text:
Why is the maximum adiabatic temperature found for an equivalence ratio a bit higher than 1?
Why does the peak adiabatic temperature usually happen for an equivalence ratio a bit higher than 1? I think it had something to do with the probability of each fuel molecule finding a corresponding oxidizer, and the fact that the molecular weight of the fuel is generally higher than that of the oxidizer.

Just a guess, but is it because nitrogen in the air is inert, and you have to "waste" a bit of the heat from combustion to heat it up?  It's a small effect, relative to the heat of combustion, so my guess is the the max temperature would be ever-so-slightly to the carbon-rich side of the equivalence ratio.

[...] $\phi = 1.1$ was chosen because the flame temperature for many fuel peaks not at the stoichiometric value but between $\phi = 1.0$ and $\phi = 1.1$ owing to lower mean specific heats of the richer products. The maximum temperature for acetylene-air peaks, for example, at a value of $\phi = 1.3$.$^{[1]}$ ## References 1. Irwin Glassman, Richard A. Yetter, Combustion, 4th ed; Academic Press: New York, 2008

I actually found the answer in this 2011 paper, here is a quote from the abstract:

The characteristic rich shifting of the maximum adiabatic flame temperature from the stoichiometric value for
  hydrocarbon/air mixtures is demonstrated to be caused by product dissociation and hence reduced amount of heat release.
  Since the extent of dissociation is greater on the lean side as a result of the stoichiometry of dissociated products, the peaking
  occurs on the rich side.


text:
The appliance of LAB in amide reduction
Lithium aminoborohydride is a versatile reducing agent. In the scheme below, I don't understand how does unhindered LAB afford an alcohol in the reduction of amides while a hindered LAB affords an amine. Why are two different products obtained and by what mechanisms do these reductions take place?

Very interesting question and not very obvious. As can be seen from the picture, LAB's reduce only tertiary amides. It is stated in the publication(1) about LAB reductions:

LAB reagents do not
  reduce primary and secondary amides even in refluxing THF.

In the same publication it is provided reason for different outcomes of reductions. They say that:

We found that the outcome of these reductions depends on
  the sterics of the amide as well as of the LAB reagent

Which means that if LAB and/or amide is sterically hindered it can't interact with nitrogen like in compound 5
In lower part of the picture is described reaction where more hindered LAB leads to amine product 9

(1)Pasumansky, L., Goralski, C. T., & Singaram, B. (2006). Lithium aminoborohydrides: powerful, selective, air-stable reducing agents. Organic process research & development, 10(5), 959-970.

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text:
Converting mol files to InChiKey in large batches
I want to convert mol files to InChiKey. I am using OpenBabel for the conversion but I have over 500 mol files and converting them one by one is too time consuming. Is there some way to convert them all at once?
Edit 1 This is a sample mol file
0187.cdx
  Sample

  8  8  0  0  0  0  0  0  0  0999 V2000
   -0.4125    0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.4125   -0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.4125   -0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
    0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.4125    0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
   -1.6500    0.0000    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0
    1.6500    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0
  2  3  1  0
  3  4  2  0
  4  5  1  0
  5  6  2  0
  6  1  1  0
  2  7  1  0
  5  8  1  0
M  END

Also I want to know is there some other format which can combine mol files into one.I think I had heard of another format like mol which can combine many mol files,but I cannot remember the name of that format
Edit 2-  What is SDF format? Can it combine many mol files into one?

Could you include the current scheme you are using? What operating system are you using? I am pretty sure there is a batch way to do things like that. (A small sample file would also be appreciated.)

@Martin-マーチン I am using V2000 and OS is windows 8.I have editted my question.

Are you converting the files in a command line or do you use a gui program?

You can store several molecules in one mol file, but I believe you have to change to the newer standard.

@Martin-マーチン I am using GUI of openbabel

@Martin-マーチン What is  SDF format.Wikipedia says it can wraps many mol files.Can it be used in my case?

I have no knowledge about mol files in general, not much knowledge about InChI other than that I read the FAQ. I cannot help you on this case, but I can try to make it easier for others to answer it for you. You can [edit] in all of the comments above, since they provide context to the question.

@LifeIsGood An SDF file is really just a bunch of MOL files one after the other. Doesn't the openbabel gui allow you to multi-select files? The alternative might be to open notepad and a command prompt and learn how to write batch files :)

If you have any sort of UNIX shell installed (such as Cygwin), I can give you a shell script that will do the trick (i.e., loop through your files and convert them) - is this an option for you?

Also, see this answer to a previous question : http://chemistry.stackexchange.com/a/9255/4665

@ToddMinehardt Ok I can install Cygwin

@gilleain I tried to write 2 mol files just after another but open babel converts just the first mol file in it.In SDF after M END of first mol file,should I start the next mol file immediately from the next line?

@LifeIsGood Sorry, I should have been clearer - there is a little more to the SDF format than just a series of mol files (there are spacer records IIRC). Probably best to do this with Cygwin and Todd's script.

@gilleain No,it worked for me.I added $$$$ after M  END and then immediately after I started the next mol files.It worked.Thanks a lot

@ToddMinehardt I figured out a way.Thanks anyway

Good to hear. Just in case you or others want a handy one-liner:  for i in $(ls *.mol); do name=${i%.mol}; babel -imol $i -oinchi $name.inchi; done

@Todd Since the question is not worded as specific to an operating system, I would encourage you to post the one-liner as an answer, with the system you tested it on. If bash, for i in *.mol; do ... should also work ;)

This is exactly why I wrote Open Babel.
obabel *.mol -oinchikey -m
As long as you're not exceeding the number of command line arguments, you're done.

I should point out that since babel / obabel will process all the filenames internally, this is faster than approaches using the shell. If you have too many files in a  directory (e.g., me), you want something like find . -name "*.mol" -print0 | xargs -0 -I % obabel % -oinchi -m

On a system where you're working in the bash shell, in your working directory where the .mol files reside, run this on the command line:
for i in $(ls *.mol); do name=${i%.mol}; babel -imol $i -oinchi $name.inchi; done
I have successfully tested this on machines running Ubuntu 12.04 and 14.04.

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text:
Predicting reaction among compounds
Suppose I have a group of organic compounds say A, B, C, D, E, F, G, H. I know their chemical formula and their names. Now there exists a reaction among these compounds but we do not know which compounds would react to give which products. It may be that A and B react to give C and D or B and D react to give F and H. If we are given these compounds, is it possible to identify a reaction that will occur among any number of these compounds to give the product which would belong to this set of compounds?
Note: If you have not understood any part of this question, please ask me before down-voting, I will edit the question and try to re-explain. Thank you. ALso I am a computer science student, so I dont have too deep knowledge in chemistry, so do help me.

So$\ldots$ your question boils down to how do we know if a reaction takes place?? Unfortunately, that's too broad.

Yes it is possible if you have a wide knowledge of organic reactions. Unless you have a question about any specific set of compounds or reactions then I have to agree with @M.A.Ramezani that this is far too broad.

Like i am talking about organic reactions occurring in the human body, Help to narrow it down, I have a very crude Idea, I don't know much about chemistry being a computer science student

At this very second there are millions of complex reactions happening in your cells. Simply saying organic reactions in body doesn't narrow it down. You can try adding what functional groups are going to be in those organic compounds.

@Rishika Reactions in the human body is still a huge field and many are highly complex due to the heavy involvement of enzymes.

okay okay, so if i know about the functional groups present in the compounds, then if these two functional groups are compatible , we can say that a reaction will occur?

It's$\ldots$ a bit more complicated than that. You can't completely decide if they'll react unless you exactly know what functional groups are and have seen a similar mechanism. Other than that, there are thermodynamics' or kinetics' considerations to be taken into account before fully answering it.

Can you please give me a link of such a functional group related mechanism?? and also the thermodynamic property link for predicting reaction??

I thought that Rishika's idea of limiting it to reactions that occur within the body was a good idea. For one thing, we know that they occur in aqueous solution at 37 C. Furthermore, I think the total number of organic reactions that occur in the body is a relatively small and well understood subset of all chemical reactions.

Please reopen the question. This is not a homework question - the person asking it is curious and a serious participant. There may be others out there that would offer answers. Comments to the effect that the chemistry is "too complex" etc. are not helpful and sound pedantic.

Maybe too broad, but intersting question. There is actually reserch on the very topic (see Bartusz Grzybowsky and http://chematica.net/, or hios paper http://pubs.rsc.org/en/Content/ArticleLanding/2012/SC/c2sc00011c#!divAbstract).

@iad22agp The question was not closed as a homework question. It was closed because it is too broad - and I agree with that. It has been flagged to be reopened and we will see how the community will decide. But I guess narrowing it down to a few functional groups and then continue this asking a few more questions is advisable. Right now it sounds to me like: Please explain organic chemistry to me.

I think your question relates to the field of combinatorial chemistry. This is an important area especially for pharmaceutical companies that want to try out lots of different structures to see if they have therapeutic potential.
Lets talk specifics. In the case of known chemistry - e.g. formation of esters from alcohols and carboxylic acids, you can identify the possible reactions. Start with a system where A, B, C, and D are different alcohols and E, F, G, and H are different carboxylic acids. Under the right conditions (acid catalyzed dehydration), the alcohols can combine with the acids all possible ways to give 16 different esters (AE, AF, AG, ... etc.). But A (an alcohol) will not combine with the other alcohols B, C, or D -- and acid E will not combine with acids F, G, or H. Under more harsh conditions, however, they could be induced to react within each group (to form ethers or acid anhydrides).
If you are talking about a random selection of compounds - especially ones with more complex structures and more functional groups, and a variety of reaction conditions, the outcome would be much less predictable.

So like you said , carboxylic acids and alcohols react, are there other such pairs that you know of? Is there a link that you can provide then It will be easy for me to understand..

@Rishika there are literally thousands of reactions that we know of, involving both organic and inorganic chemicals. I think your best bet would be to consult an introductory textbook. However, I must caution against memorising reactions as a way of learning chemistry. It is boring, difficult, and unproductive. You might want to try reading Clayden et al., Organic Chemistry.

@Rishika Or if you are only interested in the human body, find a biochemistry textbook. Most biochem textbooks will discuss some basic chemistry in the introductory chapters that will hopefully allow you to understand what is going on when they introduce the complex molecules that make up our human body.

@Rishika I think that it is difficult to just say "we have two compounds A and B - they are likely to react in the human body". The cell has lots of machinery (such as membranes and enzymes) to make certain reactions possible and others impossible, and this is a necessary result of evolution since if we indiscriminately allowed chemicals in our body to react, many of the products would be toxic to us. Therefore biochem does not focus on "what could possibly happen between all the chemicals in the body", it focuses on "what actually does happen between them".

So basically, the what can happen question is not valid, only what happens is valid

The KEGG Reaction Database, http://www.kegg.jp/kegg/reaction, is an effort to catalog all the reactions involved in known biochemical pathways - including all known enzyme-catalyzed reactions. It includes reaction classes (for example ester formation) as well as a facility for predicting reaction pathways given a reactant and optional product. I think what can happen is important, for example if a new drug is introduced into the body - we would like to predict what might happen even if we have not yet observed this process in real life.

So for such a drug based reaction, how would one find out? Do it experimentally?? And no theoretic way to find out?

Systems biology seeks to predict just what you are talking about, computationally, without having to go into the lab. If the predictions are good, you could screen a large number of drug candidates and narrow them down to a promising few. But in the end, you can't do real chemistry in a computer and you have to make the stuff and test it. That is why it costs $2 billion per new drug application to the FDA.


text:
Is density functional theory an ab initio method?
The following comment by Wildcat made me think about whether density functional theory (DFT) can be considered an ab initio method.

@Martin-マーチン, this is sort of nitpicking, but DFT (where the last "T" comes from "Theory") can be considered as an ab-initio method since the theory itself is built from the first principles. The problem with the theory is that the exact functional is unknown, and as a result, in practice we do DFA calculations ("A" from "Approximation") with some approximate functional. It it DFA which is not an ab-initio method then, not DFT. :)

I always thought that ab initio refers to wave function based methods only. In principle the wave function is not necessary for the basis of DFT, but it was later introduced by Kohn and Sham for practical reasons.
The IUPAC goldbook offers a definition of ab initio quantum mechanical methods:

ab initio quantum mechanical methods
Synonym: non-empirical quantum mechanical methods
  Methods of quantum mechanical calculations independent of any experiment other than the determination of fundamental constants. The methods are based on the use of the full Schroedinger equation to treat all the electrons of a chemical system. In practice, approximations are necessary to restrict the complexity of the electronic wavefunction and to make its calculation possible.

According to this, most density functional approximations (DFA) cannot be termed ab initio since almost all involve some empirical parameters and/or fitting. DFT on the other hand is independent of any of this. What I have my problems with is the second sentence. It states, that treatment of all electrons is necessary. This is technically not the case for DFT, because here only the electron density is treated. All electrons and the wavefunction are implicitly treated.
An earlier definition of ab initio can be found in Leland C. Allen and Arnold M. Karo, Rev. Mod. Phys., 1960, 32, 275.

By ab initio we imply: First, consideration of all the electrons simultaneously. Second, use of the exact nonrelativistic Hamiltonian (with fixed nuclei),
  $$\mathcal{H} = -\frac12\sum_i{\nabla_i}^2
                - \sum_{i,a}\frac{Z_a}{\mathbf{r}_{ia}}
                + \sum_{i>j}\frac{1}{\mathbf{r}_{ij}}
                + \sum_{a,b}\frac{Z_aZ_b}{\mathbf{r}_{ab}}$$
  the indices $i$, $j$ and $a$, $b$ refer, respectively, to the electrons and to the nuclei with nuclear charges $Z_a$, and $Z_b$. Third, an effort should have been made to evaluate all integrals rigorously. Thus, calculations are omitted in which the Mulliken integral approximations or electrostatic models have been used exclusively. These approximate schemes are valuable for many purposes, but present experience indicates that they are not sufficiently accurate to give consistent results in ab initio work.

This definition obviously does not include DFT, but this is probably due to the fact it was published before the Hohenberg-Kohn theorems. But in general this definition is still largely the same as in the goldbook.
Another point which confuses me are titles like:

"Potential Energy Surfaces of the Gas-Phase SN2 Reactions $\ce{X- + CH3X ~$=$~ XCH3 + X-}$ $\ce{(X ~$=$~ F, Cl, Br, I)}$: A Comparative Study by Density Functional Theory and ab Initio Methods"
Liqun Deng , Vicenc Branchadell , Tom Ziegler, J. Am. Chem. Soc., 1994, 116 (23), 10645–10656.

And then again we have titles like:

"Ab Initio Density Functional Theory Study of the Structure and Vibrational Spectra of Cyclohexanone and its Isotopomers"
F. J. Devlin and P. J. Stephens, J. Phys. Chem. A, 1999, 103 (4), 527–538.

Unfortunately Koch and Holthausen, who wrote the probably most concise book on DFT, A Chemist's Guide to Density Functional Theory, never really refer to DFT as ab initio or clearly draw the line. The closest they come is on page 18:

In the context of traditional wave function based ab initio quantum chemistry a large variety of computational schemes to deal with the electron correlation problem has been devised during the years. Since we will meet some of these techniques in our forthcoming discussion on the applicability of density functional theory as compared to these conventional techniques, we now briefly mention (but do not explain) the most popular ones.

But that does not really answer my question. Throughout the book they use the term only in the form of conventional ab initio theory or in combination of explicitly stating wave function and variations thereof.
In my quite extensive research about DFT selection criteria I never came about the term 'ab initio DFT'.
So the question remains:
Is density functional theory an ab initio method?

First note that the acronym DFA I used in my comment originates from Axel D. Becke paper on 50 year anniversary of DFT in chemistry:

Let us introduce the acronym DFA at this point for “density-functional
  approximation.” If you attend DFT meetings, you will know that Mel
  Levy often needs to remind us that DFT is exact. The failures we
  report at meetings and in papers are not failures of DFT, but failures
  of DFAs. Axel D. Becke, J. Chem. Phys., 2014, 140, 18A301.

So, there are in fact two questions which must be addressed: "Is DFT ab initio?" and "Is DFA ab initio?" And in both cases the answer depend on the actual way ab initio is defined.

If by ab initio one means a wave function based method that do not make any further approximations than HF and do not use any empirically fitted parameters, then clearly neither DFT nor DFA are ab initio methods since there is no wave function out there.
But if by ab initio one means a method developed "from first principles", i.e. on the basis of a physical theory only without any additional input, then

DFT is ab initio;
DFA might or might not be ab initio (depending on the actual functional used).

Note that the usual scientific meaning of ab initio is in fact the second one; it just happened historically that in quantum chemistry the term ab initio was originally attached exclusively to Hartree–Fock based (i.e. wave function based) methods and then stuck with them. But the main point was to distinguish methods that are based solely on theory (termed "ab initio") and those that uses some empirically fitted parameters to simplify the treatment (termed "semi-empirical"). But this distinction was done before DFT even appeared.
So, the demarcation line between ab initio and not ab initio was drawn before DFT entered the scene, so that non-wave-function-based methods were not even considered. Consequently, there is no sense to question "Is DFT/DFA ab initio?" with this definition of ab initio historically limited to wave-function-based methods only. Today I think it is better to use the term ab initio in quantum chemistry in its more usual and more general scientific sense rather then continue to give it some special meaning which it happens to have just for historical reasons.
And if we stick to the second definition of ab initio then, as I already said, DFT is ab initio since nothing is used to formulate it except for the same physical theory used to formulate HF and post-HF methods (quantum mechanics). DFT is developed from the quantum mechanical description without any additional input: basically, DFT just reformulates the conventional quantum mechanical wave function description of a many-electron system in terms of the electron density.
But the situation with DFA is indeed a bit more involved. From the same viewpoint a DFA method with a functional which uses some experimental data in its construction is not ab initio. So, yes, DFA with B3LYP would not qualify as ab initio, since its parameters were fitted to a set of some experimentally measure quantities. However, a DFA method with a functional which does not involve any experimental data (except the values of fundamental constants) can be considered as ab initio method. Say, a DFA using some LDA functional constructed from a homogeneous electron gas model, is ab initio. It is by no means an exact method since it is based on a physically very crude approximation, but so does HF from the family of the wave function based methods. And if the later is considered to be ab initio despite the crudeness of the underlying approximation, why can't the former be also considered ab initio?

I read and referred to that paper, too. I love that quote. I agree with you. (I still leave the accepting part for next week, to encourage more people to vote.) Would you say that the IUPAC definition should be updated, i.e. include the electron density explicitly in the last sentence?

@Martin-マーチン, it depends on how do you interpret this second sentence. In fact, I do not see any problem here with respect to DFT being ab initio method in accordance with this definition. Do we use the Schrödinger equation to treat the electrons? Yes, we do; rather indirectly, but we use it. We don't solve the Schrödinger equation, but we use it a starting point in the development of DFT: the key constituent, electron density, is defined in terms of the solution of the Schrödinger equation.

@Martin-マーチン, now, for the second part of this sentence that insists on treating "all the electrons", I again see no problem with DFT. We indeed treat all the electrons. Yes, we do so in a tricky way: throughout all the process we use only one-electron density without constructing any many-electron entity that describes our many-electron system as a whole (not like in HF where we construct the many-electron wave function out of one-electron function). But we do so because we already proved that many-electron systems can be treated in such a way: one-electron density is enough.

Yes, I had no problem with the second one, since HK clarifies that quite nicely. I was referring to the last sentence with limitations, considering that for example B3LYP is not ab initio.

@Martin-マーチン, got it. So, let us say that DFT is ab initio. Now the situation with DFA is indeed a bit more involved. From the same viewpoint a DFA method with a functional which uses some experimental data in its construction is not ab initio. So, yes, DFA with B3LYP would not qualify as ab initio, since its parameters were fitted to a set of some experimentally measure quantities.

@Martin-マーチン, however, a DFA method with a functional which does not involve any experimental data (except the values of fundamental constants) can be considered as ab initio method. Say, a DFA using some LDA functional constructed from a homogeneous electron gas model, is ab initio. It is by no means an exact method since it is based on a physically very crude approximation, but so does HF from the family of the wave function based methods. And if the later is considered to be ab initio despite the crudeness of the underlying approximation, why can't the former be also considered ab initio?

@Wildcat "basically, DFT just reformulates the conventional quantum mechanical wave function description of a many-electron system in terms of the electron density" is this really true in a sense that DFT was concieved by actually reformulating wavefunction into something else or is dft coming from just completely other angle?

The convention used by many is that ab initio refers solely to wave-function based methods of various sorts and that first principles refers to either wave-function or DFT methods with little approximation.
I can't find a citation at the moment, but I know this convention is fairly widely used in, e.g., J. Phys. Chem. journals.
The IUPAC gold book doesn't have "first principles," but Google Scholar gives over 224,000 hits for "first principles DFT".

In physics, "from first principles" is a synonym for "ab initio"; just an English equivalent of the Latin phrase. In quantum chemistry we habitually avoid using "ab initio" term with DFT since it still can potentially cause some needless terminological battles due to historical strict meaning of "ab initio" term.

But you're perfectly right, of course. In QC the convention is to avoid calling DFT methods "ab initio" and exclusively use "from first principles" term for them, while both terms can be used for wf-based methods.


text:
Are bond orders always the same in isoelectronic diatomic species?
In my textbook of chemistry it is written that:

In the Lewis description of covalent bond, the Bond order is given by the number of bonds between the two atoms  of a molecule.
   ...
  Isoelectronic molecules and ions have identical bond orders; for example, $\ce{F2}$ and $\ce{O2^{2-}}$ have bond order $1$. $\ce{N2}$, $\ce{CO}$ and $\ce{NO+}$ have bond order $3$.

I tried some more molecules and it turns out that for two diatomic isoelectronic molecules (neutral or charged) the bond orders are always the same.
My question is that :

If the bond order is always same for isoelectronic species (diatomic  molecules and ions) how can I prove it? I tried and got the following result:

For any diatomic molecule having combination of elements of same period( either period 2 or period 3) isoelectronic molecules and ions have the same bond order.

Proof: Assume the two atoms $A$ and $B$ ( atoms may be ions also) having valence electrons in their outer shell $x$  and $y$.
  Now if they combine to form diatomic molecule or ion having bond order $m$ then they must satisfy the below two equation( octet rule):
  1) $x+m=8$
  2) $y+m=8$
Now since they are along same period therefore the total no. of inner electrons( total electrons excluding valence electrons) must be equal( 2 electrons for period 2 elements and 10 electrons for period 3 elements). If we vary the atoms(or ion) $A$ and $B$ the sum of these inner electrons will obviously be constant( equal to 4 electrons for period 2 elements and 20 electrons for period 3 elements).
Therefore, For species to be isoelectronic:
  $x+y=constant$( say $k$)
  Adding equation 1) and 2) we get:
  $x+y+2m=16$ or
  $k+2m=16$ or
  *Bond order*$=m=(16-k)/2$ which is constant.  Q.E.D.

If the result given in my textbook is not true (since it looks like the fact is universal! compare $\ce{HCl}$ and $\ce{F2}$ which are isoelectronic having same bond order but elements in $\ce{HCl}$ involves period 1 and period 3 elements) then please state any counter examples.

I would not call $\ce{HCl}$ and $\ce{F2}$ isoelectronic. The first has $x=1,y=7$ and the second has $x=y=7$.

Now I have to be sure about word "isoelectronic" !  I think isoelectronic species have same number of electrons. For $\ce{HCl}$ and $\ce{F2}$  total number of electrons in these species are 18. Am I really wrong here?

I found a good link http://chemistry.stackexchange.com/questions/19065/isoelectronic-species-f2-and-hcl

I remember using a trick: total electrons =14, BO= 3.0, one electron more or less, the bond order decreased by 0.5, 13=15=2.5, 12=16=2.0.. And so on


text:
Esters of catechol - Fries rearrangement

Esters of catechol undergo the Fries rearrangement
But one product is formed more than the other.
I believe it is due to the electron withdrawing nature of the $\ce{OCOCH3}$ group which prefers the attack on meta position.I was able to write a reaction mechanism for the product which is minor in this case. But I couldn't write one for the major product. I wonder how it is formed?
Does the minor product itself rearrange to give the major product or are both of  the products formed directly in this reaction?

Your products appear to have lost an acyl group compared to the reactants. Is this intentional?

The last step is hydrolysis after which an acyl group is lost.

This Wikipedia article provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.

The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$).  Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong ortho-para director, so both ortho- and para-products are produced by the same mechanism. The $\ce{OCOR}$ group does not play a role in "directing" the substitution pattern.
The ratio of ortho/para isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic vs. thermodynamic control. Typically, the para product is the kinetic product (first formed), while the ortho product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the para-bromo product is formed in >95% yield (reference). So your specific reaction is likely being run under kinetically controlled conditions to produce the para product as the major product. As the Wikipedia link points out, the para product can equilibrate at higher temperatures to produce more of the ortho product.
Finally, just a word to address @bon 's comment.  Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (reference, p A-172) - apparently one of the acyl groups is lost in the process.

So the  mechanism is still not clear? Because i have gone through a concerted mechanism where no free acylium ion is generated.

Yes, the mechanism is not completely understood, but the reaction is not concerted and appears to involve an acylium ion.


text:
Where can we find the inventors of all chemicals?
Where can we find the inventors of all chemicals?
Discoverers of elements are easily found.

First of all, I think "discoverer" is a better word than 'inventor". Also, there are well over ten million chemical compounds previously described, and a humongous amount more are out there. I am sure that the discoverers of most chemicals have been lost to time and sheer disinterest.

@NicolauSakerNeto beat me to it. The scope of such a research and documentation task is mind-boggling.

There are >100 million substances listed in SciFinder, with 15,000 added daily. Any new ('invented') substances will most likely have their inventors listed there as a reference to the chemical literature. The discoverers of very common/simple/naturally occurring compounds is likely lost to time or don't have definitive answers.

At this moment, I have searched all websites mentioned by helpers here, but I cannot get the results. I have also tried ChemSpider.

For example,  I even cannot find the discoverer of "Digermane" in all these websites.

I think this question is quite hard.

For 'digermane' specifically, a google search leads to wikipedia which states "Digermane was first synthesized and examined in 1924 by Dennis, Corey, and Moore" with a reference to their paper. But this only points out that if the information is out there, it is not all in one place.

I doubt you can find all the information in one place, but I suspect that between the Beilstein, the Merck Index, PubChem, CAS, and patent office filings, you can find most of what you seek. That said, there will still be a healthy component of academic and proprietary findings that will prove to be hard/impossible/expensive to get.


text:
Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?
I'm wondering if the addition of a base of type $\ce{NR_3}$ (R: alkyl) to a ketone (e.g. 3-pentanone or 4-heptanone) results in the cis or trans-enolate.
I'd like to ask here about four cases:

$\ce{NMe_3}$ (small)
$\ce{NMeEt_2}$ (medium)
$\ce{NEt_3}$ (big)
$\ce{N(iPr)_2Et}$ (very bulky)

Maybe smaller bases as 1 or 2 usually give the trans-enolate, and bulkier bases such as 3 or even 4 affords the cis-enolate. (That's what I read, but I'm not able to understand why.) Is there a good explanation that rationalizes which enolate is formed, the cis or the trans?
Sterically hindered bases forms the the kinetic enolate. For deprotonation of 3-Methyl-4-heptanone I assume the enolate at the linear unbranched chain site, that's clear for me, but kinetic/thermodynamic enolate has nothing to do with cis/trans enolate or does it?

Related:
Does the addition of base to 3-pentanone give the cis or trans enolate?

With weak bases like the amines shown, the conditions are necessarily thermodynamic, so I would expect the thermodynamic Z-enolate to be favored over the E-enolate. Usually there is a trapping agent (e.g., TMSCl) available, which could change the outcome.

Why is the Z-enolate thermodynamically favoured? (Can we say the same without the carbonyl, that cis-propene is thermodynamically favoured over trans-propene?)

Tertiary amines are hardly basic enough to extract the alpha-proton of ketone (pKa ~20) to form enolate. Do you mean organolithium bases like (1) LiMe2 ...(4) Li(iPr)2 etc for a meaningful comparison with LDA or NaH? NEt3 was used to form enolate in the ACID-catalyzed enol formation. HNEt3(+) has a pKa of ~10.

@SYK Your comment is very worthful for me! Let's take the smallest Lewis Acid possible, for carbonyl activation, I would try LiI or NaI. Then I suggest a), b), c) and d) all give the cis-enolate.

I doubt LiI or NaI with a tertiary amine will deprotonate a carbonyl, so this question doesn't really make sense unless you are using some kind of boron activation (as your comment on the answer suggests), or silicon (as jerepierre suggested).

The keto-enol tautomerization cannot take place without at least a trace of acid or base, meaning that there is no direct shift of proton from alpha-carbon to oxygen or vice versa.
The two most common enolation mechanisms are:

Base-catalyzed (with organolithium reagent such as LDA (secondary amine))

In the base-catalyzed enol formation, the first step, proton extraction from the alpha-carbon, is slow. (The subsequent enolate formation is fast (due to resonance)). The base is proposed to form a chair-like transition state (TS) with the ketone, extracting the alpha-carbon proton and at the same time donating the Li to the oxygen. The resulting product is the classical lithium enolate.
Based on the structure of TS, you can then predict whether the “side chain” of the base would form a unfavorable syn-pentane interaction.
https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg

Acid-catalyzed

In the acid-catalyzed enol formnation, the first step, the protonation of oxygen and subsequent “charge re-localization” to the hydroxyl carbon, is fast. (The subsequent enol formation is slow). Since the protonation can occur from any free proton in the environment, the size of the acid plays little direct role in the stereochemistry of the resulting enolate. (However, the size of the “side chain” in the ketone reactant does).
However, there is a 3rd mechanism when the “base” is a tertiary amine (as you appear to be very interested in) but the reaction is acid-catalyzed. In this mechanism, the amine behaves like a nucleophile and a carbinolamine intermediate is detected.
See Bruice, J. Am. Chem. Soc. 1983, 105, 4982
http://pubs.acs.org/doi/abs/10.1021/ja00353a023
(Also see Bruice, J. Am. Chem. Soc. 1989, 111, 962 and 1990, 112, 7361)
In drawing this carbinolamine intermediate, you can predict a bulky tertiary amine to favor a cis-enolate due to steric hindrance. However, as suggested in the paper, "severely sterically hindered" tertiary amine can be too bulky to follow this 3rd mechanism and just catalyzes enolation via the general base-catalyzed mechanism.

Thank you very much for your answer. Probably there is a fourth mechanism which I'm more interested in than the other three: Addition of a lewis acid. a1) Bu2BOTf / (iPr)2NEt gives the cis enolate, b1) Cy2BCl / NEt3 gives the trans-enolate because of the bulky Cy. What does change when you instead use a2) Bu2BOTf / NEt3 and b2) Cy2BCl / (iPr)2NEt ? And last but not least I'd like to ask as a second question, if the addition of c) LiI and d) NaI also give the cis-enolate?

http://chemistry.stackexchange.com/questions/34153/how-many-units-lowered-pka-by-addition-of-a-lewis-acid-to-a-ketone

(−1) OP is asking about the generation of stoichiometric enolate, but you seem to be talking about keto-enol tautomerism. Plus, LDA isn't going to "catalyse" keto-enol tautomerism. Your answer seems to be confused about this: "In the base-catalyzed enol formation [...] The resulting product is the classical lithium enolate."


text:
Is enthalpy of a reaction meaningful at variable pressure?
Why should the enthalpy of a reaction be defined at constant pressure. Why can't it be defined at variable pressure?

Why should the enthalpy of a reaction be defined at constant pressure? Why can't it be defined at variable pressure?

Not sure "should" is the right choice of words: It can be defined for a reaction either at constant volume (variable pressure) or constant pressure (variable volume). You might choose one measurement over the other, but you'll get two different quantities.
At constant volume, there's no work done by the system on the surroundings, so you're just getting the change in internal energy of the system (this is typical of a bomb calorimeter lab setup); in the other case, you get the internal energy change as well as the work, and that is how enthalpy is defined (mathematically). See this Wikipedia article for a little more detailed discussion.


text:
Why does carbon monoxide have a greater affinity for hemoglobin than oxygen?
Hemoglobin is an iron-containing oxygen transport metalloprotein in the red blood cells of most mammals. Simply put, it's a carrier protein. Interestingly it doesn't carry carbon dioxide in the same way it does for oxygen. Oxygen binds to the iron atoms in the protein whereas carbon dioxide is bound to the protein chains of the structure. Carbon dioxide doesn't compete with oxygen in this binding process.
However, carbon monoxide is a very aggressive molecule. It's a colourless, odourless, and tasteless gas that is lighter than air and can be fatal to life. It has a greater affinity for hemoglobin than oxygen does. It displaces oxygen and quickly binds, so very little oxygen is transported through the body cells.
There are two equilibrium reactions of binding oxygen and becoming oxygenated hemoglobin:
$$\ce{Hb (aq) + 4O2 (g) -> Hb(O2)4 (aq)}$$
$$\ce{Hb (aq) + 4O2 (g) <- Hb(O2)4 (aq)}$$
And carbon monoxide binding equation at equilibrium:
$$\ce{Hb (aq) + 4CO (g) ⇌ Hb(CO)4 (aq)}$$
It is said
the equation is shifted towards right, generating $\ce{Hb(CO)4(aq)}$, since its bond is much stronger.
I have two main related questions:

Why is the carboxyhemoglobin bond stronger relative to that of oxygenated hemoglobin?

Why is carbon monoxide highly attracted to hemoglobin?

Does it have anything to do with the oxidation state of oxygen in each molecule? Please show some reaction equations associating $\ce{Fe}$.

Excursion into simple coordination chemistry: Bonding, backbonding and simple orbital schemes
Please refer to Breaking Bioinformatic’s answer for the MO scheme of carbon monoxide, it is very helpful. You might also look at the orbital pictures in this answer by Martin. Carbon monoxide can bind to metal centres via a σ coordinative bond where the HOMO of $\ce{CO}$ interacts with metal orbitals and also by the π backbonding, Breaking Bioinformatics mentioned. I’ll start by touching the σ bond so we can later better understand the π bond. In figure 1 you can see the molecular orbital scheme of a complex composed of a central metal ion and six ligands that donate in a σ manner exclusively.

Figure 1: Molecular orbital scheme of an octahedral complex with six σ donors around a central metal. Copied from this site and first used in this answer of mine. Metal orbitals are 3d, 4s, 4p from bottom to top; ligand orbitals are of s-type.
You will notice that figure 1 contains the irreducible representations of the orbitals beneath them. Orbitals can only interact if they have identical irreducible representations; otherwise, their interactions will sum up to zero. We can see that the metal d orbitals are split up into a $\mathrm{t_{2g}}$ set and and $\mathrm{e_g}^*$ set. This is the effect of σ bonding and why it stabilises the entire entity.
π bonding can only occur if the ligands have available orbitals of π symmetry. The 2π orbitals of $\ce{CO}$ are a nice example, but one could also simply assume a halide with its p-orbitals for the same effect. Further down in the internet scriptum I originally copied the picture from, you can see a set of two pictures that introduce p-orbitals. Twelve ligand p-type orbitals will transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, thus a further stabilising/destabilising interaction is introduced for the $\mathrm{t_{2g}}$ orbitals. Due to the $2\unicode[Times]{x3c0}^*(\ce{CO})$ orbitals being similar in energy to the metal’s $\mathrm{t_{2g}}$ and empty, they can stabilise the former well, creating a possibly strong stabilisation for the overall system. Since this is an interaction of two orbitals creating both bonding and antibonding orbitals, and since the resulting molecular orbital is filled with two electrons that stem from the metal centre this is termed π-backbonding.
The binding mode of carbon monoxide to haemoglobin
After this extensive background discussion, it may be clear that $\ce{CO}$ is generally a good ligand. For reasons I didn’t go into, a carbonyl complex of metal ions isn’t that stable, but a single carbonyl ligand will almost always be beneficial. Such is the case for the porphyrin-iron(II) system that forms the heart of haemoglobin: The central iron(II) ion is coordinated well from five directions (four from the porphyrin ring and one histidine of haemoglobin) and has a weakly bound water atom in the ground state in its sixth coordination slot, sometimes displaced by a distal histidine. Carbon monoxide can diffuse in and bind very well to this system, displacing the weakly bound water and histidine. In fact, isolated haem can bind carbon monoxide $10^5$ times better that it can oxygen.
This is something that nature could not really inhibit, because it is based on fundamental properties, but also something nature didn’t really care for, because the natural concentration of carbon monoxide is very low, and nature rarely had to deal with it competitively inhibiting oxygen binding to haemoglobin.
The binding mode of oxygen to haemoglobin
Oxygen, the $\ce{O2}$ molecule, is an absolutely lowsy σ donor — especially when compared to $\ce{CO}$. Its molecular orbital scheme is generally that of carbon monoxide except that it is entirely symmetric and two more electrons are included: They populate the 2π orbitals to give a triplet ground state. These orbitals are now the HOMOs and they hardly extend into space in any meaningful way — plus the still existant lone pair on each oxygen atoms is now at a much lower energy and also does not extend into space and thus cannot bind to a metal centre in a σ manner.
What happens here is rather complex, and the last lecture I heard on the topic basically said that final conclusive evidence has not yet been provided. Orthocresol discusses the different viewpoints in detail in this question. The diamagnetic properties of the resulting complex are, however, unquestioned and thus one must assume a singlet ground state or one where antiferromagnetic coupling cancels any spins at molecular levels. Since the ground state of haemoglobin has a high-spin iron(II) centre and the ground state of oxygen is a paramagnetic triplet, it makes sense to assume those two to be the initial competitors.
Professor Klüfers now states the steps to be the following:

Iron(II) is in a high-spin state with four unpaired electrons;
Paramagnetic triplet oxygen approaches and one of its $\unicode[Times]{x3c0}^*$ orbitals coordinates to the iron(II) centre in a σ fashion.
This induces a high-spin low-spin transition on iron and slightly reorganises the ligand sphere (pulling oxygen closer to the iron centre). We now have a low-spin singlet iron(II) centre and a coordinative σ half-bond from oxygen. We can attribute that electron to the iron centre.
Through linear combination, we can adjust the spin-carrying σ orbital and the populated iron orbitals so that an orbital is generated which can interact with the other $\unicode[Times]{x3c0}^*$ of oxygen in a π fashion.
We thus expect an antiferromagnetic coupling and an overall state that can be described best as $\ce{Fe^{III}-^2O2^{.-}}$ — a formal one-electron oxidation of iron to iron(III), reducing oxygen to superoxide ($\ce{O2-}$).
If the spin-carrying orbitals are all treated as being metal-centred, we gain a $\ce{Fe^{II}-^1O2}$ state.

(Content in the link I quoted from are in German; translation mine and shortened from the original.)
Only due to the complexly tuned ligand sphere and also due to the stabilising high-spin low-spin transition (plus reorganising) is oxygen able to bind to iron at all. The distal histidine further stabilises the complex by a hydrogen bond, alleviating the charge slightly. It is to be assumed that nature did a great deal of tuning throughout the evolution since the entire process is rather complex and well-adjusted for collecting oxygen where it is plentiful (in the lungs) and liberating it in tissue where it is scarce.
Comparison of the oxygen and carbon monoxide binding modes
The simpler picture I drew above for carbon monoxide is not correct. Inside haemoglobin, carbon monoxide also binds in an angular fashion as if it were oxygen — see bonCodigo’s answer for space-filling atom models. This is because the entire binding pocket is made to allow oxygen to bind (as I stated) thus attempting everything to make oxygen a comfortable home. Carbon monoxide is rather strained in there, its binding affinity is reduced by a factor of $1000$. However, since we started with a binding affinity difference of $10^5$, carbon monoxide can still bind $100$ times better than oxygen can.
Summary

Carbon monoxide is generally a good ligand that can bind to metal centres well.
Oxygen is generally a poor ligand.
Nature did everything to make oxygen a comfortable home in haemoglobin.
In doing so, the binding pocket became substantially less comfortable for $\ce{CO}$.
But since carbon monoxide was so good and oxygen so poor to start with, the former still binds better than the latter.
Nature didn’t really care, because individuals seldomly come in contact with carbon monoxide so the collateral damage was taken into account.

Since I was curious, I tried to find a reference for the bent CO binding. Here it is: Proc. Natl. Acad. Sci. U. S. A. 1988, 85 (14), 5062–5066. And yes, the debate about Fe(II) vs. Fe(III) in oxyHb is not yet settled. This recent article on Hb is a really nice read, summarising a lot of the work that has been done in the area: Proc. Natl. Acad. Sci. U. S. A. 2015, 112 (43), 13123–13127. Feel free to include in your answer if you wish.

Since you refer to my orbitals: I have deduced the ability of CO being a sigma donor in this answer. You might find the MO scheme in there quite useful for your first part. I'd say a job well done, you deserve more attention for this!

Is affinity of NO for haemoglobin even more than that of CO for haemoglobin?

The answer has to do with pi-backbonding.
In essence, the CO molecule has a negative formal charge on the carbon (it's neutral because of the oxygen having a positive formal charge). However, C is quite electropositive, and would like to relieve the stress caused by the negative formal charge. To relieve the stress caused by the negative charge, the CO molecule will want to bond to the iron.
How it does this is related to CO's molecular orbitals, which look like:

Note that the 4, 5, and 6 sigma orbitals, as well as the 2 pi orbital, are all antibonding. I will denote them with * from now on.
As CO stands right now, it has a bond order of three (the two lower sigmas cancel each other out, since 4* is antibonding while 3 is bonding). None of its orbitals, as they stand, can participate in bonding.
However, because carbon wants to rid itself of the negative charge, CO will do an interesting trick. It will promote an electron from the 5 sigma orbital into 2* pi. This reduces the bond order of CO from three to two, and introduces a separation of an electron pair. The picture of the 2* pi orbital is shown here, on the top:

So how is this more stable? In isolation, it is not. However, remember, we are not operating in isolation: there's an iron nearby. Most saliently, the iron has a d orbital, which looks like:

Now, the iron can put an electron into this d orbital and bond it with the pi* star orbital on the CO on the carbon end. This is hugely stable (note that we were willing to promote an electron into an unoccupied orbital and break up an electron pair to do it), especially since it relieves charge stress to do it (as opposed to molecular oxygen, which does no such thing). This backbonding is also why cyanide will stable itself onto iron.
Why isn't hemoglobin designed to bind oxygen as strongly as possible, then? Surely it would seem this would be an advantage, evolutionarily, if oxygen could bind tightly and never let go? But oxygen needs to let go: hemoglobin is a transport protein, and it needs to release oxygen into the tissues. That's why you have plenty of things that bond hemes more tightly than molecular oxygen: we need $\ce{O_2}$ in our tissues, not locked into hemoglobin forever.
EDIT: The hybridization of the iron in the protein is often (but incorrectly) assigned as $\ce{sp^3d^2}$. This is implied by the octahedral geometry the ligands around the iron take: the heme consists of an iron bound in four coordination sites by a porphyrin ring, one site by a histidine, and the last site by the oxygen the iron needs to bind.
The formula:
$\ce{Fe(II) + CO -> Fe(III)CO}$

pi-backbonding is an insight. I will search on that. Why isn't hemoglobin designed to bind oxygen as strongly as possible, then? : This definitely was not a question I have thus the last paragraph isn't necessary at all. Although the images are appreciated, more importantly, as mentioned in my question, can you write the equations for Fe and CO reactions in Hemoglobin (including the hybridization and oxidation states)?

Hybridization is kinda complicated when you're talking transition metals. That said, the equation looks like $Fe(II) + CO -> Fe(III)CO$

I need to tackle hybridization for organic chemistry, so I might as well do it together for transitions. Care enough to update your answer from that aspect?

The coordination implies that the iron needs to be sp3d2 hybridized (it's coordinated to six things, including the oxygen, and forms an octahedron).  Give me a second to put the picture together of how it fits in with the overall molecule.

@BreakingBioinformatics what evidence is there that Fe(II) is oxidized to Fe(III) upon binding CO?

@DavePhD He has done a good job by explaining most of it. Infact this was one of the main issues I face initially and also the basis to my question. I still am not able to find a credible source pointing out the exact formula for Fe, CO reaction within hemo.

Sorry, but I have to $-1$ this, even though it started out well. Reasons: 1) $\unicode[Times]{x3c0}$ backbonding incorrectly explained. There is no excitation to $\unicode[Times]{x3c0}^*$. 2) ‘hybridisation’ of iron given as complete bullshit. 3) Minor other incorrect aspects.

@Jan It started out well? Maybe the first sentence, it goes steep downhill from there. 1. CO has a formal negative charge at carbon only in one possible resonance structure. It has a positive charge, because oxygen is so much more electronegative. 2. Relieve stress? What does that even mean? 3. The HOMO of CO is not anti-bonding. According to this the BO would be one, not 3 as you stated. 4. The BO is not 3. It's somewhere between 2-3, but it's not that easy to determine. 5. " None of its orbitals ... can participate in bonding." No, CO is a sigma donor. 6. etc. pp. -> -1, sorry.

I appreciate above answer by BreakingBioinformatics. However I have been looking for the oxidation reactions involving Fe in this case. Found some useful material here. It is based on the textbook, Chemistry: Principles, Patterns, and Applications, 2007, Bruce Averill, Patricia
Eldredge.
Note : Following picture only shows the binding of Myoglobin.

Here's the text which explains some relevant and interesting details about the affinity of O2 and CO, although the main focus of the text is on primary function of the distal histidine in preventing self-poisoning of CO.

Oxygen is not unique in its ability to bind to a ferrous heme complex;
small molecules such as CO and NO bind to deoxymyoglobin even more
tightly than does O2. The interaction of the heme iron with oxygen and
other diatomic molecules involves the transfer of electron density
from the filled t2g orbitals of the low-spin d6 Fe2+ ion to the empty
π* orbitals of the ligand. In the case of the Fe2+–O2 interaction, the
transfer of electron density is so great that the Fe–O2 unit can be
described as containing low-spin Fe3+ (d5) and O2−. We can therefore
represent the binding of O2 to deoxyhemoglobin and its release as a
reversible redox reaction:
$\ce{Fe^{2+} + O2 ⇌ Fe^{3+} – O^{2−}}$ (26.8.2)
As shown in Figure 26.8.4 (the figure is not loading), the Fe–O2 unit is
bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in
CO are empty and those in NO are singly occupied, these ligands
interact more strongly with Fe2+ than does O2, in which the π*
orbitals of the neutral ligand are doubly occupied.
Although CO has a much greater affinity for a ferrous heme than does O2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O2, which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O2 and CO provides a plausible explanation for the difference in affinities. As shown in Figure 26.8.4 "Binding of O", the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO.

The existing answers explain, chemically, why hemoglobin ($\ce{Hb}$) binds $\ce{CO}$ much more strongly than it does $\ce{O2}$. In reading them, you might get the idea that nature evolved $\ce{Hb}$ to bind $\ce{O2}$ as well as possible, and simply didn't account for $\ce{CO}$.  The latter is likely true, but not the former.
In fact, nature didn't evolve $\ce{Hb}$ to bind $\ce{O2}$ as strongly as possible.  Think about how $\ce{Hb}$ operates: It doesn't merely need to bind $\ce{O2}$; it also needs to be able to release it.  Yes, there is a pH/ $\ce{CO2}$/ $\ce{2,3 BPG}$-induced conformational change in  $\ce{Hb}$ that facilitates this, but there are limits to how much the binding affinity can change as a result.  Thus $\ce{Hb}$ can't bind $\ce{O2}$ too strongly and still fulfill its biological function. Instead, the affinity of $\ce{Hb}$ for $\ce{O2}$ must be finely balanced between binding and release.
Consequently, it is not surprising that a foreign substance ($\ce{CO}$), not present in appreciable quantitites when $\ce{Hb}$ evolved, can outcompete $\ce{O2}$. The best analogy to $\ce{Hb}$'s affinity for $\ce{O2}$ is Post-It Notes.  These use an adhesive that sticks, but sticks weakly, so it can release easily when needed.

You picked on an interesting aspect: conservation of functional affinities of species  through evolution...

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text:
Hydrogenolysis of benzylic C-N bond
Why does N,N-dimethyl-1-phenylmethanamine (the left molecule in the picture) undergo hydrogenolysis of the C–N bond readily while 2-methyl-1,2,3,4-tetrahydroisoquinoline (right) doesn't react at all?

Hydrogenolysis typically involves a metal surface ($\ce{Pd, Pt}$, etc.) and hydrogen gas. In the reaction a weak bond, that is, a bond that is either strained or a bond that can generate a stabilized radical is broken. In your examples there is no strain, but a resonance stabilized benzylic radical can be generated in one case.
In the benzylic amine pictured on the left, the benzylic carbon can easily adopt a geometry such that when the p-orbital on the benzylic carbon is generated it is already aligned with the p-orbitals in the aromatic ring and is therefore stabilized while it develops.
The molecule on the right is different. I built a model and as the C-N bond starts to break and the p-orbital starts to form, the developing p-orbital is twisted 45-60° away from alignment with the p-orbitals in the aromatic pi system - it is not effectively stabilized due to conformational constraints imposed by the ring system.
Therefore this latter molecule will not undergo hydrogenolysis of the C-N bond as readily as the molecule on the left.
Here are some photos of my model:

the molecule - note the blue $\ce{N-CH3}$ nitrogen as a point of reference

conformation A - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is axial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted about (90-30) 60° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential

conformation B - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is equatorial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30-45° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted at least (90-45) 45° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential


text:
What are the active ingredients of shampoo, and how do they work?
After seeing one too many TV commercial about the fantastical properties of different high-profile shampoo products all "with documented effect" it suddenly struck me that I never actually researched what the active ingredients of a shampoo really is.
So what are the active ingredients of most shampoos, and how do they work?
Is there is any significant difference between popular brands?
Is there any significant difference between cheap and expensive brands?
Will they actually heal your hair?

Shampoo has a few common components: Surfactants, preservatives and a number of additives meant to give the product desirable properties.
When you wash your hair, you're removing debris and buildup in two steps; the polar components (such as salts and aerosols that land in your hair) are washed away by the water, and the non-polar by the surfactants.
Surfactants are molecules where one side is polar and the other is non-polar. Things with similar polarity like to group together, and the opposite happens with polar/non-polar pairs (I'll call this the 'Like attracts Like rule, it's very important to this particular problem).
These surfactants can group together with  each other to form little spheres known as a micelle if there are enough around. This level is known as the critical micelle concentration.

(source: ekshiksha.org.in)
Because the non-polar parts of the surfactants come together, the center of the micelle is also non-polar and can hold small droplets and pieces of other non-polar substances like the sebum that builds up on your hair naturally over time.
Almost all shampoos use a nearly identical set of surfactants. Common examples are sulfates (ammonium lauryl sulfate, Sodium laureth sulfate, sodium pareth sulfate, sodium lauryl sulfate) and/or polysorbates (PEG20 and PEG80).
There are also additives that thicken it up, keep it from going bad and make that rich wonderful lather that we've all come to know and love.
What you are paying for aside from the above is mainly botanical or synthetic additives meant to soothe or beautify. The best way is to look up these ingredients on a case by case basis and decide if they would benefit you. I would say "be wary"; a large number of ingredients provide virtually zero added benefit for your money. Some work extremely well for certain people; you never know until you try.
Always read your labels! If you don't know what something is, google it!

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text:
What makes cola efficient as a rust remover?
I actually tested submerging a bunch of old rusty knives blades in a cup of plain cola for a few days not expecting much. To my amazement the myth held true, coke is a really good rust remover.
So what is the chemistry behind it?

This may have been asked before.

I searched but found nothing

You're right. It wasn't asked before, but it was discussed in another question. Related: http://chemistry.stackexchange.com/questions/4732/chemically-removing-rust-without-leaving-any-unwanted-residues

Rust consists of insoluble iron oxides and hydroxides, $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Acidifying these compounds protonates the oxide and hydroxide anions to for water:
$$\ce{O^{2-} + 2H^+ \rightarrow H_2O}$$
$$\ce{OH^{-} + H^+ \rightarrow H_2O}$$
This leaves the soluble iron $\ce{Fe(III)}$, which can be washed away.
Many colas contain phosphoric acid, and are quite acidic (Coke has a pH of ~2.5 according to this science fair project), thus they can dissolve the oxides.

It's not that Fe(III) gets washed away necessarily but that Fe(III) phosphates gets formed which works afterwards as a protection layer like the oxides for e.g. aluminium.

Ah, yes. I didn't think about the phosphate ion. Good point.

It's not just a point but what happens :)


text:
Is it possible to calculate specific heat capacity of a compound based on its elements' heat capacities?
What the question says. For example, if I knew the specific heat capacities of lithium and oxygen, could I work out the specific heat capacity of lithium oxide with no further information?

There is not much connection between the heat-capacity of a compound and the heat-capacity of the elements it is made of. Heat capacity is primarily coming from the low frequency vibrations of the lattice (if it is solid), and vibration+rotation+translation if it is liquid or gas. These have no connection with the properties of the lattice etc properties of the given elements.
So short answer: no

Yes, one can calculate the specific heat capacity of the compound from its constituent elements that is just some of its constituent element's specific heat capacity. As it is known that specific heats of elements remain unchanged when they enter into compounds. However, the density/volume/mass may change and so the heat capacity.

While the empirical Neumann-Kopp method for estimating the heat capacities of compounds works reasonably well in many cases, it fails miserably in many more.


text:
In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack?
In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack ?
My approach:I considered hydroxy group as the stronger +M group.So my guess was para position w.r.t hydroxy group.But that's not the correct answer.
The answer is actually given as ortho w.r.t hydroxy and para w.r.t -CHO group!!

@yasir can you please explain why?

This is known as the ortho effect. According to it if an ortho/para and meta directing group are meta to each other, then the incoming electrophile prefers to come on ortho of meta directing group.
See the reference.

I wanted t0 know for m-hydroxy benzaldehyde

The same applies for m-hydroxy benzaldehyde too.In this case, you have $\ce{Cl}$ which is ortho/para directing meta to $\ce{NO2}$ group which is meta directing.In your case, $\ce{OH}$ which is ortho/para directing is also meta to $\ce{CHO}$ group which is meta directing.

@yasir Please edit in the complete reference. Give credit where credit is due.

So are you saying that the OP's stated 'correct' answer is wrong?

@Bon That's what i have learnt from my text.I would like someone to kindly elaborate on  ortho effect as my knowledge of it is limited.


text:
How to know which atom is the central atom in VSEPR theory?
How to find the central atom? What is the central atom in $\ce{PCl3}$
and  how to find the shape using VSEPR theory?

You don't really have to look for the central atom. Technically, VSEPR theory may be applied to any atom in the molecule, only in case of chlorines it would not produce anything useful.

You know that phosphorus has 5 valence electrons and that chlorine has 7 valence electrons ... and that you somehow have to combine them.
As chlorine is more electronegative (Allred-Rochow: Cl 2.83, P 2.06) you know that each chlorine will attract one electron from the phosphorus to get the noble gas configuration with 8 electrons in the valence shell.
Now that in turn means that there will be one single bond between the phosphorus and each chlorine, which automatically makes phosphorus the center.

In general the least electronegative atom occupies the central position in the molecule/ion. For example, in PCl3, chlorine is more electronegative than phosphorus, occupy the terminal position and phosphorus is the central atom. Hydrogen is an exception as can NEVER be the central atom. The atom with the lowest subscript in the molecular formula indicate the number of atoms of the element, in most cases, is the central atom.


text:
Why is the principal quantum number irrational?
My textbook says that for a $d$-electron, the orbital angular momentum is: $$ \frac{\sqrt{6}h}{2\pi} $$
But since:
$$ mvr = \frac{nh}{2\pi}$$
$$ n \in N$$
where $N$ is the set of natural numbers, how is the value of $n$ an irrational number?

First, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Secondly, you confused the orbital angular momentum quantum number $l$ with the actual value of the orbital angular momentum $L$.

And surely you made a mistake in your first equation: it should be $h$ instead of $n$ here, of course. Or perhaps it was a mistake (misprint) in the book which confused you.

So, first, as I said in my comment, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Both can take only integer values (positive and non-negative respectively), but for totally unrelated reasons.
Secondly, the orbital angular momentum itself is related to its quantum number $l$ by the following equation,
$$
    \hat{L}^2 \Psi = l (l + 1) \hbar^2 \Psi \, ,
$$
where $\hbar^2{l(l+1)}$ is the square of the magnitude of an electron orbital angular momentum. Thus, while the orbital angular momentum quantum number $l$ is always a non-negative integer, the magnitude of an electron orbital angular momentum $\sqrt{l(l+1)} \hbar$ can take irrational values.
For instance, for a d-electron, $l=2$, and thus, the magnitude of its orbital angular momentum is $\sqrt{2(2+1)} \hbar = \sqrt{6} \hbar = \sqrt{6} h / 2 \pi$.


text:
Trend in the melting point down the group
We know that the atomic radii increases down the group. So, less energy is required to pull out the outermost electron as we go down the group, hence,  gradually melting point decreases down the group.  The topmost element has large melting point because of its small atomic radii and so more energy is required to pull out its outermost electron.
But, it is observed that the melting point slightly increases in case of the bottom-most element of group as compared to the previous element. Why?
Examples:

Group 13 - indium - $\pu{156.6^\circ C}$ but  thallium - $\pu{304^\circ  C}$
Group 14 - tin - $\pu{231.92^\circ  C}$ but lead - $\pu{327.46^\circ C}$
Group 17 - iodine - $\pu{113.5^\circ  C}$ but astatine - $\pu{302^\circ C}$

I am curious about the melting point of astatine. The Wikipedia article lists it as the value you present, but in the body of the article it goes on to say: The bulk properties of astatine are not known with any certainty. Research is limited by its short half-life, which prevents the creation of weighable quantities. A visible piece of astatine would immediately vaporize itself because of the heat generated by its intense radioactivity. It seems to me that this problem precludes determining its melting point.

@BenNorris That could be a very good question

@santiago - I found my answer in the same Wikipedia article: Most of the physical properties of astatine have been estimated (by interpolation or extrapolation) ... The melting and boiling points of astatine are also expected to follow the trend seen in the halogen series, increasing with atomic number.

There's a counter effect in play here: van der Waals forces. As you become larger and larger, induced dipole-induced dipole interactions become stronger and stronger.

You'll notice that, in halogens (where abstracting an electron is a fairly tall order), the trend is that, the bigger your atom, the higher your boiling point, regardless of where you are in the group. The boron group loses the conflict of ease of abstraction versus van der Waals forces at around the gallium/indium leap, and for the carbon group, the conflict is lost to van der Waals forces only between tin and lead.

I agree with your explanation for the halogens but I'm less convinced by the other two. The boiling points of group 13 and 14 elements decrease down the group, which is opposite to the trend suggested by van der Waals interactions. I suspect that the increase in melting point results from the change in crystal structure and the increasing metallic character of the elements as you go down the group.


text:
Maximum Useful Work from a Process
Negative of the $\Delta G$ for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way:  $\Delta H$ is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently $-\Delta H$ energy is obtained from the system after doing expansion work. Since $\Delta S$ is the entropy created in the process, at the very least $-\Delta S$ entropy must be created in the surroundings - that is, at least $-T\Delta S$ energy must be lost as heat. This comes from the $-\Delta H$, and thus leaves $-\Delta H + T\Delta S$ to do useful work. No more work can be done than this. $-\Delta H + T\Delta S$ is $-\Delta G$, so $-\Delta G$ is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why $-\Delta G$ is the maximum possible useful work.
Now if $|\Delta H|$ is more than $|T\Delta S|$, with both being negative, then one can think of the $W_{max}$ or $-\Delta G$ as a part of the $|\Delta H|$; since $|\Delta G|< |\Delta H|$. Some part of $|\Delta H|$ goes as $|T\Delta S|$ to increase the surroundings' entropy, and the other part in doing useful work. But if both are positive, with $|T\Delta S|$ being more than $|\Delta H|$, again $\Delta G$ is negative, allowing useful work to be extracted. But now it seems as though $|T\Delta S|$ heat will be extracted from the surroundings, $|\Delta H|$ used up in the process, while the rest can be converted to work - in other words, useful work is a part of $|T\Delta S|$ - with the other part used by the process as $|\Delta H|$. Is this correct? Is $|T\Delta S|$ extracted from the surroundings actually?

"Useful work" does not include expansion work. For a proof of why $w_{\text{by,add}} \leq \Delta G$, you may want to first refer to my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work? then come back here.
In that proof, I used equalities by assuming that the process is reversible. If we drop this assumption, then by the Second Law,
$$\begin{align}
\mathrm{d}S &\geq \frac{\mathrm{d}q}{T} \\
\mathrm{d}q &\leq T\mathrm{d}S \\
dU &\leq T\mathrm{d}S - p\mathrm{d}V + \mathrm{d}w_{\text{add}} \\
dG &\leq dw_{\text{add}} \\
\Delta G &\leq w_{\text{add}}
\end{align}
$$
Equality holds if the process is reversible, or more generally, if both the initial and final states are equilibrium states. Now this seems like an obvious contradiction of the equation in the very first paragraph. The key lies in the definition of the work - whether it is done on the system, or by the system. In my proof, I have been referring to the work done on the system; however, here we are interested in the work done by the system, since that is the work that can be "extracted". The work done by the system must be equal to the negative of the work done on the system: $w_{\text{by,add}} = -w_{\text{add}}$. This leads to the final result $w_{\text{by,add}} \leq \Delta G$.
Levine's Physical Chemistry 6th ed. writes:

In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but $w_{\text{by,add}}$ is the useful work output. The quantity $-\Delta G$ equals the maximum possible nonexpansion work output $w_{\text{by,add}}$ done by a system in a constant-$T$-and-$p$ process. Hence the term "free energy". (Of course, for a system with P-V work only, $\mathrm{d}w_{\text{by,add}} = 0$ and $\mathrm{d}G = 0$ for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.

If you want a direct answer as to why your explanation isn't right, it's because you can't equate $\Delta H$ with the work done. Work done is given by $w = -\int\! p\,\mathrm{d}V$. You can only equate $\Delta H$ with the heat transferred at constant $p$. Apart from the First Law ($\Delta U = q + w$), heat and work are generally unrelated.

Yes, I understand perfectly. What I've written above seems rather nonsensical once I read your answer.

Could you also explain why the thermodynamic efficiency of a cell is given by |ΔH|/|ΔG|?


text:
Bond angles for the hydrides
I noticed the fact that all the hydrides of the elements belonging to group IV has bond angle 109.5° while in group V it varies from 107.3° for $\ce{NH3}$ to 91.3° for $\ce{SbH3}$. Similarly we see in group VI that water has highest bond angle then it significantly decreases to 89.5° for $\ce{H2Te}$. Is there any specific reason for hydrides of group IV metals to have remarkably equal bond angles, and other groups to have varying angles?

possible duplicate of Why does bond angle decrease in the order H₂O, H₂S, H₂Se?

@PH13 - while that question does answer why the variation in Groups V and VI is large, it does not answer why there is no variation in bond length in Group IV.

The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides.
According to VSEPR, electron domains (i.e. bonds and lone pairs) are arranged around the central atom in geometries that minimize electrostatic repulsion between the electron domains. These geometries also happen to be solutions to the Schrödinger equation for the set of molecular orbitals as linear combinations of atomic orbitals, and in the molecular orbital case, they arise not from electrostatic repulsion, but strictly from the mathematics of the linear combinations producing a minimum energy self-consistent set of orbitals.
VSEPR is qualitatively simpler for small molecules. All of the binary hydrides in Group IV, V, and VI have steric number four around the central atom. This means that the number of lone pairs and the number of bonding domains add up to four. These species can also be described using the $\ce{AXE}$ method, where $\ce{A}$ denotes the central atom, $\ce{X}$ denotes an atom bonded to $\ce{A}$ and $\ce{E}$ denotes a lone pair of electrons.

The Group IV hydrides all have four bonds and no lone pairs. We describe this arrangement as $\ce{AX4}$.
The Group V hydrides all have three bonds and one lone pairs. We describe this arrangement as $\ce{AX3E}$.
The Group VI hydrides all have two bonds and two lone pairs. We describe this arrangement as $\ce{AX2E2}$.

There are several factors that affect the degree of repulsion:

Lone pairs - Lone pair domains are more diffuse and have higher repulsion than bonding domains.
The nature of $\ce{X}$ - Larger and/or more electropositive atoms have higher repulsion.
Bond order - Bonds with higher bond orders have higher repulsion than bonds with lower bond orders.

Since $\ce{X=H}$ in all cases, we only need to worry about the first factor: lone pairs.
The arrangements $\ce{AX4}$, $\ce{AX3E}$, and $\ce{AX2E2}$ all have four electron domains, and thus the base electron domain geometry group is tetrahedral.
The differences in bond angle between $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2O}$ originate from the increasing number of lone pairs:

$\ce{CH4}$ has no lone pairs: the bond angles are the ideal tetrahedral bond angles of $109.5^\circ$.
$\ce{NH3}$ has one lone pair: the bond angles are contracted a little because of the grater repulsion of the lone pair to $107.8^\circ$.
$\ce{H2O}$ has two lone pairs: the bond angle is contracted further due to the repulsion of two lone pairs to $104.5^\circ$.

All of the Group IV hydrides will have perfect tetrahedral geometry due to having four bonds to the same atom and no lone pairs. Thus $\ce{CH4}$, $\ce{SiH4}$, $\ce{GeH4}$, $\ce{SnH4}$, and $\ce{PbH4}$ all have bond angles of $109.5^\circ$.

For Group V, the bond angles are:
$$\begin{array}{c|c}
\ce{AH3}&\text{angle}\\ \hline
\ce{NH3}&107.8^\circ \\
\ce{PH3}&93.5^\circ \\
\ce{AsH3}&91.8^\circ \\
\ce{SbH3}&91.7^\circ \\
\ce{BiH3}&90.5^\circ \\ \end{array}$$
Notice that there is a large decrease from $\ce{NH3}$ to $\ce{PH3}$ and then very small changes for $\ce{AsH3}$, $\ce{SbH3}$, and $\ce{BiH3}$.
You could make the argument that the larger orbitals of the heavier elements will have high repulsion than the smaller orbitals of nitrogen. This is where VSEPR breaks down. VSEPR works well for some compounds because there is significant $s-p$ mixing when constructing the molecular orbitals. This mixing need not happen for the heavier elements. One molecular orbital solution for $\ce{PH3}$, for example, indicates very little contribution from the $\ce{P}\ 3s$ orbital. We can describe the $\ce{P-H}$ bonds as predominantly $\ce{P}_{3p}\ce{-H}_{1s}$ bonds. Thus the geometry around $\ce{P}$ is determined mostly by the orientation of the $p$ orbitals.
If we look at the Group VI hydrides, we see the same trend: a large decrease from $\ce{H2O}$ to $\ce{H2S}$ and then very little change. The same argument can be made for primarily $\ce{A}_{p}\ce{-H}_{1s}$ bonding in the heavier Group VI hydrides as for the heavier Group V hydrides.
$$\begin{array}{c|c}
\ce{AH2}&\text{angle}\\ \hline
\ce{H2O}&104.5^\circ \\
\ce{H2S}&92.1^\circ \\
\ce{H2Se}&91.0^\circ \\
\ce{H2Te}&90.0^\circ \\
 \end{array}$$

VSEPR does not do a good job in predicting these geometries; it only works by chance for the second period.
For the 14th group, (the IV group in older nomenclature) things are pretty much as Ben explained them. We have a consistent number of substituents (four) and the same number of lone pairs (zero) around the central atom. Therefore, we can arrange the substituents as far away from each other as possible to generate a tetrahedric structure, and molecular orbital calculations indicate that this will lead to the lowest energy. You can predict what the molecular orbitals will look like from symmetry considerations. Changing the central atom only changes the bond length.
Once we move on to group 15, we suddenly have a lone pair on the central atom in ground state. This lone pair will occupy an s-orbital in the isolated atom, and it would be most stable to keep this lone pair in an s-atom for two reasons:

p-orbitals actually extend into a direction of space, so the overlap with a bonding partner will be larger
the s-orbital is ‘closer to the nucleus’ (in a non-quantum view) meaning that the negative charge of the electrons is better stabilised

As such, there is a general tendency across the entire periodic table to keep lone pairs in orbitals with an s-character as large as possible. The largest s-character is, of course, a pure s-orbital. Therefore, we should predict a bond angle of $90^\circ$ for all group 15 hydrides — which is well met for every case except ammonia. (This is similarly true for the group 16 ones with the exception of oxygen. One lone pair must reside in a p-orbital since only one s-orbital is available.)
So what’s wrong with ammonia and water? Well, the oxygen and nitrogen atoms are much smaller than their heavier homologues and therefore the hydrogen atoms would be much closer to each other at an ideal $90^\circ$ angle. This, however, would induce new steric stress between the hydrogen atoms, destabilising the entire structure. Therefore, the hydrogens shift out just far enough to where they are adequately far away from each other while still allowing the central atom to have as much s-character in its lone pair as possible. The sweet spot happens to be $104.5^\circ$ for water and $107.8^\circ$ for ammonia.


text:
Why do socks stink?
What chemicals are responsible for the stench emanating from socks? If it's just sweat, why don't underwear or vests stink as much? Which gas am I inhaling when smelling socks to decide whether they need a wash?

Because we do. We are just adapted to our own stink at the level prevailing over the whole body.

According to the Wikipedia page, aptly named Smelly Socks, it states that the smell

is a mixture of ammonia, fatty acids (in particular, isovaleric acid), and lactic acid.

What makes sock odor unique is that socks often pick up small amounts of organic material when walking around without shoes on. When wearing shoes, they are in a warm semi-enclosed environment with a considerable amount of sweat infusing into them (it can be quite warm and humid in shoes) particularly when exercising. A lot of the smell is due to bacterial action on the bits and pieces trapped in the socks themselves.
Interestingly, the Wikipedia article states that it is a topic of ongoing study, in part due to a possibility that the chemicals could be used in the battle against Malaria - Using smelly socks to attract mosquitoes -- and combat malaria

In addition to the answer posted by @santiago, here is some information from the following article in the UK newspaper Daily Mail:
The science behind stinky feet: Video reveals the processes that cause a nasty stench, and how to stop it
The article has a video, produced by the American Chemical Society, that explains how the bacteria that live on your skin and in your shoes eat your sweat, producing an acid by-product that causes the smells.
A snippet from the article:

The three main bacterial waste products are propanoic acid,
  methanethiol and isovaleric acid
Methanethiol makes your feet smell like sulphur and cheese and is
  produced from dead skin.
Isovaleric acid also has a cheesy aroma, and a rancid vinegar-like
  scent, while propanoic acid has a rancid sour smell.


text:
Why does the first lithium stationary state have n = 9/4?

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula
\begin{align}
E &= -2.178 \times 10^{-18} \cdot \frac{Z^2}{n^2}.\\
E &= -2.178 \times 10^{-18} \cdot \frac{3^2}{1}\\
E &= -1.96 \times 10^{-17}~\mathrm{J}
\end{align}
But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$.

I think your calculation is correct. There is an error in this textbook.  All you have to do is to switch  between $\ce{Li^{2+}}$ and  $\ce{He^{+}}$. And 'll prove my point:
$\ce{Li^{2+}}$ and  $\ce{He^{+}}$ are hydrogen-like atoms.
For these species, the ionization energy is given by the equation:
$$E_\mathrm{i}=E_\mathrm{i}(\ce{H})\times Z^2$$
For $\ce{Li^{2+}}$:
\begin{align}
E_\mathrm{i}(\ce{Li^{2+}})&=E_\mathrm{i}(\ce{H})\times Z^2\\
19.6 \times 10^{-18}&=E_\mathrm{i}(\ce{H})\times 3^2\\
E_\mathrm{i}(\ce{H})&=2.18 \times 10^{-18}  \,\mathrm {J/atom}\\
E_\mathrm{i}(\ce{H})&=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\
  &= 13.6 \,\mathrm{eV}
\end{align}
$13.6 \,\mathrm{eV}$ is well-known to be the first ionization energy of hydrogen. So, my assumption is true.
To solve the problem after correction:
 The ionization energy of $\ce{He^{+}}$ is given by the equation:
\begin{align}
E_\mathrm{i}(\ce{He^{+}})&=\frac{E_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\
E_\mathrm{i}(\ce{He^{+}})&=8.72\times 10^{-18}\,\mathrm {J/atom}
\end{align}
The energy of the first stationary state:
$$E_\mathrm{i,1}=E_\infty-E_1$$
But, $E_\infty=0$, we find:
$$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$


text:
Is ingesting water during swimming harmful?
Sometimes I feel the water in the pool is heavily chlorinated. However it is impossible for me to swim without ingesting some water. I am only a learner but even advanced swimmers do concede that they do ingest some water at times.
Could this be potentially harmful depending on the chemical content in the pool? Are there studies which advise on the extent of permissible chemicals that can be added to a pool?

Could this be potentially harmful depending on the chemical content in the pool?

Probably not, unless you managed to drink quite a bit of pool water. Worldwide, millions (hundreds of millions?) of people have collectively ingested millions (billions?) of liters of pool water - with varying chemistries - over many decades (myself included), and nobody seems to have suffered too much. I believe this falls into the "generally regarded as safe" area.

Are there studies which advise on the extent of permissible chemicals that can be added to a pool?

Numerous ones, Google away. These guidelines, from the State of Indiana, are representative of the recommended levels of various chemicals present in common swimming pools.


text:
What causes the old book smell?
I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question):

One thing I notice is that there is a distinct smell that comes from the pages of older books. (the smell is not unpleasant) - it seems that the older the book, the more distinct the smell.
What is the chemistry behind the old pages in a book smell?

I love the smell of old books, but it's probably because of association with the pleasure of reading them.

This is a very interesting question. Provided that the materials used in making papers aren't the same around the globe, this is a very broad case of study. However, a study has been conducted in which the main goal was to identify the compounds that are the cause of the smell; VOCs:[1]

Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air.

So, quoting the article:

Using supervised and unsuper-vised methods of multivariate data analysis, we were able to quantitatively correlate volatile degradation products with properties important for the preservation of historic paper: rosin, lignin and carbonyl group content, degree of polymerization of cellulose, and paper acidity.
It is a result of the several hundred identified volatile and semi-volatile organic compounds (VOCs) off-gassing from paper and the object in general. The particular blend of compounds is a result of a network of degradation pathways and is dependent on the original composition of the object including paper substrate, applied media, and binding.

The 15 most abundant VOCs present in all chromatograms were selected for further analyses: acetic acid, benzaldehyde, 2,3-butanedione, butanol, decanal, 2,3-dihydrofuran, 2-ethylhexanol, furfural, hexadecane, hexanal, methoxyphenyloxime, nonanal, octanal, pentanal, and undecane.

Strlič, M., Thomas, J., Trafela, T., Cséfalvayová, L., Kralj Cigić, I., Kolar, J., Cassar, M. (2009). Material degradomics: on the smell of old books. Analytical Chemistry 2009, 81 (20), 8617-8622.
The main link to the article might be behind a paywall, so for the interested, the relevant ResearchGate article is accessible.

What's loading weight graph? PLS Calibration?


text:
Nitrates in water
In my previous question, someones answer among other things, said that boiling the water increases the concentration of nitrates and nitrites. I searched on google and it's true, but how does that happen ? What is the reaction that leads to an increasing level of nitrates?
Im very courious because nitrates are too dangerous to be high in concentration, i would appreciate your help.

boiling the water increases the concentration of nitrates and nitrites.

Yes.

how does that happen? What is the reaction that leads to an increasing level of nitrates?

It happens because you boil away the water, and the nitrates and nitrites stay in what's left of the water, thus raising the concentration of those species in the now reduced amount of water. There is no "reaction," per se, just you are concentrating dissolved species by removing the water.

So then its mpt only the concentration  of nitrates that increases by boiling the water ?

Correct - in general if it doesn't boil off, it will stick around and get concentrated.


text:
How Does One Create A Cellulose Ingredient For Use In The Creation Of Glucose?
I am attempting to begin synthesizing glucose, however I will require a form of ball milled cellulose as a main ingredient. Having been said, what methods would be recommended to create the Cellulose? If we were to use grass clippings or cotton as our source for example, what steps would we take in order to gain the desired material from the aforementioned source?

How much equipment do you have? Do you have access to standard lab equipment? Might be really helpful to know for the answerer.

Fairly standard as comparable to high school equipment.

Cotton wool, also called "purified cotton", is inexpensive and available at pharmacies ("chemists" in the UK). I could not find the percentage of impurities (probably lignin and minerals), but it is considered one of the purest forms of cellulose commercially available, and should not need pretreatment to make glucose.


text:
What is the proper term for lines of constant volume and constant enthalpy on a thermodynamic diagram?
I know an isotherm is a line of constant temperature.
Isobar is a constant pressure line.
Isentrope is a line of constant entropy.
But what do you call a line of constant volume? An Isochore?
And what do you call a line of constant enthalpy? An "isenthalp"?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

But what do you call a line of constant volume? An Isochore?

Correct.

And what do you call a line of constant enthalpy?

The process is isenthalpic. More than one engineering/thermodynamic text I found by Googling indicates they are simply referred to as "isenthalpic lines." See this for example.

Logically, isenthalp would be the name for an isenthalpic line, but based on a quick Googling, the English world seems to be lagging the German world in this vocabulary innovation.

Agreed on all points @Curt F.


text:
Is turning graphite carbon into diamond industrially doable?
I've read somewhere on the Internet that heating graphite at 4200°K at a sufficiently high pressure would turn it into diamond.
This chart seems to support this idea:
Is that doable industrially?

The ° symbol is used for temperature scales other than kelvin.  For example, 4200K, but 70°C or 98.6°F.

It has been done since 1954 and has been a commercial success for some time. Many products now have diamond-like coatings, produced comparatively inexpensively through vapor deposition without need for high pressure.
BTW, in English "burning" means to set on fire, oxidize; I believe you mean "heating graphite", and, as your graph shows, pressure is needed as well as heat.

Thanks for your answer. Are these synthetic diamonds as valuable as the natural ones ? I mean, do they have any use in jewelry ?

There are many adds for man-made diamond jewelry, for example http://www.diamondnexus.com/pure-carbon-man-made-diamonds.html. As in synthetic sapphire, made since 1902, it is cheaper to make the artificial product (and the artificial material is also of more use in manufacture). Curiously, I believe natural quartz crystal (used in resonators etc.) are more used than the artificially grown crystals.

@DrMoishePippik Apparently natural quartz is no longer common as natural quartz has many defects and impurities. Much like as you mention for sapphire, artificial quartz is more useful for manufacturing as it can be made in defect-less shapes that are simpler to cut appropriately.

Thanks for the information, @Michael D. M. Dryden! Though I had read that Bell Labs had issues with growth of a polymorph of SiO2 taking over from the desired quartz  after hydrothermal apparatus had been cycled a few times, I should have realized that problem would be overcome.

You want to heat graphite at high pressure, not burn it - but yes, high-T, high-p will work.

Is that doable industrially?

There is a rather lengthy and informative Wikipedia article for synthetic diamond that illustrates (in part) the use of this process - HPHT, for "high pressure, high temperature"(using a high-purity source of carbon, not necessarily graphite) in industry. Note that there are several industrial processes mentioned in addition to HPHT.