text:
Mathematical Expression for Limiting Molar Conductivity
I have been fiddling around with the theory of electrolytes, specifically molar conductivity, and ways of calculating limiting molar conductivity $\Lambda^0$.
I was able to come up with the following expression:
$$\Lambda^0 = \frac{\lambda}{mT}|z_+z_-|\, ,$$
where $m$ – mass of a single ion in solution (in $\pu{kg}$); $T$ – absolute temperature (in $\pu{K}$); $z_i$ – charge of a single ion (positive or negative); $\lambda$ is a constant given by
$$\lambda = \frac{N_\mathrm{A}he^2}{k_\mathrm{B}} = \frac{hF^2}{R} = \pu{7.42e-25 S kg m2 K mol-1}\, .$$
The above formula works for some chemical species, but fails for others.
Using my formula, $\Lambda^0$ for $\ce{HCl}$ at $\pu{298 K}$ ($m = \pu{6.06e-26 kg}$) is $\pu{0.0411 S m2 mol-1}$, which agrees well with the experimental value of $\pu{0.0426 S m2 mol-1}$.
Constants in the expression for $\lambda$ should be familiar to anyone who studies physical chemistry.
Have I found a new (easier!) approach of calculating $\Lambda^0$, or is my discovery just a coincidence (or, has this already been discovered)?
If it is indeed novel, any ideas on how to get a more accurate formula?
EDIT (in response to closed post): Theory suggests that the above equation should also depend on the (dimensionless) dielectric constant (relative permittivity) of water (when dealing with aqueous solutions) which is 78.4 at 298 K, and this is also expressed in the comments.

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Did you come across your expression by guesswork, or does it have some physical motivation? The latter might present a path for further generalization. Do you notice a pattern among the species for which your expression is accurate and among the species for which it is not? If you do, that might tell you about what interactions you should consider to get a better approximation.

Actually, based on electrolyte theory, limiting molar conductivity increases with temperature. In my expression, it is inversely proportions to temperature... haven't found anything to explain this yet... but it's remarkable that the expression has the same order of magnitude and correct units nonetheless.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. I see you are an advanced $\rm \LaTeX$ user, but MathJax is a bit different:)

I actually determined the equation from guesswork, so I know of no way to verify it from theory.

Why do you have Planck's constant (h) in the equation? Is this a QM version of the classical equation?

h is indeed Planck's constant. There is no classical equation - it just so happens that the combination of constants that comprise the expression for $\lambda$ is close in magnitude to the mass of a single ion of a chemical species. In actuality, molar conductivity increases with increasing temperature. The above expression suggests an inverse relationship.

As well, the result of using the above equation is that it gives the same order of magnitudes as typical values for molar conductivity.

In your equation, the permittivity is missing. What is the relation of your equation to the usual equation of physical chemistry which uses a something similar expression for the activity coefficient?


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Finding equilibrium constant from solubility product constant

Given
  $$\ce{CuI(s) <=> Cu+(aq) + I-(aq)}\, ;\qquad K_\mathrm{sp} = 1.2 \cdot 10^{-12}$$
  write a reasonable chemical reaction that describes the decomposition of $\ce{CuI2}$ in aqueous solution, and show that this is a spontaneous reaction under standard conditions.

The answer found the Gibbs free energy change of the precipitation reaction (backward reaction of the one shown above) using the formula $\Delta G = -RT\ln{K_\mathrm{eq}}$ and substituted the value $1/K_\mathrm{sp}$ for $K_\mathrm{eq}$.
I do not understand why the inverse of the solubility constant can be used as the equilibrium constant of the backwards reaction.
Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?

Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?

Yes. The general rule actually is: For a given chemical reaction with equilibrium constant $K$:
$$\ce{aA\+bB->cC\+dD}$$
If we multiply the chemical reaction by $n$, we get $K_1$ for the following equilibrium:
$$\ce{\mathit{na} A\+\mathit{nb} B->\mathit{nc} C\+\mathit{nd} D}$$
as $K_1=K^n$.
Reversing the reaction is equivalent to taking $n=-1$, in which case we get: $K_2=\frac{1}{K}$

You must be wondering that:

Why does the new equilibrium constant $K'=K^n$?

The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction:  $\ce{A +B —> C + D}$, $K_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$
Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.
So, if we multiply our reaction with $n$, it actually can be rewritten as:
$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K_\mathrm{c}'$ will be
$$K_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K_\mathrm{c}^n$$
Hence, proved.

PS: The previous answer proved the same result, but incorrectly. That proof has now been corrected thanks to the comment by @DavePhD below.

Thank you. In regards to your bonus question, is my following logic correct: Gibbs free energy is extensive. so one of the factors on the RH side must relate to the amount of substance. R is a constant, while T depends on the average kinetic energy of particles. Therefore K must change depending on amount of substance, hence verifying your rule?

@GeorgeTian Yes, if you multiply $n$ on both sides, you'll get $n\Delta G^\circ=-nRT\ln(K)$ which is equivalent to $\Delta G'^\circ=-RT\ln(K^n)$ by properties of logarithm. Thus, for the new reaction, we have $\Delta G'^\circ=-RT\ln(K')$ with $K'=K^n$. Verified.

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Which site is more favorable for EAS in O-Acetotoluidide?
Which site is more favorable for electrophilic aromatic substitution in O-Acetotoluidide?
The answer given in my textbook says the site para to acetamide group is more favorable, however I believe the site para to methyl group should be more favorable. Here are the reasons:
1) Because of sterric inhibition of resonance, acetamide group will be slightly out of plane thereby decreasing the quality of resonance.
2) Because of hyperconjugation by 3 alpha hydrogens, electron density will increase 3 times each at sites ortho and para to methyl group. This means that methyl group due to hyperconjugation will lead to nine resonance structures as opposed to just three in acetamide.
So, because of these two reasons, I think site ortho to methyl group should be more favorable(ortho will be less favorable because of sterric hindrance).
Am I missing something or the answer given in the book is wrong?

During my entire high school, I never really took SIR into consideration as long as there weren't at least two methyl groups on both sides of the functional group (acetamide in this case). Also, the site para to acetamide is activated due to +M effect, which is generally dominant over hyperconjugation from methyl.

"Resonance effect is dominated over hyperconjugation" is just a generalisation and I don't think it should be used here. Here the hyperconjugation leads to 9 resonating structures as opposed to just three with acetamide.

Because of SIR effect, even with one methyl group, substituted benzoic acid is more acidic than benzoic acid ; substituted aniline is less basic than aniline. This means that just one methyl group can cause significant sterric hindrance and make the bulky group go out of plane leading to decrease in quality of resonance.

Why are you counting the number of resonating structures? We are not to find the stability of the overall molecule. We are to find which carbon of the benzene ring is most activated.  We've narrowed our list to (1) para to methyl (2) para to acetamide. The 2nd position is activated by +M, while the 1st position is activated by hyperconjugation. I wish to say that the major product will be formed from 2nd position (as +M dominates over hyperconjugation).

"substituted benzoic acid is more acidic than benzoic acid" yes that's true. You're right then. I'm not sure of the final answer. Sorry for disturbing.

The more number of resonating structures will cause more electron density on that atom hence making the electrophile attack to that atom more likely.

@Arishta , your arguments are on SIR effect on amide substituent. Considering stearic repulsions between amide group and methyl group , methyl group may be out of plane decreasing the quality of hyperconjugation. Which substituent effectively donates may be on a probability factor of substituent effectively in resonance with benzene ring. Thankyou

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How is aqua regia stored?
How is aqua regia stored as it dissolves all metals?

Well, for starters, try not storing it in a metal container ;-) Secondly, people don't store aqua regia for extended periods... you make it as and when you need it. On a side note, aqua regia does not dissolve all metals.

Also, make sure that there are no traces of organic matter in which you add aqua regia.

How is aqua regia stored as it dissolves all metals?

No one stores aqua regia for extended periods of time, because it decomposes rapidly, yielding volatile products that escape from the solution resulting in a loss of potency.
From the Wikipedia article on aqua regia,

Due to the reaction between its components resulting in its decomposition, aqua regia quickly loses its effectiveness (yet remains a strong acid), so its components are usually only mixed immediately before use.

Upon mixing of concentrated hydrochloric acid and concentrated nitric acid, chemical reactions occur. These reactions result in the volatile products nitrosyl chloride and chlorine as evidenced by the fuming nature and characteristic yellow color of aqua regia. As the volatile products escape from solution, the aqua regia loses its potency.
$$\ce{HNO3 + 3HCl -> NOCl + Cl2 + 2H2O}$$

The nitrosyl chloride formed is, essentially, a decomposition product of $\ce{HCl}$ and $\ce{HNO3}$. This itself, is further decomposed as,
$$\ce{2NOCl -> 2NO + Cl2}$$
$$\ce{2NO + O2 -> 2NO2}$$
The mixture itself is created by combining $\ce{HCl}$ and $\ce{HNO3}$ in a 3:1 molar ratio in glass vessels immediately before use.

Aqua regia: Why does it dissolve gold but not glass?
From what I gather from this PDF, nitric acid is a notoriously strong oxidizing agent and is quite capable of oxidizing elemental gold (a "noble metal").
$$\ce{Au(s) + 3NO3^{-}(aq) + 6H^{+}(aq) <=> Au^{3+}(aq) + 3NO2(g) + 3H2O(l)}$$
However, we wouldn't see that happening even if we plopped a gold coin in a jar of nitric acid and waited a lifetime. This is because the reaction has an extremely small equilibrium constant (I'm having trouble getting the actual value of the constant myself, feel free to give me a heads up if you know where to find it), meaning only a minuscule (almost negligible) quantity of gold actually reacts/dissolves.
So for all practical purposes, nitric acid doesn't react with gold (though strictly speaking, it does).
So you want to dissolve gold, and you know (for real) that nitric acid can do it (but not very "efficiently", if you will). What do you do?
The trick is to throw in some concentrated $\ce{HCl}$ as well.
If you remember your high school chemistry lessons on equilibrium and Le Chatelier's principle, you'll find this easy to follow: The $\ce{Cl^{-}}$ generated by the $\ce{HCl}$ complexes with the $\ce{Au^{3+}}$ and essentially "plucks" out the $\ce{Au^{3+}}$ from the earlier equation, resulting in the equilibrium shifting to the right, causing more gold to react.
$$\ce{Au^{3+}(aq) + 4Cl^{-}(aq) <=> AuCl4^{-}(aq)}$$
Now according to this Quora answer, the reason aqua regia doesn't attack glass, is because silicates (the stuff that constitutes glass) are fairly inert to the oxidizing power of nitric acid (on account of their strong $\ce{Si-O}$ bonds).

Aqua regia doesn't readily react with glass. So while you could keep it in a glass retort for years, it would be nothing like the original mixture (you be hard pressed to even call it "aqua regia" after a day, much less, after a few years).
A rather famous incident regarding the "storage" of aqua regia, was when de Hevesy dissolved two gold Nobel medals in aqua regia to hide them from Nazi Occupiers, and kept the solution in a glass retort on a shelf. He returned to collect it in about 5 years (the bottle was still intact)

When Nazi Germany occupied Denmark from April 1940, during World War II, de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck with aqua regia; it was illegal at the time to send gold out of the country, and had it been discovered that Laue and Franck had done so to prevent them from being stolen, they could have faced prosecution in Germany. He placed the resulting solution on a shelf in his laboratory at the Niels Bohr Institute. After the war, he returned to find the solution undisturbed and precipitated the gold out of the acid.

There is, however, an interesting alternate interpretation of de Hevesy's heroic deed,

 It is commonly suggested, even years after his death, that he only chose to dissolve the gold for story-telling purposes as opposed to practically. Perhaps a much simpler and efficient option was for him to bury the Nobel Prizes.

 You can also read up on how aqua regia "dissolves" gold here. And for those of you who want to know a way to easily recover the said gold, have a look at this Nurdrage video

Also, your question assumes that all metals are dissolved in aqua regia, which is apparently, not the case, as discussed under this post.

Here's a cool video! https://www.youtube.com/watch?v=EucI-BuR2Ao

Combining the two parts of your answer: what happens to a glass retort containing Nobel Prizes dissolved in aqua regia when it is left for five years?  When the components of the aqua regia react causing its decomposition, do the Nobel Prizes precipitate out automatically?  Or do the dissolved Nobel Prizes change the physical properties of the aqua regia in such a way that it can remain stable for five years?

Probably neither.  Once the gold has been oxidized and complexed to $\ce{AuCl_4^-}$ it remains dissolved regardless of what happens to the rest of the aqua regia.  You woud precipitate the gold with a reducing agent.

Don't try the video at home.  Use a lab hood and chemically resistant gloves!

@John I've edited the answer to accommodate some of that. Even Oscar's responded to the points you've raised as well :-)

I could've gotten 100 rep more today if it weren't for the rep cap XD < Shakes fist at the gods >

@paracetamol it happens to the best of us, literally!


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Which ideal gas law does a constant volume thermometer work on?
I searches online for my question and found mixed answers. I think the Ans. Should be Gay Lussac's Law as we are dealing with a case of constant volume but variable pressure and temperature. In many books the answer is given to be Charles Law. Would appreciate if someone gave me the ans. Thanks.

How many ideal gas laws are there?

@ChesterMiller Sorry for not giving the complete info. There are 4 ideal gas laws which i have been taught of. Charles Law, Gay Lussac's Law, Boyel's Law and Avagrados Law

@user516256 There is one ideal gas law.

@andselisk Arent the laws from which the ideal gas eq. Is derived also called ideal gas laws? Correct me if im wrong.

No. Those are just gas laws from which, as you said, ideal gas law is derived.

Gay-Lussac's Pressure Temperature Law.
The law simply states

The pressure of a gas of fixed mass and fixed volume, is directly proportional to the gas's absolute temperature.

As telling temperature rises, the gas in the thermometer has its pressure increased in accordance with the law. Measuring this pressure gives us the temperature as per the ideal gas equation, or by using Gay-Lussac's law directly, taking the pressure ratio against a set base temperature and pressure.

So you are suggesting the answer is Gay Lussac's Law? I know how to calculate the temp but i need the principle on which it works in. In most books and even on Wikipedia it says that it works on Charles Law rather than Gay Lussac's Law.

I think you're reading the wrong part. Volume thermometers, which I believe you're reading, work by measuring volume at a constant pressure. What you're asking about is a constant volume gas thermometer, which measures changes in pressure against a constant volume.

I didnt know there was a diffrence. I shouldve tried to frame my question better. Can you tell me about the Constand Volume Gas thermometer? On which law does that one work?

That's what I wrote. Gay-Lussac's law.


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Why nitration of biphenyl gives 2-nitrobiphenyl as major product?
In nitration reaction of biphenyl with nitric acid, 2-nitrobiphenyl was the predominant product. nitration is directed to o,p positions, but why o position is more preferred? and why nitration stops at this level? I mean why don't we get 2,4-dinitrobiphenyl?

There are really two parts to this question.  1)  Why is the Attack preferentially at the 2 position?  2)  Why does only a single nitro group get incorporated?
1)  We can think of the substrate as a substituted benzene where the substituent, the second phenyl group, contributes pi electrons by resonance.  That makes it an ortho/para director.  Then why is ortho preferred over para to give 2-nitrobiphenyl?
Draw the resonance contributions in both cases.  If you are careful, the ortho attack produces one more contribution:  the proton on the temporarily saturated C-2 can overlap with the pi electron ring on the second phenyl group, effectively becoming part of that aromatic ring.  The ortho attack is thereby enhanced by providing a lower energy transition state.
2)  Once you have a nitro group on one ring, that ring is deactivated directly and the other is also deactivated by conjugation.  Further nitration is thereby more difficult.  We may get it after all, more likely on the ring that did not get the first nitro group, under stronger conditions.  Biphenyl is by no means unique, any polynitration of an aromatic substrate faces similar problems.

So if we reduce the nitro group to amino group, that will activate both phenyl rings (which have nitro group and the other one). Then further nitration will be possible. The question her is what are the favorite positions for further nitration?Is nitration going to proceed in the same ring of amino group or in the other ring? I think nitration will proceed on o,p positions but which of them is more preferred?

This wasn't included in your original question. You may ask this as a different question.


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Why do most drugs (eg: oxycodone) have seemingly arbitrary names?
How do some medicines derive their names?
For instance, is the name oxycodone somewhat arbitrary? I am not well-versed in Chemistry (1st semester student) but does the prefix oxy- imply some chemical meaning as well as the -codone such as some sort of functional group or grouping of molecules?
I realize such nomenclature such as di- implies "two".

Note that oxy also means "sharp", and not always necessarily has to do with oxygen (it's probably not the case here, but still). There are several families and genera of living animals (snakes, birds, worms, ...) starting with oxy too, however I doubt there is a direct correlation with chemistry (Oxybelis, Oxycontin, Oxydendrum, Oxylebius, Oxytropis, Oxyura, Oxyuranus etc.).

As for other words, drugs names usually have an etymology that is not completely arbitrary. For example, the etymology dictionary etymonline.com offers this explanation for the origin of oxycodone: "from (hydr)oxy(l) + codeine". As for codeine, "coined, with chemical suffix -ine, from Greek kodeia "poppy head".

The sarcastic answer

Being a student you should do this exercise.  And if it doesn't make sense then come back and try again when you have taken organic chem.  Look at the structures for codeine, hydrocodone, and oxycodone and then look at the structures for morphine, hydromorphone, and oxymorphone.  You'll start to see part of the pattern.

@toliveira, you raised a perfectly valid point. Indeed the name oxycodone is not random at all, but has lots of Chemistry attached to, which I tried to highlight below

The reason why the generic name of most drugs have seemingly little semblance to their chemistry is in the interests of practicality; a doctor would find it easier to write sildenafil (generic name) or Viagra (brand name) in a prescription than 5-{2-Ethoxy-5-[(4-methylpiperazin-1-yl)sulfonyl]phenyl}-1-methyl-3-propyl-1$\ce{H}$,6$\ce{H}$,7$\ce{H}$-pyrazolo[4,3-d]pyrimidin-7-one (chemical name)  unless he had a weird fetish for IUPAC nomenclature .
As for how manufacturers (and the relevant regulatory bodies) arrive at those generic and brand names,  msdsmanuals.com has a take on this,

When a drug is under patent protection, the company markets it under its brand name. When the drug is off-patent (no longer protected by patent), the company may market its product under either the generic name or brand name. Other companies that file for approval to market the off-patent drug must use the same generic name but can create their own brand name. As a result, the same generic drug may be sold under either the generic name (for example, ibuprofen) or one of many brand names (such as Advil or Motrin).

Furthermore, they add,

Generic names are usually more complicated and harder to remember than brand names. Many generic names are a shorthand version of the drug’s chemical name, structure, or formula. In contrast, brand names are usually catchy, often related to the drug’s intended use, and relatively easy to remember, so that doctors will prescribe the drug and consumers will look for it by name. Brand names often suggest a characteristic of the drug. For example, Lopressor lowers blood pressure, Glucotrol controls high blood sugar (glucose) levels, and Skelaxin relaxes skeletal muscles. Sometimes, the brand name is simply a shortened version of the drug’s generic name—for example, Minocin for minocycline.

I found equally useful information on popsci.com,

As for commercial names, pharmaceutical companies are reluctant to share their strategies. These are hugely important brands, even after the patents expire--nobody knows what Sildenafil citrate is, but everyone knows its brand name of Viagra. And the odds are you know someone who insists on brand-name Tylenol instead of discount-store acetaminophen.
Though pharmaceutical companies contacted by PopSci wouldn't share their marketing secrets, you can still derive a drug's molecular and popular heritage just by studying its name. Like, for example, Tylenol. Its chemical name is $\ce{N}$-acetyl-p-aminophenol--see the -tyl and -ol in there? Then there's Viagra. It doesn't sound like sildenafil, but it certainly suggests vigor and virility--exactly what Pfizer wants you to think. Xeljanz, which was approved earlier this year after almost 20 years of research, targets a protein called Janus kinase. It works in a different cellular pathway from other arthritis drugs. That's where you get the -jan part of that name, which, to someone with a drug decoder, connotes its uniqueness.

As for oxycodone specifically, it's anyone's guess, since that information can't be found (online) easily enough.

 Note: I've italicized all generic and brand names because, well... they're generic and brand names. Italicizing actual chemical/IUPAC names is against convention, so don't get the wrong idea ;-)

PopSci mistake: "Like, for example, Tylenol. Its chemical name is N-acetyl-p-aminophenol--see the -tyl and -ol in there?" - actually, it ends in the whole enol, not just ol...

@Izkata Yup that looks like the case, good catch! :-)

I think there's a piece missing here, which is that the generic drug names are not selected by the manufacturers (as you said in paragraph 2) but ultimately (in the US) by the United States Adopted Name council, although perhaps with input from the original manufacturer.  The USAN's nomenclature is partly systematic, and this is why the generic names of the related benzodiazepine antianxiety drugs clonazepam, diazepam, lorazepam, etc. all end in -azepam.  The NLM has a (partial) list of such suffixes.

@MJD I'm not based in the US, but that's an understandable point you've raised. I've made an edit to the question (hopefully, it improves the accuracy of this post). Thanks! :-)

@flurbiprofen Well, I said "...it's anyone's guess, since that information can't be found (online) easily enough.". It was supposed to be interpreted as "I can't find that info easily enough, so until someone else posts a more precise answer about it, you'll have to make do with what's in my post" ;-) This post (and the sources I've cited) concur on the fact that generic names are (far more often than not) derived from the drug's chemical constitution or physiological function; it's just that I lack the resources/connections required to obtain oxycodone's etymology reasonably quickly. :-)

The naming of drugs is not completely a randomised process but depends on various factors most of which are linked to origins, chemistry or any modifications during the drug discovery pipeline although in old times serendipity discovery often led to trivial names being given to drugs.
While morphine seems have been named arbitrarily, the same is not true for its synthetic derivatives . In the opioid class morphine and codeine are considered prototype drugs to which other derivatives are synthesised:
Note: the additional 3-methoxy group gives rise to the -cod- stem name.
e.g. hydromorphone (a synthetic derivative of morphine prepared by the catalytic hydrogenation 7,8 double bond and dehydrogenation of morphine under acidic conditions) hydrocodone (3 methoxy version of hydromorphone)  oxycodone (14 beta-hydroxyl version of hydrocodone) and oxymorphone (4 beta-hydroxyl version of hydromorphone, analogous to the hydrocodone, oxycodone pair)

How are drugs named?

Most drugs have at least three different names by which they can be
  recognised. These are their chemical, generic (non proprietary) and
  trade (proprietary) names. Chemicals names of drugs are extremely
  cumbersome to use that they they are almost never used except by
  pharmacologists and chemists such a name for example: N-(2,6
  dimethylphenyl)-5,6-dihydro-4H-1,3-thiazin-2-amine, which is more
  commonly known as xylazine.

Every drug has several different names, and some have a very large number. This abundance of names is often a source of confusion and can indeed be dangerous. The various names have different origins and uses.

To avoid using this, a simplified name method has evolved where
  original drug manufacturers together with appropriate drug authorities derive a simplified chemical name from the full name.
  The simplified chemical name can mean little with respect of the
  chemical name since at times random extraction of pieces of the
  chemical name are used to produce a name easier to recall. e.g
  diazepam:

Drugs are named in a variety of ways. During the drug’s earliest stages of development, the first name it receives is the chemical name. The chemical name provides scientific and technical information because it is a precise description of the substance in accordance with chemical nomenclature rules established by the International Union of Pure and Applied Chemistry (IUPAC). The chemical name describes the chemical structure of the drug. Because they are long and complex, however, chemical names are rarely used in clinical medicine.
The generic name is commonly used to describe the active drug(s) in the product and is easier to pronounce and remember than the chemical name.

Morphine is derived form the Greek god of sleep, Morpheus (Morphine
  was isolated from opium in 1806 by a German pharmacist, Seturner. He
  named the compound “morphine” after the Greek god of dreams
  “Morpheus.”).

On the other hand Salbutamol derived from its chemical name which includes salicyl, butyl, amine and alcohol groups another similar example is acetaminophen or paracetamol.
Other names are derived from the nature of the drug or how it is isolated from the parent substance for example monoclonal antibodies:
because the original monoclonal antibodies (mAbs) were of mouse origin and therefore immunogenic, newer generation mAbs are genetically engineered to modify the mouse regions to appear more “humanized.” mAb nomenclature indicates “-umab” e.g adalimumab as human (produced using transgenic technology), “-ximab” e.g rituximab as chimeric, and “-zumab” e.g alemtuzumab as humanized.
Chemical name
There are precise rules for naming chemical compounds, and the chemical name defines the structure of the drug molecule. Because of this the chemical name of a drug is complex and unsuitable for use inprescribing.
Approved name (or generic)
The approved name is the official name given to the drug. It is usually a short name based on the drug's chemical structure and has the advantage that related drugs have similar names. The approved name is the one which should be used in prescribing.
Proprietary or trade name
The proprietary name is the commercial name given to the drug by the pharmaceutical company which makes and promotes it. Newer drugs are only protected by patent for a few years after their introduction so it is important for the manufacturer to firmly establish the trade name during this period so that when it is out of patent the drug will continue earning revenue. After the patent period runs out, other manufacturers can market the drug and usually introduce new trade names, resulting in the same drug marketed under many different names.

An example case study
The process for naming a marketable drug involves five steps: new chemical entity (NCE) submission and patent application, generic naming, brand naming, FDA review, and final approval.
The U.S Adopted Names Council USAN Council is responsible for creating and assigning generic names to chemicals that appear to have potential as new drugs.

The USAN Council has several criteria for a generic name. The name
  must be appropriate for the drug; short, easy to pronounce, and
  euphonic; and suitable for routine use both in the United States and
  internationally. The name cannot be misleading or confusing or imply
  efficacy or application to particular anatomical parts. The name often
  has a stem common to related drugs (e.g.,
  angiotensin-converting-enzyme inhibitors end in "pril"). The USAN
  Council attempts to minimize similarities with other drug names,

References

Names and classes of drugs. P. Lewis, Essential Clinical Pharmacology. © P. J. Lewis 1981
Veterinary Drug Use, Prescribing, Acquisition, and Pharmacy Management. Brand Name (®) Or Not?
Wilson and Gisvold’s Textbook of Organic Medicinal and Pharmaceutical Chemistry, 12th ed
Naming of drugs: Fundamentals of Pharmacology: Drug nomenclature. Gralbraith et al

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Dipolar moment between HI and HF
Which between $\ce{HI}$ and $\ce{HF}$ has a greater dipolar moment?
I think it is $\ce{HI}$ because the atomic radius of $\ce{I}$ is greater.

How much charge separation is between the atoms on each compound? What is the difference between electronegativities of H and I?  H and F?

F is more electronegative than I, how do I evalue if it compensates for the smaller distance?

One method is to look up the "rulr" for ionic character versus electronegativity difference, see the plot in https://chemistry.stackexchange.com/questions/9222/why-are-bonds-ionic-when-the-electronegativity-difference-between-bonded-atoms-i.  Use the curve to figure out where HI might be.  Which is the bigger factor now, atomic size or charge separation?

see https://chemistry.stackexchange.com/questions/31049/dipole-moment-anomaly-in-fluorine-and-chlorine-compounds-what-factors-affect-th

The dipole moment of $\ce{HF}$ is greater than $\ce{HI}$:

$\ce{HF} = \pu{1.91 D}$
$\ce{HI} = \pu{0.42 D}$

This is due to high electronegativity of fluorine. Hence from $\mu = \vec{q} \cdot \vec{d}$, the charge of fluorine is larger than iodine, but the bond length changes only a small amount.

How exactly is the charge related to the electronegativity?

Higher the charge per unit area on atomic surface more is the tendency of that atom to attract electrons. (Allred Rochow Electronegativity).


text:
Mechanism for the reaction between nitric acid and hydrochloric acid in aqua regia
After finding out that the following reaction takes place in aqua regia :
$\ce {HNO_3 + 3HCl -> 2H_2O + Cl_2 + NOCl}$,
I quickly got curious about what the mechanism for this interesting reaction would look like. Upon doing some thinking on  my own, I obtained the following:
$\ce {Step 1: HNO_3 + HCl -> H_2NO_3^ + Cl^-}$
$\ce {Step 2: H_2NO_3^+ -> NO_2^+ + H_2O}$
$\ce {Step 3: NO_2^+ + Cl^- -> NO_2Cl}$
$\ce {Step 4: NO_2Cl + HCl -> NOClOH^+ + Cl^-}$
$\ce {Step 5: NOClOH^+ + Cl^- -> Cl_2 + HNO_2}$
$\ce {Step 6: HNO_2 + HCl -> H_2NO_2^+ + Cl^-}$
$\ce {Step 7: H_2NO_2^+ + Cl^- -> NOCl + H_2O}$
I wasn't able to find any research papers online which published a mechanism for the reaction so I asked my friends to verify if the mechanism was plausible. They agreed that it was plausible and when asked which they thought was the rate-determining step, they replied Step 1 because nitric acid is acting as base. However, I disagree with them and would like to propose Step 5 since it is probably not easy for the chloride ion to abstract a chlorine cation from the intermediate to form dichlorine.
I would like to ask if anyone could find any steps in my mechanism which are unlikely to occur and also do state the reason for why they think so in their response. Also, do share your thoughts on what the rate-determining step would be.
Actually, I am most doubtful about Steps 1 and 2. I was wondering if the nitronium ion forms at all as they could just occur in a concerted manner.

I'm afraid the biggest problem here is that you're assuming there's just a single reaction there, which is not the case.


text:
Separating gold, platinum and palladium
I've dissolved all 3 metals together in aqua regia due to them being mixed together as $\ce{Pt}$ black, $\ce{Pd}$ black and gold. I isolated the gold with no problem using sodium metabisulfate, but I am struggling to separate the $\ce{Pt}$ and $\ce{Pd}$. Any suggestions? I have tried sodium chlorate for palladium isolation and ammonium chloride for platinum isolation, none had worked for some reason. Any tips would be amazing.

Palladium and platinum are in Group 10, gold in Group 11.  This is not a solution but indicates why the gold was relatively easy to get out.

What is the point of this exercise? Sounds to me as if this should be done under the supervison of some trained chemist, as with using aqua regia etc.

https://www.google.com/patents/US3024084.

I am doing it as a side project in work due to PGM waste. I am the chemist for the lab.


text:
Gas chromatography and retention time
I did an experiment of gas chromatography which contained the compounds: octane, butanone and propanol. We injected a mixture of the three compounds in the polar column and then a mixture of it in the non polar column. I have certain data which say that the most polar compound of the three is octane,then butanone and the least polar is propanol. As for the boiling points, the highest boiling point belongs to octane,then propanol and the lowest boiling point is for butanone.
I know that the lower the boiling point is, the higher the vapor pressure of the compound and the shorter retention time usually is because the compound will spent more time in the gas phase. I also know that If the polarity of the stationary phase and compound are similar, the retention time increases because the compound interacts stronger with the stationary phase. As a result, polar compounds have long retention times on polar stationary phases and shorter retention times on non-polar columns using the same temperature.
But after seeing the results, the order that these compounds appear are:
For the non-polar column:
1)air(I get it)
2)propanol
3)butanone
4)octane
I get about octane but between propanol and butanone how do we know which factor(boiling point or polarity) plays the most important role?
I get confused because if we take into consideration polarity, the retention time of butanone should be bigger but if we take into consideration the boiling point, the retention time for propanol should be bigger.

If propanol comes out first in your nonpolar column, then it's obviously the (non-)polarity which plays the bigger role. What's wrong with that?

@Karl I have to understand the concept because I have to find out the order of the compounds for the polar column and I don't get if butanone or proponal will have the bigger retention time.

If the difference in boiling points is not large enough to make a difference in the nonpolar column, then it imo surely won't matter in the polar column. Polar-polar interaction is much stronger than nonpolar interaction.


text:
Does acid and base react directly, or do they react first with water?
If an acid $\ce{AH}$ reacts with a base $\ce{B}$ in aqueous solution, in textbook they write

$$\ce{AH + B -> A- + BH+}\label{rxn:1}\tag{1}$$

But first I add acid to water and then base, so why we don't write the equations like this:

When I add acid to water:
$$\ce{AH + H2O -> A- + H3O+}$$
When I add base to the acidic solution:
$$
\begin{align}
\ce{B + H2O &-> BH+ + OH-}\\
\ce{B + H3O+ &-> BH+ + H2O}
\end{align}
$$

So the equation $\eqref{rxn:1}$ isn't true because the base doesn't react directly with the acid, but with $\ce{H3O+}$ formed by the acid reaction with water. Is what I have written true?

With strong bases and acids, only $\ce{OH-}$ and $\ce{H3O+}$ react, right, because $\ce{AH}$ and $\ce{B}$ (practically) don't exist in aqueous solution.
For weak acids and bases, that's not true, but still there the water can "transport" protons much faster between AH and B than direct exchange after diffusion. As was noted below, that's called the Grotthuss mechanism.
So yes, you are right, but the formula (1) is a valid simplification which is, for most uses, as good.


text:
Is the reaction of calcium carbide with water a Brønsted-Lowry acid-base reaction or Lewis acid-base reaction?
Consider the following reaction:
$$\ce{CaC2 + 2H2O -> Ca(OH)2 + 2C_{2}H_{2}}$$
If it can be considered as an acid-base reaction, is it a Brønsted-Lowry (B.L.) reaction or a Lewis reaction?
If it is B.L., of course it is also Lewis; but I cannot understand if it can be B.L. in this case, since 2 protons are actually passed to the double carbon atom, but there is also the $\ce{Ca}$.

Please don't use MathJax in the title field in the future. Also, you might want to read up on using the mhchem LaTex package :-)

We may regard the calcium hydroxide and calcium carbide  as essentially ionic except for the covalent bonding within the anions.  So there is no Lewis reaction involving the calcium.
That leaves the proton transfer, which can be described fully by the B-L theory.  You actually have two reactions, successively forming $\ce{C2H-}$ and then $\ce{C2H2}$.

I think it would be better to say that one can ignore Lewis aspect of this reaction, then that it lacks one.


text:
What does the M in μM-Sux stand for?
I am looking at some data regarding a study around a channel-blocking agent called suxamethonium.
The histograms of the open times of the channels at varying concentrations of suxamethonium are given. However, I am having a difficult time figuring out the dosage. Here are examples of dose listing formats:

200 μM-Sux
500 μM-Sux

etc.
I understand that μ in this context is to mean micro and I believe the suffix "Sux" is supposed to stand for suxamethonium. But what is the capital "M"? I looked it up in a scientific notation table and it stands for mega. Am I supposed to infer that this is a liquid (i.e. ML)?

$\pu{M}$ is a unit, not an SI prefix, standing for molarity, a measure of the concentration of a chemical species in solution.

I think it's useful for future purposes to refer you to the IUPAC Green Book and IUPAP Red Book which contain a lot of similar (and much more) information.

The capital M is simply an abbreviation or another way to denote mol/L, or moles per liter.
More specifically, the 'M' stands for the word 'molarity', which as a physical quantity is defined as:
$$\ \frac{\text{Amount of Substance of Solute}}{\text{Volume of Solution}}$$
Therefore, $\pu{200 μM}$-Sux means $200$ micromoles of suxamethonium per liter, or $\pu{0.000200 mol/L}$ [because the micro (μ) symbol denotes $10^{-6}$].

You are right assuming that μ is a metric prefix, and that the number denotes the concentration of suxamethonium (suxamethonium chloride, I presume). However, this is supposed to be small caps "M" (as $\rm \small M$), not capital "M" (as $\rm M$). Also note that use of the term "molarity", as well as its notation $\rm \small M$, is discouraged [1, p.27]:

The term molarity and the symbol $\rm \small M$ should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of $\ce{B}$ and such units as $\rm mol/dm^3$, $\rm kmol/m^3$,
  or $\rm mol/L$.

And, of course, $1\,\rm {\small M} = 1\, mol/L$.
Also, check out the answer to a similar question "What does the unit M stand for?"
Reference

Thompson, A.; Taylor, B. N. Guide for the use of the international system of units (SI). NIST Special Publication 811 2008.

I had never noticed that the unit is written as small caps instead of capital! :O (+1)

@LinearChristmas No one really cares, I guess, and it's easier just to type capital M (I have this sin too). However, $\rm M$ is already reserved for Mega, so amount concentration units ideally needs another notation. I also caught up to this recently, probably from one of the great answers by Loong :)

I’m sure you already know this, but technically, the prefix μ is not supposed to be italicised. But as far as I know it is not possible to have upright Greek characters in mhchem/MathJaX.

@orthocresol True; I updated it with the Unicode char;)

I cannot see the units recommend ed. Also I was unaware that M is a desuete "unit". So mol/l is the way. What are the others two?

@Alchimista If I understood you correctly, your web browser fails to render some MathJax code. Here is the screenshot:) I figured it might have to do with \pu{...} syntax, which is not properly rendered in mobile apps, so I replaced the code with the more compatible one.

@orthocresol Although a bit annoying, upright 'mu' is possible (with Unicode), for example $\unicode[Times]{x3BC} \rm \small M$ gives $\unicode[Times]{x3BC} \rm \small M$; usage in mhchem is possible with additional $$ surrounding the Unicode character. It comes in handy for some formulae, e.g. this Loong's answer.


text:
Why isn't hydration an oxidation reaction?
http://chemistry2.csudh.edu/rpendarvis/OxRed.html  states,

Addition or removal of water does not involve, by itself, an oxidation or a reduction reaction. The addition of water to an aldehyde to form a hydrate does not involve oxidation or reduction.

But in this reaction,

The oxidation states of the carbons change from
-3, -1 to -2, -2
As the oxidation states have changes, shouldn't it be considered a redox reaction?

An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron Source

You're splitting hair here. It's not instructive to categorise this reaction as (non-)redox. And the oxidation state of H-OH does not change, as it should in a redox reaction. It's some internal redox, if you insist, but I find this just pointless.

tl;dr Yes it is nominally redox, but nobody cares. This is organic chemistry, and it has its own way of doing things.

As context is always King, I will briefly mention that the original quote is about the redox chain from carboxylic acid via an aldehyde to a primary alcohol (and potentially to an alkane). In this context, the aldehyde is capable of adding water as shown in $(1)$, but this reaction does not constitute a redox process in any definition of the word (count electrons).
$$\ce{R-CH(=O) + H2O <=> R-CH(-OH)(-OH)}\tag{1}$$
In other cases, the removal or addition of water does change the oxidation states of various carbon atoms as you observed. This is not only true for the dehydration of an alcohol as in your example but e.g. also in such key reactions as the Wittig reaction or the Suzuki cross-coupling.
However, organic chemistry deals with a great number of individual synthetically useful reactions that must be classified in some way or another for humans’ benefit. In this organic chemistry classification system, only those reactions that

unilaterally oxidise a single carbon atom by means of addition of an oxygen atom or removal of hydrogen atoms; or
concurrently oxidise two carbon atoms bound to each other by the addition of oxygen or removal of hydrogen in such a manner that both oxidation states change by the same value

as oxidations. (And vice-versa for reductions. Individual exceptions most certainly will apply.) Your example is, in inorganic terms, a comproportionation which means one carbon is oxidised while the other is reduced. Not only does this not cleanly fit into ‘carbon is oxidised’ or ‘carbon is reduced’, meaning it creates a terminological mess, but it also would subsume most if not all reactions under the general redox labels which destroys the idea of classifying reactions as oxidations or reductions in organic chemistry altogether.
So: yes, in a strict (inorganic) sense it is a redox reaction, but in the interest of a more practical classification of organic reactions it is not considered one.


text:
Do most methylene groups sandwiched between a carbonyl and aryl group have a chemical shift around 3.5 ppm?
For example, take methyl phenyl acetate, we can see the chemical shifts is at 7.4 ppm, singlet, 5H; 3.6 ppm, singlet, 3H and 3.5 ppm, singlet 2H. Let us say we didn't know the identity of the compound and just its molecular formular C9H10O2.
I knew that of the 3.6 ppm, singlet, 3H signal I would get CH3-O, but I had trouble thinking about the 3.5 ppm, singlet, 2H. I later found out that the methylene group -CH2 was sandwiched between the carbonyl and the benzene ring, which made me think if this was a pattern for all methylenes sandwiched between a carbonyl and a benzene ring.

TL;DR: probably, but not guaranteed. Basically, you are asking if NMR spectra can be predicted and if the two neighbor groups are enough for predicting NMR of the metilene group. The answer for the first is "kinda, unless something weird and/or special happens, like non-obvious contact with a group capable to influence chem. shift through space (benzene ring and other non-saturated systmes, for example)". The answer for the second would probably be "yes, with a list of exceptions (strong pi-donors/pi-acceptors in benzene ring for exaple".

Sandwich has specifically chemical meaning so your post can be a bit misleading.


text:
"Sea of Electrons" - do the individual electron shells interact with each other?
I've often heard metals as having a "sea of electrons". I was wondering how the actual relationships between the electron shells of individual atoms are... do the electrons orbitals overlap the same space as each other, so the electrons are free to move or do they have to jump from the outermost orbital of their atom to another's? Is it something different entirely?

(Only) the outermost valence electrons occupy a huge set of closely-spaced orbitals (the overlapping valence and conducting band) that ideally span the whole piece of metal. All other electron shells are bound as usual to one nucleus. de:wp has a picture that i find most instructive https://de.wikipedia.org/wiki/Leitungsband

The electrons do interact with eachother, and their orbitals do intersect, but electrons in higher orbitals will be, on average, further from the nucleus of the atom. Check out this page on wikipedia:
https://en.wikipedia.org/wiki/Atomic_orbital


text:
Why is magnesium sulfate acidic? Why are my calculations showing that it is basic?
Let's say I was dissolving $\pu{1 M}~\ce{MgSO4}$ into water and I wanted to find its $\mathrm{pH}$. I would go right to the reaction of the dissociation of $\ce{Mg(OH)2}$:
$$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$$
The $K_\mathrm{sp}$ expression for this reaction is:
$$K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$$
According to wikipedia (see sources at bottom), the $K_\mathrm{sp}$ for $\ce{Mg(OH)2}$ is $\pu{5.61*10^{-12}}$
$$\pu{5.61*10^{-12}}=[\ce{Mg^2+}][\ce{OH-}]^2$$
My molarity of $\ce{MgSO4}$ will be the same as that of $\ce{Mg^2+}$, which is $\pu{1 M}$. Substitute that in.
$$\pu{5.61*10^{-12}}=[\ce{OH-}]^2$$
And then solving for $[\ce{OH-}]$ we get:
$$[\ce{OH-}]=\pu{2.37*10^{-6} M}$$
Therefore:
$$\mathrm{pOH}=5.63$$
and
$$\mathrm{pH}=8.37$$
So my final solution of $\ce{MgSO4}$ will be basic. However, on stack exchange it says that the $\mathrm{pH}$ of $\ce{MgSO4}$ is between 5.5 and 6.5. The post does not explain why, but that the general consensus is that that is the range of the $\mathrm{pH}$.
What did I do wrong?
Sources:
https://en.wikipedia.org/wiki/Magnesium_hydroxide
Is magnesium sulfate basic, neutral or acidic?

Simplified calculations which result in pH values between 6-8 (usually from extremely dilute acid/base solutions or by addtion of very weak acid/bases to water) are generally incorrect because they fail to take into consideration the autodissociation of water. Pure water starts at pH 7, and any hydroxide ions consumed by $\ce{Mg^2+}$ must cause the pH to fall below 7, as you would expect from adding an acid. Multiple previous answers here at Chem.SE touch on this point, for example here. You will have to find a slightly different equation.

Using the Ksp of $\ce{Mg(OH)_2}$ would give you the maximum concentration of $\ce{OH-}$ for a solution of a given amount of $\ce{Mg^2+}$ - it won't necessarily give you the actual concentration. (It's perfectly happy with solubility products below the Ksp value.) -- If you're getting a pH above 7.0, you have to ask yourself where the "extra" $\ce{OH-}$ is coming from. It's from water, obviously, but what's driving the "extra" splitting of the $\ce{H2O}$, and - critically - what's happening to the $\ce{H+}$ that's being generated by the auto-dissociation of water to form the extra $\ce{OH-}$?

There is an answer here for this question. The pH is depend on the concentration of $\ce{Mg^2+}$, thus why the range of pH is given.

Your answer lies not in the solubility product but in the acidity constants of sulfuric acid and the basicity constants of $\ce{Mg(OH)2}$, though the solubility product may well come into play. You would need to solve a system of 6 equations, one for each of the:

2 $\mathrm{p}K_\mathrm{a}$ values for the acid
Two $\mathrm{p}K_\mathrm{b}$ values for the base
Proton condition or charge neutrality
And as you will be groping in $\mathrm{pH}$, one to be sure you don't exceed the solubility product.

As sulfuric acid is 'stronger' as an acid than $\ce{Mg(OH)2}$ is as a base, you would expect the answer to be slightly on the "acid side" of neutral $\mathrm{pH}$.

Magnesium salts do, however, form less acidic solutions than those of most other metals.  It's an example of magnesium hydroxide being in a "gray zone" (like phosphoric acid among acids) between "strong" and "weak".

Another aspect to be taken into account is the difference between measured solubility and solubility as calculated from the solubility product. Salts like $\ce{MgSO_4}$ are always much more soluble than what can be calculated from the solubility product. I don't have the numerical values for $\ce{MgSO_4}$, but I know the corresponding values for $\ce{CaSO_4}$, which is its neighbor in the periodic table. Calculated from $\ce{K_{sp}}$, the solubility is $4.7$ mM. The measured value, obtained by titration, is $18$ mM, as stated in J. Chem. Educ. $77, 12$, Dec. $2000$, p.$1558$
This discrepancy is due to hydrolysis. An important part of the $\ce{Ca^{2+}}$ ions react with water to produce  ions like $\ce{[Ca(OH)]^{+}}$ according to the equation  $$\ce{Ca^{2+} + H_2O -> [Ca(OH)]^+  +  H^+}$$ $\ce{Mg^{2+}}$ should react the same way.$$\ce{Mg^{2+} + H_2O -> [Mg(OH)]^+  +  H^+}$$  Simultaneously, an important part of the sulfate ions are reacting according to the following equation $$\ce{SO_4^{2-} + H_2O -> HSO_4^- + OH^-}$$ And there is also the possibility that some $\ce{CaSO_4}$ ou $\ce{MgSO_4}$ gets dissolved without being dissociated. The relative extent of these two last reactions is the reason why $\ce{MgSO_4}$ gives non neutral solutions in water.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is nitric acid often used to digest any solid sample?
Why is $\ce{HNO3}$ often used to digest any solid sample?
I heard many solid samples often digested with $\ce{HNO3}$ solution because nitrate compound always dissolve in water.
Is the concept correct or not?

Could you clarify what do you mean by a sample

For example, in analytical chemistry, when we would like to analyze solid sample. It is often use HNO3 acid to digest it into the solution before.

This is pretty much correct. Nitric acid is a convenient oxidizing acid which will dissolve most of the metals (except for $\ce{Au}$, $\ce{Ta}$ and elements of platinum group), forming either oxides or nitrates; sulfur is oxidized to sulfate; phosphorus to phosphate; organics is oxidized and nitrated. This assures consistent predictable products in probe preparation.
Most oxides can be separated by filtering, and the majority of inorganic nitrates is soluble in aqueous solutions, which paves the way for the further classic analytical methods (titration, UV-Vis spectroscopy, atomic emission spectroscopy, etc.). Nitrate-ion is not prone to hydrolysis, and formed nitrates are no longer demonstrating oxidizing properties (unless they are solidified and heated), which is overall quite convenient.
Also, nitric acid is cheap and doesn't react with glass or PTFE, which allows to built various lab apparatus suitable for particular analytical method.


text:
C-13 NMR: Why are sp hybridized carbon atoms less shielded than sp3 hybridized carbon atoms but more shielded than sp2 carbon atoms?
For example, when you look at hex-1-en-5-yne, you'll find out that the terminal carbon with the triple bond has a reading of 69.7 ppm and the internal carbon with the triple bond has a reading of 86.1 ppm. This is significantly more downfield than the other sp3 carbons which have readings of 22.4 and 32.4 ppm, respectively. The other sp2 carbons have readings of 134.3 and 116.4 ppm, respectively. You can also see that the terminal carbons are more shielded, which makes me wonder why also they appear to be more electron rich?

Try to see it in terms of electronegativity of the fragments considering directionality. That should help. I e a C hybridised sp is more electronegative as seen from the single bond as compared to the opposite triple bond region, which is packed by ekectrons.


text:
Conditions for Chirality in optical isomerism
Is it necessary that any atoms which is chiral must have 4 different atoms around the central chiral atom(maybe it can be carbon if it's organic molecule) bonded to it?
More specifically:
I mean can't be just three or two different groups or atoms attached to central molecule make a molecule chiral? And if no then please explain

This video might help you - https://www.youtube.com/watch?v=JS-iAuCIexk

https://en.wikipedia.org/wiki/Allene can be chiral, or bicyclic systems, Möbius aromats can be right- or leftwound. https://en.wikipedia.org/wiki/M%C3%B6bius_aromaticity

https://chemistry.stackexchange.com/questions/77146/does-a-chiral-centre-have-to-have-3-or-4-different-groups-attached?rq=1

A nitrogen atom connected to three different groups is enough to make the molecule chiral. In this case, the lone pairs "serve the role" of the fourth group needed to make it chiral. Quote from http://research.cm.utexas.edu/nbauld/CHAPTER%2021.htm:

It is interesting to note that, since the nitrogen atom of amines is tetrahedral, such a nitrogen can be a stereocenter if it has three different R groups attached. By definition, the fourth group is an electron pair, so that all four groups are different.
However, it is observed that when chiral amines are generated, they very rapidly undergo an umbrella-like inversion to generate the corresponding enantiomer, quickly racemizing the amine. Certain amines, for which this inversion is especially difficult, can be prepared and are relatively stable as a single enantiomer.

An atom connected to three different atoms with all the four atoms in the same plane is achiral because the plane of the molecule is a mirror plane hence the molecule and its mirror can overlap.
Please also note there are cases where the chirality is not caused by a simple chirality center, such as helicenes.

Only your last sentence actually addresses the OPs question.

The question wasn't about whether non-carbon atoms can be chiral or not it was about whether you need 4 different atoms. And amines are a bad example to choose as they are often not chiral because of fast inversion at the nitrogen.

The question is about whether you need 4 different groups connected to the same atom, and the answer is "no, three is enough". Amine is an example. I know about the fast inversion and have quoted them. So why do you think it does not address the OP's question? @Karl

Didn't Werner devise a cobalt complex where the cobalt was attached to only one different ligand?  The way the ethylenediamine ligands were arranged made a configuration that could not be matched with its mirror image, and IIRC he even resolved the enantiomers!

Because at low enough temperature to avoid inversion, the free electron pair is the fourth substituent, and the geometry is very nearly a tetrahedron.


text:
Is trans-1,2-dimethylcyclopropane optically active or not?
According to my book it is optically active, but no valid reason is given. But what I found is there exists an alternating axis of symmetry, and so my conclusion is the compound would be inactive or has a meso isomer. Am I wrong?

Axes of symmetry have nothing to do with chirality unless it's an improper axis (which does not sound like what you're referring to).

First off ...
There is a difference between "chiral" and "optically active".  Something that's optically active has to have chiral molecules.  However, something can be chiral but if you have equal amounts of the mirror-image isomers, or if molecular motion can make an equal number of isomers like an umbrella-like inversion of a lone pair or an exchange of protons, then you do not see optical activity.  All we are looking at here is whether the molecule is chiral, not whether a sample of the stuff is in general optically active.
With that cleared up:
As Zhe says in the comments you have to have a mirror plane, or an improper rotation axis, or else the molecule is chiral.  A (proper) rotation axis does nothing in this regard.
Each of the carbon atoms with a methyl group on it is, in fact, a chiral center.  The four different groups attached to it are:
No 4.  A hydrogen atom
No 3.  The methyl group that's attached outside the ring
And...
No 2, No 1.  The ring going clockwise and going counterclockwise, as viewed from "above" the C–H bond, are counted as two different attached groups, as if the ring were "cut off" in either direction before rejoining the carbon atom you're looking at.
If you look at the ring attachment that way, you see that the two different directions rank 1 and 2, in some order depending on where the second carbon with the methyl group is, hence my numbering above.  So you can figure out an R or S configuration for each chiral center.  In the cis isomer you end up with one R and one S (thus mirror-image configurations), and a mirror plane does go between them making a meso rather than a chiral molecule.  But the trans isomers give both R or both S, which never allows a mirror plane or any improper rotation axis, and so they are chiral.

(Something that's optically active has to have chiral molecules... ) what about allenes, biphenyl...atropisomers?

Chiral molecules in all optically active cases.  But there can be chirality even though you don't have a specific carbon atom attached to four different groups.  You are offering examples.

Whoa, there! It's unusual to find such imprecise statements in this forum. The necessary and sufficient condition for a molecule to be chiral is that it has no improper rotation in its point group. For example, the point group of 1,3,5,7-tetramethyl-1,3,5,7-cyclooctatetraene is $S_4$ so it is achiral. To see why this is the case, any improper rotation can be written as a proper rotation about some axis followed by a reflection through a plane normal to that axis. For example inversion $i=\sigma_zC_{2z}$ or $S_{4z}=\sigma_zC_{4z}$. To superimpose $\ce{C_{12}H_{16}}$ just apply $C_{4z}=\sigma_z\sigma_zC_{4z}=\sigma_zS_{4z}$. The $S_{4z}$ operation left the molecule invariant because it's in its point group but we can see that it is now a mirror reflection $\sigma_z$ away from a realizable configuration reached via a proper $C_{4z}$ rotation. Also superimposing a molecule on its mirror image means that some proper rotation $R$ and an improper rotation $\sigma$ do the same thing: $R\psi=\sigma\psi$ so multiplying both sides by $\sigma$ we see that $\sigma R\psi=\psi$ so an improper rotation $\sigma R$ is a member of the point group of the molecule.
That is not to say that there can be chiral conformers that can interconvert; the above paragraph is just about geometry. Also, the point group of both trans-1,2-dimethylcyclopropane and trans-1,2-dibromocyclobutane is $C_2$ with no improper rotations so they are both chiral.

Since we're technical: https://doi.org/10.1351/goldbook.C01058 The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane, $\sigma = S_1$, a centre of inversion, $i = S_2$, a rotation-reflection axis, $S_{2n}$). If the object is superposable on its mirror image the object is described as being achiral. | And I don't know what the exclamation refers to…


text:
HPLC method for salicylic acid
Salicylic acid is a relatively polar compound hence I thought that normal phase would be preferred as the stationary phase of normal phase is polar. However, it is said that reverse phase would be the preferred option instead.
Can somebody explain to me why reverse phase HPLC is preferred for salicylic acid? And would reverse phase produce better graphs/results than normal phase?

Reverse phase is the standard in HPLC so, if you can use reverse phase do not complicate matters.
Normal phase usually gives more problems than reverse phase and the stationary phases degrade faster.
Reverse phase works fine and in this case you would typically use a C18 column and water and acetonitrile.
In the case of salicylic acid and, in general, all the acids and phenols, the peaks in reverse phase are normally broad and the retention times vary.
That is because there is an equilibrium between the acid and the ionized form, the carboxylate (or phenolate).
To avoid that, you have to force the equilibrium to the acid form, which is also more apolar, so the reverse phase makes even more sense.
For this, it is sufficient to add a small amount of acid to the water used in the mobile phase.
Very often, 0.1% formic acid is enough.
Sometimes it is required to use higher concentrations up to 1%.
If you need a higher $\mathrm pK_\mathrm a$, you can use trifluoroacetic acid instead.


text:
Rules to identify the most stable resonance structures
While studying resonance from this Chemistry LibreTexts article, I found the rules to follow in order to decide which structure is the most stable. Among them:

The structure with the least number of formal charges is more stable
The structure with the least separation of formal charges is more stable.

Firstly I'm not sure how to interpret the "number of formal charges" in rule 1: does it refer to the number of atoms with non-zero formal charge or to the magnitude of each formal charge? It confuses me a bit also because the total number of formal charge (the sum of them) is clearly constant (and equal to the total charge or molecule).
Secondly I don't understand which of these two rules (1 or 2) is more important and can dominate the other one.
For example two possible structures of $\ce{CNS^{-}}$ are
$$\ce{C=N=S}\qquad\ce{C#N-S}$$
The formal charge for the first one are (in order) -2, 1, 0 while for the second one are -1, 1, -1.
The first is preferable according to rule 1 (interpreting it as the number of atoms with non-zero formal charge) while the second is preferable according to rule 2 since the separation of charge is not as strong as in the first structure (I hope I'm understanding correctly rule 2).
Am I reading incorrectly the two rules or is there one that is actually "stronger" than the other?

Choose some other structure with reasonable stability. This cleared,  It seems you get the rules the right way.

Mesomeric structures aren't "stable", but more or less significant contributors. Also such few rationalisation are only supposed to prevent you from writing structures which are't significan't contributors at all.

Related.

Does this answer your question? How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?

You will encounter situations such as these when the two rules seem to contradict each other, or they confuse you.
However, when this happens, you must use another criteria to determine the best resonance structure. Specifically, you must take electronegativity into account.
What this means is that the most electronegative element should have the lowest formal charge, and the most electropositive element should have the highest formal charge.
This is because the electronegativity "rules" explained above are what are likely to happen (that is how thiocyanate is likely to appear). This makes sense because the higher EN and element has, the more likely it is to "hog" the electrons and have a lower formal charge.
When you use these rules to evaluate the resonance structures of the $\ce{CNS^{-}}$ ion, you can figure out why the first resonance structure is preferable.

Before upvoting you should consider correcting for the OP proposed an exotic structure -  without significant contributing forms -- that is not thiocyanate nor isothiocyanate.

Sorry, can you clarify?

The structures are not limiting forms for thiocyanate

Sorry, I don’t understand. If you want you can edit the answer to reflect what you’re saying feel free to do so.

There's not a stable anion with a N-C-S linkage no matter the bonds / formal charges. Thiocyanate and isothiicyanate have structures with N-C-S linkage.

@Alchimista, you meant C-N-S the first time round, I assume.

Thanks @orthocresol. Yes of course in all my comments I mean C-N-S is NOT thiocyanate nor isothiicyanate and even NOT a stable ions for which stability of limiting forms containing C with ten electrons around (!) should be discussed. :)

@Alchimista the OP didn't propose any structure with 10 electrons around carbon. I agree that the proposed linkage is uncommon, but it certainly may exist, and the discussion of the garbage rules that libretext proposed are completely independent of that.

Perhaps in editing something has  changed. Or I wanted to point out that for this discussion it would be better to choose something with a reasonable stability @Martin perhaps to cite you isn't very easy

You will encounter situations such as these when the two rules seem to contradict each other. In that case, at least one or more of those rules are wrong. Sometimes we don't understand rules as well as we thought, but if they really contradict, then at least one of them is wrong.

The hypothetical ion you discuss must have 16 electrons.
I do not see how one accommodates them in a meaningful way about a proposed structure with linkage C-N-S.
Beside that I see question marks instead of bonds, lone pairs and in case unpaired electrons (it is likely a problem of browser, I am using SE app), as the formal charges suggest you had hard time to find a way leading to something just remotely approaching a stable structure.
You ended in limiting forms with high formal charges both in number as well separation,  not to mention a C atom with 10 electrons around!
This should have pointed you to firstly ask the stability of such an anion.
As the comments to the Q suggestively already pointed out, there is little if no meaning to compare the stability of hypothetical limiting forms for a not existing unstable structure.
Besides this extreme case, an attempt to rationalised the weight of limiting forms for very unstable species based on simple rules will likely fail. Or, at least, it should be attempted just if experience show that such a structure can indeed be observed if not isolated. Likely advanced criteria and calculations would elucidate the situation.
This state, your Q has been probably dictated by the habit of writing CNS- for thiocyanate.
CNS- or less confusedly written NCS- has a structure with, in fact, N-C-S linkage for which you will easily obtain two more or less equally important limiting forms.
One with formal charge -1 on nitrogen and one with a FC -1 on sulphur.
Indeed this is an example of bond isomerism (thiocyanate and isothiocyanate for S linkage and N linkage to other fragments, respectively).
For a general answer useful in other realistic cases I point you to the first part of the A by H. Khan here in this thread.
The second part should be omitted as the author try to follow your line and ended in some "forced"assertion regarding thiocyanate ion which, as said above, is not the subject in case or identifying an hypothetical limiting form with a C -2 as the major contributor.

The down voter could provide an answer.


text:
Leaving group stability when treated with grignard
Why does $\ce{-NH2}$ not leave in $\ce{RCONH2}$ while in $\ce{RCONR2}$ $\ce{NR2}$ leaves when treated with Grignard?
I thought $\ce{NH2}$ should leave as there negative charge is more stable on $\ce{N}$ compared to $\ce{NR2}$ where the alkyl groups have a +I effect already.

I would appreciate answers instead of downvotes :)

Well, that's pretty obvious, you just don't use grigniards with compounds having even weakly acidic hydrogens.

Ikr but suppose if I use 2 eqt of grignard then why wouldnt NH2 leave ? Why does only NR2 leaves?

Grignard reagents with deprotonate amides and the RN- species is not a leaving group

$\ce{RCONH_2}$ is a protic acid, and even though it's weak the Grignard reagent is a powerful enough proton base to react.  You get deprotonation of the amide instead of nucleophilic attack on the carbonyl group.


text:
Stoichiometry question
I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.
In 50.3 gr of a compound whose formula is $Na_2X_4O_7$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.
a) What is the number of Na atoms in the sample?
I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $0.5*N_A$.
b) What is the molar mass of X?
I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.
I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:
$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$
Moreover, in a certain process 0.84 moles of $XH_{3(g)}$ were formed.
1) How many grams of gaseous oxygen were formed in the process?
I first balanced the reaction thus:
$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$
I figured that if 0.84 moles of $XH_3$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.
2) How many grams of $Na_2X_4O_{7(s)}$ reacted in the process?
I figured that 0.84/4 moles of $Na_2X_4O_{7(s)}$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.
3) What are the volumes of the oxygen and the $XH_{3(g)}$ formed in the process at room conditions?
I figured $V_{O_2}=n_{O_2}*0.082*298$ and $V_{XH_3}=n_{XH_3}*0.082*298$, where $n_{O_2}=1.26 moles$ and $n_{XH_3}=0.84 moles$
4) What is the concentration of $OH^-$ ions in the solution obtained (500ml)?
I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $OH^-$ were formed.
I'd appreciate any comments on my attempt above.

One thing I see:  when you work out the MW (or AW) of X, you assume 4 moles of X based on the chemical formula.  But, properly, you have twice as many moles of X as Na ($\ce{Na_2X_4}$), and you found just 0.5 mole of Na.  So instead of 4 moles of X you have just one mole of X making the atomic weight if X 10.8 g/mol not 2.7.  Hmmm, isn't there an element in the second row of the Periodic Table with that atomic weight?


text:
Mixing Ca(OH)2 with Ca(NO3)2
Mixing $\ce{Ca(OH)_2}$ with $\pu{0.1 M}$ concentration of $\ce{Ca(NO3)2}$.
What will actually happen in the process? What I will see with my eyes? I know that $\ce{Ca^2+}$ cation with concentration of $\pu{0.1 M}$ will be the first to react with $\ce{OH-}$, am I right? Would there be any changes in color? What will be the timing to the end point (approximately) of the reaction?

Not much is going to happen. Due to low solubility of $\ce{Ca(OH)2}$, further suppressed by the presence of calcium cations from the dissolved $\ce{Ca(NO3)2}$, you get a suspension of calcium hydroxide, which unfiltered looks like white cloudy liquid and is also known as milk of lime.

Regarding the $\mathrm{pH}$ if the solution, $\ce{Ca(NO3)2}$ is prone to hydrolysis and alone would create slightly acidic medium:
Step 1:
$$
\begin{align}
\ce{Ca(NO3)2 + H2O &<=> CaOHNO3 + HNO3} \tag{R1.1}\\
\ce{Ca^2+ + 2 NO3- + H2O &<=> CaOH+ + 2 NO3- + \color{red}{H+}}\tag{R1.2}\\
\ce{Ca^2+ + H2O &<=> CaOH+ + \color{red}{H+}}\tag{R1.3}
\end{align}
$$
Step 2:
$$
\begin{align}
\ce{CaOHNO3 + H2O &<=> Ca(OH)2 + HNO3}\tag{R2.1}\\
\ce{CaOH+ + NO3- + H2O &<=> Ca(OH)2 + NO3- + \color{red}{H+}}\tag{R2.2}\\
\ce{CaOH+ + H2O &<=> Ca(OH)2 + \color{red}{H+}}\tag{R2.3}
\end{align}
$$
(Here .1, .2, .3 stand for molecular, complete ionic, and net ionic equation, correspondingly.)
But, when you add enough $\ce{Ca(OH)2}$ (saturated solution), which is a strong base:
$$\ce{Ca(OH)2 <=> Ca^2+ + 2 \color{blue}{OH-}}\, ,$$
I'd expect the resulting solution to be basic.


text:
Weight of sodium acetate required to get to certain pH

How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$

Like solving a puzzle, I found from the asked $\mathrm{pH}$ the concentration of hydrogen cation and hydroxide anion, and wrote to myself that I need to find the final concentration of $\ce{CH3COONa}$ to seek from it the amount and then the weight.
But my problem was in all the middle steps. I understand the the concentration of hydroxide anions that I found is after the equilibrium, but I miss the point of how to use them to find the concentration on the acetic acid by given the $K_\mathrm{a}$. I'm guessing that I need to get the $K_\mathrm{b}$ from the $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$ equation. But why exactly and how then, I need to find the concentration of $\ce{CH3COONa}$?
Maybe I lack the knowledge and thus the imagination for it.
Support highly appreciated!

It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:
$$\ce{CH3COO- + H2O <=>[$K_\mathrm{h}$] CH3COOH + OH-}$$
$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\
&= \frac{K_\mathrm{w}}{K_\mathrm{a}}.\label{eq:1}\tag{1}
\end{align}
$$
At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is
$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}.\label{eq:2}\tag{2}$$
Knowing that
$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}},\label{eq:3}\tag{3}$$
one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as
$$
\require{cancel}
c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}.\tag{4}
$$
Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:
$$
\begin{align}
m &= cVM \\
  &= \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} \\
  &= \frac{(\pu{2.50 L})(\pu{82.03 g mol-1})(\pu{1.75E-5})(10^{-14})}{10^{-2\times 9}~\pu{mol-1 L}} = \pu{35.9 g}\tag{5}
\end{align}
$$

First of all thanks for your explanation, im learning a lot her in the community. I understood setps 1,2 and 4 but step 3 i`m a little bit confused. How did the Kw become $\ce{10^{-2pH}}$ in the denominator?

@Mabadai And thank you for your feedback. I tried to explain the operations behind former equation (3), hopefully now it's somewhat clearer.

pH = 0.5(pKa + logC)
pKa is log of Ka
C is the concentration
pH is Given 9.
Hence,
9= 0.5[(log1.75) - 5 + logC]
18= 0.2430 - 5 + logC
22.75 = LogC
This equation will give you Molarity.
Use molecular mass of Ch3COONa to convert it to grams
I may be wrong but I think I've used the right formula

Your answer is fine. Because i need more than "just using a formula by a small prediction". I want to understand things, and learn them from the real point of my life. Solving those questions is good for exam and so on, but i dont want to forget them after i pass the test. Thats why i`m seeking for more explanation and logic than some tricky math and equations.


text:
What is the significance of effective nuclear charge of an ion
After teaching rules of Slater, our teacher gave some examples. Some of them were to calculate effective nuclear charge on an electron, but some of them involved calculating Z* of atoms and ions.
For example, he calculated screening constant of fluoride ion as follows.
The configuration is (1s 2)(2s 2 2p 6).
Screening constant for ion (not for any specific electron) = 2×0.85 (for 1s electrons) + 8×0.35 (for 2s and 2p electrons).
I asked what is the significance of this screening constant (what is fluoride ion screening? If it were to screen a 3s electron added later, the calculation will be different). The teacher could not provide any satisfactory answer.
Is there any reason for doing this calculation? This calculation actually evaluates the screening constant for an imaginary 7th 2p electron.

It's for screening a 2p electron in fluoride.

Then it should be 2×0.85 + 7×0.35, not 8×0.35

Right. That's true. My mistake...


text:
Questions regarding the preparation and recrystallisation of methyl 3-nitrobenzoate
I have a procedure for the synthesis of methyl 3-nitrobenzoate (via nitration of methyl benzoate):

Weigh 2.0 g of methyl benzoate into a dry 50 cm3 conical flask.
Add 4 cm3 of concentrated sulfuric acid slowly to the methyl benzoate with swirling to ensure thorough mixing. Cool this mixture by partially immersing the flask in an ice-water bath.
Carefully transfer 1.5 cm3 of concentrated nitric acid into a dry test tube. Cool the nitric acid by partially immersing it in an ice-water bath before slowly adding, with swirling, 1.5 cm3 of concentrated sulfuric acid.
  Allow this mixture to cool. This is the nitrating mixture.
Using the glass dropping pipette, very slowly add the nitrating mixture (over about 15 min, 1 drop every 10 seconds) to the contents of the conical flask. Stir the reaction mixture as the addition is made. During the addition keep the temperature of the reaction mixture below 6 °C.
Once addition is complete, allow the flask containing the reaction mixture to stand at room temperature for 15 min.
Carefully pour the reaction mixture onto a small amount (approx. 20 g) of crushed ice held in a beaker.
  Stir the crushed ice throughout. Solid methyl 3-nitrobenzoate will form.
Allow the ice to melt and filter under suction. Wash the crude product with a little ice-cold water.

I have two questions:

Firstly, why does the conical flask need to be in an ice bath when adding $\ce{H2SO4}$ to methyl benzoate?
Secondly, why is ethanol used for recrystallisation?

Can't we just heat the mixture of methyl 3-nitrobenzoate after the nitrating mixture has been added?

Answer 1 - you want to keep the whole reaction cool, preferably below 6 °C, because the nitration reaction is exothermic. If it gets too hot you will lower your yield due to side-reactions. There is also the possibility of runaway reaction degrading the starting material and product. This is why you don't heat it.
Answer 2  - someone somewhere has done the comparison crystallisations and determined that EtOH is a good solvent for the recrystallisation. It is also cheap and safe for students to handle.


text:
Effect of doubling kinetic energy on phase transitions

The kinetic energy of the molecules in a sample of $\ce{H2O}$ in its stable
  state at $ \pu{–10 ˚C}$ and $\pu{1 atm}$ is doubled. What are the initial and final phases? [US National Chemistry Olympiad, 2010]

Answer: $\ce{solid -> gas}$
My attempt: Since water freezes at $\pu{0^\circ C}$ at $\pu{1 atm}$, then the water clearly begins in a solid state. In the solid state, most of the atoms are locked within a crystal of hydrogen bonds and therefore have extremely small kinetic energies. Thus, doubling the kinetic energy should not have any impact on the phase, since gases have significantly more kinetic energy than the solid/liquid phase.
Why does the solid have enough kinetic energy such that doubling it results in sublimation?

Are you given any of the heat capacities or phase transition energies? This is a good place possibly to apply the Dulong-Petit Law, which states that the molar heat capacity of most solids is $3R$. It's an underestimate for ice by about 50% though.

I strongly suspect the question is asking about equipartition, although I've already forgotten all of my phys chem.

Agree with @orthocresol For water vapor, you'd expect the average kinetic energy to be $4.5RT$ for a triatomic molecule.

@Zhe, I'm a bit rusty but I think it's $6RT$: each vibrational mode contributes $RT$, not $RT/2$. I'm assuming the vibrational levels are equally populated, when they probably aren't...

@Zhe the question is copied verbatim, no further information is given. Also, I believe that the Dulong Petit law doesn't apply to multiatomic substances well, seeing as there are more degrees of freedom in rotation.

@orthocresol Isn't it $\frac{1}{2}kT$ for each degree of freedom? There are $3N = 9$... I am also rusty, and it's been a long time since I did anything with USCNO.

@Zhe I believe the vibrational nodes in gas molecules aren't active at this temperature, so the energy should be 3RT.

So to first approximation, are we saying that the heat capacity of all 3 phases is similar? If so, then you're approximately doubling the temperature of the system by doubling the energy.

The atoms are not locked in place.   They vibrate back and forth.  They do so with plenty of velocity and thus kinetic energy, even though the vibrations switch back and forth so rapidly that the atoms do not travel a large distance in any one direction.

How would you quantify this energy of vibration, such that one can conclude that doubling this energy would be enough to cause sublimation?

Here is one way:  Calculate the kinetic energy the gas molecules would have, subtract the internal energy change associated with condensation.  This would be roughly  the latent heat of sublimation minus RT, per mole.  R = gas constant, T = absolute temperature.

I don't think there is a way to solve this without some knowledge of the equipartition theorem. The reasoning goes something like this.
In ice, the only significant method of storing energy is in bulk vibrations of the lattice. Thus, the heat capacity of ice must be at least $3RT$ because each of the vibrational modes of the lattice contributes $RT$ to the heat capacity by the equipartition theorem. In reality, the heat capacity will be larger because we ignored all intermolecular vibrational and librational modes and the intramolecular modes. If you look it up, the heat capacity of ice is closer to $4.5RT$. This does not affect our reasoning though.
Now, if we think about the gas, the heat capacity should be $3/2RT+3/2RT=3RT$ for the translational and rotational motions, then there should be a small vibrational contribution less than $3RT$. In the ideal gas situation, the heat capacity will be exactly $3RT$ so let's proceed with this assumption.
Now, note that rotational and translational energies are purely kinetic, so we have $3RT$ units of kinetic energy. In ice, half of this vibrational energy will be kinetic and half will be potential if we assume harmonic oscillators. Anharmonic effects should basically cancel out so no big worries there. Thus, doubling the kinetic energy leaves us with at least $3RT$ in kinetic energy. Therefore, the solid should turn into a gas.
I don't know how we can conclude that it won't end up as a liquid besides by arguing that water is quite structured as a liquid in the sense that any instantaneous snapshot will look like quite similar to other ones, while this is not true for gases. This basically means that translational motion and rotational motions are not important enough to scramble everything.
There may be a more elegant solution. This really doesn't take long to write down the energies for though, so this might be what they had in mind.

Note that this problem can be confusing if we think about adding energy to the system and populating modes. Instead, the problem just says that initially there is some kinetic energy and finally there is twice as much. What state can handle having this much kinetic energy? That tricked me for a bit but if it didn't confuse you then just ignore this addendum.


text:
How to convert lidocaine hydrochloride to lidocaine base?
I'm a hobbyist, no pro. I've got a formulation for a topical anesthetic I'm trying to mimic. It calls for lidocaine, ethoxydiglycol (also known as diethylene glycol monoethyl ether) and a couple of oils.
I've read that lidocaine HCl is soluble in water and insoluble in ether, whereas lidocaine base is soluble in ether, but not in water. Based on that, I'm concerned I won't be able to mix my lidocaine HCl with the ethoxydiglycol, and I need lidocaine base instead of the lidocaine HCl I've got on hand.
I'm planning to dissolve the lidocaine HCl in distilled water, then add baking soda to it until the pH drops low enough for lidocaine base to precipitate.  After that, I plan to filter the whole thing through a coffee filter, rinse with more distilled water, then evaporate the water off.  I'm hoping this will yield pure lidocaine base that I can work with.
Will it work? Are their any caveats I should know about?

related https://chemistry.stackexchange.com/questions/87644/nahco3-required-to-raise-ph-of-weak-hcl-solution-from-3-5-to-7-2-doubting-my-ca

Have you checked the maximum concentration levels of carbitol to ensure no adverse effects, for topical applications?


text:
How is the yield of a complex organic synthesis defined?
I just watched the Periodic Table of Videos episode Deadly Strychnine that describes the history of this particularly challenging synthesis. University of Nottingham chemist Rob Stockman calls the problem "an "Everest" of synthetic chemistry."
What is the definition of yield in this case? How is it calculated? Is it as simple as mass of product divided by total mass of all reactants, or something more specific and meaningful?

In theory
For simplicity's sake, let's consider a two-step sequence:

The usual terms for these are as follows: A is the "starting material", D is an "intermediate", F is the "final product", and B, C, and E are "reagents".
Why are B and C listed over the same arrow? This means that they have been added together to the starting material A, in the same flask, and together they effect the transformation of A to D.†
Now, depending on the stoichiometry of the reactions, as well as the quantities of A, B, C, and E which are added while performing the synthesis, it will turn out that one of A, B, C, or E will be a limiting reagent for the overall synthesis. Note that D cannot be a limiting reagent, as it is not an "input".
For example, let's say that the exact stoichiometry of the two reactions is as follows:
$$\begin{align}
\ce{A + 2B + 2C &-> D} \\
\ce{D + E &-> F}
\end{align}$$
Now, it is clear that if you add 1 mol of A, and if everything else is added in excess, then the theoretical maximum you can get is 1 mol of D and hence 1 mol of F. In this scenario, A is therefore the limiting reagent, because the quantity of A limits how much of F you can get.
Sadly, reactions don't ever work perfectly. You perform the synthesis, and at the end of the day you get 0.05 mol of F instead of the theoretical maximum of 1 mol. Your yield is therefore:
$$\frac{\pu{0.05 mol}}{\pu{1 mol}} = 0.05 = 5\%.$$

Identifying the limiting reagent
Usually the limiting reagent is chosen to be the most expensive, or most precious reagent you have (or else part of it goes to waste). This is most often an organic compound, and is nearly always the given starting material. So, one usually does not have to calculate the theoretical yield from every single input, as the limiting reagent can usually be identified at a glance.
MacMillan's strychnine synthesis can be used as an example here. The sequence of synthetic steps is given on SynArchive. In this case, it is reasonable to assume that the organic starting material is the limiting reagent, and that all the other "inputs" are added in excess.

In practice
In the lab, organic chemistry is done one step at a time; therefore the intermediate D is usually isolated and stored for a period of time.
If the above theoretical analysis is to hold true, then all of the D that you formed in the first step must be carried forward to the second step. i.e. if you form 0.8 mol of D in the first step, the second step must start with all of the 0.8 mol.
In practice this is never the case, due to some reasons:

Some sample must be sacrificed to obtain spectroscopic data on them; this helps the chemist know that they have made the right thing. In other words, we need to make sure that what we think is D is really D.
Some sample must also be sacrificed to small-scale experiments. Reactions often need to be tested before being performed on a full scale. If a chemist made 0.8 mol of D from the first step, and happily carried all of it forward to the second step, they would be left fuming should the second step fail, because they would have to perform the first step again. The way around this is to take a tiny amount: say 0.01 mol of D, and carry out the reaction on it. That way, if the test run works, you can then take the remaining 0.79 mol and do the reaction again; and if not, you still have plenty of D left to experiment on. Ultimately, this leads to losses in the amount of D which is carried forward.

So, the better way to do it is to measure the yield of each individual step. That is to say, measure the yield of D that you made from A; and then measure the yield of F that you made from D. If you multiply those two numbers together, you will obtain the overall yield.
For example, if you put in 1 mol of A and got 0.8 mol of D, the yield of the first step is 80%. Let's say you then took 0.75 mol of D, treated it with E, and got 0.075 mol of F. The yield of the second step is then 10%.
It looks like the overall yield is 0.075/1 = 7.5%, but in truth the yield is (80%)(10%) = 8%. Why is the true yield larger than expected? It's because you didn't carry forward all the D that you could have. In theory, if you repeated the synthesis from start to end and carried forward the entire 0.8 mol of D, you would then have gotten 0.08 mol of F, which corresponds to the correct yield of 8%.
In a paper, yields for every step are reported. This helps people identify which steps work well, and is also useful for other chemists who wish to only replicate a certain step or steps from the paper (for other purposes). The quoted yield of the overall synthesis is usually simply the product of the yields for the individual steps.

† The exact definition of a "step" has been subject to some recent controversy (1, 2), but it's not worthwhile getting into it here.

This is a wonderful answer! Thank you for taking the time to write this all out so clearly, and for choosing some interesting links as well. Reading the "what is a step" discussion will be fun as well as insightful, and definitely keep me from getting any real work done today,

Sorry for the delay accepting this thorough answer. I'll review my other Q's today as well.


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Molecular visualization software for Raspberry Pi?
I've been using our lab's code MCMD on my Raspberry Pi for fun and exploratory purposes. I'm trying to find a VMD, Avogadro, Ovito, or other equivalent that is usable on Raspberry Pi, e.g. to view .xyz or .pdb molecular dynamics trajectories, or simply single-state structures. I have found none so far; I'm using Raspbian OS.

When I tried first RPi long time ago I swear it was as easy as apt-get install openbabel  avogadro pymol jmol rasmol (there might be some extra packages you need to install), but I never really used it due to miserable performance and constant issues due to ARM CPU. Also, isn't it Raspbian?:)

Avogadro should be available from the raspbian repos: sudo apt-get install avogadro.

To search what packages are available for Raspbian: https://raspberrypi.stackexchange.com/questions/1179/does-raspbians-repository-have-a-browsable-package-list

very nice, ty all.

vmd has "support" for arm64 archetechtures, now, though not for raspberry pi. all of the config options, will set CUDA, which you will not have. I have been trying it this weekend, but it seems to fail at every turn, and there is no documentation of anyone getting VMD to work, yet. I think I came close, in a method of rewriting the installation, makefiles, config scripts, and startup scripts. VMD is kinda spaghetti code. I cant get it running on any ARM based machine, but there is code in the source files for that.


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Rate of reaction of silver salts of carboxylic acids having different alkyl groups towards Hunsdiecker Reaction
Hunsdiecker reaction follows free radical mechanism but it is given that rate of reaction with different alkyl groups (R) attached to silver salts of carboxylic acid follows the order as follows
Primary > Secondary  > Tertiary  (Rate)
But free radical stability is exactly opposite!
Primary < Secondary < Tertiary  (Stability Free Radicals)
What can be the reason for this kind of trend in this reaction?
I'm attaching the reaction mechanism for reference.

If the free radical is less stable, it will react faster.
So once the system is started, the propagation (step 4) will proceed faster for primary radicals.
That means that it is the limiting step and that step 3 is faster than 4 for any type of R.

Common knowledge holds that primary radicals form less readily than higher ones to the point where it does not happen at all. Could you elaborate why such radicals form here?

The RCOO$\cdot$ radicals are formed for all R. This radical losses CO$_2$, which is an irreversible step. Than the R$\cdot$ is formed, irrespective of which type of radical it is.

what would be yeild for f2,cl2,br2, I2

The reason behind this is that in the 3rd step the electron of the C -R bond is breaking and electron is going to each of them but if the  R is tertiary then due to the positive inductive effect the electron density gets  high in the R and as a result of this it becomes difficult for the new electron to hold by R.
In case of primary there is no such a type of situation get creates.so that's why reactivity is in the order of primary then secondary and then tertiary .