Why atomic masses aren't integers? Most of the elements have isotopes, so the atomic masses are calculated depending on the percentage of the existing isotopes. That is clear. However, what about elements that have only one isotope (monoisotopic) - like fluorine? Shouldn't the atomic mass for it be a whole number and not 18.9984?

The short answer is nuclear binding energy, which is the energy needed to disassemble an atom into its subatomic parts (or in some cases the energy released when this happens). The binding energy is a consequence of the strong and weak nuclear forces that hold atoms together.

Where does this energy come from? It comes from the mass of the nucleons! What? It is true; most atoms have less mass than the sum of their parts, and that mass defect is converted into the energy that holds them together. If you do not believe me, let us look at one atom of carbon-12, which is used as the definition of the atomic mass unit:

$$\mathrm{1\ u}=\dfrac{1}{12}\text{ of the mass of one }\ce{^12C}\text{ atom}$$

Thus, the atomic mass of an atom of $\ce{^12C}$ is by definition $\mathrm{12\ u}$. The atom is constructed from 6 protons, 6 neutrons, and 6 electrons, and the masses of those particles are:

particle   mass                     number    total mass
p          1.00727646681290     u   6         6.0436588008774    u
n          1.0086649160043      u   6         6.0519894960258    u
e          5.485799094622×10^−4 u   6         0.0032914794567732 u
                                     TOTAL:  12.0989397763600    u

The sum of the parts of the $\ce{^12C}$ atom are more than the mass of the atom! This mass defect, $\mathrm{0.0989397763600\ u}$ in this case, is the value of the nuclear binding energy when converted to energy using $E=mc^2$.

This handy graph from the Wikipedia article linked above shows that the binding energy per nucleon is not the same for all nuclei.

If it were, then we would have integer values for monoisotopic nominal masses. However, the consequences of this change in reality could be as drastic as the heavier elements (like iron) not having enough energy to hold themselves together. An additional consequence is that there would be no productive nuclear reactions (fission, fusion, or radioactivity). All nuclear transmutations are driven by conversion of nuclei with lower binding energy per nucleon to nuclei with higher binding energy per nucleon, resulting in a net conversion of mass to energy. Without a difference in binding energy per nucleon, we would not have nuclear bombs (and probably we would be better off as a civilization), but we would also not have nuclear power and radioisotopes for compound labeling, cancer therapy, and medical diagnostics. We would also not have a ready supply of deuterium for NMR solvents, and we might not be able to use variations in isotopic distributions of $\ce{H}$, $\ce{C}$, and $\ce{O}$ to determine all sorts of things like did that "locally grown" food from that restaurant you like actually come for your geographic region?


Comparing Polarity of Amino Acids I know that there are polar uncharged amino acids (serine, threonine, asparagine, glutamine, cysteine) and polar charged amino acids (the basic and acidic amino acids).
Does the charge on the acidic and basic amino acids make them more polar and hydrophilic than the uncharged polar amino acids? Moreover, cysteine is classified as an uncharged amino acid, but because it has an ionizable side chain, would it be more polar than serine, asparagine, etc.?

Does the charge on the acidic and basic amino acids make them more polar and hydrophilic than the uncharged polar amino acids?

Yes, they are less likely to be on the inside of a protein. Hydrogen bonds with water will be stronger, and interactions with ions as well.

  Moreover, cysteine is classified as an uncharged amino acid, but because it has an ionizable side chain, would it be more polar than serine, asparagine, etc.?

Cysteine and tyrosine are more hydrophilic when deprotonated. So at strongly basic pH, their properties are very different from that of serine. At neutral pH, however, cysteine is less polar than serine.


What is the diagram of the roto-inversion of $\bar 6$? Using the standard way of representation:

Here we have a rotation followed by an inversion (technically called 'rotational inversion'). So as to represent it, what I have done is:

1) Both rotation and inversion occur for the first time. The circle changes of plane:

2) Both rotation and inversion occur for the second time. The circle changes of plane again:

My issue is that I am not convinced of this, because in theory there is rotation followed by an inversion and they do not happen simultaneously. So if this is wrong, what is the correct representation and why?

Diagram for $\bar 3$

Diagram for $\bar 6$

METHOD:

NOTE: I have used comma to stress there was an inversion

Thank you

The image below gives some example of different rotation inversion operations. For the $\bar 6$ operation rotate by 60 degrees (black dot to open circle in the direction towards 3) then invert through the centre of coordinates to the opposite side. Inversion changes a point at (x, y, z) to (-x, -y, -z) thus, in the diagram point 1 becomes point 2 after the whole operation, then 2 becomes 3 and so on. Two $\bar 6$ operations moves 1 to 3 and 6 operations places it back where it started.

[The image is from 'Crystallography, An introduction ' by W. Borchardt-Ott, 3rd ed publ Springer.]


How to calculate the final concentration of calcium ions from known starting masses of carbonate and calcium ions? If $\pu{50.0 mg}$ of $\ce{CO3^2-}$ and $\pu{50.0 mg}$ of $\ce{Ca^2+}$ are present  in one liter of water, what will be the final (equilibrium) concentration of $\ce{Ca^2+}$ expressed in moles per liter?

The premise of your question is weird. $\pu{50 mg}$ of calcium ion and $\pu{50 mg}$ of carbonate ion are not equimolar. However, let's assume that they have appropriate counterions somewhere. This problem can be approached two ways.

Limiting reactant

We have a precipitation reaction between the calcium and the carboncate ions,  as calcium carbonate is nearly insoluble in water:
$$\ce{Ca^2+ (aq) + CO3^2- (aq) -> CaCO3 (s) v}$$

Convert your masses of ions to amount of substance and figure out which ion is in excess. If calcium ion is in excess, then use the amount of substance in excess to determine the remaining calcium concentration.
For example: Let's say you determined that your solution has $\pu{0.1663 mol}$ of $\ce{Ca^2+}$ and $\pu{0.1221 mol}$ of $\ce{CO3^2-}$. The maximum amount of $\ce{CaCO3}$ that could precipitate is $\pu{0.1221 mol}$, because we would run out of carbonate ion. Then there would be $\pu{0.0442 mol}$ of calcium ion left in $\pu{1 L}$ of solution for a concentration of $\pu{0.0442 M}$.

Solubility product

Calcium carbonate is not completely insoluble. It's solubility product constant is $4.8\times 10^{-9}$. Thus we can set this up as an ICE equilibrium problem. I'll use the same fake amount of substances from the previous example:

\begin{align}
\ce{CaCO3 (s) &<=> Ca^2+ (aq) + CO3^2- (aq)}\\
K_\mathrm{sp}&=[\ce{Ca^2+}][\ce{CO3^2-}]
\end{align}
\begin{array}{lcc}
\hline
                   & \text{calcium}/\pu{mol}  & \text{carbonate}/\pu{mol}\\
\hline
\text{Initial}     & 0.1663                   & 0.1221     \\
\text{Change}      & -x                       & -x         \\
\text{Equilibrium} & 0.1663 - x               & 0.1221 - x \\
\hline
\end{array}
\begin{align}
4.8\times 10^{-9}&=(0.1663-x)(0.1221-x)
\end{align}

Solving for $x$ gives you $x=0.1221,0.1663$ and only the first number makes sense. In this case the final calcium amount in solution would still be $\pu{0.0442 mol}$. The carbonate amount in solution is not actually zero. It is $\pu{1.09E-7 M}$, but that is for another day.


Hydrogen bonding in chloral hydrate Normally, chlorine doesn't form hydrogen bonds because despite its electronegativity, the size of the atom is such that its electron density is too low to form hydrogen bonds.

However, chlorine forms hydrogen bonds in chloral hydrate (2,2,2-trichloroethane-1,1-diol). What could be the possible reason for this?

As noted in the comments, the fallacy is the notion that only a select few atoms can form hydrogen bonds with protic hydrogen.  In reality, the strongest hydrogen bonds involve nitrogen, oxygen, or fluorine, but most nonmetals can form such bonds if they have an electron pair to donate to the bond.  Hydrogen bonding with chlorine is seen in the $\ce{HCl_2^-}$, which is analogous to the more familiar bifluoride ion, as well as chlorine-bearing organic compounds.


Would dissolving a salt and letting it sit for 48 hours have an effect on the solution? If one were to dissolve magnesium citrate (about $100~\mathrm{mg}$) in water (about $70~\mathrm{ml}$) and let the solution sit for 48 hours, would the solution be exactly the same as one dissolved and consumed right away?

How does one explain the difference in taste from a 48-hour solution and a 5-minute solution?

The only reactions that I would expect to occur to any significant extent with pure water are acid-base reactions between the citrate and water, where water, being amphiprotic, could act as either an acid or a base:

$$\ce{C6H6O7^2- + H2O <=> C6H7O7- + OH-}$$
$$\ce{C6H6O7^2- + H2O <=> C6H5O7^3- + H3O+}$$

Citrate, being the conjugate base of a weak acid, is itself a weak base (according to the citric acid Wikipedia page, the $\mathrm pK_\mathrm a$ for the dissociation of the second proton is 4.75, and for the third between 5.41 and 6.40). Citrate is tripotic, and in the magnesium salt citrate possesses one undissociated acidic proton, which makes it an amphiprotic molecule as well. Given that the $\mathrm pK_\mathrm a$ of water is approximately 15.7, and the $\mathrm pK_\mathrm a$ of hydronium ($\ce{H3O+}$) is −1.74, both reactions above favor the reactants over the products by many orders of magnitude, though the second reaction is favored above the first (i.e., the dibasic magnesium citrate salt should still be very slightly acidic, overall).

Acid-base reactions typically reach equilibrium quite quickly, so there should be very little change in pH and no pH-attributable differences in taste over time. If any change does occur, I would expect it to be the result of a change in concentration as the water gradually evaporates. Another possibility is that, if the water is impure (which tap water certainly is), additional side-reactions might occur. What those are would depend on the exact composition of the water as well as environmental conditions. It's possible for the magnesium to form a precipitate, for example, in combination with other anions dissolved in the water (unlikely given the quantity in your example, but possible). If the magnesium citrate is part of a commercial product containing other ingredients, then it may well be that other agents are responsible for any effects you observe.


Calculate the pH of buffer solution PROBLEM: A solution was prepared by mixing $\pu{25.00 mL}$ of $\ce{NaOH}$ $\pu{(c = 1mol/L)}$ and $\pu{10.00 mL}$ of acetic acid $\pu{(c = 2.5 mol/L)}$ in water to give $\pu{250 mL}$ of solution.

What is the $\mathrm{pH}$?

$K_\mathrm{a}=\pu{1.76\times 10^{-5} mol/L}$

Ok, so I know acetic acid will react with sodium hydroxide do form sodium acetate:

$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

initial quantities of $\ce{CH3COOH}$ and sodium hydroxide are $\pu{ 0.025 mol}$  (c*V)each

after the equilibrium is reached all base and acid have reacted, so we have $\pu{0.025 mol }$ of $\ce{CH3COONa}$

I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because concentration of acid is 0.

I got a response that I should solve it like this (after I do ICE table like I did above):

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

$[\ce{OH-}]=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$
and after I calculate $\mathrm{pOH}=5.12$ I can calculate the $\mathrm{pH}=8.88$

I dont understand the last part, why is this so? what is

$$\ce{CH3COO- + H2O <=> CH3COOH + OH- }$$

representing?

Thanks for answering! :)

I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because the concentration of acid is 0.

Yes, as commented above, sodium acetate solution cannot be considered as buffer;the result at the equilibrium , will be the same as dissolving  $\pu{0.025 mol }$ of $\ce{CH3COONa}$ in $\pu{250 ml}$ water,so you have the salt solution of $\pu{0.1M}\ce{[CH3COONa]}(c_\mathrm{\text{salt}}=\ce{[CH3COONa]_\mathrm{I}}=\pu{0.1M})$

  I don't understand the last part, why is this so? what is...

The salt ($\ce{ CH3COONa}$) is  electrically neutral substance formed by cation $\ce{Na+}$ (an acid) and  anion $\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COONa   -> Na^+_\mathrm{(aq)}   + CH3COO^−_\mathrm{(aq)}}$$
a)  Since $\ce{Na+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions $\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$ $$(K_\mathrm{b(\ce{CH3COO-})}=\frac{K_\mathrm{w}}{K_\mathrm{a(\ce{CH3COOH})}}=\frac{10^{-14}}{1.76\times 10^{-5} }=5.68\times{10^{-10}})$$

c)The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}[\ce{CH3COOH}]_\text{equilibrium}}{[\ce{CH3COO-}]_\text{equilibrium}}$$

Neglecting water autoionization and assuming :${\ce{[OH-]_\text{equilibrium}}=\ce{[CH3COOH]_\text{equilibrium}}}$, ${[\ce{CH3COO-}]_\text{equilibrium}=c_\mathrm{\text{salt}}=[\ce{CH3COONa}]_\mathrm{I}}=\pu{0.1 M}$,so:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}^2}{c_\mathrm{\text{salt}}}$$

$$[\ce{OH-}]_\text{equilibrium}=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$$
You  can calculate $\ce{[H+]}$ -without neglecting the water autoionization  and without approximation- by substituting in the following cubic equation and solving bywolframalpha

$$\
 [\ce{H+}]^3
+( K_\mathrm{a}+[\ce{CH3COONa}]_0) [\ce{H+}]^2
  -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  $$
$\ce{[H+]} =1.32658 ,\pu{pH}=8.877$


If the human body consists of 60% water why can't we put out fire with our body? I have often heard of people being burned at the stake, but if the body is 60% water shouldn't the fire just be put out?

We need to consider something. Are the people being truly burnt at the stake or are they being killed by the effects of fire? If the fire is very prolonged then their bodies will dry out and then finally be calcined to ashes.

It is an interesting (but a rather morbid) fact that cremation furnaces typically run at about 800 degrees Celsius for many hours, yet they tend to be unable to burn bodies to dust. Typically after cremation, some large fragments remain. These have to be crushed.

There are tales of spontaneous human combustion, these have been circulating for many years. I recall that the Victorians thought it was associated with extreme alcohol abuse. They thought a build-up of alcohol in the body would lead to the body being able to catch fire. A modern-day explanation is that humans wear clothing, the clothing can act as a wick.

A person dies as a result of something, then their clothing comes into contact with an ignition source. The clothing starts to burn, this heats up fat in their body. The fat melts and runs into the clothing. The clothing then acts like a candle wick and then the burning fat makes more heat which melts more fat thus turning the person into a candle.

I think that this type of human candle effect is very rare, it is likely to require a combination of several things at once. Many people who are killed in serious fires (who suffer burns as well as being exposed to hot and toxic smoke) leave behind bodies which have not been burnt to ashes.


How much water vapour can a gram of Uppsalite absorb at NTP? This site states:

  It turned out that Upsalite had the highest surface area measured for an alkali earth metal carbonate; 800 square meters per gram.

Assuming that a sphere may enclose the maximum volume for a given surface area, the radius for a sphere enclosed by $\pu{800 m^2}$ is $\pu{7.9788 m}$. It's corresponding volume is $\pu{2127.656 m^3}$ - which leads to the incredible figure of something in excess of 2000 litres of water ... per gram of dessicant.

How much water vapour can a gram of Uppsalite absorb at NTP?

One fallacy in your question is assuming that the particles 1) are spherical and 2) only accessible on the outside. The press release linked says that this material is nanoporous - the particles are perforated by pores "all having a diameter smaller than 10 nano meters". The extra surface area is inside these pores. Spheres have the lowest surface-area-to-volume ratio. Other shapes have higher ratios (in other words, more surface area for less volume). These nanoporous materials maximize the surface area to volume ration - minimizing the volume. If these particles are $\ce{MgCO3}$, then their density is $\pu{2.958 g/cm^3}$ (Wikipedia). One gram of this material has a volume of:

$$\pu{1 g}\times \pu{1 cm^3 g^-1}\times \left(\frac{\pu{1 m}}{\pu{10^2 cm}}\right)^2=\pu{3.38E-7 m^3}=\pu{3.38E-3 L}$$

The surface-area-to-volume ratio is then:
$$\frac{A}{V}=\dfrac{\pu{800 m^2}}{\pu{3.38E-7 m^3}}=\pu{2.37E9 m^-1}$$

A second fallacy is that the volume of water that can be absorbed is the same as the volume of the particle. The particle is not absorbing water per se. The water is not taken into the interior volume where the $\ce{MgCO3}$ is. That would require displacement of the magnesium carbonate, and where would it go? The water goes into the pore of the material - it adsorbs to the surface inside those pores.

You can read more about the science at the original research article:  A Template-Free, Ultra-Adsorbing, High Surface Area Carbonate Nanostructure. PLoS ONE 2013, 8 (7). DOI: 10.1371/journal.pone.0068486.


What makes these particles hygroscopic? Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment.

For cotton there is capillary action.
Hydrates like KOH attract water due to lower energy.
What are other possible reasons?
Particles like - diesel fuel, glycerol, salt, sugar and H2SO4 are also hygroscopic?
I have tried to find the reason, but I cant?

As to your question a to whether a hydrophobic compound can also be
  hygroscopic, the simple answer is "yes to a point." Organic compounds
  can elicit both behaviors simultaneously when they have varying
  functional groups with a wide enough separation. BioDiesel is an ester
  with a very short side and a long side, the short side can exhibit
  polar qualities due to the presence of oxygen while the long chain
  exhibits nonpolar qualities. The polar side is what attracts the
  water, while the long chain repels the water. Cells of living
  organisms utilize a similar method to control the amount of materials
  crossing the membrane. I hope this explanation helps. Das Nerd 20:08,
  28 October 2006 (UTC)

https://en.wikipedia.org/wiki/Talk%3AHygroscopy

The composition of petrol diesel is very different, that's why it is not as hygroscopic as biodiesel.

As far as I know hygroscopy depends mostly on polarity. The water molecule is a polar, so it likes to mix with other polar compounds, like KOH, glycerol, salt, sugar, H2SO4, etc. Apolar compounds repel water. Air consists mostly of apolar components, so when the water in the air meets with polar things like sugar, it starts to move to that phase, because it has lower energy in there.

E.g. silicagel which is a common desiccant has this structure:

where you have polar $-OH$ chemical groups on the surface, that's why it is very hygroscopic.

While silicone oil contains molecules with this structure:

where you have apolar $-CH_3$ chemical groups on the surface, which makes it not hygroscopic.


How to determine Langmuir adsorption isotherm constants and use the models I am doing some adsorption tests on bio materials and would like to test the data with Langmuir and Freundlich adsorption isotherm models. Assume these values:

adsorbent(g).   initial con (µg/L)  final con.(µg/L)
0.1                   1                0.65
0.1                   5                2.1

(Sample volume is 20ml. In both cases the dose of adsorbent was the same (0.1g) and in the first initial concentration is 1ppb and, after adsorption, is 0.65ppb.  In the second initial concentration is 5ppb and, after adsorption, is 2.1ppb.)

Using these data how can I use Langmuir and Freundlich isotherms models to get the constants and check whether my adsorption data fit the model or not.  (If any other parameter is needed you can assume some reasonable value.)

For langmuir isotherm:

Let say we are using linear langmuir isotherm.

The formula is : $\frac{1}{q_e} = \frac{1}{q_m} + \frac{1}{q_mKL}$

find the value of $q_e$. Using the formula $q_e = \frac{(C_i-C_f)V}{\text{mass}}$ of adsorbent, and then plot the graph $\frac{1}{q_e}$ versus $\frac{1}{C_f}$. It would be easier by using excel. From the graph, we can calculate the slope. The linear graph will produce $y=mx+c$ equation. m=intercept=1/qe, hence qe=1/m. C=1/qm.KL. From the C value, we can actually find KL.

From the graph generated using excel, it will shown the R2 value. This value is the important indicator to determine which isotherm model fit the system. The highest R2 value will fit the system. Let say R2 value langmuir is 0.8 and R2 for freundlich is 0.9. So, we can conclude that, Langmuir model fit your system.

For Freundlich isotherm

Plot the graph log qe versus log cf. From the graph you will get linear equation, y=mx + c and the R2 value. Compare the R2 value with langmuir's R2. Highest R2 value indicate that those model fit with the system.

Cf=final concentration
Ci= initial concentration


irreversible adiabatic process against constant external pressure? My book has these formulae for irreversible adiabatic process against a constant external pressure $p_ext$ : $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_v+(p_{ext}/p_2)R}]T_i$$. It is also mentioned that when external pressure = p2 $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_p}]T_i$$. How do I get these formulae?
I am getting confused because they are saying external pressure. In general when irreversible work don't we take into account only the external pressure and not the internal pressure? And why exactly is this so? I have assumed so far that this is because the internal pressure differs only slightly from the external pressure. Is this so? Would someone please explain this?

In this irreversible adiabetic process, only the work done on the system will be =$-P_{ext}dV$. Because in a reversible process we actually consider $P_{ext} = P_{int} + dP$, and hence we neglect the $dPdV $ term as it is negligible.  Considering this, if we go by the definition of adiabetic process and first law of thermodynamics, we will have $$0 = nC_vdT  + P_{ext}dV$$ as the heat changed in an adiabetic process is $0$. Now simply taking one term to other side and integrating we have $$nC_v \int_{T_1}^{T_2} dT = -P_{ext} \int_{V_1}^{V_2} dV$$  where $P_1,V_1,T_1$ are initial conditions, and $P_2, V_2,T_2$ are final conditions of the system.Now we put $V_1 = \frac{nRT_1}{P_1}$ and $V_2 = \frac{nRT_2}{P_2}$, so we will have $$C_v(T_2 -T_1) = -P_{ext}R(\frac{T_2}{P_2}-\frac{T_1}{P_1})$$ now taking $T_2$ associated terms to one side and $T_1$ related terms to other side we have $$T_2(C_v + \frac{P_{ext}R}{P_2}) = T_1( C_v + \frac{P_{ext}R}{P_1})$$ and thus we have our desired result as, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$


Thymolphthalein solubility I already know that thymolphthalein is soluble in alcohols, very alkaline aqueous solutions, acetone, and propylene glycol methyl ether.  My questions are what other solvents dissolve thymolphthalein, and what additives increase solubility (for example, specific surfactants)?

One expects [poly(ethylene oxide)-co-(polypropylene oxide)-co-poly(ethylene oxide)] triblock surfactants will micellize it (e.g., Pluronics); possibly Tergitols.  It looks too bulky to fit inside even gamma-cyclodextrin.

Hang a suitable side chain off the phthalic anyhydride and the product will dissolve in pretty much anything.  What does pH mean outside water?


How to determine number of alpha hydrogens in a cyclic compound? How do we determine the number of alpha hydrogens in a compound? Where should we start the naming from?
For example, bicyclo[4.2.0]oct-7-ene and bicyclo[4.2.0]oct-3-ene:

Compare the stabilities of the following alkenes

In both these structures, I need to determine their stabilities based on their hyperconjugating structures. I need to determine the number of alpha hydrogens it has, to determine the number of hyperconjugating structures it makes. Where do I start?
I am aware that stability is directly proportional to the number of hyperconjugating structures. More the number of alpha hydrogens, greater is the number of hyperconjugating structures, and hence greater the stability.

The relative stability of your two compounds is not determined by arguments based on hyperconjugation.  Rather the relative stabilities have to do with ring strain.  Placing a double bond in smaller and smaller rings increases the angle strain in the ring system.  A double bond, being roughly $\ce{sp^2}$ hybridized, would prefer to have bond angles around 120°.  This becomes difficult in 3- and 4-membered rings where the internuclear angle is 60° and 90° respectively.  Therefore molecules containing double bonds in 3- and 4-membered rings will be more strained than molecules with the double bond in a larger ring.

Edit: Response to OP's comment

No, the given solution is not the correct answer to the problem.  It is correct that structure B has more alpha (alpha to the double bond) hydrogens (4) than structure A (only 2 alpha hydrogens).  It is also true that hyperconjugation can be used to explain why alkyl groups attached to a double bond act to stabilize a double bond (see this earlier answer for an explanation), and that the double bond in B is more stabilized by hyperconjugation than the double bond in A.  However, hyperconjugation has a much smaller effect on the stability of these molecules than the ring strain effect.


how can carbon dating even work? Its at atomic level! I am a homeschooling parent, so pardon me asking stupid questions.
Everywhere I read, it is said carbon based lifeforms eat carbon and after their death that atom starts a decay.

But what I cannot understand, we dont eat or work with food at atomic levels. Yes, life forms breaks down molecules from food and use them as they wish. But that chemical process does not effect the atom structure. I dont think any life form has capacity or energy to break down atoms.

So if one atom of C14 was having a decay already before we ate it, it keeps having it while it gets digested and goes on in the same way if life is alive or dead. If another C14 was not having a decay, then it starting a decay has nothing to with the life form is alive or dead.

As Dr. Manhattan(Watchmen 2009) said atoms dont care if you are alive or dead. They stayed there at the same place before and after.

So how can a lifeform have any effect in the C14 decay at all?

Got some good answers, so I clarifying my questions:

How can we be sure that we have eaten only fresh out of factory C14? It may be already decayed by hundred or thousand of years  when we eat it. It applies to all living beings. If there is fresh C14 in the air, there is also decaying levels of C14 as old as 50K years. So when you check my c14 count, you would have a wrong answer of when I died cause origin of C14 was even before I was my great grandfather was.

You are correct in the fact carbon-14 does not care whether or not a life form is alive; however, this is irrelevant in the case of carbon dating. You may be familiar with the carbon cycle, which is, in a nutshell, the recycling of carbon through carbon dioxide. Plants are, of course, constantly using carbon dioxide from the atmosphere to generate large bio molecules that we can eat. We are also constantly recycling the carbon, ingesting it in the form of food, and exhaling it in the form of carbon dioxide, so it is clear that living organisms are constantly replacing their "old" carbon from carbon from atmospheric carbon dioxide. The key factor to carbon dating is that atmospheric carbon dioxide has a higher percentage of Carbon-14. This is caused by high energy cosmic rays which convert nitrogen-14 (the most abundant isotope of nitrogen) into carbon-14 nuclei, which in turn result in a greater percentage of our carbon content being carbon-14. When we die, we are no longer replenishing our carbon from the atmosphere (higher carbon-14 content due to cosmic rays), so the carbon-14 will gradually decay into more stable nuclei, leaving a lower percentage of carbon-14 in our remains. By using the half-life of carbon-14 nuclei, we are able to calculate the time when a certain living organism was no longer exposed to our atmosphere's high concentration of carbon-14. As a side note, carbon-14 dating is fairly accurate because the relative concentration of carbon-14 has remained constant in our atmosphere.


Substructure search with RDKit I'm trying to search for substructures with RDKit.

I would like to search for structure 1 as substructure in other structures. I don't understand why it doesn't find this structure in structures 3 and 5. I assume this is not a bug in RDKit, but more the result of my limited chemistry understanding. Something with aromaticity? Something wrong with the structures?
Can someone explain to me why it doesn't find structure 1 in structures 3 and 5?
How should the search structure look like that I will find all structures 2-5?

from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6']

pattern = Chem.MolFromSmiles(smiles_list[0])
for idx,smiles in enumerate(smiles_list):
    m = Chem.MolFromSmiles(smiles)
    print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

# Result:
# Structure 1: pattern found True
# Structure 2: pattern found True
# Structure 3: pattern found False
# Structure 4: pattern found True
# Structure 5: pattern found False

I'm not sure about why it's not matching but when I carry out substructure matches in rdkit I use SMARTS instead https://www.daylight.com/dayhtml/doc/theory/theory.smarts.html

Here is what I would have done.

from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6', 'c1ccccc1']

pattern = Chem.MolFromSmarts('[#6]12~[#6][#6]~[#6][#6]~[#6]1[#6]3~[#6][#6]~[#6][#6]4~[#6]3[#6]2~[#6][#6]~[#6]4')
for idx,smiles in enumerate(smiles_list):
    m = Chem.MolFromSmiles(smiles)
    print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

and the output

Structure 1: pattern found True
Structure 2: pattern found True
Structure 3: pattern found True
Structure 4: pattern found True
Structure 5: pattern found True
Structure 6: pattern found False

I've added a Benzene smiles to show that it's not just matching everything. The SMARTS pattern I made probably could be better I think. I hope this helps anyway.


How to prepare elemental gallium from a mixture of gallium and aluminum nitrate? How would one go about preparing and purifying elemental gallium from a mixture of aluminum(III) nitrate and gallium(III) nitrate?

If you have a mixture of aluminum nitrate and gallium nitrate, then you have to separate the aluminum nitrate from the mixture. This patent describe how to separate aluminum nitrate from a mixture of other nitrate and hence, this process can be applicable for separating aluminum nitrate from gallium nitrate.

Then, when you have gallium nitrate, just heat to produce gallium oxide.

$$\ce{4Ga(NO3)3 -> 2Ga2O3 + 12NO2 + 3O2}$$

  The thermal decomposition of nitrate gallium(III) to produce
  gallium(III) oxide, nitrogen dioxide and oxygen. This reaction takes
  place at a temperature of 110-200°C.(chemiday 1)

Then reduce gallium oxide to gallium using some reducing agent like carbon or hydrogen.

$$\ce{Ga2O3 + 3H2 -> 2Ga + 3H2O}$$

  Gallium(III) oxide react with hydrogen to produce gallium and water.
  This reaction takes place at a temperature near 700°C.(chemiday 2)

$$\ce{Ga2O3 + 3C -> 2Ga + 3CO}$$

  Gallium(III) oxide react with carbon to produce gallium and carbon
  monoxide. This reaction takes place at a temperature of 850-950°C.(chemiday 3)

You will find this youtube video of help. It shows how to synthesize gallium metal.


Effect of substituents on aniline C-N bond length Which compound has larger $\ce{C-N}$ bond length, 4-aminobenzaldehyde (1) or aniline (2)?

I think it's the second compound. $\ce{-CHO}$ is an electron withdrawing group. This means the electron density of the ring decreases. So the carbon attached to nitrogen becomes more electronegative. Hence its p-character decreases and the $\ce{C-N}$ bond length decreases.

You think right, but for the wrong reason. The inductive effect is not that great and does not quite reach that far. It is the effect of conjugated $\pi$-system that really matters. See, we may draw an alternative resonance structure for I with minus on O, plus on N, and rearranged double bonds (quinone-like ring instead of aromatic). This means the C-N bond in I is kind of "partially double" (i.e., has a noticeable $\pi$ component), hence it is shorter.

Indeed, there is some conjugation between the aromatic ring and nitrogen's lone pair in unsubstituted aniline as well. That's what makes aniline a weaker base than ammonia, and its aromatic ring more reactive towards electrophiles than benzene. It is just that the $\ce{-CH=O}$ group in para position makes that conjugation stronger yet, as we may see qualitatively by drawing the resonance structures, or (supposedly) quantitatively by means of quantum chemistry.


Will Sodium Amide react with Epoxides in the same manner as Grignard Reagents? I have been unable to find the reaction anywhere on the internet so I'm assuming the answer is no, although I would like an explanation as to why.  I can't seem to find any reason why it would not work, provided the reaction is carried out in two steps.  In the first step, H2N:- will act as a nucleophile, attacking the least sterically hindered carbon of the epoxide.  The second step would of course be an acidic workup.  As stated in the question, I would expect the same exact mechanism as the reaction of Grignard Reagents with epoxides (substituting sodium amide as the nucleophile) due to the high basicity of sodium amide and Grignard Reagents.  Like I said, I'm not expecting the reaction to be viable.  However, I would like to know exactly why not.  Im assuming it's something to do with the Hydrogen atoms of the amide anion, though I could be wrong.  Enlighten me.

It actually is possible but it works well only with simple epoxides so it is not wildly used. $NaNH_2$ is extremely basic (it can deprotonate alkynes) and too many functional groups can react with it: alkyl halides (elimination), aldehydes, ketones...

To work around that basicity, sodium azide $NaN_3$ is commonly used instead, followed by the reduction of the azido group into an amine.

For example, see this article.


Atmospheric Nitrogen gas and human body pH buffers It looks like the human body doesn't have the ability to breakdown atmospheric $\ce{N2}$, as of now $\ldots$ So an enzyme like nitrogenase (found in cyanobacteria) fixes $\ce{N2}$ and produces $\ce{NH3}$. $\ce{NH3}$ is a weak base which qualifies as a buffer component.

If the human body was able to react to $\ce{N2}$ inhaled from the air, would it negatively impact the body's ability to maintain pH levels of vascular systems/cells of the human body given the large amount of $\ce{N2}$ in the atmosphere?

One chemical equation that can be written for nitrogenase is this one:

$\ce{N_2 + 8 \ H^+ + 8 \ e^- + 16 \ ATP \longrightarrow 2 \ NH_3 + H_2 + 16 \ ADP + 16 P_{i}}$

The equation shows that not only does nitrogenase generate the basic molecule $\ce{NH3}$, it also consumes protons (i.e. acid) to form hydrogen as an inescapable byproduct of its mechanism.

So yes, in theory nitrogenase could perturb the human body's ability to buffer pH.

However, the relative rates are always important.  Human requirements for nitrogen are on the order of 80 grams of protein per day.  This translates to about 12 g of (fixed) nitrogen atoms per day, which corresponds to 0.85 mol of N.  If all of that nitrogen were to come from nitrogenase, then 0.85 × 8 ÷ 2 = 3.4 mol of acidity would be lost via proton consumption in addition to the 0.85 mol of base generated as $\ce{NH3}$, or about 5.2 mol of "base" in total.

But human pH homeostasis works even though humans breathe out about 900 g of an acid equivalent ($\ce{CO2}$) per day, or about 20 mol.

So even if nitrogenase completely replaced human nitrogen requirements in some hypothetic bioengineered human, the demands on pH homeostasis would only change by about 25%, maximum.  And considering that pH homeostasis still works even when humans are exercising vigorously, when "instantaneous" $\ce{CO2}$ production rates are far higher, I doubt even a 25% change would be too much to handle.

Thus in conclusion I don't think nitrogenase would wreck human pH homeostasis, unless its activity in the human body was increased to a level far beyond what is necessary to replace dietary N requirements.


What is the nature of interstitial hydrides? In my Chemistry book(NCERT Chemistry,Part 2, textbook for class 11, Pg 281). I found the mention of interstitial hydrides:

  Unlike saline hydrides these compounds are nonstoichiometric, for
  example $\ce{LaH_n, TiH_n}$ and $\ce{PdH_n}$, where the chemical composition is variable. Typical formulae are $\ce{LaH_{2.87}, YbH_{2.55} , TiH_{1.8}, ZrH_{1.9}, VH_{1.6}, NbH_{0.7}}$ and $\ce{PdH_{0.7}}$.

The book also said:

  Earlier it was thought that in these hydrides hydrogen occupies the
  interstices in the metal lattice producing distortion without any
  change in it's type. However recent studies have shown that except for
  hydrides of $\ce{Ni, Pd, Ce,  Ac}$ other hydrides of this class have
  lattice different from the lattice of the parent metal.

What does this mean? Please explain the nature of these compounds and type of bond present in them.

Nonstoichiometric compounds are simply compounds that cannot be represented with integer number of atoms. What the book probably is trying to say is that the hydrogen atoms do not simply fill the empty space between the atoms in the crystal. In other words, the hydrides will cause a change in the structure of the lattice. In some compounds, the atoms that are being inserted in the free space in the crystal will not cause any noticeable change in structure. The crystal may become slightly distorted. But it will retain its original shape before insertion of the new atoms. What the book is trying to say is that this not true for hydrides. Scientists used to think that this was the case, but it is not true.


How would I find R/S configuration in this case? I know about the rules to assign R/S configurations when the orientation of each bond to the central Carbon atom is given- the whole, give them a priority order based on their Atomic number/masses, move the group with the least priority as far as possible from the eye, and then draw an arrow joining the other groups in order of their priority. If the arrow goes clockwise, it's an R configuration, otherwise it's an S configuration. But how would I do the same when the orientation of none of the groups are given? The answer key says that compounds I and II have R orientation- but when I simply apply the steps above to the  compound I, I get S. How do I do this?
https://imgur.com/a/K57fCJL is the question

While assigning priority in I, $\ce{-CHO}$ is second priority group instead of $\ce{CH2OH}$ as in $\ce{CHO}$ there is a double bond between $\ce{C}$ and $\ce{O}$, so we can consider the duplicate form, i.e. $\ce{C}$ is bonded to two $\ce{O}$ atoms and $\ce{O}$ is bonded to two $\ce{C}$ atoms. So, the rotation is counterclockwise, but $\ce{H}$ is in front of the observer, so we consider the opposite rotation, i.e. clockwise. Thus it is R configuration.

Instead of using the rule that we have to consider the opposite rotation (while moving the fingers from the more priority group to the less priority), when the least priority is in front of the observer. You can visualize it this way, when you see the the compound I, drawn on paper, then $\ce{H}$ and $\ce{OH}$ is towards you and $\ce{CHO}$ and $\ce{CH2OH}$ is away from you and the rotation is counterclockwise. But when you see the same compound below the paper, the $\ce{H}$ and $\ce{OH}$ is away from you and $\ce{CHO}$ and $\ce{CH2OH}$ is towards you, but now the rotation is clockwise. So now we consider this rotation to say that it is in R configuration.

The figures are basically the Fischer projections of the corresponding molecules. The actual representation of Fischer projections is below.

Hence the 'pushing $\ce{H}$ as far as possible' thing is done when $\ce{H}$ is on dashed bond (into the screen). Just use the rules of Fischer projections and change the molecule to the form which you feel most comfortable to determine R/S configurations.


How many d-eletrons are in the outer shell of following compounds: How many d-electrons are in the outer shell of the metals in following compounds:

$\ce{ZnS}$ and $\ce{NiS}$

So sulphur does not has any $\ce{d}$ electrons and $\ce{Zn}$ has $10\ce{d}$ electrons.

But I am pretty sure that this question is more complex to answer like that.

What is considered as the outer shell? as $\ce{Zn}$ has the electrons in the $3\ce{d}$ shell but there is still a filled $4\ce{p}$ shell?

Does it also matter that it is a compound ? ($\ce{ZnS}$ not only $\ce{Zn}$). As I see no reason why they included $\ce{S}$ but I guess it has an important reason.

I am new to transition metals, any help would be really appreciated.

Sulfur does not have any "d" electrons as you said.

Before losing two electrons from the 4s sublevel, Zinc's outer sublevel is 4s (not 4p as you mentioned).  After those two electrons are lost, the outermost sublevel becomes the 3d sublevel which contains 10 electrons.

Yes, it does matter whether the atom forms a compound or not, so we can understand where the electrons are allocated/transferred.

Here is the general rule for how to consider electrons in atoms.  Electrons fill up an atom via the Aufbau principle (image attached).  However, after filling, they arrange themselves in the energy according to the their principle quantum number (1, 2, 3,...) and their sublevel type (s, p, d, f).  For example, 4s fills before 3d in zinc, but after filling, 4s gains a higher energy than 3d.


How to neutralise path weeder contamination I strongly suspect my aquarium has been contaminated by a rock sprayed with a path weeder containing Simazine, Amitrole & Ammonium Thiocyanate.

I removed all rock and made several large water changes but fish continued to die one or two a day. When the surviving fish were removed to separate tanks with identical water conditions fish are fine and not one has died for about a week. So it seems to me the chemicals have permeated the crushed coral substrate and filter media and is still leaching into the water.

I have emptied the tank completely & easily can replace the filter media but the substrate, resin decorative pieces, and plastic plants are expensive to replace and I would like to be sure I remove any residue in the tanks and piping etc. Is there anything I can wash the tank, substrate, and decor with to neutralize any remaining toxicity that would not leave a toxic residue in itself?

Would washing everything with vinegar be of any help?

Any advice from those understanding the chemical structure of these toxins and how I might rescue my prized aquarium would be greatly appreciated!

Amitrole is slightly toxic to various species of freshwater fish$^1$. However, it's quite water soluble and should not be difficult to remove from most surfaces with simple warm-water washing.

Ammonium Thiocyanate is particularly water soluble and should also readily be removed via warm-water washes. Thiocyanate ion has lower solubility with some other cations that would be present, but in general, thiocyanates are quite water soluble. Thiocyanate ion has been shown to be toxic to fish at the tens of mg per liter level, however, $^2$.

Simazine has a very low water solubility, on the order of 5 mg/L (4-5 orders of magnitude less soluble than the other compounds) but has been shown to be lethal to fish at levels on the order of its limit of solubility$^3$. I strongly suspect that residual simazine is thus to blame for the persistent toxicity to your fish.

Simazine is about 100 times more soluble in isopropyl alcohol (rubbing alcohol) than it is in water. I suggest obtaining some 99% isopropyl alcohol and scrubbing/soaking all of the decorative material that you value, as well a the tank itself, thoroughly rinsing the alcohol off with warm water, then drying completely. There is no reason to believe that vinegar would be useful in this capacity.

Anything relatively cheap should probably simply be replaced, particularly difficult to clean items like driftwood, high-surface area gravel or sand, and of course all of the filter media. My biggest concern of the items you mentioned is the crushed coral. If it were me, after the cleaning I would return everything to the tank except the crushed coral and once the water is ready, test it out with a couple fish. If things seem OK for a while, then I would set up the tank with everything including the crushed coral.

As a former aquarist, I understand how much work is involved with my suggestions. I strongly believe, however, that you have identified the culprit in the fish die-offs and that the procedure above has a good chance of resolving the issue. Please don't hesitate to ask for any clarifications in the comments below.

US EPA. 1984 (March). Amitrole: Pesticide Registration Standard and Guidance Document. Office of Pesticides and Toxic Substances, US EPA, Washington, DC.
Speyer, M. R. and Raymond, P. (1988), The acute toxicity of thiocyanate and cyanate to rainbow trout as modified by water temperature and pH. Environmental Toxicology and Chemistry, 7: 565–571. doi:10.1002/etc.5620070705
WSSA Herbicide Handbook Committee. Herbicide Handbook of the Weed Science Society of America, 6th Ed. WSSA, Champaign, IL. 1989


Spices are absorbed or adsorbed in food? When we cook food, we add several spices and condiments which enhance color, taste and smell. Are these spices and condiments just adsorbed on the food particles? Or are they absorbed? For example using pepper while cooking a piece of chicken, would it just stay on the top of the leg (adsorbed)? Or get absorbed?

The answer varies with the type of food and the type of condiment. Both are very complicated mixtures, so a same combination of food and condiment can display varying levels of adsorption/absorption depending on the region/compound being analyzed, but in general, I expect that while all of chemical/physical adsorption and chemical/physical absorption happen to some extent (cooking is quite a messy collection of physical and chemical interactions!), either physical absorption is the most prevalent, or none of them happens to the largest part of the spices.

First of all, adsorption is a surface phenomenon, which fundamentally limits how much matter can be considered adsorbed onto an object. Adsorbed films are usually very thin (probably not much more than a few nanometers thick, often a single molecule thick). Regular macroscopic objects have very little surface area relative to their volume, so it takes a comparatively tiny amount of substance to completely cover the object in a thin adsorbed layer, compared to say, activated charcoal. Thus we can likely exclude adsorption as the main factor.

It's also important to consider that most food is cooked with some amount of solvent (water or oil/fat), and in fact solvents come out of the food while it's being cooked. This means that a lot of the spices can very well end up as part of a liquid (albeit often quite concentrated) solution, which remains stuck on the surface simply due to its high viscosity and significant friction with the uneven surface of the food. I bet this is what goes on in most cases.


What does "dye content 80%" mean on Sigma Aldrich's product page for "Allura Red AC" (aka Red 40) Here's a link to the page:
http://www.sigmaaldrich.com/catalog/product/sial/458848?lang=en&region=US

Why does it say "dye content 80%"?
Shouldn't it be pure if it's a packaged powder?
Is it 20% other chemical compounds (e.g. stabilizers)?

Many dyes are sold as powders with 80-90 % purity. It is because some dyes are difficult to purify (e.g. they are difficult to crystallize). The remaining 10-20 % are impurities, such as isomers or closely chemically related side-products, but mostly they are "volatile impurities" (i.e. water). For most purposes these impurities do not interfere, so nobody bothers to remove them. Here is MSDS of Allura Red:
http://www.vidhifoodcolour.com/uploads/gallery/14067976084allura_red.pdf


Cis, trans diene reacting with trans dienophile in D-A reaction I wonder if I predicted the products correctly, because it seems to be messier than I thought - there's a pair of diastereomers... Also, I feel that the endo/exo rule does not apply here. Am I correct in my understanding? I have never seen a reaction like the one I drawn mentioned anywhere before. Hopefully someone can help me out.Thanks!

Note, there's an example in Clayden reacting cis, trans diene with DEAD, but since there's no stereocenter at nitrogen we end up just getting trans product.

Macrophage: I thought you did a fine job in analyzing the Diels-Alder reaction of the E,Z-pentadiene with fumaronitrile. To complement my comment above, I offer an illustration of the transition states of the two reactions leading to their respective products. Because fumaronitrile may add with equal facility (mirror image transition states) to either face of the achiral diene, both 2 and 4 are necessarily racemic. "Racemic" transition states represented by 1 are E-endo/Z-exo that afford racemic bis-nitrile 2. Alternatively, "racemic" transition state 3 is E-exo/Z-endo, which leads to bis-nitrile 4 as a racemate.


Horizons of a bachelor degree in chemistry If I have a bachelor degree in chemistry, what master programmes can I apply for ? In other words, are the choices limited in chemical engineeirng, biochemistry,etc? Is it possible to apply for a master degree in computer science for example ?

You would need certain prerequisite courses to get the master's degree in a different field.  For example, Boston University explains:

The Master of Science (MS) in Computer Science program is intended for computer professionals and for people who intend to move into the computer field from other areas of study (emphasis added).

So yes, you can come from a different background.

Applicants to the program are required to have a baccalaureate degree from a regionally accredited institution and proficiency in the areas covered by the following courses:
◾MET CS 201 Introduction to Programming
◾MET CS 231 Programming with C++ or MET CS 232 Programming with Java
◾MET CS 248 Discrete Mathematics
◾MET CS 341 Data Structures with C++ or MET CS 342 Data Structures with Java
◾MET CS 472 Computer Architecture

So it would be fine to switch to physics, biology, biophysics, or many other programs as long as you have the right prerequisite courses.
However, some schools only accept people wanting a PhD, and give a master's degree along the way, which you get to keep if you don't finish the PhD.


What compounds exist in the bitter peel of specific fruits? Actually, my original question was: What makes a pith of pomegranate so bitter?
Then I decided to be more generic. You'll hardly find someone that consumes pomegranate and doesn't nag about the bitter taste the white pith has! Additionally, when I was a kid, my parents taught me not to eat the bitter peel of an orange, which I later learned that is basic. So, since the evolution of these kind of fruit trees has unanimously been altogether, the chemical species that cause the bitterness of their fruits' piths can be of the same group.
I have a hunch what I seek is the "family" of phenols, but that is nothing more than a hunch. Answer in detail please.

The reason behind the bitterness of most fruits is due to presence of Tannin in them.

Tannin is a naturally occurring polyphenol found in plants, seeds, bark, wood, leaves and fruit skins.

Tannin tastes bitter,dry and astringent and you can feel it specifically on the middle of your tongue and the front part of your mouth. Besides the peels of fruit, we can feel bitterness in other food products as well like,

High-Tannin Foods =
 Tea Leaves
Walnuts, Almonds and Nuts with Skins
Dark Chocolate
Cinnamon, Clove and other spices
Pomegranates, Grapes
Quince
Red Beans

The  bitterness and astringency are caused by tannins binding with proteins in our saliva. The chemical constituent of tannins are: gallic acid , ellagic acid and their derivatives(present in hydrolyzable tannins) , catechin and its derivative (present in condensed tannins)

As for your specific question about the the pomegranate fruit pith, it contains ellagitannins, pelargonidin, punicalin, punicalagin, anthocyanins, cyanidin, ellagic acid which makes it bitter.


What is the explanation for this stereochemical product? I have this reaction with a dr = 96:4 but I can't explain the product theoretically. I suppose it is some orbital interaction?

My hyporthesis is that there is a $\pi$-$\sigma^{*}$ interaction between the alcohol-group ($\sigma^*$) that draws electrons out of the $\pi$ orbital of the double bond? I cannot formulate it really. And why the ratio is so pronounced.

No, you’re overthinking this. It’s just sterics.

The iodine will form a iodonium ion (three-membered ring with positive charge on iodine) where the double bond was. Since iodine is huge, it will come from the less sterically hindered side which is from behind the monitor plain.

The carboxylic acid then attacks the iodonium rom above closing a five-membered lactone ring (kinetically favoured against a six-membered one). The side chain carbon retains iodine pointing backwards (you can twist along the $\ce{C-C}$ bond to push the carbon into the monitor plain and iodine backwards) while the lactone oxygen coming from above means that the methyl substituent on the ring must be pointing behind the plain, again.

Also, stereochemistry on the hydroxide should be retained since it does not take part in the reaction.

Scheme 1: A not-well drawn mechanism for the reaction.


Cell walls in lath martensite - Meaning What does cell walls mean in the context of lath martensite? Are they grain boundaries?

Quote (page 66 Structure and properties of engineering alloys - Smith): In the lath martensite of low-carbon steels, there is a high density of individual dislocations and many cell walls.

Sorry for the stupid question, I don't find any image to understand it.

Thank you for your time.

According to this Wikipedia page:

  Martensite, named after the German metallurgist Adolf Martens
  (1850–1914), most commonly refers to a very hard form of steel
  crystalline structure, but it can also refer to any crystal structure
  that is formed by diffusionless transformation.1 It includes a class
  of hard minerals occurring as lath- [emphasis mine] or
  plate-shaped crystal grains.

Lath basically means the shape of a woven screen when viewed close up, or like a grid of alternating appearance.


Why are Salts Removed before Ion-Exchange Chromatography? In my organic chemistry laboratory, we used centrifugation and salting out to begin the process of purification of Myoglobin from hamurger meat. After that process, we used to dialysis to remove the Ammonium Sulfate from the salting out process to begin ion-exchange chromatography. I was wondering why we had to use dialysis to remove the salts, and why the ion-exchange chromatography couldn't just occur in the presence of the salts?

Think about what the ion-exchange material does, and the mechanism of separation by this method. The ions of interest interact with the ion exchange sites to a different degree than do the other ions or other species in the solution, thus you get separation.  What if your solution is loaded down with ammonium sulfate?  The ion exchange resins have a limited capacity to accommodate ions and all of the sites would be swamped in ammonium ions and all of the much lower concentration ions of interest thus would not be able to be separated.


Nomenclature problem, when would ring be parent? Nomenclature problem, when would ring be parent?

[(but-3-yn-2-yl)oxy]cyclopentane or 3-cyclopentoxylbutyne
I used chemsketch and it generated the former one.

The chain has a alkyne group. I thought that would make it has a seniority over the ring.
Also, the another rule said we should choose the one with more carbon as the parent chain so I am confused.

IUPAC bluebook,
P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

You have already found some relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, for the preferred IUPAC name (PIN), the ring is generally selected as the parent structure according to P-52.2.8, which also corresponds to the method described in P-44.1.2.2 (1). However, this rule applies to structures with “the same number of characteristic groups cited as the principal characteristic group”. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of characteristic groups cited as the principal characteristic group.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

For example, for compounds like 3-propylcyclohexan-1-ol or 3-cyclohexylpropan-1-ol, the senior parent structure is given according to P-44.1.1 by the principal characteristic group, which is expressed as suffix (here: ‘-ol’). The further rules P-44.1.2.2 and P-52.2.8 are not relevant in these cases.

The compound given in the question does not have any substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name).* Thus, no decision can be made using P-44.1.1. According to P-52.2.8, the ring is selecetd as the parent structure to construct the preferred IUPAC name [(but-3-yn-2-yl)oxy]cyclopentane. In general nomenclature, however, the context may favour the the chain according to P-44.1.2.2 (2), which would lead to the name 3-(cyclopentyloxy)but-1-yne.

* Note that some sources try to simplify the nomenclature rules by combining different concepts into a single one. For example, the so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order. Please do not use such sources.


Halogen bond definition I was looking for an accurate definition of halogen bonding.
I was able to find quite a few good ones, but none of them would explain if a X---H intermolecular interaction would count as a form of halogen bonding. Where X is any halogen and H is a hydrogen from another molecule. Does this count as halogen or hydrogen bonding? Or maybe as both?

Thank you,

I would say no, it is only a hydrogen bond, not a halogen bond.

To be a halogen bond, the halogen atom must accept electron density from the other member of the bond.  If the other member of the bond is a hydrogen atom bonded to a more electronegative element, I don't see have the halogen atom could be an acceptor of electron density.

If you can get access see "Halogen Versus Hydrogen" Science Vol. 321 no. 5891 pp. 918-919.  Footnote 7 there explains: "A halogen bonding donor is a species that contains an electrophilic halogen that can become a member of a halogen bond. In the literature on halogen bonding, the reader should be alert to how the words 'donor' and 'acceptor' are used. In a complex RX---B, RX is the halogen bond donor but the electron acceptor (Lewis acid); B is the electron donor and halogen bond acceptor (Lewis base)."


Best way to mix essential oils with water? I am using alcohol as an emulsifier. Of course, I can use a lot of alcohol to mix the oil and water but I do not like the color and the % of alcohol in the mix. The best mix I made was 10ml water, 0.5ml (10 drops) essential oil, 5ml alcohol because the mix looks super milky and when left alone the oil does no lt deposit. I repeated this formula again but the result is different (maybe the temperature was different, or the stirring, etc). My question is: what is the best way to stir the essential oil with the alcohol? Should I heat it? Should I stir it with an instrument other than just stirring in a beaker? All emulsifying tips are welcome!

A solution would be the best for spraying, and it would require dissolving the essential oil in an appropriate organic solvent. However, if you are concerned about the flammability, then a dispersion in water would be your best option. Basically, all you need is a stabilizing agent and a proper procedure. I am sure that you will find all the necessary information on this site. It has all the basics about emulsions, it lists various emulsifiers (stabilizing agents) and procedures. For your purpose, I think the best option would be the Bottle Method.


Why does the pH of this solution rise instead of fall when exposed to the atmosphere? I have a couple of big barrels with a $50\,\mathrm{L}$ demi-water solution of the following nutrients:

$$
\begin{array}{rl}
200~\mathrm{\mu M} & \ce{NH4NO3} & +\\
30~\mathrm{\mu M} & \ce{KCl} & +\\
10~\mathrm{\mu M} & \ce{CaCl2} & +\\
10~\mathrm{\mu M} & \ce{KH2PO4} & +\\
\\
10~\mathrm{\mu M} & \ce{Fe-EDTA} & +\\
\\
0.7~\mathrm{\mu M} & \ce{ZnSO4} & +\\
0.8~\mathrm{\mu M} & \ce{MnCl2} & +\\
0.2~\mathrm{\mu M} & \ce{CuSO4} & +\\
0.8~\mathrm{\mu M} & \ce{H3BO4} & +\\
0.008~\mathrm{\mu M} & \ce{(NH4)6Mo7O24}\\
\end{array}
$$

After mixing those nutrients, I adjust the $\mathrm{pH}$ of the barrels to $5.5$ by carefully adding $\ce{HCl}$ and continuing mixing until the $\mathrm{pH}$ stabilizes at $5.5$.

Then, when I measure the $\mathrm{pH}$in the barrels again after four days, I notice that it has risen to about $5.8$-$5.9$. This is contrary to my expectation. I'd have expected $\ce{CO2}$ from the air to be dissolved in the solution, thus lowering the $\mathrm{pH}$ rather than raising it.

Could someone explain to me what could be happening here?

At that pH, you're well outside the buffering capacity of both phosphate and boric acid, so a $\mathrm{pH}$ rise from $5.5$ to $5.9$ corresponds to a net increase in base of about $2\ \mathrm{\mu M}$.  That implies to me that it's likely something going on with one of the first five ingredients in your recipe, or the $\ce{HCl}$ added to adjust the $\mathrm{pH}$.

I can think of two scenarios:

Similar to Brinn's answer, perhaps a small amount of $\ce{HCl}$ has evaporated out of solution.
If your $\ce{Fe-EDTA}$ was added as $\ce{Fe(II)-EDTA}$, oxidation to $\ce{Fe(III)-EDTA}$ could explain the pH rise, by reducing water to hydrogen gas and hydroxide ion:

  $2\ \ce{Fe(II)-EDTA} \rightarrow 2\ \ce{Fe(III)-EDTA} + 2\ e^-$

  $2\ \ce{H2O} + 2\ e^- \rightarrow \ce{H2} + 2\ \ce{OH-}$

I don't know offhand how thermodynamically or kinetically feasible #2 is.  With all of the other metals present in solution, though, it's not unreasonable to imagine that some sort of catalysis might be possible.


The Conductivity of Electrolytes based on structure of Ionic Crystal Background
I want to run an  experiment to test the conductivity of various ionic compounds dissolved in water. I was hoping to see some sort of trend in the increase/decrease of conductivity based on my choice of salts. Except, conceptually, I don't quite get how my choice in the different compounds would affect conductivity.

Problems

I am somewhat aware of Kohlrausch's Law, but for my experiment I was not seeking to change dilution, temperature or electrode distances. What I wanted to do was to study the effect of the structure of ionic compound or charge of the ions on Conductivity.

I also wanted to compare my conductivities to some literature values.

Question
Just so this isn't a waste of time, I would like to know:

Does the structure of an ionic compound or the charge of its ions affect its conductivity, given that dilution and temperature will remain constant?
If so, how?
Is there a reliable Databse I can compare my results with?

I'm not too fussed about which ionic compounds they are, I would just like to understand how the compound itself plays a role in conductivity so I can confirm it with my experiment?
Thank you so much,

TL;DR
In a solution containing a dissolved ionic compound, to what extent is the ease of mobility of electrons or conductivity affected by either the electrophoric drag (presumably of polar water) and/or by the original structure (lattice shape) & charge of both the metallic and non-metallic ions?

It turns out, according to this paper, stated:

  The solvent-berg model which put forward the suggestion that smaller ions are strongly interacting with the nearest neighbour shell of solvent molecules. This was considered to be particularly true of cations since these are generally smaller in size than the corresponding anions and have a higher charge density. The ion essentially carries this shell of solvent molecules long enough that this leads to a larger effective diameter which lowers its conductivity to a value smaller than the conductivity of larger ions which have no strongly attached shells of solvent.

And in terms of drag from the polarity of the solvent, namely water, they showed a formula for dielectric friction:

ζDF = 3qi2(ε0 − ε∞)τD/(cri3(2ε0 + 1)ε0)

Where:
ζDF is the dielectric friction,
ri and qi are the radius and charge of the ion,
ε0 and ε∞ are the static and high frequency dielectric constants of the solvent,
c is the product of the limiting ionic conductivity and viscosity of the solvent,

So, I assume that the ionic radius will play a role in its drag by the surrounding polar solvents which will affect its mobility, and therefore the conductance.


It is said that atoms cannot be created. If so, then how did atoms get created after the Big Bang? I read somewhere that atoms cannot be created. If this is true, then how did the atoms form after the Big Bang? Also, does this mean that the number of atoms in our universe has remained the same since then? Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?

Atoms can be created and destroyed.  For example, in nuclear fusion two or more atoms make one atom.  In nuclear fission, one atom makes two or more atoms.

  how did the atoms form after the Big Bang?

The theory of this process in known as Big Bang Nucleosynthesis

Starting from protons and neutrons, light elements formed.  Later, in stars, nuclei up to iron formed through nuclear fusion.  Nuclei heavier than iron formed from supernovae by the r-process and asymptotic branch giant stars by the s-process.

  Also, does this mean that the number of atoms in our universe has remained the same since then?

No.  The number constantly changes due to fusion, fission, supernovae, creation of neutron stars, black holes and possibily quark stars.

  Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?

Currently, this process does not create or destroy atoms.  However, according to the theory of accelerating expansion, there is a senario referred to as the "big rip" in which all atoms would be destroyed.


Problem with epoxy silane (Random results in glass silanization). What are crucial points to look at? I`m trying to immobilize an antibody onto the glass surface by means of GPTOMS (glycidoxypropyl trimetoxy silane). Using the same protocol (silanization in toluene) I get different results any time, what are key points to check?
Looking through literature on immobilization, I see a lot of articles where authors use APTES (aminopropyl triethoxy silane) and only a few using GPTOMS, why? Epoxy group seems so atractive...
For toluene:

Rince slides with water
Ultrasonicate them in isopropyl alcohol for 3 min, dry them
Put for 3 minutes into piranha solution
Rinse with distilled water
Treat with oxygen plasma for 3 min
Slowly put slides into 2 or 4 % silane solution in toluene, incubate for 3 min
Rince with toluene and isopropyl alcohol.
Put them on dry heater (max temp. of ours is 97 C, most protocols say 110 C), incubate an hour

For ethanol:

Rince slides with water
Ultrasonicate them in isopropyl alcohol for 3 min, dry them
Put for 3 minutes into piranha solution
Rinse with distilled water
Treat with oxygen plasma for 3 min
Put slides slowly into 2% silane in ethanol solution (silane + alcohol + acetic acid (pH 4,5-5,5)), incubate 3 minutes
Rince slides with ethanol
Put them on dry heater (max temp. of ours is 97 C, most protocols say 110 C), incubate an hour

The triethoxysilane group has to hydrolyze first in order to bond with a silanol group of glass surface. It seems that you are cleaning the slides very well. However, I think your temperature is too low and your time of exposure is too short. The hydrolysis of ethyoxysilane group is slow.
Secondly, I guess you are posting questions on this topic for almost a year without much success. In such cases, it is always good to go back and try a control experiment and think about a new approach. The control experiment is that you try just neat glass slides and see if your antibody sticks to glass or not. Glass itself is a good adsorber. Probably you might get similar results of "fluorescence, no fluorescence and mild fluorescence". This will confirm that your silane binding procedure is doing nothing!
Second control test: Epoxy silane coated slides are commercially available. Why don't you try using them and see if your immobilization chemistry works on commercial slides or not. Schott makes commercial slides by this name: Nexterion® Slide E
Let me tell you how silanes are bonded onto silica. The reaction time is typically 10 hours under reflux conditions of toluene and other added solvents and catalysts! Sometimes a very very small amount of water is added to created polymer like bonding. There are at least several hundred papers on this topic. The surface area is silica is quite high so it is not a direct comparison with glass slides but still 3 min and room temperature is doing nothing!


How would a temperature profile in wall change if the rate of heat transfer increases Given the system below, how would the temperature profile in the wall change if the heat transfer in Fluid II was very fast?

If the heat transfer in Fluid II is very fast, then the T-profile for $\ce{T_{2∞}}$ would experience less resistance and hence have a more linear slope. However, I don't understand how this would affect the T-profile inside the wall? Wouldn't Fluid II have the same temperature and interact with the interface at the same point, meaning nothing changes within the wall?

If the heat transfer in fluid 2 is fast, the thickness of the boundary layer wall - fluid 2 is very small (the 'not so linear' part of the temperature profile on that side)

One can then approximate that the temperature of fluid 2 directly at the wall is equivalent to the bulk temperature.

So the temperature profile through the wall would be steeper, indicating a larger overall heatflux - see it as a self fullfilling prophecy: fast heat transfer will allow you to transfer heat faster


Indicators of Cu(I) or Cu(II) We performed an experiment in class where we placed a piece of copper wire in a solution of silver nitrate, we were tasked with predicted the mass of copper that should react and the mass of silver that should be formed. My question is, what are indicators of Cu(I) and Cu(II) ions?

$$\ce{2 AgNO3(aq) + Cu(s) -> Cu(NO3)2(aq) + 2 Ag(s)}$$

I need to know this in order to calculate reaction yield and efficiency (whether the product is $\ce{CuNO3}$ or $\ce{Cu(NO3)2}.$

There is very well-written chapter "The chemistry of monovalent copper in aqueous solutions" in Advances in inorganic chemistry, Volume 64 [1, pp. 220–223, DOI: 10.1016/B978-0-12-396462-5.00007-6] which extensively covers as to why copper(I) is unlikely to exist in aqueous solution and why nitrate is a poor ligand for the purpose of preserving monovalent copper:

  The redox potentials of copper in acidic aqueous solutions are as follows:

  $$
\begin{align}
\ce{Cu+(aq) + e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.521 V} \tag{1}\\
\ce{Cu^2+(aq) + e- &-> Cu^+(s)}, &\quad E^\circ &= \pu{0.153 V} \tag{2}\\
\ce{Cu^2+(aq) + 2 e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.342 V} \tag{3}\\
\end{align}
$$

  From these values, it is clear that $\ce{Cu+(aq)}$ is unstable and
  disproportionates in aqueous solutions:

  $$\ce{2Cu+(aq) <=> Cu^2+ + Cu^0(s)}, \quad K_{1.4} = \pu{(1 - 1.7) × 10^{6} M-1} \label{rxn:4}\tag{4}$$

  However, if $\ce{Cu+(aq)}$ is formed in a clean aqueous medium, the disproportionation reaction has often a relatively long induction period as reaction \eqref{rxn:4a} is endothermic due to the $\ce{Cu^0(s)}$ lattice energy.

  $$\ce{2Cu+(aq) -> Cu^2+(aq) + Cu^0(aq)}\label{rxn:4a}\tag{4a}$$

  Furthermore, the concentration of $\ce{Cu^0(aq)}$ in nonacidic aqueous solutions is limited due to the low solubility of $\ce{CuOH}/\ce{Cu2O}:$

  $$\ce{Cu+(aq) + OH- <=> CuOH(s)/Cu2O(s)} \quad K_\mathrm{SP} = \pu{1E-14 M-2} \label{rxn:5}\tag{5}$$

  Naturally, the concentration of Cu(I) species in aqueous solutions
  can be increased by adding appropriate ligands.

  […]

  …the instability of $\ce{Cu+(aq)}$ stems from the difference between the solvation energies of the two cations and the second ionization potential of copper.
  This is correct for all transition-metal ions, but as the ionization energies increase along the series faster than the solvation energies, the lower oxidation states become relatively more stable.
  In order to shift reaction \eqref{rxn:4} to the left ligands which
  preferentially bind to $\ce{Cu(I)},$ must be added to the solution.

  […]

  $\ce{Cu(I)L_n}$ complexes can be stabilized in aqueous solutions via
  employing ligands with one or more of the following properties:

  Ligands that have a lower coordination number with $\ce{Cu(I)}$
  complexes than with $\ce{Cu(II)}$ complexes; these $\ce{Cu(I)}$ complexes are stabilized due the entropy gain in reaction \eqref{rxn:7}:

  $$\ce{Cu(II)L_m + e- -> Cu(I)L_n + $(m - n)$ L} \label{rxn:7}\tag{7}$$

  Ligands that are π acids and therefore prefer $\ce{Cu(I)}$ over $\ce{Cu(II)}$ and shift the redox potential of the $\ce{Cu(II)/Cu(I)}$ couples anodically.
  Hydrophobic ligands, which often also increase the radius of
  the complex, decrease the stabilization of $\ce{Cu(II)}$ by decreasing
  its solvation energy.
  Ligands which enforce a tetrahedral coordination sphere, or
  one which approaches this geometry, on the central cation.
  Ligands that are soft bases, for example, thiols. Thus, for example, the biological copper-trafficking proteins, the metallochaperone Atx1, and the copper-transporting P-type ATPases have a highly conserved CXXC metal binding motif which binds a single $\ce{Cu(I)}$ ion via a two-coordinate site consisting of two cysteines.

  All these types of ligands stabilize $\ce{Cu(I)}$ complexes by shifting the redox potentials of the $\ce{Cu(II)/Cu^0(s)}$ and $\ce{Cu(I)/Cu^0(s)}$ couples and thus shifting the equilibrium constant of reaction \eqref{rxn:4} to the
  left.
  However, ligands might stabilize $\ce{Cu(I)}$ complexes also
  kinetically by inhibiting the disproportionation reaction by
  inhibiting the formation of $\ce{Cu^0(s)}$ via blocking the approach of two copper atoms, this is, for example, the mechanism by which
  some $\ce{Cu(I)}$ enzymes are stabilized.

  Naturally, all these ligands also increase the solubility of $\ce{Cu(I)}$ by competing with reaction \eqref{rxn:5}.

References

Inorganic/Bioinorganic Reaction Mechanisms, 1st ed.; Eldik, R. van, Ivanović-Burmazović, I., Eds.; Advances in inorganic chemistry; Academic Press: London; Waltham, MA, 2012; Vol. 64. ISBN 978-0-12-396462-5.


Why is xenon hexafluoride not stored in glass? Why is $\ce{XeF_6}$ not stored in glass containers? Isn't the reason glass is used for most acids is that it is non-reactive?

$\ce{XeF_6}$ can't be stored in glass as it reacts with the silica that makes up the glass

$\ce{2XeF_6 + SiO_2 -> 2XeOF_4 + SiF_4}$

This can proceed further to eventfully make $\ce{XeO_3}$:

$\ce{2XeOF_4 + SiO_2 -> 2XeO_2F_2 + SiF_4}$

$\ce{2XeO_2F_2 + SiO_2 -> 2XeO_3 + SiF_4}$

Dry $\ce{XeO_3}$ is highly explosive ...

That is why both $\ce{XeOF_4}$ and $\ce{XeF_6}$ are stored in nickel containers at room temperature.

These are generally considered as acid-base reactions using the Lux-Flood definition, see https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Lux.E2.80.93Flood_definition

I don't have them to hand but any decent University level Inorganic Chemistry text book such as Housecroft and Sharpe or Greenwood and Earnshaw should cover this.


Acid-catalysed Reaction of Ketone with Benzene to Make Benzyl Alcohol Benzene shows electrophilic substitution reactions. Normally, anything that can generate a carbocation shows, for example, a Friedel-Crafts alkylation reaction. I wanted to know if this reaction is also possible: taking an aldehyde or a ketone with an acid, which will protonate the oxygen atom and generate a carbocation - followed by electrophilic attack on benzene. This reaction can even go further by protonation of the alcohol and forming another carbocation for a second substitution - as I've shown in the picture below.
I couldn't find this anywhere, though it seems a simple way to make a benzyl alcohol - why isn't it more frequently carried out like other cases like Friedel-Crafts Alkylation, or Acylation, etc.?

Something almost identical happens in synthesis of DDT using chloral and chlorobenzene.
www.liceoagb.es/quimiorg/petroquimica_webquest/aad/DDT.ppt

I haven’t ever seen a reaction directly like that. To generate the alcohol you drew on your piece of paper, I would probably use phenyl lithium or phenyl magnesium bromide and add that to acetone; it is easier and faster. (Note that the reaction would stop at the tertiary alcohol and not react further under elimination of water.)

Note that the reaction to synthesise DDT starts from trichloroacetaldehyde. The three chlorine atoms draw electron density away from the carbonyl group, further strengthening the cation. I would suppose that this is not seen often because of the activation a simple aldehyde would need. Let’s face it: benzene is a very bad nucleophile.


How can I find the parent chains in these two compounds? We recently started organic chemistry in school and I decided to make a computer program that names the compounds, so far it works with alkanes that do not have branched branches and I want to make it work with compounds with double and triple bonds. I know that if there is only one double/triple bond, the answer is just the same rules as with alkanes, but it must include the 2 atoms that are bonded in that way. Also, I would appreciate it if you tell me how I decide from which side to number them.

Compound 1:

Compound 2:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the first relevant criterion to be considered in choosing a principal chain is the length of the chain.

When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. When there is a choice for the locants of multiple bonds, low locants are given first to double bonds:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

Furthermore,

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the preferred IUPAC names (PINs) for the examples given in the question are 3-ethylpent-1-en-4-yne and 3,3-diethynylpenta-1,4-diene, respectively.


Is osmosis really a special case of diffusion across a semipermeable membrane? A general idea we have about osmosis is that it is the movement of solvent particles through a semi permeable membrane from a region of high concentration to a region of low concentration. This seems correct in case of a gas. Gaseous particles spread through any container in which they are kept, which makes it seem to us as if they moved from high concentration to lower concentration. But in the case of a liquid how is this true? For example consider a bottle half filled with water, water  never moves from high concentration to lower concentration inside the bottle.

So is it wrong to explain osmosis as a special case of diffusion?

What actually makes solvent particles move from regions of higher concentration to regions of lower concentration? Is it solute attraction? Or a difference in pressure in pure solvent phase and solution phase?

I've listed a series of points about diffusion and Osmosis.

The cause of osmosis is simply diffusion: the solvent can diffuse through the membrane and the solute cannot.

Thermodynamically, we say that the reason that the diffusion occurs is that the chemical potential of the solution is at a lower chemical potential than the pure solvent and so there is a tendency for the solvent to pass into its solution and to dilute it. The osmotic pressure is the excess pressure just needed to stop that flow. If more pressure were to be applied the flow of solvent would be reversed and the solution would become more concentrated.

Osmosis occurs because there is an increase in entropy overall when the solvent dilutes the solution. There is a decrease in entropy in the pure solvent side because its volume decreases but this is balanced by an increase in entropy on the solution side as the volume increases so this is not the reason osmosis occurs. The reason is the increase in entropy that is experienced by the solute now in a larger volume or equivalently more solvent molecules in which to dissolve. This means that there are more ways of arranging solute and solvent molecules compared to that before osmosis stared, and this represents an increase in entropy. (Left unchecked this will continue until the solution is so dilute it is indistinguishable from the solvent).

When we say that a solution has an osmotic pressure of, say, 5 atm, it does not mean that the solution exerts this pressure but that if it is in contact with its solvent through a semi-permeable membrane then an applied pressure of 5 atm is needed to maintain equilibrium.

It is possible for a solute in a solution to diffuse spontaneously from regions of higher to lower concentration so that uniform concentration is reached and the solution is at equilibrium.  Naturally, diffusion does not stop at this point but is always present both for solute and solvent.

In some cases, however, diffusion can occur form lower to higher concentrations. Such a situation is that when a solute is dissolved in two immiscible solvents in contact with one another, and in which the solute has different solubilities. Solute diffusion will occur until the chemical potential in each solvent is equal which can mean diffusion into the more concentrated solution. Thus at equilibrium there will be a large difference in solute concentration but no overall flow of solute due to diffusion.


Why is water not part of the equilibrium constant? In the acid base reaction
$$\ce{NH4+ + H2O <-> NH3 + H3O+}$$

the acidity constant, which is a relation between concentrations is given by
$$K_{\mathrm{a}}=\frac {\ce{[NH3]} \cdot \ce{[H3O+]}}{\ce{[NH4+]} \cdot \ce{[H2O]}}$$

Since $\ce{[H2O]}$ is constant, it can be included inside the constant to obtain a new $K_\mathrm{a}$ which is $6.3 \times 10^{-10}$.

(Being a constant allows us to relate the molar concentration to the density of water.)

And so, even though some quantity of water reacts with $\ce{NH4+}$, but the volume of water varies in a way that $\ce{[H2O]}$ would still be constant.

What I don't understand is why the molar concentration of water is constant. Why isn't it handled the same way as the other reactant?
I understand that if $\ce{H2O}$ was the only liquid and the rest of the reactants and products were of different state (gaseous or solid), then water would exist alone in the beaker and its concentration would therefore remain constant, but this is not the case.

That was about heterogeneousness. Trying another guess; in the reaction of esterification $$\ce{{alcohol} + {carboxylic acid} <-> {ester} + H2O}$$
$\ce{H2O}$ is considered as a product which concentration is not constant, so it is counted explicitly in the equilibrium constant.

Could the answer be that $\ce{NH4+}$, $\ce{NH3}$ and $\ce{H3O+}$ are dissolved in water, while the alcohol, the carboxylic acid, and the ester are not?

I'm looking for an answer that clearly shows which volume is considered (water or the solution), because I'm confused as to why something would or wouldn't be included in the equilibrium constant.

There is no IUPAC definition of $K_a$, however, many reputable texts like Levine's Physical Chemistry define $K_a$ as:

$$\frac {a(\ce{A-})a(\ce{H3O+})}{ a(\ce{HA})a(\ce{H2O})}$$

where "a(X)" is the activity of species "X".

Then various approximations can be made, such as approximating $a(\ce{H2O}) = 1$ and approximating the activity of the solutes as the concentration of the solutes.

These approximations are only reasonable in dilute solutions.


Possible structure if a compound like AB4 is polar? I was asked the following question in JEE Mains (2nd September, Shift 1) 2020,

If a compound AB4 is polar (having non-zero dipole moment), then it's structure could be
(A) Square planar
(B) Rectangular planar
(C) Square pyramidal
(D) Tetrahedral

My thoughts

As this compound has only 4 ligands, therefore it cannot be a square pyramidal.
It can neither be a tetrahedral nor a square planar bcoz, it would be non-polar then.

So, we are left only with "rectangular planar", which I'd never heard before. How to move on?

Given all $\ce{B}$ are equivalent in terms of AXE notation, for $\ce{AB4}$ the following coordination environments are possible [1, p. 177]:

square-planar $(D_\mathrm{4h},$ SP-4);
square-pyramidal $(C_\mathrm{4v},$ SPY-4);
tetrahedral $(T_\mathrm d,$ T-4);
see-saw $(C_\mathrm{2v},$ SS-4).

Table IR-9.3 Polyhedral symbols, geometrical structures and/or polyhedra

Option (A) square planar implies a centrosymmetrical environment and such molecule cannot be polar.
The derivative of (A), option (B) rectangular planar, as well as option (D) tetrahedral also refer to the environments that posses zero dipole moment.
Speaking of (C), I think you are misinterpreting the geometry of square pyramidal coordination.
It's a square non-coplanar arrangement with a non-zero dipole moment commonly seen in period 4 transition metal porphyrins $(\ce{MN4}$ fragment) and also lead(II) oxide and tin(II) oxide $(\ce{MO4}$ fragment).
According to Brecher [2, p. 1944], the structure of bis(N,N-dimethyldithiocarbamato)lead(II) would be yet another example:

Sample depiction of a compound known to exist in square pyramidal configuration:

Since a see-saw option is not present, this leaves only one answer, option (C).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005.
Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897.


What is the pKa of the hydronium, or oxonium, ion (H3O+)? Although the wikipedia page on Hydronium indicates a $\mathrm{p}K_\text{a}$ of −1.74, I noticed in the discussion of this page that the subject seems debated (cf. http://en.wikipedia.org/wiki/Talk:Hydronium#pKa.3F) since alternative (apparently more rigorous) reasoning leads to $\mathrm{p}K_\text{a} = 0$ for $\ce{H3O+}$.

Also, many demonstrations lead to $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of 15.74 for $\ce{H2O}$ and thus, using $K_\text{w}=[\ce{H+}][\ce{OH-}]=10^{-14}$, give $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of −1.74 for $\ce{H+}$ and $\ce{OH-}$ respectively. But those demonstrations seem to make use of the concentration $[\ce{H2O}]=55.56~\text{mol/L}$ instead of the activity $a_{\ce{H2O}}=1$ (for a solvent) in the equilibrium constant of the reactions.

So it seems to me that the $\mathrm{p}K_\text{a}$ of $\ce{H3O+}$ must be equal to $0$.

Is this question still debated?

PS1:
To add details about the question, the debate in the references mentioned in the discussion on the wikipedia page as well as in the link provided by Nicolau Saker Neto bellow seems to boil down to the coexistence of two different definitions for the equilibrium constant:
- one based on the activities of the compounds
- the other derived from the 'mass action law' and which thereby uses the molar fractions of the compounds
Reading through the Callen (Thermodynamics and an Introduction to Thermostatistics) gives me the impression that the derivation from the mass action law of an equilibrium constant involving molar fractions is based on the assumption of an ideal fluid. However, it is precisely the departure from ideality that seems to justify the use of activities instead of molar fractions. Is that correct?

PS2:
Another point, which I have not looked in details yet, seems to be a debate about the possibility to apply the definition for $K_\text{a}$ to the water molecules $\ce{H2O}$ considering that they are not a highly diluted solute since they constitute the solvent. Is this a valid issue and should then the acidity of $\ce{H3O+}/\ce{H2O}$ be determined in another solvent than $\ce{H2O}$ to be compared with the other acids?

The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.

There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.

The closest IUPAC defined term is the standard equilibrium constant, which can be denoted as $K^\circ$ or just $K$.

Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:

$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$

Where $a$ is activity.

Substituting that the acid is $\ce{H3O+}$:

$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$

and of course $-\log(1) = 0$.

The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:

omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and
taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.

With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$.  For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.

However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:

New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water, which insists that

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

"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.

The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.

The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.