Why does the nitrile carbon experience more shielding than a carbonyl carbon in CNMR? In organic chemistry lecture, we were given a chart of approximate signal shift ranges for different carbon types in CNMR spectra. carbonyl carbons (C=O) have a range of about 155-210 ppm, but the nitrile group (C≡N) has a signal range of about 110-140 ppm. I understand that oxygen is a more electronegative atom, so it might pull electron density away from the bonded carbon (deshielding), but the C≡N is a more polar bond, which means the electron density spends a lot of time around nitrogen in this bond. Doesn't this imply the C≡N carbon would produce a signal more downfield than that of the C=O carbon?

From the given data ranges, it seems the opposite is true and the carbon of nitrile is more shielded despite the bond polarity.

TL;DR: It's complicated and likely has to do with paramagnetic shielding effects rather than the usual arguments based on electron density, ring currents, and so on.
Firstly, because of low-lying π* orbitals, π-bonded carbons typically have lower excitation energies than purely σ-bonded carbons and are thus more deshielded. However, due to the presence of a π-bond order term, alkynes experience less deshielding due to paramagnetic effects. So the general order of chemical shifts is: alkenes > alkynes > alkanes.
Carbonyls are more deshielded than alkenes, and nitriles more deshielded than alkynes (both due to the paramagnetic contribution), but this doesn't overturn the existing bias of alkenes > alkynes.

It is instructive to start by comparing alkenes (~120 ppm) vs alkynes (~70 ppm). There is already a large difference between those two. Carbonyls are essentially deshielded alkenes, and nitriles essentially deshielded alkynes. And so it is no longer too surprising that carbonyls come in at a higher chemical shift than nitriles.
The only question left is to ask why sp2 carbons are generally more deshielded than sp carbons. Although it's tempting to just look purely at the hybridisation and s-percentage, this doesn't explain why the pattern is generally sp2 > sp > sp3. I should first point out that the physical factors controlling 13C shifts are (usually) markedly different from those controlling 1H shifts, so any text that tries to "carry over" those ideas is likely incomplete or incorrect. The reason for that is described in further detail here, and you can see Hans Reich's excellent website for more information about the various terms that contribute to the chemical shift.
With that out of the way, I will briefly quote from Gunther's NMR Spectroscopy, 3rd ed. (p 410):

For the paramagnetic contribution, early theoretical considerations [...] led to the expression:
$$\sigma_\mathrm{p}^i \approx -\frac{1}{\Delta E} \left(\frac{1}{r_i^3}\right)_\mathrm{2p_z} \left(\color{red}{Q_{ii}} + \sum_{j \neq i} Q_{ij}\right)$$
where $\Delta E$ is a mean electronic excitation energy [...]

I have coloured $Q_{ii}$ in red because it is not in Gunther's text (I suspect this is an error). I have added it to bring it in line with an article by Pople (Mol. Phys. 1964, 7 (4), 301–306).
Note first that the paramagnetic shielding is negative, so a larger magnitude corresponds to a less shielded nucleus, or a higher chemical shift. Now, the "excitation energy" for an alkane is probably a $\sigma \to \sigma^*$ excitation, which is quite large. On the other hand, for unsaturated compounds (alkenes / alkynes), there are low-lying $\pi^*$ orbitals, so $\Delta E$ becomes smaller and $\sigma_\mathrm p$ more negative.* This accounts for why unsaturated groups have higher chemical shifts than alkanes.
It also explains why carbonyls/nitriles are more deshielded than alkenes/alkynes respectively: because a heteroatom (O/N) has a high-energy lone pair, the primary excitation becomes a $n \to \pi^*$ transition which has a smaller $\Delta E$.

[...] $Q_{ij}$ is a bond order term that originates from the presence of $\pi$-bonds. [...] The bond order term superimposes additional changes that are most pronounced for [...] the $\ce{^{13}C}$ resonances of alkynes. For [alkynes], theory yields $\sum_{j\neq i} Q_{ij} = 0$ as in alkanes, whereas for [the central carbon of] allene $\sum_{j\neq i} Q_{ij} = 0.8$.

Gunther doesn't say exactly what this term would be for a plain old alkene (not allene), but Pople writes that it is 0.44. According to Pople, the $Q_{ii}$ term (in red) is equal to 2, so the overall term in parentheses is equal to 2 for alkynes and 2.44 for alkenes. This means that an alkyne carbon experiences a smaller (less negative) $\sigma_\mathrm p$, and consequently is less deshielded than an alkene carbon.

* Note also that $\pi \to \pi^*$ excitations are symmetry-forbidden in the context of paramagnetic shielding, so the primary excitation is probably a $\sigma \to \pi^*$ or $\pi \to \sigma^*$.


Are there different types of symmetry? What factor does the symmetry of a molecule play?

For example, I know water is asymmetric so does this therefore make water polar (I know in fact that is is)? It may sound ignorant due to my lack of knowledge but considering one can draw a line through a water molecule, it appears as if if is symmetrical? Why exactly is water asymmetric and how can one tell? I have heard that it is due to its bent shape and lone pair of electrons but this is too general. Just because a shape is bent does not mean it lacks symmetry?

Finally, is there a concept to finding if a given molecule is symmetrical or asymmetrical?

Symmetry is a very broad topic, and as was mentioned in the comments there are many kinds of symmetry. However, when dealing with molecules you should focus on point group symmetry. A good place to start is this link.

But let's get back to the question:

You mention that you know water is asymmetrical and is therefore polar but at the same time you also note that you CAN draw a line through it to cut it into two equal halves. In that way, you can say water has a mirror plane and is therefore symmetric. This type of symmetry is more precisely defined as $C_{2V}$. Ammonia has  $C_{3V}$, Benzene has  $D_{6h}$, and so on and so forth.

Now, symmetry and polarity do have some relation but you can't tie them together, the case in point being the hydrocarbon $\ce{CHMeEtPr}$. From the point of view of symmetry, this is definitely not a symmetric molecule. And despite its complete asymmetry this is definitely what we would call a non-polar molecule.

So are there any properties related to symmetry? Yes - the ability for a molecule to rotate plane polarized light. We measure this and call it optical activity, and call the asymmetric molecules chiral.

Hope that helps!


An exception to the IUPAC convention for italicisation (typography) According to IUPAC, only physical quantities and variables must be italicised.

Section 8 Use of Italic and Roman Fonts for Symbols in Scientific Text of ICTNS Guidelines for Drafting IUPAC Technical Reports and Recommendations states that

  a good general rule is that quantities, or variables, can be given a value, but labels cannot.

So why is it that [X] is not written as [X] when the concentration of species X is a physical quantity?

“The overall rule” [1, p. 7] in

The overall rule is that symbols representing physical quantities or variables are italic

a keyword here meaning that not every symbol of a physical quantity should be italicized.
Chemical element symbols are always upright unless they are used to denote a locant in a name of a chemical substance [1, pp. 8–9]:

Symbols for elements in the periodic system shall be roman. […]

Symbols for physical quantities are single, or exceptionally two letters of the Latin or Greek alphabet, but they are frequently supplemented with subscripts, superscripts or information in parentheses to specify further the quantity. Further symbols used in this way are either italic or roman depending on what they represent. […]

[…]

[…] Letter symbols for elements are italic when they are locants in chemical-compound names indicating attachments to heteroatoms, e.g. О-, N-, A-, and P-. The italic symbol H denotes indicated or added hydrogen.

For example, amount concentration of nitrate can be displayed as $c_\ce{NO3-}$, $c(\ce{NO3-})$ or $[\ce{NO3-}]$, where “$c$” is italicized as a symbols for physical quantity, whereas “$\ce{NO3-}$” is upright as it is a specification with its own typography rules.
As for the general use of brackets in written text, according to Wikipedia

In English, typographers mostly prefer not to set brackets in italics, even when the enclosed text is italic.

Further, from Bringhurst' The elements of typographic style [2, p. 85]:

The sloped square brackets usually found on italic fonts are, if anything, even less useful than sloped parentheses. If, perish the thought, there were a book or film entitled The View from My [sic] Bed, sloped brackets might be useful as a way of indicating that the brackets and their contents are actually part of the title.
Otherwise, vertical brackets should be used, no matter whether the text is roman or italic: “The View from My [sic] Bed” and "the view from my [sic] bed."

Similarly, from The Chicago Manual of Style (for earlier edition see this answer on English.SE):

6.5 Parentheses and brackets in relation to surrounding text. Parentheses
and brackets should appear in the same font—roman or italic—as the surrounding text, not in that of the material they enclose. This system, though it may occasionally cause typefitting problems when a slanting italic letter touches a nonslanting roman parenthesis or bracket, has two main virtues: it is easy to use, and it has long been practiced.

Specifically for use in technical documentation and in mathematical notations, brackets could be treated as a notation for operator (like matrix delimiters) [4, p. 316], yet another reason to use roman typeface:

6.2.5.3  Special Bracket Notation for Specific Operators
[…]

Brackets only should be used to indicate chemical concentrations:
$[\ce{S^2-}];$ $[\ce{Na+}];$ and isotopic prefixes: $[\ce{^{32}P}]\text{AMP}.$

The following comparison between the symbols fro amount concentration and specific optical rotatory power illustrates the problems arising from using italicized brackets, namely uneven spacing and clashing symbols:
$$
\begin{align}
&[\ce{^{7}NO3-}] &\quad &\textit{[}\ce{^{7}NO3-}\textit{]} \\
&[α]_λ^θ &\quad &\textit{[}α\textit{]}_λ^θ
\end{align}
$$
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bringhurst, R. The Elements of Typographic Style, 3rd ed.; Hartley & Marks, Publishers: Point Roberts, WA, 2004. ISBN 978-0-88179-206-5.
The Chicago Manual of Style, 17th ed.; University of Chicago Press: Chicago; London, 2017. ISBN 978-0-226-28705-8.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.


Non-Linear Solubility Trend I am doing a lab on carbon dioxide solubility in water and I have a non-linear trend which seems to agree with the accepted trend. I am examining the change in solubility as a function of temperature. In the experiment I titrated carbonated water with sodium hydroxide and used molar ratios to determine the carbon dioxide content. The following graph shows the accepted trend:

(Graph from: http://www.rocketscientistsjournal.com/2006/10/co2_acquittal.html)

I've been thinking but, can't seem to figure out why the trend is non-linear. Is there any intuition behind this trend?

The solubility can be thought of as an equilibrium. The Van 't Hoff Equation describes the relationship between any equilibrium constant and temperature.

$lnK_2 - lnK_1 = \frac{\Delta H }{R} (\frac{1}{T_1}-\frac{1}{T_2})$ . Thus, the graph is not linear.


Finding concentration and moles given final and initial pH Questions:

How many moles of $\ce{NaOH}$ are needed to change $500~\mathrm{L}$ of solution with $\mathrm{pH}\ 2$ to $\mathrm{pH}\ 11$?
A swimming pool contains 2 million litres of water at $\mathrm{pH}\ 7.80$ and the $\mathrm{pH}$ needs to be changed to $6.80$. How many moles of $\ce{HCl}$ are needed?

Not quite sure how to tackle these types of questions.
Would I have to find the moles of hydroxide/hydrogen ions to neutralise the solution first (ie. moles of hydroxide ions needed to reach $\mathrm{pH}\ 7$) and then from there calculate the amount of hydrogen/hydroxide ions needed to get to final $\mathrm{pH}$?
Or do I just calculate the difference between the moles of ions in final $\mathrm{pH}$ with moles of ions in initial $\mathrm{pH}$?
I have tried both methods but I don't have the answer to the questions so I'm just trying to understand the concept.

Both the cases are similar, so I'll show you the calculation for the fist case only:

The $[\ce{H+}]$ ion concentration changes from $\pu{10^-2~M}$ to $\pu{10^-11~M}$. Assuming water at $\pu{25^\circ C}$, this implies the change in $[\ce{OH-}]$ ion  concentration from $\pu{10^-12~M}$ to $\pu{10^-3~M}$. Since the volume of the solution is fixed at $\pu{500L}$, the amount of $\ce{OH-}$ ions has changed from $500\times\pu{10^-12 moles}$ to $500\times\pu{10^-3 moles}$. This change is exactly the number of moles of $\ce{NaOH}$ that need to be added to this solution, since $\ce{NaOH}$ is a strong base and dissociates with $100\%$ yield.


Magnesium oxide dissolving in distilled water Magnesium oxide is a salt, and salts dissolve in distilled water, due to water being polarized. Therefore I would expect that you would find free ions, which is confirmed by the water conducting electricity. But most sources I find states that magnesium hydroxide is created through a reaction with the water. I would appreciate if someone could tell me what is correct and why the other one is false.

All of that is true, these facts do not contradict each other.

The missed part is quantification. Salts and generally ion compounds are soluble, but solubility comes in range of many orders of magnitude.

Also, some compounds, even if largely ionic, react with water.

So magnesium oxide reacts with water forming magnesium hydroxide, both being almost insoluble in water.

Oxide anion  as a very strong Lewis base immediatelly reacts with water $\ce{O^2- + H2O -> 2 OH-}$. So the only chance for metal oxides to stay being oxides is to be insoluble, with their lattice energy too high to be broken by water hydration of ions.


Synthesis Golf IX: Tezacaftor A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

The target of this Easter special round of synthesis golf is tezacaptor, which has recently been approved by the FDA as a treatment for cystic fibrosis in combination with ivacaftor:

InChI=1S/C26H28F2N2O6/c1-24(2,14-32)22-10-15-9-17(4-5-19(15)30(22)12-18(33)13-31)29-23(34)25(7-8-25)16-3-6-20-21(11-16)36-26(27,28)35-20/h3-6,9-11,18,31-33H,7-8,12-14H2,1-2H3,(H,29,34)/t18-/m1/s1

Any commercial starting material is acceptable (as previously, commercial meaning you can buy it from Sigma-Aldrich) providing it has 10 or fewer carbons
The synthesis must include a cyclopropanation (i.e. the starting materials cannot contain a cyclopropane)
The synthesis should ideally provide a single stereoisomer of the hydroxyl stereo centre, but to keep things more accessible, a racemic synthesis won't be criticised

I am really glad that you are reviving these challenge questions NotEvans.
As for the synthesis, the amide disconnection leading to fragments A and B seems like the most simplifying first disconnection:

Synthesis of fragments A and B is outlined below:

Esterification of the readily available acid with isopropyl alcohol under Yamaguchi conditions;
Demethylation (this is why I've chosen to make the isopropyl ester, it shouldn't be destroyed with $\ce{BBr3}$);
Formation of a thiocarbonate crucial for the introduction of two fluorine atoms;
JACS 2010, 132(51), 18199;
Aldol condensation of the ester with formaldehyde;
Ester hydrolysis;
Conversion of the acid to acyl chloride.

p-Nitroaniline is brominated;
Sonogashira coupling;
Gold-catalyzed hydroamination;
Fmoc protection of the indole nitrogen turned out to be necessary at this point (so it doesn't interfere with the coupling with fragment A);
Reduction of the nitro group.

Now comes the endgame:

Amide formation;
Corey-Chaykovsky reaction of a stabilized sulfure ylide is selective for enone-like systems;
Piperidine removes Fmoc;
The indole nitrogen attacks glycidol thus opening the epoxide on the more accessible site;
Reduction of the ester with $\ce{LiBH4}$ should be selective. Several equivalents are needed because of the free hydroxyl groups. My only concern is the stability of the cyclopropane, but other hydride donors could be tried.

EDIT: Here is the synthesis of the alkynyl ester needed to make fragment B:

18. Methylacetoacetate is methylated twice;

Formation of enol triflate;
Base-induced elimination of the triflate gives the ester.

  Total step count: 20

  Longest linear sequence: 12


What else other than zinc can reduce Cr(III) to Cr(II)? I successfully completed the synthesis of bis(chromium(II)hydrate)tetraacetate (aka chromous acetate) using zinc as the reducing agent. For my lab report, I was asked to suggest at least two other reagents which may used in place of elemental zinc in the synthesis of Cr2(H2O)2(u-OAc)4.

Would magnesium metal and manganese metal work since both their standard reduction potentials are more negative than Zn(s)?
And why does zinc make the more practical choice?

I couldnt find ANY literature showing this synthesis using any other reducing agent other than zinc...

Wikipedia describes the reduction of $\ce{CrCl3}$ to $\ce{CrCl2}$ with zinc metal, $\ce{LiAlH4}$$^{*}$, and $\ce{H2}$. You could use any of these methods and stir with $\ce{NaOAc}$ to form $\ce{NaCl}$ and $\ce{Cr2(OAc)4}$:

$$\ce{2CrCl2 + 4NaOAc + 2H2O -> Cr2(OAc)4(H2O)2 + 4NaCl}$$

I'm not sure if magnesium or manganese would work well, but I would assume they would do the reduction. The problem is that you want something that will reduce $\ce{Cr^{3+}}$ to $\ce{Cr^{2+}}$ but not any further. Using something with a more negative reduction potential could cause overreduction. These are good ideas, and I would certainly give credit to my students for proposing them, they just aren't mentioned specifically anywhere I'm seeing.

$^*$You would need to isolate $\ce{CrCl2}$ in the case of $\ce{LiAlH4}$ reduction (or kill off any excess LAH, at least) because you don't want to reduce the acetate added after the reduction.


Monitoring reaction progress? One of the first thing that every one learns in organic chemistry lab is how to monitor a reaction progress with TLC.

My question is what methods do we have for visualizing TLC ( UV , Iodine vapor , etc ) and each one of them is suitable for what purpose and condition ? and is there other ways like TLC (fast , easy , relatively cheap , etc) for monitoring reactions progress?

Slightly askew of what was asked, but still relevant.  TLC is not dependent on conjugation.  That reasoning suggests to me You are using UV light to visualize the TLC.  UV light is used because it is clean, quick and non-destructive.

There are destructive visualization techniques, such as vanillin stain.  In this case, after elution one dips the tlc slide into the stain, and then bakes the coated slide, and various compounds appear in vivid colours, that are normally somewhat related to the functional groups present in the molecules.  Myriad stains exist, vanillin was just my favorite when I was practicing.

(picture taken from http://www.ochemonline.com/images/thumb/5/5d/Tlc_stains.JPG/1000px-Tlc_stains.JPG)

These stains are not quite as clean or quick as UV visualization, but are still quite convenient.  There are other monitoring techniques, but nothing beats TLC for generality, cost and speed.


How much does water expand when heated, and does salt affect that? For this question, assume all things not mentioned are perfectly controlled, all at 14.7lbs of atmospheric pressure, and water is 100% pure of anything not mentioned, with no dissolved air.

Given a quantity of water that is 1 liter at 20°, what would the volume be when heated to 30°?

Secondly, the same question, except the water being heated is fully saturated with salt.

The volumetric expansion of a liquid is given by the following equation
$$\Delta V=V_0\beta\Delta T$$
where $\beta$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature.

Here's a link to a nice table containing coefficients of expansion for water, both pure and salinated.  Wikipedia shows the maximum solubility of salt in water to be around 36% (w/w).

From the first link we find that

for pure water at $25\ \mathrm{^\circ C}$: $\beta = 257\times10^{-6}/\mathrm K$

for water with 35% $\ce{NaCl}$ dissolved at $25\ \mathrm{^\circ C}$: $\beta= 297\times10^{-6}/\mathrm K$

Now you should be able to calculate the values you are interested in.


What is the name of this molecule? (C₈O₃H₁₃N) @tschoppi

Tried taking a stab at this for me and got this much:

1-(1'-oxoethyl)-2-(something-something ester)-pyrolidine

If I managed to see the structure correctly it should be Methyl 1-acetylprolinate


Why do raisins bob up and down in carbonated water? Yesterday I preformed a fun little experiment where I poured a glass of seltzer water into a glass, then dropped a few raisins into it.

I observed that the raisins would float at the top of the glass for a few seconds, then sink all the way to the bottom, then float back up, and repeat seemingly without end.

Why does this happen?

The raisin has nucleation sites on it that allow bubbles of $\ce{CO2}$ to form. The raisin is light enough to be lifted by the bubbles as they push their way to the surface. As the bubbles are released into the atmosphere, the raisin once again sinks until more bubbles form on it. This will continue until the soda water has lost the majority of its dissolved $\ce{CO2}$.


Separation and qualitative identification of cations A liquid problem may contain the following cations: $\ce{Pb^{2+}}$, $\ce{Cr^{3+}}$, $\ce{Fe^{2+}}$, $\ce{Ca^{2+}}$, $\ce{Fe^{3+}}$, $\ce{Ag^{+}}$, $\ce{Au^{3+}}$, $\ce{Co^{2+}}$. I need to devise a scheme to separate these cations individually.

I have managed to prepare the following scheme:

But, I don't know how to separate the following cations: $\ce{Cr^{3+}}$, $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$, $\ce{Co^{2+}}$.

I thought I could use ammonia or sodium hydroxide as a reagent. If I use this reagent, all the species would be precipitated, if I am not mistaken, and therefore I could not separate them. Perhaps they could be separated by using solvents and then using identification reactions (for example, I know that Co(II) precipitates with 1-Nitrous-2-naphthol). But, at first sight, I see it as very complex.

Well there is a pretty simple way, that you can do in any laboratory by standard reagents too.
React the unknown mixture with $\ce{NH4OH + NH4Cl}$
(ammonium chloride is put as a standard to prevent the precipitation of Zn, Mn, Cu and Ni hydroxides, however, you don't have those in your sample, so you can proceed with only ammonium hydroxide too)
now you will observe precipitate as

Green- confirms $\ce{Cr^{+3}}$

Reddish-brown- confirms $\ce{Fe^{+3}}$

White- confirms $\ce{Al^{+3}}$

(These are colours of the corresponding hydroxides)

Now if you have multiple salts, you carry out the following conformatory tests

For $\ce{Cr^{+3}}$, add a few dorps of strong oxidising agent in an acidified medium ($\ce{KMnO4}$/$\ce{H+}$ would do), this oxidises $\ce{Cr^{+3}}$ to $\ce{Cr2O7^{-2}}$ which gives the super famous chromyl chloride test.

For $\ce{Fe^{+3}}$, add a few drops of $\ce{CH3COONa}$ and you will initially observe a dark-red solution of $\ce{Fe3(OH)2(CH3COO)6}$, which on heating precipitates as blood red $\ce{Fe(OH)2CH3COO}$

For $\ce{Fe^{+2}}$ simply add KI, very distinct black $\ce{FeI2}$ precipitates out

For $\ce{Al^{+3}}$ add ammonium phosphate, you will see a cloudy white precipitate

You can read more here, Vogel is pretty much the most solid and trustworthy book for practical inorganic chemistry
You are welcome!


Latent heat of vaporization/evaporation It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.
But my question is, does providing energy equivalent to latent heat of vaporization to a liquid at a temperature below it's Boiling point (if you need, also assume above it's Freezing point) 'Evaporate' it?
eg. When water is kept inside a clay pot out in the Sun, we know a very small amount of water somehow seeps out through it's pores and 'evaporates' (if not vaporizes). Obviously the temperatures aren't even close to 373K(boiling point of water), but the water is spontaneously turning into steam. Because it's not boiling, but is simply evaporating which can take place any where below it's boiling point. But the point here is, 'how much' of the energy from the rest of water-pot system is entrapped in order to convert into a form that has got more energy than its predecessor.
Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.

Latent heat of vaporization/evaporation:  It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.

Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is defined for any temperature.

Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.?

Neither, neither. It is between these 2 values.
The latent heat of water evaporation at temperature $T_2 \lt T_1$ (let we take $T_1$ as the boiling point ) is equal to the  latent heat of vaporization at temperature $T_1$
plus the heat required to warm up water  from $T_2$ to $T_1$
minus the heat released to cool down vapour from $T_1$ to $T_2$.
$$\Delta H_{\mathrm{evap},T_2} = \Delta H_{\mathrm{evap},T_1} + C_\mathrm{p,water}\cdot (T_1 - T_2) + C_\mathrm{p,vapour}\cdot (T_2 - T_1)$$
where $\Delta H$ are molar latent evaporation heats at constant pressure ( called molar enthalpies of evaporation) at respective temperatures,
$C_\mathrm{p}$ are respecive molar heat capacities of water and vapour ( for simplicity considered constant to avoid integrals).
It is application of the Hess' law, which is application of the energy conservation law. It says the energy change depends only on the initial and final state, not on the way how is the final state reached. If it were not so, we could easily construct perpetuum mobile by switching between 2 states by 2 different ways.
So getting A->B directly, or A->C->D->B is equivalent, where

A is cold water
B is cold vapour
C is hot water
D is hot vapour

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature. At water critical temperature near $374 ^{\circ}\mathrm{C}$ the properties of liquid and gaseous water  converge to the same values and the latent heat of evaporation converges to zero.
If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $50^{\circ}\mathrm{C}$, then it will evaporate. If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $100^{\circ}\mathrm{C}$, then some water will not evaporate. Because the latent heat of evaporation is bigger at $50^{\circ}\mathrm{C}$, compared to $100^{\circ}\mathrm{C}$.
Technically, it is different question, as evaporation at ambient temperature takes energy from the environment, while you have to supply energy for the boiling.
Note that physicists/chemists prefer to talk about vapour, a gaseous phase of substance, that is not considered as a "permanent" gas. Steam is rather a technical/engineering/everydays term, covering pure water vapour, mixtures of vapour with gases like air, or the same with an aerosol of water or ice particles. See water phase diagram. Water vapour exists even at at mesosphere at altitude near 80 km at temperatures almost $-100^{\circ}\mathrm{C}$. Also, ice can sublimate directly into vapour, without melting.


Sulfinate R/S configuration How to determine the configuration of the attached chiral compound? It can be represented with two Lewis structures, in the one to the left, the =O substituent has the biggest priority, while in the right one, the -OMe substituent has the bigger priority. In the former case the configuration would be R, in the second S.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the trigonal pyramidal centre of the sulfinate group is converted to a tetrahedral centre using a phantom atom of low priority.

P-93.3.3.2 Trigonal pyramid
The configuration of molecules containing a trigonal pyramidal center (TPY-3) is described in a similar way to that of tetrahedral centers (T-4) described above in P-92 (see IR-9.3.4.3, ref. 12). The tetrahedral configuration is achieved by adding a phantom atom (0) to the central atom perpendicular to the base of the pyramid. (…)
Traditionally, sulfoxides have been considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom). No polyhedral symbol is used.

P-93.3.4.1 The chirality symbols ‘R/S’.
The stereodescriptors ‘R’ and ‘S’ (as defined in P-92.2) are used to indicate the absolute configuration of a trigonal pyramidal system discussed in P-93.3.3.2 (see Rule IR-9.3.4.3, ref. 12). A phantom atom of low priority, and not a pair of electrons, is used to create the tetrahedral configuration permitting the use of ‘R/S’ stereodescriptors in the manner described for tetrahedral stereogenic centers. As no locants are present, the name following the stereodescriptor is placed in parentheses, or bracket, according to the required nesting order.

The examples given for this rule in the Blue Book include ethyl (R)-(4-nitrobenzene-1-sulfinate), which confirms that this rule is applicable to the similar compound methyl (S)-methanesulfinate, which is given in the question.
Notably, an equivalent rule also exists in the current version of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book).

IR-9.3.4.3 The R/S convention for trigonal pyramidal centres
Molecules containing a trigonal pyramidal centre (TPY-3) may exist as a pair of stereoisomers. The configuration of this centre can be described in a similar way to that of a tetrahedral centre. This is achieved through notional placement of a ‘phantom atom’ of low priority in the coordination site that would create a tetrahedral centre from a trigonal pyramidal centre. The centre can then be identified as R or S by the methods described above.
The use of some bonding theories leads to the placement of a lone pair on a trigonal pyramidal centre. If this is done, the absolute configuration of the centre is also described by the R/S convention, in this case by placing the ‘phantom atom’ in the site that is occupied by the lone pair. Examples of this practice may be found in the description of absolute configurations for sulfoxides in which the alkyl substituents are different.

After placement of the phantom atom, the R/S convention for tetrahedral centres can be used. Rules for the tetrahedral configuration of elements other than carbon are given in Section P-93.2 of the Blue Book. In particular, the rule for compounds containing sulfur refers to the rule for similar phosphorus compounds.

P-93.2.5 Sulfates, sulfonates, and related compounds
Sulfates, sulfonates, and related anions are treated in the same way as phosphate anions (see P-93.2.4). (…) Sulfoxides are discussed in P-93.3.3.2.

The corresponding rule for compounds containing phosphorus reads as follows.

P-93.2.4 Phosphates, phosphonates, and related compounds
The ‘$\ce{P=O}$’ bond, as conventionally written in phosphates, phosphonates, and related compounds, is considered as a single bond, as there are already four atoms or groups in the tetrahedral configuration. Similarly, the formal arrangement of charges is not considered when determining the configuration of a chiral molecule. As the stereodescriptors ‘R’ and ‘S’ describe the entire structure, either a salt or an ester, the full name is placed in parentheses to denote the global configuration.

Therefore, the analogous ‘$\ce{S=O}$’ bond in the sulfinate group is considered only as a single bond since the tetrahedral centre of the sulfinate group with an attached phantom atom already has four groups. (Note that this is a deviation from the usual Sequence Rules for the Cahn-Ingold-Prelog (CIP) System. Using the Sequence Rules, double bonds are normally split into two bonds with duplicate representations of the atoms at the other end of the double bond.)
Thus, for the compound given in the question, the $\ce{-O-CH3}$ group ranks higher than the $\ce{=O}$ group. This leads to the configuration S and the systematic name methyl (S)-methanesulfinate.


Hydrolysis of nitriles: Amide vs Carboxylic acid I was recently studying the preparation of amides,and one of the methods listed was acid hydrolysis of a nitrile. Up until now I thought acid hydrolysis of nitriles yielded carboxylic acids. How are the methods to get amides and acids different?
I can view the amide as sort of an intermediary stage while getting to the acid,but how do you stop the reaction at amide? Do you use stoichiometric quantities of the acid? This is puzzling me.

Hydrolysis of nitrile is pH dependent. It will lead to formation of different products based on pH.
Theoretically, nitrile hydrolyze to form amide and then ammonium salt.
$$\ce{RCN ->[H2O] RCONH2 ->[H2O] RCOONH4}$$
But in practice, the reaction between nitriles and water is so slow that it is considered negligible. The nitrile is instead heated with either a dilute acid such as dilute hydrochloric acid, or with an alkali such as sodium hydroxide solution to accelerate the hydrolysis reaction. But this pH dependent reaction yields different products.

Acid hydrolysis: yields carboxylic acid

$$\ce{RCN + 2H2O + HCl -> RCOOH + NH4Cl}$$

Alkaline hydrolysis: yields sodium carboxylate salt

$$\ce{RCN + H2O + NaOH-> RCOONa + NH3}$$
To obtain amide, hydrolysis is done selectively and in controlled manner. It can be done in both acidic and alkaline medium. In acidic medium, nitriles are converted into amide in a single step process using TFA-$\ce{H2SO4}$ mixture as reagent.

In alkaline medium, it can done using alkaline solution of hydrogen peroxide (see @Light Yagami's answer).
Reference

Chemguide
Facile and Highly Selective Conversion of Nitriles to Amides via Indirect Acid-Catalyzed Hydration Using TFA or AcOH−H2SO4, Jarugu Narasimha Moorthy and Nidhi Singhal, The Journal of Organic Chemistry 2005 70 (5), 1926-1929
DOI: 10.1021/jo048240a


Why does the dipole moment order of group 15 hydrides increase down the group? The order given in my book is

  $$\ce{NH3} > \ce{SbH3} > \ce{AsH3} > \ce{PH3}$$

Phosphorous is more electronegative than arsenic and antimony. Then shouldn't the dipole moment be more in case of $\ce{PH3}?$

And why does the anomaly of $\ce{NH3}$ occur?

The query you have is a common mistake, and this is aksed as a very common question in examinations, well, I wouldn't blame you as the reason is simple but not easy to think of.

What you said about E.N. of phosphorous being more than arsenic and antimony is absolutely correct. But what you missed, is that lower down the group the direction of dipole itself reverses.

E.N. of $\ce{P}$ is $2.19$, while that of $\ce{As}$ is $2.18$ and of $\ce{Sb}$ is $2.05$

Whereas H as an E.N. of $2.20$. So hence now you can see clearly that your order is justified, as in actuality, H becomes the negative end of the dipole.

Another factor that further widens this gap is that the dipole vector in chemistry notations is from + to-, which means all vectors add up in case of $\ce{As}$ and Sb


Will dioxygen difluoride set fire to sand? I've heard that dioxygen difluoride (a.k.a FOOF) would set fire or react with nearly everything including sand ,bricks or even ice. Is this true?

It is hard to be sure as it is too reactive to test
FOOF has been made, though no sane person would want to repeat the most recent synthesis (see the paywalled article reporting it) as it involves passing oxygen a fluorine through a red hot metal tube. An earlier paper that tested its reactivity stored it as a solid at 90K to stop it from, well, exploding. Derek Lowe, in his hilarious blog "Things I Won't Work With" describes some of the (very low temperature) reactions:

And he’s just getting warmed up, if that’s the right phrase to use for something that detonates things at -180C (that’s -300 Fahrenheit, if you only have a kitchen thermometer). The great majority of Streng’s reactions have surely never been run again. The paper goes on to react FOOF with everything else you wouldn’t react it with: ammonia (“vigorous”, this at 100K), water ice (explosion, natch), chlorine (“violent explosion”, so he added it more slowly the second time), red phosphorus (not good), bromine fluoride, chlorine trifluoride (say what?), perchloryl fluoride (!), tetrafluorohydrazine (how on Earth. . .), and on, and on. If the paper weren’t laid out in complete grammatical sentences and published in JACS, you’d swear it was the work of a violent lunatic. I ran out of vulgar expletives after the second page. A. G. Streng, folks, absolutely takes the corrosive exploding cake, and I have to tip my asbestos-lined titanium hat to him.

Many other reactivity experiments were not possible because they are just too impractical with something so dangerously explosive at temperatures any normal person might want to conduct reactions at (anything much above liquid nitrogen is bad for FOOF).
So, I suspect, we don't know whether it would set fire to sand as nobody has ever been able to make enough of it to test that reactivity, or, at least, test it and survive.
FOOF's–far more benign–relative chlorine trifluoride, which is commercially available and has several uses, most certainly does set sand on fire. As rocket scientists know. This account is from John D Clarke's Ignition:

It is hypergolic with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water—with which it reacts explosively.

So, given ClF3 is far less reactive than FOOF, it seems likely that FOOF, too, would set fire to sand. But with any luck we will never know for sure as nobody will be so monumentally, suicidally insane to try making enough of it to find out.


Why doesn't 4 hydrogen and 1 carbon form a ionic bond? What is the limit of electrons that can be given away? I understand that it's a covalent bond, but to my uneducated mind isn't it possible to have the carbon give 1 electron to each of the hydrogens and form an ionic bond instead? I read this response

Why can't carbon form an ionic bond?

which kind of makes sense, which leads me to my main question. Is there a limit to how many electrons(in this case) a carbon can drop/give before it's not possible? Does it depend on how badly the other(s) elements want to take/give electrons in said reaction?

As in, it has to give up 4 electrons, would it be fine it if had to give up 2? etc, hopefully my question makes sense.

I can understand your reasoning. But we need to consider many factors. When a compound is to be formed it should be stable enough. Or else the compound breaks down. Here the electronegativity of Carbon is 2.55 and Hydrogen is 2.2 (According to  Pauling's scale). So the electronegativity difference between these two atoms is 0.35. In the table given below...,

We can see that the nature of bonds between atoms of different electronegativity. As the difference is 0.35, the bond is Non-polar. And also for removing electrons from carbon are...,
Ionisation energy number & Enthalpy(kJ/mol)
1st------------------------->1086.5;
2nd------------------------->2352.6;
3rd------------------------->4620.5;
4th------------------------->6222.7.

So total energy required is 14282.3 KJ/mol. But electron affinity of hydrogen is 72.8 KJ/mol.So four hydrogens give 291.2 energy. So totally we need 13991.1 KJ/mol of energy to form methane.

While for removing successive hydrogen atoms from methane the bond-dissociation energies are 104 kcal/mol (435 kJ/mol) for D(CH3–H), 106 kcal/mol (444 kJ/mol) for D(CH2–H), 106 kcal/mol (444 kJ/mol) for D(CH–H) and finally 81 kcal/mol (339 kJ/mol) for D(C–H). The bond energy is, thus, 99 kcal/mol or 414 kJ/mol (the average of the bond-dissociation energies). Here the formation of methane is exothermic and is spontaneous while in the above situation it highly endothermic. All these factors give us why C-H bond is covalent but not ionic. Also you can do the same for either two bonds or three bonds anything. So we can say that $\ce{CH4}$  is covalent and cannot be ionic. And two bonds also cannot be ionic.

But in October 2016, astronomers reported that the very basic chemical ingredients of life—the carbon-hydrogen molecule (CH, or methylidyne radical), the carbon-hydrogen positive ion (CH+) and the carbon ion (C+)—are the result, in large part, of ultraviolet light from stars, rather than in other ways, such as the result of turbulent events related to supernovae and young stars, as thought earlier. So we can say these ions are in existence in our universe.

I hope it helps you.
I'm new to Stackoverflow so sorry if I've made any mistake in giving your desired answer
.


Carbocation formation and stability from bromo compounds When heated,the following bromo-species all form carbocations via $\sigma$ bond cleavage. For each structure draw the corresponding carbocation and briefly describe how it is stabilised.

This is my attempt at answering the question:

I would like to know whether I have drawn the correct carbocation forms of each of these bromo-compounds and if my reasoning is sound.

Yes, both your answers and reasoning are correct. For the second compound a slight ambiguity might arise, because the $\mathrm{sp^2}$-hybridized carbons show $\text{-I}$ effect which would destabilize the carbocation, however considering the fact that there are not only  $\ce{2 Me}$— groups attached which show $\text{+I}$, they are also closer than one of the $\mathrm{sp^2}$- carbons, it is quite natural that the net inductive effect would stabilize it.


Despite being less reactive,alkanes are highly flammable I read in my book that alkanes are saturated compunds and the valencies of carbon are satisfied by single bonds. I knew that propane and butane are highly flammable and could cause explosion.
My question
If alkanes are less reactive then why they are highly flammable and could cause unfortunate explosion?

In chemistry, reactive and flammable are two different term. For example, Diethyl ether is a rather non-reactive material but it is highly flammable.
In your case, propane or butane are mostly unreactive to common chemicals, but it does react with oxygen at an elevated temperature- that is they are flammable. Under similar condition, propene or butene are more flammable.


Calcium carbonate and hydrochloric acid I am trying to solve an exercise, where a block of 605.5g calcium carbonate should be completely dissolved in 30% hydrochloric acid (w/w) and the concentration of the acid should be 3% in the end. The question is: How many g of 30% HCl-acid is needed in the beginning.

$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$

So for each dissolved molecule of $\ce{CaCO3}$:

2 molecules of $\ce{HCl}$ are used
1 molecule of $\ce{H2O}$ is produced
1 molecule of $\ce{CaCl2}$ is produced
1 molecule of $\ce{CO2}$ is produced, but leaves the solution

I got a result (about $\pu{1671.524g}$ 30% hydrochloric acid), which I verified this way:

Let $p$ be the number of dissolved $\ce{CaCO3}$ moles, which is about $\mathrm{605.5/100.0869 = 6.05}$

$\ce{HCl = 1671.524g * 0.3}$

$\ce{H_2O = 1671.524g * 0.7}$

$\ce{HCl_{end} = HCl - 2 * (p * M(HCl)}$)

$\ce{H2O_{end} = H2O + p * M(H2O)}$

$\ce{CaCl_{2end} = p * M(CaCl2)}$

$c_{end}$ = $\ce{\frac{HCl_{end}}{HCl_{end} + H2O_{end} + CaCl_{2end}} = 0.03}$

But my solution is claimed to be incorrect. Am I missing something here?

We'll assume the reaction loses only $\ce {CO2}$ from the system (although it is a exothermic reaction, we'll assume $\ce{H2O}$ produced would not be lost as vapors):
$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$

Mass of $30\%~\ce{HCl}$ needed to react with $\pu {605.5 g}$ of $\ce{CaCO3}$ $\pu {= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~ \ce{CaCO3}}\times \frac{\pu{2mol}~\ce{HCl}}{\pu{1 mol}~\ce{CaCO3}}\times \frac{\pu{36.45g}~\ce{HCl}}{\pu{1mol}~\ce{HCl}}\times \frac{\pu{100g}~\text{solution of 30%}~\ce{HCl}}{\pu{30 g}~\ce{HCl}}= 1470.04 g of 30\% \ce{HCl} solution}$.

Similarly, mass of $\ce{CO2}$ released when $\ce{HCl}$ reacted with $\pu{605.5 g}$ of $\ce{CaCO3}$ $\pu{= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~\ce{CaCO3}}\times \frac{\pu{1mol}~\ce{CO2}}{\pu{1mol}~\ce{CaCO3}}\times \frac{\pu{44.0g}~\ce{CO2}}{\pu{1mol}~\ce{CO2}} = 266.18 g \ce{CO2}}$.

According to the law of conservation of mass,

$\pu {mass of reactant = mass of products}$

Assuming no solvent left the system,

$\pu {mass of reactant + solvent = mass of products + solvent}$

$\pu{mass of reactant + solvent = 1470.04 g + 605.5 g = 2075.54 g}$

$\pu{mass of products + solvent remained in the flask = mass of reactant + solvent - mass of \ce{CO2} released = 2075.54 g - 266.18 g = 1809.41 g}$

Now, suppose extra $\pu{A g}$ of $30\%~\ce{HCl}$ added to the solution, so that final concentration is $3\%~\ce {HCl}$.

$\pu {mass of \ce{HCl} in A g of 30\% solution = A g of 30\% \ce{HCl} solution \times \frac {\pu{30g}~\ce {HCl}}{\pu{100g}~ \text{of 30% $\ce{HCl}$ solution}} = 0.3A g of 30\% \ce{HCl}}$

Thus, since final concentration of solution is $\mathrm {3\%}$,

$$\pu {\frac {(0.3 A) g}{(1809.41 + A) g} = 0.03}$$
$$\pu {0.3 A = 0.03 \times 1809.41 + 0.03A}$$
Thus, $\mathrm {A} = \frac {0.03 \times 1809.41}{0.3-0.03} = \pu{201.04 g}$

Finally, the mass of $30\%~\ce{HCl}$ initially added is $\pu {(1470.04 + 201.04) g = 1671.1 g}$.


Difference in mechanisms between formation of acetal and enamine When forming acetal from a C=O group with alcohol, the alcohol will act as a nucleophile and attack twice as the carbon is made very electrophilic twice.
I wonder in the last intermediate before forming the enamine, is it able for the 2' amine to act as a nucleophile and attack the carbon bonded to a highly electronegative N with a positive formal charge? (like the attack from the alcohol when forming acetal) Then with the help with 1 proton transfer, forming a product that has 2 RR'N?
My thought is that it may be too sterically hindered compared to the alcohol when attacking. But is it valid?

This does indeed work. One such reference includes US patent 3696080A in which propionaldehyde is reacted with dimethylamine to give the N,N-acetal as shown in the figure below.

The described reaction procedure involves cooling both reagents to $\pu{-78^\circ C}$ before addition, letting the temperature rise to around $\pu{-35^\circ C}$ at which point anhydrous potassium carbonate was added, letting the temperature rise again to about $\pu{0^\circ C}$ at which point the cooling apparatus was switched from dry ice/acetone to ice/water, and then stirring it overnight allowing the ice to melt in the process. The workup the inventors describe is again rather complex but they claim to isolate up to $\pu{73\%}$ of the bis-amino product.
SciFinder also gave a number of other references but one should serve sufficiently as a proof of concept.
It’s worth noting that $\pu{2.4 eq}$ of amine were added which probably assisted in driving the reaction to the product side.


Why is methyl α-D-glucopyranoside preferentially formed from glucopyranose in acidic methanol? According to my textbook Organic Chemistry 2e by David Klein, the formation of a glycoside from glucopyranose favors the α form (see picture below).

Why does this happen? Shouldn't the equilibrium favor the β form, where the methoxy group is in an equatorial position and has less steric interactions?

In general, if there is overlap between a filled orbital and an unfilled orbital, a stabilising interaction will take place between the two. A simple example is the resonance present in esters, where the ester oxygen lone pair (a filled orbital) overlaps with the C=O π* orbital (an unfilled orbital):

In this context, anomer refers to a diastereomer differing only in the configuration at C-1. So, the two diastereomeric products you have drawn above are often called the α and β anomers.

In the α anomer, there is one lone pair on the ring oxygen (a filled orbital) which can overlap with the C–OMe σ* orbital (an unfilled orbital):

This leads to a stabilisation of roughly 6 to 7 kJ/mol, and is called the anomeric effect.

In the β anomer, neither of the two lone pairs on the ring oxygen overlap well with the C–OMe σ*.

Of course, steric effects would disfavour the α anomer, which possesses an axial methoxy group. However, this factor only leads to a destabilisation of roughly 3 to 4 kJ/mol, and is therefore outweighed by the stereoelectronic stabilisation provided by the anomeric effect.

The rough numbers are taken from Kirby, A. J. Stereoelectronic Effects (OUP), p 16.

While the lone pair -> σ* explanation is the most commonly invoked, there are also likely to be other factors that are also present: for example, in the α anomer, the dipole moments of the ring oxygen lone pair and the exocyclic C–O bond oppose each other, which is a favourable electrostatic interaction:


Comparing the solubility of glucaric acid and mucic acid Why is glucaric acid soluble in water but not mucic acid? The chemical formulae seem to be the same so why is it that one is more soluble than the other?

Wikipedia lists the melting points of the sugar acids as:

Glucaric (or saccharic) acid: $\pu{125-126 ^\circ C}$

mucic (galactaric) acid: $\pu{230 ^\circ C}$.

The very great solubility of all the sugary materials (glucaric acid: $\pu{63 mg/mL}$1, or perhaps $\pu{912 mg/mL}$2) is due to their hydroxyl groups, which are very compatible with $\ce{H2O}$. Mucic acid, even though, it has as many hydroxyls as glucaric, is soluble only to the extent of $\pu{3.3 mg/mL}$, although it was predicted from its structure to dissolve reasonably well ($\pu{63.1 g/L}$3). (Note the differences in units; I used the same numbers and units as in the original reference.)
The likely reason for the great difference in solubility and melting point is the way the mucic acid crystal packs. This has shown up in density and x-ray crystallography4.
Looking at the structures, or playing with molecular models (typically, one molecule at a time - you never have enough atoms to make a half a dozen molecules, nor the time to rotate them all and pack them, nor the straight-line vision to see how well they pack) doesn't give you the same sense of difference that the solubility and melting point give.
Maybe this is not terribly unusual: tartaric acid(s), a simpler diacid, shows similar differences in solubility and melting point just because of the chirality of the halves of the molecule. Melting points: D or L: $\pu{171-174 ^\circ C}$; DL $\pu{206 ^\circ C}$; meso: $\pu{147/ 165 ^{\circ} C}$. Solubilities (T=?): D or L: $\pu{1.33 kg/L}$; DL: $\pu{0.21 kg/L}$; meso: $\pu{1.25 kg/L}$ (Wikipedia).
References

https://www.drugbank.ca/drugs/DB03603
https://pubchem.ncbi.nlm.nih.gov/compound/Glucaric-acid#section=Experimental-Properties
https://hmdb.ca/metabolites/HMDB0000639
The crystal structure of galactaric acid (mucic acid) at −147 °C: An unusually dense, hydrogen-bonded structure by George A.Jeffrey, Richard A.Wood, Carbohydrate Research, Volume 108, Issue 2, 16 October 1982, Pages 205-211, DOI: 10.1016/S0008-6215(00)81790-5


Does there exist a molecule of formule XαYβ where X and Y are elements that are also found in molecule XγYδ? I am not a chemist, but I will be explaining chemistry to some third-graders. I want to give the example of how different things can be made of the same parts, but arranged differently. My "real world" example is a car and a motorcycle. The car has an engine and four wheels, but we can create a different vehicle with an engine and only two wheels: we get a motorcycle. Does there exist a molecule composed of the elements $\ce{X}$ and $\ce{Y}$, where $\ce{X}$ and $\ce{Y}$ can form a different molecule when the quantities of $\ce{X}$ and/or $\ce{Y}$ atoms are different?
I was hoping to use carbon and hydrogen because the children have already heard me use those words, and $\ce{CH4}$ is a gas that will make them laugh when I tell them where it comes from. But the only $\ce{CH2}$ molecule that I could find is this, which is too complicated and unfamiliar to explain. I would prefer molecules composed of the common elements hydrogen, oxygen, carbon. I would like to avoid molecules of elements that are commonly found in their elemental state such as gold, iron, or silver as at this stage I want to reinforce the idea that we cannot see molecules with the unaided eye. I might introduce the idea that a molecule (and macromaterial) can be composed of a single element later, if this lesson goes well.
For purposes of discussion, a molecule can be defined as "a discrete body composed of atoms". There is no need to address quantum chemistry, as the target audience is a third-grade classroom.

$\ce{GeF2}$ and $\ce{GeF4}$
$\ce{GeCl2}$ and $\ce{GeCl4}$
$\ce{GeBr2}$ and $\ce{GeBr4}$
$\ce{GeI2}$ and $\ce{GeI4}$
$\ce{SnF2}$ and $\ce{SnF4}$
$\ce{SnCl2}$ and $\ce{SnCl4}$
$\ce{SnBr2}$ and $\ce{SnBr4}$
$\ce{SnI2}$ and $\ce{SnI4}$
$\ce{PbF2}$ and $\ce{PbF4}$
$\ce{PbCl2}$ and $\ce{PbCl4}$
$\ce{XeF2}$ and $\ce{XeF4}$
$\ce{XeO2}$ and $\ce{XeO4}$
Silicon dihalides ($\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$) have also all been prepared in the gas phase, but are much less stable than the tetrahalides.


Concept of Adjusted Retention Time in Chromatography Why is the relation between the adjusted retention time, retention time and dead time as follows :

tR' = tR - tM ?

To find the time that the sample spends in stationary phase (adjusted retention time), we just need to subtract the time, in which it is in the column but not in stationary phase. Right ?

But dead time is different from this idea. How does it help in finding the time spent in stationary phase ?

To find the time that the sample spends in stationary phase (adjusted retention time), we just need to subtract the time, in which it is in the column but not in stationary phase.

This is a misconception (a subtle point here). The adjusted retention time is the time an analyte spends in the column not the stationary phase. Remember that a chromatography column is made up of a stationary phase plus the mobile phase. The sample continuously partitions between the mobile phase and the stationary phase.

Dead time simply refers to the time it takes for a sample to travel from the injector to the detector without having any interaction with the stationary phase. Imagine a garden hose, when you open the tap, it take a while for the water to reach the other end. Dead time is exactly like that.

Calculating the fraction of the time spent by an analyte in the stationary phase is slightly more involved. Check this link on "Migration rate of solutes in a chromatographic separation"

https://web.njit.edu/~kebbekus/analysis/chromatography%20overheads.htm


salt hydrolysis and pH calculations I am a high school student and I am very confused in salt hydrolysis
My confusion is that: when we talk of hydrolysis of salt of weak base and weak acid, we internally assume the degree of hydrolysis of both cation and anion of the salt is assumed to be same but I think it is a very wrong assumption, why don't we simply do the pH calculation like I have done in this example? what is the problem with this?,if my approach is also correct then why doesn't it matches with the standard answer? what is the mistake in it? I have found some similar questions but no one has given the exact answer to those that's why I am asking it here please, explain it in a simplified way because I am only at high school level.

The general approach to solve acido-basic equilibrium problems is:

respecting charge balance and charge conservation
respecting mass conservation
respecting dissociation equilibrium constants for acids, bases and water itself.

Generally, it leads to a set of nonlinear equations for a set of independent variables. The application of a substitution method usually results to a polynomial equation for $\ce{[H+]}$ of the second up to the forth order. The latter is hard to solve analytically.
Fortunately, in many scenarios, there can be made justified simplifications, like

neglecting water autodissociation
neglecting concentration of one substance form wrt the other one or their sum.

That largely simplifies an equation to be solved.
E.g. to calculate $\mathrm{pH}$ of solution of acetic acid, ammonia, or sodium carbonate,
we can consider none of $\ce{H+}$, resp. $\ce{OH-}$ are created by the water autoionization and just negligible part of the compound underwent the acido-basic reaction.
E.g. for sodium acetate hydrolysis
$$K_\mathrm{b}=\frac{\ce{[HA][OH-]}}{\ce{[A-]}}\approx
\frac{{\ce{[OH-]}}^2}{\ce{[A-]}}\approx \frac{{\ce{[OH-]}}^2}{c_0}
 $$


Can photons also work as a catalyst? A lot of biological reactions need photons to work, like photosynthesis and making vitamin D in our skin. But are there any chemical reactions where photons are really just a catalyst so they are absorbed and emitted the same as the came? Or is there always a change in frequency/wavelength involved which implies perhaps that the photon is 'used'. Or is perhaps the use of the word catalyst in case of photons not very appropriate?

I do not feel that the word catalyst applies to photochemistry.

Photochemistry allows to access much higher energy states and thus reach reaction pathways that are otherwise not possible. Photons do not stabilize transition states (how would you rationalize the interaction?).

Catalysts on the other hand are focused on lowering barriers so that reactions proceed via lower energy transition states.

Because catalysts only affect the transition states, catalysts cannot affect the equilibrium position of a reaction. For example, if you had a catalyst that allowed easier conversion between (E)- and (Z)- double bonds, you could not use the catalyst to selectively provide the one that is less favorable.

On the other hand, with photochemistry, we do not establish an equilibrium because the higher energy state is not necessarily accessible via microscopic reversibility. And indeed, we can selectively use photochemistry to provide the less favorable species as is well know in the case of stilbenes.

EDIT:

Thanks to airhuff for pointing out that the requirement that a catalyst not be consumed also implies that we should not think about photons as catalysts. Hard to imagine a situation where you would absorb a photon and release one that is the same energy...


R/S chirality question Consider [(1S,2S,4S)-bicyclo[2.2.1]hept-5-en-2-yl]acetic acid:

(Source: http://www.chemspider.com/Chemical-Structure.22823922.html)

According to the chemspider nomenclature, this has three S stereocenters.  However, when I perform the CIP both manually and computationally, the center labeled ? comes up as R.  I am trying to figure out if I am misunderstanding CIP (particularly the expansion part for handling double-bonds) or if the literature is wrong in this case. I would greatly appreciate if some chemist here could work out this example to give me another data point.

According to the Cahn–Ingold–Prelog priority rules, a double bond considered as ($\ce{CCX}$), where $\ce{X}$ depends on the substitution on the double bond. In your case, $\ce{X}$ is hydrogen ($\ce{H}$). Thus, I arranged the priority as depicted in following diagram (chiral carbon in consideration is marked with asterisk):

Hydrogen is the least priority group on the chiral carbon. Among other three groups, the bridgehead carbon is a $\ce{-CH2 -}$ group, hence can be marked as ($\ce{CHH}$) on priority checking. The final two groups can be marked as ($\ce{CCH}$). I used red and blue color on one group (methine on the ring), and black on the other (double bond) to distinguish these two group. Since both has identical substitutions on first carbon, we need to moved to second carbon. On double bond, it is again ($\ce{CCH}$), but bothe red and blue carbon give ($\ce{CHH}$) priority checking. Thus, double bond has the highest priority as indicated. Therefore, assignment for this chiral center is ($S$)- as given in chemspider website.


What is the electronic configuration of Fe(II) ion? The electronic configuration of Fe is $\ce{[Ar] 3d^6 4s^2}$.
So after removing two electrons the configuration becomes:
$\ce{[Ar] 3d^6}$

But why can't the electrons rearrange themselves to give a more stable
$\ce{[Ar] 3d^5 4s^1}$ configuration?

When researching the explanation for this phenomenon, I found this:

  It is natural to question why one or two electrons are usually pushed into a higher energy orbital. The answer is because $\mathrm{3d}$ orbitals are more compact than $\mathrm{4s}$, and as a result any electrons entering $\mathrm{3d}$ orbitals will experience greater mutual repulsion. The slightly unsettling feature is that although the relevant $\mathrm{s}$ orbital can relieve such additional electron-electron repulsion, different atoms do not always make full use of this form of sheltering because the situation is more complicated than just described. One thing to consider is that nuclear charge increases as we move through the atoms, and there is a complicated set of interactions between the electrons and the nucleus as well as between the electrons themselves. This is what ultimately produces an electronic configuration and, contrary to what some educators may wish for, there is no simple qualitative rule of thumb that can cope with this complicated situation.

On a semi-related note, according to one source I found (the source sites many others as well), it is in fact the $\mathrm{3d}$ electrons that fill first, followed by the $\mathrm{4s}$ electrons. This more satisfactorily explains why the $\mathrm{s}$ electrons are first lost in the ionization of the $\mathrm{d}$-block transition metals.


How is the acid HSO₄⁻ formed? How is $\ce{HSO4-}$ formed? If Hydrogen is 1+ and $\ce{SO4}$ has 2- charge? I don't understand this. Could someone please clarify?

$\ce{HSO4-}$ look at the valencies of the hydrogen and sulphate ion, because of the numbers don't match i.e. hydrogen has a +1 charge that means it is deficient by 1 electron while sulphate has a -2 charge which means it has two extra electrons (it can also be viewed that it wants to 'give away' these two electrons) these valence numbers basically show the amount of bonds the molecules can form (since at this level of chemistry bonds can be seen as the sharing of electrons) thus the sulphate can form two while the hydrogen can only form one and in this case since it forms $\ce{HSO4-}$ that means there is one 'unhappy electron' or one that is unbonded since the bond between the sulphate and hydrogen makes hydrogen's charge neutral and takes an electron from the sulphate ion leaving the unbonded electron thus giving the molecule an overall charge of -1. Note that here I am just explaining the bonding that is occurring between the molecules but $\ce{H2SO4}$ DOES form I am just explaining the charges and values involved and also note that bonding and chemical reactions usually involve the VALENCE electrons!.

However this molecule ($\ce{H2SO4}$) will disassociate in water or solvent of some sort, and form ions as the person before me pointed out. into $\ce{H+}$ ions and $\ce{HSO4^-}$ thus forms the conjugate base of the acid ($\ce{H+}$ is the basis of most acidic reactions)

If you're confused as to why $\ce{H2SO4}$ doesn't form, it does but $\ce{HSO4-1}$ is what happens after it disassociates in solution.

I don't know which part of the question or problem is where your confusion lies, hope I answered or helped!

(I am only a first year university student if my answer is wrong please notify me but it correlates with my notes, textbooks and what I was taught back in high school)


Solubility of N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide at different pH The molecule below is a simplified analog of capsaicin. Explain in which of the following options, it should be most soluble:

  aqueous HCl (pH = 1)
  aqueous NaOH (pH = 14)
  pure water (pH = 7)

  N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide, National Center for Biotechnology Information. PubChem Compound Database; CID=606880

I chose aqueous NaOH, because then the molecule will be completely deprotonated, which will make it more soluble. Please tell me if this is correct.

I would agree that it is most soluble in NaOH.

However, the molecule will only deprotonate the OH proton not the NH proton.  I wouldn't give your explanation much, if any, credit because you aren't saying which proton would deprotonate, and you are implying the molecule will lose all protons.

Also, at low enough pH the N will protonate, i.e. it would gain a second proton.


At what point does decreasing solvent temperature cause a decrease in gas solubility? Example: At constant pressure, carbon dioxide becomes less soluble in water as temperature increases. We also know that carbon dioxide becomes more soluble in water as temperature decreases.

Does introduction of order and loss of kinetic energy of solvent molecules eventually lead to a decrease in a solvent's ability to accommodate gas molecules? It is hard for me to believe that ice cubes can hold gas more effectively than the same volume of a glass of water.

Rules such as "solubility of A in B rises with lower temperature" are only meant to be used if there is no phase transition of the participating substances.


Rules to follow while writing a chemical formula for polyatomic ions In Science, textbook for class IX, chapter $3$ [1,p.38], it states:

The rules that you have to follow while writing a chemical formula are as follows:

in compounds formed with polyatomic ions, the number of ions present in the compound is indicated by enclosing the formula of ion in a bracket and writing the number of ions outside the bracket. For example, $\ce{Mg(OH)2}$. [...]

Now my book further states that hydroxide ($\ce{OH–}$) and nitrate ($\ce{NO3-}$) are polyatomic ions. And then it says:

Formula of sodium nitrate:  $\ce{NaNO3}$

But the formula of calcium hydroxide:  $\ce{Ca(OH)2}$
My question is that even though both nitrate and hydroxide are polyatomic, why are we using ( ) in notation for latter in their symbolic representations?

What they are trying to say is that hydroxide $\ce{OH^−}$ and nitrate $\ce{NO3^−}$ ions, which are polyatomic anions react bond with cations as a group.
When $\ce{NO3^−}$ bonds to $\ce{Na^+}$ only a single nitrate anion bonds to a single sodium cation, to form $\ce{NaNO_3}$ so brackets are not needed. You can still use them, in which case it would be $\ce{NaNO_3}$ but it is redundant and current convention doesn't use it.
However, when $\ce{OH^−}$ bonds to $\ce{Ca^{2+}}$ two hydroxide anions bond to a single calcium cation, to form $\ce{Ca(OH)_2}$ so brackets are needed to avoid confusion and show the structure of the compound.
Old ways of writing it (before my time) before the conventions were standardised, such as $\ce{CaO2H2}$ or $\ce{CaH2O2}$ don't immediately show that it is a hydroxide, so would have been confusing to non-experts.
To reiterate: the brackets are used when there is more than one polyatomic ion in a compound.  So if you write down sodium hydroxide it is $\ce{NaOH}$ but calcium hydroxide is $\ce{Ca(OH)2}$


Conversion of methane to propane Is it possible to chemically convert methane into propane?

If so, what is involved and is it economical?

Yes, it can be done by mimicking the conditions in the earth's mantle.

It was shown by Alexander Goncharov and coworkers [1] by placing methane in a diamond anvil cell, applying a pressure of more than 2 GPa and heating everything to a temperature around 1000-1500 K with a laser.

Is it economical? Hardly, but very interesting anyway!

Kolesnikov, A.; Kutcherov, V. G.; Goncharov, A. F. Nature Geosci 2009, 2 (8), 566–570. DOI 10.1038/ngeo591.


Is liquid hydrogen rocket propeller pollution free? If one searches for the different fuel types used in rockets, one can find that over the years, NASA and other space agencies have used both solid fuels and liquid fuels. I can see that more and more rockets are propelled by using a combination of Liquid Hydrogen (LH) - Liquid Oxygen (LOX). Then, I assumed that the reaction would be:
$$\ce{2H2(l) + O2(l) ->[heat] 2H_2O(v)}\tag{1}$$

By looking at this reaction, one can see that the product of this reaction is water (which most probably will be in a vapor state). Now, I have three questions:

Is this reaction correct?
If the product is water, does that mean that such a combustion reaction is pollution free?
I assume that I am wrong thinking that is pollution free. What would be the polluting agent in this situation and how bad is it, compared to let's say the burning of the same amount of Kerosene?

Thank you!

I'll address you questions in order:

1) Yes, that is basically the correct reaction, although not quite balanced;  here is how I would write it:

$$\ce{2H2(l) + O2(l) -> 2H_2O(g)}$$

2) Ideally, if there were no side reactions, then yes, this particular reaction is "polution free", and the only product is water.

3) You are correct in your thinking that in the end this does not represent a net pollution-free energy source.

The reason for this doesn't lie in the reaction taking place at the rocket, but in how they produced the hydrogen in the first place. Unfortunately, there are no significant deposits of $\ce{H2}$ on our planet waiting to be pumped out and used as fuel. So we  have to make it, which requires engergy.

This means that the $\ce{H2}$ is not a net energy source, rather it is essentially acting as a battery, storing the energy from some other source. So, you can think of the actual pollutants of the $\ce{H2}$ rocket fuel as being the products from whatever process went into generating it in the first place.  This largely depends on what part of the world the hydrogen generation took place. They could have burned coal or other fossil fuel to produce the energy, or they could have built a dam to generate hydroelectric power, etc.  Any of the negative aspects of producing a particular type of energy would be considered a negative product of the $\ce{H2}$ production.


How is silica bead bonded into polystyrene? What kind of bonds silica bead forms with polystyrene? Silica is now as ~100 $\mu$m spherical particles in bulk polystyrene that is made by melting.

How can I bond the silica particles

more tightly
more loosely

to the polystyrene?

Can this be generalized to any nano/micro particle and plastics? I am thinking now metal oxides and other thermoplastics.

You can functionalize silica to add polymer "brushes" to the surface. You can also buy pre-made functionalized silica, although the downside to that is it isn't always easy to get the manufacturer to tell you exactly what is on the surface.

This isn't really going to increase the attractive interactions between the silica and the polystyrene, however. Polystyrene has a bulky aromatic ring hanging off of each monomer, and no sites that can participate in hydrogen bonding. When you add hydrophobic brushes to the surface of the silica, what you are doing is limiting the ability of the silica to aggregate due to hydrophilic interactions with itself. As a result, given enough kinetic energy (temperature for melt casting, or sonication for solvent casting) you can get the particles to disperse. Then, if you solidify the polymer quickly enough, they will be kinetically trapped in a dispersed state.

Depending on what your goal is, this could be enough for your application. However, if you want to get the particles to strongly attract to the polystyrene, you aren't going to be able to get much farther than van der Waals interactions. That's a limitation of the polystyrene, not the silica.

Other particles (gold, or carbon nanotubes, for example) will be more hydrophobic and easier to disperse without functionalizing, but you still are not going to get very strong attractive forces between the particles and the polymer.


Why is a LCAO necessary within Hartree Fock? As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.

What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?

Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.

Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set –  is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":

$$\hat{1} = \sum_i |i\rangle\langle i| \quad \Longleftrightarrow \quad |f\rangle = \hat{1}|f\rangle = \sum_i |i\rangle\langle i | f \rangle = \sum_i c_i|i\rangle \text{ where }c_i = \langle i | f \rangle$$

Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.


Why do amino acids hydrogen bond rather than ionic bond when they are forming secondary structures in proteins? When peptide bonds are formed between amino acids, electron delocalisation causes the N to be more positive and the O to be more negative. As a result, why does 'hydrogen bonding' occur to form secondary structures such as alpha helices and beta pleated sheets, rather than 'ionic bonding'?

Is this simply a case of semantics, or is it the lack of transfer of an electron that makes the distinction between the two?

(this should be a comment)
I think you are not quite understanding the concept of partial charged or charged. The N atom in the polypeptide chain (so the N in the peptide bond) is partially charged! to be in an ionic state it should have 4 bonds instead of 3.
 The peptide bond which is formed(above) just contains a N atom which is bounded to three other atoms. The oxygen atom is also partially charged but not charged! read this about H-bonding and partial charge


What type of bonding occurs in isolated silicon? Would isolated silicon engage in network covalent bonding as quartz does or would it engage in a different kind of bonding? Would germanium display similar qualities?

I am asking what is the bonding of silicon, not why silicon is different from carbon.
I am interested in isolated silicon, not silicon as a "hydrosilicon".

I believe it is network covalent, but I am not sure.

In addition to @Mithoron's comments and the Wikipedia he linked:

This is an important consideration in photovoltaics, as silicon forms the basis of the internal structure of solar technology (and image sensors). According to the PVEducation.org page Semiconductor Structure,

  Each silicon atom has four valence electrons which are shared, forming covalent bonds with the four surrounding Si atoms.

Diagrammatically, the website includes the following schematic:

The caption reads:

  Schematic representation of covalent bonds in a silicon crystal lattice. Each line connecting the atoms represents an electron being shared between the two. Two electrons being shared are what form the covalent bond.

In terms of germanium, according to the WebElements page Group 14 Elements,

  Tin and lead are very metallic although one modification of tin known as grey tin has the same diamond structure [as carbon] as does germanium and silicon.


Finding order of acidic strength for given compounds The decreasing order of acid strength of the following compounds:
The given compounds are
Water ($\ce{H2O}$)
Ammonia ($\ce{NH3}$)
Acetylene ($\ce{C2H2}$)
Triphenyl system ($\ce{Ph3CH}$)
a) $\ce{NH3}$ > $\ce{Ph3CH}$ > $\ce{C2H2}$ > $\ce{H2O}$
b) $\ce{H2O}$ > $\ce{C2H2}$ > $\ce{Ph3CH}$ > $\ce{NH3}$
c) $\ce{C2H2}$ > $\ce{H2O}$ > $\ce{Ph3CH}$  > $\ce{NH3}$
d) $\ce{Ph3CH}$ > $\ce{C2H2}$ > $\ce{H2O}$ > $\ce{NH3}$

I gave the answer to be $\ce{Ph3CH}$ > Acetylene > Water > $\ce{NH3}$. Here I have marked the triphenyl system as the most acidic because it has resonance stabilization involved and $\ce{NH3}$ as the least acidic because it has basic nature. Using these two facts I have marked the answer as above however it was wrong. What is the correct order and why?

Here, this was a simple question of checking the most acidic compound out of the four and this would have given you the answer. $\ce{H2O}$ will be most acidic as the negative charge will be formed on an oxygen compared to nitrogen and carbon in the other two cases.
However, the interesting question here is the part in the middle, what about $\ce{C2H2}$ and $\ce{Ph3CH}$? Which is more acidic?
If you take a look at the $\mathrm p K_\mathrm a$ values of the given compounds, you would notice that the values would come out to be:
\begin{array}{l|c}
\hline
\text{Compound} &\mathrm {p}K_\mathrm {a} \\
\hline
\ce{H2O} & 15.7 \\
\ce{NH3} & 38\\
\ce{C2H2} & 24\\
\ce{Ph3CH} & 31.5 \\
\hline
\end{array}
Now, this is the definitive set of values which states the order should be  $\ce{H2O > C2H2 > Ph3CH > NH3}$.

Note: Next part is speculative as I do not have a solid reason for $\mathrm pK_\mathrm a$ values being so for acetylene and the triphenyl system.
A question that may arise at this point is why is this so? Why would a triphenyl system with $3$ phenyl rings be less acidic than acetylene? What about the three rings providing resonance?
This is because in reality, only one phenyl ring provides resonance at a time. Due to the size of phenyl groups, the structure is not fully planar. Another reasoning is that the negative charge lies on an $\mathrm{sp^3}$ carbon whereas in acetylene it is on an $\mathrm{sp}$ carbon.


How can the electronic structure of pentaaquanitrosyliron be explained? The magnetic moment of $\ce{[Fe(H2O)5NO]^2+}$ is 3.87 Bohr magneton, which implies that there are 3 unpaired electrons. Since iron is in $+1$ oxidation state (due to charge transfer with nitrosyl group), it's electronic configuration is $[\ce{Ar}]\ \mathrm{3d^7}$. Having 3 unpaired electron is only possible if the 4th and 5th electron enter the higher energy orbital $\mathrm{e_g}$ (crystal field theory). The $\mathrm{t_{2g}}$ or the lower energy orbital has 5 electrons. Why do the 4th and 5th electron enter $\mathrm{e_g}$ despite a strong field ligand (nitrosyl group) being present?

As explained in Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited Inorg. Chem. 2002, 41, 4-10, the $\ce{Fe}$ is really 3+ oxidation state.

The $\ce{Fe^3+}$ is spin 5/2 and antiferromagnetically couples to spin 1 $\ce{NO-}$ to yield a S=3/2 state.


Does cyclooctatetraene also have a diradical anti-aromatic electronic state? I read that the molecular orbital diagram of cyclobutadiene predicts that it should have a diradical nature as it has two unpaired electrons in two degenerate non bonding orbitals (one in each).
Why does this diradical state exist only in cyclobutadiene? Shouldn't other anti aromatic compounds, like cyclooctatetraene, also exist in a similar diradical state?

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$).
Because of various reasons, this state never truly arises in real-life $\ce{C4H4}$ or $\ce{C8H8}$.

$\ce{C8H8}$ never adopts an octagonal geometry. See, e.g., Why is cyclooctatetraene non-planar...?
 or this image for a diagram of its conformation.

$\ce{C4H4}$ is not really square either; it is more rectangular.

So cyclobutadiene is not really special in this regard. Both $\ce{C4H4}$ and $\ce{C8H8}$ have the same status: if they were to have a regular polygon structure, they would possess a diradical electronic state.* However, neither of them have such a structure, so neither of them have such an electronic state.

* For advanced readers: even square $\ce{C4H4}$ does not really possess a diradical state either, as is explained in Cyclobutadiene - Jahn–Teller effect or not?. This is because the MO theory used to predict the diradical state is not sufficiently advanced. (Technically, the diradical state arises from Hückel theory, and the proof of its non-existence for square $\ce{C4H4}$ arises from post-HF theories, such as configuration interaction.)


Why is triptycene aromatic? I came across this wonderful compound, triptycene.

Why is triptycene an aromatic compound, even though it is non-planar? I thought planarity was a requirement for aromaticity.

The confusion arises because of a simplistic definition of aromaticity.

What matters for aromaticity is the planarity of the relevant part of the molecule. In triptycene there are three planar benzene groups joined together. Each is individually aromatic, but the whole molecule is not. Similarly, triphenyl-methane has three individually planar benzene groups, each of which is aromatic, but the molecule cannot be planar because of steric interactions between the groups. There are many other such examples of complex molecules containing an aromatic component attached to other things.


Approximated formula of acid dissociation constant $K_\mathrm{a}$ can be expressed as $K_{\mathrm{a}} = \frac{[\ce{H^{+}}] [\ce{A^-}]} {[\ce{HA}]}$, however, it can be approximated that $K_{\mathrm{a}} = \frac{[\ce{H^{+}}]^{2}} {[\ce{HA}]}$.

I usually use the approximated equation when I'm dealing with questions that don't provide [$\ce{A-}$]. What I don't understand is why/how can we approximate it to this?

To approximate to $\ce{[H+]^2}$, then $\ce{[H+]}$ and $\ce{[A-]}$ must be pretty much equal, and so this must assume that the acid completely dissociates. However, in practice, do acids dissociate to the point that $\ce{[H+]}$ and $\ce{[A-]}$ are equal? If not, then how can we approximate to $\ce{[H+]}^2$, and are there any mathematics that demonstrate this approximation?

The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.

We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.

We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.

Here is some mathematics that you requested:

  $K_{\mathrm{a}}$ can be calculated using equilibrium concentrations:
  $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$

Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)

Here is an extension to fully show how the two equate. Taken from the same textbook:

  $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$
  Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$


Given this liquid-liquid phase diagram, is the green line the nitrobenzene-rich phase? The temperature-composition diagram for hexane and nitrobenzene at 1atm. The region below the curve corresponds to the compositions and temperatures at which the liquids form two phases. The upper critical temperature, $T_\mathrm{uc}$,  is the temperature above which the two liquids are miscible in all proportions.

Am I right in assuming that the right side (green line) is the nitrobenzene rich phase? Is that always the case? Whichever one has the side with a mol fraction of 1 that side is more abundant? If you don't understand my question just tell me, and I'll clarify

Here is the same diagram with more annotation from the physical chemistry textbook by Atkins:

It shows that above the temperature $T_{uc}$, there is only one phase (the substances mix in any ratio). Below that temperature, there are two phases, with compositions shown by the solid lines. The horizontal line shows that if you try to make a mixture with a mole fraction of $a$ at the given temperature, you will get two solutions with mole fraction $a'$ and $a''$ instead. You can calculate the approximate ratio of the volumes using this technique.

  Am I right in assuming that the right side (green line) is the nitrobenzene rich phase?

Yes, because the x-axis is the mole fraction of nitrobenzene, the phase with higher fraction of nitrobenzene will be that on the right, as you say. It does not necessarily mean that the fraction is higher higher than 50% on the right and lower than 50% on the left, though.


Why is partial pressure used as a metric for oxygen in blood? I know haemoglobin basically acts as an oxygen buffer. But when we talk about non-bound oxygen in blood (or water), we're referring to the amount of O2 in solution (aqueous state). From a biochemistry point of view, the concentration seems like a good metric to me.

So why is it that medical papers report blood oxygenation in terms of partial pressure pO2 instead of concentration?

What does a partial pressure inside a liquid even mean? I tried interpreting the arterial pO2 value of 11% atm (i.e. about 11kPa). When using the formula P = CRT at 310K (body temp), I found an oxygen concentration of about 4.3 mM. This value is 20× higher than the maximum O2 concentration for water according to Henry's law. For those more familiar with DO values in mg/L, it would be 137.6 mg/L.

According to Henry's Law there is a limit to the amount of O2 that can dissolve into water at a given pressure and temperature. I know the lungs experience a slightly lower partial pressure than the atmosphere. So how can the oxygen concentration be so high? Does their metric include the oxygen bound by haemoglobin (which would be misleading at the least), or is my calculation off somewhere?

Or does a pO2 value correspond to the steady-state equivalent concentration at that pressure, in which case the arterial concentration in my example would be 0.14 mM (based on this table in wikipedia).

This similar question didn't answer my question:
What is the state of aggregation (gas, liquid) of oxygen in blood?

The equation P=CRT is the relationship for the oxygen concentration in the gas phase.  To get the mole fraction of oxygen the the liquid phase (blood), you first use the Henry's law constant.  Here is a link that can help:  https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=henry%27s%20law%20constand%20for%20oxygen%20in%20blood

If, after applying Henry's law, you end up with a concentration in units of x ml per unit volume, this result implies that x is the volume that the mass of dissolved gas (solute) would have it were an actual gas at the same pressure as that in the gas phase.


Editions of IUPAC Green Book Why does the URL https://www.iupac.org/cms/wp-content/uploads/2015/07/Green-Book-PDF-Version-2011.pdf (ending in 2011) contain the Green Book labelled 2007, and reprinted in 2008? Does a 2011 edition exist and they just uploaded the wrong file?

Same goes for the URL http://media.iupac.org/publications/books/gbook/IUPAC-GB3-2ndPrinting-Online-22apr2011.pdf (containing the exact same file).

And there is the version https://www.iupac.org/fileadmin/user_upload/publications/e-resources/ONLINE-IUPAC-GB3-2ndPrinting-Online-Sep2012.pdf with different water mark, one page more, and the front matter in a different order.

Wikipedia reads:

  The third edition of the Green Book (ISBN 978-0-85404-433-7) was first published by IUPAC in 2007. A second printing of the third edition was released in 2008; this printing made several minor revisions to the 2007 text. A third printing of the third edition was released in 2011. The text of the third printing is identical to that of the second printing.

So yes, maybe they uploaded the wrong file. But the content of 2011 and 2008 should be 100% identical.


Heat Absorption of Liquids Does same amount of heat absorbed cause the same changes of temperature in the same amount of different liquids? Explain.

No.

To answer your question we will assume certain things here.

Assume  $C_p = C_v$ for most liquids, where $C_p$ is the heat capacity at constant pressure, and $C_v$ is the heat capacity at constant volume.
Assume that $C$ is constant with respect to temperature, which is not the case in most real fluids. Here, in order to leave out the integration, we assume that it is a constant.

Then,

$Q = C\Delta{}T$

As different liquids have different heat capacities, the same amount of heat supplied to the liquid would result in a different $\Delta{}T$, the temperature change.

The higher the heat capacity, the more heat is required to raise a unit mass(or mole) of substance  by a unit temperature. Likewise, more heat has to be removed from the unit mass of substance in order to decrease the temperature by a unit temperature.

In other words, the higher the heat capacity, the more "reluctant" it is to change its temperature, for the same amount of heat supplied.


Partial reactions in redox formation of rust In the iron oxidation that results in rust, the first step is to steal 2 electrons from $\ce{Fe}$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $\ce{OH-}$ according to the reaction $\ce{O2 + 2H2O + 4e-  ->  4OH-}$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically  $\ce{4 OH-}$ comes out of the collision. I imagine that there must be some steps between the reactants and $\ce{4OH-}$. Is it that, first, $\ce{O2}$ steals the two electrons from the $\ce{Fe}$ lattice becoming $\ce{2O-}$ and then, later, each $\ce{O-}$ takes an $\ce{H}$ from $\ce{H2O}$ to form $\ce{2OH-}$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 +  4e-  -> 2O-}$$

$$\ce{2O- + 2H2O  -> 4OH-}$$

But the reaction "$\ce{O2 +  4e-  -> 2O-}$"  implies that $\ce{2O-}$ is more stable than $\ce{O2}$. Is that right?

Bringing up rust seems to indicate confusion between the concept of corrosion and the concept of oxidation. Practically, iron corrosion is the breakdown of iron as a result of oxidation reactions.

To explain corrosion, understand water can self-ionize producing hydronium, $H_3O^+$, and hydroxide, $OH^-$ as well as ionize small amounts of iron causing it to dissolve. The dissolved iron ions collide with the hydroxide ions to form iron hydroxide.  Iron hydroxide may fall out of the solution as it is not very soluble, or if the concentration is low enough the surplus of hydronium donates a proton to the iron hydroxide converting it to iron oxide.

Oxidation, which is a more abstract concept, involves any reaction which results in the loss of an electron.  Instead of considering corrosion, consider other reactions.  In photosynthesis water is oxidized because it loses an electron.  In combustion, fuel is oxidized because it looses an electron.

Chemical reactions can get extremely complex, so it may be prudent to master the language of chemistry before delving into the complexities of specific reactions.  Many reactions are described with the language of redox (Reduction and Oxidation) chemistry.