to determine relative polarity (basic procedure)? I am taking AP Chemistry, and have noticed that I can identify a polyatomic polar molecule, but struggle to determine which molecule is the most polar given a set of several polar molecules.

What would be a basic procedure/set of questions to ask myself in determining which molecule is the most polar? I hope my question is not too general...

Thank you!

Determine which molecules are polar out of the given options.
Determine the polar bond (This will be the bond causing the molecule to be asymmetrical)
The molecule with the polar bond that has the greatest difference in electronegativity is the most polar. For example a carbon-oxygen bond is more polar than an oxygen-fluorine bond because the difference in electronegativity for oxygen and carbon is greater than the difference between fluorine and oxygen.

You can find electronegativity values in the following table or a table like it, I am not sure if you are provided with one. If not you can normally judge electronegativity by the how fare an atom is from fluorine, as seen in the table below. The size of the molecule also affects polarity, for example polarity decreases with the size of alcohols, but large complex molecules should not come out at your level.


Cocrystals vs Eutectic mixtures I was reading around about the different solid states of materials and I got to a point where I have reached the multi-component solids. I was not able to find a major difference between the cocrystal and the eutectic mixture from their definitions. Do you know of any fundamental difference?

The question in another way: if you had a mixture of 2 solids that bind to each other somehow, how can you know whether this is a cocrystal or a eutectic mixture?

Many thanks,

That explains a eutectic but does not answer the question, which is a good one for solid-state chemists.
A cocrystal is a crystalline, multiple component solid with a distinct stoichiometry (molar ratio) between components where a single component is neutral overall(can exist alone without a counter ion, typically non-ionic but a zwitterion counts as a single neutral component) and the components interact by non-covalent interactions that can be partially ionic and has distinctive crystalline characterization data, i.e. to be crystalline a solid the mixed solid phase needs to have X-Ray or electron diffraction data that distinguishes it from its starting components.
So, the eutectic phase of 2 solids, or the solid mixture with the lowest melting pt, can be a cocrystal, but a cocrystal is not necessarily a eutectic. A solid mixture with no regular repeating crystal lattice and no consistent molar ratio (stoichiometry) is called a solid solution. So a eutectic phase could be a cocrystal, or a salt, or a mix of those, or physical mix (solid solution).


Why doesn't hydroxide concentration equal concentration of carbonic acid and bicarbonate in a sodium bicarbonate solution? To find the pOH of  solution of  $\pu{0.42 M}$ of $\ce{NaHCO3}$, given a   $\ce{Kb}$ of $\pu{2.4E-8}$, my solution manual does the following steps:

Reaction is similar to : $\ce{HCO3- + H2O <=> H2CO3 + OH-}$
$\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH-][H2CO3]}{[HCO3-]}}$
$\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH]^2}{[0.42]}}$ , as we have $\pu{0.42 M}$ of $\ce{NaHCO3}$ so approximately $\pu{0.42 M}$ $\ce{HCO3-}$, and $\ce{[OH-] = [H2CO3]}$
Solve equation for $\ce{[OH-]}$, giving $\ce{[OH-]} = \pu{10^{-4}}$  and hence $\ce{pOH} = 4$

What I don't understand is: Given the equation $$\ce{HCO3- + H2O <=> H2CO3 + OH-},$$ doesn't this show $\ce{HCO3} = \ce{H2O} = \ce{H2CO3} =\ce{OH-}$, as they are all in a $\mathrm{1:1:1:1}$  molar ratio in the equation.

So wouldn't $\ce{OH-}$ also just equal $\pu{0.42 M}$?

Your misunderstanding appears to be some confusion between stoichiometry and equilibrium concentrations. The chemical equation describes the stoichiometry or the ratio of the species in the reaction. In other words, each time the reaction happens, the equation describes how many of each reactant are consumed and how many are produced. In your example:

$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$

Each time this reaction occurs in the forward direction, one bicarbonate anion and one water molecule react to form one carbonic acid molecule and one hydroxide anion. In the reverse direction, one carbonic acid molecule and one hydroxide anion react to form one bicarbonate anion and one water molecule.

Since this reaction is reversible, the equilibrium constant tells you something about the extent of the reaction. In other words, what portion of the overall system is sitting on the reactant side and what portion sis sitting on the product side. Put another way, the system is at equilibrium when the rates of the forward and reverse reactions are equal. These rates are dependent on the concentrations of the species on the reactant and product sides of the equation and on rate constants.

$$\mathrm{rate_{forward}=rate_{reverse}}\\
k_\mathrm{f}[\ce{HCO3-}][\ce{H2O}]=k_\mathrm{r}[\ce{H2CO3}][\ce{OH-}]$$

The equilibrium constant is the ratio of the rate constants:

$$K=\frac{k_\mathrm{r}}{k_\mathrm{f}}=\frac{[\ce{H2CO3}][\ce{OH-}]}{[\ce{HCO3-}][\ce{H2O}]}$$

The equilibrium constant is thus a law of mass action expression for the situation where the forward and reverse rates are equal. At this state, both forward and reverse reactions are occurring following the stoichiometric ratios, but the concentrations of each species has reached a steady state.


Atomic mass unit based on other atoms Question:

  Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

My attempt:

Current standard is $1/12$th the mass of a C-12 atom. New standard is $1/9$th the mass of a Be-9 atom.

Now I understand that the only the basic unit of measurement of masses is changing. The absolute masses of the different atoms in the universe remains the same. That means the absolute mass of an He-atom in the old system and in the new system must be the same.

Now, absolute mass (of an atom in any system) = no. of "amu"s of that atom * mass of 1 amu (in that system)

We are supposed to determine the relation between the mass of 1 amu in the old and in the new system. Obviously, the one having more "amu"s would have a smaller mass of 1 amu.

And now I am stuck. I can't proceed further than this much explanation. I've looked on Wikipedia and some other questions but can't find similarities. Hints are appreciated!

UPDATE: although not explicitly mentioned, the book does NOT expect readers to look up data tables to find the answer. It expects a possible reason as to why is the Be-9/C-12 scale larger.

Question:

  Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

Answer

It is easy enough to look up the weights of $\ce{^{12}C}$ (12.00000000...) and $\ce{^9Be}$ (9.0121822) based on $\ce{^{12}C}$ scale and make a decision from that data. Thus a mass scale based on $\ce{^9Be} =9.00...$ would be heavier than an mass scale based on $\ce{^{12}C} = 12.00...$.

"Justification"

The OP has asserted that the problem is not supposed to be just a "lookup problem" but that some reasoning should be offered.

I content that the question simply asks which is larger. It doesn't ask why. Furthermore I'd assert that there isn't any "simple" reasoning by which the problem can be analyzed other than lookup.

Neutron to proton ratio reasoning

In his answer QuarkyLittleThing's reasoned that a neutron is heavier than a proton and since $\ce{^9Be}$ has a higher neutron to proton ratio (5:4) than $\ce{^{12}C}$ (6:6), it ought to be heavier.

Using the neutron to proton ratio reasoning, $\ce{^9B}$ only has 4 neutrons but 5 protons, so a $\ce{^9B}$ scale should be lighter than the $\ce{^{12}C}$ scale. But the mass of $\ce{^9B}$ is 9.0133288 u, so the neuron to proton ratio isn't entirely sound, and a mass scale based on $\ce{^9B} =9.00...$ would in fact be heavier than a mass scale based on $\ce{^{12}C} = 12.00...$.

Lower nuclear binding energy reasoning

Mithoron's made a comment on QuarkyLittleThing's answer asserting that the reason was "lower nuclear binding energy."

To properly analyze Mithoron's comment one must note that the nuclear binding energy is based on the energy differences of the actual fission/fusion of nuclei. Thus as NicolauSakerNeto pointed out on an earlier version of this answer, this is really not a "reason", but a tautology due to $\text{E}=mc^2$. So a decrease in the mass of the nucleus is proportional to an increase in the nuclear binding energy for that nucleus.


Laws of Chemical Combination I'm doing revision questions and want to double check something. The opening information reads:

  Two chlorides of iron were prepared. One was prepared by reacting iron with dry chlorine
  gas. 4.50 g of iron reacted with chlorine gas to produce 13.01 g of the chloride.
  The other chloride was prepared by reacting iron with dry hydrogen chloride; 2.80 g of iron reacted with hydrogen chloride to produce 6.35 g of the chloride.

The questions are then:

  (i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.
  (ii) Find the ratio of the different masses of chlorine that combined with 2.80 g of iron.
  (iii) Which of the chemical combination laws is demonstrated using this data?

Is that first question right? I would've thought they're asking to calculate the mass of chlorine, which you'd simply get by subtracting the mass of reactant iron from the mass of the end product.

Two chlorides of iron were prepared.

I do not read that as: $\ce{FeCl_{n}}$ was prepared by two different methods, but as:

Reacting iron with dry chlorine gas yields $\ce{FeCl_{n}}$
Reacting iron with hydrogen chloride yields $\ce{FeCl_{m}}$

Apparently, iron is oxidized in both cases.

What are the oxidants (what is reduced)?
Which of them is the stronger oxidant?
What is the stoichiometry (what are n and m)?

Could it be helpful to divide the amount of chloride produced by the amount of iron used? Would different numbers indicate which of the products has a higher chloride content?

  (i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.

Is it possible that they meant the same as indicated above, i.e.:

Calculate the amount of iron chloride that would have been produced from 2.80 g (instead of 4.50 g) iron in the first experiment and compare with the outcome of the second experiment.


Transition Metals colours and electron absorbtion I understand that when d-orbitals split some electrons can occupy the higher orbitals by absorbing frequencies of light, but do these electrons then fall back to their original ground states? If so, why? Why don't they remain in a permanently excited state?

I've heard someone mention that the electrons do indeed fall back to ground state but, you don't see that part of the emission spectra since the energy is dispersed everywhere and not just towards your eyes. This is why the colour comes from the absorbance spectra.

The reason is does not stay in the excited state is simply because it is unstable. It will gain energy and then release it as mentioned above, then repeat.


Does (chloromethoxy)ethane react in a bimolecular or monomolecular mechanism? It is known that (chloromethoxy)ethane forms stable carbonium ions based on Nobel prize Chemist George Olah research Olah's method. This would indicate that (chloromethoxy)ethane reacts through $\mathrm{S_N1}$ mechanism. However, $\mathrm{S_N2}$ mechanism is also very feasible for highly electrophile carbon.

  Which one of the mechanisms is dominant?

Organic chemistry books in this case are not very clear. For example Clayden speaks about Olah's method and neighbouring group participation (pp. 419, 970 1st ed) but also gives relative rate for $\mathrm{S_N2}$ mechanism (pp. 425 1st ed)

Probably both mechanisms can happen and it would require experimental data from reaction kinetics and computational calculations to determine which one of the mechanism is domintant.

One example of determined mechanism about neighboring group effect, also called anchimeric assistance is mustard gas (chloromethine) alkylation reaction with DNA.

Effect of chloromethine is based on its ability to block DNA replication by alkylating guanine, and after the war it has been also used as cancer drug.


Finding the weighted parameters of the vertex adjacency matrix for a derivative of azulene In my last question ( Numbering in a derivative of azulene ), I asked about the numbered of a molecule in order to find the Vertex Adjacency Matrix (VAM) of the same molecule. To explain my question properly, I will do my best to show my effort from A to Z as follows:
In general in order to find the VAM we use the formalism described by

To apply it onto the molecule I am studying, I first applied it to azulene and with agreement with the paper Quantum Interference, Graphs, Walks, and Polynomials the VAM represented by A is:

Now by having the correct numbering to my molecule (Numbering in a derivative of azulene), it also should be a 10 x 10 matrix. The difference is instead of hydrogen we now have Oxygen and hydrogen Sulfide involved in positions 1,5 and 2,6 as shown:

To find VAM the only way was to study the VAM of weighted Graphs as shown below:

Here are my questions:
1- Should I include  the parameter h only for the weighted carbons since the numbering of the molecule I am studying includes only carbon?
2- If 1 is no, how should I include the k parameters described before hand?
Any other suggestions to find VAM is appreciated.

As a comment: The matrix-representation you search emerges from graph theory, and mathematics.  The numbers drawn in the structures earlier (1 for the carbonyl carbon, 2 to the carbon adjacent to the -SH group) a synthetic chemist refers to (e.g., "let us introduce an alkyl chain in the 4-position") do not (necessarily) relate to the numbers to denote the vertices in these nets.  These are seasoned descriptions are still reviewed (2nd edition), and apparently  the Open Molecular Generator were one tool to generate them.


Residual hydroxyl groups in silicone resins Silicone resins are solutions of branched-chain siloxanes containing residual hydroxyl groups. What is meant by "residual hydroxyl groups"?

Silicones are produced by hydrolyzing chlorosilanes( in which chlorine atoms are replaced the hydroxyl groups). These hydroxyl groups undergo condensation polymerization with removal of water molecules to form polysiloxanes or silicones.

Now, residual hydroxyl groups in your question means that some hydroxyl groups has not undergone condensation yet. Solvent is evaporated and resin is heating resulting condensation of residual hydroxyl groups to form cured resinous film.


How does the wax layer of a leaf get damaged by acid rain? I know that wax is not soluble in water so I think it doesn't react with or dissolve in water-based solvent. Some sources I have consulted say that it does react but slowly. So getting confused by different opinion, I have no idea what the truth is.

Being insoluble in water does not mean something cannot react with water or water-based solvents.

Plant waxes are made up of a number of constituents; amoung them esters, alcohols, alkenes, fatty acids and carbonyls. Especially the esters can be hydrolysed by acidic conditions, creating a more hydrophilic area which water (and more acid) can use to penetrate the layer.

Another probable reason is the fact that acidic proton transfers often involve the release of heat which will make the wax layer less viscous. All these things taken together mean that slowly but surely, the wax layer will be destroyed given enough acid rain.


Galvanic cell using NaCl, Cu and Zn, why does it work? Got a paper due tomorrow in chemistry, and my friends and I are totally stuck on this:
Basically we mixed some regular salt (NaCl) into a glass of water and stuck a zinc nail and a copper nail into the solution. We connected the nails and measured a voltage of .6V. Why does this happen?

I understand galvanic cells of other types and electrolysis to a pretty OK extent, but I can't for the life of me understand what's happening at a chemical level in this reaction. I also need chemical formulas and the likes. Thanks in advance!

I want you to try and work out the details yourself, but there's one aspect of this kind of cell that's not straightforward. At one electrode, you have a reduction reaction and at the other, an oxidation, just like any other galvanic cell. In the oxidation reaction, one of the metal electrodes is oxidized to put cations in solution. (think about which metal is more easily oxidized) In the type of galvanic cell that you may be more familiar with, a metal cation in solution is reduced onto the other electrode, but in this case, the only metal cation is $\ce{Na+}$, which is extremely difficult to reduce. Therefore, some other reduction reaction must be happening. The only other reducible substance present is water.

The two reactions occurring can be found on this table, and using the potentials in the table, you should be able to work out which reactions are occurring and at which electrodes.


Why is nitric acid a stronger oxidising agent than sulfuric acid? We were taught that nitric acid can oxidize $\ce{Cl-}$ to $\ce{Cl2}$, but sulfuric acid cannot. Is this due to its smaller size, or structure, or what?

The nitrate ion is a strong oxidant whereas the sulfate ion is a milder oxidant that only shows oxidizing properties in high concentrations. The oxidizing atom in question is the oxygen atom in the acid, not the hydrogen ion. This is because hydrogen ion is not very electrophilic, which is why non-oxidizing acids cannot dissolve copper: even copper has a higher affinity to electrons than hydrogen.

In a nitrate ion, the molecular structure is inherently not very stable, as the nitrogen atom has to bear a positive charge to maintain the molecular structure. Besides, nitrogen is in itself a very electronegative element, so that makes the nitrate ion even more unstable, hence its oxygen atom readily reacts with other atoms or molecules that oxygen can oxidize, including chloride ions:
$\ce{Cl- + O -> OCl-}$
$\ce{H+ + OCl- <=> HOCl}$
$\ce{H+ + Cl- <=> HCl}$
$\ce{HCl + HOCl <=> H2O + Cl2 }$


Adding chemical equations together This is something that may seem small, but it bothers me … I mean really, really bothers me. The following issue will be expressed through the Brønsted-Lowry theory of acids and bases and specifically through the neutralistaion reaction of ammonia and hydrochloric acid.
When I dissolve hydrochloric acid in water, the following equilibrium reaction takes place:
$$\ce{HCl + H2O <=> Cl- + H3O+}\tag{1}$$
When I then add a base such as ammonia, $\ce{NH3}$:
$$\ce{NH3 + H3O+ <=> H2O + NH4+}\tag{2}$$
Adding these 2 equations should give:
$$\ce{NH3 + HCl <=> NH4Cl}\tag{3}$$

This implies that the first reaction produces hydroxonium ions which the second reaction uses up, hence the number of hydroxonium ions in the solution should remain the same as before the addition of the hydrochloric acid.

Now let's actually asses the implications of such a procedure. First of all, assume that the reaction has gone to completion. This implies that when I mix water with hydrochloric acid, I will get hydroxonium and chloride ions produced. Subsequently, when I add the same number of moles of ammonia as hydrochloric acid, this will react with all of the hydroxonium ions present in the solution, as is shown in reaction (2). One problem though; this isn't a full reaction. If I add the same number of moles of ammonia as hydrochloric acid, this time there will be more hydroxonium ions than there were in the solution before addition of the hydrogen chloride (this what what I guess will happen, and this is the center of my problem). So how do I become sure that the hydroxonium ions cancel out when the equations are added?

When you write a chemical equation, you don't care whether the reaction has gone to completion or not. On a deeper level, none of them do. There is always an equilibrium, and there is always some starting compound left. To find out how much is that "some", you need to know the equilibrium constant and make certain calculations.

The written equation does not tell you that each and every molecule in the sample should react exactly like that. This is simply not what it is all about. It does, though, tell you another thing: if you got such-and-such amounts of products, you must have lost such-and-such amounts of the starting compounds. This is simply the conservation of mass, and there is no going around it.


What advantages does the mole have over counting large numbers using SI prefixes? I'm trying to understand why chemists use the mole unit instead of just counting and using SI prfixes to simplify the big numbers with units such as zetta- or yotta-molecules (yotta=10^24).

Here's what I've found so far, but it doesn't seem to be enough to make moles particularly important:

Reasons for moles:

You can easily approximate the number of atoms from the mass of a sample because the mass number of an isotope approximately equals the mass of 1 mole of atoms in grams.
It's a historical convention that would be too expensive or uncomfortable to change now.
You can use millimole and micromole instead of having to learn several more obscure SI prefixes like exa and peta. EDIT this reason is added from Jan's answer
Shorter descriptions, eg "mole of carbon" vs "yotta atom of carbon" (from matt_black's answer).

Reasons against moles:

For accurate calculations (beyond about 3 significant figures), the relationship between mass number and mass in the first reason above breaks down. So this is a risky thing to use.
It's a additional concept and set of facts that chemists have to spend effort learning but which describes human convention rather than nature.
We need extra conversion factors such as 1/mol which people often neglect and Faraday's constant used in Q = n(e-) x F [1] which wouldn't exist without moles.

Non-reasons for moles:

You can easily calculate the number of atoms from the mass of a sample because the atomic mass of an element equals the mass of 1 mole of atoms in grams. This relationship only exists because of the special units commonly used for atomic mass. If periodic tables listed atomic mass in grams (or perhaps yoctograms), then we could do the same calculations just as easily without moles.
In practice, we can't measure the number of molecules so we have to measure mass or volume instead and should therefore count them using a unit that's defined in terms of mass. I don't think the precise details of how a unit is defined matter for practical purposes. If you have 1 litre of an ideal gas, you still have to do calculations to find out many moles it contains just as you would to find how many molecules it contains. There is even a proposal [2] for SI to redefine the mole to be independent of the mass of any substance, indicating that keeping the definition isn't very important.
Chemists would make mistakes with those big numbers. They'd use "yotta" in the same way they use "mole", not doing calculations with the actual number it represents and not being at risk of other types of error. (from matt_black's answer) However it would be more complex replacing millimole with zetta because you'd have to remember that a zetta-atom is a milli-yotta-atom.

Here's a similar question but it's mixed in with the idea of measuring number rather than mass or volume -
Why do people still use the mole (unit) in chemistry?

This other similar question mainly addresses the uncommonness of quantities as big as 1 mole in other areas of life -
The mole is used extensively in chemistry, why not elsewhere?

[1] http://www.ausetute.com.au/faradayl.html

[2] https://en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units

Of course, it’s convention and has been so for a good century. And there is no real reason why this convention (and not a different one) happened in the first place — it is entirely conceiveable to define an ‘imperial mole’ so that the atomic mass of an element is equivalent to the same mass in ounces. 1 imperial mole of carbon atoms would then be twelve ounces of carbon atoms.

So there are basically two questions in this:

Why replace a large number with a unit just meaning ‘this certain large number’?
Why is the presently defined mole a good choice for this unit?

The thing about large numbers is that they are large. Almost everybody who grew up in a metric country can name at least three SI-prefixes: kilo, centi and milli. Thanks to IT, many people now also know mega and giga (and maybe tera), even if they don’t use metric units at home. But tera only gets you to $10^{12}$. We need some $10^{21}$ for moles.

I often work with milli- or micromoles of substances in my research. In plain numbers, that’s $10^{20}$ or $10^{17}$ — I don’t know those prefixes and thus would have to learn an entirely new subset. With the mole, everything one uses in the lab will nicely fall into something between nano and kilo.

It also helps to have a single unit there. Molar mass is expressed in grams per mole, concentrations in moles per litre and many more. If there were no unit, it would be simple grams per 1 or 1 per litre — precisely the reason why some people prefer to use rad or some other way to show radians rather than just writing the number. If the unit is there, you’re unlikely to forget it, you know if your calculations are good and more. If there was no name for this unit, it would have to be invented.

So if the mole didn’t exist, it should be invented for simplicity.

The good thing about the size definition of the mole is, as noted above, that it brings everything into one general range. Whether it’s mass, volume, concentration or amount, every unit is going to be prefixed by only a small subset of the SI-prefixes: kilo, milli, micro, maybe nano. Thankfully, those are the ones that are most used in everyday life, too (excluding nano and maybe micro).

It doesn’t really matter where one ends up. If the mole had originally been defined imperially,[1] that would be fine, I wouldn’t have memorised $12.01\,\frac{\mathrm{g}}{\mathrm{mol}}$ for carbon but $340.48\,\frac{\mathrm{g}}{\mathrm{mol}}$. That would create significantly larger molar masses, but that shouldn’t be a problem it should only mean that nano is more prevalent.

The thing about the definition of the unit is that it doesn’t bother $99.9\,\%$ of the scientists working with the unit. Seconds are defined (by SI) according to a number of transitions occuring in some weird isotope that I wouldn’t even know how to measure. I would explain a second by saying it’s the 86400th fraction of a day if someone asked me. Same meaning, different exactness. If the moles are soon defined by mere counting rather than weighing atoms then so be it. Nothing will change for me in practice. (Maybe the fourth digit of a molar mass but I don’t really count those.) So as long as a new definition doesn’t break anything, we’re good to carry on.

Notes:

[1]: With imperially, I meant to assume the following definition using imperial units:

  One mole is the number of atoms in exactly $12.00~\mathrm{oz}$ of atoms of the carbon isotope $\ce{^{12}C}$.


Why does bromide not react with water when methyl ammonium bromide is dissolved in water? In the following reaction, the salt $\ce{CH3NH3Br}$ is formed:

$$\ce{CH3NH2 + HBr -> CH3NH3Br} $$

In water this dissociates completely into $\ce{CH3NH3+}$ and $\ce{Br-}$

In my chemistry book, it says that the $\ce{CH3NH3+}$ dissociates further in water into $\ce{CH3NH2}$ and $\ce{H3O+}$ (meaning that $\ce{CH3NH3Br}$ an acidic salt) according to the following:

$$\ce{CH3NH3+ + H2O <=> CH3NH2 + H3O+}$$

Fair enough, but why does the $\ce{Br-}$ not do the same and dissociate in water?

$$\ce{Br- + H2O <=> HBr + OH-}$$

In this way, the dissociation of the $\ce{CH3NH3Br}$ salt would not have so much of an effect on the pH, because the quantity $\ce{OH-}$ ions has also increased?

The key is comparing $\mathrm{p}K_\mathrm{a}$ values. A good resource for this is the Evans $\mathrm{p}K_\mathrm{a}$ table. Unfortunately, the table does not contain $\ce{CH3NH3+}$, but in the section protonated nitrogen there is a value quoted for $\ce{C2H5NH3+}$ — which is chemically similar enough to methyl ammonium to assume the $\mathrm{p}K_\mathrm{a}$ values to be essentially identical.

\begin{array}{lr}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}\\ \hline
\ce{EtNH3+} & 10.6\phantom{0}\\
\ce{HBr} & -9.00 \\
\ce{H3O+} & -1.7\phantom{0} \\ \hline\end{array}

Notice how the $\mathrm{p}K_\mathrm{a}$ values differ by almost $20$ logarithmic units. $\ce{HBr}$’s $\mathrm{p}K_\mathrm{a}$ value is even lower than water’s by $7$ to $9$ logarithmic units depending on the source used. Thus, while the amine exhibits extensive acid/base chemistry in water, bromide will not be protonated to any noticeable extent. Or you could say that your final equation will be strongly favouring the reactant side.

I’ll also point you to a recent answer of mine wherein I calculated the concentration of $\ce{HF}\ (\mathrm{p}K_\mathrm{a} \approx 3.17$, so notably weaker than $\ce{HBr}$) in a 1 molar solution of $\ce{NaF}$. The case is very comparable to your’s.


Can aniline be made with the method described in Dr Stone? In Dr Stone, the main character makes aniline by rinsing coal tar with HCl, then mixing in ethyl acetate. See the image below. Is a method like this possible? What is occurring here?

Believe it or not, aniline was actually isolated from coal tar way back in 1834. A chemist called Friedlieb Runge isolated aniline along with phenol from coal tar. He discovered that aniline produced a beautiful blue color when treated with chloride of lime (calcium hypochlorite). He named it cyanol/kyanol and that was the birth of the first aniline dye but it was made a dye by the exhaustive work of another chemist, William Perkin.
Perkin distilled the components of coal tar like benzene, aniline, toluene, anthracene, phenol, naphthalene and other aromatic hydrocarbons. He first worked around toluene to make allyl toluidine which he then transform to quinine by oxidizing it with a mixture potassium dichromate and sulfuric acid. He was left with a useless reddish-black powder. He then took aniline and repeated the experiment. He was left with a black gooey substance. Disappointed by the results, he took the test tube and washed it off but then came out a beautiful looking purple residue which hung off at the wiping cloth. He named the dye "mauveine". Perkin was so amazed by the result that he started working with other distilled products hoping to make other dyes$\ce{^{[4]}}$ . His discovery revolutionized the dye industry. Perkin proposed the following statement:

Benzene, toluene, and other components of coal tar were colorless
because they absorbed ultraviolet light undetectable by the human eye.
But if those aromatic hydrocarbons were treated with an acid to create aniline or another amine, after some additional steps the
newly synthesized molecules very efficiently absorbed light particles
from specific wavelengths in the visible spectrum.

So, Senku (the main character of Dr. Stone) being well-versed in chemistry knows what he's talking about.
Notes and references

https://trc-leiden.nl/trc-needles/materials/dyes/aniline
https://www.scientificamerican.com/article/toms-river-excerpt-on-aniline-dye/
https://www.cavemanchemistry.com/cavebook/chtar.html
Magentas from toluene, reds from anthracene, pinks from phenol, and indigos from naphthalene.
Dr. Stone is an amazing anime. I recommend fellow readers to check it out.


Acidity of ester and benzoic acid Can the oxygen in an ester participate in resonance donation? I think so, although the donation is slight. The hybridization is correct. However, resonance withdrawal by the ring via the oxygen doesn't seem possible as my picture depicts. Therefore, because of resonance donation from the oxygen in the conjugate base of the ester, would that make benzoic acid the stronger acid? It seems that the resonance donation would destabilize the conjugate base of the ester.

Can the oxygen in an ester participate in resonance donation?

Sure, structure III in the following diagram is a valid resonance structure.

  However, resonance withdrawal by the ring via the oxygen doesn't seem
  possible

No, that's possible too.  It might not be a major contributor, but structure V is a valid resonance structure.

  would that make benzoic acid the stronger acid?

I think you're asking which is the stronger acid, benzoic acid or the half-ester of carbonic acid. Resonance structures like III and VII will stabilize both the acid and the corresponding carboxylate anions.  One molecule doesn't appear to have a more resonance stabilized starting acid or a more resonance stabilized carboxylate anion than the other.  So resonance arguments don't seem to suggest that one molecule should be more acidic than the other.  This conclusion, that resonance factors do not control the acidity in these two compounds, can also be arrived at by noting that the starting acids and particularly the carboxylate anions are analogous to the allyl MO system which has a node on the central carbon atom in the HOMO ($\ce{\Psi 2}$). The node effectively  disconnects the substituent (-OR or phenyl) from the MO, meaning that the substituent cannot interact via resonance with the MO.

Inductively one can argue that the electron withdrawing oxygen in the half-ester would destabilize the already positively polarized carbonyl carbon in the starting acid and then also provide some inductive stabilization of the electron-rich carboxylate anion, more so than a phenyl group would.  So perhaps, based mainly on inductive arguments, one could argue that the carbonic acid half-ester would be more acidic than benzoic acid.

I couldn't find any pKa values for carbonate half-esters, but we can compare the pKa of

carbonic acid 3.6

to

benzoic acid 4.2

The values are close, as expected - even in the right direction, but that's probably just luck.


Why is gas pressure measured in mmHg? Pressure can be measured in $\pu{mmHg}$ which is equal to the pressure exerted by a “column of mercury” of a height expressed in $\pu{mm},$ at $\pu{0 °C}$ under Earth’s normal gravity.

Does the diameter of a tube that contains the mercury affect how high the mercury will rise? Why isn’t pressure measured in $\pu{mmHg^3}$ or some other unit of volume of mercury?

History and because of how pressure works

The original reason why mm of Hg are used is because of how chemists first started to measure and understand atmospheric pressure. The idea of the vacuum and atmospheric pressure is surprisingly new. Otto von Guericke demonstrated the power of atmospheric pressure by showing that two horses could not separate a brass sphere containing a partial vacuum only in the 17th century (remember a vacuum doesn't suck the atmosphere pushes).

This kicked off a vast amount of experimentation. Around the same time scientists like Torricelli and Pascal realised that they could measure atmospheric pressure by filling a closed tube with mercury and inverting it in a bath of mercury (you can do this with any fluid but mercury is about 13 times denser than water so the tube only needs to be a meter or so long; with water it needs to be ~10m long which is an inconvenient length in an indoor laboratory). The height of the mercury in the tube is a measure of the external pressure as a vacuum is formed above it in the closed tube. This simple instrument was used by a friend of Pascal to show that atmospheric pressure drops with height when he took a barometer up France's Puy de Dome mountain.

The diameter of the tube is irrelevant as pressure is weight/area (do the calculations and tube diameter cancels out).

We now call a tube like this a barometer. And the height the mercury rises to is a convenient measure of atmospheric pressure and can be read off using a simple ruler. We could use mm of water, but would need a much longer tube and an unusually inconveniently sized ruler. Hence why mercury is the fluid and a convenient way of measuring pressure is mm of Hg.

We tend to use modern SI units for pressure these days, not least because the wide use of mercury has been banned and calculations in consistent nits are far easier.


Finding formula of hydrated potassium chromium(III) sulfate through experimentation I'm in Year 13 and this is a past paper question which I don't have a mark scheme for and I'm confused on the chemistry.

  Hydrated potassium chromium(III) sulfate has the formula $\ce{KCr(SO4)2.xH2O}$

  An experiment using heating can determine $x$.

  Give a step by step description of how you would carry out the experiment

  Include all measurements taken

  Outline how you would use your results to find $x$.

  You are provided with solid $\ce{KCr(SO4)2.xH2O}$ and apparatus usually found in a school laboratory

I think mostly I want to know how would begin with this type of question, like what part of my knowledge do you think this is testing me on because I don't know where to start.

All the clues you need to solve the problem have already been given. A key clue is the second line; the word heating. What do you know about heating and how it affects water content? Use this as a starting point.

Thenceforth, the question’s solution should suggest itself.


What are some of the simplest chiral molecules in biological systems I have understood generally for a while there is a significant preference in "life" for so-called "left-handed" (chiral) molecules, though I'm reading things now that say "life" sugars are right-handed, while "life" amino-acids are left-handed.

I'm wondering: how simple do these molecules get (particularly the molecules that are in intelligent life, if there happen to be any that are exclusive to intelligent life)?

Don't get hung up on the "left-handed" and "right-handed" designations for amino acids and sugars. Those are just conventions for distinguishing one mirror image (enantiomer) of a chiral molecule from another. It just so happens that based on a particular way of drawing (Fisher projection) amino acids and sugars that it appears that the configurations are "left-handed" or "right-handed."

As for how simple chiral biological molecules get, the simplest ones that I can think of are alanine or lactic acid. I'm not sure what you consider "intelligent", but these are likely to be found in all life (as we know it).


Why is it that the least reactive metals are the best electrical conductors? Silver, Gold and Platinum are amongst the best conductors of electricity, but also the amongst the most unreactive. Since electrical conductivity depends on the number of delocalized electrons (along with resistivity), if an element gives up its electrons readily (making it highly reactive), then shouldn't that mean it will be a great electrical conductor? Why isn't Cesium the best conductor?

One of the reasons why non reactive metals are good conductors is that they are good at staying as metals. Most metals react with the atmosphere to form oxides. And the majority of oxides are insulators or semiconductors. Of course there are few exceptions to this rule.

Also, just a note: calcium and iron have better conductivities than platinum.


What are the conditions for using skeletal replacement ('a') nomenclature? The Blue Book 2013 in P-63.2.4.1 gives “1-methoxy-2-(2-methoxyethoxy)ethane” as the IUPAC preferred name for $\ce{CH3OCH2CH2OCH2CH2OCH3}$ and comments,

  Skeletal replacement (‘a’) nomenclature [method (4)] generates preferred IUPAC names, when the conditions for using this type of nomenclature are met (see P-15.4); otherwise substitutive nomenclature must be used.

It then names $\ce{CH3OCH2CH2OCH2CH2OCH2CH2OCH3}$ as “2,5,8,11-tetraoxadodecane.”

My question is why skeletal replacement nomenclature was used in the latter case but not the former—why the conditions for using skeletal replacement nomenclature are not met by the former molecule.

P-15.4 states that

  “Skeletal replacement (‘a’) nomenclature is used in two ways:

  “ (a) …

  “ (b) to generate simpler names for heteroacyclic structures than those formed by following the principles of substitutive nomenclature, for example, in naming polyamines, polyethers, etc., by replacing carbon atoms in corresponding acyclic hydrocarbons” (89).

The former molecule seems like an instance of a polyether, and the name “2,5,8-trioxononane” seems simpler than the given preferred name, so it seems that this molecule meets the this criterion.

What factor led to the use of substitutive nomenclature where replacement nomenclature seemed to apply?

Rule P-51.4, not P-15.4, specifies the conditions under which skeletal replacement nomenclature generates preferred IUPAC names.

  P-51.4.1.1 Skeletal replacement (‘a’) nomenclature rather than substitutive or multiplicative names must be used to generate preferred IUPAC names for acyclic structures when four or more heterounits are present in an unbranched chain containing at least one carbon atom and when none of the heteroatoms constitute all or part of the principal characteristic group of the compound.

The former compound has only three oxygen heteroatoms, not enough to require the use of replacement nomenclature.


Why are ionic reactions very fast in aqueous medium? (No, I am not talking about why Ionic Reactions are faster than Covalent reaction.)

This question has me stumped. I mean, at first it looks obvious; water is polar in nature, but what exactly does that help with?  Also, can ionic reactions even happen in non-aqueous mediums?

Water has a permanent dipole character, which means it has a slight positive and negative charge permanently in it (due to electronegativity difference). So, in case of an ionic reaction, it helps in easier solvation of ions by pulling them with its charged centres (positive centre attracts negative ions and negative centre attracts positive ions)
Aqueous is a term specially for water as a solvent. Any polar solvent can support ionic reactions.


Classical Least Squares in UV-Vis Absorbance Experiment How do I find the minimum number of standards needed to calibrate a spectrophotometer using CLS? We never talked about CLS in class so I'm not sure of my approach. Would 1 standard solution per analyte be enough? We'll be plotting absorbance vs. concentration then finding the residuals from the best fit line.

That depends on the nature of the analyte and matrix. Ideally (Beer's Law) A should be linearly proportional to concentration but that may not be the case. As an example of this in the water industry barium chloride is often added to samples to determine how much sulfate ion they contain. Spectrophotometers are used to determine how much barium sulfate is in suspension. The relationship between A and sulfate concentration is not linear in this case.

What you should do is make up a series of standards spanning the range of concentrations you are interested in, measure the absorptions and plot A vs. c. Now start fitting curves, starting with linear and progressing to higher order until doing so does not make the residuals look any noisier. That gives you the order of the response curve you need and the number of standards required is one more than the order.

Ideally, the matrix wont interfere and Beer's law applies. In that case when c = 0, A = 0 and you only need one standard.


Why are square planar coordination compounds with four different ligands optically inactive? Why do square planar coordination compounds of type $\ce{[Mabcd]}$ not show optical activity, although they contain 4 different ligands (i.e. chiral central metal atom)?

Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from Wikipedia):

A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, results in the same molecule[...] For tetrahedral molecules, the molecule is chiral if all four substituents are different.

So for 4-coordinated carbon atoms which are tetrahedral having four "ligands" actually establishes optical activity. But in the case of a square planar metal complex you always have a mirror plane, namely the one made up by the central atom and the ligands:

And as a result you don't get optical activity. But it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited.


Kolbe Schmitt followed by methylation The first step in this synthesis is a Kolbe Schmitt process, but why is two equivalents of base needed? Also, what is the purpose of the two equivalents of methyl iodide? I assumed that there will be the protection of the phenol (to create a methyl ether), which would justify the two equivalents of base while converting the carboxylate into a methyl ester. However, the base is shown in separate steps, which makes me doubt if base would be available in the second step. Also, another problem is that if you follow the sequence (2 Grignard reagents used to convert the ester to a tertiary alcohol, followed by dehydration, followed by hydrogenation of the double bond just formed, followed by a Friedel Crafts acylation), you get to compound R, which only has 13 carbons, when if you had methylated the phenol, you would have 14. Does this mean that the hydrogenation somehow deprotected the methyl ether? Even more confusing, there is methylation occurring from R to S, which tells me that the phenol was protected at some point. How is this possible?

EDIT: Here is the structure of the final product. The problem was about derivatives of natural products from Salvia plants, but I could not find a total synthesis of this compound anywhere. Also, there is no easy name for it so I couldn't search it well. If you can find the source, please let us know. Note that there are other steps in the synthesis that I didn't include in this question.

Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol O undergoes thermal dehydration catalyzed by the acidic phenolic hydroxyl group. Steps P --> Q are straightforward. Step R --> S methylates the phenol hydroxyl and excess KOH insures that any ester that is formed is saponified. Clemmensen reduction of S and remethylation afford T.
Addendum: The OP's comment reveals that T is a carboxylic acid, not a methyl ester. Presumably the Clemmensen reduction effected some demethylation which is the reason for the last methylation step.


How can I easily explain to a layperson the difference between types of "contains" I had a conversation with somebody who had somewhat confused ideas about food. He thought that "all food contains sugar" and when I gave a bottle of oil as the counterexample, he said the glycerol is the sugar there.

Now let's forget for a minute that glycerol is not a sugar. The point is, glycerol is indeed a "building block" of a triglyceride, and glucose is a "building block" of glucose. But still, if you are talking about a process which needs sugar, starch won't work unless you cleave it off first. Just like starch won't spontaneously conduct electricity or emit beta radiation even though it's chock full of electrons.

Given that we have somebody with a very basic level of knowledge, and who is already trying to wrap his head around several confusing concepts at once, what would be an intuitive way to get him to really grok that, just because a given atom or group is present in the structure of a molecule, it does not mean that a substance "contains" this atom or group in the same way that a wet sponge contains water? And that without going off on a long tangent of theory?

I would say something like this:

All matter -- that is, all substances -- are made of atoms.  The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.

Chemists refer to the "types" of atoms as "elements".  For example, vegetable oil and table sugar (sucrose) are both composed of three types of atoms only: carbon, hydrogen, and oxygen.

The arrangement of the atoms in the two materials differs drastically.  (You could refer to structure diagrams at this point if your friend is interested.)  First, oil has much less oxygen and slightly more hydrogen per carbon atom.  Second, the way the atoms are connected -- their arrangement -- is very different.

Chemists sometimes give a name to a group of similar arrangements of atoms.  For example, "sugars" are compounds that (a) have carbon, hydrogen, and oxygen atoms in ratios that are approximately $\ce{CH2O}$, and (b) often form ring structures, as seen in table sugar, glucose, and corn starch.  But the key words are "approximately" and "often" because some molecules that chemists call sugars don't satisfy these rules.  Biochemists interested in metabolism sometimes refer to molecules that can "easily" be converted into "true" sugars as sugars.

Naming in chemistry is complicated by the fact that chemists use words for different purposes: to indicate metabolic similarity, to indicate structural similarity, or to indicate similarity in reactivity.  Those different purposes sometimes lead situation where chemist B says that what chemist A calls a "sugar"  isn't really a sugar.  But that's just an argument about humans' imperfect way of naming things, not about chemistry.  There is a lot of wiggle room in the words "approxiately", "often", "similar", and "easily".  So if we focus on metabolism only, the glycerol part of the vegetable oil arrangement (i.e. structure) is a bit like a sugar.  But it is dissimilar from sugars in terms of both structure and reactivity.

Again, what ultimately matters is (a) the type and (b) the arrangement of the atoms in a substance.  Those two things are together necessary and sufficient for a complete description of all substances.  Whether to call particular arrangements of particular atom types "sugars" or not is something people can argue about.


Safest Methylating Agent I am planning on doing some synthesis in the near future and due to safety purposes I am looking for a methylating agent that is both effective and safe. The substrate is a deprotonated amine and the solvent is water. I understand that DNA methylation is involved in the toxicity of the compounds, but there should still be ways to minimize risk. Ideally, the compound should be a powder that is soluble in water. Can any ideas that work for this substrate and solvent be thrown on the table?

I was planning on using methyl iodide solely as a last resort.

What you are looking for does not — to the best of my knowledge — exist and likely won’t ever exist.

Methylating agents are usually classified as toxic as they are usually able to penetrate cells and nuclei and methylate DNA bases — although they will also methylate most nucleophiles on the way there. This can lead to all sorts of things will the shadow of cancer overarching it all. Off the top of my head there are a few methylating agents that might not penetrate the skin such as methyl Meerwein salt. These, however, are so reactive that they just react before they reach the skin.

Secondly, methylating agents will likely never work in water. Water itself is a nucleophile and a (Brønsted) acid. If you use an electrophilic methylating agent such as methyl iodide or methyl Meerwein salt, water is able to nucleophilically attack these to generate methanol and then dimethyl ether, thereby deactivating your methylating agent. If you choose to use a nucleophilic methylating agent such as methyllithium instead, water will supply a proton to liberate methane (and lithium hydroxide in the case of methyllithium) instead. So practically no methylating agent will survive in aquaeous solution.

Finally, your proposed reaction pathway is skewed from the beginning: you suggest using a deprotonated amine in water as a reactant. However, the $\mathrm{p}K_\mathrm{a}$ of amines is typically around $30$, depending on the substituents. ($\mathrm{p}K_\mathrm{a}(\ce{LDA}) \approx 35; \mathrm{p}K_\mathrm{a}(\ce{LiHMDS}) \approx 25$; source: Evans $\mathrm{p}K_\mathrm{a}$ table.) On the other hand, the $\mathrm{p}K_\mathrm{a}$ of water is $14$ or $15.7$, depending on the source you consult. This is a difference of $10$ to $20$ logarithmic units depending on the type of amine — meaning that amines are practically never deprotonated in aquaeous solution.[1]

If you mistyped and are using a deprotonated amide, things won’t get that much better ($\mathrm{p}K_\mathrm{a}(\ce{RCONH2}) \approx 25$).

Therefore, you will likely have to give up your plan and go for a more traditional methylation method.

[1]: Note you might argue that the free base, $\ce{R3N}$ is enough, so all you want to do is lower the $\mathrm{pH}$ of water to below $\approx 10$ so that the protonated form $\ce{R3NH+}$ is neglegible. However, at these $\mathrm{pH}$ values, water is also deprotonated to a non-neglegible amount and a hydroxide ion is a better nucleophile than a free amine base.


How exactly do you measure the free energy of a reaction? It is to my understanding that ΔG can be positive or negative given whether the reactant is exothermic or endothermic.  How would you go about measuring the energy (in joules) of a reaction using the heat of the reaction?

How would you go about measuring the energy (in joules) of a reaction
  using the heat of the reaction?

You can measure the heat of a reaction (the enthalpy of reaction, $\ce{\Delta H}$) using a calorimeter, but the enthalpy of the reaction is only one part of the free energy of the reaction ($\ce{\Delta G}$). $\ce{\Delta G}$ is related to $\ce{\Delta H}$ through the following equation $$\ce{\Delta G = \Delta H} - T \Delta S$$ where $\ce{T}$ is the temperature and $\ce{\Delta S}$ is the change in entropy for the reaction.

To use a calorimeter you put your reactants into the calorimeter and let the reaction proceed.  The calorimeter is thermally isolated from its surroundings, so any heat given off (exothermic) or absorbed (endothermic) during the course of the reaction can be accurately measured.  Knowing the moles of reactant you put into the calorimeter and the heat change in joules (using the measured $\ce{\Delta T}$ and the known heat capacity of the calorimeter) you can determine $\ce{\Delta H}$ in joules/mole.  Here is a link that provides more detail about calorimeters along with some problems that are worked out in detail.


What is the oxidation state of C in this lewis structure? What is the oxidation state of $\ce{C}$ in this lewis structure?

I know how to find it: based on the number of electrons leaving/coming to the most/least electronegative element. But it s the $\ce{OH}$ that gets me confused...

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Foreword: As mentioned in the OPs comments, electronegativity and oxidation states are atom properties, not group properties.

Comparing Electronegativities

This table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{C} & 2.55\\
\ce{O} & 3.44\\
\hline
\end{array}$$

Hence, the following applies:

$$\chi_{\mathrm{Pauling}}(\ce{H}) < \chi_{\mathrm{Pauling}}(\ce{C}) < \chi_{\mathrm{Pauling}}(\ce{O})$$

Looking at heteroatomic Bonds ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$)

Since the oxygen atom has a higher electronegativity than the carbon atom, it attracts the bonds atoms more, resulting in an polarized bond. Presuming one would apply pulling force to the two atoms in a $\ce{C-O}$ or $\ce{C=O}$ bond until the atoms separate, it is assumed that the bond cleavage is heterolytic, thus allocating the bonding electrons to the oxygen atom.

The same applies to $\ce{C-H}$, $\ce{O-H}$, or any other heteroatomic bond ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$).

Putting the jigsaw puzzle together

Now, having a look on the complete structure of the molecule in question, and applying the aforementioned rule to each bond:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$

With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (Never forget the lone pairs).

This will lead to the following oxidation states for each atom:

$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H-1} & 1 & 0 & +1\\
\ce{H-5} & 1 & 0 & +1\\
\hline
\ce{C-2} & 4 & 2 & +2\\
\hline
\ce{O-3} & 6 & 8 & -2\\
\ce{O-4} & 6 & 8 & -2\\
\hline
\end{array}$$

The structure with assigned oxidation states:

The oxidation state of the carbon is $\mathrm{+II}$.


How do I import CSVs to OpenChrom? My end objective is to compare GC outputs of samples in two groups using PCA. I have bene given the data in Excel format but have saved to CSV.

Whilst I can locate CSV files through OpenChrom, they all appear blank when I do so.

Having spend a lof of time trying to find the information, I cannot find the relevant info about how my CSV (or Excel) file should be formatted for OpenChrome.

I have columns for retention time (labelled "RT") and area (labelled "Area").
I have tried a simple file with two columns (RT and Area) for one sample. I have also tried files with multiple samples and columns, both with and without sample no. headers. (see pictures).

Whatever option I try, the charts in OpenChrom appear completely blank. I don't know if it's the data itself, the headings, the process I'm going through, not a clue what's going on to be frank.

Lastly, I would like to apologise that this may be a basic/naiive question for the forum. I've essentially been dumped with the work of running PCAs comparing GC data sets with little to no experience in doing so.

Thank you.

Oh my God! I hope I help you in time. I just spent hours dealing with the same issue, and I finally figured it out. Basically, OpenChrom will only read CSVs formatted in a very specific way, so you will have to change your .csv files to meet these requirements:

The first three columns MUST be have the headers 'RT(milliseconds)', 'RT(minutes) - NOT USED BY IMPORT', and 'RI', exactly as I have written them.
The second column ('RT(minutes) - NOT USED BY IMPORT') does not need any values; as its name suggests, it is not used by the import, but it still needs to be there! You will have to convert your times to milliseconds to populate the first column. The third column ('RI') is for retention indices, if you do not have them you must populate the entire column with zeros (0.0).
Every subsequent column must have an integer header representing the mass of the signal (e.g. '176' for m/z 176).
No trailing comma at the end of the line! E.g. a line of values should look like

  1,2,3,4

  5,6,7,8

and NOT

  1,2,3,4,

  5,6,7,8,

In Excel, the only way for my files to save like this is if I save them as CSV (MS-DOS).

I hope this helps!


"The centre of positive and the centre of negative coincide" What does that mean? I'm doing AS level chemistry right now and my teacher told us that when a molecule is non polar , the centre of positive and the centre of negative coincide , I don't really get what he means by that .

Each molecule has negatively charged electrons, and positively charged nuclei.

If the centroid of the electrons' charges (electron probability density) is different than the centroid of the nuclear charges, then the molecule will have a permanent electric dipole moment.


Determine and draw planes and axes of symmetry of a molecule using software (VESTA or Mercury) I designed the $\text{MoS}_2$ molecule (bilayer), with polytype 2H in the VESTA. My goal is to draw the plans and axes of symmetry of the bilayer set. According to the information in the literature, this molecule belongs to the $\text{D}_{3d}$ point group. How can I draw and determine the planes and axes of symmetry in VESTA?

These are the parameters I used in VESTA:

Mercury Attempt:

I exported the .vesta data to a .cif file.

Table with space and point groups for $\text{MoS}_{2}$ (taken from https://iopscience.iop.org/article/10.1088/0953-8984/28/35/353002/ampdf) (see 2H phase for $\text{MoS}_{2}$):

I do not know VESTA well enough to provide an answer for this program.

However because your data are about a crystal structure I suggest to export your model in the .cif format.  CCDC's freely available version of CCDC's Mercury is able to read this file type, and add to its visualizations the symmetry elements of the unit cell (Display -> Symmetry Elements).  There will be a new menu where you may choose among the symmetry elements the program is able to recognize, and how they are displayed:

For example only the centres of inversion:

The export options in Mercury include bitmap .png, ray-tracing PovRay .pov, and .vrml / .stl.

Given lattice constants and space group $P6_3/mmc$, do not forget to compare your model with the literature known models for example in the freely accessible COD about experimental data (entries #1010993, #1011286, #1531960, #9007660, #9009144); and consider contributing to its sibling TCOD about theoretical data and MPOD.


Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets? Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets? I know for sure it has to do with the van der Waals force and the fact that the molecules are huge, but why? Why does a big molecule break down before the van der Waals forces break?

Thermosets are liquids or soft solids which irreversibly 'set' or 'cure' when heated. This 'setting' produces covalently  bonded cross-linkages between adjacent molecular chains, resulting in in their mechanical rigidity.

Thermoplastics too are formed from linear polymer chains, but their molecules typically have very long chains with high molecular weight, with very few cross-linkages. The weaker van der Waals bonds which link adjacent molecules in thermoplastics are easily broken when heated, allowing the chains to 'slide' past each other, thereby giving them the property of 'softening' when heated.

  Why does a big molecule break down before the van der Waals forces
  break?

In the case of polytetrafluoroethylene (PTFE), we have a very long, linear polymer chain of carbons covalently bonded with fluorine. PTFE is quite unique because it doesn't have any hydrogen atoms. Instead, it has these large highly electronegative fluorine atoms.

The bonds within the PTFE molecule are very strong (C-C bonds are 607kJ/mole and C-F bonds 552kJ/mole, much stronger than the typical C-H bonds found in other plastics such as polyethylene).  In effect, the PTFE molecule is like a 'rod' of carbons bonded to a surrounding 'blanket' of fluorine atoms (arranged in a slight helical twist).

Although the van der Waals bonds which form between adjacent the adjacent linear 'rod' molecules of PTFE are much weaker than the C-C and C-F bonds within the molecules, these van der Waals forces are able to combine in an additive manner, so the long straight PTFE 'rods' are able to be 'packed' together in a hexagonal arrangement, a bit like pencils in a box. This results in PTFE having a relatively high density and melting point of $327^o C$.

But when PTFE melts, its extremely long molecular chain have a high probability of entanglement, resulting in a extremely high melt viscosity. Although it's melt viscosity varies with the shear stress applied and temperature, in the temperature range between $360^o C$ and $380^o C$ and shear stresses of $0.1MPa$, commercial samples of the granular polymer display viscosities of the order of $10^{10}Pa.s$. This is exceptionally high compared with most other engineering thermoplastic melts. In fact, at atmospheric pressure, it hardly seems to 'melt' in the conventional sense of 'becoming a flowing liquid', although it becomes slightly softer, it forms a near-clear gel with very high elasticity.

Even below it's melting point, PTFE will start to decompose at $260^o C$, releasing toxic compounds through sublimation. When heated above $360^o C$ thermal decomposition is more rapid and the PTFE degrades noticeably. As a consequence, PTFE can not be processed by many 'conventional' thermoplastic techniques, such as injection molding. Instead, it is processed in a finely powdered or granular form, using modified powder metalurgical techniques such as cold-molding, sintering, compression and extrusion, to fabricate rods, tapes, wire, tubing, sheeting and other profiles which are typically suitable for further machining.

Above $480^o C$ PTFE thermally decomposes in air to release a mixture of vapours and particulates, including carbonyl fluoride, hydrogen fluoride, tetrafluoroethylene and other fluorinated hydrocarbon products, including a small amount of some highly toxic species such as perfluoroisobutylene.

The precise nature of the reactions responsible for this decomposition is still subject to research, but it appears likely that it is initiated by random-chain incisions in the PTFE chain, followed by depolymerization.


Reversibiliy and equilibrium My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".
The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $\ce{Tsys=Tsurr}$ at every instant of the process.
Atkins-Physical Chemistry, page 77 states, while explaining the entropy change of the surroundings: "furthermore, because the temperature of the surroundings is constant whatever the change, for a measurable change,$\ce{\Delta}Ssurr=\dfrac{q_{surr}}{T_{surr}}$".
But then this treatment seems to contradict the fact that the system and surrounding are in thermal equilibrium! Since $\ce{Tsys=Tsurr}$ implies that $\ce{Tsurr}$ is not constant as long as the process isn't isothermal.
I therefore believe that my first statement is incorrect.If it isn't incorrect,then what am I misinterpreting?

When a system is in thermodynamic equilibrium, there is no tendency for spontaneous change. In other words everything stays the same, the composition, temperature, pressure do not change over time, etc.
Therefore:
Tsurr is constant
and
Tsys is constant.
If you make an infinitesimal change, the system will tend to revert to the equilibrium.
So there is no issue.
When you are talking about a reversible process in thermodynamic terms, you're talking about a system which can be taken from its initial state to another state, and then back to the initial state without any net change to either the system or the surroundings.
The idea being that you make an infinitesimal change to some aspect of the surroundings, the system changes to a new state which is again in equilibrium with the surroundings, lather, rinse, repeat. Of course it would take infinite time for the process to be completed.


Stritations in a discharge tube I am reading about A cathode ray discharge tube and I still do not fully understand why that as you keep on lowering the pressure in the chamber it creates striations, Faraday's dark space and Crooke's dark space and why it does not just stay as a constant stream of light? Also why does the gas have to be at a low pressure in the first place for the constant column of light to appear?

To verify my questions are:

Why when you lower the pressure to a certain extent you can see striations, Faraday's dark space and Crooke's dark space and why doesn't it just stay as a constant column of light?
Why does the pressure in the chamber have to be very low for you to be able to see a constant column of light?

The most important question is why do we need low pressures? If you recall kinetic theory of gases, there is a concept of mean free path. This quantity roughly tells us the distance a particle would travel before colliding with another particle. If you want the electrons to travel from one end to the other, you need to provide a large mean free path to them. How can one obtain that? Simply by removing "matter" from the cathode ray tube. Practically this is achieved by vacuum pumps. Although nobody gives importance to vacuum pumps, it was this technology which allowed great progress in the field of cathode ray experiments and hence understanding matter.

Now, the "glow" phenomenon is rather a spectroscopic problem. Electron don't glow green/pink etc, it is the matter which starts to glow when it interacts with electron. The glow can originate from several sources such as recombination phenomenon, fluorescence of the glass, a fluorescent zinc sulfide screen placed inside, or even Xrays. When the pressures are moderate, other phenomena like which you quote as striations, appear. This arises from accelerated electrons exciting the gas particles and then losing energy (dark space). This video explains some of it. Cathode ray striations. However it is not a trivial phenomena. It took ages to understand it.
The observation of striations indicate that there is enough gas (low vacuum) in the discharge tube.


Pressure turning out to be zero? I was studying for my freshman chemistry course, and ran into a mind-boggling contradiction.

We know, at equilibrium, entropy is at a maximum:
$$\left(\frac{\partial S}{\partial V}\right)_U=0$$

From the first law

$$dU=TdS-pdV$$

which leads to

$$\left(\frac{\partial U}{\partial S}\right)_V=T$$

From the cyclic rule

$$\left(\frac{\partial U}{\partial V}\right)_S\cdot\left(\frac{\partial V}{\partial S}\right)_U\cdot\left(\frac{\partial S}{\partial U}\right)_V=-1$$

$$\left(\frac{\partial U}{\partial V}\right)_S=-\left(\frac{\partial U}{\partial S}\right)_V\cdot\left(\frac{\partial S}{\partial V}\right)_U$$

Using the result ($\left(\frac{\partial U}{\partial S}\right)_V=T$) it gives

$$\left(\frac{\partial U}{\partial V}\right)_S=-T\cdot\left(\frac{\partial S}{\partial V}\right)_U$$

But we know that $\left(\frac{\partial S}{\partial V}\right)_U=0$, so

$$\left(\frac{\partial U}{\partial V}\right)_S=0 \tag{1}\label{eq1}$$

So far so good.

Writing the total differential of $U$

$$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV$$

and comparing it to

$$dU=TdS-PdV$$

We get:

$$\left(\frac{\partial U}{\partial V}\right)_S=-P$$

Now I'm in real trouble, because equation $\eqref{eq1}$ says this quantity is zero, and even in this general series of thermodynamic equations, I have reached the conclusion that

$$P=0\,\,\,\,\,\,\text{(?!)}$$

I would appreciate if someone could help identify what the fundamental problem is? I've checked and rechecked my calculations and assumptions, but I'm sure I'm missing something.

The problem begins with your first statement:

$$\left(\frac{\partial S}{\partial V}\right)_U=0$$

Because in fact:

$$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$

which is clearly non-zero (except in limit of zero pressure or infinite temperature).

More generally, being at equilibrium does not equate to a partial derivative being zero!  Rather, being at equilibrium equates to a partial derviative being defined (i.e., they are only defined at equilibrium).  Thus, whether or not a partial derivative equals zero has nothing to do with equilibrium, but rather with the specific nature of the system. And, most typically, partial derivatives are non-zero; it's only for certain specialized cases that a partial would be zero.

Consider, for instance, the total differential for $U=U(T, V,n)$:

$$\mathrm{d}U =  \left(\frac{\partial U}{\partial T}\right)_{\!V,\, n_i}\,\mathrm{d}T+\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}\,\mathrm{d}V +\sum_i \left(\frac{\partial U}{\partial n_i}\right)_{\!T,\, V,\, n_{j\ne i}}\,\mathrm{d}n_i,$$

where the last term allows for chemical reactions and/or matter flow into or out of the system.

Note that state functions ($U, H, A, G, S...$) are only defined at equilibrium, and thus these partial derivatives are likewise only defined at equilibrium.  What the above partial derivatives tell us, then, is how the state function (in this case, $U$) changes as we infinitesimally change the system from one equilibrium state to a new equilibrium state by varying $T$, $V$, or one of the $n_i$.

Thus, far from being zero at equilibrium, these partials typically need to be non-zero at equilibrium, unless the state function is independent of the change in the corresponding independent variable, which only applies in special cases.

One such special case is the internal energy of an ideal gas, which depends only on $T$ and $n$. Hence, for a single ideal gas, or a mixture of ideal gases:

$$\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}=0$$


Why at constant pressure and temperature Gibbs energy change of a process can be negative? $dG < 0$, for a spontaneous reaction at constant Temperature $T$ and Pressure $P$, this means that at contant $T$ and $P$, $dG$ can be negative
$dG = VdP -SdT$, this means that at constant $P$ ($dP =0$) and constant $T$ ($dT = 0$), $dG = V\cdot 0-T\cdot 0 = 0$. At constant $P$ and $T$ according to this equation $dG$ is always zero.

Statement $1$ contradicts statement $2$. Why does this happen?

The relation $dG = VdP - SdT$ implies that the Gibbs free energy of the system depends only on the two variables $T$ and $P$, e.g., as in the case of a single-component gas consisting of a fixed number of molecules. For such systems, the macrostate is completely determined if $T$ and $P$ are given; nothing can change further.

On the other hand, suppose that there is an additional parameter $X$ describing the system and that $X$ is not fixed. Then, according to the second law of thermodynamics, the system, being at constant $T$ and $P$, self-adjusts $X$ in such a way that $G$ is minimized.

In case there exist multiple species that can undergo a chemical reaction, $X$ could be a measure of how far the reaction has proceeded. The condition $\partial G/\partial X = 0$ gives the well-known law of chemical equilibrium.


Difference between Sildenafil and Sildenafil citrate The data I found seems to show no difference between Sildenafil and Sildenafil citrate. But if they are the same things, why Pfizer use Sildenafil citrate instead of Sildenafil as their product Viagra?

In many pharmaceuticals (and many recreational drugs) the choice of form is determined by the convenience of the manufacturing and delivery

Many drugs (not just sildenafil) are manufactured as salts or simple derivatives of the active molecule. For example, the anti-inflammatory steroid fluticasone is usually delivered as the propionate or furoate esters and the recreational drug cocaine is usually delivered as the hydrochloride salt. In all this cases the active ingredient is exactly the same in the body.

But there are two factors that matter as well as the active ingredient.

One is how easy it is to obtain a convenient form that is pure and easy to make. Amines, for example, are much less prone to produce nice crystals than their salts (sildenafil and cocaine are amines). Since the industry needs to have a good route to a verifiably pure compound and since crystallisation is a good way to obtain pure compounds, they often experiment with different salts to find one that gives a satisfactory end result. This is convenient for the manufacturer and makes no difference to the active ingredient which is determined by the pH of the biological environment.

The second reason is the ease of delivery of the drug. The physical form and chemistry of the substance or salt do matter for the bioavailability. If the drug is taken as a pill, it has to be well absorbed in the digestive system (which means it should, usually, dissolve quickly in the stomach). Salts are more likely to be rapidly soluble than simple amines (though this depends on the form as much as the fact they are a salt though the form is more controllable if you have nice crystals to start with in manufacturing).

The illegal drug cocaine is an interesting case. It is easy to maufacture it as a salt and it is commonly available as the hydrochloride. The hydrochloride dissolves in nasal linings giving a high. But some prefer a faster, more intense (and also, unfortunately, more addictive) high obtained by using the vaporised freebased amine. This is easy to make by reacting the hydrochloride with household baking powder. But it can't be sniffed in that form (as a powder), it has to be heated and vaporised for effective absorption. The vapour is absorbed more rapidly directly into mucous membranes, making a faster hit. But this isn't as convenient as sniffing a finely ground powder and would not be a convenient way to distribute legal pharmaceuticals (imagine the police attention if you had to heat a spoon full of fluticasone to relieve your blocked nose).

So, in short, convenience of manufacture and delivery frequently lead to the ideal form of a drug being a salt and not the pure active ingredient.


How can metals like magnesium and sodium burn underwater? Please answer with as much detail as possible. I can't find the answer anywhere else and I need it for an essay.

Any websites I have found have conflicting answers.

One said that they don't actually burn underwater, rather than separate hydrogen from oxygen and burn at the surface. Another actually said that they do burn underwater because they take the oxygen source from the water. I do not have a specific definition for "burn" (the question itself is my assignment- write an essay answering that question!), so I know about as much as you do concerning the wording of the assignment.

I do know that alkali and alkaline earth metals are generally very reactive with water, and usually result in exothermic reactions.

Your last sentence was the key.  Metals like sodium and magesium react with water according to the following equations
$$\ce{2Na + 2H2O -> 2NaOH +H2 + heat}$$
$$\ce{Mg + 2H2O -> Mg(OH)2 +H2 + heat}$$
Both reactions produce hydrogen gas and a lot of heat.  As the hydrogen gas bubbles out of the water it mixes with atmospheric oxygen. The intense heat ignites the hydrogen-oxygen mixture and the mixture continues to burn (possibly explosively depending upon the quantities involved) as long as hydrogen continues to be generated.


Why can't small trans-cycloalkenes exist? Why when a ring has 8 carbon's and contains a double bond the trans version is very unstable to exist, for instance a molecule like cholesterol?

Because you're stuck in the ring structure.

The trans isomer has branches that, in their natural state, would want to rest antiparallel to one another. But in double bond 6 membered ring structure (like in cholesterol), these bonds would need to be twisted a lot so that they could sit in the octagonal ring structure. Given how a double bond is rigid and will already resist rotation, this theoretical trans isomer will have such a high energy requirement for it to exist that it will simply refuse to do so.


Is Gibbs Free Energy of formation equal to Gibbs Free Energy of transition state? Is activation energy equivalent to Gibbs Free Energy of transition state as related by Eyring equation?
$$E_a=\Delta ^\ddagger G \, \, ?$$

Is Gibbs Free Energy of transition state defined by the Gibbs Free Energy of formation?
$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G \, \, ?$$
If not, which are the conditions for which such an approximation (second equation) would apply?

The Arrhenius equation

$$k=A\exp\left(-\frac{E_a}{RT}\right)$$
places all of the $T$-dependence in the exponential factor. The pre-exponential factor is not assumed to be temperature-dependent. By contrast, the Eyring equation

$$k=\frac{\kappa k_\mathrm{B}T}{h}\exp\left(-\frac{\Delta ^\ddagger G^⦵}{RT}\right)$$

which is justified on the basis of statistical-mechanical principles, displays a a temperature-dependence in the pre-exponential factor. So the exponential terms are not expected to be equal, at least theoretically, unless the pre-exponential factor in the Eyring equation becomes T-independent by some coincidental cancellation of terms.

As regards the second equation, I would define the free energy of the transition state more accurately as

$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G(\mathrm{ts})-\Delta _\mathrm{f} G(\mathrm{ini}) $$


Transformation of acyl group to carboxyl group in biphenyl system with two acyl groups at para position How do I perform this reaction?

I have prepared 4,4'-acylbiphenyl and want to make one acyl group into $\ce{-OH}$ and other into $\ce{-COOH}$ group.

Is it correct that you are aiming to synthesize 4-(4-hydroxyphenyl)benzoic acid [58574-03-1]?

The target compound is available in one step from biphenyl using potassium peroxydisulfate $(\ce{K2S2O8})$ as an oxidant, Palladium(II)-trifluoroacetate, $\ce{Pd(OCOCF3)2}$, as a catalyst, and formic acid as the carbon source for the $\ce{-COOH}$ group. The synthesis was published in
 Org. Lett., 2004, 6, 2437-2439 (DOI).

EDIT

Limited resources and experience will probably rule out Stille-type coupling reactions too.

Alternatives pretty much depend on which 4,4'-disubstituted biphenyl are available.

A methoxyarene can be converted to the corresponding phenol by $\ce{BBr3}$
or the cheaper hydrobromic acid. The $\ce{-OCH3}$ group resists oxidation, is a +M substituent and is not acidic. All this might be helpful (see below).

In order to introduce the $\ce{-COOH}$ substituent, two approaches seem to be possible in an undergrade lab:

oxidation of a methyl group by permanganate
Grignard reaction of a bromoarene with carbon dioxide


Figuring out the volume of air trapped inside a bottle of water I am doing a chemistry lab report on figuring out the volume of air trapped inside a bottle.  Here is the procedure:

Fill a larger container or tub about half full with tap water
Fill a 2 L plastic soft drink bottle with tap water.  Put the cap on loosely
Carefully turn the bottle upside down without spilling any water
Put the bottle into the large container of water so that the mouth of the bottle is underwater.  Remove the cap underwater
Feed the flexible tubing under the water so that one end goes inside the bottle
Put your straw into the other end of the tubing.  Do not share straws
When it is your turn, exhale into the straw to collect the air of one normal breath
With a marking pen, mark the volume of air on the soft drink bottle
Figure out the volume of the air trapped inside the bottle.  Record this volume

How might I do number 9? My proposed procedure is:

With a marking pen, mark the new volume of air on the soft drink
bottle
Record this volume, which will be smaller than the volume found in
step 8 as some of the water will have been displaced by the breath
Subtract the new volume of air from the old volume of air to find
the volume of the air trapped inside the bottle
Record this volume

Thanks!

Displacement! Let the container that you have bottle in be filled to the brim inside a bigger container. Then when you breathe into the bottle, the amount of water that is displaced will be the same as the volume of the air.