text:
Do solids affect equlibrium?
After doing a calculation on $K_{C}$, I was told that solids do not affect the calculation.
This is confusing me. Is there a specific reason why solids do not affect the equilibrium shift?
Thanks.

Why, they do, only in a binary manner. Liquids and gases may have different concentrations or activities (10, 2, 1, 0.1, 0.005...), but a solid is either there (1) or not there (0).

Loose generalizations make me nervous. Let's be more specific and note that for example the $\mathrm{K}_{\mathrm{sp}}$ doesn't depend on how much of the solid phase is present. If however you're using activated charcoal to absorb something from solution, then how much charcoal you add does change the equilibrium.

@IvanNeretin - while in 'pure' chemistry you may be correct, the rather trivial extension to materials science and the thermodynamics of solids (alloys, etc.) suggests otherwise. There, of course, one has varying concentrations in the solid state, and have to apply the correct thermodynamics to obtain the equilibrium concentrations leading to the correct phase diagrams.

@Jon_Custer - That was why I was trying to bound the remark. If you want to talk about the solid phase in a liquid metal mixture using a phase diagram then equilibrium has other ramifications.


text:
Can hydrazine be protonated twice?
First, $\ce{N2H4}$ is a base when it reacts with water, it accepts a proton.
$$\ce{N2H4 + H2O -> N2H5+ + OH-}$$
Can I follow up the reaction with this?
$$\ce{N2H5+ + H2O -> N2H6^{2+} + OH-}$$

It's not that easy to shove in the second proton. You'll need a strong acid to do that. Water will not suffice.

What does it depend on?

Having two positive charges that close to each other would cause the second proton to be very acidic. Only a very weak base would resist protonation by that dication.

Yes, it is possible, but only to a very small amount.
The $\mathrm pK_\mathrm a$ of the hydrazinium (2+) ion in water is $-1$ (CRC Handbook of Chemistry and Physics, Ed. D.R. Lide, CRC Press, Boca Raton, FL, 2005), so the equilibrium
$$\ce{N2H5+ + H2O <<=> N2H6^2+ + OH-}$$
is almost completely on the left side.

It is the difference in $\mathrm{p}K_\mathrm{a}$s of the two acids that determines the equilibrium position. You can't know where it lies just by knowing one of the values.

Chemical suppliers sell hydrazinium sulfate or $\ce{N2H6^2+\,SO4^2-}$. So yes, you can protonate it twice; however, as the other answer mentions you need a strong acid. Sulfuric acid is strong enough as can be seen.

Probably it isomerizes to $\ce{N2H5+HSO4-} $ in water.  Crystal lattice energetics may serve to stabilize the dicationic form in the solid state.

when one proton is gained by hydrazine the l.p of other nitrogen will not be avialable due to the strong en value of $\ce{NH3^+}$. So biprotonated hydrazine is rare. We can get it by using strong acid.

source: Stack Exchange

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sort: votes


text:
Nickel(II) + borohydride debenzylation reaction mechanism
I've been looking at this reaction for cleavage of benzyl esters. I've also looked at the original paper and some others, but haven't been able to find any mechanism on how this reagent does hydrogenation.
It also works with Cbz-protected amines/amides/alcohols. With palladium/platinum, the transition metal does oxidative insertions into the X-Bn bond, but I'm not sure nickel could do this insertion. If it could, I would expect this reagent to debenzylate amines, amides and ethers in addition to esters, carbamates and carbonates.
I was wondering if it is known how this reagent debenzylates esters or if anyone has any insight into the mechanism.

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text:
Why is sulfuric acid used in the Jones oxidation of alcohols?
When oxidising primary alcohols, $\ce{K2Cr2O7}$ is used because it's a useful & readily available oxidising agent.
Why does the sulfuric acid need to be added?
Does it have to do with the acidity of the resulting dichromate solution? How would this facilitate the oxidation?

Low pH raises oxidation potential, AFAIK.

It protonates the chromic acid to form an oxonium ion, increasing positive charge at the chromium center.
Using $\ce{H^+}$ typically protonates the electrophile, making it more electrophilic (more electron deficient = more electrophilic), just like with acid-catalyzed esterification. In fact, you're making a chromate ester in this reaction.
See the first step here in this image (and the rest is related, also):
Image source

Also, it apparently helps break the dichromate ion into two chromic acid molecules.

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text:
What happened to the authors of 'Modern Quantum Chemistry"?
This is kind of a silly question, but my google-fu has failed me, and I can think of no better place to ask it:
What ever happened to the authors of the classic "Modern Quantum Chemistry", Attila Szabo and Neil S. Ostlund? I notice that they were faculty at IU and University of Arkansas, respectively, but see no mention of their retirement or anything else. Are they still with us today? If so, what are they up to?

Not really a silly question but very broad.  I will let it stand in the off chance that someone is a colleague.  The chances of that are fairly slim, though.

If I remember correctly,  Neil S. Ostlund moved to the University of Waterloo and then founded HyperCube Inc., known for HyperChem, at the same location.

I just googled "Attila Szabo" and landed on an NIH page here.

Ostlund also recently started Chemical Semantics which is a startup claiming to bring semantic web technologies to computational chemistry.

@ToddMinehardt - I came up with this search as well, but  found no mention of the book in his list of publication/works, so assumed it was not him.

@GeoffHutchison - Very cool. I had no idea this existed. I guess that answers my question then.


text:
What is the most stable oxide of francium?
I read that the heavier alkali metals, like potassium, rubidium, and cesium all prefer to form superoxides. Since francium is the heaviest alkali metal so far, I assumed it would follow the same trend as the previous alkali metals, forming $\ce{FrO2}$. However, I was told that the most stable oxide of francium is $\ce{Fr2O}$. Why would this be the case?

You know that Fr is itself unstable?

@Mithoron Yes, I'm aware that Fr decays quickly, but I suppose this is somewhat of a hypothetical question.

https://en.wikipedia.org/wiki/Francium says this superoxide would have more covalent character, the cited source doesn't say much more.

All Alkali metal oxides below sodium are superoxides. This is because of the bigger alkali metal cations' extremely low charge density and the high polarizability of superoxide relative to peroxide and regular oxide.
What this means in the real world, though, I do not know. I don't think Francium itself is stable, so what can we really say about the stability of its oxides?

Was that the correct answer to the local exam? I also chose the superoxide.

@YunfeiMa The correct answer is just the oxide $\ce{Fr2O}$

Francium Oxide(Fr2O) is an ionic compound composed of Fr+ and O^2- ions. To balance the negative charge in the compound, 2 molecules of francium must be present for every 1 oxygen, meaning the compound must be Fr2O

The claim is that heavier metals will combine with the superoxide anion  $O^-_2$

Perhaps the question made a distinction between Oxides and Superoxides  (considering oxide  to be the formal name of the  $O^{2-}$ ion)

Related: http://chemistry.stackexchange.com/questions/32062/isolating-radium-oxide

With a half-life of just 22 minutes for the longest-lived isotope francium-223,
Wikipedia says that due to intense heat of radioactivity it vaporizes itself. This is because as more and more atoms of francium clump together, the collective heat produced increases tremendously.
As the article states that $\ce{FrO2}$ is quite covalent, it couldn't withstand the intense heat that Fr atoms emit together from within $\ce{FrO2}$ and thus $\ce{Fr2O}$ which is more stable is favorable over $\ce{FrO2}$. This is the same reason we couldn't get an observable chunk of francium element in nature as of yet, either in free state or in combined state.
Look up this list of inorganic compounds that also mentions the existence of $\ce {Fr_2O}$.

Please don't post links to the mobile versions of the websites (unless there is a really good reason to do so). Note isotopes are usually denoted as "francium-223" or "$\ce{^{223}Fr}$", but not as "Fr-223" (this notation is often used for labeling and counting atoms).

@andselisk The simple notation Fr-223 is widely used in radiochemistry and radiation protection (e.g. by ICRP); probably even more widely used than the somewhat unwieldy notation $\ce{^223Fr}$.

FrO2 woukd not convert to Fr2O. The radiant energy would blow the atoms completely apart. Francium samples need a big heat sink, which is usually supplied by making a highly dilute solid solution in the corresponding caesium compound.

@OscarLanzi May I say that the environment in francium favors the formation of $\ce{Fr_2O}$ over $\ce{FrO_2}$ instead of saying that it converts?

The environment in francium favors the sample being blown apart by radioactivity. Period.

What else can be said in regard to the query of the question? So much heat should form oxide after all even though it would blow away much of it.

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text:
Why are gold-sulfur bonds so stable?
In my analytical chemistry book, I have seen that molecules' electrical conductance has been measured by putting sulfur-containing groups on them and using atomic force microscopy with a gold surface. They say that this works very well because of gold and sulfur's special affinity for one another. I know that thiols and disulfides will be very attracted to gold, forming very stable bonds. Why is this?

Because they form really stable $\ce{Au4^{2+}}$ cation. see this and this.

I found one more paper although less attractive - http://library.lanl.gov/cgi-bin/getfile?rc000062.pdf


text:
Using baking powder to make baked food crispy
The recipe I use for chicken wings achieves a fried texture by dusting the wings with baking powder and then baking at $220^\circ\pu C$ ($425^\circ \pu F$). The chicken is also dusted with salt and other seasonings, but I assume they're not chemically relevant.  What chemical process is going on, here?
I'd assumed it was some kind of reaction between the baking powder and the skin of the chicken, though I had no idea what. Then, one time, there was a stray boneless, skinless chicken thigh in with the wings I bought, and that also became crispy on the outside when cooked this way. I've also heard of people using the same technique on oven fries.

Check the baking powder to see if it has cornstarch. I suspect that is the secret.

@MaxW It does, as does a different brand that I've used in the past. But your question suggests that not all baking powder contains cornstarch, and the recipe doesn't specify baking powder that does contain it. Also, cornstarch is widely and easily available -- why specify baking powder if it's really the cornstarch, which constitutes at most a third of the powder by weight, that's needed? (And what's the chemical process involved in the cornstarch making things crispy?)

The Charlie Brown's Thanksgiving episode of Binging with Babish claims that the baking powder "lower[s] the temperature at which the Maillard reaction occurs."  Wikipedia doesn't quite back that up but says that the Maillard reaction "is accelerated in an alkaline environment (e.g., lye applied to darken pretzels [...]), as the amino groups ($\ce{RNH3+ -> RNH2}$) are deprotonated, hence have an increased nucleophilicity", which amounts to much the same thing.

Raaaaandomly found the answer to my question, three-and-a-half years later.

Not exactly an academic source, but a few articles on Serious Eats suggest that baking powder's gas releasing reaction forms small bubbles of protein on the surface that add surface area that can quickly dehydrate (a prerequisite for crispiness) and solidify.
It's also a pretty common ingredient in things like batter for fried chicken, where I imagine the expanding bubbles do a similar thing—allow easier removal of water from the batter by increasing the surface area so it can crisp up.
The pH change may also help break down the skin, if applied in advance.

It's not applied in advance in this case -- toss food in powder, put in baking dish, put in oven -- and it also works on meat with no skin and, allegedly, on potato, which doesn't have nearly as much protein.

The acid could definitely help break down the skin increasing its permeability to water - in other words the reacted skin drys out faster thus gets crispy.


text:
Why can one double bond exist in some resonance structures but two can't?
I am doing some biology and have not done chemistry in ages. I have two concepts I am struggling with.

When drawing Lewis structures, am I right in assuming that the central atom will always require covalent bonds to fulfil its octet before the outer atom's octets are filled?
For example when drawing lewis structures, the central carbon atom should be 'given' enough covalent bonds or lone electrons to have a full octet before lone electrons are allocated to the outer atoms in the structure.
I do not get why, for example, $\ce{HCO2-}$ can have resonant structures with one double bond (but not two double bonds) between the carbon and oxygens. (please see picture attached)

In your drawn example, the red-boxed structure requires the carbon atom to have five bonds, which, to channel my secondary school teacher, says "yuck, I don't like that".
This is a case of breaching the octet rule which has to with how the orbitals hybridise. Five full bonds on a carbon is not "allowed". However three full bonds and two partial bonds is common in both resonance and transition structures, and is in fact a much more accurate representation of the molecule you're drawing.
Don't forget too, that the location of the negative charge matters, and I can't see it in any of your drawings.

1) the rule of an octet must always be satisfied. That being said,a carbon can have a maximum of 4 covalent bonds.
2) Think of a resonance bond as the bonds being kind of 'smeared' between all the atoms, much as shown in the figure as the dotted line between two oxygens and carbon. In that way, the rule of an octet is never broken.


text:
Why is Avogadro's statement true that at STP conditions all matter have same number of atoms or molecules?
As atomic number increases, atomic size, number of electrons and shells also increase. So due to higher atomic size and maximum number of electrons, there should be maximum repulsion forces between atoms in a state of matter as the atomic number increases. Then how is it that at STP conditions all states of matter have equal number of atoms as stated by Avogadro?

You've mixed so many concepts that I'm not sure how to offer an explanation that would make sense to you. All in all Avogadro's constant gives the number of atoms (or molecules) in a mole of an element (or compound).  https://en.wikipedia.org/wiki/Avogadro_constant

Quiye simply, Avogadro said no such thing.

It is not true, nor did Avogadro claim it was true, that all states of matter have the same number of atoms in STP conditions.
You are probably confusing this with the related result that most gasses have approximately the same number of molecules per unit volume in STP conditions. This holds because, in the gas phase, attractions between molecules are typically weak, and the distances between molecules are typically much larger than the molecules themselves.


text:
Why isn't thallium triiodide stable?
We know that stability of higher oxidation state (+3) decreases and stability of lower oxidation state (+1) increases, so thallium is most stable in its +1 oxidation state due to the inert pair effect and paired s electrons.
In the compound $\ce{TlI3}$, since $\ce{I3}$ exists as a linear molecule $\ce{I3-}$  so the oxidation number of $\ce{Tl}$ is +1 which is stable and the compound should exist but I have read in many of the books that it does not exists. Why is it so?

Which books? Thallium Tri-Iodide,Tl+(I3)-, does exist. Its Thallium (III) Iodide,  Tl3+ (I-)3 to my knowledge, does not. Interestingly the [TlI4]- ion does, and contains Tl3+.

Note that, according to IUPAC nomenclature, the name tris(iodide) must be used for compounds containing $\ce{3I-}$ rather than triiodide, which is used for $\ce{I3-}$. Therefore, correct names for $\ce{TlI3}$ are “thallium tris(iodide)”, “thallium(III) iodide”, and “thallium(3+) iodide”. Correct names for $\ce{Tl(I3)}$ are “thallium triiodide(1−)”, “thallium(I) (triiodide)”, and “thallium(1+) (triiodide)”.

Thallium triiodide does exist  and it is formulated as $\ce{Tl+I3-}$ and not $\ce{Tl^3+(I^{-})3}$. Thus $\ce{TlI3}$ is a thallium(I) compound and contains the triiodide ion, $\ce{I3^{−}}$. This is confirmed by considering the standard potential which indicate that $\ce{Tl(III)}$ is rapidly reduced to $\ce{Tl(I)}$ by the iodide.
$$\begin{align}
\ce{Tl^3+ + 2e− &-> Tl+} & E^\circ &= +1.252~\mathrm{V} \\
\ce{I2 + 2e- &-> 2I-} & E^\circ &= +0.5355~\mathrm{V} \\ \hline
\ce{Tl^3+ + 2 I- &-> Tl+ + I2} & E^\circ_\mathrm{cell} &= +0.717~\mathrm{V} > 0 \\ \hline
\end{align}$$
(data from Wikipedia).
$\ce{TlI3}$ can be prepared by the evaporation of stoichiometric quantities of $\ce{TlI}$ and $\ce{I2}$ in concentrated aqueous $\ce{HI}$, or by reacting $\ce{TlI}$ with $\ce{I2}$ in ethanol.


text:
What causes the yolk of a frozen egg to be more spreadable instead of fluid?
If you freeze an egg the yolk of it gets more spreadable instead of it was fluid before. It appears that this state of the yolk is perm. So what chemical reaction causes this?
The question is of course after the egg had been melted again.

Can you rephrase this question? It's not clear what you mean by "spreadable" or "perm".

with spreadable I mean that the yolk had become more solid and perm means that it stays in that solid shape and not becomes fluid again after it is melted for a while.

OK, but neither of that is an English word.

Spreadable: with the capability of being spread. Perm: a contraction of 'permanent', most often used in relation to permanent hair colouring. They are definitely both English words even if you didn't know them.


text:
What is the mechanism of synthesizing hydroxybenzotriazole?
I have big problem with this reaction.

Is this nucleophilic substitution? Cl is removed by hydrazine? It's my only idea, I don't know if this mechanism below is correct for this example:

This is 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. Please [edit] in your full reasoning or thoughts on this.

related http://chemistry.stackexchange.com/questions/23482/benzotriazole-to-o-phenylenediamine

It looks pretty reasonable to me. I'm not positive, but I like it.

The second picture presents mechanism of o-nitrophenylohydrazine with KOH/H2O....so my reaction would have the same mechanism?

Yes it looks plausible except two things 1) Your conditions above don't have base present. 2) In the top-right corner of the second picture I would have an intermolecular proton transfer instead of a 4-membered intramolecular transition state.

So what it looks like, when I'm using EtOH and hydrazine? I must do it on laboratory...

Given that you've shown my proposed mechanism, yes, I think that's correct. http://chemistry.stackexchange.com/questions/26770/amine-addition-to-nitro-group/26880#26880


text:
Are labile tetrahedral [Mabcd] complexes optically active?
In the above mentioned complex, there is a chiral centre at the metal, and hence it seems the compound must be optically active.
But my course notes say that the ligands keep on changing their their positions at a very fast rate and as a result such complexes get converted into their mirror images and are hence optically inactive.
I am unable to reconcile my understanding of the stereochemistry with this statement in the notes. How can such complexes be optically inactive?

What's unclear about it? Yes, technically the molecule is optically active, and would have (R) and (S) isomers. But you can't separate them, because (R) changes to (S) and back pretty quickly (or not so quickly, depends on the metal, but whatever).

Please see my edits & change anything if I've misrepresented the situation or your confusion.

In a labile complex, ligands are detaching and attaching to the metal centre at extremely high rates. This means the solution will racemise quickly.  Therefore, even if the complex is truly optically active, you have equal proportions of each enantiomer and thus equal rotations of the polarised light plane - in opposite directions.
The end result is that rotation cancels and so none is observed.


text:
Gaussian: Relaxed scan with modredundant optimization and dummy atoms
I would like to perform a relaxed scan in Gaussian09 on my molecule using the modredundant optimization. Because I want to scan over two bond lengths simultaneously, I defined some dummy atoms which are close to the real atoms whose bond lenghts should be scanned. However, this approach does not work, because I can't properly index the dummy atoms. The Gaussian09 documentation says (emphasis mine):

[Type] N1 [N2 [N3 [N4]]] S nsteps stepsize [[min] max]]
N1, N2, N3 and N4 are atom numbers or wildcards (discussed below). Atom numbering begins at 1, and any dummy atoms are not counted.

How can I then perform a scan referencing a dummy atom? I don't want to do the scan using the z-matrix optimization because it does not converge.
Here is my input file:
%nprocshared=16
%mem=5gb
%chk=4MeAminoButanal
#P HF/6-31G(d) opt=(modredundant) scrf=(pcm) nosymm

4MeAminoButanal dummy scan

0 1
 C     0.000000     0.000000     0.000000
 N     0.000000     0.000000     1.451901
 C     1.325038     0.000000     2.055476
 C     1.991320     1.377897     2.032970
 C     1.229182     2.449098     2.809397
 C     1.309917     2.335179     4.306919
 X    -0.248373    -0.404008     1.611583
 X    -1.041414    -0.131882     1.066579
 H     1.218960    -0.323776     3.083767
 H     1.990264    -0.714465     1.564774
 H     2.993872     1.281489     2.436811
 H     2.106328     1.711600     1.006802
 H     1.602469     3.443240     2.566292
 H     0.178120     2.455789     2.536445
 H     0.708887     3.065770     4.854632
 H    -1.016528    -0.113770    -0.358506
 H     0.375985     0.943036    -0.379032
 H     0.606031    -0.798850    -0.431982
 X     1.641992     1.939406     4.607116
 X     1.311963     2.318294     5.471711
 O     1.974063     1.543637     4.907309
 H    -0.496745    -0.808017     1.771266

B 7 19 S 60 -0.05
B 22 7 F
B 2 7 F
B 6 19 F
B 21 19 F

When I run it, Gaussian09 obviously complains with

The following ModRedundant input section has been read:
   Invalid atom number.
   B 7 19 S 60 -0.05

This question might be better suited at [scicomp.se], but it is not off-topic here. You can indeed not use dummy atoms with a relaxed scan via modredundant, because they are not existent in that coordinate system. There is also no option, where this would be possible. Can you explain, what you are actually trying to do. To me it appears, that you are moving the middles of the CO and NH bonds closer together. If your input would work, then it would very likely not produce what you intended. (Btw. You are using too much proc and mem for HF.)

@Martin-マーチン I do not agree with you: this problem is very specific for a QChem program, so it is very unlikely that Computational Science people can say much about it, while many Chemist have experience with the given software. However I fully agree with you that OP may want to do something where dummy atoms are most probably not needed at all.

@Martin-マーチン That is exactly what I'm trying to do. How would you do it?

@Greg From our [tour]: Don't ask about... Computational questions ... . Other than a technical aspect of the program Gaussian which is used to calculate molecules it does not have any bond to chemistry. [scicomp.se] has the appropriate tag and most of the program users that are active here are (somewhat) active there. If you want to talk about this, we should probably go to meta.

@shrx I actually meant what you want to investigate. What is the chemistry behind this approach. How I would go about it, you can see in my answer. I guessed what you want to do though.

@Martin-マーチン If you think this question is better suited for another SE site feel free to migrate it.

Disclaimer: This answer concerns the use of Gaussian 09. While the ModRedundant interface is retained and therefore the same restrictions also apply in newer versions, there is now a newer way of using internal coordinates. See my other answer for this option.

Like I posted earlier in the comments, you can indeed not use dummy atoms with a relaxed scan via the ModRedundant interface, because they are not existent in that coordinate system. Internal redundant coordinates are generated only from actual atoms. I tried to find an explanation of that, but failed. I believe the reason for that behaviour is, that dummy atoms do not have basis functions and are therefore not included in the wave function.
In this case you need alternative means to what you actually want to achieve. That of course depends on what you would like to model. From the input you provided, I can see, that you are trying to move the middle points of the carbon-oxygen and nitrogen-hydrogen bonds closer together, possibly to simulate the carbon nitrogen bond formation and proton transfer. While I seriously doubt this will happen in one step, this is at least one way to try to go about it.
If this method had worked, I believe you would have gotten a different than expected result. There are some more issues, that you might have not considered with your approach.
You might think that the following two lines keep the nitrogen hydrogen bond distance fixed. They do not.
B 22 7 F
B 2 7 F

Because the bond angle A 22 7 2 is still flexible, and the bond length B 22 2 is still flexible, this can possibly do whatever it wants. The same does hold for the carbon oxygen bond.
B 6 19 F
B 21 19 F

If I am correct with my assumption, then what you want look at is the carbon nitrogen bond formation. Why not take this internal coordinate and scan it? It is easy and it should pretty much do what you like. The following input should take you there. Note that I changed the method to pm6 to massively cut down computer time - and even with that, it took to long for my patience. (I did also not include any trailing lines about resources - see my notes below.)

#P pm6
opt=(modredundant)
scrf=(pcm)
nosymm

4MeAminoButanal scan

0 1
C        0.000000     0.000000     0.000000
N        0.000000     0.000000     1.451901
C        1.325038     0.000000     2.055476
C        1.991320     1.377897     2.032970
C        1.229182     2.449098     2.809397
C        1.309917     2.335179     4.306919
H        1.218960    -0.323776     3.083767
H        1.990264    -0.714465     1.564774
H        2.993872     1.281489     2.436811
H        2.106328     1.711600     1.006802
H        1.602469     3.443240     2.566292
H        0.178120     2.455789     2.536445
H        0.708887     3.065770     4.854632
H       -1.016528    -0.113770    -0.358506
H        0.375985     0.943036    -0.379032
H        0.606031    -0.798850    -0.431982
O        1.974063     1.543637     4.907309
H       -0.496745    -0.808017     1.771266

B 6 17 F
B 2 18 F
B 2  6 S 60 -0.05

Also note, that I did not check that the N-C bond length will become shorter than an Ångström, so the last steps take forever to converge and it will die in the last step running out of cycles. Run it with something more reasonable.
The calculation itself produced a somewhat reasonable result. The following animation cycles through all optimised structures. It was made with ChemCraft.

Some more notes on your input:

%nprocshared=16 is quite a number you are requesting for a Hartree-Fock calculation. Consider that parallelisation is not linear. While it became better throughout the years, I guess 4 threads is the most reasonable for this system size. You can go higher, I'd not recommend to go above ten though.
%mem=5gb is probably too much. I doubt that any of these processes will exceed 200MB of memory, probably a lot less.
%chk=4MeAminoButanal You should consider adding the .chk ending in all cases. Gaussian is known to mess things up when there are non-ordinary characters like _-. in a file name.
It is maybe a good idea to not write the whole route section in one line. Consider this as improving the readability of your input.  You can cluster similar commands (like pop statements) in one line, etc..
In previous versions, Gaussian might have had trouble reading in anything beyond the  80th char. I cannot reproduce this bug anymore and it's been a long time since I was told about it, so this should not be an issue. However, a well structured input file can help you finding possible syntax errors that result from typos.

Gaussian sometimes has (still) trouble reading in anything beyond the 80th char.  Wow, I never knew this...  Can you offer up an explicit example? I'd love to see this behavior.  Great answer by the way.

@LordStryker Thank you and happy to see you. I could not reproduce this bug anymore, so I am fairly certain they fixed it. Gaussian used a fixed width of 80 char in the input (and still uses that as output for readability, but not everywhere) and break longer lines, apparently that could have destroyed some keywords. My former supervisor told and showed us, but that might have well been the 98 or 94 version. I'll fix that statement.

I just realised, that this bug may have resulted from preprocessing of the input file, inserting spaces within a keyword, which the produces a syntax error. It might not be gaussians fault after all...

Indeed I'm trying to do the C-N bond formation and simultaneous proton transfer. I tried doing it the simple way like you did, to just move the C and N atoms closer together. However, this does not lead to the desired proton transfer, as evident in your animation. I provided my input file as an example, but I am interested in a general way to do simultaneous scan over more than one variable with the modredundant optimization in Gaussian. For now, I have resorted to a not yet fully-functional python script that simulates a multi-variable scan. I can provide the code if there is interest.

@shrx Well technically you are scanning only one parameter in your example input. You can do a multidimensional scan by providing more than one scan command in the modredundant section. I am still not sure what you are actually trying to achieve, because I believe that there is no true concerted transition state.

This is the result of my simultaneous scan over the C-N and O-H distances (animation made with molden): http://i.imgur.com/NjhSJyN.gif

@shrx I really don't want to go any deeper into this discussion here. If you want to talk about it, feel free to find me in chat. I answered your question to the best of my knowledge and I am very sorry that I could not help you with your problem.

@Martin-マーチン no problem, I have already upvoted your answer as I think it is quite informative, even if it doesn't lead to the desired result. If I have further questions I will contact you in chat.

Disclaimer: This answer concerns the use of Gaussian 16. While the ModRedundant interface is retained and therefore the same restrictions also apply to it as mentioned in the other answer. However, there is now a newer way of using internal coordinates as explained below.

The new magic is provided by the generalised internal coordinates (GIC), which provide awesome new ways of defining input. It is invoked via the geom or opt interface and the option specified is GIC. That alone doesn't let you perform scans, it just changes the internal use of coordinates and builds them. It has to be combined with the usual ModRedundant, or the new AddGIC (GIC would then be redundant) options.
The latter invokes the reading of a new input section after the molecule definition, just like in Gaussian 09. However, if you read through the info page, you will quickly notice, that many more options are possible, including mathematical operations. In this way you can define intermediate anchors which you may use as substitutes for dummy atoms. Dummy atoms in the geometry definition still don't count, and are completely superfluous with this approach.
The following input will perform the scan that you initially intended. It will break close to the end, as a four-membered ring will form and with two sides and the mid-points distance fixed, it simply runs out of cycles. You may want to adopt the input according to your needs. I'll explain a little more below the input-file.

#P PM6
OPT(MaxCycle=100,Loose)
SYMMETRY(none)
GEOM(ModRedundant,GIC)

4MeAminoButanal scan

0 1
C        0.000000     0.000000     0.000000
N        0.000000     0.000000     1.451901
C        1.325038     0.000000     2.055476
C        1.991320     1.377897     2.032970
C        1.229182     2.449098     2.809397
C        1.309917     2.335179     4.306919
H        1.218960    -0.323776     3.083767
H        1.990264    -0.714465     1.564774
H        2.993872     1.281489     2.436811
H        2.106328     1.711600     1.006802
H        1.602469     3.443240     2.566292
H        0.178120     2.455789     2.536445
H        0.708887     3.065770     4.854632
H       -1.016528    -0.113770    -0.358506
H        0.375985     0.943036    -0.379032
H        0.606031    -0.798850    -0.431982
O        1.974063     1.543637     4.907309
H       -0.496745    -0.808017     1.771266

xCO(inactive)=XCntr(6,17)
yCO(inactive)=yCntr(6,17)
zCO(inactive)=zCntr(6,17)
bCO(freeze)=B(6,17)
xNH(inactive)=XCntr(2,18)
yNH(inactive)=yCntr(2,18)
zNH(inactive)=zCntr(2,18)
bNH(freeze)=B(2,18)
distCOandNH=sqrt[(xCO-xNH)^2+(yCO-yNH)^2+(zCO-zNH)^2]*0.529
distCOandNH(NSteps=30, StepSize=-0.10)

The first line of the GIC input defines the x-coordinate of the centre of the CO bond, and the following two do the same for y and z. They are set to inactive as we only need them as an intermediate anchor for the distance. They will therefore not show up in the output.
The fourth line freezes the CO bond length. I decided to include that because it is part of your initial set-up. After running the calculation I think it should be relaxed though.
The four lines after that define the same as above for the NH bond.
The second to last line defines the distance between the centres. The multiplication with 0.529 converts from bohr to angstrom. That is basically only necessary because it makes the following operation, the actual scan in the last line, easier to specify.
This new interface therefore provides many wonderful opportunities, and possibilities for the most exotic scans. However, it also lets you define certain values to be printed in the output, for much easier post-processing. I can only recommend reading the manual and making a lot of use of this new system.
In order to not leave you hanging dry, here is the resulting scan animated:

Thank you for your answer. What if one wants to perform the equivalent rigid scan ? I could not find a straightforward solution without employing Z-matrix ...

@gluuke I don't think there is an easy straight forward way to do a rigid scan without a z-matrix. Of course you can created all resident coordinates yourself and then freeze the majority, but that is probably as extensive a task as creating a sane matrix.


text:
Henry's law Numerical
I got very basic doubt. Henry law's Give relation between concentration vs Partial pressure.  So here is the Numerical Of my textbook

**Now My question is, why concentration is Considered only in Grams? shouldn't it be in Molarity, Mole fraction or Strength(gm/Vol) etc? **

In Other textbook, They simply used formula to solve it.
mass = Henry's Constant * pressure.

Frankly I think you're right. It is a sloppy problem statement. The extra "$\mathrm{L}^{-1}$" should be there.  !@#$%^& the formula for the answer uses c for concentration not m for mass of solute. For the saturated solution the volume of the solution doesn't make any difference. However for the unsaturated solution if the total volume is 1 liter or 10 liters it does make a difference.

Thank you for reply. I didnt get you. How Volume is linked upto saturation?

The concentrations of the gases in solution should be $\text{ g L}^{-1}$ not just $\text{ g}$.

For this example, the volume of the solution doesn't matter. It's a thought experiment, doesn't say what sovent is used, and you could replace ethane by helium or miraculane.
As long as you talk about the same compound, mass is proportional to molar quantity. Plus the units cancel out in the calculation anyway.
Think of the hypothetic experimentator as a guy who only has one glass beaker, without graduation. ;)

SO it is assumed that (Volume of) container is same in both cases? But most of the text are simply using formula ( mass= Kh*P)

Have you understood the principle behind Henry's law? You sound like you just want to learn this by heart for an exam. ;) That formula you cite is constructed for this special case, and the way they give it in your second textbook is totally inadequate.

Yes I understand the principle. Personally I think , if I put concentration as  mass/Volume. Volume will cut out each other as they are same in both cases, but i checked 4-5 textbooks, All of them are using formula (m=k.p) which according to me is blunder!!   .... it is actually not for exam.  I am just curious guy who cant proceed Till the point I am 100% satisfied with the basic concept.

What textbooks are those?

Most of the Indian Author's relate henry law with mass per unit volume. I dont know the reason.

They are implicitly assuming that the number of moles of ethane in solution is very small (in both cases) compared to the number of moles of the solvent.  So, in both cases, the concentration of ethane and its mole fraction are essentially proportional to the mass of ethane in the solution and the amount of solvent is the same.

Hi i have added the Screenshot of other Solution as well. How they came up with that relation?

That formula works but is only useful for this special case with a constant, unknown volume, and the calculation should never be done this way, calculating an unnecessary quantity like "kH" in between. It's not even a Henry's constant, that should of be concentration per pressure. I'd try to get rid of that book.

I agree totally with @Karl.  They basically assumed what I indicated in my answer, but with no explanation of what they were doing, or their assumptions.  And, as Karl said, that $K_H$ is not even close to what  we conventionally define as the Henry's law constant.

What if we take Concentration in  GRAMS/VOLUME and perform the experiment in same Flask (same Volume).  Wouldn't in this case mass will be proportional to partial pressure?  ( without any assumption)

To have the same volume, you would have to change the amount of solvent.  If that's OK with you, that's fine with me too.  As long as the amount of the solute is very small compared to the amount of solvent, it won't matter significantly whether you work with concentration or mole fraction.

Henry's law is usually expressed using molar concentrations, which makes more sense, because the partial pressure is proportional to activity of the solute, strictly speaking. But as long as you stay at low concentations, you could use any measure, vol-%, mol/kg, g/l, etc., Henry's law will still work acurately.

If in the calculation, both concentration are multiplied with the relative molecular mass of ethane, which is as good as multiplying by 1. The quantities to use on the concentration variables will be masses.


text:
How does an energy input actually change a molecule?
I'm reading an online biochemistry textbook for fun, and a question that has bothered me on chemistry before popped up again. The question here is not specific for the chemical process described, but I needed an example:
By adding energy to a molecule of formic acid ($\ce{CH2O2}$), it kicks out an O and becomes formaldehyde ($\ce{CH2O}$). More energy makes it methanol ($\ce{CH3OH}$), and more energy makes that methane ($\ce{CH4}$).
I get that energy allows the molecule to become a different molecule of a higher energy state; it essentially stores the energy as new or altered chemical bonds. What I never understood is why? It is natural for the molecule to try to go to a lower energy state, releasing energy in the bonds, as it does under oxidation/"burning". So why doesn't it just immediately release the added energy, rather than store it in bonds?
What prevents the formaldehyde from saying "I see your energy, but I'll just do what's easier for me and release some of my own, and become formic acid"? After all, no matter how much energy I use to blast a block of stone into the air, it's going to fall down again, unless someone up there catches it. What's "catching" the higher-energy molecule and preventing it from "falling back" into the more stable, lower-energy molecule?
The answer seems like it might be leaning on atomic theory more than chemistry, but I thought I'd give it a shot here.

To use your "throwing a rock up" analogy: if you throw it hard enough, it will reach orbit or escape the planet's attraction. After that happens, there is a time period in which two different sections of a molecule may move so that if-and-when the electron does fall back, it's not into the same place.

It is about transition energy barriers.
Imagine a mountainous landscape with lots of valleys of different depths. Stable molecules are large balls than sit in the bottoms of the valleys. Add some energy and the ball might rise up the valley floor but it will fall back. Add enough energy and the ball can rise high enough to get over the barrier to the next valley and will fall into a different place with a different depth (and the analogy falls down as it also becomes a different ball). So, whatever the relative depths of the valleys, there is a big barrier to getting from one valley to another. Even when one valley is much lower than another (so moving from one to another releases energy) you still have to put energy in to cause the transition.
Chemical reactions are often controlled by those transition barriers between one molecule and another rather than by the absolute energy difference between the molecules. Methane and oxygen will burn to give various compounds and will release energy, but there is a big barrier to starting the burning process. This means that the deoxygenation reactions you describe are possible and won't spontaneously reverse even though they would release energy if they did.

That makes a lot of sense! One thing, though: When it reaches the peak, what prevents the "ball" from rolling back the side it came from? Is there some kind of chemical momentum (bad analogy, my sincerest apologies, but...) or is it a matter of chance (which would create a lot of other interesting questions...)?

@HenryStone (1) Yes, sometimes the "ball" does roll back down the side it came from; see transition state theory. (2) Approximately speaking, yes, there is a kind of chemical momentum; see reaction dynamics. (3) Yes, there is an element of "chance" (technically, of statistics) to the progression of a reaction.

@HenryStone adding to Brian's comments: biological systems are often even more complex as enzymes often direct reactions down a small hard-to-find path between one valley and another (in some cases by altering the landscape, though we risk getting lost in analogy land if we go to far with this idea).

Heh. Enzyme catalysis as terraforming.

It is also important to mention (and this is especially true in organic reactions like your example, but also other chemical reactions),  that when we have a vessel with reactants and we heat it to get product, most often than not, you will get a bunch of different products and not just the one you wanted to make initially. And also, often there will be unreacted material. This happens precisely for the reason of ball being able to roll into different valleys and also because of kinetics. Really, there are a lot of factors to think about.
So if you bring enough energy to transform formic acid to methanol, there still might be some molecules changing into the aldehyde.
This is not to say that reactions with the singular product don't occur, they are just much more rare. It all depends on energies of a reactant, product and particularly on energy of transitional state and how they relate to each other (or in analogy, the relative distances between hills and valleys)


text:
Looking for solubility kinetics of zinc chloride hydroxide monohydrate
$\ce{Zn5(OH)8Cl2.H2O}$ is a crystalline solid insoluble in water. Often used as nutritional supplement. Preparation and chemistry of zinc chloride hydroxide monohydrate is known. I am looking for some additional data such as solubility or solubility kinetics basically in anything. I wasn't able to find any valid data.

Guess it would dissolve in any strong acid or alkali.

There's 500 pages report in German by the Gesellschaft für Anlagen- und Reaktorsicherheit (GRS) titled  Entwicklung eines thermodynamischen Modells für Zink, Blei und Cadmium in salinaren Lösungen. Around page 283, there a bit on Simonkolleite.

@KlausWarzecha Thank you a lot! I will take look at it.


text:
l- versus d-glucose
I recently purchased Dextrose from a local bulkfood store. I want to use it instead of table sugar. My concern is that it may be l-glucose instead of d-glucose. According to Wikipedia l-glucose is too expensive to produce but l-glucose produces laxative effects which is what I'm experiencing when I consume this dextrose.
Any tests I can do to check if it's truly d-glucose?

related http://chemistry.stackexchange.com/questions/42458/taste-of-sugars-unnatural-enantiomers

The canonical experiment is to measure the optical activity (rotation of plane polarized light) by a solution of the substance. Dextrose, unsurprisingly, is dextrorotatory (rotates light to the "right"). l-glucose is levorotatory (rotates to the "left"). Two things to keep in mind: There are different d and l nomenclatures. A compound labeled "D-" is not always dextrorotatory.) Also, optical activity is only definitive if you've already established the compound is d-/l-glucose. If it's fructose ("levulose") or some other carbohydrate, the normally occuring isomer could be the levorotatry one.
Given that you likely don't have a polarimeter handy, another easy experiment you can do is to add some yeast to a weak solution of the sugar. Yeast can readily use d-glucose, but should have a difficult time fermenting l-glucose. The reason for this is that the enzymes which catalyze the utilization of glucose are chiral, and will only match the typical d- configuration. l-glucose won't fit, like a left foot doesn't fit well into a right shoe.
All that said, it's so exceedingly unlikely that you were sold l-glucose that it boggles the mind. First off, there's the labeling concern. "Dextrose" specifically means the d- isomer, so anyone selling l-glucose as dextrose is committing fraud. Secondly, the price differential is astronomical. Spot checking, from a chemical supplier l-glucose costs over \$50 per gram (so ~\$400 per teaspoon). In comparison, d-glucose of a similar grade from the same supplier is around \$0.04 per gram (so \$0.30 per teaspoon). With over a thousand-fold price differential, someone along the way would recognize if they were accidentally selling l-glucose.

Ok, then I should be sound assuming that I'm consuming d-glucose. I am going to try the yeast trick eventually. Thank you R.M.


text:
How much CO2 is produced by plant litter?
I made measurements of CO2 formed during decomposition of dead plant material using gas spectrometer. It gives the results of CO2 in ppm. How do I convert it to $\large \rm \frac{mg~CO2}{\text{1 g plant matter}}$ per day? I know that it has something to do with volume of my jar and and mass of plant litter, but that's all.


text:
Why so few fluid phases?
Taking a look at the phase diagram of water

By Cmglee - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=14939155
I see ice phases as much as up to 'Ice XI', but only one 'liquid' and one 'vapor' phase. Why is this? Either there are many ice phases, or too few liquid phases relatively.
The latter seems to be the case as it seems to be consistent with other substances. No matter what number of solid phases you get, there is usually only one liquid and one vapor phase.
Why are there so few common$^1$ fluid phases? What is the physical explanation for this?

1: For the sake of not complicating the answer too much, please ignore more exotic phases such as mesophases etc. If, though, your answer benefits from relating to them without confusing the reader too much, by all means $\ldots$

To me, there are too many fluid phases. Molecules in a solid are ordered, like soldiers in a formation - no wonder you can arrange them in a multitude of ways. But when there is a complete chaos, when they are moving, bouncing, turning this way and that, never standing still - how come it's possible to have two different manners of doing that? Come to think of it, they are not all that different, because critical point and stuff. But still they are different, and that bothers me.

The difference between liquid and gaseous is short range order, in the case of water resulting from hydrogen bonding. Nothing to worry about. ;-)

And if the the intermolecular potential is purely repulsive (i.e. no short range attraction) you only get 1 fluid phase.

Even with water there's possibility of additional liquid phases, supercooled one https://phys.org/news/2013-07-liquid-liquid-phase-transition-low-temperature.html and "ionic water" at high temp. and pressure.

The solid (crystalline) state is defined by structure, for which there can be many. Liquid means loss of long range order, unless you talk about liquid crystalline, where orientational, but not positional long range order, is partially retained. In the gas phase, also short range order is lost.
So, in the liquid or gaseous state, phases can only differ by the molecular structure of the constituents. For water, that leaves only one stable possibility. Sulfur has a few different molecular modifications, but they're hard to separate and they tend to interconvert, so they're not really phases.
There are a few liquids with two phases which differ not in structure but in quantum states. Helium I and II are an example, para and ortho hydrogen another.

And binary (and higher) mixtures can exhibit miscibility gaps, although that isn't completely the same thing.


text:
If ionic compounds are strongly bonded, why are acids with a more polar central atom the stronger types of acids?
If the central atom is more polar, it pulls electron density away from the H+, making it more detachable. But why, then, are ionic compounds (which are even MORE polar than acidic covalent compounds) held tightly together?

The acidity trend goes to the right and down. There is the electronegativity effect and the size effect, but the size effect is more important!

Ionic compounds make a network of electrostatic attraction. Molecules have covalent intramolecular bonds which are very strong, but each molecular unit is weakly attracted to one another (relatively) due to lack of ionic attraction.  So, essentially, in the solid form, each $\ce{NaCl}$ has more attraction to another $\ce{NaCl}$ than does each molecule to another in a molecular system. That does not, however, translate to each $\ce{Na+}$ ion of an $\ce{NaCl}$ unit being more tightly bound to a $\ce{Cl-}$ ion than two atoms of a covalent bond are to each other.
The idea that ionic bonds are "stronger" than covalent bonds is confusing to early chemistry student (and maybe more so to chemists, but it's tough to convey concisely what is actually occuring) but, for some reason, seems to be nearly ubiquitous in general chemistry courses. It is much easier for compounds like water to stabilize $\ce{Na+}$ and $\ce{Cl-}$ than to dissociate something like paraffin wax into ions in order to solvate the molecules as ions.
So I guess what I'm trying to say is: the ions of ionic compounds aren't usually held more tightly together than two atoms of a covalent bond.
As for the idea that it is a higher electronegativity difference in $\ce{H-A}$ bonds that allow for higher acidity, that can be disproven by looking at $\ce{HF}$ vs $\ce{HI}$. $\ce{HF}$ has a higher electronegativity difference but is a weaker acid. This is because $\ce{I-}$ is larger, so it can spread the negative charge out, essentially stabilizing the negative charge on its own. $\ce{HF}$ produces $\ce{F-}$, and because fluorine is more electronegative than iodine, that may be expected to be a stronger acid than $\ce{HI}$. However, because $\ce{F-}$ is much more of a point-charge rather than a diffuse charge like $\ce{I-}$, it has to be stabilized by solvation effects to a larger extent.
So what causes an acid to be more acidic is the thermodynamic stability of the resultant conjugate base. Because the charge in $\ce{I-}$ is more stabilized by the ability to spread out its own charge, it more readily donates $\ce{H+}$ than does $\ce{HF}$.


text:
If d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block?
If I understand correctly, the shielding effect of d- (and f-) electrons seems to be much poorer than those of s- and p-electrons, due to the fact that they are less penetrating, have less electron density near the nucleus and so on (as can be seen from radial distribution functions).
However, trends for ionization energy, electron affinity, and electronegativity are MUCH more gradual than for s and p-blocks. This means that the effective nuclear charge ALSO increases much more gradually in the d-block, as confirmed by Slater and Clementi values. If d-electrons were poor shielders, surely this would mean that Z-eff would be much higher?
Doesn't this mean that, say for period 4, 3d-electrons are much more effective at shielding the 4s-electrons, than the 4s-electrons are at shielding 4p-electrons? Why does this happen?
Is there another factor at play? Is it because we are adding electrons to an "inner shell" that the shielding becomes more effective? Is Z-eff = Z - S an incorrect guideline to use?

Your question is similar to what I had in mind like a week back.
I know some textbooks do state a contradiction — that d-electrons have a poor screening effect, but then quickly use the explanation that due to increasing screening effect of nuclear charge from left to right in d-block elements (due to more electrons coming in the d-orbitals), trends are gradual.
Let me clear your mind of this false statement. Remember: it is not the screening effect that leads to a gradual change in trends across d-block elements, but  rather the repulsion between d-electrons and outermost s-electrons.
The follwing example from the 1st series of d-block elements should make this clear to you.
     As nuclear charge increases, electrons come in the 3d-orbitals, which is not the outermost. The outermost is the 4s orbital, already filled up with 2 electrons.
Now, from left to right in this period, the increased nuclear charge will tend to hold on to the outermost 4s electrons tighter, but the d-electrons have an opposite effect. They repel these 4s electrons, and hence net force felt by the 4s electrons decreases. So, atomic size is increases more than expected, and hence it is easier to remove these electrons too (less ionisation energy)
Hence, the increase in ionization energy, reduction is size, etc. is much more gradual while traversing from left to right in a period.

I don't get what you are trying to say. The screening effect arises precisely because of electron-electron repulsions, so "3d electrons repel 4s" and "3d electrons screen 4s" mean exactly the same thing.

orthocresol: no they dont. screening effect does arise out of electron repulsion, but that doesn't mean those two statements convey the same meaning :'D


text:
Inferring bond breakage in a reaction from InChI strings of reactans?
I'm currently confronting a problem:
I would need to enumerate all chemical bond breakages that occur during a given chemical reaction, including the type of bond and the atoms participating in the bond. This is straightforward to do by visual inspection if one has, say, the 2-D structures of the reactants and products of a given reaction.
My problem is that I have a list of several thousand reactions, and I need a computer to do the job for me. To this end, I got the InChI (and SMILES) representations of all molecules participating in each reaction. At first, I thought that it would be rather easy to write a program that identifies and enumerates the broken bonds in the reactants (given the InChI strings), but by now I'm starting to become desperate.
Perhaps someone knows any software able to do this?

There is the Reaction Decoder Tool (RDT on github) that can - from the README - do:

Reaction Annotator (Extract Bond Changes, Identify & Mark Reaction Centres)

EDIT: It potentially could have InChI support, but a quick glance at this file suggests that it currently doesn't...
In fact, that would require the CDK/RDT to implement the rInChI format
The input formats includes reaction smiles which should be possible to generate from your smiles strings. Looks like you just jam a ">" inbetween reactants and products with a "." separating the molecules in each set.
which sounds like what you want, I think. Also it should be able to handle those volumes as it has been developed to work with reaction databases at the EBI.
(Full disclosure : I worked on the project for a little while :) but it does have InChI support  - via the CDK).

Yeah, sounds promising! I'll take a look at that. Thanks a lot for the suggestion.


text:
Conversion of 4-nitrotoluene to 2-bromobenzoic acid

In the above conversion why doesn't bromine add to methyl group? Is the first reaction following free radical or polar reaction (Friedel-Crafts). If it's following free radical mechanism then Br should add to CH3, if it's F-C substitution then why there is no catalyst when NO2 is deactivating group?
What would be the change if I first do seconds step (i.e reduction to amine) and then bromination? Would it be wrong to do bromination at last?
Is the intermediate (2-bromo-4-nitrotoluene) the only major product of first step?

Too many questions in one! 1) You are right, ring substitution needs a Lewis acid catalyst. Without the catalyst nothing is going to happen. For free radical substitution of the methyl group you will need either UV light or a radical initiator. 2) and 3) You need to look up directing effects of substituents. 4) Yes.

At what temp or conditions was the bromine reacted in the first step? The methyl group may require a higher temp or specific conditions to be activated I'm only guessing tho , good luck.

@Swastik        okay no problem

I hope my answer will clarify your doubts @physics

Sorry but I would like to answer your questions from last to first,
(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place
(2) -COOH  is deactivating meta directing group (due to presence of  >C=O) so friedel-crafts bromination would take place at meta position.
(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2
(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced  one of the -H at ortho (to -CH3) of benzene.
(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)

preference of -CH3(activating) OVER -NO2(deactivating) will take place - I doubt this. $\ce{NO2}$ is a very strongly deactivating group whereas methyl is only weakly activating.

The directing effects of the methyl and nitro groups in p-nitrotoluene reinforce each other so that's not a problem in this specific case, but as bon said, the strongly deactivating nitro group is likely to exert a greater influence on the regioselectivity.

[for your reference}(http://www.utdallas.edu/~biewerm/17.pdf)  is the site which will answer your question.You can refer "Benzene and its derivative" chapter of any standard organic chemistry books like J.D. March or Paula Yukarinis or P.Bahadur  or O.P Tandonor Solomons(name of eminent chemistry book writer)   @ bon,orthocresol


text:
Why is the melting point of magnesium anomalous?
The melting and boiling points for the Group II elements are as follows:
$$\begin{array}{ccc}
\text{Element} & \text{Melting point / }\mathrm{^\circ C} \\
\hline
\ce{Be} & 1289 \\
\ce{Mg} & 650 \\
\ce{Ca} & 842 \\
\ce{Sr} & 769 \\
\ce{Ba} & 729
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., p 112)
As you can see, the overall trend is decreasing, but magnesium is an exception. Just like the alkali metals, melting point should decrease for the heavier elements of the group because of the weaker bonding. Why is magnesium different?
The other trends for this group, such as density, atomic radius, and first ionization energy follow the expected pattern with no exceptions.

I am guessing it might be to do with the structure of the metals. Be and Mg are hcp, Ca and Sr ccp, Ba and Ra bcc.

As a further comment, counting on trends in the periodic table will lead you in many wrong directions. Expect the unexpected.

Mg has it's special properties, but the m.p. of Be is the one that is actually out of line, as are many properties of the elements in the second period. Plus Be lies nicely on the diagonal in the periodic system, where some properties are metallic, and some not.


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Strontium is more common than sulfur?
I was really surpised to see that strontium is apparently more common than sulfur in the earth's crust, since it occurs in what I thought were pretty rare minerals.
Sulfur which seems to be everywhere such as in volcanoes, in the atmosphere as sulfur dioxide, in coal deposits, and is even found elementally in large lenses.
Sulfur seems to be ubiquitous. For example, fool's gold, iron pyrite is iron sulfide and it can be found in practically any stream bed. I would be hard pressed to find strontium in my local stream beds. And, of course, this sulfide is only one of many commonly sulfides.
Is strontium really more common than sulfur? It seems implausible. What is the basis for this claim?

Related question on Physics SE: http://physics.stackexchange.com/questions/1918/why-the-distribution-of-elements-on-earth . Also, the Earth Science SE might be a better place for this question. Note there is a large distinction between elemental distribution in the crust, and in other layers of the earth.

This section of a Wikipedia article on the abundance of the elements  has a nice graph of the mass abundance over the universe https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe and this section has a nice table of overall abundance on earth You can look at both mass abundance and abundance by number of atoms https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Earth.27s_detailed_bulk_.28total.29_elemental_abundance_in_table_form

The overall earth abundance of S is 6350ppm by weight, and for Sr 13 ppm.

It worth to distinguish abundance from "easy to find". Many elements can be found in enriched sources even if they are rare in average. E.g gold is a very rare element, yet it is relatively to mine as it is concentrated in given geological formations.

The primary reason for this and other seemingly implausible features of element distribution is this: some elements are concentrated, and some are scattered. Strontium is much like calcium (which is way more abundant) and can hide as a tiny admixture in its minerals, while sulfur is pretty much on its own.
Just think of the rare earth elements. Their name is in fact a misnomer; they are not rare at all, just very scattered and thus difficult to come by, hence the relatively late discovery. On the other hand, silver and gold, which are thousands times less abundant, were known since antiquity (or probably before that).
Also, you see yourself that there are several columns with the same name "Abundance in crust", and the data for the same element vary quite significantly between the columns. This is due to the fact that finding an element's content in the crust is not easy at all. Wait, don't they say explicitly that the numbers are but estimates?

I'm sure that you're exactly right. I was an analytical chemist specializing in x-ray fluorescence. All Ca containing compounds will have the other Group 2 elements as well. (Really a wonderful confirmation of utility of the periodic table.) So massive amounts of $\ce{CaCO3}$ in the crust means a fair amount of Sr in the crust as well.


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Enthalpy change defined at constant pressure only?
Is change in enthalpy defined only at constant pressure?
I know that $Q-W=\Delta U $, and $Q$ at constant pressure equals $\Delta H$ (enthalpy change).
Can $Q$ (heat given to the system) be used interchangeably with $\Delta H$ in the first equation?
Is $ \Delta H =nC_p\Delta T$? How?

Is change in enthalpy defined only at constant pressure?

No. The definition of $\Delta H$ is $\Delta H = \Delta U + \Delta (PV)$  Simple as that.

I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy
  change).
Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No.  Only if the applied pressure is held constant during the change.

Is del(H) =nCp del(T)? How?

No.  Only for an ideal gas.  For real gases, this equation is not correct outside the limit of ideal gas behavior.  For an ideal gas, $\Delta U=nCv\Delta T$ and $\Delta (PV)=nR\Delta T$, so $\Delta H=n(C_V+R)\Delta T$.  And then $C_p=(\partial H/\partial T)_P=C_v+R$, so $\Delta H=nC_P\Delta T$.
For a real gas, liquid, or solid,
$$dH=nC_PdT+\left[V-T\left(\frac{\partial V}{dT}\right)_P\right]dP$$
Note that the term in brackets is zero for an ideal gas.

Why we say that for chemical reactions at constant pressure $dH=dq$ at constant pressure? It is true that $$dH=Tds+Vdp+\sum_i μ_i dn_i$$ Only the middle term cancels out so $dH \neq dq$.

For a closed system at constant pressure with no mass entering or leaving, irrespective of whether there is chemical reaction occurring, the 1st law tells us the $Q=\Delta H$.  Good luck with integrating the equation that you have written, given that s is a function of the chemical composition also.

I thought about this some more, and I have an additional answer for you.  The equation you wrote describes the change in enthalpy between two closely neighboring (differentially separated) thermodynamic equilibrium states.  For a chemical reaction at constant temperature and pressure, the summation represents the change in Gibbs free energy, which is zero at equilibrium for all differential variations in the number of moles of the various species.  So, under these circumstances, you are left with dH=TdS.

how did you derive the terms in square brackets?

@Buraian This derivation is in every thermo book.

Is change in enthalpy defined only at constant pressure?

No, enthalpy-change is not meant only for isobar processes.

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No, not always. Only when the non-compression work done by the system is zero, then only $\delta Q$ be interchanged with $\mathrm dH\;.$ See my answer to this ChemSE post.

Is del(H) =nCp del(T)?

Yes, indeed.

How?

Check my answer to this ChemSE post.

As you said $dH=\delta H$ only when non-compression work done is zero. Does this mean that the formula $$dG_{T,P}=-TdS_{total}$$ is only valid when the non-compression work done is zero?


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Identifying a polyprotic acid using titration
I have the following question from a high school chemistry lab:

If an unknown acid has more than one equivalence point, is there a reason to expect a specific $K_\mathrm{a}$ ($K_{\mathrm{a1}}$, $K_{\mathrm{a2}}$ or possibly $K_{\mathrm{a3}}$) to be a better value to identify the acid?

I understand that if an acid has multiple equivalence points, it must have more than one hydrogen disassociating. As we titrate with a strong base, we will get an equivalence point for each hydrogen. Similarly, we will get a $K_\mathrm{a}$ value for each hydrogen, but why would one of these values be better than another for identifying the acid? Would we expect one to be more accurately measured than the other(s)?

No. For instance if Ka1 and Ka2 are close then the two dissociations are muddled and you don't get a good value for either. So if Ka3 is much different then it would be the better one to determine. If Ka2 and Ka3 are close, but Ka1 is much different than Ka2, then Ka1 would be the best. So it depends...

@MaxW What if there are only two equivalence points? Is there still an advantage of picking one over the other?

Yes. You pick whichever one works better. For $\ce{H2SO4}$ for example the first ionization is so favored that you'd never be able to see it in aqueous solution.


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Why is pyridinium chlorochromate selective compared to chromate, dichromate etc.?
One can use PCC (pyridinium chlorochromate) to oxidise an alcohol selectively up one level - to an aldehyde/ketone, without further oxidation to a carboxylic acid.
Why is pyridinium important to this selectivity?
I understand that usually PCC is dissolved in dichloromethane and not in anything like THF, and this prevents the formation of a hydrate, which can then act "like" an alcohol in it's oxidation to a carboxylic acid, but I can't figure out why it's important that pyridinium is used.

Though pyridinium chlrochromate (PCC) is a salt, the pyridnium cation is easily dissolved by a wide variety of organic solvents. As such, the chlorochromate anion then becomes dissolved in the organic solvent, and can oxidize $1º$ and $2º$ alcohols present in solution to carbonyls. Oxidation of carbonyls to carboxylic does not occur when using PCC for the very reason you said yourself—no water is present to hydrate the carbonyl species to its geminal diol to allow for further reaction.
Other benefits of PCC include that it is not particularly hygroscopic, is commercially available, and can be stored for some time, though other methods (Swern oxidaiton, Moffatt oxidation, Dess-Martin Periodinane) are generally preferred as the reaction workup for PCC can be tedious and the chromium(IV) species produced is toxic.


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Mechanism of the oxidation of alcohols with KMnO4
Many oxidising agents, like chromate, dichromate, iodine in $\ce{NaOH}$ etc. seem to work via ester formation and elimination. For example, chromic acid will react with the $\ce{-OH}$ of alcohol to form a chromate ester, and then this will undergo an E2-like elimination, with a proton abstracted from carbon and loss of leaving group from oxygen.
However it seems that manganate works via a different mechanism. I think this because here:
http://www.masterorganicchemistry.com/2015/05/21/demystifying-alcohol-oxidations/
It says in a footnote at the bottom:

"..main exception you’ll encounter is $\ce{KMnO4}$, which likely proceeds through a C-H abstraction/internal return type mechanism followed by collapse of the hydrate to give the new carbonyl."

I don't quite understand this brief summary of the mechanism, could someone expand on it? I don't quite understand what "internal return type mechanism" is refering to - I kind of understand it refers to something a little like the "intimate ion pair" in the $\mathrm{S_Ni}$ mechanism, but haven't been able to find online a more expanded mechanism - many places seem to show it like the case of chromic acid above, and so I am slightly confused as to what mechanism it really follows.

March 7th ed. (p 1436) seems to imply that it is a ester formation followed by elimination, the same as it is with CrO3/H2SO4. Unfortunately, I am not quite sure what the phrasing above is referring to.

@orthocresol I have seen ester formation followed by elimination for manganate, but the footnote seems to imply that manganate is somehow different in it's mechanism, but I couldn't understand what the phrasing of it meant either.

$\ce{KMnO4/OH-}$ Oxidation

$\ce{H2CrO4}$ Oxidation

The two main differences are

$\ce{H2CrO4}$ Oxidation occurs in acid, $\ce{KMnO4}$ oxidation occurs in base
The source of the nucleophile in the removal of the metal ester. In $\ce{KMnO4}$ oxidation, the $\ce{Mn}$ ester itself extracts the $\ce{H+}$ from the alcohol carbon, while in $\ce{H2CrO4}$ oxidation the nucleophile is the solvent.

Are you sure about your first difference? I am very sure manganate oxidation can take place in both acid and base, depending on how harsh you want the conditions to be.


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The equilibrium constant of a reaction only involving pure solid or liquid
How to calculate the equilibrium constant of a reaction only involving pure solid or liquid.
For example reaction
$$\ce{S(s) + Hg(l) -> HgS(s)}$$
or
$$\ce{S(s) + Mg(s) -> MgS(s)}$$

Do a quick search for "equilibrium constant solids", there are several questions which are of help to you.

Concentration of  pure solids and liquids is conventionally taken to be 1. Since reaction involves only solids and liquids the equilibrium constant will be 1


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Is the rate determining step the step with the largest Ea?
I've seen some controversy on this question while doing a brief search. For example, this SE answer quotes Wikipedia and says that the RDS is the step with the largest $E_a$. However, this UC Davis page says that

In the potential energy profile, the rate-determining step is the reaction step with the highest energy of transition state.

My textbooks don't explicitly mention whether the RDS is the step with the largest $E_a$ or the step with the highest energy transition state. When I was doing a past exam, one of the questions was on this topic, and the correct answer assumed that the RDS is the step with the highest energy transition state. However, I've always thought that the RDS is the slowest step, and the slowest step has the highest $E_a$. So which viewpoint is actually correct?

Anslyn & Dougherty p 361: "The step of the reaction whose rate determines the observed rate of product formation is called the rate determining step (rds). It is commonly the chemical step that involves the highest energy transition state." (emphasis mine)

The operative word in the Anslyn & Dougherty quote is commonly. Dogmatic statements in chemistry always bother me. More often than not there are exceptions.

But really the previous question that you linked has the answer in a comment. I don't have the time to write up an answer but the summary is that neither of your proposed approaches is complete. The reference is J. Chem. Educ., 1981, 58, 32 cc. @MaxW

@orthocresol I can't read the article you linked because it says I do not have a subscription to the publication. Can you post a summary of (or quote) what it says?

Here's the conclusion http://i.imgur.com/zuqtOHm.png (RLS = "rate-limiting step") Sorry I would really read it more carefully and write an answer if I had time but not now. Hopefully someone else can

Yes, the rate determining step is the largest energy difference between any starting material or intermediate on a potential energy diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction.
The transition state with highest absolute energy may not necessarily correspond to the rate determining step, because if it is transforming from another high energy state, then little additional energy is required to reach this (low $E_a$).

See the Anslyn & Dougherty quote pointed out orthocresol above. Is and is commonly are very different. There are no exceptions to is.

@MaxW I saw it, and I don't see any disagreement. "Commonly," "not necessarily"-- neither are absolute statements.

Also as a cool side note, Anslyn is currently my professor for organic chemistry. He is an immensely smart guy.

You said "Yes, the rate determining step is the largest energy difference..." I think it should be "Yes, the rate determining step is commonly the largest energy difference..."

That is how the RDS is determined though. It does not have to include the highest energy transition state if there is an intermediate following it with a higher following $E_a$. I think that is where it is important to qualify the statement with "commonly."

Aren't there reactions that are thermodynamically favorable but that don't happen because of a steric hindrance? So you use a catalyst to get the proper steric orientation, not to lower the activation energy.

I link to an article entitled "The Rate-Determining Step is Dead. Long Live the Rate-Determining State!" in which the authors provide examples of reaction mechanisms where the RDS is the transition state (TS) that has the highest free energy as well as a case where that breaks down (refer to Section 2.3). They also show an example of a case where both the step with the highest $E_{a}$ and TS with the highest free energy is not the RDS (refer to Section 4.1).

In addition, one thing I think that's missing from the discussion here is that, if the RDS is the "slowest" step, presumably that is in reference to the rate, not the rate constant. Since the rate of that step is a function of the rate constant as well as concentration terms, I would argue that it is not always wise to refer to a step as being slowest solely by looking at its $E_{a}$.

I couldn't fit this as a comment, so I'll post it as an "answer." This are just my ponderings.
For everybody, this is the problem: (it's not that clear, but the vertical axis is Gibbs Free Energy).

The correct answer according to the problem source is III-IV, even though the first and second transition states have higher $E_a$ s from their respective "valleys."
The most relevant equation here is $k = Ae^{-E_a/RT}$. This is a high school level competition, so this is purely a conceptual problem involving that equation.
As we all know, as the activation energy of an elementary step increases,  the rate of that step slows down as the fraction of molecules with the requisite kinetic energy decreases. ($e^{E_a/RT}$ ) is the fraction factor: the fraction of particles that have equal to or greater than $E_a$.) Fraction of what, though? I believe that's where the answer lies.
My reasoning in favor of III-IV this: even though $E_a$ for III-IV may not be as high as I-II or II-III, already only a small fraction of molecules out of the total even made it to state III, due to the high potential energy at state III. Thus, when we apply the fraction factor again to that fraction via the Arrhenius equation, the final fraction to of particles that have enough energy to progress from III-IV is indeed the smallest.
To put it mathematically (and this is a purely mathematical argument, not chemical), the fraction of the particles that have enough energy to progress from III-IV is the fraction of total particles that are at state III, multiplied by the fraction of those particles at state III that have the energy to go from there to IV. Multiplying the two fraction factors, adding the exponents, the net "minimum energy" is the sum of the energies involved in getting to III and the $E_a$ of the III-IV reaction, giving the "absolute energy" of the transition state between III and IV, which is the highest.

But for any reaction, the fraction of particles that pass through the last step is always the smallest since it needs to have passed through all the previous steps (and each step multiplies another "fraction factor" onto it), right? Also, if II-III has the largest $E_a$, that step would be slowest since a small fraction of the particles that reach II have enough energy to go to III. But of the particles that make it to III, a larger fraction of them have enough energy to go to IV, so III-IV progresses faster. So wouldn't the RDS be II-III since a reaction can't be faster than the slowest step?

@Yunfei Ma Does the energy difference of I to III-IV transition state matters? In order for reactants to go to products they must posses kinetic energy greater than the III-IV (with respect to I) transition state energy. So why we don't take into account this?

From the Wade textbook on kinetics and activation energy: the rate-determining step, is the step which leads to the transition state with the highest overall energy.
If there is an early step with a low transition state, the intermediate will accumulate (the forward and reverse reaction of the first step achieve equilibrium), and the overall reaction will depend on the later step with the higher transition state.


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Does all organic matter originate from organisms?
The organic matter in the atmosphere, in the ocean, and on the land is very diverse. Does all of this organic matter originate from organisms?

Your question is at the threshold of the greatest mystery in the universe. It sort of depends on how complex a molecule needs to be to be "organic." So is there a God that created organisms, or did organisms evolve from some natural process?

@MaxW No need for philosophy or theology in this question.

The type of molecules you are asking about are called natural products. The Earth did not coalesce from the debris of our solar system's previous sun with life on it, so clearly some conditions must have been present on prebiotic-Earth for organic molecules to form from other molecules. The Miller-Urey experiment actually demonstrates this in a very eloquent way by simulating the conditions of prebiotic-Earth. From Wikipedia:

The experiment used water ($\ce{H2O}$), methane ($\ce{CH4}$), ammonia ($\ce{NH3}$), and hydrogen ($\ce{H2}$). The chemicals were all sealed inside a sterile 5-liter glass flask connected to a $500\: \mathrm{mL}$ flask half-full of liquid water. The liquid water in the smaller flask was heated to induce evaporation, and the water vapour was allowed to enter the larger flask. Continuous electrical sparks were fired between the electrodes to simulate lightning in the water vapour and gaseous mixture, and then the simulated atmosphere was cooled again so that the water condensed and trickled into a U-shaped trap at the bottom of the apparatus.

This would have effectively simulated the conditions you would find on prebiotic Earth. Vast, boiling seas in an atmosphere of low molecular weight gases. Constant lightning storms in the atmosphere ionized these gases and acted provided the energy needed for chemical reactions to take place.

After a day, the solution collected at the trap had turned pink in colour. At the end of one week of continuous operation, the boiling flask was removed, and mercuric chloride was added to prevent microbial contamination. The reaction was stopped by adding barium hydroxide and sulfuric acid, and evaporated to remove impurities. Using paper chromatography, Miller identified five amino acids present in the solution: glycine, α-alanine and β-alanine were positively identified, while aspartic acid and α-aminobutyric acid (AABA) were less certain, due to the spots being faint.
In a 1996 interview, Stanley Miller recollected his lifelong experiments following his original work and stated: "Just turning on the spark in a basic pre-biotic experiment will yield 11 out of 20 amino acids."

Sure many of the more complicated natural products were produced by living organisms, but they all originate from the molecules formed in prebiotic Earth's atmosphere. Petroleum & natural gas products are only the decomposed molecules of ancient algae and zooplankton, after all.
Additionally, there is $\ce{CH4}$ in Earth's mantle has been there since Earth first formed, meaning it is totally abiogenic.


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Transmission, absorption, and reflection of light
My understanding is: when light hits an object, it can do one of three things: transmit through the object, be absorbed into the object, or reflect from the object.
However, I have also learned that when light hits atoms, it can excite electrons, leading to another emission of light as the electrons go back down to their ground states.
The first paragraph, while intuitive, does not seem to be logically explained by the second. Can someone please help me understand the first in terms of the concepts of the second?
Specifically, is absorption of light exciting an electron with more energy than necessary, and some energy remains in the atom, or is the entire energy of the photon absorbed?
Is reflection just hitting the atom with a photon that is just the right amount of energy to get it excited, so no net energy is gained?

Not quite right. Light can be transmitted, absorbed or scattered. Reflection is a particular form of scattering. https://en.wikipedia.org/wiki/Scattering

While I can't explain the phenomenon of light transmission, I can say a few things. The energy levels correlating to physical states and the gaps between the states are quantized. This an all or nothing reaction. The frequency of light must exactly match the energy (E=hv) needed for the excitation event (electron is pushed into a higher orbital, a vibrational mode to expressed, etc) for absorption to occur. When the light frequency is mismatched and does not absorb, it will reflect. The frequency that induces excitation is theoretically the frequency of emission, but phenomena such as intersystem crossing may create multiple paths by which the excitation energy will disperse. Hope this helped.

If light is neither absorbed nor scattered, it is transmitted through the substance. This occurs when the energy from the light is transferred into the substance, and the vibrations of the electrons are passed on through the substance and re-emitted on the other side. Different arrangements of atoms in substances can cause these vibrations to propagate differently through different substances. This is the cause of Snell's law.
Different substances absorb specific wavelengths of light. This is determined by the differences in energies of the different atomic or molecular orbitals. The energy of the photons promote electrons from the ground state to an excited state, and is then dissipated as vibrational energy as the electron returns to the ground state. All other light that does not match these energy differences are cause electrons to vibrate on a much smaller scale for a short amount of time before re-emitting the light, thereby scattering it.
White light consists of all the wavelengths of the visible spectrum. When all wavelengths are scattered by a substance, it appears white. When all visible light is absorbed, it appears black. The dissipation of all this energy absorbed by black substances explains why black cars get hotter in the sun than white cars.

When only some wavelengths of light are absorbed, the resulting scattered wavelengths constitute what our eyes perceive as color.
It's also important to note that visible light isn't the only part of the electromagnetic spectrum that can interact with matter. The spectrum ranges from high frequency gamma rays and x-rays to low frequency microwaves and radio waves. If you were wondering if there was anything particularly special about visible light, check out this physics post!

Hmm, the way I learned it was that light is emitted when an electron falls from an excited state to the ground state. However, you are saying that if a photon exactly matches the energy difference necessary to excite an electron, the light will instead be absorbed. Is this correct? Also, What exactly happens in the process of scattering? Does scattering mean that the original photon hitting the object never excites an electron in the first place?


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Predicting dissociation energy of diatomic iodine computationally
In one of my classes we used ORCA with three different functionals: BP86, B3LYP, and M06-L, all with the basis set def2-SVP. In addition to calculations we did it experimentally by a Deslandres table and a Birge-Sponer plot. The experimental values for the most part match the values given in NIST, however for the three functionals my $D_{0}^{"}$ value is twice as big as the value given in NIST, and I don't know why. Thanks!

Welcome to chemistry.se! Feel free to take the [tour] to learn more about this site. In order to be actually able to help you, it would be good to if you could [edit] the values you have obtained/ researched into the question. Please also share what your thoughts are on the matter.

Are you making a distinction between $D_{0}$ and $D_{0}^{"}$?


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Benzenesulfonyl chloride dehydration of an amide to give nitrile
1) I am unable to understand how benzenesulfonyl chloride acts as a dehydrating agent in dehydration of amides? What is the driving force for losing chloride and taking the oxygen from the amide?
2) Why I can't use other dehydrating agents like $\ce{P4O10}$, $\ce{H2SO4}$, or $\ce{CaCl2}$?
3) How can I estimate what is suitable dehydrating agent for a particular reactant?
4) Is it necessary to use only those dehydrating agents that undergo chemical change? Or can I use dehydrating agents that adsorb water like silica gel?

Similar http://chemistry.stackexchange.com/questions/33742/what-makes-a-compound-a-dehydrating-agent

One key detail is that this reaction is performed in the presence of base. Typically pyridine is the solvent. Amides are nucleophilic at oxygen, explained by the resonance structure shown. Displacement of chloride by the amide oxygen gives the O-sulfonylated intermediate. After neutralization, the O-sulfonylimidate is set up for an E2 elimination across the C-N bond to give the nitrile.

In addition to sulfonyl chlorides, $\ce{P4O10}$ and $\ce{POCl3}$ are common reagents for this reaction. In general, the reagent should be electrophilic and be able to act as a leaving group.

The sulphur of the sulphonyl group is pretty electrophilic due to the two oxygens and one chlorine atom attached to it. So the nitrogen oxygen from the amide attacks on this sulphur atom, displacing the chlorine atom on it. The $\ce{Cl-}$ ion takes a proton from the amide with it and forms $\ce{HCl}$. The nitrogen then pushes its lone pair towards the former carbonyl carbon, causing the benzenesulphonyl group along with the former carbonyl carbon to take off, leaving a triple bond between the nitrogen and the carbon. The benzenesulphonate anion then picks up the proton left on the nitrogen, giving the products $\ce{R-C#N}$ and $\ce{Ph-SO3H}$.
Generally $\ce{P4O10}$ is a very strong dehydrating agent, so must be used where dehydration is difficult, or more dehydration than done by normal dehydrating agents is required. $\ce{H2SO4}$ is a good dehydrating agent, but probably isn't good enough for this particular reaction. It is difficult to predict which dehydrating agent must be used for any arbitrary reactant (atleast to my knowledge).
$\ce{CaCl2}$ or silica gel are physical dehydrating agents (more accurately drying agents) and cannot be used for chemical dehydration reaction like the above reaction.
Edit: My original mechanism was shown to be wrong by @jerepierre and @orthocresol, and hence i have corrected it above.

Do you have a reference for the proposed dehydration? I doubt that mech is correct.

I don't have a reference :/ But your mechanism does seem to make more sense...but I am not able to find concrete reference for the oxygen acting as a nucleophile, and not nitrogen. That is the part where I went off on a wrong track, otherwise your mechanism easily follows.

@jerepierre It seemed intuitive to me initially that the nitrogen would act as a nucleophile....that is the reason for me proposing the above mechanism.

The most nucleophilic part of an amide (aka the HOMO) is the carbonyl oxygen. The nitrogen lone pair is tied up by delocalisation into the C=O π*.

@orthocresol Okay. Thank you for correcting me! I have corrected the mechanism in my answer now!


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Benzene and proton in 1H NMR
The chemical shift in benzene is around $\pu{7.27 ppm}$, which is lower than, for example, aldehydes. However, the magnetic field due to the pi electrons are in line with the applied magnetic field $B_0$, resulting in $B_{eff}>B_0$. Whereas in aldehydes, the hydrogen is simply "less shielded". So shouldn't the $\ce H$ atoms in benzene (or alkynes etc.) have even greater chemical shifts than species that are deshielded "normally" through electron withdrawing substituents?
Also, somewhat of a side question: What kind of chemical shift would we see for a proton ($\ce{H+}$) in a proton NMR, since it is completely deshielded? Could this theoretically be measured?
(Following the previous logic, shouldn't benzene have a higher chemical shift than a naked proton?)

The aldehyde is both deshielded by reduction in electron density about the carbon it is bonded to through electron withdrawal and by the magnetic anisotropy of the double bond. Although a benzylic hydrogen is slightly more deshielded than the hydrogen attached to a two electron pi bond like a carbonyl or an alkene, it does not also have electron density withdrawn from its adjacent carbon. Generally electron withdrawal through resonance is more important than inductive electron withdrawal, so the difference in chemical shift between an aldehyde and an alkene will be greater than the difference in chemical shift between the proton proton geminal to a halogen in  a halogen-substituted alkene and the proton in its unsubstituted analogue. All that is to say the additive effects of resonance electron withdrawal and magnetic anisotropy make an aldehyde proton more shifted than a benzylic proton.


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Comparing strengths of Hydrogen Bonds
What kind of a bond is the Hydrogen bond? Is it electrostatic forces of attraction or more like a partial coordinate bond?
What pair of atoms would provide us the strongest Hydrogen bond? If we consider electrostatic forces then HF and HF should be strongest. But how about HF and NH2?

http://chemistry.stackexchange.com/questions/35488/is-hydrogen-bonding-a-type-of-dipole-dipole-interaction and https://en.wikipedia.org/wiki/Bifluoride


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Why is DBU the most popular base with an amidino group? Does the basic strength have something to do with the ring size being 7 for one of the rings?

Structure of DBU
DBU is used as a very strong base(which is due to resonance stabilization of amidinium cation) with steric bulk. Is any other compound used instead? If not, why not?

DBN is used, too.

Caprolactam is $38/kg from Aldrich. Acrylonitrile is $55 for two liters. My guess: it's cheaper to make than similar ring sizes. I know valerolactam is significantly more expensive than $38/kg.

What does the price of lactams have to do with price of amidines ?

@gluco Caprolactam + acrylonitrile = starting materials for DBU. Imagine caprolactam with a nitrogen in place of the oxygen and three more carbons, then cyclized.

DBU is a popular amidine base because it can be deemed to be of lower nucleophilicity than cheaper amidines e.g. formamidine.  It has been sometimes erroneously labelled as "non-nucleophilic" but seeing as it reacts with carbon dioxide and sulfonyl halides and anhydrides,  this cannot be supported. As has been pointed out, the price of bulk chemicals such as caprolactam (made from cyclohexanone which in turn comes from phenol) and acrylonitrile makes DBU accessible.
The 7-member ring structure isn't special per se but assists with steric bulk to reduce nucleophilicity of the free base and increase the nucleophilicity of the counter anion when acting as a (large) amidinium ion. This occurs through a reduced association of the ion pair.
For cost weighed up against basicity,  tetramethylguanidine is better value as an organic base if the extra strength is required.

I actually just saw an article that suggested that DBU isn't even of low-nucleophilicity, per se, but rather that it just dissociates much in the way pyridine will.

I was referred to an article about 16 years ago that allegedly showed that where DBU reacted with carbon dioxide, DBN did not. I can't place the article unfortunately.