Migratory aptitude in pinacol pinacolon rearrangement Out of ortho-methoxyphenyl and para-methoxyphenyl which one has more migratory aptitude in pinacol-pinacolon rearrangement?

Groups with higher electron density have a greater tendency to migrate. So, o-methoxy phenyl should be the answer but due to steric factors p-methoxy phenyl can always be an equally likely answer.  It can only be found out accurately by experimentation and considering other reaction conditions.


What are all the structural isomers of a tertiary amine with the sum formula C6H15N? According to the solutions there exist 7 different structural isomers 'containing 3 degree amines' for the molecular formula  $\ce{C_6H_{15}N}$

I tried and made a few, here are those

I have been trying since an hour and all I made were the repetitions of these 5 only. Could anyone suggest me the remaining 2?

Please note that this is just a visualization of the comment given by LDC3!

Here are the two missing isomers:


Angle of Light Source for Absorbance Spectroscopy I apologize if this question is stupid, but have been puzzled by it for a few days. Say I have a spectrometer measuring in the visible light range and we're looking to measure the absorbance on a solid material. In lab examples I've seen and read, the light source is placed on one end of the sample and the spectrometer on the other end. so the light passing through goes into the spectrometer. But what if you have a sample around 3 feet away from you. Could you place the light source next to the spectrometer and focus it on the sample then use a collimating lens in an attempt to see what was absorbed or not?

Once again, I apologize if this is a stupid question, I feel like I am missing some fundamental points of this process here.

Yes you could do what you propose provided that the light is properly collimated so that it all reaches the detector. It is quite common to place mirrors inside the sample housing of a spectrometer to  take the light out into the lab, through a sample (usually a weakly absorbing gas ) and then use a mirrors to get it back into the spectrometer. Multi-pass cells can also be used to do this.

In atmospherical studies very long path lengths (100's metres) are sometimes used to identify trace amounts of various species. Of course if you do something like this you have to make sure extraneous light does not reach your detector.


Is there an edible, colour-less, taste-less, odour-less, water-soluble, cheaply-producable gel, solid or other substance? Suppose, for example, that some oatmeal is fixed in gel, or some other substance. A portion of the substance-with-oatmeal is then removed and added to cold milk or some similar dairy product, where it quickly dissolves, yielding a standard bowl of milk and oatmeal. Is there such a substance?

Ideally, the substance should be unobtrusive to the eating experience. This also means that it should not dilute the milk significantly. Otherwise, some taste, colour or odour may be permissible, provided it does not clash with the taste of the meal.

The substance should also not soak the oatmeal, which is why it may not be a liquid.

Are you talking about maltodextrin? It's a solid powder assembled from glucose units with no taste smell or color. It's commonly used as a carrier; it's used to dilute the actual sweet -flavored molecule in Splenda.

If you're dealing with Hot oatmeal, you could add methyl cellulose to gel it up. Upon mixing with cold milk it will melt, acting like reverse jello; solid when hot and liquid when cold.

You asked a rather strange question, I'd recommend instant oatmeal to prevent a possibility of bacterial/fungus growth when storing oatmeal for extended periods of time.


White powder observed after boiling water in electric kettle for many weeks I used my water kettle (metal ones, not the plastic ones) to boil my water. Usually, if I did not consume all the water in the kettle and the water was cold, I usually topup extra water to the kettle and turn it on to boil.

So, after many weeks of boiling in this way, I notice that there are white powder outside the kettle. (I notice that when the water is boil, some water was splashed to the outside of the kettle and I left those water splashed out to dry as time goes)

Is the white powder actually called fluoride? How do I tested it to verify if the white powder chemical is safe or not?

It's likely that your home is supplied with "hard" water (that is, water with a high mineral content). These dissolved minerals typically include calcium or magnesium ions (from dissolved calcium bicarbonate and magnesium bicarbonate, respectively). Heating or boiling your "hard" water in a kettle will soften it by precipitating the calcium carbonate as its solubility decreases with increasing temperature.

This (calcium carbonate) could be the white powder you have observed after evaporation. Hard water should not pose any health risks, and in fact can add some extra calcium and magnesium to your diet. If it's still concerning to you, there are hundreds of products available that can soften your water before you use it, just google "water softener" and you'll be overwhelmed.

http://en.wikipedia.org/wiki/Hard_water


Why do reusable heat packs have a limited time frame before the reaction becomes irreversible? I have two "rechargeable" heat packs which from my understanding operate on the phase change
$$ CH_3COONa_{(aq)} \rightleftharpoons CH_3COONa_{(s)} + \Delta H.$$
Triggering the crystallization releases heat, and leaving them in boiling water returns it to its supersaturated form. The packs say if they are left in solid form for more than two months, they may not be able to return to aqueous form. I left one solid for many months, and sure enough, I couldn't quite redissolve the crystals. It gets 90% dissolved, but the remaining crystals just trigger the forward reaction as the pack re-cools, even after multiple attempts. What mechanism prevents this?

The sodium acetate that shows this behavior is not anhydrous $\ce{NaCH3COO}$ as you write. It is the hydrated salt $\ce{NaCH3COO·3H2O}$. The anhydrous salt does not show the phenomena of surfusion. Only the hydrated form does it. If you keep the solid form of the hydrated salt a long time, I think it will slowly loose water by evaporation through the plastic bag. So it will be partly transformed into anhydrous acetate. And by heating the whole system, it will never melt entirely. So it may never return to its supersaturated form. It is necessarily to add a few drops water to get supersaturation again.


Can strong acids be amphoteric? There are amphiprotic (or amphoteric) substances, like $\ce{H2O}$ and $\ce {HCO3-}$, which act both as an acid and base.

If you mix a strong acid like $\ce{HCl}$ ($\mathrm{p}K_{\mathrm{a}} = -6.3$) with an even stronger acid, such as perchloric acid ($\ce{HClO4},~\mathrm{p}K_{\mathrm{a}} \approx -10 $) can it act as a base?

Well an acid can act as a base when there is a stronger acid present, But if you look at the molecular level acidity depends on how well the negative charge is balanced. so out of the two molecules present the one which can disperse(stabilize) the negative charge to a greater extent will act as acid.


What Can Dissolve Polypyrrole? I am looking for a solvent for polypyrrole. I read M-cresol and chloroform can be used, but I am not entirely sure.

Polypyrrole is quite a rigid polymer (one of the reasons it's a conductive polymer) and it's generally not considered particularly soluble without being derivatized. It seems to be possible to dissolve it somewhat in polar organic solvents if it's been polymerized with a large anion like dodecylbenzenesulfonate, basically reducing the interactions of the chains of polymer, but it loses some of its conductivity when cast into a film. One group has apparently managed to make it water soluble by polymerizing oxidatively with horseradish peroxidase in the presence of a sulfonated polystyrene counterion, which keeps the chain growing linearly along the length of the polystyrene, apparently allowing it to be soluble.


How can I properly calculate the isoelectric point (pI) of amino acids? The following amino acid is called lysine. I was asked to calculate its isoelectric point, with the given $\mathrm pK_\mathrm a$ values.

I've searched a lot, and the most helpful post that I found was How do I calculate the isoelectric point of amino acids with more than two pKa's?
According to orthocresol's answer:

Since the $\mathrm{pI}$ is the $\mathrm{pH}$ at which the amino acid has no overall net charge, you need to average the $\mathrm pK_\mathrm a$ values relevant to the protonation/deprotonation of the form with no net charge.

Let's call the ends $e_1, e_2$ and $e_3$ (from left to right).
Approach $\#1$

deprotonate $e_3$ (i.e., carboxyl group)
deprotonate $e_1$ or $e_2$ [neutral point]

So, $\mathrm pK_\mathrm a$'s of $e_1$ and $e_2$ are relevant.
$$\Rightarrow \mathrm{pI} = \frac{10.53 + 8.95}{2} = 9.74$$
But, is there some limit to number of protonations/deprotonations or some procedures to follow?
For instance,
Approach $\#2$

deprotonate $e_1$
deprotonate $e_3$ [neutral point]
deprotonate $e_2$ and protonate $e_1$ [neutral point]

This time, $\mathrm pK_\mathrm a$'s of $e_3$ and $e_1$ are relevant. But, the calculated $\mathrm{pI}$ isn't correct.
So, how can I validate the approaches?

Problem source: FIITJEE study material
$\mathrm pK_\mathrm a$ and $\mathrm{pI}$ values table for amino acids: https://www.anaspec.com/html/pK_n_pl_Values_of_AminoAcids.html

Isoelectric point of an amino acid is the $\mathrm{pH}$ at which the molecule carries no net charge[1]. It can be calculated by the average of the relevant $\mathrm pK_\mathrm a$ values as you have mentioned.
Your confusion seems to stem from choosing the relevant $\mathrm pK_\mathrm a$ values. For this we should refer to the titration curve of the amino acid.
For a neutral amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the carboxylic acid and the amino group.
For a basic amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two amino groups. The $\mathrm pK_\mathrm a$ of the carboxylic acid group is not relevant.
For an acidic amino acid[3]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two carboxylic acid groups. The $\mathrm pK_\mathrm a$ of the amino group is not relevant.
Here are examples for all three cases:

References:

Wikipedia

Titration curves for neutral and basic amino acids

Titration curve for acidic amino acids


Which is the correct definition for metamers (or metamerism)? I have heard two definitions of metamers.

Compounds having the same molecular formula but different number of carbon atoms on either side of the functional group.
Compounds having the same molecular formula but different alkyl groups on either side of the functional group.

Which one of these is correct?

Are pentan-2-one and 3-methylbutan-2-one metamers?

There is no IUPAC definition for metamers or metamerism in the gold book (not even one that makes it obsolete), and Wikipedia doesn't even have a real article about it, see Metamerism.

  In chemistry, the chemical property of having the same proportion of atomic components in different arrangements (obsolete, replaced with isomer). In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups) on either side of functional group ( i.e., -O-,-S-, -NH-, -C(=O)-) are called metamers and the phenomenon is called metamerism.

I cannot find any authoritative source for this, but there is a question discussing it on our platform: What is metamerism? Also related: Are methyl n-propyl ether and methyl iso-propyl ether metamers?

Any of the two definitions cited in the question match the general definition of isomers.

  isomer (DOI: 10.1351/goldbook.I03289)
  One of several species (or molecular entities) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

More particularly, they match the definition of constitutional isomers.

  constitutional isomerism (DOI: 10.1351/goldbook.C01285)
  Isomerism between structures differing in constitution and described by different line formulae e.g. $\ce{CH3OCH3}$ and $\ce{CH3CH2OH}$.

Metamers/ metamerism is an archaic, ambiguous, and deprecated term in chemistry. Don't use it.


Are there any free electrons in metal vapor? Metals contain free moving electrons, in both the solid and liquid state. Does this happen when it is in vapor form?

Interesting question. There are free electrons and ions in a metal's gas phase, but even at rather high temperatures, they are relatively rare. The formation of free electrons is given by the reaction:

$$\ce {M_{(g)} → M^+_{(g)} + e^{-}_{(g)}}$$

This is an endothermic reaction for all atoms and molecules, and the energy involved receives a special name: the (first) ionization energy. Comparing a chart of ionization energies for the elements, one finds that almost all elements require over $ 500\ kJ/mol$ (or about $5\ eV/atom$) to ionize. Even at $5750\ K$, the temperature of the surface of the Sun, the average thermal energy of a particle is about only about $0.5\ eV$.

Also, a large part of the electrons that did manage to free themselves from their parent neutral atoms would actually be consumed by other neutral atoms, in a different reaction:

$$\ce {M_{(g)}  + e^{-}_{(g)} → M^{-}_{(g)}}$$

This reaction is slightly exothermic for most elements and molecules, and the energy associated with it also receives a special name: the (first) electron affinity.

Calculating the equilibrium concentrations of each species at a given temperature is possible but is a bit of work. However, for most metals, you can expect to find significantly less than one free electron per ten thousand neutral atoms at $5750\ K$, and the relative amount of free electrons decreases exponentially with temperature.


How do I find the theoretical pH of a buffer solution after HCl and NaOH were added, separately? This question is about the theoretical pH of a buffer solution.

  Calculate the theoretical pH values expected for a $\pu{200mL}$ buffer solution containing a 1:1 ratio of acetic acid and sodium acetate (their concentrations are $\pu{0.05M}$), following the addition of;

  $\pu{10mL}$ of $\pu{0.2M}$ $\ce{HCl}$ and
  $\pu{10mL}$ of $\pu{0.2M}$ $\ce{NaOH}$.

  $K_\mathrm{a}$ of acetic acid is given.

So I just found the $\mathrm{p}K_\mathrm{a}$ for the Henderson equation.

Attempt for (1):

$\ce{[HCl]}$ = $\pu{0.2M}$ and $V=\pu{10mL}$/$\pu{0.01L}$ thus $n = 0.002 $

$0.002 \times \pu{0.2L} = \pu{0.01M}$

ICE table:

$$
\begin{array}{cccc}
&\ce{H3O+ + &CH3COONa <=> &CH3COOH + &H2O}\\
\text{I}&   x& 0.05&0.05\\
\text{C}& -0.01&-0.01&+0.01\\
\text{E}& x - 0.0&0.04&0.06\\
\end{array}
$$

$\mathrm{pH} = 4.74 + \log (0.06/0.04) = 4.926$

Although in our experiments the actual pH was 4.17 after $\pu{10mL}$ of $\ce{HCl}$ was added and the pH after 10 mL of NaOH was 4.98

Attempt for b:

$\ce{[NaOH]} = \pu{0.2M}$ and $V = \pu{10mL}/\pu{0.01L}$ thus $n= 0.002$

$0.002 \times \pu{0.2L} = \pu{0.01M}$

Is this how I would properly set it out?

$$
\begin{array}{cccc}
&\ce{CH3COOH + &OH- &<=> &CH3COONa + H2O}\\
\text{I}& 0.05&x&&0.05\\
\text{C}& -0.01&x-0.01&&+0.01\\
\text{E}& -&0.04&&0.06\\
\end{array}
$$

$\mathrm{pH} = 4.74 + \log(0.04/0.06)= 3.74$

So I feel like I have done these the wrong way because it makes more sense but I don't know why or how.

First I want to point out a couple things.

When you add more liquid to a solution, the concentration of the solute is going to decrease. Therefore, after the addition of 10 mL of HCl, the concentration of sodium acetate and acetic acid will no longer be 0.5 M since you diluted the solution a little bit. When calculating the concentration after dilutions, keep in mind the amount of moles is still the same, so just use the formula $\ce{M1V1 = M2V2}$.
Also, sodium acetate is a soluble salt (see solubility rules), so in the net ionic equation, don't write out the sodium as it is a spectator ion. The same applies for HCl. It is a strong acid, so it completely dissociates in water.
For this problem, using the Henderson-Hasselbalch equation is not really the right way to do it. When you are using this formula, keep in mind that it is only good for calculating pH when you already know the equilibrium concentrations of the acid/conjugate base. Here's a derivation of the formula if you're interested that will probably help clear things up a bit.

I think you mostly understand this point last point since you did made an ICE table before you applied the equation. However, in the context of this problem, making an ICE table will actually lead you to the pH value without requiring the HH equation. You'll see why.

Alright; now we can actually get to how to approach the problem. First, you're going to want to look out for a reaction. When solving these problems, being able to identify a reaction will make it 10 times easier.

How do you identify a reaction? Well, look at the reagants in this problem: $\ce{HCl}$, which is essentially just hydronium and chloride, sodium acetate, which as mentioned earlier is essentially just sodium and acetate, and acetic acid. Acids react with bases, so using that, it isn't too hard to determine what exactly happens in the solution.

You actually got the net ionic equation for the reaction correct (without the Na, however), but you didn't solve the ICE table correctly.

Actually, we're going to start with simple stoichiometry as a reaction is occurring. To simplify the process, we're going to assume the reaction reacts to completion, and then reverses until it reaches equilibrium.

For this, just use limiting reactants and such to calculate the final concentrations of $\ce{CH3COOH}$ and $\ce{CH3COO-}$.

As I said earlier, we're going to first assume that the reaction reacts to completion, and then goes back on itself and reaches equilibrium. We're at the second part, the equilibrium part. This is where an ICE table comes in.

First, write out the equilibrium reaction. This is pretty simple; it is just the acid dissociation reaction. It's going to look something like this: $$\ce{HA<->H+ +A-}$$

For your ICE table, you didn't really set it up properly. Remember these three things:

You usually won't be writing any unknowns for your initial concentrations. Everything should be known.
You usually won't be writing any numbers for your change. It's an equilibrium. You don't know the exact amount it is going to react to reach equilibrium. That's why we write out the ICE table, so we can calculate how much needs to react for the system to reach equilibrium.
Don't use the Henderson-Hasselbalch equation. Use the $K_a$ and its definition to calculate the concentration of $\ce{H+}$, which leads to your pH.

I know I pretty much just wrote a wall of text, but hopefully that helps nonetheless.


How does the mixing of two gases increase entropy? I have learnt that $\mathrm{d}S = \mathrm{d}Q_\mathrm{rev}/T$, where $Q$ is the heat energy absorbed by the system. Now consider an insulated compartment with two chambers, separated by a valve. Each chamber consists of a gas at a particular temperature and volume. Now, the valve is screwed open. We know that the entropy of the isolated system will increase as the gases have more freedom.

But, starting with the assumption that the walls were insulated, where did the heat energy come from?

How did the entropy increase?

Here is a reversible process to do what you are asking for (i.e., unmixing the gases reversibly).  You have a cylinder containing the two gases already mixed.  The cylinder has a piston.  Connected to this main cylinder are two other cylinders joined to it through semipermeable membranes.  Each membrane allows one of the gases to pass, but not the other.  You gradually advance the piston in the main cylinder while, at the same time moving the other pistons outward in the other two cylinders so that, at any time, the pressures of the pure gases in the attached cylinders are equal to their partial pressures in the main cylinder.  The change in entropy for this adiabatic reversible process is zero.  This is the first step.  Now, in the second step, we close off the semipermeable membranes, and compress each of the pure gases in the two cylinders isothermally and reversibly to the original total pressure that was present in the main chamber.  So, if $p_1$ was the partial pressure of one of the gases in the main chamber and $p_2$ was the partial pressure of the other gas in the main chamber, the total pressure of the pure gases in their chambers after completing step 2 will be $p_1+p_2$.  The change in entropy for the 2-step process we described is minus the entropy of mixing.  So, if the process is done in reverse, the entropy of mixing will be positive.


How can I prepare two phosphate buffer of different concentrations but give same pH? How is it that buffers of different compositions can have the same pH? For example, it is possible to prepare 0.01 M phosphate buffer of pH 7.0 and 0.1 M phosphate buffer of pH 7.0? How?

I used Henderson-Hasselbalch equation, but still failed. No way is working. Is it something related to volume? Can it be shown mathematically? Can anyone show it mathematically that it is possible?

For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram below:

In this figure, the green trace is the $\ce{H2PO4-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14.$ The blue trace is the $\ce{HPO4^2-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14$. So, at $\mathrm{pH} = 7,$ the blue curve is below the green curve, and the ratio of alpha fractions is $10^{-0.2}$, i.e., about $0.631.$ This is directly from the Henderson-Hasselbalch equation. Since all the phosphate species are in the same buffer volume, this is also the ratio of the respective concentrations.

The difference between the $\pu{0.01 M}$ phosphate buffer and the $\pu{0.1 M}$ phosphate buffer is that the $\pu{0.1 M}$ buffer is ten times higher in both $\ce{HPO4^2-}$ concentration and $\ce{H2PO4-}$ concentration. So the ratio is the same as in the $\pu{0.01 M}$ buffer and the $\mathrm{pH}$ is the same for the two buffers. Of course, the more dilute buffer has lower buffer capacity, but that is another issue.


Why exception is noted in the solubility of lead salts? There are many exception in the solubility of lead salts like:

Lead sulfate is insoluble in cold water whereas most of the sulfates are soluble in cold water.
Lead chloride is also insoluble in cold water but is soluble in hot water. Why?
Carbonates are generally insoluble in water but they dissolve in dil. $\ce{HCl}$ except lead carbonate.

What property of lead causes such type of exceptions?

Lead sulfate is insoluble in cold water whereas most of the sulfates are soluble in cold water.
  Lead chloride is also insoluble in cold water but is soluble in hot water. Why?

Lead ion is a borderline acid in the context of hard-soft acid-base theory. Water is a hard solvent. Hard-hard and soft-soft interactions tend to be favorable; mixed interactions are not as favorable. Remember, solvation entails not only breaking bonds within the salt but also creation of bonds between ions and solvent.

Hot water likely weakens the lead ion-anion bonds.

  Carbonates are generally insoluble in water but they dissolve in dil. $\ce{HCl}$ except lead carbonate.

Many carbonates are soluble in acid because the acid can protonate the carbonate portion. Singly protonating carbonate ion creates hydrogen carbonate, or bicarbonate, and bicarbonate salts tend to be soluble, likely because of its reduced negative charge density. Doubly protonating the carbonate portion creates a neutral entity - carbonic acid. This is rather unstable and tends to disproportionate to carbon dioxide in water. A common qualitative test for carbonates is to acidify the (unknown) salt and if bubbling is observed, the salt is likely a carbonate salt.

The interesting thing about lead ion is that many of its salts are insoluble in water. This suggests to me that there is a) an especially strong bond between lead ion and its counterions and/or b) an inability of water to effectively solvate lead ion. The latter is discussed above. The former, I'll leave to someone more familiar with the chemistry of metals.

(source)


Are there any sulfides, sulfates, carbonates that cannot be dissolved by aqua regia? I was looking into these specific minerals: arsenopyrite, pyrite, chalcopyrite, barite, calcite, dolomite. I investigate issues with tellurides not showing expected gold concentrations.

Chemical Book reports that barium sulfate (barite, or baryte) is

slightly soluble in concentrated sulfuric acid and soluble in carbonate alkali metal solution in which it is converted to barium carbonate; it is insoluble in other kinds of acid or base.

It sounds like the barite is your culprit.  If you try grinding it with an eye towards physically separating the denser gold from it, the Mohs hardness of the mineral is 3-3.5, so there appears to be no issue with the physical grinding process.


Is cycloheptatriene aromatic or not? Is cyclohepta-1,3,5-triene aromatic?

I think it should not be, because even though the π-system possesses six π-electrons, the conjugation is not complete around the ring.

You are correct. The conjugation does not extend all the way around the ring, so the molecule is not aromatic. In addition, the single methylene group ($\ce{CH2}$) sticks out of the plane of the rest of the molecule.

(On the other hand, removing a hydride ion $\ce{H-}$ from the $\ce{CH2}$ group leaves the tropylium ion $\ce{C7H7+}$, which is planar and aromatic, with resonance structures putting the positive charge on any of the seven carbons.)


Ionization energy and Electron Affnity Which of the following statements correctly interprets the relationship between ionization energy, atomic radius  Electron affinity when comparing two atoms.

The correct choice as per my instructor is that when atomic radius increase ionization energy decrease and electron affinity becomes less positive.

I do understand that as the atomic radius increase it becomes easier to remove and an electron due to shielding from multiple orbitals. Hence Ionization energy is decreased as the atomic radius increase. Additionally, electron affinity is an exothermic process.I am not sure how  electron affinity (addition of an electron) related to ionization energy and atomic radius. Please clarify. Thank you.

For example ,when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released).

Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.

When an electron is generally added to a neutral atom, i.e first electron affinity, energy is released; thus, the first electron affinities are negative. However, when an electron is added to a negative ion, i.e. second electron affinity, more energy is required. Thus, more energy is released to add the electron into an ion because the negative ion has to force the electron to go into its electron orbital; thus, the second electron affinities are positive.

First Electron Affinity (negative energy because energy released):

$\ce{X(g) + e- -> X- (g)}$

Second Electron Affinity (positive energy because energy needed is more than gained):

$\ce{X- (g) + e- -> X^2- (g)}$

This is because you are practically trying to forcefully add an electron to it.

Coming to your question: Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.

One might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table.

As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately.)

The electron affinity is a measure of the attraction between the incoming electron and the nucleus — the stronger the attraction, the more energy is released. The factors which affect this attraction are exactly the same as those relating to ionization energies — nuclear charge, distance and screening.

The increased nuclear charge as you go down the group is offset by extra screening electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about.

So,

as atomic radius increases, the nuclear attraction on outtermost electron decreases — this leads to decrease in IE — but the shielding also increases — less energy is released when an electron is added to the outside orbital — decreasing the EA.


How does Aloe Vera leave a cool feeling on your skin? Why does it evaporate fast? If you have ever grown Aloe Vera and applied it to your skin, it feels a bit slimy. However, after it evaporates it has a cool feeling on your skin.

Also, why does the Aloe Vera evaporate so fast? What is it made of, apart from water that makes Aloe Vera do this?

I am assuming Hand Sanitizer and Aloe Vera are not directly connected.

The leaf of aloe vera is composed of three layers: 1) An inner clear gel that contains 99% water and rest is made of glucomannans, amino acids, lipids, sterols and vitamins. 2) The middle layer of latex which is the bitter yellow sap and contains anthraquinones and glycosides. 3) The outer thick layer of 15–20 cells called as rind which has protective function and synthesizes carbohydrates and proteins. So overally , it consists of about 99% water and remaining are vitamins, enzymes, minerals, sugars, lignin, saponins,etc.

So what makes inner layer gelly ?? Well, the innermost part of the leaf is a clear, soft, moist and slippery tissue that consists of large thin-walled parenchyma cells in which water is held in the form of a viscous mucilage consiting of cellulose, hemicellulos and acetylated mannans.

As for you question, as most of it consists of water(99%) the cool feeling might has been felt due to its evaporation but some of its remaining constituents are absorbed in the skin. Mucopolysaccharides present in it help in binding moisture into the skin.Its one of the important medical applications is that it could potentially be used as a
penetration enhancement agent for the transdermal delivery of drugs and intestinal absorption of coadministerd drug.link

Saponins that are the soapy substances are found in the gel of aloe vera and have cleansing and antiseptic properties link so you can say that hand sanitizer and aloe vera are somewhat connected.


What is the safe level of energy output for hydrogen combustion at home? With confinement, my son would like to do some home experiments for a class. So far, he has a good setup that seems sound for the electrolysis of water. A later part of the experiment would combine back the $\ce{O2}$ and $\ce{H2}$ to get back water vapour.

What amount of atmospheric pressure $\ce{H2}$ and $\ce{O2}$ can be safely combined, at home, without it turning to a fireball and/or a news event?  From Googling, I found 12,749 joules per litre of hydrogen. What I'm asking is:

How does dissipating 12K joules look like in a plastic bucket? Is this a case of needing only gloves and eye protection, or does it need further protection? Will it be just some heat, just a small flame, or a McGyver-type explosion?

Also, will $\ce{H2}$ and $\ce{O2}$ at atmospheric conditions instantly react, or would this be a progressive reaction?

First things first: hydrogen gas and oxygen gas do not spontaneously react but need activation. Activation can be provided by a flame, an electric spark or similar.

That said, here is the important part. The reaction will not turn into a fireball but even small volumes will explode with pretty loud bangs. It’s not the loudest bang known to chemistry instructors but it’s certainly something to be aware of – you don’t want your neighbours ringing the police because they think a shooting has been going on.

If you’re going to try this, I recommend starting with the smallest volumes possible, not bigger than a shot glass. My chemistry teacher in high school used a glass cylinder approximately the size of a small (200 ml) water glass. That gave an even stronger bang but the cylinder itself remained unharmed, probably because there was a large opening on one side.

However, the experiment is unlikely to show you much because the water vapour is obviously gaseous and will dissipate rather than being visible anywhere. It would almost be more instructive to identify the two gases as hydrogen (burns when brought to a flame) and oxygen (ignites glimming wood) by the standard procedures used in high school (at least in mine).

Tl;dr: I suggest you don’t try igniting a 2:1 mixture unless you’ve seen it performed recently and know what you’re doing.


Nomenclature of spiro compounds Does the nomenclature of spiro compounds follow the same pattern as that for bicyclo compounds? That is, in absence of functional groups do we start form the largest chain to the smallest chain. What would be the case if there were functional groups?

No, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), numbering of spiro compounds starts in the smaller ring.

  P-24.2.1 Monospiro alicyclic ring systems

  Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of heteroatoms, double bonds, or functional groups. However, low locants must be attributed if there is a choice.


Adiabatic expansion of gas If a real gas is adiabatically expanded against constant pressure, then which of the following will definitely increase?
Compressibility factor (Z) or Entropy (S)
Correct answer: Entropy
My answer: Compressibility
My Reasoning :
$Z=\frac{PV}{nRT}$
In adiabatic $TV^{γ-1}=\text{constant}$
Therefore, $T$ ∝ $\frac{1}{V^{γ-1}}$
$γ-1$ is greater than zero always. So we can say is volume increases, temp decreases. Therefore Z should increase as its numerator increases and denominator decreases. Entropy for isobaric adiabatic has $\ln\frac{T_2}{T_1}$. Since $T_2$ is less than $T_1$, entropy decreases. What am I doing wrong ?

First note that the process is an irreversible adiabatic expansion. If it were a reversible one, the gas has to expand against a varying pressure because pressure of gas decreases during expansion and this varying gas pressure must be matched with external pressure in all infinitesimal steps of expansion. "The gas expanded against constant pressure" ⇒ the gas must have a pressure which is always higher than the constant external pressure during the expansion ⇒ irreversible.
Will entropy definitely increase?

ΔSsur = 0 as q = 0. Irreversible ⇒ ΔStot > 0. ΔStot = ΔSsys + ΔSsur = 0 + ΔSsys = ΔSsys. So ΔSsys > 0, so entropy of system will definitely increase.

Will compressibility factor definitely increase?

In adiabatic irreversible expansion all p, V, T changes. If Z is to increase, the increment in pV must be greater than the increment in T, or the magnitude of decrement in T must be greater than that in pV, which cannot be sure will happen as all p, V, T change.

What did you do wrong?

First of all, TV equation is for perfect gas doing reversible adiabatic expansion only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change. All p, V, T will change. You said as volume increases, temperature decreases, but pressure of gas will decrease too.


Naming tetrahalogenated benzene ChemSketch generates the IUPAC name of the compound above as "2-bromo-1-chloro-5-fluoro-3-iodobenzene" but I have a different answer.

I know that I have asked a related question before as well. The key takeaway from orthocresol's answer was that:

  first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants

I have applied the same here: the prefixes ordered in alphabetical order are "bromo, chloro, fluoro, iodo". Now, I have to choose a locant set that minimizes the locants. A possible locant set is "1-bromo-2-chloro-4-fluoro-6-iodo" which correctly locates the substituents. Also, the locant set "1246" is lower than "2153", as noted in this answer. So, according to me, the preferred IUPAC name should be "1-bromo-2-chloro-4-fluoro-6-iodobenzene".

What exactly am I doing wrong here?

The locant set ‘1,2,4,6’ is not lower than ‘2,1,5,3’.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the lowest locant set is defined as follows:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)

Especially note that the locants are cited in order of increasing value when they are compared.

Thus, the locant set ‘2,1,5,3’ is regarded as ‘1,2,3,5’, which is lower than ‘1,2,4,6’.

Therefore, the correct name is 2-bromo-1-chloro-5-fluoro-3-iodobenzene.


Why flame color of salts is determined by metal If a pure metal is heated up in a flame, we see it giving the flame a characteristic color, e.g. $\ce{Na}$ burns yellow.

If a salt, e.g. $\ce{NaCl}$ or $\ce{Na2SO4}$, is heated up, resulting color to a greater degree still is that of the pure metal. Although the color does change depending on the anion, this effect is usually not much pronounced.

Why is that the case? Shouldn't ionization happen to anions as well, causing electrons (including those pulled from the cation) to "fall back" and give off photons? In other words shouldn't we see a mix of two characteristic colors?

Not all flame colours are determined by the metal, some anions also colour flames (see the linked question.)

Also, not all metals impart a notable flame colour (platinum, for example, is often used as a substrate material to do flame tests because it doesn't cause interference: small amounts of the substance of interest are deposited on a platinum wire which is then inserted into the flame).

What actually causes colour are the electron transitions where excited electrons fall back to lower energy levels emitting light in the process. Flames cause enough excitation that this process is common. Some metals have strong electronic transitions that match wavelengths of visible light. The bright yellow characteristic of sodium come mostly from electrons falling from the 3p to the 3s levels in sodium atoms and have a wavelength around 589nm. This is the same bright yellow that is visible in low pressure sodium street lighting (and is caused by exactly the same electronic process).

Other metals have different electronic structures and their emission spectra therefore have different colours. But not all metals have bright lines in the visible (the strongest lines in the spectrum of mercury, for example, are in the UV so won't give the same notable visible colour in a flame test). Many metals do, though, which is why flame tests are often useful. And even when you can't see the colour, instruments can, so spectroscopy is a useful analytical tool even when your eye can't see much.


Is there a hydrophobic gas? I am looking into the subject of water repelling and have been led to the hydrophobic subject area. I have two questions regarding this subject.

Is there a hydrophobic gas ?
If a hydrophobic liquid is heated to a gas does it loose its water repellent qualities.

Thanks for the help guys

The trouble with the question is that hydrophobicity is not a meaningful concept in a gas.

The concept is mostly useful when characterising how surfaces behave. Typical hydrophobic compounds don't interact strongly with water (e.g. non polar hydrocarbon waxes) and tend to disrupt water's hydrogen bond network. So water doesn't "wet" the surface and tends to form droplets that seem to be trying to minimise the contact area with the surface. The contact angle that a droplet makes with the surface is used as a measure of static hydrophobicity (if the droplet spreads out because the surface likes to interact with water, the angle is low, if the surface reels water the angle is high).

It is less clear that describing liquid phases as hydrophobic is meaningful, though it is not uncommon in casual usage. Compounds that would be hydrophobic as solids will probably exist as liquids that don't mix with water or can form micelles where they cluster together in very small droplets in the water phase. The bulk effect that matters is that hydrophobic compounds don't mix with water and form separate phases. The hydrophobicity of the molecules describes why the separation happens, but the effect is immiscibility.

None of this matters in the gas phase. The point of gases (compared to solids or liquids) is that the bulk properties are not determined by molecular interactions. The molecules of a gas don't interact much and are, to a good approximation, interchangeable whatever the individual molecules are made of. Gas equations describing gas behaviour can effectively ignore the molecular interactions.

What this means is that if you take a compound that would be hydrophobic as a solid and immiscible with water as a liquid, it will mix quite happily with water vapour. So a hydrophobic compound can be a gas, but a gas can't be hydrophobic.


What toxic gases are emitted during the combustion of potassium ferrocyanide? I have read that Hydrogen Cyanide is emitted if Potassium Ferricyanide is combusted, but would this be the same case for Potassium Ferrocyanide? I am preparing to perform some case hardening on steel using a compound that contains 30-50% Potassium Ferrocyanide, so I need to be sure what measures are needed for protection from the fume.

Potassium ferrocyanide cannot burn, because it is the product of a combustion. When meat or blood is overheated, carbonized and burnt, the remaining ashes contain potassium ferrocyanide. And this product was initially recovered by washing these ashes and evaporating the obtained solution. In German, this compound is called Blutlaugensalz, which means "salt from washing blood" This is even the way this product was originally discovered in the old times of beginning chemistry.
So potassium ferrocyanide resists high temperature, and will not produce any $\ce{HCN}$ by heating or by combustion. The only way of producing HCN out of it, is to heat it with diluted sulfuric acid, according to the equation : $$\ce{K4Fe(CN)6 + 3 H2SO4 -> 2 K2SO4 + FeSO4 + 6 HCN}$$This reaction is difficult to carry out properly, because, if the sulfuric acid is more concentrated, it will destroy $\ce{HCN}$ according to  $$\ce{H2SO4 + 2 HCN + 2 H2O -> (NH4)2SO4 + 2 CO}$$


Can vicinal dihalide produce conjugated alkene in strong base? In Klein's Organic Chemistry (3rd edition), p. 410, we learn that vicinal dihalide can produce alkyne in strong base. And on Reaxys I see many reactions of this sort. For example:

However, why isn't the conjugated alkene, which should be more stable, the major product under such conditions?

Somebody once gave me an explanation, to the effect of "the hydrogen geminal to the halide is more acidic, hence more easily eliminated." However, Smith's Organic Synthesis (4th edition), p. 121 says the following:

  [In the following reaction] $\mathrm{H_a}$ is more acidic than $\mathrm{H_b}$ since it is attached to the less substituted carbon. The base should react with
  the more acidic hydrogen atom, which would lead to the less substituted alkene, 3-ethylhex-1-ene, but that is not the major product.

This suggests that we cannot base our logic on the acidity of the hydrogen atoms.

Comprehensive Organic Transformations (2nd edition) p. 467 gives examples of converting 2,3-dibromobutane to buta-1,3-diene.

On the other hand, Reaxys has many examples of producing alkynes from vicinal dihalides, even when potassium t-butoxide is used.

According to Carey's Advanced Organic Chemistry (5th edition, volume A), p. 554, the direction of E2 elimination is determined by whether the mechanism is more similar to E1 or to E1cb. If it is similar to E1, then the thermodynamically more stable product dominates; if it is similar to E1cb, then the direction is determined by the more easily removed hydrogen atom.

On page 556, Carey attributes the Hoffman rule of t-butoxide to its stronger basicity, not to its bulky size, and states that stronger bases drive the mechanism toward the E1cb side.

Therefore I think the problem might be understood this way: If a strong base such as $\mathrm{NaNH_2}$ is used, then the elimination direction is determined by which hydrogen atom is the easiest to remove. Since bromine acidifies its geminal hydrogen, the alkyne product dominates. If a weak base is used, the diene product might predominate.


Are a meso compound and its mirror image isomers? How are meso isomers even isomers? By definition, meso isomers are compounds that are superposable on their mirror images, but doesn't that mean that the mirror image is just the same compound?

Can someone explain the concept of meso isomers and diastereomers to me?

meso-compound
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.[1, 2]
isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.[1, 3, 4]

Your notion is correct in so far as the meso-compound and its mirror image are not isomers of each other since they are actually the same compound.
For example, meso-tartaric acid (preferred IUPAC name: (2R,3S)-2,3-dihydroxybutanedioic acid) can be superimposed on its mirror image “(2S,3R)-2,3-dihydroxybutanedioic acid”; i.e. both structures actually describe the same compound. Thus, they are not isomers of each other. (And hence, “(2S,3R)-2,3-dihydroxybutanedioic acid” is not a correct IUPAC name.)

However, two further stereochemical configurations are possible, namely (2R,3R)-2,3-dihydroxybutanedioic acid and (2S,3S)-2,3-dihydroxybutanedioic acid.

These structures are non-superimposable mirror images of each other, i.e. they are enantiomers, and the above-mentioned meso-tartaric acid is a diastereomer of both of them.
Therefore, there are a total of three isomers of tartaric acid:

(2R,3R)-2,3-dihydroxybutanedioic acid,
(2S,3S)-2,3-dihydroxybutanedioic acid, and
the meso-compound (2R,3S)-2,3-dihydroxybutanedioic acid.

References
[1] Basic terminology of stereochemistry (IUPAC Recommendations 1996). Pure Appl. Chem. 1996, 68, 2193
[2] meso-compound. In IUPAC Compendium of Chemical Terminology, (Gold Book)
[3] Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994). Pure Appl. Chem. 1994, 66, 1077
[4] isomer. In IUPAC Compendium of Chemical Terminology (Gold Book)


Why does solvent particles move towards solution when semi permeable membrane is added? When a SPM is added in between solute and solution the water molecules move towards salt water even though the concentration is same both sides. Why does this happen? According to SPM's perspective, concentration (of water) at salt water side is lower than that at fresh water side but how does that change the fact that the overall concentration is same both sides? My book says it depends on vapour pressure. If that's the case then osmosis should take place without SPM because it depends only on vapour pressure?

Here what happens is that the fresh water is a pure solvent and a pure solvent always has a greater vapor pressure than that of a solution. So due to the greater vapor pressure on water's side, the molecules of water will experience a force in salt water's direction as the pressure is less in salt water side. Therefore the water will flow towards the salt water.


Chemistry Experiments at Home I'm looking to set up a small chemistry lab in my garage for doing demos. I think it would be good for a freshman student I'm mentoring and also as a hobby. Does anyone know where I can buy apparatus (like flasks, pipettes, heaters) and chemicals?

If you live close to a research university, I would recommend you to look up if they have a surplus store. Often times when professors retire a lot of the lab supplies get sent to surplus, where they are sold very cheap. You can also build equipment out of basic stuff you can find at any thrift shop (pressure cookers, blenders, etc).

As for chemicals, it really depends on what you are interested in doing. Metals are all around you. You can take some pennies, dilute hydrogen peroxide, any carboxylic acid (vinegar, citric acid, amino acid supplements, etc.), and stick the mixture into a microwave for a few seconds. The copper will form complexes with the carboxylic acid you add. Different acids will give you different colors. Leave them standing for a bit and you will get crystals.

If you get to know the plants around you, you will have access to a huge array of naturally occurring small molecules and enzymes. Some palm trees have a high concentration peroxidases, onions and garlic undergo some pretty cool chemistry which results in the formation of pigments (blue, green, pink), there's pH indicators, capsaicin, etc.

Buying pure chemicals for organic synthesis is usually tricky, and for many of them you need to be affiliated with a research institute.

There are several YouTube channels with chemistry experiments which mentions where they obtained the chemicals. I recommend 'NurdRage'.

Good luck!


Is there a chemical treatment for pipes to reduce or eliminate leaching of lead into tap water? Given events in Flint, Michigan, where a switch to acidic river water in the city water supply resulted in leaching of lead into tap water from the insides of pipes, I wonder -- has anyone has heard of a chemical treatment for pipes that would reduce or eliminate further leaching of lead into the water?

In some cases the Langlier Saturation Index can be adjusted so a small carbonate scale barely forms on a pipe to prevent leaching.  The opposite may have happened to the index making the water corrosive when the water supply was switched.

This adjustment must be done by qualified professionals because if this index is too high the pipes will clog.


What does the symbol ϕ mean in skeletal formulae? I was given this formula by my A-level teacher and was curious as to what it means. It seems to be the Greek letter phi. I have searched online and have not been successful. Any help would be greatly appreciated.

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the Greek lowercase letter phi $(\phi)$ may be used to represent a phenyl group.

  It is acceptable to use the Greek lowercase letter phi $(\phi)$ to represent a phenyl group. That abbreviation has a long history, and ambiguity is unlikely since that letter is rarely used for other purposes in chemical structure diagrams.

Note that this traditional use is only considered “acceptable”, i.e. not “preferred”.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name for the hydrocarbon parent ring component is the retained name chrysene. The indicated numbering of this polycyclic hydrocarbon is is the result of the application of the specific criteria used to number fused ring systems:

Therefore, the complete name for the entire compound given in the question is 1,3,7,9-tetraphenylchrysene.


Thiol prefixes for nomenclature I've seen multiple prefixes for naming thiols: mercapto-, sulfanyl-, and -thio. Is there a rule for when specific ones are used? And are they used for IUPAC and/or common naming?

Mercapto- and sulfanyl- are both prefixes for use with the $\ce{-SH}$ group (known as a thiol), while thio- is used to denote the "sulfur" equivalent of an oxygen-containing functional group.

More details from the IUPAC Blue Book:

Mercapto- is mostly abandoned, and only used as a prefix for -SH
groups. This quote is from the older edition of the Blue Book. "When -SH is not the principal group, the prefix "mercapto-"
is placed before the name of the parent compound to denote an
unsubstituted -SH group."
Sulfanyl- "In these [1993] recommendations,
the prefix "sulfanyl-" is preferred to "mercapto-" which was used in
previous editions of the IUPAC Nomenclature of Organic Chemistry."
Thio- "In nearly all its applications "thio" denotes replacement
of oxygen by sulfur."


What is the identity of the unknown compound using GCMS and NMR? I've been asked to identify an unknown compound by performing several analytical techniques on it.

The compound was a white powder and had a boiling point of 65 degrees celsius.
I have the GC mass spectrum as well as both Carbon-13 and Proton NMR spectra (note that the compound was dissolved in chloroform for NMR analysis).
So far I think it contains an aromatic ring because of the chemical shifts for the NMR spectra but that's all I have so far.

These are the images of the spectra (Fig 1. GC and Mass spectrum, Fig 2. Proton NMR spectrum, Fig 3. Zoomed in Proton NMR spectrum, Fig 4. Carbon NMR spectrum, Fig. 5 Zoomed in Carbon NMR spectrum):

Hints:

If you counted carbon signals in $\ce{^{13}C}\mathrm{~NMR}$, you'd find the compound has at least 10 $\ce{C}$ atoms.  Thus, it's indicated boiling point is incorrect, unless it's determined at reduced pressure. If so, you need to give that pressure.
You are correct about saying your compound should be aromatic. Did you count how many aromatic hydrogens are there? At least, by doing that, you may able to speculate some framework for your compound, as a beginning point.
Now, if you are careful, you see no aliphatic carbon signals in your $\ce{^{13}C}\mathrm{~NMR}$. Then, what is the signal at $\mathrm{\delta~1.6}$ in your $\ce{^{1}H}\mathrm{~NMR}$ doing there?
Now, go to $\mathrm{GC/MS}$. In there, your parent peak is almost looks like a doublet. Do you know why? Can you recall what element have the two stable isotopes about 50:50 natural abundance with $2~m/z$ units? If you don't remember, then subtract each of your parent peaks from your base peak in your $\mathrm{MS}$. Now, you almost have the empirical formula of your compound. Now it's matter of analyzing peak patterns in $\ce{^{1}H}\mathrm{~NMR}$ and fragmentation pattern in $\mathrm{MS}$.


Nucleophilic Substitution Reactions help Which halogenoalkane reacts fastest with sodium hydroxide?

A. 1-iodobutane
B. 1-chlorobutane
C. 2-chloro-2-methylpropane
D. 2-iodo-2-methylpropane

Iodine has a larger radius, thus less electronegative making it a better leaving group. And since OH is a protic substance it favors Sn1, which is a tertiary, so the answer must be d... but it says that the answer is b.. how????

Yes iodine's good leaving group.  the leaving ability is best described by the Gibbs energy change that occurs.

So since chlorine is a poor  leaving group so that make the reaction slower as less  energy is available.
So I think  that you are right so far.


Activation energy and temperature It is said that activation energy does not change with temperature.

But I am unable to figure out why.
If we increase the temperature, the kinetic energy of the molecules will increase and they will need less extra energy and hence lesser activation energy to overcome the threshold energy barrier.

When I surfed the net, at most of the places it has been explained that increasing the temperature increases the no. of molecules having more energy and hence reaction rate is more but activation energy is same. This makes some sense but I still don't get why the activation energy doesn't decrease as the KE of an individual molecule should also increase on increasing the temperature.

The two distribution plots shown here are for a lower temperature $T_{1}$ and a higher temperature $T_{2}$. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of $E_{a}$ that are shown.

It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.

Activation energy seems to be more of a material or reaction pertinent . Say a reaction has $10J$ as $E_{a}$ to proceed but has $20 J$ kinetic energy - I would say the the additional energy is transferred/carried on by the products after reaction is complete. So having more K.E shouldn't affect the character of a reaction but only its kinetics .


Predicting the product of a hydrogenation reaction The question is to predict the nature of the product(s) of the above reaction, that is if the product(s) form a racemic mixture, meso, diasteromers, or are structural isomers.

I am aware of the fact that addition of $\ce{H2}$ is a syn addition. So it should form a meso compound. However the answer key gives it as a mixture of diastereomers. I am not quite sure if the reaction is strictly selective and some anti products are also formed.

To answer and understand this, let’s first consider the products of the hydrogenation ($\ce{H2; Pd/C}$) of 1-methylidene-4-methylcyclohexane (the same reactant except the methyl group is in para-position). This compound is symmetric and therefore achiral. We can draw the methyl group pointing upwards iwthout loss of generalisation. The hydrogenation is syn-selective and it can occur from the top or the bottom. We expect a $1:1$ mixture of the former methylidene group pointing upwards (the syn-isomer) and pointing downwards (the anti-isomer). Both are achiral (the plane of symmetry is retained in the reaction) but not identical; therefore, they must be diastereomers.

Now let’s move the methyl group back into the 2-position. By moving the methyl group, the plane of symmetry is lost; it suddenly makes a difference whether the methyl group points upwards (S) or downwards (R). For each of the two isomers, we can again hydrogenate from both above and below; we can expect two anti and two syn isomers at first approximation: we enter one stereocentre into the reaction and we form an additional one making two stereocentres in total and thus 4 theoretically possible isomers.

In explicit forms, starting from the (2S) isomer the syn-product could be called (1R,2S) and the anti-isomer would be (1S,2S). From the (2R) isomer, another syn (1S,2R) and anti (1R,2R) isomer is expected. Only when taking a closer look at the products, it becomes obvious that the two syn isomers are in fact identical because they can be transformed into each other by rotation; and that they are achiral because they generated a new plane of symmetry. Thus, the syn isomer is a meso isomer.

Putting this together, the hydrogenation products are two enantiomeric anti products and a meso (syn) product. These two classes are diastereomers of each other, thus ‘a mixture of diastereomers’ is the only correct answers.


How to achieve regioselective (base-catalyzed) enolization of a carbonyl group in a diketone? One can convert a ketone to an enolate, either base- or acid-catalyzed.

Let's say you now have a diketone. You choose to treat this compound with base (e.g. hydroxide anion, alkoxide anion, LDA/THF, NaH).

How can one control the regioselectivity of this enolization reaction such that only one specific carbonyl group is converted to an enolate group?

As @Zhe and @Zenix have mentioned, you have to hope that there is a difference in the rate of enolate formation to differentiate the two ketones.

To do this cool a rigourously dried solution of the diketone and a trapping reagent ($\ce{Me3SiCl}$ if you're doing an anhydrous workup otherwise TIPS-Cl or MeI) to $\mathrm{-75 \, ^\circ C}$ and slowly add 1 eq. of LDA keeping the reaction mixture cold. This will trap the first formed enolate (kinetic enolate). Allow to warm, dilute with hexanes, filter (or wash with aq if you used TIPS or MeI), concentrate and analyse.

If this gives a mixture then you try making the more stable (thermodynamic enolate) and trap that. This means forming the enolate without the trapping compound ($\ce{Me3SiCl}$, MeI or TIPS-Cl), giving it time to equilibrate then adding your trapping compound. This might be done with NaH in dry DMF, left to stir at $\mathrm{-30 \, ^\circ C}$ for an hour then the trap is added, worked up and analysed.

Even then with this example I think you'll get a mixture, but you don't know until you try. If you do get a mixture then you have to rearrange your synthetic pathway so that you don't have this problem - you'll probably have a different one but that's the  way it goes!


Numbering for the IUPAC name of this benzene derivative My first guess was 2-chloro-4-nitrotoluene, but this was not among the choices for this question. My teacher says we will begin numbering from Cl, as halogens have higher priority than a nitro-group. However, this compound is named 2-chloro-1-methyl-4-nitrobenzene on PubChem.
I am confused. Please explain where should I begin numbering from.

The main goal of IUPAC names is to keep them simple and understandable by anyone. So, you basically want to keep numbers the lowest possible. Hence, you shall start numbering where it will give the lowest numbers for your substituents.
For instance, if you began numbering from the carbon bonded with the chlorine, your molecule would have been named 1-chloro-2-methyl-5-nitrobenzene or 1-chloro-6-methyl-3-nitrobenzene. Or, if you chose your numbering to begin from the other group, you'd have named it 3-chloro-4-methyl-1-nitrobenzene or, even worse, 5-chloro-4-methyl-1-nitrobenzene. With these examples, you can notice that the simplest name, i.e. the one that has the lowest numbers, is the one PubChem gave you : and that's the reason one should start numbering by the methyl.


Molecular orbitals of the enolate ion The reactivity of the enolate ion is most frequent explained with reference to its system of $\pi$ molecular orbitals. The image below taken from Wikimedia Commons depicts this system of $\pi$ molecular orbitals. $\ce {C}$-alkylation is favoured when the orbital interaction with the $\ce {HOMO}$ of the $\pi$ system is dominant while $\ce {O}$-alkylation is favoured when the electrostatic interaction with the negative charge built up on the oxygen atom is dominant. This is the case because the coefficient of the $\ce {HOMO}$ is the largest on the terminal carbon atom while being more electronegative, oxygen naturally bears greater negative charge.

The explanation provided by Fleming (2010) on p. 160 is that in the lowest-lying $\pi$ molecular orbital, the oxygen atom exerts the strongest polarising effect on the electron distribution while for the second highest $\pi$ molecular orbital, the polarisation is in the opposite direction. I understand that these MO diagrams are grounded in quantum chemistry theory but is there any way that we could intuitively understand why the coefficient of the $\ce {HOMO}$ is indeed the largest on the terminal carbon?

Update

After discussion with some friends, I have more thoughts I would like to share. There is this connection between resonance structures and MO diagrams. For example, the MO diagrams of the allyl cation and allyl anion feature a node in the second MO, informing us that there is no sharing of the charge with the central carbon atom. This information is also reflected when we draw the two resonance structures. Can we perhaps, extend this connection a little further and say that the first MO for this enolate $\pi$ system reflects the charge distribution in the resonance structure with negative charge localised on the oxygen atom while the second MO reflects the charge distribution in the resonance structure with the negative charge localised on the carbon atom?

Image source

Reference

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.

One possible way to intuit the result is to remember that all of the orbitals are normalized. Thus, the contribution of a single AO to the three MOs must add up to 1, and the contribution of the three AOs in each MO must also add up to 1. (This is a bit of an oversimplification, since the normalization is of the square of the wave function, but conceptually it's the same idea). So if a particular AO contributes a lot to one MO, it is proportionally less represented in the others. Also, the other AO's have to contribute less to the first MO in order for it to have the right magnitude.
In MOs composed of AOs from atoms of differing electronegativity, the lower energy MOs have a larger contribution from the more electronegative atom (and therefore smaller contribution from the less electronegative atoms). The higher energy MOs must therefore have a smaller contribution from the electronegative atom.
As a very simple first example, consider the $\sigma$ and $\sigma^*$ MOs of a polar diatomic molecule such as HF. The valence F AO is much lower in energy than the hydrogen orbital, so the bonding MO has a large contribution from F and very small contribution from H. In fact, the MO is nearly indistinguishable from the fluorine AO, as evidenced by the bond being nearly completely ionic. In order for the normalization to be correct, the antibonding orbital must be the reverse - a very small contribution from F and large contribution from H.
A slightly more complicated example is a triatomic like H2O. Here is a drawing of the lowest three valence MO's for XH2, when X is more electronegative than H:

(This is a portion of Figure 7.3 of Albright et al. Orbital Interactions in Chemistry).
You can see that again the electronegative atom (O) is overrepresented in the lowest energy MOs, at the expense of the H atomic orbitals. Therefore, in the high energy orbitals, we have to compensate with the reverse, an overrepresentation of the H orbitals (as seen in the 3a1).
Now getting to your question of C-C-O, again the O is most electronegative, so it contributes most to the lowest energy orbital. That means it has less to contribute to the higher energy orbitals, where the carbon orbitals are the bigger contributors.
This overly simplistic approach is harder to apply to the difference between the two carbons, for which you really have to incorporate the actual math of the atomic orbital contributions, but maybe it at least helps with the concepts.


What's the meaning of wavefunction being finite? I know ,in quantum mechanics it's important that the wavefunction has to be finite everywhere , including infinity. It's a criteria for acceptable wavefunction . What exactly does this statement mean ? And how it's associated with square integrability ?

A broader point regarding functions that are finite:
A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.
This is a generalisation of the comments written above, not specifically quantum related, but a more general point on mathematical functions.


Ionic equilibrium between multiple ions This is a question from IIT-JEE 1993:

  For the reaction

  $$\ce{[Ag(CN)2]^- -> Ag^+ + 2CN^-}$$

  the equilibrium constant at $\pu{298 K}$ is $\pu{4e-10}$. The silver ion concentration in a solution which was originally $\pu{0.10 M}$ in $\ce{KCN}$ and $\pu{0.03 M}$ in silver nitrate is?

This is what I've done:

$$
\begin{align}
[\ce{K^+}] &= 0.1 \\
[\ce{CN-}] &= 0.1 - 2x \\
[\ce{Ag+}] &= 0.03 - x \\
[\ce{NO3-}] &= 0.03
\end{align}
$$

$$\frac{(0.03-x)(0.1-2x)^2}{x} = 4\cdot 10^{-10}$$

But this gives $x = 0.03$, hence, $[\ce{Ag+}] \approx 0$.

The answer key says $[\ce{Ag+}] \approx 7.5\cdot 10^{-18}$. Where am I going wrong?

In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)_2^− \approx 0.0300}$ and $\ce{CN− \approx 0.0400}$. Then you solve for [Ag+]. So:

$4\times10^{-10} = \dfrac{\ce{[Ag^+][CN^-]^2}}{\ce{[Ag(CN)_2^-]}}$

or rearranging for $\ce{[Ag^+]}$

$\ce{Ag^+} = 4\times10^{-10} \dfrac{{\ce{[Ag(CN)_2^-]}}}{\ce{[CN^-]^2}} = 4\times10^{-10} \dfrac{0.03}{0.04^2} = 7.5\times10^{-9} $

Checking $7.5\times10^{-9} \ll 0.03$ so the assumption that the reaction essentially went to completion is valid. in other words:

$0.0300 - 7.5\times10^{-9} \approx 0.300$


Will glycerol and citric acid react? If yes, under what conditions? I would like to know whether the reaction will take place if glycerol and citric acid are merged? In particular, what will happen if both substances are heated up to high temperature (+100°C)?

As @Mithoron said, it will form esters. Actual name is Citric acid esters of tri-glycerides or citroglycerides.

  (source)

  Where at least one of R1, R2 or R3 represents a citric acid moiety,
  one represents a fatty acid moiety and the remainder may represent
  citric acid, fatty acid or hydrogen.

  Obtained by esterification of glycerol with citric acid [...] acid;
  consists of mixed esters of citric acid and edible fatty acids with
  glycerol; may contain minor parts of free fatty acids, free glycerol,
  free citric acid and mono- and diglycerides; may be wholly or
  partially neutralized with sodium hydroxide or potassium hydroxide.

  They are white to ivory coloured, oily to waxy material. They are used
  as stabilizer, emulsifier, dough conditioner, antioxidant synergist.

According to this site:

  The influence of various heating methods without catalysis to prepare
  polyesters from citric acid : glycerol blends were studied. In the
  presence of short-term microwave treatments, i.e., 60 s at 1200 W,
  blends of glycerol and citric acid invariably formed solid amorphous polyesters. Fourier transform infrared spectroscopy showed that
  citric acid and glycerol blends can form highly stable polymers
  composed of ester bonds. The glycerol citrate polyester polymers
  exhibited the least degradation in water, more in acid solutions
  (0.1–1.0M HCl), and the most deterioration in strong alkaline
  solutions (0.1–1.0M NaOH) after 72 h soakings. Polyesters of glycerol
  and citric acid were studied with differential scanning calorimetry
  and thermal gravimetric analysis. The polyesters were found to be
  thermally stable (up to 313°C).

A noble application of glycerol + citric acid esters or citroglycerides is in the making of a biodegradable thermoset polymers.

1. Source 1

  A new biodegradable thermoset polymer, was made from simple,
  economical, biocompatable monomers without the need for a catalyst.
  Glycerol and citric acid, non-toxic and renewable reagents, were
  crosslinked by a melt polymerization reaction at temperatures from
  90-150°C. Consistent with a condensation reaction, water was
  determined to be the primary byproduct. The amount of crosslinking was
  controlled by the reaction conditions, including temperature, reaction
  time, and ratio between glycerol and citric acid. Also, the amount of
  crosslinking was inversely proportional to the rate of degradation. [...]

2. Source 2

  The notable increase in global consumption of plastics and their long
  residence time in the atmosphere show the great need for biodegradable
  products. In this project, we developed biodegradable polymers based
  on citric acid and glycerol. Their synthesis was carried out at
  different conditions of constant temperature and concentration and
  they were synthesized using an economically viable method.
  Characterization was undertaken using: Acid number, FTIR spectroscopy,
  refractive index, viscosity, impact analysis, tensile test, hardness,
  calorimetry, the % humidity (oven method with recirculating air), and
  density determination, along with qualitative evidence to corroborate
  their biodegradability. These elaborated polymers were mixed with
  medical grade PVC formulation, obtaining a hybrid polymer, showing
  their mechanical properties were changed.


Confusion regarding specific conductivity Is there any difference between the terms specific conductivity and conductance. If yes, please explain.

Let's take a wire inside cable for example. That wire's conductance is just the inverse of the resistance this cable is making when electricity passes through it. This depends on things like the length, what is made of, the maximum/minimum width of certain areas of the wire, etc.

On the other hand, the conductivity of that wire is a direct property of the material it's made of. If the wire is made of copper for example, it'll have (at the same temperature) the same conductivity as any copper wire in the world (because they are all made of copper), but not the same conductance (because they are not physically equal)

I hope this helps, I'm sorry if I was wrong!


Calculating volume of gas produced For the following reaction, what volume of $\ce{SO3}$ can be produced from $4.2\ \mathrm{L}$ of $\ce{O2}$ (measured at the same temperature and pressure), assuming an excess of $\ce{SO2}$?
  $$\ce{2SO2(g) + O2(g) -> 2SO3 (g)}$$

I demonstrate what I tried here but I have no idea what to try here, anyone have any suggestions on what to do?

TL;DR There is twice  the amount of $\ce{SO3}$ produced compared to oxygen, thus the volume will be twice as much as oxygen. Therefore $8.4$ litres of $\ce{SO3}$ will be produced.

Since $\ce{SO2}$ is in excess, the amount of $\ce{SO3}$ produced will be twice as the amount of moles of oxygen. So if there is originally x moles of oxygen gas, 2x moles of $\ce{SO2}$ is produced. It is important to note that temperature and pressure remain constant throughout the reaction.

Note one of gas laws state, the same number of moles of any gaseous substance will have the same volume at constant temperature and pressure. So x moles of $\ce{SO3}$ will hold the same volume as x moles of oxygen gas, so 4.2 litres.

Now there is also another gas law which states that the volume of a gas is directly proportional to the number of moles of the substance. So 2x moles of $\ce{SO3}$ will have twice the volume of x moles of $\ce{SO3}$. Therefore the volume of 2x moles of $\ce{SO3}$ will be 8.4 moles.

Therefore the answer is $8.4$ litres


What are the other ways (if they exist) that atoms can gain or lose electrons that aren't bonding? I know that electrons can be isolated through human intervention from their atoms. My question rather concerns if electrons are ever lost or gained naturally. I recognize that this is a stupid question but I couldn't find the answer on the Internet.

My question rather concerns if electrons are ever lost or gained naturally.

Yes, they can!

Here are two examples of natural processes by which atoms can gain or lose electrons "naturally":

Electron Capture

In the natural radioactive decay process known as electron capture the nucleus captures an inner-shell electron, K- or sometimes L-shell. This results in a proton being converted to a neutron, and the emission of a neutrino. The atom has lost an electron, but conveniently a proton as well, so it remains neutral but with a deep vacancy.

Auger Process

If the vacancy is filled by a higher shell electron falling into it this can result in the emission of a photon, or in the emission of yet another electron. That emission is referred to as the Auger process.

See Electron capture; Reaction_details for examples.

 Source


Singlet or triplet carbene On reaction of diazomethane with ethene a carbene is formed as a intermediate . This carbene then attacks on the double bond and form cyclopropane . However I read that both singlet and triplet carbene can react in this reaction but according to me since the carbon of carbene is overlapping with two different orbitals hence each electron of carbene must be in different orbital hence must be triplet. I can't find a reference to this on the internet, perhaps you all could help me .

Diazomethylene is singlet therefore carbene is formed in the singlet state.
singlet carbene is highly reactive therefore it goes through reaction probably before intersystem crossing (depends on competitive speeds). this is why singlet state of carbene is usually invisible. Triplet carbenes react like radicals unlike singlet state (besides some aryl carbenes) therefore the products are different. you can look up Skell-Woodworth for reaction of different spin states of carbenes.
based on the products you can find which spin state of the carbene has been involved in the reaction.


Electrolysis of molten nitrates, sulphates and carbonates We know that nitrates, sulfates and carbonates doesn't or hardly oxidized by anode in the aqueous solution.

What will be produced on anode in the electrolysis of molten salts of these ions?

A common theme is disproportionation of the anion, and potential reactivity with the electrodes.

Nitrates

Electrochemical behavior of nickel electrodes in molten potassium nitrate with increased basicity, Electrochimica Acta, Vol. 36, No. 3/4, pp. 65S-660, 1991:

  $\text{Ni}$ electrodes in a $\ce{KNO_3}$ melt [...] some ionic species of nickelite ($\ce{NiO_2^{2-}}$) and nickelate ($\ce{NiO_2^{-}}$) type

Voltammetric investigations of the oxygen electrochemical systems in molten sodium nitrate at 420°C, Electrochimica Acta, Vol. 33, No. 4, pp. 545-549, 1988:

  the potential range is limited anodically by the electrochemical oxidation of nitrate ions into nitrogen dioxide and oxygen: $\ce{NO3-} \to \ce{NO2} + \frac{1}{2}\ce{O2} + \ce{e-}$ and cathodically by the electrochemical reduction of nitrate ions into nitrite ions: $\ce{NO3-} + \ce{2e-} \to \ce{NO2-} + \ce{O^2-}$

Carbonates

Electrochemical formation of carbon nano-powders with various porosities in molten alkali carbonates, Electrochimica Acta, Vol. 54, pp. 4566–4573, 2009:

  reduction of carbonate ions: $\ce{CO3} + \ce{4e-} \to \ce{C} + \ce{3O^2-}$ [and reduction of metal ions, along with oxidation of carbonate ions:] $\ce{2CO3^2-} \to \ce{2CO2} + \ce{O2} + \ce{4e-}$

Sulfates

Polarization measurements on solid platinum-molten sodium sulphate-sodium chloride interfaces, EIectrochimica Acta, Vol. 23, pp. 381-388, 1978:

  the limiting reactions [...] are suggested to be the $\ce{SO3}$ reduction to $\ce{S^2-}$ and $\ce{O2-}$ and the $\ce{SO4^2-}$ oxidation to $\ce{SO2}$ and $\ce{O2}$


Why is trimethylsilanol a stronger acid than tert-butanol? The Wikipedia article about trimethylsilanol says the following:

  Trimethylsilanol is a weak acid with a $\mathrm{p}K_\mathrm{a}$ value of 11. The acidity is comparable to that of orthosilicic acid, but much higher than the one of alcohols like tert-butanol ($\mathrm{p}K_\mathrm{a}$ 19).

Why is it so? A silicon atom is more electropositive than a carbon atom, therefore the conjugate base should be more destabilized by the +I effect of the $\ce{Si(CH3)3}$ group compared to the $\ce{t-Bu}$ group.

The involvement of d-orbitals in the stabilisation of silanol anions is minimal. The ions $\ce{H3CO-}$ and $\ce{H3SiO-}$ were investigated by Hopkinson and Lien,1 who concluded that:

  Silicon is a more electropositive element than carbon and consequently the silyl group might be expected to be an electron donor. As such the silyl group should stabilise cations and destabilise anions. However, ab initio molecular orbital calculations show both the α-silyl and β-silyl to strongly stabilise anions and to weakly destabilise cations. Inclusion of d orbitals on  silicon does not significantly alter these conclusions. This contradictory behaviour can be largely understood in terms of n $\to$ π* delocalisation from the lone pair to π* of the silyl group.

These π* orbitals are antibonding orbitals derived from linear combination of Si 3p and H 1s. A partial MO diagram for the $\ce{H3Si}$ fragment is provided here (from the same paper):

Delocalisation of the oxygen lone pair into the $\pi_x^*$ and $\pi_y^*$ orbitals therefore leads to stabilisation. The authors write that the π* orbitals of the $\ce{H3Si}$ fragment are lower in energy than the corresponding π* orbitals of the $\ce{H3C}$ fragment, and hence the stabilisation is greater for silicon than it is for carbon.

A good test for this is to examine how the Si–H bond lengths change upon deprotonation. If the delocalisation is into Si–H antibonding orbitals, then the bond lengths should increase upon deprotonation. In contrast, if the delocalisation is into Si 3d orbitals, then there should be no change. Upon deprotonation, the Si–H bond length increases from 1.473 Å to 1.528 Å, which indicates it is the former.

However, there isn't a thorough explanation of why the energy levels are so. I would surmise that there are two factors: one is the fact that Si–H overlap is poor, compared to C–H overlap. Therefore, Si–H antibonding orbitals are lower in energy than C–H antibonding orbitals, and can act as better acceptor orbitals. Another factor (and one commonly taught in textbooks) is that since Si is more electropositive, the π* orbitals have a larger coefficient on Si; the π* orbitals of the $\ce{H3C}$ fragment have a smaller coefficient on C.

There is also an issue of why the stabilisation is larger in the anion than it is in the neutral molecule: if both the anion and the neutral molecule were equally stabilised, then there would not be any difference in the acidity, which is essentially a measure of the relative stability of the two species. It may be easily explained - at least partly - by the fact that the lone pairs in the anion are higher in energy than those in the neutral molecule, and consequently act as better donor orbitals.

Now, this explanation can be easily extended to the methylated compounds $\ce{Me3CO-}$ and $\ce{Me3SiO-}$. Due to the poorer acceptor capability of Si–C antibonding orbitals, we expect the factors above to play a smaller role in the preferential stabilisation of the silicon anion. As such, the difference in $\mathrm pK_\mathrm a$ between $\ce{X3COH}$ and $\ce{X3SiOH}$ should be smaller when $\ce{X = Me}$ compared to when $\ce{X = H}$.

Is there a way to test this? A crude analysis is as follows. I will take the value of $\pu{27.6 kcal/mol}$ given by Hopkinson and Lien1 for the reaction

$$\ce{H3SiO- + H3COH -> H3SiOH + H3CO-}$$

which corresponds to a $\mathrm pK_\mathrm a$ difference of 20 units when $\ce{X = H}$. This is obviously not rigorous, but serves as a good preliminary benchmark. Now, if Wikipedia's sources are to be believed, then when $\ce{X = Me}$, the $\mathrm pK_\mathrm a$ difference is decreased to a mere 8 units - which is in line with our prediction.

The topic of negative charge stabilisation by silicon has been studied much more thoroughly in the context of carbanions, because they are synthetically useful species (e.g. Peterson olefination). In this context, Schleyer et al.2 also conclude that d-orbital participation is negligible. The myth of d-orbital participation in main group bonding has been thoroughly debunked decades ago,3 so it is regrettable that it is still being taught nowadays.

References

Hopkinson, A. C.; Lien, M. H. A theoretical study of the effects of silyl and methyl groups on acid-base reactions at oxygen, nitrogen and carbon. J. Mol. Struct.: THEOCHEM 1983, 92 (1–2), 153–166. DOI: 10.1016/0166-1280(83)80065-7.
Schleyer, P. v. R.; Clark, T.; Kos, A. J.; Spitznagel, G. W.; Rohde, C.; Arad, D.; Houk, K. N.; Rondan, N. G. Structures and stabilities of α-hetero-substituted organolithium and organosodium compounds. Energetic unimportance of d-orbital effects. J. Am. Chem. Soc. 1984, 106 (22), 6467–6475. DOI: 10.1021/ja00334a001.
Reed, A. E.; Schleyer, P. v. R. Chemical bonding in hypervalent molecules. The dominance of ionic bonding and negative hyperconjugation over d-orbital participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. DOI: 10.1021/ja00160a022.


What are those... crystals? I have been trying to make muratic acid from salt and phosphoric acid (didn't know that it's made at 150 °C). After waiting a few days, a third of the time it was sealed by placing a container over it, I wanted to check if some HCl was created, by adding peroxide and etching a small PCB, after 2 days I noticed small white crystals and the solution turned green.

What are those crystals and why is the fluid green?

Likely the green solution contains copper(II) phosphate, $\ce{Cu3(PO4)2}$. "It is commonly encountered as the hydrated species Cu2(PO4)OH, which is green."

Try crystallizing it!

Libethenite, $\ce{Cu3(PO4)2}$, crystals.

You have $\ce{H+}$, $\ce{Na+}$, $\ce{Cu+}$ and/or $\ce{Cu++}$, $\ce{OH-}$, $\ce{Cl-}$ and $\ce{PO4{^-}{^3}}$ in solution. You could identify the white crystals by color (white, so probably no Cu in them) shape, birefringence, and solubility; they might be $\ce{NaCl}$ or $\ce{Na3PO4}$. The less soluble should crystallize first, of course.


Confusion on electrode potentials Standard electrode potential values can be used to judge the reducing ability of a metal. When I was introduced to this topic a simple picture was painted: you put a metal into a solution of its ions. An equilibrium is set up. You do so for another metal. If you now hook up the two electrodes and measure the voltage, you can tell which one is 'more negative' hence more electron donating.
I was taught like it is covered here :
  https://www.chemguide.co.uk/physical/redoxeqia/introduction.html

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals?

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.

That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:

$$\ce{Ag+ + 1/2Mg -> Ag + 1/2 Mg^2+}$$

has the same cell potential as

$$\ce{2Ag+ + Mg -> 2Ag +  Mg^2+}.$$

(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).

  How has the voltage helped us determine the difference in the tendencies of these two metals?

What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.

You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.


Temperature of chlorine gas at given pressure and volume Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of $\ce{Cl2}$ has is $\pu{9.22 L}$ at $\pu{1124 torr}$ and $\pu{24 ^\circ C}$. At what temperature will the volume be $\pu{15.00 L}$ if the pressure is $\pu{8.76e2 torr}$ ?

I found the moles of the sample to be $\pu{0.559 mol}$. However, now I'm stuck at the last part of the question. Can I use $\pu{0.559 mol}$ to solve for the temperature when the volume is $\pu{15 L}$ and the pressure is $\pu{8.76e2 torr}$?

Assuming at these conditions it's safe to use ideal gas law ($pV = nRT$) for chlorine, and given the same gas sample ($n = \text{const}$, hence $nR = \text{const}$), let's express the unknown temperature $T_2$:

$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \to \quad T_2 = \frac{p_2V_2}{p_1V_1} \cdot T_1$$

Generally, I would recommend not to do intermediate calculations, especially if you are not asked to do so. Construct the mathematical expression first, convert the units if necessary, and plug in the values when you are done. Note that in this case you only need to convert temperature units.

   $$T_2 = \frac{\pu{876 torr} \cdot \pu{15.00 L}}{\pu{1124 torr} \cdot \pu{9.22 L}} \cdot \pu{297 K} = \pu{377 K}~\text{or}~\pu{104 ^\circ C}$$


How to calculate the number of geometrical isomers in allenes? Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$

If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of cis-trans isomers is possible in compounds with a single double bond.

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.

Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.


Three general questions in silanization of a glass surface For silanization, I put a glass slide horizontally and pour the silan solution on it to cover the glass surface. Then I leave it for a specific time and then wash with water and acetone.

I have three questions:

Is this method correct or I should necessarily deep coat the glass plate into a container of silan solution (because in the 2nd method, there are larger number of available agents than can react with glass or have better accessibility to the glass surface)?
How durable is the silanized glass? i.e. after silanization, when I want to clean the glass surface from reaction products, can I rub something onto the surface to help clean it better or the treated layer of glass might be damaged? (rubbing with cleanex, etc)
For getting maximum hydrophobicity, can I use Octadecyltrichlorosilane>90% directly on glass surface (for how much time?) or I should dilute it?

Is this method correct or I should necessarily deep coat the glass plate into a container of silan solution (because in the 2nd method, there are larger number of available agents than can react with glass or have better accessibility to the glass surface)?

Your method is fine, anything method that covers the surface with solution will work.

  How durable is the silanized glass? i.e. after silanization, when I want to clean the glass surface from reaction products, can I rub something onto the surface to help clean it better or the treated layer of glass might be damaged? (rubbing with cleanex, etc)

In my experience hydrophobic coatings are durable, but will wear due the the small surface thickness.  My recommendation is to use rinsing only, if you can and use abrasive cleaning when you have to.

  For getting maximum hydrophobicity, can I use Octadecyltrichlorosilane>90% directly on glass surface (for how much time?) or I should dilute it?

I find that hydrophobic solutions typically make  very effective coatings nearly instantaneously.  You can use it directly which should only require a few seconds maybe a whole minute to work.  If you want to conserve octadecyltrichlorosilane you can dilute it. I have had 5% solutions also work nearly instantaneously, but if you want the most hydrophobic surface you should wait 1-5 minutes.


Equation for salt formation for ammonium oxalate? I'm having trouble finding out the salt formation equation for $\ce{(NH4)2C2O4}$

I thought it was $\ce{NH3 + H2C2O4}$ but that doesn't seem to balance out.

$\ce{HC2O4-}$ seems to work, but are ions allowed in salt formation/neutralization formulas?

Where am I wrong? First of all, note that the ammonia is dissolved in water. So there is an ion that's involved in the reaction, not the ammonia itself. And second, the charge of the ion involved requires a different coefficient for ammonia. Notice that the $\ce{H+}$ cations in Hydrogen oxalate are spectator ions.

The formation of Ammonium Oxalate will be a precipitation reaction. (Or some may put it an acid-base one) Answer these in order to reach a final net ionic reaction:

What are the ions that are present in the final product (ammonium oxalate)?

   The ions involved are ammonium ($\ce{NH4+}$) and oxalate ($\ce{C2O4^{2-}}$).

Notice the charges of those ions:

   Ammonium has a 1+ charge (that's written correctly as +, but of course, I wanted to emphasize) and oxalate has a 2-.

The other ions are spectator ions. These two ions actually form the precipitate. But the final formula (where your misunderstanding rises from) is

   $\ce{(NH4)2C2O4}$

You can now write the net ionic equation and get it balanced.