Evaluation of a synthesis from aspirin to paracetamol from this Nile Red video Recently the YouTube chemist Nile Red performed an interesting conversion from aspirin to paracetamol Aspirin to Acetaminophen - Part 1 of 6: Extracting Aspirin from Pills using the following steps:

He performed the following steps:

1) Extracting aspirin from pills (95%)

2) Hydrolysis of the ester (86%)

3) Decarboxylation (51%)

4) Nitration of phenol (6%)

5) Reduction of nitro group (58%)

6) Condensation reaction with acetic anhydride (18%)

Final yield: 0.26%

As you can see, the yield is quite discouraging, particularly the nitration of phenol. The synthesis that he planned was quite linear, first removing functional groups and then adding them. This unfortunately did not take full advantage of the functional groups present.

I feel that this is a good opportunity to reflect how we can carry out the reaction efficiently, and I will illustrate it with the picture below:

1) I feel that nitrating the aspirin directly would have been more effective.

Since the carboxylic acid is 3-directing and the ester 2,4-directing, this would lead to two nitration positions as shown by the asterisks. However, the site to the left would be more ideal as the site to the right is blocked by steric hindrance from the bulky ester group. The combined efforts of both functional groups would also increase the reaction yield quite considerably. Since this is done in the presence of concentrated nitric acid and sulfuric acid, it is not surprising that a hydrolysis can occur here when we add water, thus allowing us to do a one-pot reaction.

2) Now, decarboxylation should be carried out. Due to the powerful electron-withdrawing nature of the nitro group, the $\ce{C\bond{-}C}$ bond should be weaker than the original synthesis, giving us a better yield as well. This is important as the close proximity of the two groups allows for intramolecular hydrogen bonding, thus making decarboxylation more difficult than usual.

I am wondering how much the yield would be affected if it were synthesised in this method. Of course, it is impossible to theoretically come up with a value and can only be done experimentally, but it would be good if I can come up with some estimates. Let me know what you guys think. Thanks!

You are absolutely correct that $6~\%$ yield is not acceptable for a synthesis — especially if it is a reaction as simple as the nitration of phenol. A quick search in SciFinder gives numerous reaction conditions that allow for up to $98~\%$ yield of the para-substituted product, depending on which catalysts and sets of conditions you want to use. One of the simplest sets of reagents is probably Ducry’s and Roberge’s Angewandte paper which, however, uses a microreactor. While their best selectivity is $1.2:1$, their yields are way above $12~\%$ (which would be the corresponding yield for $6~\%$ of the para-product in a $1:1$ mixture).[1]

Having an acetate group rather than a hydroxy group greatly modifies the properties of that particular oxygen atom. I synthesised a phenol and the corresponding acetate and the chemical shift of the ortho carbon changes by $7.5~\mathrm{ppm}$ upon acetylation while the meta carbon remains essentially unshifted. Therefore, I wish to challenge your assertion that acetate be ortho and para-directing — please back this with appropriate references. However, I will acknowledge that an acetate group is less meta-directing that a carboxylic acid; therefore the preferred reaction site should still be ortho/para to the acetate.

Whatever the selectivity, I do not agree that nitration of acetylsalicylic acid or salicylic acid would be more efficient than nitration of phenol. Remember that the reaction you are trying to perform is an electrophilic aromatic substitution $\mathrm{S_EAr}$. For an aromatic species to be attacked electrophilicly, it should have a high electron density. The carboxyl and acetate groups are both electron-withdrawing and therefore hinder the reaction. It should be harder and require harsher conditions to nitrate a benzoic acid than a phenol. The choice of steps is therefore better in the originally proposed way, in my opinion.

I will agree that decarboxylation of a meta-nitrobenzoic acid should be easier than of salicylic acid. Since the yield of carboxylation is already rather acceptable already, I do not think that switching the order would provide any significant advantage here.

Finally, it should also be rather easy to increase the yield of the final-step acetylation. $18~\%$ needs a lot of work.

Reference:

[1]: L. Ducry, D. M. Roberge, Angew. Chem. Int. Ed. 2005, 44, 7972. DOI: 10.1002/anie.200502387.


Homemade filter paper I used normal paper from my notebook to perform filtration. The filtration is 100% perfect, but takes incredibly long to happen, because the paper is too dense.

I'm not planning to buy filtration papers. They're cheap, but I don't filter something every day, so they'd be laying around useless. I would prefer to be able to use some household item as a filtration paper.

One idea I had was to apply some substance on normal writing paper to make it softer.

I have tissues which are very soft, but these are too soft and will leave paper chunks in my filtrate.

Do you have any idea what should I use for filter paper, or how could I alter normal paper to suit my needs?

Apparently, the particles of your solid are either very small, or / and not crystalline.  This may complicate filtration, though this is one of the reasons filter papers used in the chem labs are sold in different grades of paper hardness.
A practical alternative to separate such a slurry from the liquid phase would be to substitute filter and funnel by a centrifuge.  With some luck you still find some with a manual crank (with conic tubes, about $\pu{20 mL}$ each). Based on own experience, this often is faster than setting up and then performing a suction filtration over a glass filter. On small scale, it is simpler to wash the slurry, and clean up the tools, too. Just keep the opposite tubes balanced and don't mind the surprised gaze of your colleagues around.

(credit)


Why doesn't pH = pKa1 in the buffer zone for this titration? Below is an image from my textbook representing the titration curve for the titration of $5.00\ \mathrm{mL}$ of $0.010\ \mathrm M$ $\ce{H2S2O3}$ with $0.010\ \mathrm M$ $\ce{KOH}$, which has $\mathrm pK_\mathrm{a1} = 0.6$ and $\mathrm pK_\mathrm{a2} = 1.74$.

The Henderson–Hasselbalch equation implies that at the first half equivalence point, $\mathrm{pH} = \mathrm pK_\mathrm{a1}$, and at the second half equivalence point $\mathrm{pH} = \mathrm pK_\mathrm{a2}$, because there are equal changes of the acid and conjugate base.

This means that at the first buffer region b, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a1}=0.6$ and at the second buffer region d, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a2} = 1.74$. However, this is clearly not the case with the diagram.

What am I missing here? Is it because I have to account for the autoionization of water in the dilute solution? How would I do so?

Your acid dissociation constants are $10^{-0.6}$ and $10^{-1.74}$ whereas your thiosulfate species have only a total molar concentration of $10^{-2.00}$. When the dissociation constants of an acidic solute are greater than the solute concentration the solute is rendered like a strong acid by dilution.  If you have 0.01 M of an acid with a $pK_a$ of 1.74, and you use the equilibrium relation to determine how much of the acid is dissociated, you find that in fact most of the acid is dissociated.

The proposed use of the H-H equation then does not work because this equation assumes that aqueous hydrogen or hydroxide ions are only a small part of the total solute concentration.  When the solute becomes like a strong acid (or strong base) as above, it floods the solution with aqueous hydrogen or hydroxide ions and the assumption needed for the H-H equation fails.


Help understanding how "steric effects" are distinct from "electronic effects"? @jakebeal's excellent answer to Why do animal cells “mistake” rubidium ions for potassium ions? includes the following passage:

In the case of potassium versus sodium, which are both very important in biochemistry, a recently discovered mechanism in sodium-potassium pumps uses binding properties to grab both, then atomic size (via steric hindrance) to distinguish sodium versus potassium as described nicely in the answers to this question.

Wikipedia's Steric effects says:

Steric effects are nonbonding interactions that influence the shape (conformation) and reactivity of ions and molecules. Steric effects complement electronic effects, which dictate the shape and reactivity of molecules. Steric repulsive forces between overlapping electron clouds result in structured groupings of molecules stabilized by the way that opposites attract and like charges repel.

Since "the way that opposites attract and like charges repel" sounds "electronic" to me, why does one say that "Steric effects complement electronic effects" instead of simply being electronic effects?

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds
All the normal physical interactions we experience are arguably electronic. When you touch your desk, you feel force because of interactions between the molecules of the desk and the molecules of your hand because the molecules of one interact "electrostatically" with the molecules of the other.
But your hand and your desk are not bonded together.
When a noble gas condenses into a liquid, it has a finite volume. The atoms become liquid because there are weak attractive forces (not usually described as "bonds") but have a fixed volume because the strong forces between their filled electron shells repelling each other prevent the atoms packing together any more closely. This second effect is very similar to the force stopping you hand penetrating your desk.
Arguably this is all electronic interactions.
But, if you are a chemist, it is worth drawing a line because it makes it much easier to talk about the properties of molecules. Electronic interactions to a chemist are usually interactions transmitted via bonds in a molecule.
So a molecule like fluorescein (shown below) has an extended pi-system so the oxygens at either end of the three fused rings can interact via bonds (and the overall molecule has electronic transitions that make it a useful fluorescent dye).

But in a molecule like this substituted biphenyl the electronic interactions might drive the molecule to be planar (so the two pi-systems can communicate) but the steric size of the 4 methyl groups prevent that because the methyl groups would come too close in 3D space to allow a planar configuration.

So the molecule will not have two aromatic rings in the same plane but will have significant angles between the rings. A "space filling" view (below) shows the methyl groups take up too much space to allow the rings to be coplanar.

Chemists think of the forces trying to make this molecule planar as "electronic" as they are transmitted via the bonds but the forces preventing this as being "steric" interactions. They are more like the forces preventing your hand from penetrating your desk than they are alike the forces holding the molecule together.
It is a useful distinction even when the forces are internal to a molecule.


VO2 unexpected conduction $\ce{VO2}$ is a weird metal which becomes a conductor at temperatures higher than $\pu{67 ^\circ C}$, but below $\pu{67 ^\circ C}$, it is an insulator!

How is this possible anyway? This is against solid state theory and how are we supposed to create band gaps with it?

It's not really the same vanadium dioxide.  The 67°C temperature corresponds to a phase change.  Compare with the phase change in solid tin.


Major Product of Alkene in Bromine Water At school, we are taught that in bromine water, the following equilibrium is present:
$$\ce{Br2 + H2O \rightleftharpoons BrOH + HBr}$$
Therefore, when you add an alkene to bromine water, you don't get bromination, but rather the $\ce{BrOH}$ reacts instead and you get the following reaction (lets say for ethene for example):
$$\ce{C2H4 + BrOH -> C2H5BrO}$$
However I have looked at a few organic textbooks such as Clayden and they don't mention anything about $\ce{BrOH}$ forming in bromine water and then adding across the double bond. It says that the product of the above reaction will be $\ce{C2H4Br2}$.

In reality, both reactions probably both occur to some extent (also, for the first reaction, wouldn't you expect some of the $\ce{HBr}$ to add across the double bond as well?), but which one is the most prominent, or in other words, what is the major product?

@orthocresol mentioned the main points in his comment but I would just like to expand a little more. The main electrophile in bromine water is $\ce{Br2}$ and this is what does the initial attack on the double bond, forming a cyclic bromonium ion. The bromonium ion is a strong electrophile and so it will react with any nucleophiles that are around. Since the water molecules will far outnumber bromide ions in the solution then the major nucleophile is water, even though it is not as good a nucleophile as bromide. The nucleophile attacks in an $\mathrm{S_N2}$ reaction, leading to a halohydrin product, after loss of a proton. This mechanism also explains the trans stereochemistry observed in the product.

As @orthocresol mentioned, the equilibrium constant for the formation of $\ce{BrOH}$ will be very small and so $\ce{BrOH}$ will be negligible as an electrophile. This is explained in a bit more detail in @Jan's answer here.


A problem regarding the first law of thermodynamics 2 moles of an ideal gas are in a closed system with a temperature of $25\ \mathrm{^\circ C}$.
The gas is cooled under constant pressure and afterwards heated gas under constant volume.
During the cooling process the gas has lost $1000\ \mathrm J$ of heat to the environment.
It is known that the molar heat capacity of the gas is $C=25\ \mathrm{J/(^\circ C\ mol)}$
What is the work done by the system during the cooling and heating combined?

Official answer: $333\ \mathrm J$.
The problem: I understand how to get to this solution, however I don't understand why an alternative method yields an incorrect answer. Here's what I tried:
First, during the heating process there is no work done because the volume is constant.
$Q=-1000\ \mathrm J$
We know that:
$\Delta E=\frac32\cdot n\cdot R\cdot\Delta T$
Now, $Q=n\cdot C\cdot \Delta T$, therefore:
$-1000=25\times2\times\Delta T$ and therefore $\Delta T=-20$
So $\Delta E=\frac32\cdot n\cdot R\cdot\Delta T=\frac32\times2\times8.314\times(-20)=-498.84\ \mathrm J$
By the first law of thermodynamics: $\Delta E=Q+W$, therefore $-498.84=-1000+W$ and $W=501.16\ \mathrm J$
Could anyone please tell me where is my mistake?

The question didn't specify which process the given molar heat capacity was specified for. So we will assume that it is molar heat capacity for the first process. The first process is isobaric so $$q=nC_p\,\Delta T$$
You calculated $\Delta T$ correctly. The only mistake you did was assume that the gas was monoatomic. The expression for $\Delta U$ you wrote was $\Delta U=\frac32nR\,\Delta T$. This expression would be correct if this was a monoatomic gas. But we don't know that do we?
The correct expression to use here is $$\Delta U=nC_V\,\Delta T$$
It is given that molar heat capacity for first process (constant pressure) is $C_p=25\ \mathrm{J/(^\circ C\ mol)}$ and  we know that $$C_p=C_V+R$$
So you can calculate the value of $C_V$ and use it to get the value of $\Delta U$ which I calculated to be $$\Delta U=667.2\ \mathrm J$$
Plug this into first law of thermodynamics and you will get the correct answer.
Edit: I used the sign $\Delta U$ for Change in Internal Energy intead of $\Delta E$


Why May K Expressions Be Manipulated As Such? Why is the above massaging of various K expressions valid? Something in me says "Hess's Law." But then doesn't that deal with enthalpy? What's the true reason?

Your intuition is correct. The reason equilibrium constants from different reactions can be strung together and multiplied comes from a generalization of Hess' law which covers Gibbs' free energy changes (apparently also called Bordwell thermodynamic cycles). Enthalpy is a thermodynamic function with a special property; it is a state function. That is, enthalpy changes in cycles are determined solely by the initial and end states of a system, and not the trajectory in which the initial state travels to the end state (as happens for functions such as work $W$ and heat $q$). Any state function can be manipulated in the way enthalpy is used for Hess diagrams. Thus, by adding reactions, you can add their respective reaction free energy changes $\Delta_rG$.

Let us now take two reactions, with free energy changes equal to $\Delta_rG_1$ and $\Delta_rG_2$. There is a relationship between a reaction's free energy change and its reaction quotient $Q$, namely:

$$\Delta_rG = \Delta_rG^o + RT\ ln\ Q$$

, where $^o$ indicates standard conditions. For reactions 1 and 2, we therefore have:

$$\Delta_rG_1 = \Delta_rG^o_1 + RT\ ln\ Q_1$$
$$\Delta_rG_2 = \Delta_rG^o_2 + RT\ ln\ Q_2$$

Adding the reactions, we obtain a reaction 3 where the same formula applies:

$$\color{\red}{\Delta_rG_3} = \color{\navy}{\Delta_rG^o_3} + \color{\green}{RT\ ln\ Q_3}$$

However, since we can add the free energy equations for reactions 1 and 2, reaction 3 is also represented by:

$$\Delta_rG_1 + \Delta_rG_2 = \Delta_rG^o_1 + \Delta_rG^o_2 + RT\ ln\ Q_1 + RT\ ln\ Q_2$$
$$\color{\red}{\Delta_rG_1 + \Delta_rG_2} = \color{\navy}{\Delta_rG^o_1 + \Delta_rG^o_2} + \color{\green}{RT\ ln\ Q_1 Q_2}$$

Clearly, we have $Q_3 = Q_1 \times Q_2$. Considering the equilibrium conditions of all reactions ($\Delta_rG = 0$), then the reaction quotients are defined as the equilibrium constants $K$, and similarly one finds $K_3 = K_1 \times K_2$.


Why can wood burn all the way to charcoal instead of just creating a passivating charcoal layer? I know burning to charcoal is actually incomplete combustion because charcoal can also burn. When a strike anywhere match burns, it burns all the way to charcoal but the charcoal doesn't further burn. How can all of it burn to charcoal? According to https://en.wikipedia.org/wiki/Pyrolysis, the heat causes pyrolysis which undergoes flaming combustion generating more heat. Once some charcoal has already formed, why doesn't it just passivating the wood blocking further endothermic pyrolysis. I know wood is porous but at the small scale, should't it's surface become passivated by charcoal blocking further pyrolysis? Does it occur for a similar reason that iron can continue rusting after it already formed a rust layer by producing a porous rust layer? Could nonporous single crystal cellulose also burn to charcoal? If so, can you give me a detailed mathematical description of how it would continue burning to create a thicker charcoal layer and how iron can just keep on rusting?

Charcoal doesn't passivate and the results you see depend on the physical structure of the fire

The difference between a match and a larger wood fire has little to do with the formation of charcoal and more to do with the circumstances of the reaction. And the charcoal isn't really passivating in the reaction on a match.

Wood consists of a complex mix of chemicals as well as the core cellulose, which is a carbohydrate. Some of the ingredients, like the sap and other volatile components, burn easily. Cellulose burns in multiple steps one of which involves pyrolysis where some components are driven off quickly leaving mostly charcoal. Charcoal itself will burn but it takes more energy to ignite and therefore more time to get going. It isn't so much that charcoal passivates, it is just that it is the last component to burn.

The difference between a match and a larger wood fire isn't about passivation but about the scale of the fire. There isn't enough energy in a match head to drive combustion all the way so only the easy-to-burn components ignite (unless you hold the match upside down so the flame spreads up the matchstick when the heat from the tip is enough to cause much more extensive combustion). In a larger wood fire there is enough energy in the flame to cause more complete combustion. So the volatile components burn quickly but leave enough heat for the charcoal to burn as well. Since heat penetrates charcoal very well, the volatile components under the charcoal will tend to ignite before the charcoal itself (it isn't doing much to passivate the wood underneath). In a badly organised fire the overall heat may never be enough to set off the charcoal.

So the issue isn't passivation just a reflection of the order the components of wood burn and the physical structure of the fire. Charcoal isn't passivating, just the last component to burn and, if the fire isn't well constructed, there may never be enough heat to burn it. Much more of a match will burn if you drive the flame along the stalk by holding it upside down and a well organised fire will burn the whole way to ash.

Also controlled pyrolysis (which is used to create charcoal deliberately) relies on controlling the amount of air available to the work being pyrolysed. If there were enough air, the charcoal would burn but the conditions for pyrolysis involve restricting the air so the end-product cannot burn. Charcoal doesn't passivate anything when there is enough air (which is why you can cook on a barbecue).


Predicting product of amine elimination From section 'Nitrogen Containing Compounds'

I spent quite a lot of time here but could not figure out the answer. It's not mentioned in my school text book either. The solution given in my book is:

I'd like to know:

Is the answer provided (c) correct?
How do I know that heating that compound in the presence of a base results in the formation of that product? We've got plenty of cases in our textbook (Tollen's Test, Fehling's Test and the aldol reaction, to name a few) where we heat organic compounds in the presence of a base, but none of them go by what happened in this question.

In chemistry, you don't solve things like that; you just recall the right class of reactions (Hofmann elimination in this case) and apply that knowledge. You can hardly expect to rediscover that from the first principles. Knowledge about other classes of organic compounds and their behavior in the presence of a base is not particularly helpful either. It is not until much later that those seemingly random pieces of data start to fit into a bigger picture.

As for the mechanism, it is all there in your second picture. Indeed, that $\ce{OH-}$ would readily snatch a proton from an ammonium cation $\ce{NH4+}$, only there is no ammonium, it is a substituted ammonium, and we don't have any H linked directly to N. So we go for the next best thing and grab a proton from some carbon in $\beta$ position, simultaneously with the formation of a double bond.

Note the reversed regiochemical preferences as compared to Zaitsev-type elimination, which has a different mechanism, usually involving the appearance of a carbocation. (Not that it matters much in your case, since we only have one option anyway.)


Is sulfuric acid or hydrochloric acid stronger? Of my belief, sulfuric acid might be a little stronger than hydrochloric acid. Because even though they both are strong acids, one mole of sulfuric acid produces two times as much $\ce{H+}$ as one mole of hydrochloric acid.

Am I wrong?

When using $\mathrm{p}K_\mathrm{a}$, one typically does not consider multiple dissociations for polyprotic acids, as the acidity of the conjugate base ($\ce{HSO4-}$ in the case of sulfuric acid) can and should be measured (or calculated) separately. Using that metric, $\ce{HCl}$ is stronger (at least per Wikipedia's acid strength page). One needs to keep in mind that $\mathrm{p}K_\mathrm{a}$ is solvent-dependent, and that the values typically given are for relatively dilute solutions. Acidity can sometimes vary widely with concentration ($\ce{HF}$ being a notorious example). It should also be noted that experimentally measuring acidity for very strong acids is actually quite difficult for numerous reasons.

On the other hand, if you evaluate acid strength by, say, the $\mathrm{pH}$ of the resulting solution, then you'd need to take subsequent dissociations into account. Using the crude approximation that an aqueous "strong acid" dissociates completely, then any diprotic "strong acid" is going to be stronger than a monoprotic one for equal concentrations (assuming a non-zero $K_\mathrm{a}$ for the second dissociation). Of course, this is not accurate, and a proper calculation would take into account not only $K_\mathrm{a}$ values for all dissociations, but also water's auto-ionization. I suspect that $\ce{HCl}$ would still be stronger, given that the $K_\mathrm{a}$ for the dissociation of $\ce{HSO4-}$ is fairly small, but one would need to actually crunch the numbers to be sure.


Are the C-C bond lengths different in 2-methyl propene When we were first taught resonance it was with the example of benzene and I myself justified it by thinking that the double bond oscillates because no two carbon atoms are special and hence there is no reason there will be a double bond between any two and not any other two. I now know that's not true. But just to avoid any latent confusion, can you verify my reasoning is wrong by answering this question? By my method, all three bonds will have same length and bond order 4/3

The cases of benzene and 2-methyl propene are different.

In benzene, all carbon atoms are equivalent. One can also imagine the electron resonance.

In case of $\ce{CH3-C(CH3)=CH2}$ , the electrons have no place to move to create a delocalization cloud.

Additionally,  it would have to switch a hydrogen atoms between 3 atoms,
or create highly unstable $\ce{C=CH3}$

Such molecule would be highly reactive.


How to make sense of the crystal systems names I cannot fix the names and structures of the seven crystal systems in my head. There's a thousand books explaining their geometry and it's totally fine and understandable, no problem. But what do the names mean, etymologically?

Cubic
Tetragonal
Orthorhombic
Monoclinic
Hexagonal
Rhombohedral or Trigonal
Triclinic

OK, cubic is simple enough, hexagonal too. I know that "-eder/-hedral" means "side" and "-gon" means "angle". Why is a system with three different edges and angles triclinic? What are the rationales behind each of the names?

A rhombus has sides of equal length. Orthorhombic has only right angles (orthogonal), but the sides a,b,c have different lenghts. So why "rhombic"?

Monoclinic and triclinic

Ivan (in the comments) is spot on about these two. Among all the systems, these are the only ones with angles that are not fixed by symmetry, so you actually have to measure them.

Rhombohedral or Trigonal

These are two different settings of the same lattice. The unit cell can either be a prism (with angles of 90, 90 and 120 degrees, blue lines in the figure) or a rhombus (with all three angles equal, and all three sides equal, black lines in the figure).

In general, there are infinite choices for unit cells (you just need three translation vectors that are not linearly dependent, so pick an origin on a lattice point and three other lattice points and its a unit cell). The conventional choices usually have more of the symmetry elements along the axes, face diagonal or body diagonal, and keep the volume of the unit cell fairly small.

Orthorhombic

This used to be called pseudo-orthorhombic.

Again, you have a choice of unit cell, and the more uncommon one is a rhombohedral prism. When you switch from all right angles to a rhobohdral prism setting, face-centered turns into primitive and primitive turns into face-centered.

Here is a quote from Wikipedia:

  In the orthorhombic system there is a rarely used second choice of crystal axes that results in a unit cell with the shape of a right rhombic prism;2 it can be constructed because the rectangular two-dimensional base layer can also be described with rhombic axes. In this axis setting, the primitive and base-centered lattices swap in centering type, while the same thing happens with the body-centered and face-centered lattices.

Historically, the first data available to crystallographers (and minerologists) were the shapes of crystals. Orthorhombic crystals can have faces that are at 90 degree angles or not, depending on the kinetics of growth. Pictured below are three crystal forms with underlying orthorhombic symmetry, with the 2nd one showing is a shape with (+/- 1, +/- 1, +/-1) faces labeled orthorhombic pyramid:

Cubic, Hexagonal and Tetragonal

See OP's comments for those.

History

The German Wikipedia article on crystal systems has the terms Bravais used in 1886:

Assemblages terquaternaires (cubic)
Assemblages sénaires (hexagonal)
Assemblages quaternaires (tetragonal)
Assemblages ternaires (trigonal)
Assemblages terbinaires (orthorhombic)
Assemblages binaires (monoclinic)
Assemblages asymétriques (triclinic)


Need reactants for a rocket I am doing a project in chemistry at the moment to build a rocket out of materials that can be easily bought and that react strongly to create thrust.
I was wondering if anyone knew a good chemical formula for my rocket fuel? I know lots of good fuels and reactants used in amateur rocketry however commercial fuels have been banned from the project.
Any help would be appreciated.

Put some potassium nitrate $\ce{KNO3}$ plus powdered charcoal and sulfur in a mortar. Wet it carefully and grind the whole until you get a black homogeneous paste. Let it dry overnight in the mortar. The obtained black gun powder may be detached from the mortar with a wooden spoon. It is a safe rocket fuel. The proportions of the powders are defined by stoichiometry. They must allow the chemical reaction : $$\ce{2 KNO3 + 3 C + S ->  K2S + 3 CO2 + N2}$$
I have done it many times with my students. It's a good exercice of stoichiometry. Never had an accident ! And remember : Never ! Never grind dry powders ! Grind as wet powders as possible ! In case of doubt add more water ! Pastes are even a better choice for grinding purposes !


Finding equilibrium constant for the reaction between non-stoichiometric mix of gaseous components $$\ce{CH3OH(g) + NOCl(g) <=> CH3ONO(g) + HCl(g)}$$
The volume of the container is $\pu{433 cm^3},$ $T = \pu{50 °C}.$ Methanol was added until the pressure was $\pu{50.1 mbar},$ then $\pu{0.059 g}$ of $\ce{NOCl}$ was added. At equilibrium the partial pressure of $\ce{NOCl}$ is $\pu{27.6 mbar}.$
Calculate $K.$

I assumed that $\pu{50.1 mbar}$ was the partial pressure of methanol (I do not know if this is correct) and calculated the initial partial pressure for $\ce{NOCl}$ from the ideal gas law:
$$p = \frac{nRT}{V} = \frac {(\pu{0.059 g}/\pu{65 g/mol})\times 8.314\times 323}{\pu{4.33E-1}} = \pu{5.63 Pa} = \pu{5.63E-5 bar}$$
Then
$$
\begin{array}{cccc}
&\ce{CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)} \\
&\pu{0.0501 bar} && \pu{5.63E-3 bar} && 0 && 0 \\
&\pu{0.0501 bar} - x && \pu{0.0276 bar} && x && x\\
\end{array}
$$
When I then solve for $x$ I get a negative value which isn't correct. Is $\pu{50.1 mbar}$ supposed to be the total pressure and if yes, how do I use that then? The correct answer for $K$ is $1.33.$

As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$:
$$
\begin{array}{lccccc}
&\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\
&\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) && 0 && 0 \\
&\text{C} && -x && -x && x && x \\
&\text{E} && p_0(\ce{CH3OH}) - x && p_0(\ce{NOCl}) - x && x && x
\end{array}
$$
You are interpreting given conditions wrong.
The problem explicitly provides you with the initial partial pressure of methanol $p_0(\ce{CH3OH}) = \pu{50.1 mbar}.$
As for the initial partial pressure of nitrosyl chloride, it can be found applying the ideal gas law:
$$
\begin{align}
p_0(\ce{NOCl}) &= \frac{m(\ce{NOCl})}{M(\ce{NOCl})}\frac{RT}{V}\\
  &= \frac{\pu{0.059 g}}{\pu{65.46 g mol^-1}}\frac{\pu{8.314 J mol^-1 K^-1}\times\pu{323 K}}{\pu{4.33E-4 m^3}} \\
  &\approx \pu{5590 Pa} \\
  &\approx \pu{55.9 mbar} \tag{1}
\end{align}
$$
Note that methanol and nitrosyl chloride reaction mix is apparently not equimolar.
Also, the total pressure is not relevant for obtaining the solution.
Do keep in mind though it is not the initial partial pressure of the first component because the second gaseous reagent has been added afterwards.
The unknown $x$ can be found trivially since you are given the equilibrium partial pressure of nitrosyl chloride $p(\ce{NOCl}) = \pu{27.6 mbar}$:
$$
\begin{align}
p_0(\ce{NOCl}) - x = p(\ce{NOCl}) \quad\implies\quad x &= p_0(\ce{NOCl}) - p(\ce{NOCl}) \\
  &= \pu{55.9 mbar} - \pu{27.6 mbar} \\
  &= \pu{28.3 mbar} \tag{2}
\end{align}
$$
Finally we can write an expression for $K_p$ (by the way, here $K_p = K_c$ since $\Delta n = 0)$:
$$
\begin{align}
K_p &= \frac{p(\ce{CH3ONO})\cdot p(\ce{HCl})}{p(\ce{CH3OH})\cdot p(\ce{NOCl})}\\
  &= \frac{x^2}{(p_0(\ce{CH3OH}) - x)\cdot p(\ce{NOCl})}\\
  &= \frac{(\pu{28.3 mbar})^2}{(\pu{50.1 mbar} - \pu{28.3 mbar})\times\pu{27.6 mbar}} \\
  &\approx 1.33 \tag{3}
\end{align}
$$
Don't forget to do dimensional analysis to check yourself and never omit units in your calculation.
Also, keep an eye on significant figures: the number for the molar mass you used was too sloppy; instead of two figures $(\pu{65 g mol^-1})$ you had to go with four $(\pu{65.46 g mol^-1}).$


How does solution which electrodes are immersed in effect electrode potential? When we make an electrochemical cell, we dip the electrodes in the salt solution. How exactly does the electrode potential change as we dip the electrodes in the salt solutions?  Would different salt solutions change the effect of the electrode potential?

The short answer is that we want a cathodic reaction so we can use the metals with a positive standard reduction potential (because 0 is for $\ce{H2} $ reduction).

In reality however to really find a suitable electrolyte cyclic voltammetry is typically used and electrolytes with suitable electrochemical windows for the cell reaction are chosen.

The electrochemical window is the voltages where the current is zero in the picture above. If the wrong electrolyte is chosen then a full redox reaction happens(with the electrolyte participating) rather than half reactions at each electrode negating the need for the external circuit and creating heat


Are Cubic Equation of states applicable to Liquids? I was just going through a graph of isotherm given by cubic equation of state and noticed that isotherms passes trough the subcooled liquid region. Are these equation applicable to liquids?

Cubic equations of state are summarized here.  They can, indeed, describe both the liquid and gaseous state.

Under subcritical conditions, a given temperature and pressure will give three roots.  The largest is the vapor volume, the smallest is the liquid volume and the middle one is an unstable state that must exist along with the vapor and liquid roots.  By using appropriate integrated forms we can compare the free energies associated with these roots and thus identify which is more stable.  Or, we can identify when the free energies match and thus predict vapor-liquid equilibrium.  The catch?  The equations are of only limited accuracy, especially for liquids.  Some are better than others; for more details the reader can start with the Wikipedia reference given above (https://en.m.wikipedia.org/wiki/Equation_of_state).


Is the existence of chirality center equivalent to chiral? To make my question more clear, I want to ask two questions:

Is it possible for molecule to be achiral even though molecule has chirality center?

Is it possible for molecule to be chiral even though molecule doesn't have chirality center?

I thought of these questions because the definitions of chiral center and chiral do not have a one to one correspondence.
A chiral center is defined by carbon having four different ligands around it and chiral is defined by a compound being non-superimposable with its mirror image. They are quite related, but I think there will be counterexamples that satisfy one condition but not satisfy the other one.

The answer to both questions is yes.

In regards to the first question, meso compounds like tartaric acid have multiple chiral centers, but are not chiral overall because they have a mirror plane.

For the second question, a compound can be geometrically chiral without having a chiral center. A cool example class of compounds is helicenes, which are helical arrangements of aromatic rings. These compounds have axial chirality, where the stereoisomers are the left and right handed versions of the helix.


Urban air pollution and the smell of freshly cut lawn grass? Recently, all the lawns in the apartment complex that I live in have had their lawns cut - giving that pleasant smell of freshly cut lawn.

Now, I am aware that the smell arises from a chemical distress signal from within the plant itself - are the chemicals from freshly cut lawn contributors to urban air pollution?

Related: How to determine source of air pollution

The information page from the Compound Interest page of the same name describes it well (below).

A summary of the main points are that grass already emits small amounts of VOCs (volatile organic compounds) - mostly comprising of 6-carbon atom based aldehydes and alcohols.

When the lawn is cut, the main smell is from (Z)-3-hexenal, at which only 0.25 parts per billion can be sensed (smelled) by humans - this compound is short lived, hence why the smell is a freshly cut smell.

As a little twist, in research presented in the paper Air Pollution and the Smell of Cut Grass, the authors suggest that a significant proportion of aerosol VOC's in urban air pollution actually originate from the release of these chemicals when lawns are cut - and given the sheer number of manicured lawns in an urban centre, this starts to make sense.

Some of these released chemicals are highly photoreactive. The flux of VOCs is seasonal as well in many places, contributing to seasonal variations of VOC-based photochemical smog.

Within the article, VOC flux measurements taken in the urban area of Melbourne Australia indicate that 6% of the flux arises from uncut grass; however, this flux doubles when the lawn is cut. The combined flux is comparable to industrial sources.

It is suggested in the article, that

  within an
  urban airshed, during the growing season, about one third of the emissions of
  photochemically reactive VOCs come from grass and cut grass.


Molecular Orbital Theory and No. of bonds The order of filling of molecular orbitals $\ce{O2}$ by MOT is this :

But from it, how can I deduce that there is one sigma and one pi bond in an oxygen molecule ?

Overall bond order equals number of electrons that occupy bonding orbitals minus numer of electrons that occupy antibonding orbitals divided by two.

Now you can apply the same concept indivudually:

For s molecular orbitals, both electron pairs "cancel out".

For p molecular orbitals, you have 2 electrons in sigma bonding orbitals and none in sigma antibinding orbitals, which means (2-0)/2 = 1 sigma bond.
Also, you have 4 electrons within pi bonding orbitals and only 2 electrons within pi antibonding orbitals, which means (4-2)/2 = 1 pi bond.
This yields the result that you expected.


What chemical file formats are compatible with 3D printing? I would like to print some molecules rather than building them using a kit. What file formats work?

None of the chemical file formats are typically accepted by 3D printing services. The reason is that chemical file formats are fundamentally different from those describing 3D objects.

In the simplest case, a chemical file format describes a list of atoms and their positions (plus possible metadata). An example is the XYZ format, which stores exactly that. Other “chemical objects” which can also be described in some chemical file formats include bonds, protein structure, crystalline structure information (unit cell, symmetry, occupancy of each atomic position, …), etc.

3D file formats, on the other hand, describe a series of objects. In the simplest case (e.g., the STL format or the Wavefront .obj file format), the object itself is not stored, but an unstructured triangulated surface delimiting the object. In those, you can't describe a sphere, but only a tesselation of the sphere.

So, you need a conversion step to create a 3D representation of your chemical structure. Chemical file formats don't typically contain representation information: you need to pick the one you want (CPK, liccorice, protein cartoon, …) and use software to create a 3D scene. VMD, for example, allows export to STL format, which is suitable for 3D printing.


Why is copper a better conductor than iron? Metals are good conductors as the electrons in their outer shell are loose and can plunge out of the atom with the application of the slightest force (voltage).

Silver is the best metallic conductor, then comes copper.

Why is one metal better at conduction than the other one?
What determines the difference in conduction between 2 metals?
Why is silver a better conductor than copper, and why is copper a better conductor than iron if all of them have loose electrons?

This is, ultimately, a question on solid state physics rather than chemistry. Further, the OP indicates that they are in high school, which kind of limits the depth of the answer that might be useful to them. However, I will try to make a simple, yet detailed answer.

As atoms are brought closer and closer together, their electron clouds overlap and interact. When a crystalline solid is formed, one abandons consideration of the atomic energy levels. Instead, one seeks out the Bloch functions to describe the electonic states - the Bloch functions explicitly consider the symmetries of the crystal. Further, they represent extended electron states, those that extend across the entire crystal. They represent the allowed (energy and momentum) combinations of electrons.

For a metal, the highest Bloch states occur as a band of states which is only partially filled. Unoccupied states lie just above states with electrons. To move an electron about, it needs just a little bit of energy, and off it can go. (Fully filled bands do not allow conduction - for any electron going one way there is one going the other way. This needs a good grounding in solid state physics to fully grasp).

OK, in general then the conductivity should just be the number of electrons near the line between occupied and un-occupied. Kind of. This line is called the Fermi surface, and things get more complicated quickly. The shape and connectivity of the Fermi surface impact conduction dramatically. Lets take a deeper look at copper and iron.

Copper is a simple noble metal with a beautiful near-ideal Fermi surface (see Ashcroft and Mermin's Solid State Physics book, or C.Y. Fong et al., Comparison of band structures and charge distributions of copper and silver). It turns out that the Fermi surface of copper is almost exactly what one would expect from the free electron sphere. Electrons can move easily in any direction, the surface is fully connected, life is good.

Iron is much more complicated. Following along from J. Callaway and C.S. Wang, Energy bands in ferromagnetic iron one finds two major differences. First, because iron atoms indeed have individual magnetic moments, the overall band structure is split into two different spin states for the electrons. So, each spin state is separated from the others, reducing how each can move (relative to copper, were such splitting does not exist). Second, and more importantly, the Fermi surface looks nothing like a free electron sphere. Instead, for iron the Fermi surface is not fully connected, comprising instead of multiple 'pockets' of electrons that do not communicate directly with the other pockets (and each spin state has a different set of pockets). To move electrons around, they have to 'jump' from one pocket to another through scattering. This makes it much harder to move the 'free' electrons in Fe around to get net conductivity.

An analogy would be the difference of crossing a stream using either (1) a nice smooth bridge (copper) vs (2) jumping from rock to rock (iron). The band structure for copper just makes it so much easier, so the resistivity is much lower.


Differentiating between Prins reaction / ene reaction experimentally The Prins reaction – stepwise addition of an alkene to a carbonyl compound – leads to the same product as a concerted ene reaction. This is illustrated here with tin tetrachloride as a Lewis acid, which (as far as I know) lowers the activation energy of both pathways:

How can these two mechanistic pathways be differentiated experimentally?

One thing that I can think of right now would be the stereochemistry of the alkene formed - presumably in a concerted pathway, the cyclic transition state (envelope-like, if I remember correctly?) the substituent R' would be positioned equatorial or pseudo-equatorial. However, I'm not confident enough to undertake an analysis of the stepwise reaction.

I'm open to hearing about other possibilities, e.g. kinetic studies or spectroscopic studies. I suppose the free carbocation in the Prins reaction could lead to a more negative coefficient $\rho$ in a Hammett analysis, compared with a stepwise mechanism?

I'd also be glad if somebody could provide an example in the literature, if it exists. (I confess to not having looked this up - been very busy.)

The traditional Prins reaction was run in aqueous acid (H2SO4, HCl) with formaldehyde. The carbocation formed by addition of the alkene was trapped by a nucleophile, or if E1 elimination occurred, the most stable alkene formed. The Lewis acid-catalyzed ene reaction is intramolecular without nucleophile capture and the less stable alkene is formed, at least when an isopropenyl group is present in the reactant. Two examples follow:

Malonaldehyde 1, which has two enantiotopic aldehyde groups, reacts via transition state 2 to produce alcohol 3 as the major product. The cyclopentadiene ring is in effect a t-butyl group that prefers being equatorial in the transition state. Note the exocyclic double bond as opposed to a more stable endocyclic position. The hydroxyl group is axial. Aldehyde 4 follows the same type of pattern forming axial alcohol 5. If an open transition state were operable, the equatorial alcohol would be anticipated.
[PS: You can Google much of this information.]

1) J. Am. Chem. Soc., 1990, 112, 2149-2158

2) J. Am. Chem. Soc. 1984, 106, 718-721


How does the redox of a battery happen? I tried to understand how does a battery work. After doing some research, I knew chemical reactions inside are key. But when I came across the reactions inside a zinc carbon battery, almost all websites say "Zinc is oxidized and gives out two electrons".

Why does this reaction happen in the first place? Is it the result of the reaction between the electrolytes and the zinc? And I'm confused what exactly do electrolytes do because I saw they are described as "catalyst". I only understand they allow the flow of ions but prevents the flow of electrons.

Thank you and if what I say above is wrong, please correct me.

The zinc-carbon cell uses a zinc anode and a manganese dioxide cathode. It is manganese dioxide that takes part in the reaction, not carbon. Carbon is added to the cathode to increase conductivity and retain moisture(catalyst). It also contains a layer of ammonium chloride. Following is the overall reaction.
$$ \ce{Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + ZnCl2 (aq) + 2NH3(aq)} $$
This is why the half reactions are as follows,

Anode: $\ce{2Zn -> Zn^{2+} + 2e–}$

Cathode: $\ce{2NH4+ + 2MnO2 + 2e– -> Mn2O3 + H2O + 2NH3}$

An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. Here, ammonium chloride solution acts as the electrolyte immobilized in the form of a paste. Basically its a chemical medium for flow of charge.


Relationship between normality and molarity I found this formula for calculating the normality from the molarity:
$$\text{Normality} = n \cdot \text{Molarity}$$

with
$$n = \frac{\text{Molecular weight}}{\text{Equivalent weight}}$$

But in one book I found that if equivalent weight is equal to half of the molecular weight then
$$2N = M$$

Under the given condition, I am confused if the above statement is true.

Am I following a factually correct book for solving problems on volumetric analysis?

The normality of a solution is the molarity of the solution multiplied by the number of equivalents per mole:

$$\mathrm{N} = \mathrm{{eq\over mol}}\cdot\mathrm{M}= \mathrm{{eq\over mol}}\cdot\mathrm{{mol\over L}}=\mathrm{{eq\over L}}$$

Example: Calculate the normality of a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$. There are 2 equivalents per mole, from the reaction

$$\ce{H2SO4 -> 2H+ + SO4^{2-}}$$

which yields 2 moles of $\ce{H+}$ for every 1 mole of $\ce{H2SO4}$.

Using the relation given above, we find that a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$ is 2 $\mathrm{N}$:

$$\mathrm{{2\;eq\over mol}}\cdot\mathrm{1\;M}=2\;\mathrm{N}$$

One resource you might find useful in order to check your math is Sigma-Aldrich's Normality & Molarity Calculator. Additionally, there are details relevant to the definitions of molarity and normality that you might find clearer that the ones you've conveyed in your question.


Facial selectivity of iodonium formation in an iodolactonisation I am trying to understand the stereochemistry of iodolactonisation of this cyclohexene. I can't work out if the iodonium ion would form on top or the bottom.

There is $\ce{CH2COO-}$ and $\ce{Me}$ pointing up, and $\ce{CH2OTBS}$ and $\ce{Me}$ pointing down. My guess would be iodine attacks from the top because there is a bulky $\ce{TBS}$ group pointing down as well as a methyl, however there is a semi-big $\ce{CH2COO-}$ group pointing up right next to the double bond. Would that have more of an effect as it is nearer to the double bond?

Here, it does not matter which face is preferred. The formation of the iodonium ion is reversible and it can be formed on both faces. However, the iodonium ion can only be attacked by the carboxylate if it is formed on the opposite face from the carboxylate, which is the bottom face in your diagram. Attack on the top face is said to be unproductive because it cannot lead to iodolactonisation.


Where does the additional hydrogen in the Fischer depictions of glucopyranose come from? My book shows this figure of Fischer formula of D-glucose:

I don't understand this figure, and I wonder why the upper carbon atoms in both the right and left formulas have 5 bonds.

And from where comes this hydrogen atom in the upper left side of the figure (attached to α-D-glucopyranose).

When turning D-glucose to Fischer formula, it seems there's an additional hydrogen atom, isn't it? Why and how does this happen?!!

Could someone explain how to get from the aldehyde form of glucose (the middle structure) to the β-D-Glucopyranose and the α-D-Glucopyranose?

Throw away that book, you have found a very terrible error. Okay the rest of the book may still be good, but at least black out that picture or stick a correct version of the picture over the top.

In the image below, you can see four depictions of α-D-glucose taken from Wikipedia (where a full list of authors is available). From left to right these are: Fischer projection, Haworth projection, chair depiction and ‘flat circle’ (a depiction which follows the guidelines for 2D representations of 3D structures).

As you can see, there is no additional hydrogen on the anomeric carbon and it is not pentavalent.

The mechanism of ring-closure is also mentioned in sufficient detail on the corresponding Wikipedia page to which I hereby point you. Basically, the 5-hydroxy group attacks the $\ce{C=O}$ π system as a nucleophile. Proton transfer will lead to the cyclic product. (In actual aquaeous solution, the carbonyl oxygen will be protonated immediately from a nearby water molecule and one will also take away the hydroxy group’s excess proton, so the two are not the same.)


How do CO2 Fire extinguishers procure their CO2 How do CO2 Fire extinguisher manufacturers get such a large amount of Carbon Dioxide? Do they get it through a chemical reaction or burning hydrocarbons?

They just buy it from companies producing CO2.
Actual sources may be sequestration from combustion (e.g. power plants) or fermentation (e.g. ethanol production) processes.


How to make foggy and cloudy plastic fully transparent? I used sandpaper of different grits to sand the surface of plastic. The surface is smooth but it looks dull after sanding. I read that one could use toothpaste to remove the foggy look but it did not work. I also tried vinegar with baking soda. When the plastic was inside the vinegar+baking soda bath, it looked transparent and shinny. However, when I took it out to wash it with water to clean the smell, it looked shinny and transparent when being washed but it became foggy and cloudy again after washing. I put it back to vinegar+baking soda bath. It became transparent again but when I used a clean microfiber cloth to clean it, it became foggy again. Looks like some kind of thin film on it. Putting it in vinegar+backing soda bath for about 8 hours did not help. What caused the foggy/cloudy look? Would you please tell how to I make the plastic fully transparent?

The tiny scratches on the plastic surface reflect and diffract light in different directions, so edges of objects behind the plastic become distorted, and the clear areas of the plastic are less clear.
There are several ways to make the plastic more transparent.

Use a finer grit sandpaper - really really fine: 1500 grit or finer, if you can get it. Plastic abrasive sponges are more gentle and can be used wet, so may create smaller, less deep scratches.

Use toothpaste to "sand" the plastic in the first place. You have to get the paste to be pasty thick, not watery thin. It will reduce scratches, but if you don't put deep scratches on in the first place, it will be easier to shine up a plastic surface. Toothpaste will reduce the shininess of a perfect surface, but usually will improve an ordinary surface. There are commercial pastes for cleaning and shining stovetops that are more abrasive, and you can make your own with sodium bicarbonate (no vinegar) - but you have to give it some energy, some elbow grease, and it takes a while. You could use a buffer wheel on a power drill.

Filling in the tiny scratches with a liquid helps even out the surface, but water evaporates and the effect does not last. An oil (e.g., silicone) or petroleum jelly (very thin layer) might last long enough for you - days, maybe weeks.

Depending on the plastic film, you might be able to heat it slightly to allow the scratches to deform slightly and become slightly wavy instead of sharp valleys and projections. This will require a lot of experimentation to get enough heating without distorting the plastic beyond your requirements. Hot air or hot smooth metal, or perhaps hot water or steam might work.

The least likely method to work is using a solvent to smooth out the surface. You might get some immediate effect that looks like an improvement, but any solvent that dissolves the very top surface will also diffuse into the plastic and when it evaporates, you will have a plastic that is disturbed even deeper, and will be foggier. Working this approach further suggests that you could use what would be better described as a plasticizer, a high molecular weight liquid that is compatible with the plastic, but does not migrate (out, usually, but in here), buff the surface and quickly clean the plastic to remove all the solvent/plasticizer. Now, since you probably don't have access to different kinds of plasticizers, you could try some vegetable oil plus a little of a more powerful solvent, like a few percent of alcohol, or mineral spirits, or even acetone, and you could add some sodium bicarbonate for a little abrasion, and use a buffer wheel.


What's this glassware? Can anybody tell me what this glassware with a stem, bulb and long tube is?

The glassware shown is a drying tube. The bulb holds cotton or glass/quartz wool while the tube part holds a drying media such as anhydrous calcium sulfate to keep moisture out of the vessel during moisture sensitve reactions.


Recrystallizing Sodium Acetate from solution I made sodium acetate using the following reaction:

$$\ce{NaHCO3 + CH3COOH ->CH3COONa + H2O + CO2}$$

The resulting solution of sodium acetate isn't pure, it has impurities. I want to crystallize the sodium acetate to extract the pure crystals. I crystallized some if it in the fridge, sodium acetate has a higher freezing point than water but I would like to know the technical side of it. Do temperature and saturation affect the crystallization and how can I crystallize sodium acetate reliably?

The Purification of Laboratory Chemicals by Perrin and Armarego states about sodium acetate this procedure:

  Cryst[alise]d from acetic acid and pumped under vacuum for 10h at 120$^\circ$.
  Alternatively, cryst[alise]d from aqueous EtOH, as the trihydrate. This material can be converted to the anhydrous salt by heating slowly in a porcelain, nickel or iron dish, so that the salt liquefies. Steam is evolved
  and the mass again solidifies. Heating is now increased so that the salt melts again. (NB: if it is heated too strongly, the salt chars.) After several  minutes, the salt is allowed to solidify and cooled to a convenient
  temperature before being powdered and bottled (water content should now
  less than 0.02%).


Why does Pb normally have an oxidation number of +2? Why does $\ce{Pb}$ have an oxidation number of $2^{+}$ while carbon and other elements in the same group have an oxidation number of $4^{+}$?

Furthermore, why doesn't carbon also have a $2^{+}$ oxidation number, as can't it also give up its two electrons in the p-suborbital to have a full s-suborbital? Why does it prefer to gain four electrons instead to fill the p-suborbital?

Similarly, why can't $\ce{Pb}$ gain four electrons just like carbon instead of giving up two?

Finally, doesn't this mean that $\ce{Pb}$ forms ionic bonds? Why doesn't carbon do the same?

Why can't lead gain four electrons? The lower you go on the periodic table, the lesser you will find elements that have a more non-metallic character (so with this I mean elements that will have a lower tendency to gain electrons). Just visualize a lead atom with a very small nucleus and a large area surrounding the nucleus, filled with electrons. Now, the more you go away from the nucleus, the weaker the positively charged nucleus can 'pull' on the outer , negatively charged, valence electrons. Elements that haven't got a very tight grip on these valence electrons, will lose their valence electrons. However, elements with a small atomic radius (such as Fluorine) will have such a tight grip on its valence electrons, that it will even be able to snatch electrons from other atoms.

Of course the electron configuration of Radon is stable, because it has fully filled orbitals and it has a noble gas configuration, but you need to keep the nuclear forces in mind. This you can also see by looking at the electronegative value of lead : it is 1.87 (Pauling scale). Now compare it to carbon, which has 2.5. The lower you go in the same group, the lower the electronegativity goes and the more metallic character you will get. Elements with more metallic properties (such as lead) will mostly form ionic bonds and elements with more non-metallic properties (such as carbon) will mostly form covalent bonds. There exists a rule to predict (however it isn't always correct) if a compound will be covalently/ionically bonded : if the difference in electronegative value is higher than 1.66, then you have an ionic bond and if it is lower than 1.66 then you have a covalent bond.

Why does lead have an oxidation number of 2+ while carbon and other elements in the same group have an oxidation number of 4+? Just to correct you on this : lead does have an oxidation state of 4+ and elements above such as carbon and tin also have 2+ as an oxidation state. Just look at the electron configuration : the outer $\ce{s^2 p^2}$ orbitals will be able to lose 2 or even 4 electrons, or even gain 4 electrons. In fact these are all possible oxidation states of the elements in the carbon-group:

   Carbon 4, 3, 2, 1, 0, −1, −2, −3, −4
   Silicon 4, 3, 2, 1 -1, -2, -3, -4
   Germanium 4, 3, 2, 1, 0, -1, -2, -3, -4
   Tin  4, 3, 2, 1, -4
   Lead   4, 3, 2, 1
   (source : wikipedia)

Please do note the tendency to have more positive oxidation states, the lower you go in the group. (the reason why? see the beginning of the answer) Note : not all of these oxidation states are (very) stable.

Also have a look at this picture:

Notice the metallic (shown by red) character increases the lower you go on the periodic table.


Solid and Liquid Equilibrium I understand why solids and liquids are not included in the $K_{eq}$ expressions. However, I'm wondering what the $K$ value for a reaction involving only solid or liquid reactants and products looks like. Theoretically, it'd be 1 no matter what since $1^n = 1$. Is this right? Also, do all such reactions go to completion?

You could calculate an equilibrium constant at a given temperature for a reaction from the standard free energy change ($\Delta G^{\circ}$) for any reaction using:

$\Delta G^{\circ} = -RT \ln K$

$\Delta G^{\circ}$ can be calculated for reactions using tabulated standard free energies of formation ($\Delta G_{f}^{\circ}$) using the following equation:

$\Delta G^{\circ} = \sum{\Delta G_{f}^{\circ} \mbox{(Products)}} - \sum{\Delta G_{f}^{\circ} \mbox{(Reactants)}}$

If the reaction only contains solids and liquids, then K will still depend on the ratio between the product and reactant concentrations.  However, these concentrations will not change as a reaction moves towards equilibrium, which implies that the reaction quotient ($Q$) remains constant


Is the valency of carbon satisfied in graphite? Graphite has hexagonal parallel planes. In a hexagonal structure the carbon atom has three bonds. Since the valancy of carbon is 4 is the valency satisfied in graphite?

In the hexagonal graphite structure the carbon atoms are $\ce{sp^2}$ hybridized, just like in benzene.  In graphite, this p-orbital is used for bonding just as it is in benzene, resulting in an extended pi system in the graphite structure.  Graphite is really just a large number of benzene rings annelated together to form a continuous, ring structure.  Just as carbon satisfies its valence requirement of 4 in benzene (3 sigma bonds and 1 pi bond), carbon's valency requirements are satisfied the same way in graphite.


Make pyridine from furfural Furfural $(\ce{C5H4O2})$ has been used as a feedstock for synthesis. Furfurylamine $(\ce{C5H7NO})$ is made from furfural.
Given that industrial manufacturing ammonia is matured, how about combining furfural, ammonia and hydrogen to directly synthesize pyridine? Is the following reaction realizable?
$$\ce{C5H4O2 + NH3 + H2 -> C5H5N + 2 H2O}$$
Both furfural and pyridine are significant organic compounds. If one compound can be made from the other, the cost of the other can drop. So far is how pyridine itself is being manufactured remained an industrial secret? Substituted pyridines are covered in many articles and online information.
Wikipedia pyridines production indicates that

Chichibabin synthesis makes dihydropyridine, a pyridine derivative from acrolein, acetaldehyde and ammonia and byproducts are observed.
Bönnemann cyclization of acetonitrile and two molecules of acetylene gives 2-methylpyridine, a substituted pyridine. Following dealkylation gives pyridine.

In essence, substituted or derived pyridines were first made in multisteps and they are further processed to the parent pyridine.
Another possible route would be using the aza variety of the Achmatowicz reaction on furan.

Hydrogen replaces Ts() and removing of hydroxyl and alkyl make the parent pyridine.
( removing the hydroxyl makes the parent pyridine.

Still, substituted furfurylamines $(\ce{C5H7NO})$ is needed at the beginning. In other words, my question was about if as few steps as possible making the parent pyridine from a prevalent starting material such as furfural is realizable.

I do not know whether the following is a real answer or not but Wikipedia 3-methylpyridine article shows

3-Methylpyridine is produced industrially by the reaction of acrolein with ammonia:
:$$\ce{2 C3H3O + NH3 -> 3-CH3C5H4N + 2 H2O}$$ This reaction also affords substantial amounts of pyridine.

Wikipedia Nicotinonitrile article

A colorless solid, it is produced by ammoxidation of 3-methylpyridine:[2]. Nicotinonitrile is a precursor to the vitamin niacin.

In other words, the question could become one of separating pyridine from 3-Methylpyridine.


Is a quinone ring aromatic? I'm trying to work out which are the aromatic ring systems in the drug aklavinone:

I am trying to follow Hückel's rules that the compound must be cyclic, be planar, possess a p-orbital on every atom in the ring, and have a π-system with $4n+2$ electrons.

I see that the benzene rings are obviously aromatic. But does the quinone ring (in red) also count as part of the 'aromatic ring system'?

Strictly speaking, once we have anything more than a single cycle without conjugated side chains, the Hückel's rule does not really apply, and the very definition of aromaticity becomes somewhat muddy. But that's not an answer they expect from you, anyway. So just point to the first and third rings and call them aromatic, and that would be it.

Why not the second? Well, because the quinone fragment is not aromatic. Its bond length are different from each other. These $\ce{C=O}$ bonds are always double, and the $\ce{C-C}$ bonds next to them are always single; you can't draw a resonance structure where that would be otherwise.


How to acurately measure pH in heterogeneous solution? Following my experiment, I adjust pH of $\ce{Ca3PO4}$ from acid to base (about 11)with ammonia solution by using pH meter for precipitating $\ce{Ca3PO4}$. I waited and found pH wasn't constant. In addition, I tried to wash a pH probe and be placed in the solution again but pH was changed from last value even if I didn't add more acid or base.

The $\mathrm{pK}_{a3}$ of phosphoric acid is 12.319 according to Wikipedia. So at a pH of 11 most of the phosphate species would be in the form of $\ce{HPO4^{2-}}$. Thus as $\ce{Ca3(PO4)2}$ precipitates the solution will become more acidic. So an unanswered question here is how fast would equilibrium be established in the solution. Normally once precipitation starts, then equilibrium is established fairly quickly.

Thus it seems likely that equilibrium would depend a lot on how well the solution was stirred, thus swirling the ppt in the solution rather than having it sit at the bottom...

Also the solution is basic enough that $\ce{CO2}$ from the atmosphere will be forming carbonate species which would also slowly make the solution more acidic. (not sure how much drift in pH that you are observing...).


Separation of Binary mixture? I will be given a binary mixture containing two unknown compounds(the mixture could be either liquid or solid- I don't know yet). The plan of the experiment is to identify the two compounds. So my primary concern is to separate the binary mixture and purify the individual compounds(then the subsequent classification tests/NMR/IR analysis should be straightforward).  I am thinking of doing an acid-base extraction but don't know how to since I don't know any physical properties of the compounds. Any suggestions would be appreciated.

N.B.: the advice below mostly applies to organic compounds. If you're dealing with inorganic salts, or solutions thereof, the approach will be very different.

For a solid mixture, I would take an IR first. That should allow you to rule out potentially useless chemical tests by identifying the functional groups present in your mixture without wasting any of the sample. This would also give you some indication whether acid/base extraction is likely to be of any use. After that, it seems logical to test solubility in a wide range of solvents and at different temperatures. If you can find one that dissolves your mixture only partially at low temperature, but completely at high temperature, then you can likely separate the components immediately by recrystallization. Judicious use of a centrifuge can help if the crystals form a fine suspension.

In the case of a liquid mixture, I would again take an IR initially. In some cases, one component of the mixture can be chemically separated, and IR results would again be useful in determining the viability of that approach. Ketones and aldehydes, for example, will react to form semicarbazones that can be recrystallized and then identified by melting point. As far as physical methods of separation, standard are distillation and fractional freezing.


Potassium carbonate and water/alcohol/toluene? Recently I did a distillation. At a certain point an azeotropic mixture of alcohol, water and toluene evaporated and was caught in a beaker with anhydrous potassium carbonate. I had to shake it and then pour it back into the reaction mixture (while filtering out the solid).

Why did I have to catch it in a $\ce{K2CO3}$-mixture?

(The product was diethyladipate)

The synthesis you are referring to is outlined here. Potassium Carbonate acts as a drying agent as you said as the later part of the distillation requires a specific temperature to remove the toluene and alcohol.

The water comes from the concentrated sulphuric acid solution added with apidic acid. The residue is also rinsed with alcohol which is a strong indicator that water needs to be removed.


Which of the five main branches of chemistry has the most labwork? Budding chemistry student here.

I am aware of the so called "five branches" of chemistry, namely organic, inorganic, bio, physical and analytical.

Which of these paths lead to work with the most lab time?

I suspect physical chemistry would be fairly academic, considering it's links to physics and maths. Which is also very appealing.

All the five so-called branches of chemistry require heavy use of the brain. Modern chemistry is very interdisciplinary. At the research level, the distinction among these branches becomes blurry. Everyone is doing some part of each branch. Traditional inorganic and organic chemistry both require a lot of wet work and synthesis and characterization.
If you ask which branch of chemistry has the highest level of automation, and most of the work is done by instruments-it is analytical chemistry. Many modern physical chemists or theoretical chemists rely on in silico techniques (=computers, modelling, simulations etc). You may not be doing a single real experiment (if your not lucky). Computer is your virtual lab. As a hardcore theoretical chemist, finding a job afterwards in the field, may be difficult. My acquaintance switched to computer programming after a PhD, which was a wise decision after remaining jobless for a year. He learned programming languages in the mean time. In any case, all fields are good. It all depends on your passion.


Inductive effect and hyperconjugation - one elephant, different parts? Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon - i.e. methyl groups donate negative charge for example. Inductively we can argue that carbon is of greater electronegativity than hydrogen, and so carbon withdraws negative charge from hydrogen, and thus negative charge is "donated" to whatever that the methyl is attached to.

From a hyperconjugative effect we can argue that there is some interaction between the C-H orbitals and some antibonding orbital ... how is this supposed to be stabilizing if it involves an antibonding orbital?

Are the inductive effect and hyperconjugation both different ways of
  looking at the same phenomenon

I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around.  I tend to separate the two effects as follows:

inductive effects are associated with electron movement through sigma bonds due to electronegativity differences
resonance effects are associated with electron movement through p orbitals due to electronegativity differences

For example, we often say that a methyl group stabilizes a double bond (carbon-carbon or carbonyl) or a carbocation.  The methyl group is roughly $\ce{sp^3}$ hybridized while the double bond carbon or carbocation carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow from $\ce{sp^3}$ to $\ce{sp^2}$ orbitals.  The direction of this electron flow also indicates  that $\ce{sp^2}$ orbitals are more electronegative than $\ce{sp^3}$ orbitals.  How can the $\ce{sp^3}$ methyl group shift electrons to the lower energy, more electronegative $\ce{sp^2}$ carbon?  It can do it inductively (through sigma bonds) or through resonance (through p orbitals).

I've added a picture showing resonance structures involving hyperconjugation, just to be clear as to what hyperconjugation involves.


What is the dipole moment of the most stable conformer of 1,1,2,2-tetrafluoroethane? I want to know about the dipole moment of this molecule. I think the molecule has no dipole moment, but a book says me that it has a dipole moment.

The dipole moment is 2.45 Debye. The molecule is dipolar as you have asymmetrical electron distribution around both the carbon atoms.

Source: Complete Structure of Gauche 1,1,2,2-Tetrafluoroethane Determined by Microwave
Spectroscopy, J. Phys. Chem. A, 2000, 104, 9489-9493.


Why do dithiins have 8 π-electrons? It is mentioned on Wikipedia that dithiins have 8 π-electrons. Why is this so?

I was thinking it would suffice for just one of the lone pairs on one of the sulfur atoms to participate in resonance delocalization and hence give dithiin a total of 6 π-electrons, but why does the other lone pair on the other sulfur delocalize as well to give a total of 8 π-electrons?

TL;DR Ab initio HF molecular orbital studies of the potentially antiaromatic $8\pi$-electron systems of distorted and planar 6-membered rings revealed that conformational rigidity (calculated or measured folding energies) is most likely the best extension to a $(4n+2)\pi$ criterion. 1,4-Dithiins experience large energy decreases when permitted to fold, and when no external strains are applied, there is no reason for sulfurs to unevenly donate electronic pairs.

Crystal structure of 1,4-dithiin [1] is indeed suggesting a non-planar molecule with the fold angle between "red" and "blue" planes (both plotted on the mirrored $\ce{-S-C=C-S-\,}$ fragments) of $137.14^\circ$:

Same goes for the structure of (1,4)-Dithiino(2,3-b)-1,4-dithiine [2] with the dihedral angle of $131.03^\circ$:

  The chair conformation of the molecule prevents conjugation, and the C-C bond distances clearly corresponds to normal unconjugated bonds, (simple double bond = 1.337(6) Å) so the molecule is non-aromatic in the solid state.

Theoretical calculations ($\Delta_fH^\text{MNDO}$) suggest that both molecules remain bent not only in crystal structure, but they also are flexible in gas phase and in solution [3].

At this point one might think that geometrical conformation dictates the aromaticity or the lack thereof, which is in my opinion an example of violation of the law of cause and effect. Under certain "strains" 1,4-dithiins can exhibit planar structure, like in case of 1,4-dithiine-tetracarboxylic N,N'-dimethyldi-imide acridine [4], where $\pi$-stacking causes a coplanar arrangement of both molecules:

  The intermolecular distances ... are shorter than the van der Waals distance (3.40 Å). This suggests that $\pi-\pi$ interactions between molecules ... exist in these parts and the molecules form a paired molecular group. The dihedral angle between the least-squares planes of molecules ... is $2.3^\circ$}.
  Distances between paired groups are larger than the van der Waals distances... The stacking of paired molecules is in accordance with the poor electrical
  conductivity of this complex.

  Anti-aromatic 1,2-dithiins display properties opposite to those of 1,4-dithiins, whose dications show aromatic stabilization. Unlike other antiaromatic compounds, the 1,2-dithiin derivatives, with eight $\pi$-electrons ..., appear in natural products synthesized by composite plants. The Dewar resonance energies of nonplanar 1,2-dithiins are close to zero. The aromatic 1,4-dithiin dication is generated by two-electron electrooxidation of the neutral 1,4-dithiin. The aromaticity was established by NMR measurements and theoretical calculations.

Conformational criterion for aromaticity and antiaromaticity has been extensively investigated for a number of $8\pi$-electron systems [5], and for 1,4-dithiin the energy required to force it into planar geometry is less than $\pu{2.7 kcal/mol}$ (max. value of 3-21G folding energy), allowing for a high degree of conformational flexibility. This seems to be one of the most recent methods of quantitative determination of both flexibility and aromaticity, as authors propose:

  Relative to planar reference geometries of maximum symmetry, folding of aromatic and antiaromatic systems results in large endothermic and exothermic changes, respectively. Aromatic compounds are planar and strongly resistant to
  folding, whereas antiaromatic systems exhibit large energy decreases when permitted to relax (fold) from planar geometries to nonplanar, minimum-energy conformations. According to this criterion, systems ... which have
  shallow folding energy curves, are nonaromatic.

Of course, $6\pi$-aromaticity can be achieved via electrooxidation of non-aromatic 1,4-dithiine, which leads to a planar dication which exhibits aromatic character [6]:

Bibliography

Howell, P. A.; Curtis, R. M.; Lipscomb, W. N. Acta Cryst, 1954, 7 (6–7), 498–503. DOI 10.1107/S0365110X54001582.
Varma, K. S.; Sasaki, N.; Clark, R. A.; Underhill, A. E.; Simonsen, O.; Becher, J.; Bøwadt, S. Journal of Heterocyclic Chemistry 1988, 25 (3), 783–787. DOI 10.1002/jhet.5570250314.
Bock, H.; Roth, B.; Lakshmikantham, M. V.; Cava, M. P. Phosphorus and Sulfur and the Related Elements 1984, 21 (1), 67–77. DOI 10.1080/03086648408073129.
Yamaguchi, Y.; Ueda, I. Acta Cryst C, 1984, 40 (1), 113–115. DOI 10.1107/S0108270184003218.
Podlogar, B. L.; Glauser, W. A.; Rodriguez, W. R.; Raber, D. J. J. Org. Chem. 1988, 53 (9), 2127–2129. DOI 10.1021/jo00244a059.
Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Chem. Rev. 2004, 104 (5), 2777–2812. DOI 10.1021/cr0306790.


Oxidation of acetone using SeO2 I was trying to look-up the mechanism of reaction of organic compounds with SeO2, but I am not able to understand it. I would really appreciate if someone can predict the end product with the mechanism.

I am getting 2-oxo-propan-1-al (I hope the nomenclature is correct), as a product. But I have a doubt, will it react further, because it could still exist as a enol (2-Hydroxy-prop-2-en-1-al), which can further undergo the reaction with SeO2.

Well I guess that is not a perfect solution but I think you will improve it easely if your are studying organic chemistry. I"ll try to make a beautiful picture with chemdraw when i'll have time.

Remember that you can always draw an enol when you have a ketone. If you have no idea at first with the ketone, try it with the enol.


How did early chemists measure concentrations and purity? My chemistry teacher loves going back through the history of famous chemists. This got me wondering how these chemists would first determine the concentration of a sample before they had any other chemicals of known concentration to compare against?

What modern methods are there for determining concentrations and purity without other chemical standards to compare against?

Let me explain my question a little more clearly with an example.

If a historical scientist starts their own laboratory and wanted to do a titration, how would they start? They might produce some hydrochloric acid, or some barium sulfate, and want to confirm its concentration/purity. However, this would require they already know the concentration of something else, which they don't. So it seems a little paradoxical. How did they ever start performing analytical experiments?

Not everything comes as a solution. Rather, the historical chemist would have started of with solid substances. These were purified according to ‘established procedures’ (distillation, recrystallisation, dissolution and precipitation) which worked quite well. Then, one of these substances would have been used to make a stock solution — and because it is solid you can simply weigh it.

Now you might assume that the historical chemist had no knowledge on how two substances would react — which is likely true. However, once he knew the composition of one type of solution, he was able to express the other in equivalents: ‘this solution is able to reduce two equivalents of the permanganate solution’ (using the first example that jumped to my head; not sure whether that is actually a likely thought).

Going via the diversion of analytical chemistry (is this a compound or an element?), it was established that a law of multiple proportions existed: if two elements form more than one compound, the mass ratio of one of those elements is always a simple integer ratio. This was easily extended to reactions: if something reacted with 2 equivalents of A but 1.5 equivalents of B, then maybe it wasn’t the concentration I assumed but I should multiply by two?

Over time and more data points, the current system would have been established.


Diazo transfer to 1,3-dicarbonyl with tosyl azide What is the mechanism for the above reaction?

This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.

Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.

The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:

In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!

Anyway, the product above is simply a resonance form of the diazo product you want.

Reference

Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc. 1990, 112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.


Does the salt in the ocean act as a binding agent on sand? When I walk on the beach, there appears to be an upper layer of the sand that is 'crusty'.

This appears to be the opposite of what you see in sand in a dessert setting, where the individual particles are not bound together, and blow to form 'dunes'.

I'm trying to figure out what causes the sand to bind together like this.

To me the crystalline properties of salt could cause this. (Ie the way that salt binds together to form crystals could cause this salt-binding that leads to 'crustiness'.

My question is: Does the salt in the ocean act as a binding agent on sand?

Good observations!

Seawater contains salts in small amounts such as such as $\ce{LiCl}$ and $\ce{MgCl2}$ that are quite hygroscopic, which could keep the sand wet longer.

There is also dissolved organic matter (http://www.eoearth.org/view/article/154471/) in seawater that might act as a hydrating agent or as a wetting agent, helping to hold onto the water and to bond it to the sand.

You might try an experiment comparing the appearance of (washed) sand after being wet with plain water, with $\ce{NaCl}$ salt water of the same concentration as seawater, and with seawater itself to see if there is a difference. Though this seems like a casual test, this could be of significance in cases where it's necessary to drive on a sandy shore.


What is the freezing point of sulfuric battery acid? As we all know, the freezing point of absolute sulfuric acid is 10 °C. The average lead-acid battery (sulfuric) uses ~30% sulfuric acid, and has a freezing point much lower than 10 °C.

What constituents contribute to this effect? What is the diluent used in these batteries? What would the freezing point be?

The general effect you're talking about is freezing point depression, one of the basic colligative properties. There are several flavors of lead-acid batteries, so there are different additives that can modulate the freezing point.  The actual composition of the electrolyte varies with the charge state of the battery, so that means the freezing point also varies with how charged it is. I think all of the lead-acid batteries use water as a diluent (it will always be present as it's a species in the electrochemical equilibrium.) From the Wikipedia article on "lead-acid battery":

  Due to the freezing-point depression of the electrolyte, as the battery discharges and the concentration of sulfuric acid decreases, the electrolyte is more likely to freeze during winter weather.


What kind of isomerism does Ni(PPh3)2Cl2 have and what is their IUPAC name? I know that there are 2 isomers that $\ce{Ni(PPh3)2Cl2}$ have and they exist in equilibrium. I am tempted to name them cis- and trans- but it does not sound right.

A metal phosphine complex, dichlorobis(triphenylphosphine)nickel(II) exists in two isomeric forms (blue and red) based on how you prepared it (Wikipedia). The blue isomer, which is paramagnetic can be prepared by treating hydrated nickel chloride ($\ce{NiCl2.6H2O}$) with triphenylphosphine in alcohols or glacial acetic acid. When allowed to crystallize from chlorinated solvents (e.g., dicloroethane (Ref.1), dicloromethane (Ref.2)), the tetrahedral isomer converts to the square planar red isomer, which is trans and diamagnetic:

References:

Benedetto Corain, Bruno Longato, Roberto Angeletti, Giovanni Valle, “trans-[Dichlorobis(triphenylphosphine)nickel(II)]•$\ce{(C2H4Cl2)2}$: a clathrate of the allogon of venanzi's tetrahedral complex,” Inorganica Chimica Acta 1985, 104(1), 15–18 (https://doi.org/10.1016/S0020-1693(00)83780-9).
A. S. Batsanov,  J. A. K. Howard, “trans-Dichlorobis(triphenylphosphine)nickel(II) bis(dichloromethane) solvate: redetermination at $\pu{120 K}$,” Acta Cryst. 2001, E57, m308–m309 (https://doi.org/10.1107/S1600536801008741).


Why is iodine the heaviest bonded element in circumstellar/interstellar space? As I understand, the heaviest bonded element found in space to date is iodine, in the form of iodide in sodium iodide. This is the only element heavier than iron on this list, and that makes sense to me because the nuclear fusion inside stars only continues until iron. The only way that heavier elements are formed is when a star is near death and the energy is so high in the center of the star that more nuclear fusion occurs, and then the elements are flung into space.

If this is true, then wherever they found the sodium iodide should be traces of more heavier-than-iron elements. Why then hasn't an element such as, say, bromine been found? Especially when it behaves so similarly to iodine.

This had stumped me for a while and information on the topic is sparse, so I'd just like some input on if it may be a limitation on observation of space or an astrochemical scenario where somehow iodine is more likely to form bonds in space than its heavier-than-iron counterparts

I disagree with the assertion that Sodium Iodide has been found in circumstellar or interstellar space.

Sodium Iodide is not on the list Molecules in Space kept updated by University of Köln.

Instead it is likely that someone misunderstood an article such as Interstellar NaI absorption towards the stellar association ARA OB1, where NaI does not mean sodium iodide, but instead means neutral atomic sodium.

Also, LaO (lanthanum oxide) has been found in stellar atmospheres, and lanthanum is heavier than iodine. See table 5 here: http://www.ucl.ac.uk/phys/amopp/people/jonathan_tennyson/papers/311.pdf

as well as the 1948 article On the Occurrence of Lanthanum Oxide in S-Type Stars.


Colours in Coordination Compounds I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion. It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$. When $\ce{H2O}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{15}$, though in the spectrochemical series, ammonia is far ahead of water (Source-Triump chemistry ,target publications).
So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as $\ce{Ni}$ is in +2 oxidation state.
The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region (PYQ- AIIMS 2005) the options are  $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(NH3)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ and    $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ <  $\ce{[Ni(NH3)6]^2+}$ .
I used $E=hc/\lambda$. So, E is inversely proportion to wavelength. Therefore the ideal answer should be $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ <  $\ce{[Ni(NH3)6]^2+}$ . As $\ce{[Ni(H2O)6]^2+}$ has a better stability constant, more energy is required to break it. Hence a shorter wavelength than $\ce{[Ni(NH3)6]^2+}$ , whose stability constant is less. This less energy is required, hence a longer wavelength. But the correct answer is $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(NH3)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ according to the AIIMS . Please explain.

I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion.

(I assume that SFL = strong-field ligand and WFL = weak field ligand.)
No, this is not correct. Ammonia is neither a true strong-field ligand nor a true weak-field ligand. Instead, it can form high-spin and low-spin complexes depending on the properties of the metal. Thus, it is best to term it a medium-field ligand.
It is also not true that all metal(III) ammine complexes be low-spin while all metal(II) ammine complexes be high-spin. However, among the many different factors that influence the spin state of a complex, it is true that a higher oxidation state tends to favour (note the wording) a low-spin configuration.

It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$.

It would be very weird if the stability constants of all $\ce{[M(NH3)6]^2+}$ complexes have the same value. Think about it: the metals are different while everything else is the same. There has to be variation. The same is true for the aquacomplexes.

So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as Ni is in the $\mathrm{+II}$ oxidation state.

You can’t say much about the stability of the two a priori. What you can say is that both of these complexes should be relatively inert (although not as inert as, say, $\ce{[Co(NH3)6]^3+}$) from a simple CFSE estimation.

The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region

This has absolutely nothing to do with the stability of the complexes or their stability constant; only with their electronic structure and the energy difference between the orbital an electron is excited from and the orbital an electron is excited into. Furthermore, there has to be an error in $\ce{[Ni(NO2)6]^2+}$ – I find it hard to see a stable complex involving six stable radicals around it.
As for the order the question asks for, this is where the spectrochemical series actually comes in and in handy. Determine which orbitals are involved, what effect a greater field split has on these, find the ligands on the spectrochemical series, determine their relative field splits and translate that all into an photon wavelength. You will soon notice that the wavelength of the hexaaquanickel(II) complex must be longer than the wavelength of hexaamminenickel(II) which gives you your answer by inferring.


Mono-oxidized and dioxidized tryptophan I'm translating a Russian text in which there is the following diagram of tryptophan degradation:

Further on in the text, hydroxytryptophan is called "the mono-oxidized form" (моноокисленная форма) and dihydroxytryptophan is called "the dioxidized form" (диокисленная форма). Is this correct and would this be understood by the readers if I translate it as-is?

First, you can use Google Ngram viewer to check out how often and in which context proposed words and their combination is used/has been used in the literature.

Second, I feel that "mono-oxidized" is overly redundant and just "oxidized" will deliver. Even in Russian you would rarely say so ("моноокисленный" or "единожды окисленный" is usually seen as an unnecessary complication in most cases). For "dioxidized" there are plenty examples when it is used in a similar context both in inorganic and bio chemistry, so it should a safe usage from what I see.

There is also an another option to say "two-stage oxidation" ("окисленный по двум ступеням") instead of "dioxidized" if you want to underline that the process has been carried out step by step.


Is acetate a weak base due to its resonance structure? I just read that acetic acid is an exception to the general idea that the conjugate base to a weak acid is a strong base.  The example says that the conjugate base of a weak acid will only be strong if the acid is a weaker acid than water.

What I'm curious about though, is if acetate is a weak base due to its resonance structure.  Does the resonance cause acetate to be more stable than hydrogen acetate?

Is this true in the more general case where some substituent can bond to the atoms involved in the resonance?

Thanks.

Consider this generic acid-base reaction in water:
$$\ce{HA <=>[K_\text{a}] A- + H+ }$$

The (thermodynamic) dissociation constant $K_\text{a}$ depends mainly on two things:

The stability of the product $\ce{A-}$
The instability of the starting material $\ce{HA}$

The higher one of those factors is, the higher the dissociation constant. So in effect, what a chemist states is the following:

The carboxylate anion is the conjugated base of the carboxylic acid, whose strength is determined by the stability of said carboxylate ion and the instability of the acid. It may be reasoned that carboxylic acids would be much less strong if it was not for the additional stability via conjugation.


Reacting Grignard reagent with amides Propene in the presence of $\ce{HBr}$ and $\ce{H2O2}$ gives A.
A on reaction with $\ce{Mg}$ in dry ether gives B.
B on reaction with acetamide, $\ce{CH3C(O)NH2}$, gives C.

In the above question, I got A as 1-bromopropane and B as propylmagnesium bromide. As for C, I got
$$\ce { CH3-CH2-CH3 + CH3C(=O)NHMgBr}$$
The ratio of moles of the Grignard reagent vs. acetamide is not specified. Should I add an additional propyl Grignard and arrive at $\ce{CH3-CH2-CH3}$, $\ce{CH4}$, and $\ce{ CH3C(OH)N(MgBr)2}$ ?
I need help with figuring out the product in the reaction of B and acetamide.

In the reaction that you have mentioned above, the reaction essentially stops after the reaction has taken place in a 1:1 ratio. That is because the electrophilicity of the polar bond of the acid derivative has reduced sharply so it cannot undergo further reaction.
However, had the nucleophile been $\ce{RLi}$ instead of $\ce{RMgX}$ ($\ce{RLi}$ is more ionic and hence a better nucleophile), further reaction would've been possible, although it provides the product in low yield. R attacks C=O resulting in the formation of $\ce{R-C-O- Li+}$ (The other groups remain as they were- the only difference being that it is $\ce{NHLi}$ instead of $\ce{NHMgX}$)


Standard Hydrogen Electrode Could someone, please, explain to me how the standard
Hydrogen Electrode works and how it is used to measure electrode potentials? Also, why does it have a potential value of zero?

The SHE (Standard Hydrogen Electrode) is a standard equipment for measuring the relative electrode potential, as the process of measuring the actual potential is quite difficult. It consists of a platinum foil dipped in 1.0M $\ce{HCl}$ in the presence of $\ce{H2}$ gas at 1 bar of pressure.

Now when the SHE is connected to another half-cell, either oxidation or reduction may occur in the SHE.

Suppose we have to find the electrode potential of a half cell represented by "A" Now, on setting up the cell and connecting a voltmeter we get the standard potential(Estandard) of the cell i.e.Estandard = EA - EH2
Now as our our aim is to just find EA, we take EH2 to be equal to 0
Therefore,
Estandard = EA - 0
or,
Estandard = EA

The value of Estandard is taken from the voltmeter.
It should be noted that the electrode potential calculated using this method is relative(to the electrode potential of hydrogen). It isn't the absolute value.
Also, the potential for the SHE is taken to be 0 for the simplification of the process.