text:
How to label atoms or groups "backwards" in ChemDraw
I'm trying to draw L-tyrosine with the hydroxyl part of the phenol as HO instead of OH as it appears in the ChemDraw image. Everytime I type in HO, it automatically corrects to OH. I have the settings in ACS Document 1996 style. Is there anyway to make the format render my "backwards" type?

I have version 14.0.0.117 and it works as expected: "HO" instead of "OH". It should automatically rotate the labels.

I have 15.0.0.106 and I don't think it worked when I had version 14 either.

The problem with the question is, that it becomes a software issue, that might have a very specific solution. Support might be the best option there. However, as it is working with my version and you have a newer one, I suppose it should work with yours, too. You can find me in [chat] to talk about it. Have you ticked "File > Preferences > Building/Display > Automatic Atom Label Alignment"? Have you tried entering "OH" instead?

Okay, so apparently typing it in as "OH" works.

If you type "HO". Click on it. Go to Text>flush right and it will also fix it.

ChemDraw is a piece of software that attempts to be as intelligent as possible. In many cases, that is truly helpful, in others it can be annoying.
When labelling an atom in text mode, ChemDraw will automagically assume a label from left to right (i.e. in standard North American/European writing direction) and interpret the label as such. If it finds that the atom has bonds to the right but none to the left, as tyrosine’s hydroxy group has in your example, it will group the label into ‘sensible’ fragments[1] and reverse them. This makes sense if you think of rotating a molecule: If you had drawn tyrosine rotated by $180^\circ$, you would have wanted to add OH on the rightmost atom, not HO. If you then manually rotate the entire molecule within ChemDraw, you will notice the OH flipping into an HO as soon as it graphically makes sense to do so.
Note that if you just hover over the atom in lasso mode or bond adding mode and then hit o (lowercase!) it will automatically add OH (or HO, depending on the orientation) at that position. Hitting o again will remove the additional hydrogen (this holds true for all chemical elements that are attached to single letters regardless whether the atom is attached to a lowercase letter (i.e. o) or an uppercase one (B for boron)).

[1]: Consider OTBS: ChemDraw automatically recognises TBS as an abbreviation for the $\ce{TBS}$ group so when reversing, it becomes TBSO. However ChemDraw does not standardly recognise the triethylsilyl group $\ce{TES}$, so when reversing OTES it arrives at SETO. The preferred way would be to add an abbreviation to ChemDraw.

Note that you can view the known nicknames using File>List Nicknames. You can define your own nicknames using Structure>Define Nickname. Also note that ChemDraw breaks atom labels (not nicknames) into tokens that consists of an uppercase letter followed by any numbers or lowercase letters. Therefore, $\ce{-OH}$ is flipped to $\ce{HO-{}}$, whereas $\ce{-Br}$ is not flipped to $\ce{rB-{}}$.

@ortho How write you are in assuming a righting error there … ;)


text:
Value of mixing ratio and rate constant in Chapman mechanisn
In the Chapman mechanism (which tells about the activities of ozone in the stratosphere), there is a equation for the lifetime ($\tau_{\ce{O}}$) of an oxygen atom, $\ce{O}$ against conversion to ozone, $\ce{O3}$, through a certain reaction is given as
$$\tau_{\ce{O}} = \frac{1}{k_2 C_{\ce{O_2}}n_a^2},$$
where $k_2$ is the rate constant of a reaction involving consumption of $\ce{O2}$ to form $\ce{O3}$,
$$\ce{O + O2 + M -> O3 + M},$$
 $\ce{M}$ is a third body present during the reaction. $C_{\ce{O2}}$ is the mixing ratio of $\ce{O2}$ (which is given as $0.21~\mathrm{mol/mol}$), $n_a$ is the air number density.
Source: Daniel J. Jacob: Introduction to atmospheric Chemistry. 1999, To be published. Chapter 10.
I am confused regarding this mixing ratio of $\ce{O2}$. Shouldn't be this value in the stratosphere different from that in troposphere?
Also, regarding $k_2$, nothing is mentioned about how and where this was measured. I mean if the value of $k_2$ is derived through laboratory experiments, then how reliable is this value for the reactions happening in the stratosphere?

The calculations you refer to are only valid within the "low-pressure limit". 2. The steady state approximation fails, as it overestimates the density of the ozone layer by at least a factor of two. 3. It is explicitly stated that "in the lower stratosphere, a steady state solution [...] would not [...] be expected [...]". 4. The kinetics of certain reactions are probably not so much influenced by gravity as some of the other necessary simplifications, so conducting the experiments in a lab probably yields very reliable results as opposed to the theory.

The entire stratosphere lies within the "homosphere", where the composition of gases is dominated by mixing processes and the atmosphere is thus homogenous.  This reaches up to about 100 km, whereas the stratosphere doesn't go any higher than about 50 km.

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text:
Is magnesium sulfate basic, neutral or acidic?
A question in my college test was whether $\ce{MgSO4}$ is neutral, acidic or basic. I was told to solve this kind of problem by hydrolyzing any ions that do not "come from" a strong acid or a strong base, and if nothing is hydrolyzed, then the salt is neutral. So by this logic, as $\ce{Mg^2+}$ comes from $\ce{Mg(OH)2}$ and $\ce{SO4^2-}$ comes from $\ce{H2SO4}$, I answered "neutral".
My professor marked it as wrong, and said that the salt $\ce{MgSO4}$ is basic. How can you explain this?

Hint: weak bases are usually insoluble.

Basic makes no sense at all. The only way for the solution to be basic is for $\ce{SO4^{2-}}$ to hydrolyze to $\ce{HSO4^{-}}$. The pKa2 for sulfuric acid is 1.99, so that isn't going to happen in a solution of $\ce{SO4^{2-}}$.

@MaxW Suppose you start with neutral water, so that initially hydroxide and hydronium ions are equal. You then add the sulfate anion, the equilibrium constant for the following equation is $K_b = K_w/K_a$ so that $pKb = 12.01 $:  $ \ SO4^{2-} + H2O \implies HSO4^{-} + OH^{-}$ Although you are correct that the reverse direction here is favored, bisulfate is a weak enough acid for the forward direction to be non-negligible. In the process of reaching equilibrium, there will be a generation of hydroxide ions. Since the solution was initially neutral, the final solution is slightly basic....

… This analysis ignores the contribution of the cation. In this instance the two oppose each other, Mg works to make it acidic whereas sulfate attempts to make it basic. Then you have to do a computation to see which one "wins". Had it been sodium sulfate the answer WOULD be basic because sodium hydroxide disassociates completely beneath its saturation limit and thus it would be analogous to a "strong acid" in that the reverse direction in which a sodium ion bonds with a hydroxide ion to form sodium hydroxide is negligible and can be fairly said not to occur.

Looking at the following references, it seems that most agree that a water solution of $\ce{MgSO4}$ is slightly acidic.
"The pH of an aqueous magnesium sulfate solution is related to the molarity of the MgSO4. Typically, the pH is between 5.5 and 6.5 due to magnesium's affinity for hydroxide ion (OH-). As the sulfate goes into solution, hydroxide anions associate with the magnesium, increasing the relative ratio of H+ to OH-. This shift results in more acidic solutions."
"The pH of a pure magnesium sulfate solution is approximately 6.4"
"The pH of [magnesium sulfate] hydrates is average 6.0 (5.5 to 6.5)."
"pH: Neutral to litmus"
Why not perform the experiment in class, using an accurate pH meter and/or indicators (see the pH indicator chart in Wikipedia for suggestions).

+1 for nice answer and all the references. // I'm not so sure that the pH meter experiment would be illustrative. A very low amount of impurities could shift pH acidic or basic. So how do you obtain salt of "sufficient" purity?

"The pH of a pure magnesium sulfate solution is approximately
6.4, a pH distinctly more acidic being obtained by adding e.g. 1 wt% FeSO4 • H2O
(pH: 4.6) or 1 wt% KH 2PO 4 (pH: 3.2)." There is no evidence given for the pH of about 6.4, and they seem to think that adding $\ce{KH2PO4}$ makes the pH more acidic perhaps through some interaction, which is weird.

The wikipedia article no longer makes a claim for the pH of dissolved hydrates.

Pubchem: "Efflorescent crystals or powder; bitter, saline, cooling taste; density: 1.67; pH 6-7; soluble in water (g/100 ml): 71 @ 20 °C, 91 @ 40 °C; slightly soluble in alcohol; its aqueous soln is neutral; ". So they say the pH is between 6 and 7, but then say that the aqueous solution is neutral.

Aqueous solution of magnesium sulfate is going to be slightly acidic, and there is no need to even go to the lab to test it. $\ce{MgSO4}$ is a salt formed by a weak base and a strong acid (both dibasic), so there is a two-step hydrolysis of magnesium cation.
First and primary step is the formation of a basic salt:
$$
\begin{align}
\ce{2 MgSO4 + 2 H2O &<=> (MgOH)2SO4 + H2SO4} \tag{R1.1}\\
\ce{2 Mg^2+ + 2 SO4^2- + 2 H2O &<=> 2 MgOH+ + SO4^2- + 2 H+ + SO4^2-}\tag{R1.2}\\
\ce{Mg^2+ + H2O &<=> MgOH+ + H+}\tag{R1.3}
\end{align}
$$
Second minor (negligible) step:
$$
\begin{align}
\ce{(MgOH)2SO4 + 2 H2O &<<=> 2 Mg(OH)2 + H2SO4}\tag{R2.1}\\
\ce{2 MgOH + SO4^2- + 2 H2O &<<=> 2 Mg(OH)2 + 2 H+ + SO4^2-}\tag{R2.2}\\
\ce{MgOH+ + H2O &<<=> Mg(OH)2 + H+}\tag{R2.3}
\end{align}
$$
Here indices .1, .2, .3 denote molecular, complete ionic, and net ionic equation, correspondingly. As a result of the hydrolysis reaction, some excess amount of hydronium cations appears, therefore, the the medium is expected to be acidic.

$\ce{SO4^2-}$ is actually going to make the pH basic because $\ce{HSO4-}$ is a weak acid (not $\ce{H2SO4}$, only the first hydrogen ionizes completely). By itself aqueous $\ce{Mg^2+}$ will not form magnesium hydroxide in water. But with the $\ce{OH-}$ from the $\ce{HSO4-}$ is will form a suspension with some dissolved and some not so dissolved, but still the solution is basic.

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text:
Predicting Extent of Exothermic Reactions
Sodium Acetate is widely used in hot packs due to the fact that it produces an exothermic reaction when it reacts with water.
I decided to see if other sodium salts, such as sodium bromide, sodium sulphate, etc. would produce similar exothermic reactions in hopes of finding a potential substitute for sodium acetate.
Is there a reason why some of these sodium salts would react with water to produce a larger enthalpy change than others? Is this predictable given the molecules' properties and/or geometry? Can enthalpy change at all be predicted given every property we currently know about these molecules?

Difficult to predict.

It is also worth noting that the exothermic behavior of sodium acetate and water is the exothermic crystallization of a supersaturated solution of sodium acetate in water. Go look up hot ice


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Gases produced by pyrolysis of cellulose
I heated cotton in a sealed container (with a small hole) over a natural gas flame. Some gases and smoke were produced. What would they probably be? I can come up with some guesses based on the composition of cellulose: $\ce{CO2}$, $\ce{CH4}$ or possibly other hydrocarbons, $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, however I do not know which of those they are. Obviously, soot ($\ce{C}$) was also formed, due to the visible smoke particles.

During pyrolysis, organic compounds are thermally decomposed in the absence of oxygen. The pyrolysis products are classified into categories based on their physical state of existence: char (solid), bio-oil (liquid) and non-condensable gases (gas). The relative proportions of these three product fractions significantly vary depending upon the process conditions, as is shown in the table below.
$$ \small
\begin{array}{lcc}
\hline
\text{Pyrolosis Technology} & \text{Residence Time} & \text{Heating Rate} & \text{Temperature} & \text{Char} & \text{Bio-Oil} & \text{Gases} \\
\hline
\text{Conventional} & \text{5-30}\ \mathrm{min} & \text{<50} ^\circ \mathrm{C\ min^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<35%} & \text{<30%} & \text{<40%}\\
\text{Fast Pyrolysis} & \text{<5}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<25%} & \text{<75%} & \text{<20%}\\
\text{Flash Pyrolosis} & \text{<0.1}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{650-900} ^\circ \mathrm{C} & \text{<20%} & \text{<20%} & \text{<70%}^{[1]}\\
\hline
\end{array}
$$
The exact compositions of the products of cellulose pyrolysis at different temperatures can be seen below.
$$ \small
\begin{array}{lcc}
\hline
\text{Products} & \text{Peak Temp,}\ 500 ^\circ \mathrm{C} & \text{Holding Temp,}\ 400 ^\circ \mathrm{C} & \text{Peak Temp,}\ 750 ^\circ \mathrm{C} & \text{Peak Temp,}\ 1000 ^\circ \mathrm{C}\\
\hline
\ce{CO} & \text{0.99%} & \text{0.25%} & \text{15.82%} & \text{22.57%}\\
\ce{CO2} & \text{0.3%} & \text{1.45%} & \text{2.38%} & \text{3.36%}\\
\ce{H2O} & \text{3.55%} & \text{6.49%} & \text{8.72%} & \text{9.22%}\\
\ce{CH4} & \text{0%} & \text{0%} & \text{1.11%} & \text{2.62%}\\
\ce{C2H4} & \text{0%} & \text{0%} & \text{1.05%} & \text{2.18%}\\
\ce{C2H6} & \text{0%} & \text{0%} & \text{0.17%} & \text{0.28%}\\
\ce{C3H6} & \text{0%} & \text{0%} & \text{0.70%} & \text{0.80%}\\
\ce{H2} & \text{0%} & \text{0%} & \text{0.36%} & \text{1.18%}\\
\ce{CH3OH} & \text{0.25%} & \text{0.21%} & \text{1.03%} & \text{0.98%}\\
\ce{CH3CHO} & \text{0.01%} & \text{0.05%} & \text{1.58%} & \text{1.7%}\\
\text{tar} & \text{16.37%} & \text{83.35%} & \text{59.92%} & \text{49.12%}\\
\text{char} & \text{83.63%} & \text{6.17%} & \text{3.32%} & \text{3.91%}\\
\text{other} & \text{0.19%} & \text{0.16%} & \text{2.14%} & \text{1.78%}\\
\text{total} & \text{105.25%} & \text{98.36%} & \text{98.8%} & \text{99.86%}\\
\hline
\end{array}
$$
The holding time for each of these reactions was $30\ \mathrm{s}^{[2]}$. As shown in the table, $\ce{CO}$, $\ce{H2O}$, and $\ce{CO2}$ are the major gaseous products, with $\ce{H2}$ and hydrocarbons being produced in considerably smaller proportion.

$^{[1]}$ Patwardhan, Pushkaraj Ramchandra, "Understanding the product distribution from biomass fast pyrolysis" (2010). Graduate Theses
and Dissertations. Paper 11767.
$^{[2]}$ Hajaligol, M. R.; Howard, J. B.; Longwell, J. P.; Peters, W. A. Product Compositions and Kinetics for Rapid Pyrolysis of Cellulose. Industrial & Engineering Chemistry Process Design and Development Ind. Eng. Chem. Proc. Des. Dev. 1982, 21, 457–465.

Impressive. Do you make the tables by hand or is there some tool for making them?

@CopperKettle  I did have to enter the data manually, but the formatting was taken from this meta post. Digging around the sandboxes can yield some really neat $\LaTeX$ tools :)

As usual, the most interesting part (nearly everything that gives color and smell) is also the hardest to analyze, so it is just swept under the rug and called "tar".

@IvanNeretin On page 30 of the first source, there is a detailed analysis of exactly what this tar is. This question isn't about tar though, so I didn't mention it.

Wow. Now that's very impressive indeed.

This is really nice, with the graphs and all. Thanks (and the bounty goes to you, when I can award it.)


text:
Chirality of carbon and R/S naming when enatiomeric groups attached
when checking chirality for a tetrahedral, let's take carbon as an example, will two enantiomers or diastereomes of a group attached to same carbon cause chirality at that carbon, I mean are they considered different for comparing valencies and what about r/s naming , in cis trans and r/s, which enantiomer or geometrical isomer have more priority(supposing one chirality centre each) have more priority in cip?
I am not clear on how to apply method of plane of symmetry here

R takes priority over S.

Also related: IUPAC name for 1,2,3-trichlorocyclopropane?


text:
Why is this structure impossible?
I'm asked the following:

Explain why the structure $\ce{H-C=C-H}$ doesn't make sense chemically.

Not only am I just looking for a check up of my answer, but am I making any sense in what I'm saying here? Am I mixing up or misunderstanding concepts here?
My answer was:
The hydrogen and carbon atom form a covalent bond. Hydrogen has only 1 electron in its valence shell, so it 'needs' 1 of carbon's electrons to complete it. Now there are 3 electrons left to take in the carbon atom. But this isn't enough to make a double bond with the other carbon atom: we're one electron short.
Does this sound about right?

There are three electrons left on each carbon. Your assumption will lead you astray.

Could you elaborate a bit on this? Does a single bond in a structure have a 'higher priority' than a double bond?

My comment has nothing to do with "priority" it is about the the total number of electrons in play for bonding.

But the 3 electronscin the carbons are there because the hydrogens take one of each, right?

@Aperion We are just counting the total electrons available for bonding. The order you count them in doesn't matter so you are free to deal with  simpler bonds first and then see what is available for the rest. But you have to count the total correctly or you will not get a sensible answer.

OK, got it. Cheers! If you make an answer of it, I can accept it as such

@Apeiron You can also always write up a self-answer. That is encouraged on Stack Exchange ;)

Your analysis was a good start, but then it dropped. You need to consider four electrons on each carbon atom. Each of the two carbons can then form a single bond to one hydrogen atom using one of its electrons. Each of the carbon atoms then uses two of the remaining electrons to form the double bond.
Thus, each of the carbon atoms has one extraneous electrons; the structure would be a diradical structure. It should better be represented e.g. like this: $\ce{H-C^.=C^.-H}$.
It is possible for these two radicals to be alongside each other — see for example the structurally strongly related dioxygen molecule which is a triplet (diradical) in ground state and not adequately described in simple Lewis structure formalisms. However, it is highly unlikely for two carbons without additional constraints to form a diradical like that; forming a triple bond ($\ce{H-C#C-H}$) would be much better energetically (for carbon!).

Shine light on acetylene $\to$ get $\ce{H-C=C-H}$ $\to$ profit?

@orthocresol If only relaxation were not a thing ;P


text:
Can I convert elemental nitrogen to nitrous oxide?
Can I convert nitrogen to $\ce{N2O}$, to conduct barking dog reaction, where it's needed as oxidizer?

In short: no, you can't. Molecular nitrogen is insanely hard to convert to anything.

'Just' run an oxygen-nitrogen plasma, accelerate the ionized species, mass separate for $N_{2}O$ and voila... So, yes, you can but whether it is useful or not is a different question. As @IvanNeretin suggests, starting with $N_{2}$ is typically a bad (hard) thing to do.

That reaction seems like something I would avoid, too.

A more well-known way of making $\ce{N2O}$ is the thermal decomposition of ammonium nitrate: $\ce{NH4NO3 -> N2O + 2H2O}$.

There are several ways to fairly safely make $\ce{N2O}$ in the laboratory that don't involve the difficulties associated with trying to react molecular nitrogen.  The following reactions all are mentioned in this Wikipedia article which I suggest reading, along with the references.  Depending on the quantity you are making, I'd be very cautious about using the thermal decomposition of ammonium nitrate as this can kind of blow up on you (literally). Of course every reaction has it's own risks and it's up to you to decide what you are most comfortable with.
I think the direct oxidation of ammonia over a manganese dioxide-bismuth oxide catalystis an intriguing method$\mathrm{^{(1)}}$
$$\ce{2 NH3 + 2O2 -> N2O + 3 H2O}$$
This also seems like a fairly straight-forward route:
$$\ce{2 HNO3 + 8 HCl + 4 SnCl2 → 5 H2O + 4 SnCl4 + N2O}$$
And this seems to me a bit of a more complex synthyses:
$$\ce{2 (NH2)2CO + 2 HNO3 + H2SO4 → 2 N2O + 2 CO2 + (NH4)2SO4 + 2H2O}$$
Again, I'll say molecular nitrogen makes a lousy starting material for most syntheses, including $\ce{N2O}$.
1) Suwa T, Matsushima A, Suziki Y, Namina Y (1961). "Synthesis of Nitrous Oxide by Oxidation of Ammonia". Kohyo Kagaku Zasshi, Showa Denka Ltd. 64: 1879–1888.


text:
Naming compounds with an OX group
How are compounds which have a bond between an oxygen atom and a halogen atom named according to IUPAC rules? For example this compound:

I looked in many sites, but couldn't find how such compounds are named.
Also what is their seniority compared to alcohols, esters, ketones, and carboxylic acids? Or are they always a prefix?

These are esters, however this exact case is problematic. Also it wouldn't be stable.

The preferred IUPAC name (PIN) of the compound given in the question is butan-2-yl hypofluorite. It is named as ester of hypofluorous acid. In the seniority order of classes, esters are ranked between anhydrides and acid halides.
The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-67.1.3.2 Esters of mononuclear noncarbon oxoacids
Esters of mononuclear noncarbon acids are named in the same way as esters of organic acids (…). Alkyl groups, aryl groups, etc. are cited as separate words, in alphanumerical order when more than one, and followed by the name of the appropriate anion. Partial (acid) esters of polybasic acids are named by citing alkyl groups, aryl groups, etc. as separate words, in alphanumeric order if more than one, followed by the word ‘hydrogen’ (with the appropriate multiplying prefix, as necessary) also cited as a separate word, and the name of the appropriate anion. Salts of partial (acid esters) are named by citing the name of the cation before the name of the organic group; remaining acids groups are denoted by the word ‘hydrogen’ as described above. Structural specificity for esters of chalcogen analogues of mononuclear noncarbon oxoacids is provided by the appropriate italic element symbols O, S, Se, and Te, prefixed to the name of the group, as needed.

But why is it an ester and not an ether?

@indjev99 Ethers have the general formula $\ce{R-O-R'}$, in which $\ce{R}$ and $\ce{R'}$ are substituent groups derived from organic parent hydrides.

The are covalent hypohalogenites. Few are known to exist, such as methyl hypochlorite and tert-butyl hypochlorite. The one you have shown would be called sec-butyl hypofluorite, and is so unstable it likely cannot be isolated.
Due to the powerful oxidizing properties of these groups, such compounds are so unstable that they explosively decompose into toxic fumes, so you wouldn't have to worry about naming a multi-substituted compound. It would be difficult enough to form a compound with more than just a hypochlorite group without having severely altered the other functional groups, and any molecules that did form would react so violently with one another that it would probably be considered little more than a transition state.

Note that the retained prefix ‘sec-butyl’ was still contained in the 1993 IUPAC recommendations, but it is no longer recommended in the 2013 recommendations.


text:
Finding vendor information programmatically
I have a large collection of structures (>10,000) that I've gathered from literature searches and I would like to find out which of those can be obtained from vendors.  I know that eMolecules and ChemSpider provide such information, but getting that information for 10,000 structures in daunting.
On eMolecules, I can load a list of SMILES, search the databases, and download the results.  The download process is not very robust, though, and I have to reconstitute a lot of pieces of information.
ChemSpider provides additional information regarding vendors, and I've tried using the ChemSpiPy API, but I consistently get an error from the ChemSpider server.  Obviously something has changed with respect to the server that the API doesn't account for.
Are there other suggestions for where/how I might gather such vendor details for a very large collection of structures?

https://www.cas.org/content/chemical-suppliers

I have no problem with the ChemSpider server. Did you request an API token for ChemSpider? http://www.chemspider.com/AboutServices.aspx

You can try molport, they are an intermediate. Just ask them directly, it will be done in a jiffy.
Or if you like you can do it in knime and use any API (like molport's API), which is much more fun, but for just a one-time thing a bit complicating.

This is really more of a comment than an answer.


text:
Regioselectivity in radical chlorination of a carboxylic acid
From an online textbook, the first step in the preparation of an amino acid:

Can we be sure that the chlorine will connect exactly at that spot, in the alpha-position? Won't we get an acyl chloride instead? Or a molecule where Cl is connected in the beta-position, or further down the line?

By what mechanism would you expect to get the acid chloride? The $\sigma _{C-H}$ bond $\alpha$ to the carbonyl is going to be the weakest bond in terms of homolytic cleavage.

@SendersReagent could you please tell why that sigma bond will be the weakest one, generating a free H atom.

Sure, I'll put it in an answer since it will go into more detail than I feel is appropriate for a comment.

This turned out to be a pretty nice thread. Three answers approaching the problem in a different way each and equally useful. Favoriting.

@PhMgBr Agreed, and it's a perfect example of "Theory can only get you so far, what actually happens in the lab is most important."

I am going to say: no, this does not work in practice. In fact it does not even work in theory.
In simple theory you would say that that the α C-H bond is the weakest because the resulting radical is stabilised by conjugation into the C=O. But under radical conditions, we don't observe α-chlorination in the lab. Is there a better way to explain this?
The chlorine radical
The problem is twofold. Firstly, the chlorine radical, $\ce{Cl.}$, is much less stable and therefore extremely unselective in radical chlorinations. In physical organic chemistry, we quote something called the Hammond postulate, which says that in this case, the transition state for hydrogen abstraction resembles the starting material more than the product. Therefore, the stability of the product does not significantly affect the activation energy, and the rate, of the hydrogen abstraction.
On top of that, it is what we would consider an "electrophilic radical"; it has a very low-energy SOMO, and therefore has a tendency to abstract a more electron-rich hydrogen atom. Essentially, it means that the radical behaves more like an electrophile than a nucleophile. In the transition state for hydrogen abstraction, the chlorine wants to have a significant build-up of negative charge on itself (remember, the product of hydrogen abstraction is going to be $\ce{HCl}$). It therefore goes for the hydrogen that is most willing to give it this negative charge, and that is the most electron-rich hydrogen atom.
An example
Consider the radical chlorination of propionic acid, i.e. $\ce{R} = \ce{CH3}$ in your diagram above.

The α-hydrogen, compared to the β-hydrogen, is comparatively electron-poor because of the electron withdrawal by the $\ce{-COOH}$ group. So, it turns out that the rate of abstraction of the α-hydrogen is much slower: $k_\alpha/k_\beta = 0.03$ (Moody & Whitham, Reactive Intermediates, p 12). Here's an article which discusses the selectivity observed in the chlorination of propionic acid under different conditions: Tetrahedron, 1970, 26, 5929.
As written in Carey & Sundberg 5th ed., Part A, p 1022:

Radical chlorination shows a substantial polar effect. Positions substituted by EWG [electron-withdrawing groups] are relatively unreactive toward chlorination, even though the substituents are capable of stabilizing the radical intermediate. [...] Because the chlorine atom is highly reactive, the reaction is expected to have a very early TS [transition state] and the electrostatic effect predominates over the stabilizing effect of the substituent on the intermediate. The electrostatic effect is the dominant factor in the kinetic selectivity of the reaction and the relative stability of the radical intermediate has relatively little influence.

As an example, they use the chlorination of butyronitrile, which is the same as propionic acid except that the EWG is now $\ce{-CN}$ instead of $\ce{-COOH}$:

As @Loong mentions in the comments, in the specific case of propionic acid, there is a further problem: there are three β-hydrogens for it to abstract, and only two α-hydrogens. Therefore, even in the absence of everything else I have described, we would expect there to be 1.5 times as much chlorination at the β-position. This certainly does not help at all with the selectivity! The importance of statistical factors is most clearly seen in the radical chlorination of simple alkanes, where all the hydrogens are pretty much similar. I shan't go into detail here, but you can look at this answer (by Loong again), and there are examples on Clayden 2nd ed., pp 987-8.
The solution
If you wanted to selectively chlorinate propionic acid at the α-position, you would use a polar mechanism that goes via the enol. The paper above has some examples of conditions that work. All of them involve suppressing the radical reaction by adding radical scavengers.

The paper you cited uses $\ce{Cl2}$ under dark conditions. I'll look at it further, but can it be applied directly if the OP is referencing photolytic conditions?

@SendersReagent I don't feel like there would be a difference whether the $\ce{Cl.}$ is generated photolytically or via radical initiators. Once generated there is no reason to think that hv would make it prefer the alpha position.

This paper might contain the information, but I can't read French: http://onlinelibrary.wiley.com/doi/10.1002/bscb.19520610705/abstract Reaxys says "Irradiation.im UV-Licht". I doubt the result is any different.

Hmm... Actually, good point. I was thinking of diazo cleavage, which would lead to singlet or triplet carbenes. Would the $\sigma _\ce{CH}$ be weakened by photolytic conditions? Or is this an "all or nothing" situation?

@SendersReagent Well, I'm not 100% sure on whether the $\sigma(\ce{C-H})$ is affected. Although I personally don't see a reason for it to be, I don't profess to know everything so if there is I'd be happy to hear it :)

Note that the reaction of the chlorine radical is also affected by a statistical factor since there are one acid-H, two α-H, and three β-H atoms in propionic acid.

Yes, we can :-D
Note the $h\nu$ in the reaction. It is a radical reaction. Irradiation will cleave the $\ce{Cl-Cl}$ bond and yield chlorine radicals ($\ce{Cl*})$. These will abstract hydrogen atoms (not protons) from the carboxylic acid to form a carbon centered radical (and $\ce{HCl}$). The favoured and position for this radical is the $\alpha$ carbon, where the radical is stabilized by interaction with the $\ce{-C=O}$ bond.

Update
Chances are even better in the presence of reagents that would facilitate the formation of an acid halide, as mentioned in the question. Catalytic amounts of phosphorus might be sufficient. This type of transformation, usually performed with bromine, is known as the Hell-Vollhardt-Zelinsky reaction.

Debatable. Alkylradicals are electron-deficient, and carboxygroup is in general electron-withdrawing. I would ask a reference for this claim.

@permeakra In this case, the carbonyl wouldn't be electron withdrawing (as far as resonance is concerned). You would have $\pi _{C-O} \ce{->} \text{SOMO}_C$, I think?

The $\sigma _{C-H}$ bond $\alpha$ to the carbonyl is going to be the weakest bond in terms of homolytic cleavage.

The $\ce{C-H}$ bond is weakened due to hyperconjugation. Hyperconjugation goes beyond the "intro organic" explanation of "adjacent $\ce{C-C}$ bonds stabilizing carbocations." Essentially, electron-rich $\sigma$ bonds, or bonds where the electrons are fairly central within the bond (in this case, the $\ce{C-H}$ bond) will donate electron density into an adjacent empty or partially-empty $\pi$ or $\pi ^\star$ orbital. This will strengthen the $\ce{\sigma _{C-C}}$ bond of the two adjacent carbons, but will weaken the $\ce{\sigma _{C-H}}$ by reducing electron density in its $\sigma$ bond and will also slightly lower the bond order of the carbonyl due to donation into $\pi ^\star _{C-O}$.

In the image below, you can see the molecular orbital diagram for hyperconjugation between a $\ce{C-H}$ bond and an adjacent alkene.

When looking at a carbonyl, this picture is essentially the same, except the orbital on the right would be ${\pi ^\star _{C-O}}$. Essentially, in the $\ce{H-C-C=O}$ system, the $\ce{H-C}$ and $\pi ^\star _{C-O}$ bonds are weakened/lengthened and the $\ce{C-C}$ bond is shortened and strengthened.

Because some electron density is lost from $\ce{\sigma _{C-H}}$, it becomes weaker and is, therefore, the easiest bond to break by the resultant chlorine radical.


text:
N-O bonds in Nitric acid
The wikipedia article on nitric acid mentions the following structure of $\ce{HNO3}$ in which the two $\ce{N-O}$ bonds (other than that of $\ce{N-OH}$) are of different length i.e., $119.9\,\pu{pm}$ and $121.1\,\pu{pm}$.

But it also mentions that $\ce{HNO3}$  molecule is a resonance hybrid of - (only major contributing ones)

I can't understand how can both be simultaneously correct since the resonance hydrid should have same two $\ce{N-O}$ bond lengths (other than $\ce{N-OH}$ one). How is it correct or is the source itself wrong?

The difference is very minor, such minor differences may be caused by multiple very weak effects rarely significant and consequently rarely discussed and explained. Generally it is expected to have unequal bonds (i.e. not transformed into each other by symmetry transformation) to have slightly different length. Specifically, at least 2 effects may be involved here: 1) effect of charge distribution.  Significant polarity of O-H bond makes resonance hybrids unequal in energy, so one has higher impact 2) the unconjugated lone pair of OH oxygen can conjugate with only one another N-O bond.

In addition to permeakra's comment, I want to add the following thoughts:

1) How were these bond lengths determined?

Were these values determined computationally? If so, what is the standard error in the computational method uses? In other words, is the difference between the two values within the error of the method, and thus the difference is not significant. Additionally, a computational method operates on single molecules "in the gas phase", and there might be just enough interaction between the hydrogen atom and one of the other oxygen atoms. This interaction would also explain the disparity in bond angles:

Were they measured experimentally, for example from x-ray crystallography? If so, then dissymmetry arising from a preferential crystal packing arrangement can lead to differences. In particular, in the solid state, the position of the hydrogen atom can make a big difference in preferential hydrogen bonding, either intermolecularly or intramolecularly. For units like the following hyrogen-bond dimers, the hydrogen bonding could lead to dissymmetry between the N-O bonds.

2) Bonds vibrate

They do, even in the solid state. Thus, an x-ray crystal structure has error its measurements as well. Again, is this difference in bond lengths within the experimental error?

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


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Acidic strength of organic acids. Substituted PhCOOH acids
Could someone please explain the order of acidity of ortho, para and meta substituted benzoic acids with the substituent as methyl group?
Methyl group operates +I and +R effect, I believe.

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Related: Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid?

The ortho substituted benzoic acid is most acidic, followed by the meta substituted and finally para substituted benzoic acid.
The methyl group, when at ortho position, causes steric inhibition of resonance, which essentially means that the $\ce{-COOH}$ group goes out of the plane of the benzene ring (due to steric hindrance from the $\ce{-CH3}$ group). Due to this, the benzene's $\pi$ electrons can no longer be transferred to the central carbon of the $\ce{-COOH}$ group by resonance, hence increasing the positive charge on it. This makes the hydrogen of the $\ce{-COOH}$ group more acidic.
When the methyl group is on meta position, it operates by +I, thus reducing its acidity. When on para position, it operates by both +I and +R (actually +H, which means hyperconjugation effect), hence decreasing the acidic strength even more.


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Slips System of BCC
We have been taught that there are 48 slip Systems in BCC. I need the miller indices of each slip plane and direction.
Out of 48 I got miller indices of 12 systems of $\{110\}\ \left<111\right>$ and 12 systems of $\{112\}\ \left<111\right>$ , but I am not getting remaining 24 slip Systems of $\{123\}\ \left<111\right>$.
I don't know how to find all 24 planes of $\{123\}$ family of planes.If anyone know how to do it, it will be helpful

This is simple enumeration of the possibilities. There are more because there are no repeats of indices that are equivalent.

How to find all 24 planes of $\{123\}$ family of planes?

For any of the $\left<111\right>$ directions, we have $3!$ i.e. $6$ $\{123\}$ slip planes. And we have $4$ $\left<111\right>$ slip directions for each plane. So in total $6\times 4 = 24$ slip systems.

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Why do I have to use a 1:1 ethanol:water solution to hydrolyse aspirin?
I understand that water itself would not work as aspirin is not very soluble in water but Why not just use 100% ethanol?

Because you can't hydrolyze something without water.

Hydrolysis means to break something apart (e.g. chemical bonds) with water.

If there's no water, there's no hydrolysis.

Also the ethanol is needed to solubilise the aspirin which is not very soluble in water.


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Does NaCl reduce the surface tension of water?
Does NaCl reduce the surface tension of water? And why?
For example soap does it, and I was wondering if NaCl does the same.

The surface tension of NaCl solution is greater than that of water.

Gibbs adsorption isotherm tells us that (roughly speaking) the decrease in surface tension is proportional to the surface excess concentration of the solute.
Positive surface excess means that the solute tends to concentrate on the surface rather than stay in the solution. The surface concentration has no physical limit and can go arbitrarily high. In soap and other surfactants, it is hundreds of times higher than the concentration in the solution, therefore their effect on surface tension is negative and very strong.
Negative surface excess means that the compound prefers to stay in the solution rather than on the surface. It has a physical limit, though: the concentration can't be lower than 0. Therefore the effect of such compounds on surface tension is positive, but small (not noticeable with naked eye). This is the case with $\ce{NaCl}$ and pretty much all "normal" inorganic salts.


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Why is platinum an excellent catalyst in comparison with similar metals?
What is special about platinum structure which actively participate in catalytic reactions while similar metals with similar atomic structures cannot.

Indeed, similar metals with similar structures ($\ce{Pd,;Ir}$) are just about as good in many catalytic reactions, so $\ce{Pt}$ is not all that exceptional.

@IvanNeretin Righto. The whole cluster of them has a name: Platinum group metals


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How is the association/ dissociation constant derived?
I came across the following formula to find the association of a solute in particular solvent. Suppose the concentration of the solute given in water (or in some other solvent in which it does not associates or dissociates) is $C_1$ and that in some other solvent ( like benzene ) is $C_2$ then the association factor $n$ is determined by $$K=\frac{C_1}{{C_2}^\frac{1}{n}},$$
where $K$ remains constant.
On what concept is this formula based on? Can we derive it? Is it an empirical formula ?

Please don't use MathJaX in titles

Why, it is just like any other equilibrium constant.

This looks a lot like a normal equilibrium constant. There is very little background information provided, but I assume that this formula applies to an experimental procedure in which the same total amount of the solute is dissolved into each solvent and the concentrations of free (dissociated) solute are measured. In this case that $C_1$ and $C_2$ should be the other way round. In the following, I assume that $C_1$ is the measured concentration in benzene and $C_2$ is the measured concentration in water.
In order for the formula to work we also need to assume that in benzene (or whatever the other solvent is), the solute associates almost completely.
With these assumptions the derivation of the formula is practically identical to the standard derivation of the formula for an equilibrium constant.
Let's call a single molecule of the solute A and a complex of $n$ associated molecules B.
The association reaction then is
$$n\ce{A <=>[k_1][k_2] B},$$
where $k_1$ and $k_2$ are the rates of association and dissociation reaction respectively.
The speed of the dissociation reaction is proportional to the concentration of B, so we can write $k_2=b\ce{[B]}$.
The association reaction requires $n$ units of A coming together and from probability theory we can tell that the probability of that happening is proportional to the concentration of A multiplied by itself $n$ times, so we can write $k_1=a\ce{[A]}^n$.
At equilibrium the rates of association and dissociation reactions are equal, so we can write:
$$k_1=k_2 \Rightarrow a\ce{[A]}^n=b\ce{[B]} \Rightarrow \frac{b}{a}=\frac{\ce{[A]}^n}{\ce{[B]}}$$
Now we raise both sides to the power of $\frac{1}{n}$ in order to get
$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{\ce{[A]}}{\ce{[B]}^{\frac1n}}$$
Now, $\ce{[A]}$ is the concantration of dissociated solvent in benzene, so we can substitute $\ce{[A]}=C_1$. On the other hand $C_2$ is the concentration of the solute in water, which is the same as the concentration of total concentration of both associated and dissociated forms in benzene, so we can write $C_2=\ce{[A]}+n\ce{[B]}\approx n\ce{[B]}$ if $\ce{[A]}$ is much smaller than $\ce{[B]}$ as we assumed in the beginning.
By substituting we can write
$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{C_1}{\left(\frac1n C_2\right)^{\frac1n}} \Leftrightarrow \left(\frac{b}{na}\right)^{\frac1n}=\frac{C_1}{C_2^{\frac1n}}$$
Because $n$ is independent of $C_1$ and $C_2$, we can substitute $K=\left(\frac{b}{na}\right)^{\frac1n}$ and we get
$$K=\frac{C_1}{C_2^{\frac1n}}.$$


text:
How to determine what organic molecule has the highest boiling point?

Compare and answer which one has the highest boiling point:

$\ce{CH3CH2CH2CH3}$ (butane) ... [$\pu{−1}$ to $\pu{1^\circ C};\ 30$ to $\pu{34^\circ F};\ 272$ to $274\ \pu K$]
$\ce{CH3NH2}$ (methylamine) ... [$−6.6$ to $−6.0\ ^\circ\pu C;\ 20.0$ to $21.1\ ^\circ\pu F;\ 266.5$ to $267.1\ \pu K$]
$\ce{CH3OH}$ (methanol) ... [$64.7\ ^\circ\pu C\ (148.5\ ^\circ\pu F;\ 337.8 \pu K)$]
$\ce{CH2F2}$ (difluoromethane) ... [$−52\ ^\circ\pu C\ (−62\ ^\circ\pu F;\ 221\ \pu K)$]

My effort:
At first, without referring to factual data (but the factual data has been provided in parenthesis, see above-options, Source: wikipedia.org), I reached the following conclusion:

Butane appears to have the lowest boiling point because there is only Van-der-Waals forces, which are weaker than dipole-dipole interactions, present in the rest of molecules.

Next I would compare between methylamine and methanol because they both have dipole-dipole interactions and most importantly that these molecules associate themselves with hydrogen bonds, but as oxygen is more electronegative than nitrogen, hydrogen bonds formed within methanol molecules would be stronger than those formed between methylamine molecules. So methylamine takes the 3rd position.

Lastly, it becomes easy for me to differentiate between methanol and difluoromethane, because according to my knowledge the latter has got fluorine which is more electronegative than oxygen, present in former. Moreover difluoromethane has got 2 fluorine, which add weight to my above argument in the way that there will be even more extensive hydrogen bonding within the molecules.

Inferences and comparison with the source's answer
Answer: Methanol
I am full of doubts as to why difluoromethane is not the answer. Revisiting my arguments again and again, I always ended up with the same conclusion i.e., mine answer would always contrast with the source's answer.
Afterthoughts & Deductive Conclusions
A few days after extensive thinking and arguing myself, I ended with the following reason, which seemed plausible to my brain. It may be possible that the because of: the presence of two fluorine atoms, small and compact nature of carbon-fluorine bond (according to bond length - bond strength - bond dissociation enthalpy relation/ theory), and the overall small and virtually spherically shape (But, in fact, it is tetrahedral geometry) of difluoromethane; the two fluorine atoms might be finding themselves in a position, in which they cannot simultaneously form hydrogen bonds with the hydrogens of other molecule. My above explanation can also imply that they might be hindering into each other's hydrogen bonds (reason can be same as mentioned in list in the previous sentence).
This reasons and explanation thus explain why the boiling point of difluoromethane falls below that of methanol. Remember that in methanol, there is less compact nature of carbon-oxygen (similar size of orbitals of carbon and oxygen as well as bigger size of oxygen atom as compared to that of fluorine), a single carbon-oxygen bond is weaker than a single carbon-fluorine bond (based on electronegativity difference and concept of dipole-development & interactions), and the last that is a molecule of methanol is less symmetrical (3-H and 1-O) as compared to difluoromethane, which is more symmetrical (2-H and 2-F).
Final conclusion and additional Info.

Solution1: $\ce{C4H10 < CH3NH2 < CH3OH < CH2F2}$ (in increasing order from left to right)
Solution2: $\ce{C4H10 < CH3NH2 < CH2F2 < CH3OH}$ (in increasing order from left to right)
Solution3: None (it's should be totally based on your own conclusions and knowledge, backed up with info. from  appreciable sources)

It's simpler than that. Hydrogens that sit on carbon just don't form hydrogen bonds.

@Ivan That is a crude approximation, that is not true in many cases. It is, however, known that such hydrogen bonds are weaker.

Yes, but a crude approximation is just what we need here.

So, should we consider the 'crude approximation' of another Hydrogen atom's interaction in our analysis.

There are commonly three types of intermolecular forces quoted when it comes to melting and boiling points: dipole-dipole interactions, London interactions and hydrogen bonding. Hydrogen bonding is by far the strongest force of them all. In all cases except for very rare ones, for a hydrogen to take part in hydrogen bonding, it must be bonded to $\ce{O, N, F}$ — one of the three very electronegative atoms.
In difluoromethane, there are no hydrogens bound to $\ce{O, N, F}$ — they are both bound to carbon. It is mainly carbon that experiences fluorine’s electronegativity and by the time the inductive effect reaches the hydrogens it has been substantially lowered. On the other hand, methanol and methyl amine both have hydrogens bonded to the electronegative atom — nitrogen or oxygen — so these compounds actually take part in hydrogen bonding. Since we have two compounds that participate in hydrogen bond networks it is safe to assume that these will be the top runners.
As you mentioned, methanol’s hydrogen bonds should be slightly stronger since oxygen has a higher electronegativity. I think your reasoning as to why methanol is stronger than methyl amine is sound enough.
Difluoromethane’s principal intermolecular forces will be dipole-dipole interactions. However, you should also note that the presence of fluorine atoms strongly reduces the underlying London interactions, since $\ce{C-F}$ bonds are more localised and fluorine’s lone pairs are much less disperse than most other carbon-heteroatom bonds — this is due to fluorine’s very high electronegativity. Thus, while we have significant dipole-dipole interactions, we are somewhat lacking London interactions in spite of the rather large molecular weight of $52$.
This is important when comparing it to both methyl amine $M_\mathrm{r} = 31$ and butane $M_\mathrm{r} = 58$. Butane, given its larger electron count ($26$ bonding electrons versus $14$ bonding and lone pair electrons in methyl amine) and its very non-spherical shape can give rise to much stronger London interactions, especially since all its bonds are almost or entirely unpolar $\ce{C-H}$ or $\ce{C-C}$ bonds. A general rule of thumb is the heavier a molecule, the stronger the London interactions and they quickly catch up with what dipoles or hydrogen bonds can do.
So putting it all together:

Difluoromethane has very weak London interactions but some dipole-dipole interactions. It is expected to have a rather low boiling point.
Methyl amine can form hydrogen bonds but is also rather small and thus lacking London interactions. It is less polar than methanol giving rise to lesser dipole-dipole interactions. Put it somewhere in the middle.
Butane is significantly larger in volume than all the others and also has much more electrons that can take part in London interactions. Thus, London interactions are predicted to be rather strong but nothing else is there.
Finally, methanol of course forms strong hydrogen bonds, is a good dipole and can take part in London interactions well. A clear winner.

I wouldn’t know how exactly to rank butane and methyl amine but I would put difluoromethane at lowest and methanol at clearly highest boiling point. This is consistent with the experimental data.


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Copper chloride color in electrolysis of salty water
I tried a simple water electrolysis experiment at home, with $\ce{NaCl}$ as electrolyte, and a $12 \,\rm{V}$ battery.
My wires were made of copper and $\ce{H2}$ bubbles were only forming at negative terminal (cathode). On the positive side (anode), nothing was happening.
After $3$-$4$ minutes, water started to change into a greenish-blue color and I realized that it must be copper reacting with chloride ions. After $15$ minutes, my solution was light blue with exceed $\ce{NaCl}$ at the bottom, after taking the wires out the solution started to turn into a yellow color in about $10$ minutes.
What was the yellow solution and why didn't cathode produce any oxygen?
I'm also worried that the yellow solution might be chlorine because I later boiled the solution.

Related: Why is chloride oxidised instead of water in copper chloride electrolysis?

The color of the solution is probably due to the presence of blue copper(II) ions ($\ce{Cu^2+}$) and greenish tetrachlorocuprate(II) ions ($\ce{[CuCl4]^2-})$.
The change in color comes from the equilibrium
$$\ce{Cu^2+ + 4 Cl- <=> [CuCl4]^2-}$$
Low concentrations of chloride favor the formation of the blue copper ions. High concentrations of chloride favor the formation of the greenish tetrachlorocuprate ions.
If a significant amount of chlorine had been created you should have observed bubbles at the anode and the typical chlorine smell.

I wouldn't expect any significant amount of chlorine (or oxygen, for that matter) to appear in this setup. Dissolving the anode is so much easier.

Well I didn't observed any bubbles at the anode but I think it was chlorine I was smelling.(smelled just like swimming pools) I've just noticed bright green color around the anode, I'm assuming that is tetrachlorocuprate(II), after few hours the solution now looks almost clear yet huge amount of blue and yellow precipitation are floating at the bottom. I'm leaving it outside waiting for the water to evaporate. If the blue precipitate is copper (II) or copper(II) chloride ions then what is the yellow precipitated?

Copper(II) chloride is readily soluble in water, so I would not expect it to precipitate. I guess that the precipitate rather contains copper(II) hydroxide.

Copper is not an inert electrode especially at 12 volts. Even platinum electrodes can dissolve at that voltage. If something is happening at the cathode something must be happening at the anode if current is flowing.


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How is purine aromatic?

If we count the total number of π electrons, there are 10. But to apply Hückel's rule to this molecule, we should count only the electrons of the outer periphery and there are 8 of these. How can the molecule be aromatic?

Hückel's rule applies to monocyclic systems only.

Draw the other tautomer where the six-member ring does not have the double bond in the 'inner'side of the 5-6 ring junction.  That will get you a 10-electron system.  Or live with Ivan's comment and accept two aromatic systems that are fused together.

Consider this structure of purine:

The ring on the left contains two nitrogen heteroatoms and is aromatic due to π electron conjugation. Having six π electrons, it obeys Hückel's rule.
For the ring on the right, it contains four π electrons in bonds and the lone pair on the NH group takes part in conjugation. Hence the total number of electrons taking part in conjugation is six, which makes it aromatic by Hückel's rule as well.
Since both the rings of purine are aromatic, the compound is aromatic.
For more information see: Ciesielski, A.; Krygowski, T. M.; Cyrański, M. K.; Balaban, A. T. Defining rules of aromaticity: a unified approach to the hückel, clar and randić concepts. Phys. Chem. Chem. Phys. 2011, 13 (9), 3737–3747. DOI: 10.1039/C0CP01446J.

It does not obey Hückel's rule. It is not monocyclic, and you must not apply it to each ring independently. Hückel's rule is very strict and can only applied to a handful of systems.

@Mart Clarification: Did you mean "must apply" or  "must not apply"? Thanks!

@paracetamol I meant not. Hückel's rules apply only to mono cycles

@Mart Follow-up question: I see, so to employ Hückel's rules the entire system must be a monocycle, but I should not apply the rules to individual cycles in a polycyclic system? (Trying to see if I understood this right) Thanks again!

@paracetamol Hückel's rules are guidelines. They have been used for situations as generalisations, which were not originally covered by the set they were derived from. If used for these occasions they might lead to wrong conclusions. In general: not all aromatic compounds obey Hückel's rules, but all compounds that do obey them are aromatic.

@Martin-マーチン But, consider the case of biphenyl. We apply Huckel's rule independently to the two different Benzene rings and conclude that it's aromatic.

@DJKoustav I fail to see your point. These are two linked monocycles. The inherent property of the phenyl moiety, possessing aromatic character, will not be disturbed much by adding another (in this case also aromatic) substituent. If you conclude anything from Hückel's rules, you are stretching the limitations of the theory (beyond the useful). It is an observation, with a more rigorous theoretical background. It has very limited predictive power, and only if you exactly understand, why you can apply it. Use it for explanation and rationalisation, not prediction.


text:
"Exactly Equal" and "At Least" in electron excitation
When we examine IR spectra, we see troughs corresponding to absorption at exactly a specific frequency that corresponds to the energy needed to stretch certain bonds (although translational motions and intermolcular forces can broaden the accepted frequency).
Coordination compounds absorb light at exactly the frequency corresponding to the crystal-field splitting energy.
However, when we talk about energy from photons required to break a  bond, we say "at least" the frequency instead of "exactly" in the examples above. Take $\ce{Cl-Cl}$ for example, we say that the photon must have at least a frequency of 607 THz (or wavelength of no more than 496 nm). There is still a definite energy involved here: 242 kJ/mol to promote an electron from a bonding MO in the molecule to an anti-bonding MO.
As another example, when we talk about ionizing radiation, we say that there is at least a certain amount of energy needed to ionize a compound. For example, my textbook says that a photon needs "at least 1216 kJ/mol" to ionize water. There is still a definite energy level involved: bringing the bonding electron from its negative potential energy MO to 0.
In all of these examples, definite energies were involved. Why is it that sometimes we say that that a photon needs to have exactly the energy needed, and other times the minimum energy needed?

When you ionize or break something, the resulting particles (an electron and an ion, or maybe two atoms, or whatever) would fly away from each other with arbitrary speed. Their kinetic energy would absorb any excess energy of the photon. This is the "at least" case.
On the other hand, when you excite some particle (molecule or whatever) in such a way that it remains in one piece, then you can't send it flying, because the momentum conservation forbids that. This is the "exactly equal" case.
Surely this is a simplification; in fact, a photon may excite a molecule using only a part of its energy, while keeping the rest of it to itself (Raman scattering), but that's another story.


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Why Don't Sulfur Atoms Hybridize?
I am doing some chemistry problems with given answers, and supposedly the $107^{\circ}$ and $92^{\circ}$ bond angles in $\ce{H_2O}$ and $\ce{H_2S}$, respectively, is due to the fact that"$\ce{O}$ uses $\ce{sp^3}$ hybrid orbitals for bonding, $S$ uses its $\ce{3p}$ orbitals."
There is quite a lot of stuff on the Internet pertaining to this exact problem. For example, the first answer on this page says that sulfur doesn't have as much electron repulsion because it's bigger, so it doesn't need to hybridize to lower its nergy. That answer assumes that hybridization in oxygen and sulfur atoms requires energy like it does in oft-used carbon, but I think that is wrong: since oxygen and sulfur don't need to promote any $ns$ electrons to $np$ orbitals (they already have filled $p$ orbitals), they can simply "mix" what they already have. Thus, why doesn't S just hybridize anyway to save a little bit of energy, and consequently have tetrahedral bond angles?
Another answer involves lone pair repulsion: perhaps both atoms hybridize, but the lone pairs on $S$ repel the $H-S-H$ bond more. I simply don't think this can have such a large effect. Even more, the other $H-X-H$ bond angles for Group 6 elements (selenium and tellurium) are around $90^{\circ}$, so it cannot be due to increased lone-pair size, as Se and Te have even bigger ones.
Because of the anomaly in $\ce{H_2O}$'s angle ($107^{\circ}$) versus the other ones (all around $90^{\circ}$), I think that the non-hybridization is more likely to be correct, so why don't sulfur atoms hybridize?

consider $\ce{SO4^{2-}}$

In principle, hybridisation is only a mathematical tool to understand bonding better. Hybridisation always results from a geometry. It is never cause for a certain geometry.

http://chemistry.stackexchange.com/q/14981/16683

see: http://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h?rq=1

@Martin-マーチン I agree with you absolutely on that and that is why we cannot predict the hybridisation state of atoms in molecules without first observing geometry and other molecular characteristics. However, it is possible to deduce whether orbitals can hybridise. According to Mingos, an indicator for whether different orbitals can hybridise is the difference in their values of the most probable radius (r max). For the period 2 elements, Mingos says that it is due to the similarity in r max values of the 2s and 2p orbitals which allow for hybridisation.


text:
Why triiodide ion does not form ionic bond with cation having +2 charge?
In general, I found that triiodide ion generally forms ionic bond with cations having +1 charge like  $\ce{NH4+I3-}$, $\ce{K+I3-}$, $\ce{Tl+I3-}$. But, I have never seen triiodide ion forming ionic bond with cations having +2 or +3 charge like  $\ce{Ba^2+(I3^{-})2}$, $\ce{Pb^2+(I3^{-})2}$, $\ce{Bi^3+(I3^{-})3}$. If properly written, it would  look like $\ce{Ba(I3)2}$, $\ce{Pb(I3)2}$, $\ce{Bi(I3)3}$. These compound has never existed and would not likely to exist. But what is the reason behind this?
Recently I went through this question:-

Why isn't thallium triiodide stable?

One of the  commenter  said:-

Thallium Tri-Iodide $\ce{Tl+(I3)-}$ does exist.  Thallium (III) Iodide, $\ce{Tl^3+(I^{-})3}$ to my knowledge, does not.

I was quite surprised at first seeing that $\ce{TlI3}$ exist because I knew that due to inert pair effect, thallium will form Thallium(I) compounds like $\ce{TlCl}$, $\ce{TlBr}$,  $\ce{TlI}$ etc. but I  realized that the compound contains  $\ce{Tl+}$ and not $\ce{Tl^3+}$. But what about the corresponding $\ce{Pb^2+}$ and $\ce{Bi^3+}$ ions? Why does not they form triiodide compounds?

Why triiodide ion does not form ionic bond with cation having +2 charge

It does. Specifically, quick googling for "bis-triiodide" gives out of the blue links like this: poke, poke, so the question is not 'why no bivalent triiodides', but 'why only super-large bications can form triiodides.'
The answer, I suspect, lies in charge density of the cations and their ability to form strong ionic bonds.
It is well known, that energy of ionic crystal lattice strongly increases with decrease of the size of ions involved. Triiodide ion, being quite large, is not very good for forming strong ionic lattices, forming only very weak ionic bonds. Iodide ion, on the other hand, is much better for that.
The energy of formation of triiodide anion from iodine molecul and iodide anion is very low. Thus, triiodie will break to form a crystal lattice with higher formation energy at first chance. So, for a solid triiodide to form, an ion with very low charge density is required, so the energy of crystal lattice would not benefit significantly from triiodide anione dissociation. This rules out any 'naked' cation with charge +2 or more and actually most 'naked' cations with charge +1. On the other hand, by covering dication in a shell of organic molecules strongly bound to the cation, we gain a very large cation, and a solid triiodide can form.
A very similar effect guards temperature of decomposition of solid carbonates, for example, as the decomposition involves breaking of a larger carbonate dianion into a much smaller oxide dianion.

They are some seriously big compounds :)

The problem boils down to competing stabilization factors. As discussed here, an isolated triiodide ion, or any trihalide ion ($\ce{X_3^-}$), is more strongly stabilized than the separate monatomic halide ion plus the neutral halogen diatomic molecule ($\ce{X^- + X2}$). But in practice, except for plasmas, ions are generally not isolated; they are accompanied by counterions or the oppositely-charged ends of solvating dipoles. These are more favorably attracted to a compact, monatomic ion and so tend to pull the triiodide ion apart. Even where the triiodide is stable, such pulling apart commonly occurs to some extent. From Wikipedia:

In ionic compounds, the bond lengths and angles of triiodide vary depending on the nature of the cation. The triiodide anion is easily polarised and in many salts, one $\ce{I−I}$ bond becomes shorter than the other. Only in combination with large cations, e.g. a quaternary ammonium such as $\ce{[N(CH3)4]^+}$, may the triiodide remain roughly symmetrical.1

A table from the Wikipedia article, citing 2 for the methanol solution, renders results for some singly charged cations. In effect the anion is not purely a triiodide ion, but an intermediate between triiodide and dissociated iodide plus $\ce{I2}$:

When we try a doubly charged cation, the electrostatic attraction is stronger and with it, there is an even greater tendency to pull the proposed triiodide ion apart completely. Attempting to bind triiodide ion with $\ce{Mg^{2+}}$ or even $\ce{Ba^{2+}}$ is likely to produce instead the more tightly packed monatomic iodide salt plus $\ce{I_2}$. Cutting down the electrostatic attraction, or equivalently making the anion more isolated, by using a large multiatomic counterion offers the best chance for creating a bis-triiodide salt.
Cited Reference

Atkins; et al. (2010). Inorganic Chemistry (5th ed.). Oxford University Press. p. 431. ISBN 9780199236176.

Heo, Jun; Kim, Jong Goo; Choi, Eun Hyuk; Ki, Hosung; Ahn, Doo-Sik; Kim, Jungmin; Lee, Seonggon; Ihee, Hyotcherl (26 January 2022). "Determining the charge distribution and the direction of bond cleavage with femtosecond anisotropic x-ray liquidography". Nature Communications. 13 (1): 522. https://doi.org/10.1038/s41467-022-28168-0. ISSN 2041-1723. PMC 8792042. PMID 35082327.


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Are Daniell cells rechargeable (reversible)?
I know that the Daniell cell is a primary cell and the reaction occurs as:
$$\ce{Zn -> Zn^{2+} + 2e^-}$$
$$\ce{Cu^{2+} + 2e^- -> Cu}$$
However, I do not know if we can reverse the reaction by applying a potential greater than $1.1~\mathrm{V}$. Also, please give the reason if it is not
P.S.: Please do not confuse this question: Why can't a primary cell be recharged?
I asked it as neither the above link nor the internet could give me satisfactory answer.

Daniel Cell is not a reversible cell. You can read the exact reason in Levine's Physical Chemistry book. I think it has something to do with the membrane separating the two halves of the cell.

The Daniell cell is not rechargeable. According to this Wikipedia article:

In the Daniell cell, the porous barrier cannot prevent the flow of
  copper ions into the zinc half-cell. Hence, recharging (reversing the
  current flow by an external source of EMF) is impossible because, if
  the zinc electrode is made to become the cathode, copper ions, rather
  than zinc ions, will be discharged on account of their lower
  potential.

A Daniell Gravity cell is reversible.  I just did the experiment and it worked.  I set up a cell with copper powder at the bottom and a zinc electrode at the top and filled the cell with zinc sulphate solution.  I applied 3 volts at 0.5 amps for a half hour and noticed that zinc plated onto the zinc electrode and copper was turning into a layer of blue copper sulphate solution at the bottom of the cell.  The cell produced 1.1 volts at .2 amps.
See attached photo:

In your image I can not see how you close the electrical circuit for your electrochemical cell. How did you do it?

My students usually find in the laboratory that if the $\ce{Cu^2+}$ ions get in contact with a Zn electrode, they will start corroding right away. Guess your device relays on the very slow diffussion of the copper(II) ions during the experiment, but eventually the  ions will reach the Zn electrode. So we can not say the cell is rechargable.

Hi PAEP,  the copper powder at the bottom has a copper wire going to it and connected to the power supply as does the zinc.  You are correct that the copper plates out in the zinc.  It falls off back to the bottom so it will just turn back into copper sulphate when I recharge the battery.  The one trick is that I need to scrape the zinc electrode clean before recharging.  A selective anion exchange membrane between the top and bottom of the cell would take care of the  unwanted diffusion of copper.


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Space Group names
In space groups, how do know if which axis/planes the glide/mirror/rotations are perpendicular to.
For example in Pnma,
how do we know that a-glide plane is perp to c-axis, n-glide plane perp to a-axis, and mirror plane is perp to b axis?

@IvanNeretin So for Ama2, we would have _m(⊥a),a(⊥b),2(c⊥)?

2 is an axis, not a plane, hence it is parallel to c.

@IvanNeretin Thanks, if you want to post those comments as an answer I'll accept it.

That's quite simple, if we are talking of orthorhombic groups. The elements corresponding to the cell directions $a,b,с$ follow in the same order as $a, b, с$ in the alphabet. Here by "corresponding to" we mean "perpendicular to" in case of a plane or "parallel to" in case of an axis.
So this is what $\rm{Pnma}$ consists of:

glide plane $n\,(\perp a)$,
mirror plane $m\,(\perp b)$,
glide plane $a\,(\perp с)$.

Another example: $\rm{Ama2}$ has

mirror plane $m\,(\perp a)$,
glide plane $a\,(\perp b)$,
axis $2\,(\parallel с)$.

Yet another example: $\rm{P222_1}$ has rotation axes 2 along $a$ and $b$, and a screw axis $2_1$ along $с$.
Things get hairy when we reach the tetragonal groups, but that's another story.


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Minimizing Gibbs Free Energy for Multiphase Equalibria
I'm a geologist and don't have a strong background in chemistry or thermodynamics.  Any help on solving this problem is much appreciated.
I'm trying to minimize Gibbs free energy to get initial estimates of Henry Coefficients ($K_i$) for a flash calculation.
Several publications presented the equation for Gibbs free energy for multiphase systems as the equation in the photo below.  My question is relatively straight forward: how do I calculate the chemical potential of each component (methane, ethane, propane, water) as a function of temperature?
Also, I have Matlab scripts to calculate fugacity. However, it calculates a fugacity for each phase (gas and liquid). I'm not sure how to calculate the reference fugacity of each component while knowing that part of each component is in two phases?
From there, I can use simulated annealing in MATLAB for cycle through potential compositions to evaluate the minimum.
$n_i$ = mols of component $i$
$\mu_{i0}$ = chemical potential of reference state
$f_{i0}$ = fugacity of the reference state
$f_{ij}$ = fugacity of component $i$ in phase $j$
$n_{ij}$ = moles of component $i$ in phase $j$


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Why are bromine oxyanions uncommon?
Sodium hypochlorite is used in bleach, calcium hypochlorite is used in public swimming pools, ammonium perchlorate is used in solid rocket fuel.
Sodium periodate is used in the cleaving of syn-vicinal diols, and is generally useful as an oxidizing agent.
Bromates are water contaminants, and perbromates weren't even successfully synthesized until 1968, but only through radioactive decay. Though alternate methods for their production exist today, what makes bromine oxyanions so unstable when chlorine and iodine oxyanions are widely used in chemistry?

Related but not duplicate:http://chemistry.stackexchange.com/questions/31113/middle-row-anomaly-in-periodic-table

I can only really speak for bromine(VII). Bromine(V) is pretty common and I'm not entirely sure what's the deal with Br(I) readily disproportionating to Br(V) + Br(-1). (The tendency of Br(I) to undergo disproportionation explains why we don't see it very commonly, but why it has this tendency I'm not so sure.)
This is an example of what is sometimes called the "alternation effect", or "d-block contraction" in p-block chemistry.
Due to the intervention of the ten 3d electrons, which are poorly shielding, the 4s and 4p electrons of the elements Ga through Kr are more tightly bound than one might expect. These elements show a distinct reluctance to take on their group valency. In $\ce{BrO4-}$, the higher-than-expected energy cost of ionisation to form Br(VII) is not fully compensated for by covalent bond formation (the strengths of which decrease linearly down the group). A selection of other observations that can, to various extents, be attributed to this are:

the anomalously high electronegativity of germanium (2.01), compared to silicon (1.90) and tin (1.96)
the anomalous instability of $\ce{AsCl5}$ (decomposes above $-50~\mathrm{^\circ C}$) even though $\ce{PCl5}$ and $\ce{SbCl5}$ are well-known,
the fact that nitric acid oxidises sulfur to $\ce{H2SO4}$, but selenium only to $\ce{H2SeO3}$

In a sense, it is exactly the same as the inert pair effect, which arises partly due to the lanthanide contraction - i.e. the intervention of the fourteen 4f electrons, which shield 5d, 6s and 6p electrons poorly. The difference is that the inert pair effect is even more pronounced because of significant relativistic stabilisation of the 6s electrons. Here's an insightful discussion on the topic: Why does tin form tin (II) compounds?


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α versus β D-ribopyranose
I understand that sugars are named α/β according to the arrangement of the OH on the anomeric C with respect to the hydroxymethylene group (Ex. glucose), but I'm a little confused how you determine α/β when you don't have that group (Ex. ribopyranose, below).

α-D-ribopyranose
 β-D-ribopyranose
Thanks!

Draw the compound in its linear form.

@Joel I don't see how that would matter? It's the same exact compound in it's linear form...

Yes it is but it's easier to determine its isomerism in its linear (straight chain ) form.

It's actually not the hydroxymethelene group that determines the α/β status of a sugar, instead it's the relative orientation of the anomeric hydroxyl to the stereocenter which determines the absolute stereochemistry (D/L) of the sugar. This just so happens to match up with the same side/opposite side heuristic for the common six carbon pyranoses, but that's a quick rule of thumb, not the official definition:

α (alpha), β (beta): Relative stereodescriptors used in carbohydrate nomenclature to
  describe the configuration at the anomeric carbon by relating it to
  the anomeric reference atom. For simple cases the anomeric reference
  atom is the same as the configurational reference atom. Thus in
  α-d-glucopyranose the reference atom is C-5 and the OH at C-1 is on
  the same side as the OH at C-5 in the Fischer projection.

IUPAC Gold Book

The configuration at the anomeric centre (that derived from the
  carbonyl carbon) is denoted alpha- (α-) or beta- (β-) by reference to
  the stereocentre that determines the absolute configuration. In a
  Fischer projection, if the substituent off the anomeric centre is on
  the same side as the oxygen of the configurational (D- or L-) carbon,
  then it is the α--anomer. If it is directed in the opposite direction
  it is the β-anomer.

CAZypedia - includes diagrams and examples, including for five carbon pyranoses.


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How can chlorine have an oxidation number of 3 in chlorite?
I am studying redox reactions right now and I'm a bit confused about something:
$\ce{ClO2^-}$ is a compound ion and in a compound ion the sum of oxidation state of participating ions has to be equal the charge on compound since $\ce{ClO2^-}$ has a whole charge of $-1$ the oxidation state sum of $\ce{Cl}$ and $\ce{O2}$ must be $-1$, right? Oxygen has an oxidation state of $-2$ so $\ce{O2}$ will have
\begin{align}
2\cdot -2 &= -4\\
-4 + x &= -1\\
x &= -1 + 4\\
x &= 3\\
\end{align}
Chlorine has to have an oxidation state of $+3$.
But if you look at the periodic table chlorine is in 7th period which means it has 7 electrons in valence so it gains one electron while reacting.
Getting oxidized by definition means losing electrons so an oxidation state of $+3$ means chlorine is losing 3 electrons. I just can't get my head around why would chlorine do that? Or am I calculating the oxidation state wrong?

It is indeed in the +3 oxidation state. Keep in mind, though, that that does not mean chlorine lost three electrons. That polyatomic ion is not made up of smaller ions.

First and foremost, please remember that an oxidation state is simply a book-keeping method that assumes that all bonds between atoms are ionic in nature.
In the chlorite ion, we know that the bonds between chlorine and oxygen are covalent in nature.
Therefore, when we say that chlorine has an oxidation state of +3 we do not mean that the chlorine atom has transferred three electrons to the oxygen atoms, as that would imply that the reality of the nature of the bonds between oxygen and chlorine is ionic.

Chlorine can indeed have an oxidation state of +3. Infact, it has multiple oxidation state to a maximum of +7. This is because chlorine is able to engage its d-orbital by hybridization, in which an electron in a lower energy level of may be an s-orbital or any other orbital below the d-orbital, reaches an excited state where it is promoted to a higher energy level which is the d-orbital. The d-orbital has an orbital orientation of 5. I hope this helps.

OK, I'll bite.  What is an orbital orientation of 5?

The fuck you talking about


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How does Carbon Dioxide have a degree of freedom of 6?
Carbon Dioxide has a degree of freedom of 6, yet from what I understand, it is a linear molecular shape. Therefore, the number of axis of linear movement is 3 (x,y,z) but there are only two axis of rotational movement since the center of mass essentially does not rotate on one of the axis due to its linear shape. Yet I'm pretty sure my textbook is saying that carbon dioxide has a degree of freedom of 6 when that should be wrong. Can someone explain?

There are 3N degrees of freedom (trans+rot+vib) for all molecules, regardless of their geometry, so maybe you want to specify what types of degrees of freedom you are looking at. For carbon dioxide, there are 3 translational, 2 rotational, and 4 vibrational degrees of freedom. http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Statistical_Mechanics/Equipartition_Theorem

Thank you, but why did my textbook say it only had 5? I just looked it up on wikipedia and the formulas and degree assignments dont match up. And some places say the same thing as my textbook and some say different. I am so confused.

Look, I can't speak for the author of whatever textbook you're using. At least give some context.. Which degrees of freedom are you trying to consider? Exactly what is the problem? Are you trying to calculate the constant-volume heat capacity using equipartition or something? If you are, then the vibrational modes don't contribute to the heat capacity (they aren't excited at normal temperatures), so you only have 3 trans + 2 rot = 5 degrees of freedom for a heat capacity $C_{v,m} = 5R/2$.

Thanks, that last part was actually what I needed. Thanks for clearing it up!

No problem. It's always a good idea to be clear exactly what you're talking about - if the term "degree of freedom" is not qualified any further, there are always $3N$ of them, including all trans/rot/vib ones. Depending on the specific application, some of them may not be relevant, so just make sure you know what's going on.

Yup! I'll keep that in mind, although my question was actually pretty hard to phrase.

There are always 3N total independent degrees of freedom for a molecule, where N is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the x, y, and z directions.
Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.
The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have 3N total degrees of freedom, but you can set aside 3 of them as translation of the whole molecule in each of the x, y and z directions, leaving (3N-3) degrees of freedom.
Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the x, y, and z directions. But for linear molecules like $\ce{CO2}$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the 3N total. So while for non-linear molecules there are (3N-3-3) = (3N-6) degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are (3N-3-2) = (3N-5) degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for any of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that none of the atoms move due to the "rotation".
So since $\ce{CO2}$ has three atoms and is linear, it has 3*3 - 5 = 4 degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes. There's different ways you can decompose them, but the most useful one is 1) the symmetric stretch (both oxygens going out and in at the same rate while the carbon sits still), 2) the asymmetric stretch (the carbon going back and forth while the oxygens sit more-or-less still), and 3&4) two different out-of plane bending modes. (If the molecule is aligned along the x-axis, one each where the carbon moves in the y and z directions.)
That's it for $\ce{CO2}$: any movement you make - any position you put the atoms in - can be thought of as a combination of those nine movements (three global translations, two global rotations, and four internal vibrations/bending).

By the way. if that still isn't making sense with respect to your textbook, it would help to edit your question to include an exact quote of the sentence/paragraph you think is in conflict.


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Can you derive a formula for any internal energy of a gas as long as you know the degree of freedom?
I have the following formula that I derived from studying my textbook that generalizes to all ideal gases:
Monoatomic molecular gases ($\ce{He, Ne}$ etc.)
Degree of freedom: 3
$\ce \frac{3}{2}nRT = U$
Diatomic molecular gases ($\ce{O_2, N_2}$ etc.)
Degree of freedom: 5
$\ce \frac{5}{2}nRT = U$
Polyatomic molecular gases ($\ce{CH_4, H_2O}$ etc.)
Degree of freedom: 6
$\ce \frac{6}{2}nRT = U$
As you can see here, we observe that a general trend for the formula for the internal energy of a gas is:
$\frac{\text{degree of freedom}}{2}nRT$ = U
Is this correct?

Yes it is, look up "equipartition theorem".

It's correct for ideal gases.

cool! thank you orthocresol and chester miller


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Confusion with degrees of freedom, some say CO2 has 9, others say 6
My textbook doesn't consider vibration modes when determining the degree of freedom. Some other sources I have found don't either. But wikipedia, other sources, and even other users here apparently use the vibrational part to determine the degree of freedom. What is going on here? Can someone explain why it is so different?

If you have other information to add to the question it's usually a better idea to edit the previous one instead of posting a new one.

Are you referring to the "possible duplicate" or is there a question of mine that is very similar to this one

Also my questions are different because the other one wanted to know why CO2 had a degree of 6 when I thought one of the possible rotational axis was actually not a degree of rotation

$\ce{CO2}$ has 4 degrees of freedom for vibrations - the rule is $3N - 5$ for a linear molecule.  There are 3 translational and 2 rotational degrees of freedom - the other 4 are vibrational, of which 2 are degenerate. See this page for a more complete explanation.


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Is this mechanism for the formation of a tosylate correct?
I've been given this mechanism by my lecturer for the formation of a tosylate. It claims that NEt3 is needed to remove the hydrogen from the O+ in the second step. However surely the Cl- from TsCl would remove the H and no NEt3 is needed?

No, $\ce{Cl-}$ is a much worse base than $\ce{NEt3}$.

No, $\ce{Cl-}$ will not remove the $\ce{H+}$ because it is not basic enough. The $\mathrm{p}K_\mathrm{a}$ of $\ce{Cl-}$'s conjugate acid ($\ce{HCl}$) is $-7$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NEt3}$'s conjugate acid ($\ce{HNEt3+}$) is $10.75$ (source). This means that $\ce{HCl}$ almost completely dissociates in solution, and is very unlikely to form $\ce{HCl}$ molecules. $\ce{NEt3}$, on the other hand, is much more basic and will easily pick up the $\ce{H+}$.
Also the mechanism is incorrect. A lot of the arrows are wrong. See here (pyridine serves the same function as $\ce{NEt3}$):

pKa of Et3NH+: 10.75 in water I'm fairly sure that website got it mixed up. All the pKb values cited (including the anilines, pyridine etc.) are actually pKaH values

This sulfur-oxygen bond movement is... eh.


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Solubility of EDTA in water
Why is EDTA ($\ce{C10H16N2O8}$) poorly soluble in water?
I am interested in analyzing the hardness of water in a sample due to the presence of $\ce{Ca^2+}$ cations. I attempted to do this by titrating with ethylene diamine tetraacetic acid ($\ce{C10H16N2O8}$) however these attempts have proven to be futile.

There are all sorts of standard procedures for doing a total water hardness. You can also do Ca only. Did you use a standard procedure (if so which one?), or did you try to make something up?

I've done EDTA titration of hard water. We did it in the class I used to teach. What procedure are you using? Are you using EBT? To answer your question, EDTA isn't typically considered poorly soluble in water. I believe more is going on here.

I ran into the same problem myself when I bought (foolishly, it turned out) $\ce{H4EDTA}$ instead of $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ for some of the experiments for my doctorate.  It really confused me, because carboxylic acids are generally quite soluble and, with four of the things, you'd think it'd dissolve right away.  But, no.  There are even amines in there, too, but they apparently don't help much either.
Based on the abstract of a 1973 article by Ladd and Povey1, my guess is the apparent poor solubility of $\ce{H4EDTA}$ is actually a kinetic effect, due to a network of intra- and inter-molecular hydrogen bonds present in the solid (emphasis added):

The [EDTA] molecule exists in a cis conformation and possesses two-fold symmetry, the diad axis passing through the central $\ce{C-C}$ bond. The nitrogen atom is protonated, and takes part in a bifurcated intramolecular hydrogen bond. A short $\mathbf{\left(2.46~Å\right)}$ hydrogen bond links the oxygen atoms of adjacent molecules.

I don't have the necessary thermodynamic data at hand to argue with certainty, but my guess is that the activation energy for solvent penetration into this hydrogen-bond network is rather high, and thus the rate of solvation is slow even though the overall process is (presumably) exergonic $\left(\Delta G < 0\right)$.
Reference: Ladd, M. F. C.; Povey, D. C. Crystallographic and spectroscopic studies on ethylenediaminetetraacetic acid (edta). J. Crystal Molec. Struct. 1973, 3 (1), 15–23. DOI: 10.1007/BF01270899.

An option would be to dissolve EDTA through $\mathrm{pH}$ with an alkaline hydroxide, the cation of which will not form complexes, in effect re-creating $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ in vitro. The kinetics aren't great but much better than for the dissolution of the tetravalent acid if I remember correctly.


text:
Why are some organic reactions run at elevated temperatures?
We are using a water bath at $70\ ^\circ\mathrm{C}$ for different reactions to identify oxygen-containing organic compounds, such as the Tollens reaction, the iodoform test and hydroxamic acid test.
What is the function of heat in these reactions? Does the heat act as a catalyst?

Many if not most organic reactions are actually run under reduced temperatures. My personal lowest so far was the Masamune aldol which required $- 95~\mathrm{^\circ C}$.

Though they can produce a similar effect (faster reaction rate), heat doesn't act as a catalyst.
Reaction rate is dependent on how often reagents can overcome the activation energy of a reaction. Therefore, a lower activation energy, $E_\mathrm{a}$, means a faster reaction.
A catalyst lowers the activation energy of a reaction by producing lower energy transition states. This is how catalysts speed up reactions. A generic version of this is shown in the image below. Transition states occur at the local maxima of the graph in this image.

Heat, on the other hand, increases the average kinetic energy of the molecules within a system (i.e. most molecules are moving more quickly). By doing this, the chances of two particles colliding and having enough energy to react are higher because the particles will move faster. Therefore, they will collide more often, and when they do, they will have more "momentum" (so to speak) to get over this energy hill. By making it so more of the reactant particles can overcome this activation energy, heat also speeds up reactions.

Take a look at either the Eyring equation or the Arrhenius equation. The Eyring equation is:
$$k = \frac{k_{\mathrm B}T}{h}\exp{\left(-\frac{\Delta G^{\ddagger}}{k_{\mathrm B}T}\right)}$$
Notice that for higher values of $T$, the rate constant increases, implying that the reaction occurs faster.


text:
Which procedure(s) will allow a student to differentiate between solid sodium sulfate and solid sodium sulfite?
I. Make solutions of each and look for a precipitate when added to 0.10 M Ba(NO3)2.
II. Add crystals of each to 0.10 M HCl and watch for bubbles.
III. Make solutions of each and test with a pH indicator.
The answer is II and III only. I understand how III works, but I don't see how sulfite and sulfate can be differentiated by adding HCl; don't both get reduced to SO2 bubbles?

What's getting oxidized, then?

Sodium sulfite and hydrochloric acid partially react to $\ce{SO2}$ which is gaseous and thus will form bubbles if not dissolved immediately.
When sodium sulfate is added to hydrochloric acid it will dissolve. The solution then contains sulphate and hydrogensulphate ions. No gaseous product will be formed.


text:
Sulfuric acid's effect on rocks
I am no chemist, nor did I pick up much understanding of chemical reactions in school. But I would like a basic understanding of how "reactive" and "potent" sulfuric acid is on rock; i.e., whether or not, and if so, how fast it would dissolve or otherwise affect the rock. Now, I know there are various types of rocks, and I'm sure this question could be too broad if not narrowed to some extent. So let me give some specifics.
My questions
Assuming a highly concentrated sulfuric acid (98%; but weaker, if perchance that would be more potent against the particular rock):

Will the sulfuric acid affect (dissolve or otherwise) it at all (and the ideal concentration to do so)?
If it affects it, then:

how fast would it dissolve/otherwise a solid piece of rock (not powder), say 1 cubic inch (though I'm looking for rate answers, so whatever the evidence gives as a rate works)?
Is rate affected by pressure (i.e., the rock "sitting" in the acid versus a pressurized spray of the acid on it)?
How much faster(?) would it dissolve it if the rock was powdered?
What are the general products/results left after the effect?

Igneous Rocks

Basalt
Gabbros, which apparently has a large amount of pyroxene and plagioclase
Granite

Sedimentary Rocks

Limestone
Dolomite

Metamorphic Rocks
(Perhaps the chemically diverse make-up of these rocks makes answering not feasible.)

Marble (perhaps reacts similar to limestone, "Geologists use the term 'marble' to refer to metamorphosed limestone"?)
Slate

While I have no problems with you giving in your answer the chemical reactions involved, I desire at least a "laymen's" summary of what occurs to each.
(Also, feel free to re-tag if there are some viable tags I was not aware of.)

Limestone is calcium carbonate so it will be dissolved into hydrogen carbonate and then water and carbon dioxide. Presumably this would leave behind calcium sulfate, which is gypsum.

Related for a minor part of the question (not duplicate): http://chemistry.stackexchange.com/questions/47049/why-does-caco3-react-with-hcl-but-not-with-h2so4/47050

This is flagged as too broad, and I'm afraid I have to agree with the close flag.

This question probably cannot be answered correctly without conducting any experiment.
However, I would not expect that any of the rocks given would be dissolved by highly concentrated sulfuric acid. None of them has a composition that may completely result in the formation of soluble salts when reacting with sulfuric acid.
Presumably (hot?) dilute sulfuric acid may be able to dissolve parts of the rocks if they are powdered. E.g. $\ce{Na2O}$, $\ce{K2O}$, $\ce{MgO}$, and up to a small amount $\ce{CaO}$ and $\ce{Al}$.
Hydrochloric acid would do better since it forms soluble salts not only with Na, K, Mg, Fe, and Al but also with Ca. E.g. it could almost completely dissolve limestone, dolomite, and marble.

I agree about an "experiment," but that was why I asked here;  somebody must have experimented with sulfuric acid on stone and reported results in a journal/report. My hope is that somebody with access to such info summarizes the results in an answer. My understanding is that sulfuric acid does at least affect limestone and marble, but as Klaus commented on marble, needs to have a resulting protective layer dealt with to continue dissolving.

I don't think that someone tried to do serious research on dissolving the given types of rock in sulfuric acid. In my opinion it just does not make sense. But I am quite sure that there exist result about other aspects of the recaction with sulfuric acid. E.g. effects on the rock surface, or long time effects of exposure to acid rain, which also contains sulfuric acid.


text:
Boyle's law applied on mixture of gases
Q1. If $\pu{1.00L}$ of oxygen gas at $\pu{1.00 bar}$ pressure is combined with $\pu{1.00L}$ of nitrogen gas at $\pu{1.00 bar}$ pressure in a $\pu{2.00L}$ vessel, what is the total pressure in the vessel?
a) $\pu{0.50 bar}$
b) $\pu{1.00 bar}$
c) $\pu{2.00 bar}$
d) $\pu{4.00 bar}$
Q1. If $\pu{1.00L}$ of oxygen gas at $\pu{1.00 bar}$ pressure is combined with $\pu{1.00L}$ of nitrogen gas at $\pu{1.00 bar}$ pressure in a $\pu{1.00L}$ vessel, what is the total pressure in the vessel?
a) $\pu{0.50 bar}$
b) $\pu{1.00 bar}$
c) $\pu{2.00 bar}$
d) $\pu{4.00 bar}$
My thought process:
I know that Boyle's law states that the volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure, so in these cases it is safe to assume that Boyle's law can be used. For Q1, would the answer be simply 2.00 bars (c)? Instinct tells me it is, but I can't seem to explain why. For Q2, I'd think the answer is 4.00 bar (d) because the volume is halved, so the pressure should be doubled? Again, I'm not too sure though. Any help would be much appreciated.

you have both answers wrong.
(1) b - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 2 liter container. So the pressure stays the same.
(2) c - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 1 liter container. Since the volume is halved the pressure must double.


text:
Stereoselectivity differences in Lindlar's catalyst and Na/NH3?
Why does using the Lindlar's catalyst form cis-alkenes whereas $\ce{Na/NH3}$ forms trans-alkenes?

Related: Hydrogenation stereochemistry-Pd,Pt, Ni catalysts

In Lindlar's catalyst, it is a matter of the physical way in which the molecules react. Firstly, $\ce{H2}$ dissociates into atomic hydrogen and binds to the catalyst, and the alkyne coordinates to the metal surface. Since the surface of the catalyst can be thought of as a vast plane, all carbon to metal bonds must be facing down onto the plane. The hydrogen atoms then get inserted in place of these carbon to metal bonds, and the result is syn-addition.
In the $\ce{Na/NH3}$ reduction, when the electrons are first added to the triple bond and the $\mathrm{sp}$ hybridization is broken, the groups bonds on either side the triple bond repel each other into the lowest energy conformation, the trans position. This is why the $\ce{H}$ atoms being donated by the $\ce{NH3}$ add through anti-addition.


text:
Why is there syn-addition in this bromination of an alkene?
In my book I read that halogenation of alkenes results in anti-addition. Is this example showing syn-addition?
The example:

In this example, it is impossible to tell from the product whether the mechanism was syn- or anti-addition. Let's look at the two possible products. After anti- or syn-addition of bromine, two compounds are produced that look different. However, facile bond rotation around the C1-C2 bond shows that they are in fact the same compound.

How would one go about proving that anti-addition is actually occurring in 4-methyl-1-pentene. One solution would be to use an isotopically labeled variant as shown below. Starting from the deuterated version of 4-methyl-1-pentene, we can draw the anti- and syn-products again. Similar rotation around the C1-C2 bond makes it clear that the products are not the same but instead are diastereomers. It would be possible to distinguish these products by some analytical technique (1H NMR) and show that the anti-product is formed selectively.

No, this is not syn-addition. Your textbook is being lax in showing the stereochemistry of the prodcut (wrongfully so). The true stereochemistry of such a product should be written:

or

The question is asking about distinguishing syn- vs. anti-addition (diastereoselectivity). This answer addressing the enantioselectivity (or lack thereof).