Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution? If I were to attempt to nitrate Pyridine (essentially benzene but with one carbon replaced with Nitrogen) would I get substitution at the Ortho-, Meta- or para position? Also why is this? I can only think of an inductive effect that the Nitrogen might have but I can't think of more factors and I also can't think whether it will be directing towards o-,m- or p- nor why. Can you help?

The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.

Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons.  Below are the resonance structures we can draw for pyridine.  They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects.

This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).

If we look at the resonance structures that we can draw for electrophilic attack at the ortho

meta

and para positions,

we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen.  Therefor, meta-substitution is preferred in electrophilic attack on pyridine

So electrophilic attack on pyridine will produce meta-substituted pyridines.  Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation.  The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred.  In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.


Fundamental forces behind covalent bonding I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other?

Also, am I right to think that "repulsion occurs when atoms are too close together" comes from electrostatic interaction?

I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other?

The role of Pauli Exclusion in bonding

It is an unfortunate accident of history that because chemistry has a very convenient and predictive set of approximations for understanding bonding, some of the details of why those bonds exist can become a bit hard to discern. It's not that they aren't there -- they most emphatically are! -- but you often have to dig a bit deeper to find them. They are found in physics, in particular in the concept of Pauli exclusion.

Chemistry as avoiding black holes

Let's take your attraction question first. What causes that? Well, in one sense that question is easy: it's electrostatic attraction, the interplay of pulls between positively charged nuclei and negatively charged electrons.

But even in saying that, something is wrong. Here's the question that points that out: If nothing else was involved except electrostatic attraction, what would be the most stable configuration of two or more atoms with a mix of positive and negative charges?

The answer to that is a bit surprising. If the charges are balanced, the only stable, non-decaying answer for conventional (classical) particles is always the same: "a very, very small black hole." Of course you could modify that a bit by assuming that the strong force is for some reason stable, in which case the answer becomes "a bigger atomic nucleus," one with no electrons around it.

Or maybe atoms as Get Fuzzy?

At this point some of you reading this should be thinking loudly "Now wait a minute! Electrons don't behave like point particles in atoms, because quantum uncertainty makes them 'fuzz out' as they get close to the nucleus." And that is exactly correct -- I'm fond of quoting that point myself in other contexts!

However, the issue here is a bit different, since even "fuzzed out" electrons provide a poor barrier for keeping other electrons away by electrostatic repulsion alone, precisely because their charge is so diffuse. The case of electrons that lack Pauli exclusion is nicely captured by Richard Feynman in his Lectures on Physics, in Volume III, Chapter 4, page 4-13, Figure 4-11 at the top of the page. The outcome Feynman describes is pretty boring, since atoms would remain simple, smoothly spherical, and about the same size as more and more protons and electrons get added in.

While Feynman does not get into atoms how such atoms would interact, there's a problem there too. Because the electron charges would be so diffuse in comparison to the nuclei, the atoms would pose no real barrier to each other until the nuclei themselves begin to repel each other. The result would be a very dense material that would have more in common with [neutronium[(http://en.wikipedia.org/wiki/Neutronium) than with conventional matter.

For now I'll just forge ahead with a more classical description, and capture the idea of the electron cloud simply by asserting that each electron is selfish and likes to capture as much "address space" (see below) as possible.

Charge-only is boring!

So, while you can finagle with funny configurations of charges that might prevent the inevitable for a while by pitting positive against positive and negative against negative, positively charged nuclei and negatively charged electrons with nothing much else in play will always wind up in the same bad spot: either as very puny black holes, or as tiny boring atoms that lack anything resembling chemistry.

A universe full of nothing but various sizes of black holes or simple homogenous neutronium is not very interesting!

Preventing the collapse

So, to understand atomic electrostatic attraction properly, you must start with the inverse issue: What in the world is keeping these things from simply collapsing down to zero size -- that is, where is the repulsion coming from?

And that is your next question:

  Also, am I right to think that "repulsion occurs when atoms are too close together" comes from electrostatic interaction?

No; that is simply wrong. In the absence of "something else," the charges will wiggle about and radiate until any temporary barrier posed by identical charges simply becomes irrelevant... meaning that once again you will wind up with those puny black holes.

What keeps atoms, bonds, and molecules stable is always something else entirely, a "force" that is not traditionally thought of as being a force at all, even thought it is unbelievably powerful and can prevent even two nearby opposite electrical charges from merging. The electrostatic force is enormously powerful at the tiny separation distances within atoms, so anything that can stop charged particles from merging is impressive!

The "repulsive force that is not a force" is the Pauli exclusion I mentioned earlier. A simple way to think of Pauli exclusion is that identical material particles (electrons, protons, and neutrons in particular) all insist on having completely unique "addresses" to tell them apart from other particles of the same type. For an electron this address includes: where the electron is located in space, how fast and in what direction it is moving (momentum), and one last item called spin, which can only have on of two values that are usually called "up" or "down."

You can force such material particles (called fermions) into nearby addresses, but with the exception of that up-down spin part of the address, doing so always increases the energy of at least one of the electrons. That required increase in energy is a nutshell is why material objects push back when you try to squeeze them. Squeezing them requires minutely reducing the available space of many of the electrons in the object, and those electrons respond by capturing the energy of the squeeze and using it to push right back at you.

Now, take that thought and bring it back to the question about where repulsion comes from when to atoms bond at a certain distance, but no closer. They are the same mechanism!

That is, two atoms can "touch" (move so close, but no closer) only because they both have a lot of electrons that require separate space, velocity, and spin addresses. Push them together and they start hissing like cats from two households who have suddenly been forced to share the same house. (If you own multiple cats, you'll know exactly what I mean by that.)

So, what happens is that the overall set of plus-and-minus forces of the two atoms is trying really hard to crush all of the charges down into a single very tiny black hole -- not into some stable state! It is only the hissing and spitting of the overcrowded and very unhappy electrons that keeps this event from happening.

Orbitals as juggling acts

But just how does that work?

It's sort of a juggling act, frankly. Electrons are allowed to "sort of" occupy many different spots, speeds, and spins (mnemonic $s^3$, and no, that is not standard, I'm just using it for convenience in this answer only) at the same time, due to quantum uncertainty. However, it's not necessary to get into that here beyond recognizing that every electron tries to occupy as much of its local $s^3$ address space as possible.

Juggling between spots and speeds requires energy. So, since only so much energy is available, this is the part of the juggling act that gives atoms size and shapes. When all the jockeying around wraps up, the lowest energy situations keep the electrons stationed in various ways around the nucleus, not quite touching each other. We call those special solutions to the crowding problem orbitals, and they are very convenient for understanding and estimating how atoms and molecules will combine.

Orbitals as specialized solutions

However, it's still a good idea to keep in mind that orbitals are not exactly fundamental concepts, but rather outcomes of the much deeper interplay of Pauli exclusion with the unique masses, charges,  and configurations of nuclei and electrons. So, if you toss in some weird electron-like particle such as a muon or positron, standard orbital models have to be modified significantly and applied only with great care. Standard orbitals can also get pretty weird just from having unusual geometries of fully conventional atomic nuclei, with the unusual dual hydrogen bonding found in boron hydrides such as diborane probably being the best example. Such bonding is odd if viewed in terms of conventional hydrogen bonds, but less so if viewed simply as the best possible "electron juggle" for these compact cases.

"Jake! The bond!"

Now on to the part that I find delightful, something that underlies the whole concept of chemical bonding.

Recall that it takes energy to squeeze electrons together in terms of the main two parts of their "addresses," the spots (locations) and speeds (momenta)? I also mentioned that spin is different in this way: the only energy cost for adding two electrons with different spin addresses is that of conventional electrostatic repulsion. That is, there is no "forcing them closer" Pauli exclusion cost like you get for locations and velocities.

Now you might think "but electrostatic repulsion is huge!", and you would be exactly correct. However, compared to the Pauli exclusion "non-force force" cost, the energy cost of this electrostatic repulsion is actually quite small -- so small that it can usually be ignored for small atoms. So when I say that Pauli exclusion is powerful, I mean it, since it even makes the enormous repulsion of two electrons stuck inside the same tiny sector of a single atom look so insignificant that you can usually ignore its impact!

But that's secondary, because the real point is this: When two atoms approach each other closely, the electrons start fighting fierce energy-escalation battles that keep both atoms from collapsing all the way down into a black hole. But there is one exception to that energetic infighting: spin! For spin and spin alone, it become possible to get significantly closer to that final point-like collapse that all the charges want to do.

Spin thus becomes a major "hole" -- the only such major hole -- in the ferocious armor of repulsion produced by Pauli exclusion. If you interpret atomic repulsion due to Pauli exclusion as the norm, then spin-pairing two electrons becomes another example of a "force that is not a force," or a pseudo force. In this case, however, the result is a net attraction. That is, spin-pairing allows two atoms (or an atom and an electron) to approach each other more closely that Pauli exclusion would otherwise permit. The result is a significant release of electrostatic attraction energy. That release of energy in turn creates a stable bond, since it cannot be broken unless that same energy is returned.

Sharing (and stealing) is cheaper

So, if two atoms (e.g. two hydrogen atoms) each have an outer orbital that contains only one electron, those two electrons can sort of look each other over and say, "you know, if you spin downwards and I spin upwards, we could both share this space for almost no energy cost at all!" And so they do, with a net release of energy, producing a covalent bond if the resulting spin-pair cancels out positive nuclear charges equally on both atoms.

However, in some cases the "attractive force" of spin-pairing is so overwhelming greater for one of the two atoms that it can pretty much fully overcome (!) the powerful electrostatic attraction of the other atom for its own electron. When that happens, the electron is simply ripped away from the other atom. We call that an ionic bond, and we act as it if it's no big deal. But it is truly an amazing thing, one that is possible only because the pseudo force of spin-pairing.

Bottom line: Pseudo forces are important!

My apologies for having given such a long answer, but you happened to ask a question that cannot be answered correctly without adding in some version of Pauli "repulsion" and spin-pair "attraction." For that matter, the size of an atom, the shape of its orbitals, and its ability to form bonds similarly all depend on pseudo forces.


Water as a "Harmful Acid" Apologies for the question which is far below the brilliance of the minds on this page. I have an A-Level in Organic Chemistry but I'm tying myself in knots with this one.

I'm a magician looking to create a Russian Roulette effect with water and acid (sort of!) the actual secret is that both solutions are water.

Is there a readily available metal; or substance that looks like metal, that would make the water behave like an acid to convince an onlooker that it is a potentially life threatening acid.

As a bonus question – is there a way to cheat the litmus test so that one solution (also water!) turns it red, the other reacting neutrally (that's the easy one!)?

If you are using water alone, then sodium lead alloy reacts as sodium would but, with the lead added, it's much safer.  But you can have a wider variety of options if, as the comments suggest, you use a dilute acid like vinegar (reacting that, for instance, with zinc).  Wear eye protection no matter what you do.


Why is the enthalpy of solution for NaCl 3.88 and not 4.1 kJ/mol? Based on Hess' law, the molar enthalpy of solution is equal to the sum of the enthalpies of formation of products minus reactants. In the case of dissolving table salt ($\ce{NaCl}$) in water, those reactants are $\ce{NaCl (s), Na+ (aq)}$, and $\ce{Cl- (aq)}$, with enthalpies for formation of $-411.2$, $-239.7$, and $-167.4\ \mathrm{kJ/mol}$, respectively.

So, based on Hess' law, the enthalpy of solution should be $4.1\ \mathrm{kJ/mol}$. However, the actual enthalpy of solution in this case is $3.88\ \mathrm{kJ/mol}$ according to multiple sources. The discrepancy appears to be that the enthalpies of formation for $\ce{Na+}$ and $\ce{Cl-}$ don't exactly equal the combined enthalpy of formation for $\ce{NaCl (aq)}$.

I am at loss for why there is this small offset. Is it due to only partial dissociation of $\ce{NaCl (aq)}$ while in solution? And, if so, wouldn't that have a larger effect? Or is the effect slight because there are some inter-molecular forces at play that only minimally affect the total enthalpy of solution? It could be that my numbers are wrong, but a cursory search on the internet matches the values for $\ce{Na+}$ and $\ce{Cl-}$ in the textbook I have on hand.

Your approach is correct. The small discrepancy of the results when using parameter values from different sources could be caused by inconsistent standard states (temperature and pressure), or simply by measurement uncertainty (in this case, the resulting difference would not be considered significant).

By way of comparison, I found the following parameter values for a temperature of $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ and a pressure of $p=1\ \mathrm{bar}=100\ \mathrm{kPa}$.

Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})=-411.2\ \mathrm{kJ\ mol^{-1}}$ [1]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})=-240.1\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})=-167.2\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of solution $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})=3.88\ \mathrm{kJ\ mol^{-1}}$ [3]

Calculating the standard molar enthalpy of solution from the given values for the standard molar enthalpy of formation yields

$$\begin{align}
\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})&=\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})+\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})-\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})\\[6pt]
&=-240.1\ \mathrm{kJ\ mol^{-1}}-167.2\ \mathrm{kJ\ mol^{-1}}+411.2\ \mathrm{kJ\ mol^{-1}}\\[6pt]
&=3.9\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

Considering the number of significant digits, there is no significant difference between the calculated result of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.9\ \mathrm{kJ\ mol^{-1}}$ and the corresponding literature value of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.88\ \mathrm{kJ\ mol^{-1}}$.

References

[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[2] “Thermodynamic Properties of Aqueous Ions”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[3] “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Does 2,2-dimethylpropan-1-ol give a positive Lucas test? I learnt that the Lucas test involves the formation of a carbocation and it gives a positive test based on the stability of the carbocation formed, and hence primary alcohols do not show turbidity (condition for positive test).

Now consider 2,2-dimethylpropan-1-ol. It is a primary alcohol and I was taught that it will not give a positive Lucas test. But my doubt is, why won't a methyl shift take place making it stable and hence give a positive result?

No. It does not.

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions

In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.

Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test


Clarification on IUPAC nomencalture This is 1,6-dimethyl-cyclohexene. Why cannot this same structure have the IUPAC name 1,2-dimethyl-cyclohexene. In the given picture the locant sum comes to 7 but in my case it comes to 3, implying that my answer is in accordance with the rules. Aren't the alkyl groups supposed to be given the lowest number as per lowest locant sum rule?

Please do clarify. Thanks!

Between alkenyl and alkyl groups, alkenyl groups (double bonds) have seniority over simple alkyl substituents groups such as methyl and ethyl groups (See the new version of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Prepared by Henri A Favre & Warren H Powell); Royal Society of Chemistry, Cambridge, UK, December 2013, ISBN: 978-0-85404-182-4:

Briefly, it suggests "When several structural features appear in cyclic and acyclic compounds,"  low locants (numbering) are assigned to saturated/unsaturated groups in the following decreasing order of seniority: (a) low locants are given to ‘ene’ and ‘yne’ endings; (b) low locants are given first to multiple bonds as a set and then to double bonds; and (c) detachable alphabetized prefixes, all considered together in a series of increasing numerical order.

Thus, ene-group on cyclohexene get the highest priority such as cyclohex-1-ene (or, 1-cyclohexene). Since the double bond has the highest priority, the substituents on  it (if any) should get the next priority such that it should have been 1-methylcyclohex-1-ene (or, 1-methyl-1-cyclohexene), if there are no other substituents. But in your molecule, there is an another methyl group present. Therefore that has the lowest priority. Since 1,2-numbers belongs to the double bond as mentioned above, that methyl can have been only 6 with numbering. Thus, the compound is:  1,6-dimethylcyclohex-1-ene (or, 1,6-dimethyl-1-cyclohexene).

Note that 1-methylcyclohex-1-ene has higher priority over 2-methylcyclohex-1-ene (lowest locant rule within double bond). Therefore, naming the molecule as 2,3-dimethylcyclohex-1-ene (or, 2,3-dimethyl-1-cyclohexene) is also not possible within IUPAC rules.


Raman Spectra of Toluene This is the raman spectra of Toluene I obtained while working at my lab using a raman spectrometer with laser of 532 nm wavelength. Well, I am a physics student, but I want to know how information about vibrational modes and functional groups are inferred from the raman spectra, let's say from this spectra of Toluene. It would be of great help, since I am very curious about this spectroscopy method. I know physics part of raman spectroscopy, but really struggling with its application part.

Chemists have tables and tables and tables that correlate functional groups to numbers.

For example:
http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html

If we see a number on the graph, a certain type of functional group may be present.  The sharp one at 3056 tells me aromatic hydrogen.  The 2919 is the benzylic hyrodgens.  Since I'm a synthesis guy, knowing the exact types of vibrations occurring there was never important to me.

Be warned, chasing every peak is normally quite frustrating.  Vastly more important is normally the overall look for comparison with a known compound, which is sort of a molecular finger print.


Delocalization of π-electrons in nitrate ion Benzene and nitrate ion are given in my textbook as examples for the delocalization of π-electrons. Benzene, due to symmetry of its resonating structures, is simple enough. We assume that σ-electrons are localized and π-electrons are delocalized in the ring.
Each carbon atom promotes one electron from its $\mathrm{s}$ orbital to the empty $\mathrm{2p}$ orbital. It hybridizes two $\mathrm{p}$ orbitals with the $\mathrm{s}$ to form three $\mathrm{sp^2}$ orbitals which it then uses to form three σ-bonds, two with other carbons and one with a hydrogen.
The remaining unhybridized $\mathrm{p}$ orbital then sticks out above and below the plane of the ring. The electrons from all the six unhybridized $\mathrm{p}$ orbitals of the six carbons are then delocalized above and below the plane of the ring.
However, when I try to apply a similar reasoning to the nitrate anion, problems arise. The resonance Lewis structures of nitrate ion are:

Now, assuming again that only the π-electrons are delocalized, we would expect that only two electrons are delocalized (since there is only one double bond). But my textbook claims that each atom is $\mathrm{sp^2}$-hybridized, and as in benzene, one unhybridized $\mathrm{p}$ orbital per atom sticks out above and below the plane of the molecule.
Assuming this to be true, I presumed that two electrons would be localized and tried to arrange the remaining 22 electrons into a Lewis structure with $\mathrm{sp^2}$ hybridized atoms. I couldn't find any viable structures with only three lone pairs around each atom.
What is the exact mechanism for delocalization of electrons in nitrate? Is there a general scheme to this mechanism that applies to all such similar molecules such as $\ce{CO3^2-}?$

Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.

Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp2 hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. D6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)

The nitrate ion $\ce{NO3-}$ is what I like to call a Y-aromat since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene.
The resonance structures, however, look much more complicated.

In quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. D3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene.

Since the carbonate anion $\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.

* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another.
For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\ce{O\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that.


Ammonia for cleaning - what is the interaction between ammonia and oils? Household ammonia is recognized as being an excellent cleaner for grease and oil. My question is, how does this work? Does ammonia merely put the oils into a suspension, like a detergent does? Or is there a more significant chemical reaction between the two - along the lines of how enzymes affect oils?

Common food fats and oils usually contain some free fatty acid1. The alkalinity of household ammonia serves to form a soap which will emulsify the main body of the fat/oil, if enough elbow grease (i.e. vigorous agitation with a scrub brush or wash rag) is applied. Other alkalis do as well - better, if you consider breathable ambient air to be desirable.
Some brands of commercial household ammonia already contain fatty acid soaps and exhibit a cloudy appearance.
It also seems to me that a dilute solution of washing soda ($\ce{Na2CO3}$) would clean quite well, and ammonia should do no better. But when the odor of $\ce{NH3}$ disappears, and the oily/fatty surface is now clean, it is as if there are now two indicators of cleanliness: no odor and no oiliness, therefore the surface must certainly be clean, because all evidence of the cleaning process has been removed. It must be a psychological thing, like foaming with detergents.
Reference:

Analytical approaches for the assessment of free fatty acids in oils and fats
S. A. Mahesar, S. T. H. Sherazi, Abdul Rauf Khaskheli, Aftab A. Kandhro, Siraj uddina, Anal. Methods, 2014 ,6, 4956-4963, DOI: 10.1039/C4AY00344F


Can the human body withstand the heat from setting plaster? There was an incident a couple of years ago where a teenage girl during an art lessen got her hands stuck in a bucket with plaster. As the plaster set, the heat from the chemical process caused her fingers to melt (it says in the article that the temperature reached 60 °C; I am not sure how this can melt your fingers, but this seems to be the case). As a result, she had to have all but two fingers amputated. The lesson we seem to learn is that sticking your hands into a bucket with plaster is a bad idea.

At the same time, plaster casting the human body is something that people do all the time, both in art and for medical purposes. Of course, the layers are thinner in this case. But in this video (which is erotic, sorry about that, but it was the only video example I could find), a woman puts her entire body into a bucket of plaster and seems to get away with it just fine.

So, what are these people doing differently than in that school incident? Could they be using a different kind of plaster that generates less heat?

The thermal effect depends on the amount of water used. If you mix plaster of Paris with 0.6 times its amount of water (in weight), the temperature will not go higher than 37°C.


How can an Organic chemist know the chemical formula of a natural product she isolated from an organism? As a non scientist I several times wondered how can an Organic chemist successfully isolate a single molecule which a natural product of an organism;
Whether a plant-part/mushroom-part/algae-part or a coral-part or even from an animal or human organ in autopsy (neurotransmitter/hormone and so forth).

I also wondered how could she go further to know the natural product's (exact?) chemical formula and give such newly discovered molecule a name.

Given that molecules are such small entities, way smaller than the smallest uicellular organism and are hard to be photographed uniquely with current microscopal technologies (AFM/STM); it is unclear to me How could she know if a plant part contains a certain molecule (say caffeine or nicotine) so precisely that she could tell:

  This is what I will now name Caffeine → not Theobromine
  or any other Methylxantine → just Caffeine, pure and exact and unequivocal.

Practical examples for my problem are:

How did organic chemists isolated THC from various Cannabis plant-parts and gave it a formula and a name, although today it is known that Cannabis plant parts has tens of cannabinoids?
How did organic chemists isolated Ganodermadiol in some Ganodrma mushroom part/s and gave it a chemical formula and name and claimed it is similar to steroidal hormones?
How did they know they weren't wrong?

My question

How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

There are two levels of answers. One is historical and one is modern. Historically, determining the chemical formula for had been a trivial job for most small or medium sized molecules. You do a combustion analysis, determine how much carbon dioxide, and water were produced. By doing other classical element detection tests, you would determine if it had halogens, nitrogen and sulfur. These were the most common approaches. The determination of molecular weights was done by classical physical chemistry experiments.

Today you would still do combustion analysis and elemental analysis as a first step and determine the CHNS ratios. Then you run a mass spectrum of the compound, and assess the molecular weight of the compound.

Determining a structural formula may take a long time and others have already mentioned in the comments. Today advanced NMR techniques, high resolution mass spectrometry, X-ray of crystals (if they exist), all help in structure determination. These things have become routine now.

Nobody routinely determines the molecular structure by AFM or STM as far as I know.


What's the difference between RHF and UHF? I am doing a course on electronic structure methods and am somewhat puzzled by the content (I have a physics background). From what I understand Hartree-Fock is a method where you only use the Slater determinant for the ground state of occupied spin orbitals. Then configuration interaction (CI) is an extension of this where you include unoccupied spin orbitals for your Slater determinants. But if Hartree-Fock considers the ground state energy, then how can you have unrestricted Hartree-Fock?

The names RHF and UHF refer to the invariance of Hartree-Fock wave functions with respect to certain symmetry operations. Fukutome has worked out the details. Also, Stuber gives the details in his PhD work. I.e., a rRHF ( = real Restricted Hartree-Fock) wave function is invariant under the $\hat{S}^2, \hat{S}_z, \hat{K}, \hat{\theta}$, where $\hat{S}$ is the spin operator, $\hat{K}$ the complex conjugation operator and $\hat{\theta}$ the time reversal operator. rUHF is invariant only w.r.t. the operators $\hat{S}_z$ and $\hat{K}$. So the operators form a group and the labels rRHF, rUHF, $\ldots$ are used to group wave functions.

Configuration Interaction (CI) is the interaction of a predefined set of Slater-Determinants (SD) or Configuration State functions (CSF) with each other. I.e., one forms a CI wave function
$$
\Psi_\mathrm{CI} = \Phi_0 + \sum\limits_{r}^{a} \Phi_{r}^{a} + \sum\limits_{rs}^{ab} \Phi_{rs}^{ab} + \ldots
$$
$\Phi_0$ is the reference wave function and $\Phi_r^a$ denotes the wave function constructed from the reference by exciting one electron from orbital $r$ to orbital $a$. The choice of $\Phi_0$, however, is up to you. Oftentimes a rUHF solution is chosen. But similarly, rRHF is possible.

Ground state in your nomenclature just refers to the lowest energy HF solution of a given symmetry. But in some cases, the wave function can be unstable w.r.t. the loss of some symmetries. E.g., you can calculate an rRHF wave function for the disscociated $\ce{H2}$ molecule and you will find it to be a (local) minimum. Using this wave function as a starting point for an rUHF calculation, you will quickly discover the local rRHF minimum to be a local rUHF maximum. Hence an optimization will lower the energy. This is closely connected to the terms singlet or triplet instability. For a more complete discussion please refer to books as, e.g., Molecular Electronic-Structure Theory by Helgaker, Jørgensen and Olsen.

In the mentioned case of $\ce{H2}$ dissociation it is also quite logical. As indicated already by others, RHF implies one set of spatial orbitals for all electrons of $\alpha$ and $\beta$ spin or, to put it differently, double occupation of each molecular orbital (MO). Triplet states cannot be represented with an RHF wave function! But while dissociating $\ce{H2}$ it is unreasonable to assume that the electron spins on the atoms stay paired forever. One would rather expect both atoms to be completely identical at infinite distance and thus to end up with two hydrogen atoms instead of a $\ce{H+H-}$ system. And this is what happens when you use UHF.

The complex versions of HF become interesting especially when dealing with relativistic effects. Due to the Pauli matrices the Hamiltonian becomes complex and hence the MOs become complex as well. By this the wave function is not symmetric w.r.t. to $\hat{K}$ anymore and hence belongs to a different group in the scheme of Fukutome and Stuber.


Replacement for Centrifuge in Calcium Content Determination Basically, I'm performing a lab investigation to determine calcium content in different milk sorts. Since my school didn't have any EDTA, I'm using Sodium Oxalate to form a precipitate with Calcium Oxalate. However, the source I have says that the best way to make these two things react is by centrifuging, a tool which my school lacks. How could I do this without centrifuging for 5 minutes on 1500 RPM? Can I just wait or heat up increase the rate of reaction? Any tips?

If your "source" claimed that a centrifuge increased the reaction rate, then you need to find a better source. (I admit that it is possible that the compressive work a centrifuge does causes heating which for high speed centrifuges can be significant, but it isn't likely to be a factor here with small analytical amounts at relatively low speeds, <300 g's, imho) Without some calibration, you really need to find a method you can follow closely. If you have some knowns, say some calcium phosphate and some dietary calcium supplements, then you may be able to validate any deviations from the given method by showing that both highly insoluble salts and highly bioabsorbable calcium (usually calcium carbonate, acidified with HCl to the Ca++ ion) both can be analyzed with the modified method. (I also would guess that the centrifuge wouldn't "push down" other suspended (colloidal) material in the milk (which might either also contain calcium or improve the precipitation rate or amount of the calcium oxalate).)


How can the solid state structure of cyclopentadienyllithium be explained? In the solid state, LiCp adopts a polymeric multidecker structure where Li atoms are sandwiched between two Cp rings (shown in this diagram from Organometallics 1997, 16 (17), 3855–3858):

If we consider the η5-binding mode of the Cp rings, then it seems that the number of electrons around Li is more than eight. However, Li can't extend its octet. How can the structure and bonding be explained?

This summary identifies several calculations indicating that covalent interactions between lithium and cyclopentadienyl moieties do occur; the interaction is mostly but not purely ionic.

Draw the occupied $\pi$ orbitals of the cyclopentadienyl groups and, for comparison, draw the $s, p_x, p_y, p_z$ orbitals on lithium with the $z$ axis along the length of the chain.  Observe:

• The lithium $s$ and $p_z$ orbitals overlap the lowest energy cyclopentadienyl $\pi$ orbitals.

• The lithium $p_x$ and $p_y$ orbitals overlap the bonding degenerate pair on the cyclopentadienyl ring.

With these overlaps, we can construct three lithium-cyclopentadienyl bonding combinations per cyclopendienyl group, one corresponding to each cyclopentadienyl $\pi$ bonding orbital.  We then occupy these orbitals with six electrons per ring.  In effect we have extended the aromatic coupling of the cyclopentadienyl ion into the third dimension with the lithium orbitals.

Heavier alkali metals do not work as well because their orbitals are too diffuse, but they could overlap larger rings.


How to explain regioselectivity in nucleophilic aromatic substitution I’m trying to rationalise the major regioisomeric product of the following reaction of an arylchloride in the presence of sodium amide:

I have predicted that the para-substituted product is formed as the major product because the negative charge in the meta position in the carbanion intermediate is more strongly stabilised by the electron-withdrawing CO2NMe2 group:

Is this observation based on inductive effects correct? Can the carbanion be stabilised mesomerically by delocalizing the negative charge through the π-bond framework?

In the benzyne mechanism, mesomeric effects are not involved as when protons are removed, the electrons are situated in orbitals perpendicular to the pi-system, in non-bonding orbitals (which makes sense since the electrons come from a sigma bond). Therefore the electrons left over are not delocalised, which is why such a strong base is required. This also means regioselectivity is explained by inductive effect, which acts through the sigma bonds. The amide is not involved in any -M stabilisation in this case.
In the first part of the mechanism, the most acidic proton is abstracted, which will be either of the identical protons either side of the Cl, as this provides the most stabilisation by inductive effect. This is correctly shown in your diagram. Deprotonation at the carbons adjacent to the amide is less favourable and leads to no reaction due to lack of leaving group anyway.
The point where the alkyne is attacked is determined by both inductive and steric effects; and optimisation of both simultaneously usually occurs in this mechanism. What I mean by this is that in the major product above, the negative charge is situated as close to two -I substituents as possible, so is stabilised, and at the same time the added NH2 is furthest from the other substituent, reducing steric strain.
In the minor product, the negative charge has less -I stabilisation, and more steric strain since the substituents are now closer together.
Of course this only applies if it progresses through the benzyne mechanism. As @Waylander hinted at, a likely possibility is aromatic nucleophilic addition-elimination since you have an electron withdrawing substituent para to the leaving group. This just so happens to give a product identical to the major product of the benzyne route.


Why are calcination and roasting processes carried out at a temperature just below the ore's fusion temperature? I had read about the two important methods used for the concentration of an ore - Roasting and Calcination. I have a doubt regarding why both the methods are carried out at temperatures not above the ore's melting point.
After some research, I have been able to conclude that one of the reasons is the formation of complicated slag (i.e., difficult to remove from the ore) that tends to form if the ore is molten.
Could someone please explain what exactly happens so that a complicated slag is formed? Also, are there more reasons behind keeping the temperature below the ore's fusion temperature (any specific reasons pertaining to roasting and calcination individually)?

The purpose of roasting a/o calcination is oxidation a/o decomposition of original ore to form metal oxides.
It is desirable to keep the intermediate product solid to obtain porous product. This allows easy access for further reduction by carbon monoxide by coke or charcoal reduction process. If oxides were in form of a solidified melt, the reduction arrangement would be problematic.
Additionally, some ores or oxides have too high melting point to be practical or easily possible to melt them.


R, S Configuration and Branching - Which Path? Which path do I take when we have a branched chiral center? None of the textbooks I've consulted adequately covers this or even provides an example problem in which there was branching.

Consider the above:

1) Chiral carbon with the following attachments H, an carboxylic acid group, a methyl group, and then the ring.

2) H gets the lowest priority.

3) The methyl group gets the second lowest priority since it has just one carbon and then hydrogens; the other paths have either more carbons or oxygens!

4) Between the carboxylic acid group and the path along the ring, we have an oxygen and a hydrogen going for the carboxylic acid group. But which path do I take? After comparing the carbon in the carboxylic acid group to the carbon on the ring, do I compare the double-bonded oxygen to the next carbon on the ring or the single-bonded oxygen? How does one choose in cases of branching such as these?

No,  you have a carbon figuratively attached to 3 oxygen atoms vs.  a carbon figuratively attached to 3 carbon atoms; the former has higher priority.


Do SN1 reaction always result in Racemisation In the following reaction the products formed are not enantiomers but are diastereomoers.So there is no racemisation occured.So is it always correct to say that SN1 result in racemisation?

This doesn't qualify as racemisation, it is epimerization. Racemization would require both stereocentres to invert to form the enantiomer and - as you rightly suppose - the stereocentre bearing the methyl group is untouched by this reaction.

It is also unlikely to lead to a 1:1 mixture of products since the two diastereomers (and the transition states leading to them) are different in energy, so under either thermodynamic or kinetic control one would expect a non-1:1 mixture. This contrasts with the case of a molecule undergoing true racemization in which the two enantiomeric products will be equal in energy so will be formed in equal quantity.

Your title also poses an interesting and more general question - do SN1 reactions always lead to racemisation even in molecules containing only one stereocentre? I was winding up to write an answer to this question, before finding the outstanding answer previously given by NotEvans here - well worth a read. (hint: the answer is - not necessarily).


Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry When being tested in the spectrometer why is it vital that the atoms in the same sample are given the same electrical charge e.g all are c+1 or all are +2 etc

This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.

There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.


Type of analysis for water What other analysis suitable for water as general? Right now, I have turbidity, pH, Total Soluble Solid and color.

Hardness
TC Total Carbon

TIC Total Inorganic Carbon
TOC Total Organic Carbon
VOC Volatile Organic Carbon

COD Chemical Oxygen Demand
BOD Biochemical Oxygen Demand


Why is propenoic acid more acidic than p-methoxybenzoic acid? Why is propenoic acid more acidic than p-methoxybenzoic acid?

I looked online for their $\mathrm{p}K_\mathrm{a}$ values and found them to be $4.37$ and $4.25,$ respectively. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20.$ Why does the $\ce{-OMe}$ group make such a big effect on $\mathrm{p}K_\mathrm{a}?$

Why is propenoic acid more acidic than p-methoxybenzoic acid?

This is propenoic acid, $\mathrm{pK_a}=$4.25

This is p-methoxybenzoic acid, $\mathrm{pK_a}=$4.47

An electron withdrawing group increases the acidity of a carboxylic acid. It disperses negative charge by inductive/resonance effect and stabilizes the carboxylate ion. Thus, propenoic acid (pKa 4.25) is more acidic than p-methoxybenzoic acid (pKa 4.47).

However the molecular structures are totally different. It seems to me to be comparing apples and oranges! To me such hand waving comparisons between molecules are only meaningful if the two molecules are similar.

  The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20.$ Why does the $\ce{-OMe}$ group make such a big effect on $\mathrm{p}K_\mathrm{a}?$

This is a more interesting question.

\begin{array}{|c|c|} \hline \hline
\text{acid} & \text{pka}  \\ \hline
\text{2-Methoxybenzoic acid} & 4.09   \\ \hline
\text{3-Methoxybenzoic acid} & 4.10  \\ \hline
\text{Benzoic acid} & 4.19  \\ \hline
\text{4-Methoxybenzoic acid} & 4.50  \\ \hline
\end{array}

So the phenyl group has about the same effect on the acidity of a carboxylic acid as observed with an ethylene group. However the substitution of other function groups, and their position on the ring can increase or decrease the pKa of the acid.

Data sources:
2-Methoxybenzoic acid

3-Methoxybenzoic acid

Benzoic acid

4-Methoxybenzoic acid


What happens in the attempted diazotisation of 1,2-diaminobenzene? Usually, $\ce{NaNO2}$ and $\ce{HCl}$ are the reagents used to generate an aryldiazonium chloride from an aniline. However, in this compound there are two amine groups adjacent to each other, so it seems unlikely that a diazonium chloride would form. Is it possible for the two amine groups to react during the reaction and form some cyclic product?

Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride:

However, in addition to substitution reactions by loss of $\ce{N2}$, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are diazo couplings, where the nucleophile is an activated arene. However, other nucleophiles, especially intramolecular nucleophiles, can attack diazonium cations. The other amine is poised to do this:


2-methylpropene to tert-butyl alcohol Reaction Mechanism Can someone explain what reaction is going on here?
I looked in my list of mechanisms learnt: I find nothing related!

It is an acid-catalyzed hydration of the double bond. First, the double bond is protonated to yield the tert-butyl cation which is most stabilized by hyperconjugation (protonation at C2 would form a less stable primary cation). The cation is captured by nucleophilic attack of water, which after deprotonation yields tert-butanol and the regenerated proton.

$$\ce{(H3C)2C=CH2 + H3O+ <=> (H3C)3C+ + H2O} \tag 1$$

$$\ce{(H3C)3C+ + H2O <=> (H3C)3C-OH2^{+}} \tag 2$$

$$\ce{(H3C)3C-OH2^{+} + H2O ->~ (H3C)3C-OH + H3O+} \tag 3$$


How to use Kohlrausch's law with ionic equivalent conductances? My text says that for an ionic compound like AxByif ionic equivalent conductance of Ay+ is a and that of   Bx- ion is b, then the equivalent conductance for AxBy would be a/y + b/x
Now my question is:
as the unit of equivalent conductance is S cm2 (gm equivalent)-1 so if i multiply it by a quantity x with unit (gm equivalent) mol-1, then i will get a quantity with unit S cm2 mol-1. This should be the Molar conductance of the ion granted that the value of x was correct. Now for example in Al2(SO4)3 for Al3+ ion the value of x would be 3 because one mole of Al3+ ions would contains 3 gram equivalents. So if equivalent conductance of Al3+ ion was a, then its molar conductance would be 3a and similarly for SO42- ion with equivalent conductance b, the molar conductance would be 2b. Now from all that i can tell that the molar conductance of Al2(SO4)3 would be, 2(3a) + 3(2b) = 6a + 6b
Now from the formula of equivalent conductance and molar conductance
(Molar conductance)/n = Equivalent conductance
where n is the valence factor for desired compound.
So in our case n=6 for Al2(SO4)3

∴ equivalent conductance = (6a + 6b)/6 = a + b
but according to my text it would be a/3 + b/2
Tell me where am i going wrong.
And the result I ended up on works for some questions and the thing that my text provided works for some questions, and then there are some questions in which nothing works. I am really confused... And sorry for any grammatical atrocities. EDIT: I am just trying to ask that when equivalent conductance of two ions are given to me as a and b, then what would be the equivalent conductance of the compound to which these ions belong? Would it be a + b or would it be a/(charge on first ion) + b/(charge on second ion) or is it something totally else?

You textbook writer has explained it in a very complicated way. I think you are confused by the concept of equivalent. Gram equivalent is an obsolete unit now but it is still used in conductance measurements.

First understand the definition of gram equivalent weight for a salt (and it applies to salts only). Acids/ Bases/Redox agents have different definitions.

Gram Equivalent weight of a salt or an ion = gram molecular weight/charge on the ion

This is the definition of gram equivalent.

From the context of your book, ionic equivalent conductance is same as "molar conductance" of aluminum ion. Here a and b refer to molar conductance. This is the mistake you are making. Thus, a 1 Molar Al3+ solution has 3 gram equivalents of Al/ 1 L or it is a 3 N solution (N = Normal).

So if you are given ionic equivalent conductance of Al2(SO4)3 (at infinite dilution) then your textbook is right = (ionic equiv. conductance Al)/ charge on Al + (ionic equiv. conductance sulfate)/ charge on sulfate = a/3 + b/2

By the way, equivalent conductance can only be added at infinite dilution. Whatever you see in tables are usually given at "infinite dilution".


Ultraviolet acid base indicator Is there a chemical(s) that provides good discrimination between absorbance curves for different pHs in the UV light spectrum (100-400nm?)

There are some good examples in Multiwavelength Spectrophotometric Determination of Acid Dissociation Constants Pharaceutical Research Volume 16, Issue 3, pp 374-381.

Benzonic acid (pka = 3.98), phenol (pka 9.73), phthalic acid (pkas 2.70 and 4.86), nicotinic acid (pKas 2.1 and 4.63) and p-aminosalicylic acid (pKas 1.79 and 3.58) would be good in the ~220nm-300nm range.


What is this colorless ferric solution? I made a solution of iron acetate by placing steel wool in one litre of acetic acid (vinegar $5\%$) and adding a few tablespoons of hydrogen peroxide of questionable age. The solution turned a nice brown almost instantly upon shaking.

I left it overnight and it lost its colour while forming a white precipitate:

I added some more peroxide (new, good peroxide this time), and the moment the peroxide hit the clear an colorless solution it looked like dark red-brown food-coloring! Unfortunately I could only get out my camera fast enough to make the following photo:

I squeezed the air out of the steel wool before the above picture, which caused it to sink (noted this so you don't think something of its sinking). Added one-fourth cup hydrogen peroxide and shook:

With a flash from the top:

Held up to the light:

The result seems to be ferric acetate (III), since ferric acetate (II) is light brown in color (according to the Wikipedia link)--possibly like the light brown in the first/top photo. Ferric acetate (III) is supposed to be a red-brown color (again according to the Wikipedia link).

It is the clear an colourless solution in the second picture that I am not certain as to its composition. Any assistance will be appreciated in identifying that solution.

I would hypothesize that the white precipitate is either acetate or carbonate (more likely carbonate). I would guess that the clear solution occurred as a result of the solution interacting with either the steel wool or the air. I took the steel wool out after making the final dark red solution and it has been 2 days now without it diminishing in colour, so it is likely not the air that was interacting with the solution. Perhaps, if I placed the steel wool back in the solution, it might go clear again. I'll post again in a day or two.

I don't believe you have any ferric in the colorless picture since you have a metallic iron which would reduce the ferric to a ferrous. I believe you have a ferrous (Iron(II)) acetate characterized as an off-white precipitant.


Is the smallest or largest diameter important for the color of a molecule? I learned that phenol absorbs longer wavelengths than benzene because the electron cloud is bigger.

But what is about Penta-1,2,3,4-tetrien (1) and 2,2,-Dimethylpropane (2)?
(1) has the largest diameter but (2) has the smallest. Which one absorbs the longer waves?

The size of a molecule does not correlate with the color of a molecule.

Color depends upon which wavelengths of light are absorbed and which are transmitted by a molecule.  Visible light typically ranges from around 390 to 700 nm.  Light with a wavelength between approximately 635 to 700 nm appears red to the human eye.  If we have a molecule that absorbs all visible light below 635 nm, the molecule will appear red; it absorbs all visible light except for the red wavelengths, hence only the red wavelengths will be reflected or transmitted and be available for the eye to detect.

What determines which wavelengths of light are absorbed by a molecule?
This is controlled by the energy separation (energy gap) between the highest energy orbital that contains an electron (the HOMO or highest occupied molecular orbital) and the lowest energy orbital that an electron can be promoted to (the LUMO or lowest unoccupied molecular orbital). If this energy gap falls in the visible range, then the molecule will appear colored to the eye when the molecule absorbs light at the wavelength that corresponds to the energy gap (e.g. when a photon is absorbed that promotes an electron from the HOMO to the LUMO) .

Molecules like 2,2-dimethylpropane contain only sigma bonds.  The energy gap in typical sigma bonds is very large meaning that it takes a lot of energy to promote an electron from the HOMO to the LUMO.  The amount of energy required typically falls in the ultra-violet region, outside the visible range; therefore these molecules are usually colorless since they transmit all visible wavelengths.

Molecules containing pi bonds have smaller HOMO-LUMO energy gaps than molecules with just sigma bonds.  If there are enough pi bonds and they are conjugated then the gap can be small enough that light in the visible range can be absorbed and make the molecule appear colored.  Here is a diagram showing how the energy gap decreases as we increase the length of our conjugated pi system.

Even with 1,3,5-hexatriene the light absorption at 258 nm is still outside the visible  range, so the molecule will not be colored.  But as we continue to extend the conjugation, eventually the molecule will absorb light around 400-500 nm, the blue light range.  As the molecule absorbs blue light it will appear yellow. In addition to pi conjugation, certain substituents such as the nitro group and the carbonyl, to just name two, can also serve to extend the conjugation path in a molecule.

The molecule you asked about, penta-1,2,3,4-tetraene, looks roughly like this.

If we examine it closely, we'll notice that the orbitals really look like those in two distinct (orthogonal) 1,3-butadiene molecules.  As the chart above shows, 1,3-butadiene does not absorb in the visible range.  Nonetheless, because penta-1,2,3,4-tetraene does have an extended pi system it will have a smaller HOMO-LUMO gap than 2,2-dimethylpropane and will absorb light at longer ultra-violet wavelengths (lower energy) than 2,2-dimethylpropane.


Application of the Woodward-Hoffman rules to a [14+2] cycloaddition I am just doing some pericyclic practice problems and I am just trying to understand the Woodward Hoffman rules.

The attached picture is a problem I attempted that was checked against the answers but I am confused how the symmetry is allowed.

When I add the (4n + 2) and 4n together I get 4. Doesn't n have to be an integer and the sum an odd number for the symmetry to be allowed?

EDIT: I copied the answer below with incorrect stereochemistry. Please refer to the second image for the correct stereochemistry

EDIT: Correct stereochemistry below

General rules for cycloadditions

Before getting into the actual Woodward Hoffman 'analysis', one can quickly check any proposed cycloaddition reaction against the following table of general results.

The table is essentially an overview of the Woodward Hoffman rules that doesn't require any consideration of orbitals/orbital symmetry. It can often be useful to check against a table like this to ensure you've done the real Woodward Hoffman analysis correctly.

\begin{array}{ccc} \hline
{\text{Number of }\\\text{Electron Pairs}} &
  \text{Photochemical} &
  \text{Thermal}\\[2ex]\hline
\text{even}~(4n)    &
  \text{suprafacial-suprafacial}  &
  \text{suprafacial-antarafacial} \\[1ex]
\text{odd}~(4n + 2) &
  \text{suprafacial-antarafacial} &
  \text{suprafacial-suprafacial}  \\ \hline
\end{array}

From your example, we can count $14 + 2$ electrons in the π system, giving a total of $16$ which is a $4n$ integer (where $n = 3$). We also know from the reaction scheme given that the cycloaddition is taking place under photochemical conditions. We can, therefore, conclude that the reaction is allowed photochemically if both components react suprafacially.

The actual Woodward Hoffman analysis

Note: I'm doing this for the thermal $[14 + 2]$, firstly because its a little more interesting in terms of orbital overlap, but secondly to give you the opportunity to do the analysis properly for your problem. The approach taken here is the one taken by Ian Fleming in his Molecular Orbitals and Organic Chemical Reactions book.

First: Draw a diagram of the reaction showing the orbitals involved. In this case we are interested in the π system of the heptafulvarene and the π system of the tetracyanoethene.

When drawing a diagram, no orbital phases are needed- this is already taken into account by the generalised Woodward Hoffman rule. (Also note that its not important how you define the components, the 14π system could be considered as a series of seven 2π systems).

Second: Consider how each component is reacting. In a cycloaddition, each component may either react suprafacially (the reaction takes place on the same side of the π system) or antarafacially (the reaction takes place from opposite sides of the π system) (also note that although this is theoretically possible, it is geometrically impossible in some cases due to the inability to get into the correct orientation for orbital overlap).

Third: Apply the Woodward Hoffman rule, as given below.

  A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

In the example above, the 14π component is taken to be reacting antarafacially (denoted π14a), and the 2π component suprafacially (denoted π2s). When the Woodward-Hoffman rule is applied, this gives an odd number of components, making the reaction thermally allowed.

We could, however, also have drawn the reaction as having the 14π component reacting suprafacially (denoted π14s), with side on overlap of the 2π component reacting antarafacially (denoted π2a). Both are equally valid.

From that,you should now be able to consider your photochemical reaction.  One slight difference exists in the formulation of the Woodward-Hoffman rules for photochemical reactions, taking account of the excited state that a component exists in.  Namely, the total is now even, not odd, for an allowed reaction.

  A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is even.

Further reading

All of the diagrams were taken from Fleming's Molecular Orbitals and Organic Chemical Reactions, student edition, 2009. The book contains many more examples and a slightly more rigorous explanation of the Woodward Hoffman rules.


hydrogen environments of benzophenone The hydrogen in $\ce{C4}$, $\ce{C6}$, $\ce{C10}$ and $\ce{C8}$ belong to same hydrogen environment.
The hydrogen in $\ce{C3}$, $\ce{C1}$, $\ce{C11}$ and $\ce{C13}$ belong to the another hydrogen environment.
Hydrogen in $\ce{C12}$ and $\ce{C2}$ belong to the once again, another hydrogen environment.

$\ce{C3}$, $\ce{C1}$ , $\ce{C11}$  $\ce{C13}$, $\ce{C12}$ and $\ce{C2}$ all have two hydrogen on the neighboring carbon atoms.
So my question is why aren't they classified as one hydrogen environment? Instead, why are the categorized under two different environments?

Whether environments are the same or not is based on symmetry considerations and not the number of hydrogens attached to the adjacent carbon atom.

Molecules and atoms within a molecule can be of three types:

identical - they can be interconverted by simple rotations or translations.
enantiotopic - they can be interconverted by reflection in a mirror or about a center of symmetry
diastereotopic - they cannot be interconverted.

Molecules or atoms within a molecule that are found to be identical are said to be in the same environment.  Molecules or atoms that are identical, or in the same environment, must be identical with respect to all chemical and physical properties.  These properties would include chemical reactivity, spectroscopic environment, etc.

In benzophenone, I can rotate the benzene ring about the $\ce{C5-C7}$ bond. This rotation interconverts $\ce{C4}$ and $\ce{C6}$.  If I told you to look away and I performed this rotation (and we removed the numeric subscripts), there is no way that you could tell if the original carbon or the other carbon was now in the $\ce{C4}$ location.  These two carbons (and their attached hydrogens) are therefore said to be equivalent by the above definition.  The same applies to $\ce{C1}$ and $\ce{C3}$, they are also equivalent to each other, or in the same environment.

I can also rotate the molecule around the $\ce{C7=O}$ carbonyl axis.  If I do this I can interconvert $\ce{C4, C6, C8, C_{10}}$.  This same rotation also (separately) interconverts $\ce{C1, C3, C_{11}, C_{13}}$, and also (separately) $\ce{C2}$ and $\ce{C_{12}}$.

Therefore, these 3 sets of carbons

$\ce{C4, C6, C8, C_{10}}$
$\ce{C1, C3, C_{11}, C_{13}}$
$\ce{C2, C_{12}}$

are different (diastereotopic) from one another (I can't interconvert $\ce{C4}$ and $\ce{C_{12}}$, for example).  So we have 3 different chemical environments, but within each environment there are several identical atoms. The same methodology and reasoning applies to the hydrogens attached to these carbons.  They also exist in 3 different environments, and within each of these 3 environments there are several equivalent hydrogens.


How can one synthesise a conjugated alkyne chain? Let's take the conjugated alkyne chain (e) below, taken from Wikipedia:

How would one typically go about synthesising such a conjugated chain with a series of alkyne functions?

A carbenoid Fritsch-Buttenberg-Wiechell (FBW) rearrangement of a substituted dibromoolefinic precursor is used to generate a lithium acetylide, and subsequent trapping with carbon-based electrophiles provides a wide range of di- and triynes. The lithium acetylide formed from the FBW reaction can also undergo transmetalation to provide zinc, copper, tin, or platinum acetylides [1]. The paper described the preparation of poly-ynes of up to 6 triple bond units.

Image source: organic-chemistry.org

References

Luu, T.; Morisaki, Y.; Cunningham, N.; Tykwinski, R. R. One-Pot Formation and Derivatization of Di- and Triynes Based on the Fritsch−Buttenberg−Wiechell Rearrangement. J. Org. Chem. 2007, 72 (25), 9622–9629. https://doi.org/10/ffbmwt.


Is this a correct way to write a chain reaction? I don't know much about chain reactions and how I understand them, they are just a series of chemical reactions where the products of the preceding reaction act as the reactants for the succeeding reactions.
Now, let's assume that $\ce{A}$ and $\ce{B}$ are two compounds that chemically react under certain conditions to give $\ce{X}$ and $\ce{Y}$. This chemical reaction can be written as:
$$\ce{A + B -> X + Y}$$
Now, let's also assume that $\ce{X}$ and $\ce{Y}$ react under the very same conditions to give $\ce{M}$ and $\ce{N}$. This chemical reaction can be written as:
$$\ce{X + Y -> M + N}$$
Now, can the series of these reactions be written as follows :
$$\ce{A + B -> X + Y -> M + N}$$

Your scheme is one of sequential reactions. The chain reaction has feedback so that one to two reactions occur repeatedly until the reactants are all consumed.
The general scheme is shown in the diagram where $R\cdot$ represents any radical species ( H or Cl in Maurice's answer) and $\alpha$ has to be at least 1 for species $R\cdot$ to continue the chain reaction.


What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH? Here's all the info I have and my attempt in solving the question:

The chemical equation for the titration:

$$\ce{C4H4O6 + 2NaOH -> Na2C4H4O6 + 2H2O}$$

Here are the 1st and 2nd $\mathrm{Ka}$ value for tartaric acid:

$$K_\mathrm{a1} = 9.2 \times 10^{-4}, K_\mathrm{a2} = 4.3 \times 10^{-5}$$

Molarity of tartaric acid = $5 \times 10^-4$

Standard solution of $\ce{NaOH }$ was made using $\pu{0.98 g}$ of solid $\ce{NaOH}$ dissolved in $\pu{250 ml}$ of water. The molar mass of $\ce{NaOH}$ is $\pu{39.997 g/mol}$.

This is what I have done so far.

n (C4H4O6) = cv = (5 × 10-4 M) (25ml ÷ 1000L) = 0.02 mol

n (NaOH) = m ÷ M = (0.98g) ÷ (39.997 g/mol) = 0.0245 mol (4dp).

C (NaOH) = n ÷ v = (0.0245 mol) ÷ (250ml ÷ 1000L) = 0.098M

N (NaOH) at equivalence point = cv = (0.098M) (0.204L) = 0.02 mol

C4H4O6(aq)+ 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l)

I 0.02         0.02           0            -
C - 0.02      - 0.02       + 0.02          -
E   0           0            0.02          -

C (Na2C4H4O6) = n ÷ v = 0.02 ÷ (0.025L + 0.25L) = 0.0727M (4dp)

Na2C4H4O6(aq) + H2O(l) → C4H4O4Na2^+(aq) + OH^-(aq) below is ice table, stackexchange won't allow me to seperate the values.

I  0.0727          -           0              0
C   - x            -         + x             + x
E 0.0727 – x       -          x               x

Kb = Kw ÷ Ka = (1 × 10-14) ÷ (9.2 × 10-4) = 1.087 × 10-11

Kb = [C4H4O4Na2+] [OH-] ÷ [Na2C4H4O6] 1.087 × 10-11 = x2 ÷ 0.0727 x = √ ((1.087 × 10-11) (0.0727)) x = 8.89 × 10-7

pOH = - log [OH-] = - log (8.89 × 10-7) = 1.9

pH = 14 – pOH = 14 – 1.9 = 12.1

For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$

For the solutions of acid salts, we can with advantage use the formula

$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

that can be derived from equilibrium

$$\ce{2 HA- <=> H2A + A^2-}$$

For the 2nd equivalence point there is  $\ce{Na2A}$ solution, which is hydrolyzed with equilibrium

$$\ce{A^2- + H2O <=> HA- + OH-}$$

With neglecting OH- from water hydrolysis, $[\ce{HA^-}] = [\ce{OH^-}]$

$$\begin{align}
K_\mathrm{a2}&=\frac{[\ce{H+}][\ce{A^2-}]}{[\ce{HA-}]}\\
\mathrm{p}K_\mathrm{a2}&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{HA-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{OH-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -\mathrm{pOH}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -14 + \mathrm{pH}\\
\mathrm{pH}&=7+\frac12\left(\mathrm{p}K_\mathrm{a2} +\log{ c_\mathrm{\ce{Na2A}}}\right)\\
\end{align}$$

The molarity of $\ce{NaOH}$ solution $c_{\ce{NaOH}}=\frac{m}{M\times V}=\pu{0.09801  mol/L}$

The volume of $\ce{NaOH}$ solution for given volume of tartaric acid solution

$$V_{\ce{NaOH}}=V_{\ce{H2A}}\times \frac{2\times c_{\ce{H2A}}}{c_{\ce{NaOH}}}\\=0.0102\times V_{\ce{H2A}}$$

The final concentration

$$\begin{align}
[\ce{Na2A}]&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +V_{\ce{NaOH}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +
0.0102  \cdot  V_{\ce{H2A}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]}{1+ 0.0102}\\
&=5\times 10^{-4}/1.0102\\
&=\pu{4.9495\times10^{-4} mol/L}\\
&\overset{\mathrm{rounding}}=\pu{4.95\times 10^{-4} mol/L}\\
\end{align}$$


Which reactions occur during the oxidation and dissolution of bromide/bromine? I'm trying to understand the fate of bromide anions during and after they undergo anodic oxidation. I have a solution of 94% DMF/6% water (by volume) with 0.1 M tetrabutylammonium  bromide (TBAB) as the supporting electrolyte. I've used this electrolyte to reduce a cathode at -2.5 V or less v. the Ag/AgCl reference electrode. During the reaction, the solution near the platinum counter electrode (the anode) turned yellow. I would assume this is due to the reaction:
$$\ce{2Br–(aq)-> Br2(l) + 2e– }$$

The next day, I noticed that the yellow liquid in the solution had completely disappeared. I'm assuming that the bromine had dissolved into the solution into $\ce{Br-}$ ions again. However, I'm not 100% sure of the chemical reaction associated with this. Also, would there be any side products associated with the dissolution? Or, do we end up with the species we initially started with in terms of bromine?

I ask because I would like to be able to reuse the electrolyte multiple times and so want to figure out exactly how the electrolyte changes after the reaction. At the cathode, there will be reactions such as the hydrogen evolution reaction which will make the solution more basic. However, I'm not sure what happens with the bromine.

A possibilty is that elementary bromine formed -- causing the the yellow color -- then solubilized in water yielding bromine water.

(source)

It does not dissolve well (2.8% ref -- but do not forget the high density (about 3.10 g/mL) of liquid bromine here). It then may disproportionate into then colorless hypobromit and bromide

$\ce{Br2 + H2O -> Br- + BrO- + 2H+}$ (ref.)

especially under light, forming an acidic solution.  The hypobromit may act as an oxidizer.

To draw an analogy, chlorine gas does dissolve into water, too (ref, ref).

Proceed with caution since bromine is corrosive, though.


Software for simulating NMR spectra I am wondering is there any open source software that predict NMR spectra by giving the chemical shift?

Something like Spinach library http://spindynamics.org/group/?page_id=12 (Matlab is not free, so...).

There is still the old WINDNMR programme developed by Hans Reich: https://www.chem.wisc.edu/areas/reich/plt/windnmr.htm which apparently should still work on modern versions of Windows.

It is not quite as sophisticated as Spinach, but if all you want is a 1D spectrum with various multiplets, then it is probably good enough.

If you have the computational resources, then there is software developed by Stefan Grimme's group https://xtb-docs.readthedocs.io/en/latest/enso_doc/enso.html, but you can't specify shifts manually, you need to generate them via a quantum chemical calculation using ORCA. Technically, all of these are free to use, but obviously it is not always practical. But in any case, if you are interested, the relevant publication is Angew. Chem. Int. Ed. 2017, 56 (46), 14763–14769. It should be possible now to run all the steps using just one ORCA input file; I suggest looking in the ORCA manual for more information.

For what it's worth, if you have the programming knowledge required, it's easy enough to write a (e.g.) Python script that is capable of simulating two- or three-spin (maybe four-spin) systems. At the end of the day, an NMR experiment is "just" a bunch of unitary operators acting on density matrices, i.e. lots and lots of matrix multiplication. For example, you could take a look at some of my MATLAB code here which simulates the first FID of a sensitivity-enhanced HSQC experiment. It is fairly trivial to port this to Python, since numpy provides you with pretty much every function you might need.

[Disclaimer: the code will not necessarily stay there forever.]

The only problem is that this scales exponentially with the number of spins, so unless you do some serious optimisation of the code (like that done in Spinach), it becomes intractable very quickly. But if your molecule can be "decomposed" into several spin systems which are small, then the overall spectrum is just the sum of the spectrum of each spin system, so you might be able to get away with simpler code. I've never thought about how you might do that, although my guess is that you would want to construct a Hamiltonian matrix, then pass it to a function which would call itself recursively if the matrix can be written in block-diagonal form.


About undocumented conventions(?) in PDB files ATOM 22 N  GLN L 3 23.945 -2.849 106.974 1.00 15.00 N
ATOM 23 CA GLN L 3 23.125 -2.052 106.036 1.00 15.00 C
ATOM 24 C  GLN L 3 22.898 -0.648 106.616 1.00 15.00 C
ATOM 25 O  GLN L 3 22.613 -0.491 107.825 1.00 15.00 O

In every amino acid there are two O atoms; which one is described by the (in this example) fourth entry? (I guess it's the one that's next to the next amino acid's N, but then where are the coordinates of the other one or should this be obvious?)

ATOM 26 CB   GLN L 3 21.754 -2.682 105.749 1.00 15.00 C
ATOM 27 CG   GLN L 3 21.577 -3.299 104.379 1.00 15.00 C
ATOM 28 CD   GLN L 3 20.956 -4.676 104.431 1.00 15.00 C
ATOM 29 OE1  GLN L 3 20.940 -5.348 105.476 1.00 15.00 O
ATOM 30 NE2  GLN L 3 20.496 -5.137 103.279 1.00 15.00 N
ATOM 31 H    GLN L 3 24.339 -2.372 107.727 1.00 0.00  H
ATOM 32 HE21 GLN L 3 20.637 -4.559 102.496 1.00 0.00  H
ATOM 33 HE22 GLN L 3 20.049 -6.007 103.275 1.00 0.00  H

What's this entry's nr.31 supposed to describe? I thought that an entry without remoteness indicator (and branch designator) might be bound to the previous atom, but in this case the N's adjacent H atoms are already dealt with with entries 32 and 33.

Also, this doesn't quite follow the convention(?)* of increasing the branch designator by one each time a new branch emerges—is this usual when H atoms' positions are given as well?

* like here:

ATOM 1661 CG  TRP A 214 24.705  7.905 6.526 1.00 50.03 C
ATOM 1662 CD1 TRP A 214 23.435  7.531 6.857 1.00 49.65 C
ATOM 1663 CD2 TRP A 214 25.087  8.914 7.475 1.00 51.52 C
ATOM 1664 NE1 TRP A 214 22.999  8.246 7.947 1.00 51.68 N
ATOM 1665 CE2 TRP A 214 23.993  9.098 8.350 1.00 52.47 C
ATOM 1666 CE3 TRP A 214 26.251  9.675 7.670 1.00 52.68 C
ATOM 1667 CZ2 TRP A 214 24.019 10.027 9.407 1.00 54.73 C
ATOM 1668 CZ3 TRP A 214 26.279 10.600 8.724 1.00 53.86 C
ATOM 1669 CH2 TRP A 214 25.167 10.761 9.581 1.00 54.84 C

First, your PDB data seems to have some formatting issues, in that the columns are in the wrong places.  I'm not sure if it's actually messed up, or if it simply got messed up when you posted it.

"In every amino acid there are two O atoms":  True for a single amino acid; not true for an amino acid residue.  I presume you're referring to the two oxygens on the carboxyl group.  However, one of these is lost when the amino acid forms an amide linkage with the next amino acid.

"What's this entry's nr.31 supposed to describe? ":  I opened the PDB in PyMol (after I corrected the format), and it seems to be the amide hydrogen

Regarding the branch designator -- I've never heard of such a thing in a PDB file, and I don't see any mention of it in the format description.  Could you clarify?

ETA:  I've tried to post the correctly formatted PDB data, but it seems to be messing up the formatting.  I'll try to figure out how to do that.
Here's an image of the residue


Entropy and specific heat capacity I have seen the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ where $C_p$ is the molar heat capacity at a constant pressure. I understand that this assumes that the temperature range is sufficiently small that the constant pressure heat capacity does not vary significantly over it.

My question pertains to the derivation of this expression. One of the first steps is substituting $q=C_pdT$ into $dS=\frac{q_{rev}}{T}$. I am not 100% sure as to why these enthalpies are equated. Is it because the heat exchanged by a system at a constant pressure is always reversible heat, or is this not true?

If the above were true, it would explain to me why I never see an expression involving constant volume heat capacity. If the above is not true, could someone also explain why suh a form does not exist?

Finally, is the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ only applicable where the pressures are the same at the initial and final temperatures? On the one hand, the derivation assumes a constant pressure in the infintesimal changes from the initial to final temperature (and of course uses a constant pressure heat capacity!); however on the other hand entropy is a state function. So I would think that if one could find how entropy varies wit pressure at a constant temperature, and the change from the final to the initial state involved both a temperature and pressure change, then I would think that the above expression can be used to find the change in entropy due to the temperature change only, and then the additional factor from the pressure could be used?

Thank you in advance.

EDIT: For the second part, from having a quick think about the reversible heat enxchange in and isothermal process, I think $S(T_2,p_2)=S(T_2,p_1)+R\
 \ln(p_1/p_2)$. So then one could construct a Hess-like cycle to go from one temperature and pressure/volume, to another temperature and pressure/volume. The use of the cycle and step-wise calulcation works because entropy is a state function I think?

Hopefully I answered the second part of my query (I would appreciate if someone could verify that it is correct), although I am still not sure about why the heats used are the same in the derivation, and whether the expression really does refer to the same pressure at the initial and final temperatures, even though the process in getting from these two temperatures can involve pressure changes (again, thinking about $S$ being a state function, so the process itself when changing states does not matter).

It is not true that heat exchanged at constant pressure is always reversible.
But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2.  It is of no further use.  You instead need to focus exclusively on the two end states.  And, in order to apply the entropy equation, you need to devise a reversible path between these exact same two ends states.  Any reversible path will do, because they will all give the same value for the entropy change.
In the case of your example, you can choose a constant pressure reversible path for convenience. So, for such a path, from the first law we know that $\mathrm{d}H = \mathrm{d}q$. But, now how do we figure out a reversible path between the two states in which the system temperature is changing reversibly along the path?  If we contact the system with at constant temperature reservoir held at the final thermodynamic equilibrium temperature $T_2$, that path would not be reversible because there would be a finite temperature difference between the system and the reservoir over most of the path. But, now, suppose we contact the system, not with a single constant temperature reservoir, but with a sequence of constant temperature reservoirs, each at a slightly different temperature over the range $T_1$ to $T_2$. This would guarantee that, over the entire path, the system would differ only slightly from the reservoir it is currently in contact with. So, in this case, $\mathrm{d}q_\mathrm{rev} = \mathrm{d}H = C_p\,\mathrm{d}T$, where $\mathrm{d}T$ is the change in the system temperature over the present increment of heat exchanged. We use the equation
$$\mathrm{d}H = C_p\,\mathrm{d}T$$
because of the precise definition of constant pressure heat capacity in thermodynamics:
$$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$
So now, for the change in entropy, we have:
$$\Delta S = \int{\frac{\mathrm{d}q_\mathrm{rev}}{T}} = \int_{T_1}^{T_2}\frac{C_p \,\mathrm{d}T}{T}$$
So the key thing to remember here is that, if you want to determine the change in entropy for an irreversible process, you first need to determine the initial and final end states (say, using the first law of thermodynamics) and then devise a reversible path between these same two end states to calculate the integral of $\dfrac{\mathrm{d}q_\mathrm{rev}}{T}$ for that path.


What mass of water is present as a liquid when equilibrium is established? $1.00~\mathrm{g}$ of water is introduced into a $5.00~\mathrm{L}$ flask at $50~^\circ\mathrm{C}$. What mass of water is present as a liquid when equilibrium is established? Also you know $92.5~\mathrm{mmHg}$ (vapor pressure).

  $0.083~\mathrm{g}$
  $0.41~\mathrm{g}$
  $0.59~\mathrm{g}$
  $0.91~\mathrm{g}$

I need thorough explanation for each step because I'm new to this. This question is in the gases section of the book, so I think I need to apply the ideal gas law, but I don't know.

With the help from the comment I came up with the following solution:

$$
\begin{align}
\frac{92.5}{760}~\mathrm{atm} \cdot 5.00~\mathrm{L} &= n \cdot 0.08206 \cdot 323~\mathrm{K}\\
n &= 0.023~\mathrm{mol}
\end{align}
$$

Since $1~\mathrm{mol}$ of $\ce{H2O}$ is $18~\mathrm{g}$, the answer is $0.414~\mathrm{g}$, or (2) $0.41~\mathrm{g}$.

The given temperature of $T=50\ \mathrm{^\circ C}$ and the given pressure of $p=92.5\ \mathrm{mmHg}$ approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of $T=50.000\ \mathrm{^\circ C}$ is $p=92.647\ \mathrm{mmHg}$, which could be rounded to $p=92.6\ \mathrm{mmHg}$ and not $p=92.5\ \mathrm{mmHg}$; in the opposite direction, however, the corresponding saturation temperature for a given pressure of $p=92.500\ \mathrm{mmHg}$ is $T=49.968\ \mathrm{^\circ C}$, which could be rounded to $T=50\ \mathrm{^\circ C}$.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of $T=50\ \mathrm{^\circ C}$).

Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions.

The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour.

The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about $\rho=988\ \mathrm{kg\ m^{-3}}$, the given mass of $m=1.00\ \mathrm g$ corresponds to a maximum volume of $V_\text{liquid}=1.01\ \mathrm{ml}=0.00101\ \mathrm l$, assuming that no water has evaporated. Considering that the total volume is given as $V=5.00\ \mathrm l$ (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected.

Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred.
$$1\ \mathrm{mmHg}\approx 133.3224\ \mathrm{Pa}$$
You might want to convert the given values to coherent SI units, or find a value for the molar gas constant $R$ that is expressed in a suitable unit.

The following data are given:

Temperature $T=50\ \mathrm{^\circ C}=323.15\ \mathrm K$
Pressure $p=92.5\ \mathrm{mmHg}=12332.322\ \mathrm{Pa}$
Volume $V=5.00\ \mathrm l=0.00500\ \mathrm{m^3}$

The value of the molar gas constant is $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$.[source]

The amount of vapour $n$ in the container may be estimated using the ideal gas law
$$\begin{align}
p\cdot V&=n\cdot R\cdot T\\[6pt]
n&=\frac{p\cdot V}{R\cdot T}\\[6pt]
&=\frac{12332.322\ \mathrm{Pa}\times0.00500\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times323.15\ \mathrm K}\\[6pt]
&=0.022949674\ \mathrm{mol}
\end{align}$$
(Note that this intermediate result is given with an excessive number of digits. It is deliberately not rounded in order to avoid round-off errors in subsequent calculations.)

Since the molar mass of water is $M=18.01528\ \mathrm{g\ mol^{-1}}$, the corresponding mass of vapour is
$$\begin{align}
m&=M\cdot n\\[6pt]
&=18.01528\ \mathrm{g\ mol^{-1}}\times0.022949674\ \mathrm{mol}\\[6pt]
&=0.413444803\ \mathrm g\\[6pt]
&\approx0.41\ \mathrm g
\end{align}$$
(Note that this final result is rounded to two significant digits. The number of significant digits is estimated in view of the fact that the mass of water $m=1.00\ \mathrm g$, the volume $V=5.00\ \mathrm l$, and the pressure $p=92.5\ \mathrm{mmHg}$ are given with three significant digits, but the temperature $T=50\ \mathrm{^\circ C}$ is only given with two significant digits.)

This result is the mass of vapour $m_\text{vapour}$ in the container, whereas the question is about the mass of liquid water $m_\text{liquid}$, which can be calculated from the given total mass of water $m_\text{total}=1.00\ \mathrm{g}$.
$$\begin{align}
m_\text{total}&=m_\text{vapour}+m_\text{liquid}\\[6pt]
m_\text{liquid}&=m_\text{total}-m_\text{vapour}\\[6pt]
&=1.00\ \mathrm g-0.41\ \mathrm g\\[6pt]
&=0.59\ \mathrm g
\end{align}$$

Therefore, the correct answer is option 3) $0.59\ \mathrm g$.

By way of comparison, precise engineering calculations for water and steam usually do not rely on the ideal gas law but use so-called steam tables. We may use such steam tables to check the accuracy of our estimate that has been based on the ideal gas law.

For example, for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium, we find the following values in REFPROP – NIST Standard Reference Database 23, Version 9.0:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988.00\ \mathrm{kg\ m^{-3}}=988.00\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.083147\ \mathrm{kg\ m^{-3}}=0.083147\ \mathrm{g\ l^{-1}}$

Furthermore, the database includes various other thermodynamic and transport properties, which are not required for the following calculations.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium can be obtained using the input “water boiling at 50 °C”:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988\ \mathrm{kg\ m^{-3}}=988\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.08315\ \mathrm{kg\ m^{-3}}=0.08315\ \mathrm{g\ l^{-1}}$

Note: The indicated values for STP are not in accordance with IUPAC recommendations.

We know that the mass balance in the container is
$$m_\text{total}=m_\text{vapour}+m_\text{liquid}\tag{1}$$
where $m_\text{total}=1.0000\ \mathrm g$ is the total mass of water (liquid and vapour) in the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Since density is defined as
$$\rho=\frac mV$$
the corresponding volume balance can also be expressed in terms of mass:
$$\begin{align}
V_\text{total}&=V_\text{vapour}+V_\text{liquid}\\[6pt]
&=\frac{m_\text{vapour}}{\rho_\text{vapour}}+\frac{m_\text{liquid}}{\rho_\text{liquid}}\tag{2}
\end{align}$$
where $V_\text{total}=5.0000\ \mathrm l$ is the total volume of the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Solving the system of the equations $\text(1)$ and $\text(2)$ yields the solutions

$$\begin{align}
m_\text{vapour}&=\frac{\rho_\text{vapour}\cdot\left(V_\text{total}\cdot\rho_\text{liquid}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=\frac{0.083147\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times988.00\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.41569\ \mathrm g
\end{align}$$

and

$$\begin{align}
m_\text{liquid}&=-\frac{\rho_\text{liquid}\cdot\left(V_\text{total}\cdot\rho_\text{vapour}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=-\frac{988.00\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times0.083147\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.58431\ \mathrm g
\end{align}$$

These results confirm that the above-mentioned estimates $m_\text{vapour}\approx0.41\ \mathrm g$ and $m_\text{liquid}\approx0.59\ \mathrm g$, which have been obtained using the ideal gas law and neglecting the volume of the liquid water, are reasonable and sufficient in order to answer the question.