What exactly are electron states? From Tro's Chemistry: Structure and Properties [1, p. 93]:

2.5 Quantum Mechanics and the Atom
As we have seen, the position and velocity of the electron are complementary properties—if we know one accurately, the other becomes indeterminate. Since velocity is directly related to energy (recall that kinetic energy equals $\frac 1 2 mv^2),$ position and energy are also complementary properties—the more we
know about one, the less we know about the other. Many of the properties of an element, however, depend on the energies of its electrons. In the following paragraphs, we describe the probability distribution maps for electron states in which the electron has well-defined energy, but not well-defined position. In other words, for each of these states, we can specify the energy of the electron precisely, but not its location at a given instant. Instead, the electron’s position is described in terms of an orbital, a probability distribution map showing where the electron is likely to be found. Since chemical bonding often involves the sharing of electrons between atoms (see Section 4.2), the spatial distribution of atomic electrons is important to bonding.

The textbook says that electrons can exist in two states simultaneously. As I went further, I saw that an electron could exist as both a particle and a wave. So I thought that this must be the states that I was introduced to earlier, until I was learning about quantum mechanics.
In the text above, the writer was talking about electron states with well defined energy. From what I know the electron state that has well defined energy would be one with well defined velocity. And the velocity of an electron is associated with it's wave nature.
So, aren't we supposed to be dealing with electron as waves only? Or what exactly does the writer mean by the electron states?
Reference

Tro, N. J. Chemistry: Structure and Properties, 2nd ed.; Pearson: Hoboken, NJ, 2018. ISBN 978-0-13-429393-6.

In QM, properties that can be measured (such as energy, spin, position, etc) are each associated with an "operator". Every observation will always return an eigenvalue for the relevant operator. Eigen values are values for $\lambda$ for which the equation $$\hat{O}\Psi = \lambda\Psi$$ is true, where $\hat{O}$ is the operator of interest and $\Psi$ is the wave function of the entity of interest.
For example, let's consider electron "spin", by which we typically mean specifically the component of the spin in the z-direction, given by the operator $\hat{S}_z$. For the electron, this operator has two possible eigenvalues, which we abbreviate as $+1/2$ and $-1/2$. An electron having one of these values of spin is described as being in the eigenstate (or more simply just "state") associated with the given eigenvalue.
So we can say that having $+1/2$ spin and having $-1/2$ spin are two possible states of the electron.
What makes QM different from classical mechanics is that measuring an electron to have $+1/2$ spin does not mean that it was necessarily in the $+1/2$ spin state prior to our taking the measurement. An electron can (and often does) exist in a super-position of states, meaning that it simultaneously has both $+1/2$ and $-1/2$ spin, but our measurement causes it to collapse into one state or the other.
As you described in your question, some pairs of operators have the property that you cannot measure a defined value for both at the same time. Position and momentum are the most famous pair, since they are the example for the Heisenberg uncertainty principle, but many other pairs behave similarly. Energy and position are one such pair. So if we measure energy precisely, we cannot know where the electron with that energy exists. Since energy levels are the basis of the Schrodinger equation and define much of the behavior of atoms, chemists tend to treat electrons as if they are in a state in which the energy has been measured and other properties are unknown.
To address your point of confusion - "wave" and "particle" are descriptions of the behavior of electrons, most useful when understanding the concept of positional uncertainty, since "position" is most associated with particle behavior. When we learn about the Heisenberg uncertainty, we learn about experiments in which electrons behave as waves when their position has not been measured and like particles when their position has been measured, so it's easy to see why you associate wave and particle behavior with velocity and position, respectively, but those are just convenient examples to show the concept of duality. Operators generally should not be thought of as being associated with one description or the other, nor should we think of an electron as switching between particle and wave. It is better just to think of it as having specific position in space or not, and whether or not it has specific position it is neither a wave nor a particle, but rather just a quantum mechanical entity.
The results of any experiments we perform with electrons are predicted very well by the mathematical system that has been developed for quantum mechanics, and some of that math has parallels with classical wave behavior and some with classical particle behavior and some with neither of them. Beyond that, we have very little understanding of what an electron actually is and how its behavior accords with our understanding of the physics of the macroscopic world.
ADDENDUM For completeness, I'll add that in addition to $\hat{S}_z$, two other important operators that measure properties of electrons that can be known at the same time as energy is known are $\hat{L}^2$, the operator for the magnitude of total (quantum) angular momentum and $\hat{L}_z$, the z-component of that angular momentum. Thus, we can theoretically have an electron for which all four properties are fixed at one time, and we define the state described by the four eigenvalues by representing those eigenvalues with the "quantum numbers" (simplified forms of the eigenvalues) with which you are probably familiar:
n is the quantum number which represents the eigenvalues of the energy operator $\hat{H}$ (ie the Hamiltonian, which we associate with a specific shell)
l is the quantum number which represents the eigenvalues of the operator $\hat{L}^2$ (which we equate with s,p, d etc orbitals)
ml is the quantum number which represents the eigenvalues of the operator $\hat{L}_z$ (which we equate with the axial orientation of orbitals)
ms is the quantum number which represents the eigenvalues of the operator $\hat{S}_z$ (which we equate with the above-described spin orientation)


How do we actually know the chemical structure of a compound? If a compound is placed before you and you are asked to determine what it is and what it's structure is, how do you actually do that?

This might be a baby question for most of you but I am not from chemistry background and so this amazes me!

The easiest way is to inquire where the compound originated, observe the compound with generic senses (color, texture, phase-gas/liquid/solid) and then make educated guesses based on that! If there remains any doubt as to the chemical or if that info is not available then a systematic process of ruling out the most likely possibilities is conducted. The single largest category in chemistry would be organic or carbon containing compounds. They may be analyzed using NMR to determine their structure with good accuracy. Any decent sized chemistry department will have an NMR available. For inorganic compounds, some simple acid-base tests could narrow the possibilities.

With the most coveted and expensive tools, it is easy to tell what elements compose a sample. I once visited a forensic laboratory where they had a scanning electron microscope that could display individual grains of gunshot residue (invisible to the naked eye)for which police swab the hands of suspects. The particular SEM I saw in action had an x-ray source connected to it, which allowed any point including an individual microscopic dot of suspected gunshot residue to be analyzed for its elemental composition. The guy there clicked on any part of the SEM image to display what element it was made of... completely automated!

One of the longest laboratory experiments I did in school was determining the percentage of iron in a mixture of iron and clay. It took 12 hrs and used only wet chemistry from 100 yrs ago.

There are many techniques that could be used but it really comes down to which one is most promising or available.


Why IUPAC name this amine is written as 2-methylpropan-2-amine and not as 2,2-dimethylethanamine? Why IUPAC Nomenclature of following amine is written as 2-methylpropan-2-amine and not as as 2,2-dimethylethanamine?

The basic principle in nomenclature is to look for the longest carbon chain: here it is the chain with 3 carbon atoms so the suffix prop; since it is an amine connecting to the carbon 2 it gives the suffix propan-2-amine.
As you have a $ \ce {CH_3} $ group grafted in position 2 of the main carbon chain, you must add the prefix 2-methyl
So you have 2-methylpropan-2-amine.


What are introductory level books on chemical engineering? I'm currently enrolled in a chemical engineering course in my college, and in my first year, I'd like to get a nice introductory book on that field of study.
I'd prefer that the book covers several topics, but less in-depth analysis of each.

Unfortunately the big-list in the Resources for learning Chemistry post does not yet cover this topic yet. With respect to the meta discussion the answer to this question will be amended to that list (and this question can be closed as duplicate after the fact).

I'm studying chemical engineering for more than two years and what I can tell you is that there is not really a lot of "good" books you may read as an introduction.
You have to understand that Chemical Engineering contains both "Chemical" and "Engineering". The first one refers to chemistry whereas the second refers to the way to make something. A chemical engineer who works at Metalor (which makes precious metal bars amidst other things) really has not the same problems as another one working at Air Liquide (which produce liquid gas).

Now if you are looking for what we call the basics, any chemical engineer must be good at, fluid mechanics, thermodynamics and kinetics, mathematics and informatics are a good start.

Fluid mechanics because in any plant, whatever you produce, you need to be able to size pipes, reactors, etc. It's also a must if you encounter a two-phase flow;
Thermodynamics because when you are sizing somethings you must ensure safety and security so you need to know what properties the molecules involved in your plant have. Kinetics because this is important to know the rate of a reaction and a lof other stuff related to catalysts etc.;
Mathematics because you can't do the first three without it;
Informatics because this is pretty hard to perform all the correlations and models by hand when sizing equipment and knowing how to use the common software (Aspen, Matlab) plus having some skills in coding a few lines in either Fortran/C/C++/Python is always a must. Especially if your company does not have the budget to pay the licences and Excel is limited in many ways;
Finally, mathematics and informatics are a strong requirement to design process and safety control loops efficiently and test them with data from the process before implementation.

If you speak French, the book "Le génie chimique à l'usage des chimistes" by Joseph Lieto gives a pretty good overview about Chemical Engineering. How to make a balance (of mass or energy or momentum), basic operation units (distillation and crystallisation), mass and heat transfer, thermodynamics, kinetics and process control.

The best reference I have for chemical engineering is the McGraw-Hill Chemical Engineering Series which contains more than you should know as a student. After you will need to get strong knowledge in many areas of chemistry (electrochemistry is sometimes a big plus) because as an engineer and a chemist especially if you are looking for sizing operation, you need to think about details (Production practices, HSE, Quality), which are not in the books, unfortunately.


Kinetics of converting a water + heavy water mixture to HDO Imagine an initial 50%-50% mixture of normal ($\rm H_2 \rm O$) and heavy water ($\rm D_2 \rm O$).

I think, simply mixing them, the result will exponentially converge to an equilibric state, where about a half of the solution is $\rm{HDO}$.

What is the time constant with what the system is going to the equilibric state?

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

  Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.


Molpro - Connection between atomic term symbols and system symmetry Let's say I have a carbon atom with the electron configuration $1s^22s^22p^2$. We know, that using atomic term symbols we can describe the following states:

$${}^3P_0, {}^3P_1, {}^3P_2, {}^1D_2, {}^1S_0$$

Now I'd like to compute their energies with Molpro. But how should I specify those terms using just symmetry point groups?

I saw something similar using state-averaged multi computation from several different symmetries. But, I don't know, how to derive them just from the term symbol?

My attempt

I was thinking about writing the state wavefunction as the linear combination of microstates and specifying each of then separately.

I.e. let's have the state ${}^1S_0$. According to this answer it can be described like this:

$$|L = 0, M_L = 0\rangle = \frac{1}{\sqrt 3} \left( |m_{l1}= 1, m_{l2} = -1\rangle + |m_{l1}= -1, m_{l2} = 1\rangle - |m_{l1}= 0, m_{l2} = 0\rangle \right)$$

I presume, that I can choose any of the available point groups in Molpro for 1 atom. Let's say I'll choose $D_{2h}$ point group.

Then the orbitals' symmetry can be described like this:

$1s$ ($A_g$)
$2s$ ($A_g$)
$2p_x$ ($B_{3u}$)
$2p_y$ ($B_{2u}$)
$2p_z$ ($B_{1u}$)

According to this, the symmetry of the state $\left| m_{l1} = 1, m_{l2} = -1 \right>$ should be
$$(A_g)^2 \otimes (A_g)^2 \otimes (B_{3u})^1 \otimes (B_{1u})^1 = B_{2g}$$

and the corresponding wavefunction specification in Molpro: wf,6,6,0.

The state $\left| m_{l1}=-1, m_{l2} = 1 \right>$ posses the same symmetry, i.e. $B_{2g}$ and so the Molpro wavefunction specification stays the same wf,6,6,0.

Finally, the state $\left| m_{l1} = 0, m_{l2} = 0 \right>$ has the symmetry

$$(A_g)^2 \otimes (A_g)^2 \otimes (B_{2u})^2 = A_g$$, i.e.
the wavefunction will be specified like wf,6,1,0 and multi will be called
with both of the wavefunctions together

{multi;
 occ,2,1,1,0,1,0,0,0;
 core,1,0,0,0,0,0,0,0;
 wf,6,6,0;
 wf,6,1,0;
}

Is this the correct approach?

You are trying to use the wf card to pass individual configurations. This is wrong in 2 ways:

The wf card does not specify which configurations to use, it specifies the symmetry of the wave function (or electronic state).
You don't need to tell Molpro which configurations to use. It will figure this out on its own based on the symmetry of the wave function you are requesting.
All you need is the IRREP of the $^1S_0$ state, which is totally symmetric.

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,1,0;
 }

However, this will give you different a state than expected. Add expec2,lxx,lyy,lzz; and check $\langle L^2\rangle$ in the output. It should be 6, telling us the $^1D_1$ state is lower in energy than $^1S_0$. So we need to request more states of that symmetry (use the state card). Also, we should make sure the $^1D_1$ is described appropriately, as it might negatively affect the targeted $^1S_0$ state. So you should add all of the 5 degenerate $^1D_1$ components. I leave that as an exercise for you (see also below).

Some further note about your configuration $(a_g)^2\otimes(a_g)^2\otimes(b_{3u})^1\otimes(b_{1u})^1=B_{2g}$ (Note that the symmetry of orbitals is commonly denoted by small letters, while capital letters are used for electronic states.)

Only IRREPs of same symmetry mix, so this one will not contribute to $^1S_0$.
It is breaking the spherical symmetry of the atom. This can only happen if there is some external perturbation, e.g. a magnetic or electric field.
It is however, one component of a degenerate (different) state. But you need to consider all the degenerate components to preserve symmetry. This is an issue of the symmetry reduction from $O_h$ to $D_{2h}$. For example you can calculate the $^3P$ state with

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,4,2;
  wf,6,6,2;
  wf,6,7,2;
 }

Check that you get 3 exactly degenerate $p$ orbitals, and 3 exactly degenerate electronic state energies.
If you would request each state separately, by running multi 3 times with only one wf card each, you will get 3 degenerate state energies, but the $p$ orbitals won't be degenerate.


What is known about liquid carbon? The wiki tells me that if you heat carbon at atmospheric pressure it eventually turns directly into a gas without being liquid first.  At what pressure can you make liquid carbon?  Has anyone actually ever studied liquid carbon?  Surprisingly a web search did not answer these questions.

The carbon phase diagram shows that liquid carbon is only achievable at high pressure (~100 atm) and high temperature (~4500 K). This means that liquid carbon probably has very limited applications, and not many will be researching it. Funding is hard to come by when the applications of the research are not apparent.

Image source


Acetone Enolate-HaloBenzene Reaction A reaction I am trying to perform says to use a strong base like Sodium Amide with Acetone and Iodo-Benzene as reactants with DMSO as the solvent to form Phenylacetone. The main idea is to use the Acetone Enolate to form Phenylacetone when the Halogen is removed from the Halobenzene.

Can I just put NaOH in Acetone with a Halobenezene?

I could be wrong but I think this reaction works by forming benzyne, which then undergoes enolate attack:

I don't think hydroxide is basic enough to do this, so no I don't believe your proposed reaction conditions will work. Is there any reason you can't just use NaNH2, or BuLi, LDA etc.?


Place cooling bath apparatus in freezer to speed up the process? As part of an at-home lab, my chemistry class is creating cooling baths to measure the freezing point of different solutions. We have to determine the molar mass of the solutes in these solutions by observing the freezing point. The cooling bath is constructed from ice and salt inside a 500 mL beaker, into which are placed test tubes of solutions (one at a time). A thermometer sticking out of the test tube hypothetically measures the freezing point. The problem is that the ice bath is taking hours to freeze the solution.
To speed up this process, can I just stick the whole apparatus in the freezer and check it every 10 minutes or so to see the temperature at which the solutions freeze?
Will the freezing point of salt water and calcium chloride water (the solutions in question) be radically different if frozen in an ice bath vs. frozen inside a kitchen freezer?

The freezing point remains the same, but its measured value may depend on the way of measurement.
In case the solution freezing point is too close to the ice/salt 3:1 bath temperatures, it may be the reason for too long freezing time. The bath ice/$\ce{CaCl2 . 6 H2O}$ bath 1:2 could help, if applicable.
Note that fast freezing may lead to biased value, especially, if supercooling occurs. E.g. a PET bottle with water can get supercooled in a freezer deep below the freezing point.


Why does the earth have more radon in some locations? I am not a chemist so I'm not going to give specific amounts of radon; different countries use different units. My question: why do some areas have more radon? For instance, Australia has a spreadsheet of all locations and their radon levels. I've looked it up and where I used to live, nearby, has half the listed Radon levels of where I live now. Wikipedia was a bit general to me for American examples:

  Iowa has the highest average radon concentrations in the United States due to significant glaciation that ground the granitic rocks from the Canadian Shield and deposited it as soils making up the rich Iowa farmland.

So it is rock or soil types or weather or something else that causes the different amounts?

Radon nuclides are short-lived members of the natural radioactive decay chains of thorium and uranium. On a time scale that would be relevant for geology, you should better ask why some locations have more uranium or thorium. The distribution of radon concentrations in the ground air is a result of the distribution of uranium or thorium in the ground.

The main reason for the inhomogeneous distribution is igneous differentiation.  The original liquid magma might have been rather homogenous. But when the magma cooled down, fractional crystallization started. This removal of mineral precipitates also changed the composition of the remaining melt. Uranium and thorium ions are incompatible, i.e. they do not fit well in the crystal lattices of the silicate minerals that precipitate first. Therefore, they were accumulated in the remaining melt. Primary deposits of uranium and thorium minerals (such as pitchblende) were formed where the last parts of the cooled magma solidified.

Furthermore, secondary deposits were formed when uranium or thorium minerals were dissolved, transported and precipitated in ground waters. (This includes the nicely coloured and fluorescent uranyl ($\ce{UO2^2+}$) minerals.


How can there be concentration and pressure terms simultaneously in the Nernst equation? In the Nernst equation we are taught that the reaction quotient can have pressure terms and concentration terms simultaneously. According to my knowledge this does not happen anywhere else, so why can it happen here?

This topic often causes confusion.

The most correct form of the Nernst equation uses thermodynamic activity of the substances:

$$E = E^{\circ} - \frac{RT}{zF} \ln\frac{a_\mathrm{red}}{a_\mathrm{ox}}$$

For diluted solutions, we can replace the activity of a substance with its concentration in the solution. But how do we express the activity of a gas?

This is basically the question of how to formulate the equilibrium constant of the redox reaction. There are actually multiple solutions to this. The problem is that all of them are correct.

Solution 1: Express activity of the gas as a function of partial pressure.
\begin{align}
pV &= nRT \\
\Rightarrow p &= \frac{n}{V}RT = cRT\\
a(\mathrm{gas}) &\approx c(\mathrm{gas}) = \frac{p(\mathrm{gas})}{RT}
\end{align}

Solution 2: Just use the partial pressure itself:
$$a(\mathrm{gas}) = p(\mathrm{gas})$$

Solution 3: Use the partial pressure divided by a reference pressure:
$$a(\mathrm{gas}) = \frac{p(\mathrm{gas})}{p(\mathrm{reference})}$$


Why does heptane have an odour I read that alkanes are 'colourless and odourless'. Heptane is an alkane. Why does heptane smell like heptane? Impurities?

Some higher-molecular weight alkanes definitely have an odor, but they aren't as intense as most of the compound types you'll find stinking up an organic lab.

"The odor threshold for hexane is 130 parts per million (ppm), with a faint peculiar odor reported." ---http://www.epa.gov/ttnatw01/hlthef/hexane.html
"[Heptane is a] colorless liquid with a mild gasoline-like odor." ---http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf
"Alkanes have relatively high odor detection threshold (37 to 50 mg/m$^3$ for dodecane) as compared to other VOC groups such as sulphides (0.0003 to 0.16 mg/m$^3$  for
dimethylsulphide), phenolics (0.022 to 4 mg/m$^3$  for phenol), and nitrogen heterocycles (0.0004 to 0.0008 mg/m$^3$  for skatole) (O'Neill and Phillips, 1992)" ---http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf

According to the following paper, there are apparently at least two olfactory receptors  in rats that respond to alkanes; one accepts alkanes with areas around 3.5 nm$^2$, and another  responds to hydrocarbons with areas of about 5 nm$^2$. The authors note that response to cycloalkanes is higher than for normal alkanes, so the shape matters too.

"Concanavalin A reveals olfactory receptors which discriminate between alkane odorants on the basis of size." ---http://www.ncbi.nlm.nih.gov/pubmed/2803264


Oxidation of fructose with conc. HNO3 I recently studied carbohydrates. I was taught that fructose on oxidation with strong oxidizing agents like concentrated $\ce{HNO3}$ yields glycolic acid and tartaric acid along with trihydroxy glutaric acid. Though I am confused why glycolic acid is formed instead of oxalic acid (since the terminal $\ce{-OH}$ should also have been oxidized) I am also satisfied that oxidation of the keto group in fructose breaks it off into 2 carboxylic acids and yields a 2 carbon(glycolic) and a 4 carbon(tartaric) containing acid.
But I am very much unable to figure out the formation of trihydroxy glutaric acid by oxidizing fructose since its a 5 membered carboxylic acid and in no reaction I know did I ever come across such breakage of ketones in 6=5+1. Some sources even cited the formation of trihydroxy butyric acid along with oxalic acid where I couldn't understand why tartaric acid wasn't formed by oxidation of the terminal $\ce{-OH}$ group.

Sir Arthur7: From your comments you appear to know how to liberate CO2 and a proton by pushing arrows as illustrated by the blue arrows in generic α-oxoacid 1. You are correct in discrediting acyl anion 5 for two reasons. First, it is unstabilized and secondly, its protonation would provide an aldehyde, not a carboxylic acid. So why not push the red arrows in 1 to generate an acyl cation equivalent? The downside is that you don't want hydride as a leaving group. The solution is to replace the acidic hydrogen of 1 with a species that likes electrons, e.g., a high oxidation state metal or a hypervalent, non-metallic atom. Certainly, the nitrogen of HNO3 fits the bill where M = NO2 in structure 2. In this instance, nitrogen is n = 5 in structure 2 and 4. Bear in mind that for the purpose of electronic bookkeeping, transformation of the acyl anion 5 to a acyl cation 2 is a 2-electron oxidation while species M undergoes a 2-electron reduction. ---continued---

Moriarity, et. al.,1 have effected this oxidative decarboxylation with the hypervalent iodine species, iodosobenzene 7. The path of the red arrows within the brackets provides a step-by-step version of their mechanism using α-oxoglutaric acid 6.2 Notice that mixed "anhydride" 8 is the operational equivalent of species 2. The authors propose the formation of cyclic intermediate 9 that collapses to 12 with loss of CO2 and a proton. [Note: Structure 10 is drawn for clarity. Collapse of 9 to succinic acid 12 is proposed as concerted.] The collapse of structure 8 might be thought to follow the course of the green arrows but because the reaction is conducted in anhydrous dioxane, it is unlikely that an acylium ion is formed directly because it is apt to form succinic anhydride. See examples studied in the chart below.

The Moriarity paper describes the iodosobenzene oxidation of mandelic acid to benzaldehyde and CO2. See if you can now write a mechanism for this reaction.

1) R. M. Moriarty, S. C. Gupta, H. Hu, D. R. Berenschot, and K. B. White, J. Am. Chem. Soc., 1981, 103, 686. https://pubs.acs.org/doi/10.1021/ja00393a040
2) α-Oxoacid is preferred over α-ketoacid. The term keto (C=O) overcounts carbons. Oxo = (=O).


Finding the pressure equilibrium constant of the reverse reaction At $1000~\mathrm{K}$, $K_\mathrm p = 1.85$ for the reaction.
  $$\ce{SO2 + \frac12O2 <=> SO3}$$

  What is the value of $K_p$ for the reaction
  $$\ce{SO3 <=> SO2 + \frac12O2}$$

Now, since the task gives us $K_\mathrm p$, it seems strange to ask for it again with the equilibrium put the other way around. Does this make sense? Is it a typo? Either that or my understanding is flawed. It would make more sense to ask for $K_\mathrm c$.

The chapter has this equation: $K_\mathrm p = K_\mathrm c(RT)^{\Delta n}$, I think this is relevant given the temperature is available to solve it.

Source: task 15.21 in the 11th edition of Chemistry: The Central Science, and the answer is suppose to be $0.541$.

They are actually just looking for the equilibrium constant for the reverse reaction.

$$K_p(-\xi) = K_p^{-1}(\xi) = \frac{1}{1.85} = 0.541$$


Why is phosphorous acid diprotic and not triprotic? In H3PO3 only only 2 hydroxide groups exist, instead of 3. If there is 3 hydroxide group, the phosphorous will have 1 lone pair electrons, which will still satisfy the valence electron of Phosphorous.

The two isomers you're referring to both exist and are in equilibrium:

When only one or two protons are removed, resonance stabilizes the anions formed, e.g. the $\ce{HPHO3^-}$ anion:

The trianion (shown below) doesn't have the benefit of resonance stabilization. This makes removing the third proton much harder.


Does glass adsorb methanol to an exceptional extent? Does (borosilicate) glass adsorb methanol to an exceptional extent? We are doing vapor density measurements of organic solvents using a microbalance and a sealed glass sphere. Other polar molecules such as acetone present no problems, but we are getting anomalous results for methanol for some reason.
[Alternatively, does it absorb less than other solvents?]

Exceptional is subjective term, and can't be responded to sensibly.  But more/less can be discussed.  This is somewhat complex.  The isotherm is an equilibrium between adsorption and desorption.

Methanol is small, and when adsorbing to glass, presumably through hydrogen bonding, a single molecule will do very little to block other adsorption sites.  Other, larger, molecules might 'lay down' on the surface or have branched structures that may inhibit their peers from adsorbing, giving smaller surface coverage.  However, long linear molecules will tend to orient, and there is an energetic favorability to forming a monolayer for long linear alcohols (and thiols, carboxylates, amines, silanes, etc).

In these highly ordered linear systems there is an attractive force of roughly 0.2 kcal/mol per methylene unit.  Since methanol has the fewest number possible, if a methanol molecule desorbs it can (almost) freely leave the molecular layer, as opposed to longer chain molecules.  When you begin to discuss molecules that are more topologically diverse than simple linear chains, this attraction is significantly weakened, and as such those molecules may also (almost) freely desorb.

In essence the methanol isotherm has maximum kinetic lability, and as such the coverage will happen reasonably quickly, and be fairly complete.  At equilibrium I would expect methanol to have apparently higher coverage than any branched molecule, and before equilibrium, I would expect methanol to have more coverage than long-chain analogs.

Please note this is on a per molecular basis, and the mass equivalents of methanol will be smaller than any other molecule to be compared with (save water).


Explanation of the orbital energies resulting from Jahn-Teller distortion of octahedral complex using MO theory I am confused about the energies of the orbitals after Jahn-Teller distortion of an octahedral complex. Take elongation for example.
To my understanding, since the ligands on the $z$ axis move away, there is less orbital overlap with the $d_{z^2}$ metal orbital, leading to weaker orbital interactions, leading to the antibonding orbital being lowered in energy.
There are two things I don't get: 1. The lowering of $d_{xz}$ and $d_{yz}$ and 2. the raising in energy of the $d_{x^2-y^2}$ and $d_{xy}$ orbitals.
1. The $d_{xz}$ and $d_{yz}$ don't seem to have the correct symmetry to sigma bond with any $p$ orbitals of the ligands. They are nonbonding in a perfect octahedron, so how can the be lowered in energy after distortion? Maybe they only appear lower relative to the $d_{xy}$?
2. Are they not actually raised in energy, only appearing higher relative to the $d_{z^2}$ and $d_{xz}$ and $d_{yz}$ orbitals? Or are the equatorial bonds shortened so there is a stronger interaction, raising the $d_{x^2-y^2}$? If so, then why is the $d_{xy}$ orbital raised?
Is there an explanation for the orbital energies using MO theory? If you could clear up my confusion that would be great. Thanks!

I disagree with your statement that

"The dxz and dyz don't seem to have the correct symmetry to sigma bond with any p orbitals of the ligands. They are nonbonding in a perfect octahedron"

Consider the case of a halide ligand like Cl-, which has filled p orbitals perpendicular to the axis of the M-Cl sigma bond. If these are aligned with the coordinate axes, they can each interact with proper symmetry with one of the off-axis t2g orbitals. In the axial position, these orbitals match up in symmetry with dxz and dyz.
In the perfect octahedron, all three t2g orbitals are affected equally (each interacts with the four ligands in its plane), so whether or not the ligand has p orbitals (and therefore pi bonding) does not qualtitatively change the splitting diagram. It just changes the magnitude of the splitting.
When the octahedron is distorted, however, the interactions with each of the t2g are no longer the same in magnitude. Just as with the sigma bond, as the axial ligand moves away from the metal center, the pi bonding interaction also weakens, so the d orbitals involved in pi bonding (which as you noted are anti-bonding orbitals) lower in energy.
The reason that the dxy appears to be raised in energy is likely just an artifact of centering the orbitals on the average energy in both cases. Alternatively, the diagram could be based on the assumption that as the axial ligands move away from the metal, the equatorial ligands move closer, increasing the strength of the sigma and pi bonding of those ligands. In the pi case, this would raise the energy of all three t2g orbitals because all of them can engage in pi bonding with equatorial ligands, but the dxy would go up the most since it can pi bond with all four equatorial ligands, whereas the other two can only pi bond with two of the four equatorial ligands.
UPDATE: Here's an image from some old lecture notes showing the difference in convention of alignment between the "MO" view and "d-orbital view" in which the "d-orbital" diagram recenters the energies around the average, even though only the dxz and dyz MO (or more properly the antibonding orbitals primarily composed of them) energies actually changes:


Diluting acetic acid to obtain a solution of pH 5 What percent of a solution needs to be acetic acid for a particular $\mathrm{pH}$?  I have 100% acetic acid, and want $1~\mathrm{L}$ of solution with a $\mathrm{pH}$ of 5. I've done some research, but I still do not understand how to solve this problem.

This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?

It is necessary first to find the molarity of the acetic acid. On PubChem, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.

If we have 1 L of acetic acid, then

$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$

We know that,

$$ \text{pH} = -\log{\ce{[H3O+]}} $$

which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our RICE table,

$$\ce{HC2H3O2 + H2O -> C2H3O2- + H3O+}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & x \ \text{mol} & - & \text{0 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -1 \cdot 10^{-5}\ \text{M} & - & 1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\text{End conc.} & x - 1 \cdot 10^{-5}\ \text{M} & - &  1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\end{array}

$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$

$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$
$$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$
$$2.75 \cdot 10^{-10} = 0.0000175x$$
$$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$

Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.

Two assumptions,

Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration.
Autoprotolysis of water does not significantly contribute to pH.


According to Gaussian's Modredund description, there is no such type as "O" anymore in Gaussian16.
Hence, the input deck is rejected. Gaussian16 provides generalized internal coordinates, for which more complex expressions are possible.


How to predict amine donor in biochemical reactions (whether glutamine or aspartate)? In biochemical reactions amine $(\ce{-NH2})$ transfer is done in multiple pathways (e.g. urea cycle, purine biosynthesis etc.).
In many of the situations glutamine donates amine group (e.g. xanthosine monophosphate → guanosine monophosphate). In few instances, however, aspartate donates amine $(\ce{-NH2})$ group (e.g. inosine monophosphate → adenosine monophosphate).
This seems to be very random to me. Is there any pattern in such amine transfer reactions? Can we generalise it in some way?

There are several very general concepts that you have to know about in order to see a pattern in biochemical reactions driven by enzymes. A quick answer is given at the very end. Amongst many concepts to understand, two main concepts are as following:
Enzyme are classified with a specific enzyme commission (EC) number, which is a type of ID number based on catalytic function. The first number groups the enzyme into an overall group. In example, enzymes that can transfer functional groups are placed in EC.2. Within EC.2 you can then subdivide into more specific functions, e.g. transaminases that transfer nitrogen containing groups have the the EC number of 2.6, and within this group again you can find enzyme that specifically transfer amine groups. Those enzymes will then have the EC number of 2.6.1 (aminotransferases), and you can then further specify into families, e.g. aspartate aminotransferases EC.2.6.1.1.
Enzymes folds within specific functional families is evolutionary conserved and highly specific. The three-dimensional shape of enzymes within a specific family, e.g. aspartate transaminase; EC 2.6.1.1, is very conserved (the amino acids sequence identity will also be very high)
and enzymes that are grouped into the same family will have highly similar overall structure and thus very similar mechanisms of catalysing reactions. In contrast, enzymes in completely different families, with different foldes and different amino-acid sequence will have completely different catalytic mechanisms.
To predict how some enzyme will catalyse a reaction, you need to find what enzyme it is related to. Here it will also be important to remember that structure is more conserved than sequence. A good starting point is to do a BLAST search with the protein sequence that you are investigating, note the you could also do a blast search against the PDB database to find related published protein structures.
Continuing with the exsample of amino-transferases, you can even subdivide the enzymes into where the catalytic reaction will take place. In example, you have alpha-aminotransferases or omega-aminotransferases, depending on the location on the substrate that the amino will we translocated to and from.
To answer your question:
Now that we covered some very general concepts about enzymes, it is important to know that their structure and sequence are specifically designed to accommodate a particular substrate. In other words, some enzymes will take aspartate as substrate and donate its amine group - while other type of enzymes will take glutamine as substate - and donate its amine group.

As a side-note. In proteins, the amine-group in aspartate would be part of the protein backbone (amide bond / peptide bond), and the side-chain (no amine group) would therefore be the only accessible part of this amino acid. It is therefore a good idea to specify if amino acids are substrates or catalytic residues (i.e. part of the enzyme) when asking about biochemical reactions.


Lennard-jones cut-off I am using a Lennard-Jones potential in a computer simulation and I was wondering if i should place a cut-off for low distances since the potential blows up quickly   making the calculations quite difficult.
I am trying to simulate basic molecules ( few atoms ) , calculating bonds stretching , bond angle bending and lennard-jones interactions and trying to minimize the total energy to get an optimal structure prediction .

It's hard/impossible to pinpoint the reason for your problems without seeing the  code and knowing which terms you use. If you feel like sharing it, then you should probably head over to Computational Science and make a post there.

A few general things come to mind:

Check that you have $\sigma$ and $\epsilon$ values inserted in the correct order.
Check the correctness of the combination rules you are using. It seems the force field you linked previously contains noncovalent terms from OPLS/AA. Make sure that the rest of the FF is compatible with OPLS/AA.
Check the units you are using, and make sure they are all the same. Probably won't make your simulation blow up, but it's still worth to take a look, especially if you are using parameters from different FFs.


Neutralization reactions I saw that most neutralization reactions produce water while some do not.
How can I know if water will be a product in a given neutralization reaction?
I initially thought that water is a product only when an acid reacts with an alkali (producing hydroxide ions) but there apparently exist some counterexamples like the reaction between ethanoic acid and copper oxide.

Neutralizations of any acid with oxides, hydroxides and carbonates always produce a salt and water.
The only neutralizations that do not produce water are reactions of an acid $\ce{HX}$ with ammonia $\ce{NH3}$ or with organic derivates of ammonia, like amines. These reactions produce an ammonium salt $\ce{NH4X}$ and no water
$$\ce{NH3 + HX -> NH4X}$$ If the acid is $\ce{H2Y}$ or $\ce{H3Z}$, the neutralization reactions are $$\ce{2  NH3 + H2Y -> (NH4)2Y}$$ $$\ce{3 NH3 + H3Z -> (NH4)3Z}$$ It never produces $\ce{H2O}$. If one atom $\ce{H}$ from $\ce{NH3}$ is replaced by an organic radical $\ce{R}$, a primary amine $\ce{RNH2}$ is obtained and the neutralization reaction produces also an ammonium salt $\ce{RNH3X}$ $$\ce{RNH2 + HX -> RNH3X}$$ Similar reactions can be obtained with secondary and tertiary amines. All these reactions produce salts and no water.


Can decarboxylation of carboxylic acids be effected by bases other than soda lime? What about other bases such as Group 1 or Group 2 hydroxides? Does it have to do something with the mechanism?

I don't understand the mechanism either:

What is NaOH supposed to with the acetate ion? I thought the acetate ion would be produced when NaOH reacted with acetic acid after which the reaction would proceed as follows:

Decarboxylation can be effected by pretty much any base strong enough to produce the anion plus a high enough temperature. I've used DBU to effect an aromatic decarboxylation.
The role of the $\ce{NaOH}$ is this reaction is to capture the $\ce{CO2}$ as sodium carbonate.


Naming rules and acidity of Cl2CHCH2COOH How do we name $\ce{Cl2CHCH2COOH}$ ?

Is seeing branched chain our number 1 priority so we name it

1,1-dichloro-3-propanoic acid?

also

Which is more acidic $\ce{Cl2CHCOOH}$ or $\ce{Cl2CHCH2COOH}$ ?
I feel it should be the first one as there is more −I effect but not sure.

1- 3,3-dichloropropanoic acid. Please see http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/

2- The first one  is more acidic, as the inductive effect, due to chlorine atom decrease fast with the distance. (So, your answer is correct!).


Will the dextrorotatory and laevorotatory enantiomers always have a set of stereocenters with opposite absolute configuration? I was faced with a question where the skeletal structure of of D(+)-glucose was given and I was asked to give the right structure of L(-)glucose. I noticed that the correct answer was the one where all the stereocenters had opposite absolute configuration of those of D(+)glucose.

But my question is: Will the dextrorotatory and laevorotatory enantiomers always have a set of stereocenters with opposite absolute configuration?

Precisely, these are based on the configurations proposed by German chemist Emil Fischer, who established the structures of many sugars.

  Emil Fischer elucidated these configurations for the aldohexoses in
  1896. According to the Fischer convention, D sugars have the same absolute configuration at the asymmetric center farthest removed from
  their carbonyl group as does D-glyceraldehyde. The L sugars, in
  accordance with this convention, are mirror images of their D
  counterparts,

Similary,the  Fischer projection for glucose  is shown below.

It is very important not to confuse these enantiomers with epimers   (two sugars that differ only in the configuration around one carbon atom e.g D-glucose and D-mannose, which differ only in
the stereochemistry at C-2)

References

Voet and Voet Biochemistry
Lehninnger Principles of Biochemistry
Biochemistry (Campbell and Farrel)


What are the factors affecting osmosis (bio)? Asking from the blood barrier angle. The wikipedia states that glucose, urea, and $\ce{Na/K}$ count. Can there be any other factors, like $\mathrm{pH}$, temperature, hydraulic pressure on the cell walls? Why do minerals not play a role? Because there are only traces of them?

Osmotic flow is the movement of a substance (typically water), across a semi-permeable membrane, from the side on which the chemical potential of the substance (water) is higher to the side on which it is lower.  I.e., just as temperature provides a gradient for the flow of thermal energy, chemical potential provides a gradient for the flow of matter.

Thus anything that affects the chemical potential of water will affect osmotic flow.  This includes dissolved solutes, pH (which of course comes from a particular dissolved solute—H$^+$), temperature, and pressure. For instance, increasing the solute concentration decreases water's chemical potential, while increasing the pressure increases it.

Indeed, that's the basis of osmotic pressure—water will flow toward the side with the greater solute concentration (activity) until the pressure on that side increases enough such that the reduction in water's chemical potential due to the presence of the dissolved solutes is exactly negated by the increase in water's chemical potential due to the increased pressure.


What is the mechanism of action of anaesthetics? A range of very different compounds are used in medicine as anaesthetics. They don't seem to have much in common chemically but they all seem to keep people asleep while medics are doing nasty things to them.

Inhaled anaesthetics include a variety of fluorocarbon gases such as halothane and sevoflurane plus barbiturates like thiopental, oddballs like propofol and even xenon:

What do these heterogeneous compounds have in common and how do they work?

This answer is a general answer, based on some research.

The mechanisms behind anaesthetics is not entirely understood, according to the article Mechanisms of Anesthesia: Towards Integrating Network, Cellular, and Molecular Level Modeling (Arhem et al. 2003), primarily due to, as you alluded to in your question, the fact that anaesthetics come from a very diverse group of chemicals.

In the fragmentary data presented in the article, the conclusions the authors made were that

  general anesthetics modulate the activity of ion channels, the main targets being GABAA and NMDA channels and possibly voltage-gated and background channels, thereby directly or indirectly hyperpolarizing neurons in thalamocortical loops, and thereby disrupting coherent oscillatory activity in the cortex. In our view, it does not seem unreasonable that the ultimate target is NMDA channels.

There is also a suggestion that the unconsciousness that results is in a way like sleep, an adaptive response to the chemical input.

Interesting perspectives are provided in the article Why anesthetic mechanism research
has failed, and what to do about it (Hammeroff), where the author suggests, research is needed in investigating anaesthetic mechanisms by

  Look closely at the quantum nature of anesthetic binding by van der Waals
  London forces.

In terms of local anaesthetics, the slides included in the presentation Local Anesthetics: Overview state that the mechanism here is due to

  Ionized form of weak base
  blocks $\ce{Na^+}$
  channels
  by binding to an internal
  sequence involved in channel inactivation


How to cheaply extract iodine from an alcoholic solution? I have an alcoholic solution of iodine. My goal is to extract the pure iodine without using rare or expensive compounds. How could I do this?

Alcoholic solution of iodine

What firstly comes to mind is that you meant "iodine tincture".
This question must have been frequently asked, since answers are available out there with different degrees of reliability. You must first follow the safety instructions:

Do your experiment somewhere that the noxious fumes won't be able to do harm to anyone. It's best to work under a fume hood.
You're gonna be dealing with strong acids. They're very corrosive when in contact with skin, metals or anything you can imagine (except glass).
Iodine tends to sublime. If not contained wisely, you'll get irritated skin, along with damages to your eyes (See about iodine hazards here).

In short, what you should do is:

Pour the tincture in a beaker, or any container. The important thing to note is to have "control" over what's happening in there.
Add water to the solution. The volume of the water you add should preferably be equal to that of tincture.
Add $\ce{HCl}$ to the solution, $\approx \pu{5-7 ml}$.
Add bleach to the solution, preferably equal or more than that amount of $\ce{HCl}$ you added.
Wait...wait...wait. For a relative amount of time, you should leave the solution for the reactions to happen.
You've past the hard part! Congratulations! Now, filter the dilute solution with a coffee filter.

Additional tips:

Iodine crystals sublime, so you shouldn't leave them in open air for long.
You should at least be wearing appropriate eye and skin protection. If you don't want them, do not do the experiment!

Wikihow has additional tips in case of any failures.
Sources:
http://www.wikihow.com/Extract-Iodine
http://www.sci-spot.com/Chemistry/iodine.htm


Recover Solder (Sn-Pb) from wires I have began to do an "art project" which utilizes solder as main material.

While I know that solder is cheap and easily found in multiple shops, I have set my mind on using recycled solder from out of use PCB's.

I've recovered a somewhat pure amount of material by using solder suckers, but a significant amount is trapped in my DIY desoldering braid which is simply salvaged wires. Those wires might be copper or some grey wires salvaged from a PC power supply.

What I have been able to come with is:

Some video I watched used (IIRC) $\ce{HCl}$ to dissolve solder from PCB's. My main issue is that I can't easily procure high concentration $\ce{HCl}$. My second issue is the acid reacting with the copper and maybe rendering the solder not usable (if this turns out to be a non-issue, I could figure a way to procure the acid).
I have been thinking of pulverizing the solder covered wires using a coffee grinder then heating the powder to the meting point of the solder, mechanically separating the two.

I wanted to ask if any of the two ways above would work without specialized equipment, or if there was a third way.

Unfortunately the reason solder works is that it partially dissolves copper as shown in Figure 1 to form $\ce{Cu3Sn}$ and $\ce{Cu6Sn5}$ phases (also known as bronze) shown in Figure 2. Because of this you cannot simply melt solder off the solder wick to recover it as the tin will continue to dissolve the copper forming bronze. There is also the issue of needing rosin in your recovered solder to be useable. This 1960's video from NASA is an excellent video for learning more about the soldering process.

Figure 1. Reactions of the Copper-Solder Interface

Figure 2. Copper-Tin Binary Phase Diagram

To address the acid method. Physically, it could work (I would not recomend doing so with leaded solder) as most ROHS compliant electronics are $95\%+$ tin solders with either silver or antimony added as alloying agents to prevent tin whiskers from forming. That said it economically dissolving in acid then precipitating then reducing tin in a way which does not also produce copper would be a tedious and time consuming process. You are far better off just buying solder which does not cost that much and has rosin needed to flux the solder.


pH of an aqueous solution with acetic acid and sodium acetate An aqueous solution of $\pu{1L}$ contains $\pu{0.2 mol}$ of acetic acid and $\pu{0.2 mol}$ of sodium acetate ($K_\mathrm{a} = 1.8 \times 10^{-5}$). Compute the $\mathrm{pH}$ .

I'm trying to help my little sister in solving this exercise... however, a lot of years passed away since the time I attended the school...
Our (unsuccessful) attempt:
$\ce{CH3COOH <=> H+ + CH3COO-}$
$K_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{x\cdot x}{0.2} = 1.8 \cdot 10^{-5} \rightarrow \mathrm{pH} = 2.72$
Then
$\ce{CH3COONa -> Na+ + CH_3COO-}$
$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$
\begin{align}
K_\mathrm{b} = \frac {K_\mathrm{w}}{K_\mathrm{a}} &= 5.56 \cdot 10^{-10} \\
 &= \frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}  = \frac{x^2}{0.2} \\
 \implies \mathrm{pOH} &= 4.97 \\
\implies \mathrm{pH} &= 9.02
\end{align}
Hence, $\mathrm{pH}_\text{solution}= (9.02+2.72)/2 = 5.87$
However, the result should be $4.74$.

Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:
$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\ {\rm M}}\right)$$
which reduces to
$$\pu{pH} = 4.74$$


Formation of Lactone from 5-amino-decanoic acid? As expressed in the title, I am trying to figure out how to form a lactone from 5-amino-decanoic acid. The thing that comes to my mind as a start is the attachment of the amino group ($\ce{NH2}$) to the carboxyl carbon and then immediately break the double bond of this carbon with oxygen … But the process gets more likely to form a lactame, where my exercise is asking a Lactone. Can anyone help?

I doubt your book is asking you to convert this molecule into a lactone. This would be very hard to do. It could, however, be easily converted into a lactam. 5 and 6 membered lactams are formed by simple dehydration. For your γ-lactam, this can be achieved by heating the amino carboxylic acid, forming the thermodynamically favourable, stable 6-membered lactam. Similarly, lactones can be be interconverted by adjusting the pH.

IF you really wanted to convert the amine into a lactone, it would involve some pretty uncommon reagents. This blog post mentions that pyrylium salts can be used to make a primary amine into a leaving group, which would, presumably, allow you to substitute a hydroxyl group, thereby giving you access to the lactone.
http://curlyarrow.blogspot.hu/2009/09/how-to-turn-amine-into-leaving-group.html


What is the chemical formula for the synthesis of Hydroxyapatite (Ca10(PO4)6(OH)2)? This might be a basic question but I am a little confused.
I have some lecture notes that says calcium phosphate ($\ce{Ca3(PO4)2}$) reacts with calcium hydroxide ($\ce{Ca(OH)2}$) to form Hydroxyapatite ($\ce{Ca10(PO4)6(OH)2}$).
So, the chemical formula (even when unbalanced) should be:
$$\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca10(PO4)6(OH)2}$$
I can't figure out why $\ce{Ca10(PO4)6(OH)2}$ is the product of $\ce{Ca3(PO4)2}$ and $\ce{Ca(OH)2}$ and I am having trouble balancing it (see below). On Sigma Aldrich's website, it says the linear Formula for hydroxyapatite is $\ce{3Ca3(PO4)2·Ca(OH)2}$. I copied and pasted this directly. I assume the dot between the calcium phosphate and calcium hydroxide means 'plus'.
On the Sigma Aldrich's website, the calcium phosphate is written/balanced as $\ce{3Ca3(PO4)2}$, whereas the lecture notes I have are just $\ce{Ca3(PO4)2}$. Even with the three in front as the coefficient, I get $\ce{9Ca + Ca -> Ca10}$, $\ce{(PO4)2}$ and $\ce{(OH)2}$ which is $\ce{Ca10(PO4)2(OH)2}$, instead of $\ce{Ca10(PO4)6(OH)2}$. If I use the formula in my lecture notes, I get $\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca4(PO4)2(OH)2}$, which is even more off.
Should it be $\ce{3(Ca3(PO4)2)}$ instead of $\ce{3Ca3(PO4)2}$ so the coefficient 3 can be applied to the subscript 2 on the $\ce{PO4}$ to give $\ce{(PO4)6}$?
What am I missing?

First, let me state the standard formula of hydroxyapaptite:

The term "apatite" applies to a group of compounds  with a general
formula in the form $\ce{M10(XO4)6Z2}$, where $\ce{M^2+}$ is a metal
and species $\ce{XO4^3-}$ and $\ce{Z-}$ are anions. The particular
name of each apatite depends on the elements or radicals M, X and Z.
In these terms, hydroxyapatite (HAp) has the molecular structure of
apatite, where M is calcium ($\ce{Ca^2+}$), X is phosphorus
($\ce{P^5+}$) and Z is the hydroxyl radical ($\ce{OH-}$). This is
known as stoichiometric hydroxyapatite and its atomic ratio Ca/P is
${1.67}$. Its chemical formula is $\ce{Ca10(PO4)6(OH)2}$, with $\ce{39 {%}}$ by
weight of $\ce{Ca}$, $\ce{18.5 {%} P}$ and $\ce{3.38 {%}}$ of $\ce{OH}$.

The anion radical can be replaced with fluorine/chlorine to become fluoroapatite/chloroapatite. This formula can also be written as $\ce{3[Ca3(PO4)2]· Ca(OH)2}$ which indicates three moles of calcium phosphate and one mole of calcium hydroxide which is basically the same as the Sigma-Aldrich's formula. The confusion arises since they have omitted the brackets, you didn't considered calcium phosphate as single entity for which you had a wrong formula. Brackets are generally not given but you have consider them as a single entity or you can even use brackets conveniently for your purpose. But do note that this formula doesn't prove that it is a basic salt but is actually a product of polymerisation:

According to Bassett, the solid existing in stable equilibrium at $\pu{25 °C}$., with solutions of a range from faintly acid to nearly pure
lime-water, is oxy- or hydroxyapatite, but, although its composition
may be indicated by the formula $\ce{3Ca3(PO4)2.Ca(OH)2}$, it is not to be
regarded as a basic salt, but rather as the salt of a polymeride of
phosphoric acid, $\ce{H11P3O13}$ (or $\ce{3H3PO4.H2O}$), with one hydrogen atom
un-neutralised (like orthophosphoric acid or pyrophosphoric acid). The slowness of
formation of hydroxyapatite favours the view that it is produced by
polymerisation.

Reference

Hydroxyapatite-Based Materials: Synthesis and Characterization
By Eric M. Rivera-Muñoz, 2011, DOI: 10.5772/19123
http://calcium.atomistry.com/basic_calcium_phosphates.html


Reaction mechanisms for organohalides from alcohols For the syntheses of organohalides from alcohols, what would plausible reaction mechanisms be for the three following reactions?

$$
\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}
$$

$$
\ce{R-OH + PCl5 -> R-Cl + POCl3 + HCl}
$$

$$
\ce{R-OH ->[red P][Br2 or I2] R-X}
$$

Nucleophilic substitutions, regardless whether proceeding via $S_N1$ or $S_N2$ require a decent leaving group. $\ce{HO-}$, however, is a lousy leaving group!

To overcome this, several methods exist:

Perform the reaction under acidic conditions, where the alkanol is protonated. Your leaving group now is $\ce{H2O}$. In addition, the corresponding anion of sulfuric acid as a bad nucleophile.
React the alkanol with an acid chloride to form an ester and replace the ester group with a halide. Consider $\ce{PCl5}$ to be an acid chloride of phosphoric acid.
Form the acid halide in situ by reaction of phosphorous and iodine.


Reaction between 2 hydrogen halides I know that hydrogen halides can react with each other to produce halogen halides.

As an example here is a reaction between $\ce{HCl}$ and $\ce{HF}$:
$$\ce{HCl + HF -> H2 + ClF}$$

As for the mechanism it is like this (taking into account that chlorine is hypervalent but fluorine isn't):
$$\ce{HCl + FH -> HCl\bond{...}F + H. -> H. + ClF + H. -> H2 + ClF}$$

I know that both of these are acids and that $\ce{HF}$ is weaker than $\ce{HCl}$. However with the halogen bond the negative charge on the fluorine (partial negative after the formation of the halogen bond) is more delocalized. This stabilizes the $\ce{H-Cl\bond{...}F}$ intermediate and makes the dissociation easier for the hydrogen and harder for the fluorine. And yes the hydrogen radical is very reactive but that is exactly why $\ce{H2}$ forms in this process.

But I could see some problems with this.

It would have to be in a vacuum for it to not be explosive(most reactions involving fluorine or fluorine compounds are explosive because of fluorine's high electronegativity (same reason reactions with $\ce{O2}$ burn things))
If it is in air $\ce{O2}$ could react very violently with both the chlorine and the fluorine giving you these products:
$\ce{ClO}$, $\ce{ClO2}$, $\ce{ClO3}$, $\ce{ClO4}$, $\ce{Cl2O}$, $\ce{Cl2O2}$,
$\ce{Cl2O3}$, $\ce{Cl2O4}$, $\ce{ClO6}$, $\ce{ClO7}$, $\ce{OF2}$, $\ce{O2F2}$, $\ce{ClO2F}$, $\ce{ClO3F}$
Chlorofluorocarbons could also form from reactions with carbon based compounds in the air like methane and $\ce{CO2}$.

And this is all because the halogens are very reactive.

Similar things would be true for other halogen fluorides.

So is there a more practical way to form halogen halides than reacting $\ce{HX}$ with $\ce{HY}$ where both $\ce{X}$ and $\ce{Y}$ are halogens?

The reaction you are proposing is very unlikely to happen under standard conditions. Taken average bond energies from UC Davis ChemWiki,[1] the approximated heat of reaction would be $\Delta_\mathrm{R}H = 307~\mathrm{kJ\,mol^{-1}}$.
$$\ce{HCl + HF <=> H2 + FCl}$$

The reaction mechanism you are proposing is also not quite feasible. If we would ignore the energetics of this reaction, and we speak hypothetical about this reaction, we would still have to consider an activation step. What we also need to consider is the high affinity of fluorine towards electrons and therefore have to exclude anything that would enhance the heterolytic bond cleavage of hydrogen fluoride. Hence this reaction can only take place in gas phase in an inert atmosphere.
A feasible starting point would be the thermolytical or photolytical cleavage of one of the bonds.
$$\begin{align}
\ce{HCl &->[\Delta T/ h\nu] H + Cl}\\
\ce{HF  &->[\Delta T/ h\nu] H + F} \\
\end{align}$$

Then we would need to consider propagation of the reaction.
$$\begin{align}
\ce{HCl + H/Cl/F &->{} [ HCl.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/Cl(H/Cl/F) + H/Cl}\\
\ce{HF + H/Cl/F &->{} [ HF.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/F(H/Cl/F) + H/F}\\
\end{align}$$
And then finally recombination of the remaining free radicals.

We are basically looking at the reverse mechanism of the reaction of hydrogen and chlorine.[2]

The practical approach is given in orthocresol's answer. Interhalgen compounds, i.e. halogen halides, are formed by reacting the pure halogen gasses. For the reaction of chlorine and fluorine we find the approximate enthalpy of reaction to be $\Delta_\mathrm{R}H = -113~\mathrm{kJ\,mol^{-1}}$.
$$\ce{F2 + Cl2 <=> 2FCl}$$
Side products also include $\ce{ClF3}$ and $\ce{ClF5}$. The reaction itself is exothermic and ignoring entropy should proceed spontaneously.
$$\begin{align}
\ce{Cl2 &->[h\nu] 2Cl}    &&\text{(start)}\\
\ce{Cl + F2 &-> ClF + F}  &&\text{(propagation)}\\
\ce{F + Cl2 &-> ClF + Cl}\\
\ce{F + Cl &-> ClF}       &&\text{(stop)}\\
\end{align}$$

All reactants are highly reactive and may decompose explosively when other reactants are involved. Even in diluted systems these compounds react with almot everything, see also Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?

Conclusion

The formation of interhalogen compounds via their hydrogen halides is energetically not feasible. They can be synthesised from their pure compounds under moderate conditions.
The mechanisms for these reactions can only develop in a closed system as radical chains, which need an initiation step.

Appendix

Average bond energies (Bond dissociation energies; BDE) from UC Davis ChemWiki:
$$\displaystyle\begin{array}{lr}
&\text{BDE}/\mathrm{[kJ\,mol^{-1}]}\\\hline
\ce{HCl} & 427\\
\ce{HF}  & 565\\
\ce{H2}  & 432\\
\ce{FCl} & 253\\\hline
\ce{F2}  & 154\\
\ce{Cl2} & 239\\\hline
\end{array}$$
A. L. Marshall, J. Phys. Chem. 1925, 29 (7), 842–852.
Carl M. Furgason and John W. Moore, J. Chem. Educ. 1943, 20 (1), 41.


Why do solutions become saturated? Why can a solvent dissolve only a particular amount of solute?
If we add more solute to the solution, the number of solute particles in contact with water increases. So rate of dissolution should increase, and subsequently, the solubility product should also increase. But why do we have a constant value of solubility product?

P.s. An explanation at the particle level would be appreciated

Take a slightly more general problem of miscibility, all that in thermodynamic equilibrium. In your question you mix up kinetics and equilibrium, and additionally in a wrong way. Equilibrium is the most important part.

If you put two liquids together and shake them well, than in general, they do not mix. So you will observe a boundary between, say, diethyl ether (upper layer) and water (bottom layer).
If you separate them and analyze their composition, you will find, that the ether fraction contains little bit of water (1.5g / 100 mL) and that the water fraction contains little bit of ether (6g / 100mL). Knowing this, if you now take 100 mL of water and mix it with 5g of diethyl ether, you will observe, that no boundary is formed, and all ether has dissolved in water.

As another example, if you shake ethanol with water, they mix in any ratio.

That's observation, now the rationale. The particles of pure liquid interact with each other, in attractive way (otherwise you will have no liquid, but gas). Now you mix two pure liquids together (A + B) and look, if some of the particles of A can easily sneak into the B. If their behavior is reasonably similar, they can do it and interact with particles of B also in attractive way. The attraction should be comparable or greater, than within pure A. That is the enthalpy part.
But as we are dealing with free energy, there is also the entropy, which you have to take into account. On one hand, entropy favors mixing, as you add lot of accessible states. On the other hand, if the attraction of B to A is very strong, it starts to organize the molecules of A around itself, so they are not as free to move as they were in pure A, and limits the miscibility.

For the solubility of solids, it is technically similar, the A is solid and the interaction which hold sit together is the lattice energy.

As a side note, if you ask about solubility, miscibility and such, you should refrain from ideal approximations, as they do not predict such phenomena. Their calculation from pure substance properties is therefore terribly difficult, and physical chemistry tables are the only reasonable answer.


The type of H-NMR spectra observed for S=PF2H and other inorganic molecules similar to this I was hoping to learn more about the types of splitting observed in inorganic molecules such as S=PF2H. I am familiar with the typical splitting patterns associated with the study of organic molecules studied by C and H NMR. We know that H and F have a spin of 1/2, in reading about the effects of fluorine in on H NMR, it appears that fluorine relaxes in a similar way to protons when pulsed by a radio frequency. Since the two nuclei have the same spin but are different types of nuclei altogether and impact each other according to the Nuclear Overhauser Effect, what kind of signals would be given by an H NMR for the molecule? Would we see signals corresponding to fluorine given that it has the same spin as the proton?

The everything OP and  orthocresol discussed in the comment session is in this reference book (Ref.1). I include its PDF file (Ref.1) for benefits of OP and future readers. An example of coupling is depicted in the following figure (from Ref.2):

This spectrum is qualified as other inorganic molecules similar to $\ce{S=PF2H}$ molecule. What I want to show is alike different coupling constants in $\mathrm{^1H \ NMR}$ due to the molecular structure, bond distance, etc., the coupling constants in $\mathrm{^{31}P }$ and $\mathrm{^{19}F \ NMR}$ are also changed as well. For instance, $^1J_\ce{HH}$ is $17$-$\pu{20 Hz}$ in $\mathrm{^1H \ NMR}$ organic compound. However, $^1J_\ce{PH}$ is $\pu{836 Hz}$ for this particular molecule ($\ce{PF2H(^{15}NH2)2}$) while $^2J_\ce{PH}$ is $\pu{11.8 Hz}$ for same molecule (c.f., avarage $^2J_\ce{HH}$ is $\approx \pu{7 Hz}$ for a hydrocarbon with more than 2 carbons).
This particular spectrum consists of 90 peaks due to additional coupling of $\ce{P-^{15}N}$ coupling. If the compound is not labelled with $\ce{^{15}N}$, the number of peaks would reduced to 30 peaks, a relatively simplified spectrum, but still complicated coupling pattern of doublet of triplet of quintet.
References:

Michael O’Neill, In Assigning Inorganic NMR Spectra: a programmed approach to solving NMR problems in main group chemistry; Copyright © 2019 Michael O’Neill (ISBN-13: 978-1-0864-12017) (PDF).
David W. H. Rankin, Norbert W. Mitzel, Carole A. Morrison, In Structural Methods in Molecular Inorganic Chemistry; First Edition, John Wiley & Sons: Chichester, United Kingdom, 2013 (ISBN-13: 978-0-470-97278-6).


Reaction mechanism in fluoride adsorption to aluminum oxide Despite a fair amount of research (excluding non-open access journals, to which I have not got access), I cannot seem to find an explanation of the process that takes place when $\ce{F^-}$ adsorbs to activated alumina ($\ce{Al_2O_3}$).

I do know that the process involves both chemisorption (monolayer formation) and physisorption, but since the latter is explained solely by van der Waals forces, I am primarily looking for an explanation of the former.

EDIT: Does adsorption take place only at the aluminium ions? If not, does that mean that all aluminium compounds with oxidation state +3 will be equally good adsorbents?

EDIT: I have found an article (https://link.springer.com/article/10.1023/A:1012929900113) which proposes a mechanism that involves aluminium-fluoride complex formation. If this is the case - why is this even classed as an adsorption reaction at all, and not just a normal chemical reaction? And what role do the $\ce{OH^-}$ ions play? Do they keep the aluminum in the ionic lattice of the alumina?

Besides the presence of Al and O atoms on an Aluminum Oxide surface, there are also H atoms, which are bounded to the solid by OH bonds, - O --- H+ . Direct evidence for the presence of H atoms, and OH bonds, come from XPS or FTIR (grazing incidence) data. Indirect evidence comes from several sources, but especially from the reactivity of the surface to some compounds, like Silanes, where Si --- O --- R groups react readily with OH. Based on what is said above, I´d say the best explanation for the F– anion interaction with the Aluminum Oxide surface is a chemisorption process, through an indirect process,  bonding of the F– anion to the  H+ of the surface, forming an -O --- H+ --- F– , anion (F–) – to – dipole (+H  --- O– ) complex structure. A real solid – state Hydrogen bonding.


Periodic Trends of Zeff and Electronegativity The effective nuclear charge $Z_\mathrm{eff}$ increases from left to right and from top to bottom. Can you explain why it increases from top to bottom?

Also can we explain the periodic trend of electronegativity using the shielding effect?

EDIT:
The calculation of $Z_\mathrm{eff}$ for Li would be 1 (3 protons − 2 core electrons). For Na is 1 (11 protons − 10 core electrons). So far it doesn't seem to me that $Z_\mathrm{eff}$ should increase from top to bottom. Considering the fact though that for Na there are more electrons shielding the only valence electron than Li as well as Na's electron is relatively more distant than that of Li, the first should be experiencing less $Z_\mathrm{eff}$. This is not true however. As far electronegativity is concerned I can't come up with a neat explanation.

The Slater shielding rules give more quantitative information about the amount of shielding.

Put simply, valence electrons (as you go across a row) do not shield very well, roughly 35% of a full electron. So when you go from B to C to N, you keep increasing the nuclear charge by one proton, but the electrons don't fully shield the nucleus.

Core electrons, as suggested in comments and other answers do shield fully, since they're closer to the nucleus.

So yes, the shielding for Li and Na are similar. As you mention, the valence electron in Na is further from the nucleus, and that does effect the ionization energy and electronegativity as you suggest.


Why are group 1 elements called alkali metals and group 2 elements are called alkaline earth metals? I was studying the periodic table. My teacher said that

  Group 1 elements are called alkali metals because their oxides are soluble in water and group 2 elements are called alkaline earth metals as their oxides form in the earth and are water soluble.

What did she mean by "oxides"? Aren't alkali metals bases which are soluble in water? Then what is the term "oxides" doing here?
What is the difference between "alkali" and "alkaline"? Why do we call one of them "alkali metals" and the other "alkaline earth metals" when the concept is the same?
Why are the bases of group 1 and group 2 elements soluble in water? What is the reason?

The description of the reasoning of the naming is … let’s call it simplified.

First, concerning your first point. Alkali metals are not bases by themselves, they are metals. Metals typically cannot be classified in the original Arrhenius or extended Brønsted-Lowry acid/base classification. The cannot dissociate to liberate either $\ce{H+}$ or $\ce{OH-}$ because they only consist of metal atoms. And they cannot act as a donor or an acceptor for protons — one because of the lack of protons and two because of the lack of favourable lone pairs.

It is only when we oxidise these metals in some way or another that they form a species that can interact with water in the acid-base formalities. One possibility for oxidation is simple burning, i.e. reacting with atmospheric oxygen. Unfortunately, only lithium is nice enough to ‘do what you might expect’, hence why I’m using that as an example in the following equations:

$$\begin{align}\ce{4Li + O2 &-> 2 Li2O}\tag{1}\\[0.4em]
\ce{Li2O + H2O &-> 2 LiOH}\tag{2}\end{align}$$

As we can see, the oxide we generated, lithium oxide, can react with water to form a hydroxide which dissociates into $\ce{OH-}$ ions — an Arrhenius base. We can also say that the oxide anion is capable of accepting protons donated by water — a Brønsted-Lowry base. Therefore, lithium oxide (and also lithium hydroxide) are the bases while lithium metal is not.

The second question is easily answered. Alkali is a noun and alkaline is an adjective. Thus, alkali metals is a compound word formed by two nouns while alkaline earth metals is also a compound word formed by two nouns where the first noun is further specified by an adjective.

The third question is difficult to answer. There is not really an easy or understandable explanation; it all boils down to churning numbers and finding out that the dissolution is favourable, i.e. the corresponding equation represents a decrease in Gibbs free energy: $\Delta G < 0$.

We can break dissolution of solids into two general processes:

breakdown of the ionic lattice, i.e. liberation of individual ions from a solid

$$\ce{NaOH(s) -> Na+(s) + OH- (s)}\tag{3}$$
dissolution of the individual ions

$$\begin{align}\ce{Na+(s) &->[H2O] Na+(aq)}\\[0.3em]
\ce{OH-(s) &->[H2O] OH-(aq)}\end{align}\tag{4}$$

The first of these steps requires energy because you are breaking bonds (this is always unfavourable). It is typically represented by $\Delta_\text{latt} H^0$ — the standard lattice enthalpy.

The second process is exothermic because the individual ions are now again somewhat balanced by the polar water molecules. It is typically represented by $\Delta_\text{solv} H^0$.

Both processes are favourable from an entropic point of view.

Whether the entire reaction $(5)$ happens, depends on the difference between lattice and solvation enthalpies. If the lattice enthalpy is too high to be offset by solvation enthalpies and the entropy effects, an ionic compound will not dissolve. For alkali metal oxides and alkaline earth metal oxides, the lattice enthalpy is typically low and thus they dissolve.

$$\ce{NaOH (s) -> Na+ (aq) + OH- (aq)}\tag{$3+4=5$}$$

The actual reason for their names is in another castle beyond the scope of this answer. At first glance, I would forward you to the corresponding Wikipedia articles.


Why are wave functions orthogonal? It seems that all wave functions studied in physical chemistry are orthogonal (e.g. particle in a box, hydrogen atomic orbitals). Does this come about because we purposefully make them orthogonal, or are they derived that way naturally? Can there be useful wave functions that are not orthogonal?

In general, orthogonal wavefunctions are much easier to treat. In some cases they appear naturally, but usually, the orthogonality is imposed as a constrain while constructing the wavefunction.

For example, if you construct electronic wavefunction in the atomic orbital basis, you try to construct the orthogonal basis. This guarantees that the AOs are linearly independent. (Implication, not equivalence). Would you fail to fulfill this, the solution might still be possible, but much more difficult.

If you manage to solve the eigenvalue - eigenvector problem, the solutions are by definition orthogonal to each other. This is the case for the examples you provided.


Ripening of bananas? So, obviously the ripening of bananas involves changing it’s chemical composition. For example, ripe bananas have a much higher amount of sugar.

It is also known that reaction rates are generally increased with increasing temperature, yet bananas ripen faster in a refrigerator.

Why do bananas ripen faster in the fridge, even though it’s colder? And is there any difference (except ripening time) between ripening a banana in the fridge, and having it ripen in room temperature?

I like bananas, but I almost went bananas doing some research for this question!

Let me post some quotes - too much info to post the sources:

1) Is it safe to store my bananas in the refrigerator?
Dr. Brecht: “Never, never, never put bananas in the refrigerator!

2) The answer, far from dissuading me from keeping bananas in the fridge, only confirms my habit. Banana-eaters are divided into two fairly equal camps: those who like the delicate flavour of the green (unripe?) banana and those who prefer the quite different strong flavour of the yellow (over?)ripe banana. Using cold to inhibit the enzymes which aid the ripening process is just what is needed by the 'greens' but should be avoided by the other camp.

3) If they're kept at a cold temperature, the enzymes that enable them to ripen are inhibited. And as those enzymes become inactive, other enzymes operate more efficiently. Some cause cell damage, while others (browning enzymes) cause the skin to blacken. Bananas do best if stored at around 10°C.

4) Bananas require gas, that is ethylene, to trigger the ripening process.

5) Cooler temperatures also slow down the chemical ripening process. Once bananas are ripe to your likening, store them in the fridge.

6) De gustibus non est disputandum

So it seems that bananas do change because of the cold, but the process is not exactly ripening. It does involve a chemical change, which is so complex that it can be predicted and analyzed, but is not useful as a guide to developing a tasty banana for a specific person. Too much variation in the people who eat bananas! I did an experiment some time ago, wondering if the economy of buying green-yellow bananas (with noticeably thick, firm skins) was significantly less than buying bananas with exactly the color for immediate eating (deep yellow for me), where the skins were noticeably thinner, softer. How much weight loss do I incur and pay for, but don't get the benefit of? Not much, it seems, about 10% of the weight of a banana, maybe 15% if you got a green one and let it go almost brown.

So banana color may not be the whole story: enzyme activity occurring at different temperatures may affect taste without being "ripening" - a banana may go from underripe to overripe without ever being exactly ripe! The process of marketing, from green to yellow, is described in some detail, and interestingly, ripening bananas develop heat (and I would guess, ethylene also), so the recommendation was to unpack the banana boxes quickly when they get to the grocery store to reduce the temperature (and C2H4) so the bananas keep longer.


how many calories per unit of sweetness between table sugar and aspartame? How many calories per unit of sweetness between table sugar and aspartame?

Aspartame 4 calories per gram source
Sucrose 3.9 calories per gram source
But you need far less aspartame than sucrose to get the same level of perceived sweetness


Why does LiAlH4 reduce an amide to an amine, but only reduce a ketone/aldehyde to an alcohol? Here are two different reactions in which lithium aluminium hydride, $\ce{LiAlH4}$, reduces a carbonyl group:

Why does $\ce{LiAlH4}$ completely remove the oxygen from the amide, but only reduces the carbonyl group of a ketone/aldehyde to a hydroxyl group?  Is the difference due to the fact that $\ce{H2O}$ is added in the first reaction (i.e. if the aldehyde/ketone reduction had a second step where water was added, would the carbonyl group be reduced to a methylene group)?

You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ and $\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 38$.)

(Don't worry about how I've drawn the metals coordinating to oxygen - the exact details aren't important and sometimes aren't even fully understood.)

In the aldehyde reduction, there is no lone pair to assist in the expulsion of oxygen. So the intermediate alkoxide just sits there, until you add water or acid:

The aqueous workup serves two purposes: firstly to reprotonate the product (for example, with the aldehyde reduction, your final product is an alkoxide ion and adding water gets you the alcohol), and secondly to destroy any unreacted $\ce{LiAlH4}$. In fact, the aqueous workup is also carried out in the reduction of the ketone or aldehyde - we often just omit it because it's so common and it's tiresome to have to write it on every reaction we do. So even though it's not there, it's still implied.

If you wanted to reduce the carbonyl group of a ketone/aldehyde to a methylene group, the three most well-known methods available are the Wolff-Kishner reaction, the Clemmensen reduction, and the Mozingo reduction.


Repeating unit of poly lactic acid (PLA) Which repeating unit for poly lactic acid (PLA) is correct, picture 1 or picture 2?

 and

I read somewhere that in polymerization, acids give $\ce{OH}$ and alcohol gives $\ce{H}$, creating water. According to this explanation, picture number 2 is correct but I have seen on websites like Wikipedia that picture 1 is correct. So, which is the correct picture and why?

In fact, both pictures show the same polylactic acid, only different disconnection points have been chosen:

The red disconnection (or your picture 1) is preferred because you can immediately see that it is a polyester.

Edit:

Your teacher is right in that, during the polymerization, the $\ce{-COOH}$ group loses $\ce {-OH}$ and the $\ce{-OH}$ group loses $\ce {-H}$ to form $\ce{H2O}$.
Therefore, your teacher prefers this arrangement:

If we draw it the other way, it are the wrong $\ce{-OH}$ and $\ce{-H}$ at the end of the chain, in the sense that it are not the ones that will give water:

However, note also that in either case, the part between brackets is not formed by identical fragments.
The atoms are indeed the same, but since not all are connected to the same atoms, they are not all chemically equivalent. There are different functional groups.
There are acid, alcohol and ester groups; mainly ester of course.
So in reality, we are writing $\ce{[...]_n}$, but those $n$ are not all the same and thus it is not correct.

If you want to depict the whole chain including the polymer chain and the ends that can continue polymerizing, the most correct way is to draw the end monomers explicitly.
At that point, you can place the brackets where you want, since the structures are absolutely equivalent.


How is the trend of acidity in the sequence: phosphinic, phosphonic, phosphoric acid explained? How is the trend of increasing $\mathrm{p}K_\mathrm{a}(1)$ value in the sequence:
phosphinic $(1.24)$, phosphonic $(2.00)$, phosphoric acid $(2.21)$
explained (the image was taken from Housecroft Inorganic Chemistry)?

The stability of conjugate bases is indeed a good way of estimating the relative acidities of different compounds.

However, in the case of the three compounds you've brought up, it is not sufficient. Taking the conjugate bases of each acid, one sees that each has a doubly bonded oxygen (oxo) and a deprotonated hydroxy. These are the only relevant groups as far as resonance is concerned.

Note that phosphorus-bound hydrogens cannot delocalize the negative charge, and neither can phosphorus-bound hydroxyl groups.

Given this, all three acids actually have the same resonance stabilization. Any resonance structure where a hydroxyl $\ce{O}$ donates an electron pair to the phosphorus is insignificant because it creates increased charge, likewise any resonance structure where the hydroxyl $\ce{O}$ receives an electron pair is insignificant because oxygen already has a complete octet in its hydroxyl form.

Since the resonance argument only tells us that their $\mathrm{p}K_{\mathrm{a}}$'s are similar (looking only at first $\mathrm{p}K_{\mathrm{a}}$) we can use the argument that the additional hydroxyl groups reduce the polarization of the $\ce{O-H}$ bond. This can be demonstrated by looking at the resonance structures of the acids (in neutral form). The hydroxyl $\ce{O}$'s in the neutral form can donate electron density to the phosphorus; however when there is only one hydroxyl group, all the positive charge is localized there, while when there are two or three, the positive charge is delocalized across three hydroxyl groups.

Note that in the case of the neutral acid, it is ok to have resonance where $\ce{O}$ donates to $\ce{P}$ because $\ce{P}$ does not gain a $2-$ charge as it would with the conjugate base.


Can enamines be formed from α,β-unsaturated ketones? This question was a result of discussion form here: α,β-unsaturated carbonyl compounds and alkyl addition

The reaction below (condensation of cyclohexenone with a secondary amine) does not yield any results on a literature search, even when I only search for the product by itself (unless my search was not good enough):

Can enamines actually be formed from α,β-unsaturated ketones? If not, what prevents it from being possible?

Stork and Birnbaum have reported (Tetrahedron Lett. 1961, 10, 313) the formation of dienamine 2 from octalone 1. Alkylation with methyl iodide leads to α-alkylation of the dienamine. Buffer conditions are required for hydrolysis of the dienamine to octalone 4.


Can the number of intermediates for a reaction be more than the total number of transition states? I'm getting results from a transition state guess software, in which the number of intermediates are two and there's only one transition state for a reaction, is that possible?

Let's assume your reactive complex (reactants interacting) are optimised, i.e. its gradient equals zero, and your potential energy surface is continuous.

Then any direction is either uphill or constant (otherwise the optimisation hypothesis would not apply).
There are three possibilities: (a) going from reactants to intermediate gives a transition state which has higher energy than both reactive complex and intermediate; (b) reactants and intermediate lie on a plateau where energy does not change; or (c) one hypothesis is wrong.

So yes, for barrierless processes.


Need to dissolve glue placed intentionally in a lock Someone intentionally sabotaged the lock on the front door of my house by putting glue inside it apparently, I have no idea what type of glue it is. Is there any chemical substance I can use to dissolve the glue instead of having to replace the entire lock?

Also, would heating the lock work?

I took a look at this, the heating option wasn't possible since they're talking about paper, but I was wondering if it'll work before trying any of the chemicals mentioned.

Please consider that I don't have much experience in Chemistry while answering, thank you for your time.

To paraphrase the link you mentioned,
Without knowing the specific glue, you should try different solvents to see if it dissolves the glue:

Start with a mild, non-polar solvent like paint thinner.
If that doesn't work, try a polar solvent like acetone.
If that doesn't work, try tetrahydrofuran (PVC pipe solvent)

All of these solvents are flammable and pretty nasty if you get them on you. To minimise risk, use small quantities at a time, minimise the time that the lid is off the container. The fumes shouldn't be too much of a problem if you're outside, but still take care.


Why does Li get darker in a glove box? I'm having some issues with lithium getting darker in a glove box. Please see attached photos, Li gets dark overnight.

My sensors for $\ce{H2O}$ and $\ce{O2}$ are showing 0 ppm. I also checked oxygen and moisture contents using chemicals in case sensors were bad, and looks like there is no oxygen and moisture in the glove box. Just to make sure, I also replaced the nitrogen tank with argon (there are some reports in the literature saying Li reacts with nitrogen even at room temperature and gets darker in the presence of trace levels of moisture).

Does anyone had this problem before? Any recommendations would be of great help.

Lithium can react with nitrogen giving a dark product.  https://en.wikipedia.org/wiki/Lithium


How can I calculate the energy of a solution by knowing all of it's elements? How can I calculate the energy of a solution by knowing all of it's elements?

For example:

If I have 2 liters of salting water $\mathbf{NaCl_{(aq)}+H_2O_{(l)}}$ contains 1.5 g of salt, how can I determine the internal energy stored in it?

Since you ask specifically for internal energy, would it be that you already have some ideas?

So what is internal energy? It is the energy due to the interaction among the species plus the energy due to the interaction between the species and any external field. That is, the energy it takes to "assemble" the system.

Then for gaseous species, it would be mainly the bonding energies of the molecules. Those are usually inferred from experiments. And you can look up tables in textbooks and IUPAC website.
For ionic crystals, you would reason this way:  first ionize the species and them find the Columb interaction among them in crystal. Ionization energies are measured. Columb interaction is probably calculated and is a nightmare to calculate.
The further problem is from NaCl(s) to (aq) where the solvation involves quite a bit of interactions between the ions and water. For the case of Na+ Cl-, water molecules do not form regular cages around them, even though we CAN study the ideal case of regular cages. I think there has been some minimal models which can possibly guide a simulation study for the solvation energy. Though I don't think there has been any simulation study or any theoretical study that is sufficient to predict solvation energy.  I can be wrong. On the other hand, direct measurement should at least be able to measure the heat change during solvation.

So in digression, I don't think you can find the complete energy too well.


How to prevent dripping when using a Pasteur pipette with solvents like diethyl ether? I'm trying to use a Pasteur pipette for pipetting diethyl ether and regardless of the angle I hold the pipette at, the diethyl ether will automatically start dispensing itself (I am unsure of what this phenomenon is called). I'm sure it's not the rubber bulb since I've been using the same one for pipetting other solvents that don't do this. I've also tried using a different rubber bulb, but to no avail.

Is there any way to prevent this from occurring? If not, is there anyway to mitigate the flow rate of the drips?

Usually I avoid pipetting diethyl ether because it can be a bit annoying, but recently I've realized there is a very simple technique which markedly increases ease of transfer (though admittedly I've been using it mostly for DCM). All you have to do is saturate the interior of the pipette with the vapour of the low-boiling solvent before performing the transfer.

The problem arises because the low-boiling solvent will quickly (but not instantly) evaporate into the partial vacuum inside the pipette after pulling the solvent. If you try to transfer the liquid while it's still evaporating significantly, the large volume change from the phase transition will push the liquid beneath through the pipette exit.

So how do you get around this problem in practice? Using the pipette, simply pull and dispense the solvent a few times in succession over 5-10 seconds inside the outgoing vial. Then pull what you need and transfer. In my experience this greatly mitigates dripping. If you're not too slow between transfers, you can get away with doing this only once, at the start.

Other things which can be done to help:

As mentioned in the comments, avoid dragging the filled pipette over a  large distance. Keep the outgoing and ingoing vials right next to each other, at about the same level.
Pull a little bit of air into the pipette, keeping it between the exit and the solvent. This gives you a small time buffer before it drips out.
If you have a steady hand, keep very gently pulling air while you move the pipette between vials, matching the speed at which the liquid tries to drip out. This can be somewhat fiddly, so train to get a feel for it.
Don't fill the pipette to its maximum capacity on each transfer; do more transfers, each with less volume.
There's always a drip left on the very tip of the pipette, and that can easily fall. To avoid it, contact the remaining drop onto one of the outgoing vial's walls above the surface of the liquid, then pull a little bit of air into the pipette, sucking the last droplet along and leaving the tip dry.
Practice.