text:
Abstraction of a proton in cycloheptatriene
So, whenever there is abstraction of H+ by an acid-base reaction, I see that the proton on the sp3-hybridized carbon is taken. Why is this if you take into account that vinylic hydrogens are more acidic?
Also, on a sidenote, what is the hybridization state of a carbon in a C=C double bond where the vinylic proton was removed by a base?

Are you sure you don't mean transfer of hydride, i.e. oxidation to the tropylium cation?

Actually Abel, the tropylium cation is drawn. But the anion with the abstraction of the proton only on the sp3 carbon is also shown (Cycloheptatrienyl anion). It was an exercise where we had to say whether the compounds were aromatic or not. Could we make such compound from acid-base reactions?

The pKa for a vinylic proton is ~ 43, the pKa for propene is ~ 40, and the pKa for 1,4-pentadiene is ~ 35.  Consequently it's not the vinylic protons that are the most acidic, it's the allylic ones.  Now the question whether tropylium anion is aromatic/antiaromatic is complicated.

Abel, are you saying that an acid would rather rip apart a hydrogen from the sp3 carbon in the propene?

If you think about it, is CH3:- not a much stronger base than CH2=CH:-? So, it is the CH2=CH2 that is the stronger acid.

In the allyl anion the charge is delocalized over all three carbons.  In saturated hydrocarbons it's confined to a specific carbon.

Okay, so in this particular molecule (Cycloheptatrienyl anion), the abstracted hydrogen will always come from the sp3 carbon.

And in the case of propene, the hydrogen will come from the sp3 too. Why? Because although the double bond is more acidic if we look at it individually, the possibility of removing the hydrogen from the sp3 carbon adds resonance, which further stabilizes the compound than abstracting an H from the double bond.

Please say yes if all my reasoning is correct.

I think you are correct in saying this,


text:
Interpretation of TD-DFT results
For the first time I'm doing TD-DFT calculations (wB97XD functional) in Gaussian 09 for an open-shell system and the results look like the hell of a mess for me. The molecule is rather big, so I started with just 10 excited states to see how it would work.
First off, the ground electronic state is $^2A_2$, orbital 453 is HOMO and 454 is SOMO. Now to the excited states. In the output I found, for example, the following piece that looks strange to my eyes

Excited State   3:  2.139-B2     0.9431 eV 1314.65 nm  f=0.0816 <S**2>=0.894
453B -> 454B       0.98895
453B <- 454B       0.17190

The first line is fine: it says that the third excited state is $^2B_2$ (taking spin contamination into account).
The second line is also fine: it tells that the determinant corresponding to excitation from HOMO to SOMO has the largest coefficients in the CI expansion.
But what is the meaning of the third line?

Time-dependent DFT can be used to predict excitation energies through a linear-response formulation.
In this Gaussian result, beyond the first line, you are looking at the largest coefficients in the configuration-interaction (CI) style expansion.
(It's not strictly CI, but the implementation of time-dependent HF or RPA is essentially the same for TDDFT or Tamm-Dancoff Approximation TDDFT.)
Anyway, since you're performing a calculation on an open-shell system, the formulation for the state will include all one-electron excitations AND all one-electron de-excitations. That state has a de-excitation from 454B into 453B (i.e., the SOMO will fill the hole left by the excitation). Now you may think "but that doesn't accomplish anything." Remember that Gaussian only prints the dominant coefficients.
For this reason, Rich Martin created the Natural Transition Orbitals method as a nice way to express the excitation, rather than a set of coefficients and the ground-state orbitals.

Oftentimes, there is no dominant configuration in the list of excitation amplitudes, thereby making a straightforward interpretation of the excited state difficult. This is particularly unsatisfactory when attempting to determine the qualitative nature of an excited state. An additional complication in the DFT case is that in principle all orbitals in DFT but the HOMO are devoid of physical significance. However, chemical intuition is built on the orbital construct, and a simple orbital interpretation of "what got excited to where" is important.

This is available in Gaussian from the Population keyword and in other codes.
Incidentally, there's a nice review of TDDFT in: "Progress in Time-Dependent Density-Functional Theory" Annual Review of Physical Chemistry
Vol. 63: 287-323.

I spent almost the week reading about RPA here and there, and as far as I understand your are right: "de-excitations" (as well as "excitations") are involved in the wave function expansion in RPA. And I simply did not know that.

@Wildcat I don't think I've ever seen a case where there's an excitation and a deexcitation from the same states, but I think NTOs would help to understand what's going on. The "minor" contributors are probably important.

NTO seems to be quite interesting, but so far I decided to do calculations within Tamm-Dancoff approximation in which "de-excitations" are neglected.

I'd do some benchmarks if I were you. Things may have changed, but when I benchmarked TDA vs. TDDFT, full TDDFT won. Granted, that was on closed-shell systems, but if possible, I'd test out some experimental systems.

Sure. I'm doing some benchmarks right now. Luckily we have experimental UV-vis spectra for the molecules in question, so I can benchmark TDA approximation vs. full TD-DFT, as well as different functionals/basis sets/ECPs combinations.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What is the difference between absolute and relative configuration in stereochemistry?
Can you help me understand the difference between absolute and relative configuration?

Let's say that we have synthesized a new compound, C(1)(2)(3)(4), where (1)-(4) represent 4 different groups attached to the central carbon, so the molecule is chiral.
Let's say that we find a method to separate the two enantiomers present in the new compound and that we placed the enantiomers into separate test tubes.
Next we might take each enantiomer by itself and place it in a polarimeter.  We would find that the material in one test tube rotates light in the (+) direction and the material in the other test tube rotates light in an equal, but opposite direction (-).
At this point, all we know is that we have two enantiomers that, as expected, rotate light in equal and opposite directions.  We don't know which test tube contains the "R" enantiomer or which contains the "S" enantiomer.  We just know their relative configuration with respect to each other, they are mirror images of each other.
To find out which is "R" and which is "S" we would either need to 1) put a crystal of one of enantiomers in an X-ray diffractometer and get a three-dimensional picture of the structure of the molecule from which we could see whether it is the "R" or "S" enantiomer or 2) carry out a series of chemical transformations where the stereochemical consequences of each step are known (e.g. does each step proceed with inversion or retention of configuration at the chiral center) and convert it to a compound of known absolute configuration
Let's say we pick one of our test tubes and are able to convert that enantiomer to a chiral compound C(A)(B)(C)(D) where we know that R-C(A)(B)(C)(D) rotates light in the (+) direction.
We find that the C(A)(B)(C)(D) we created from our C(1)(2)(3)(4) enantiomer rotates light in the (-) direction; therefore we have produced the "S" enantiomer of C(A)(B)(C)(D).
For simplicity, let's say that all of the reactions we used in our conversion of C(1)(2)(3)(4) into C(A)(B)(C)(D) are known to proceed with retention of configuration.  Then we know that the test tube of C(1)(2)(3)(4) used in our reaction contains the S-C(1)(2)(3)(4) enantiomer. Now we know the absolute configuration of our sample.

EDIT: d, l and D,L
The lower case "d" and "l" can be used interchangeably with "+" and "-" respectively.  The letters stand for dextrorotatory and levorotatory.  They simply indicate how the plane of polarized light will be rotated when a sample of the material is placed in a polarimeter.  So l-2-butanol is the same as (-)-2-butanol.  Use of "d" and "l" is discouraged because it can create confusion with "D" and "L".
Upper case "D" and "L" are somewhat archaic.  Way back when, the two enantiomers of glyceraldehyde were separated, purified and placed in a polarimeter.  One enantiomer rotated the light in a dextrorotatory manner; this enantiomer was called D-glyceraldehyde.  The other enantiomer rotated light in the levorotatory direction and was referred to as L-glyceraldehyde.  If you take what ever chiral molecule you are studying and convert the chiral carbon to glyceraldehyde, and it rotates light in the "+" direction, then you've made D-glyceraldehyde, and if all of your reaction steps involved retention of configuration (see above) then you would have the "D" enantiomer of your material. The terms "D" and "L" relate whatever chiral center you are studying to the chiral carbon in glyceraldehyde.

You should mention that the (+) isomer can be labeled D (for dextrorotatory).

@LDC3 I just wanted to keep it focused on the OP's specific question.

But it does point out that the D and L labels are applied to relative enantiomers. In fact, they can sometimes be applied to a series base on one compound (all natural amino acids are labeled L even though some have dextrorotatory rotation).

Thank you very much, that was a very useful answer. Just a question regarding the comments. Does D- (capital D) and capital L- mean dextrorotary  and leverorotary? I thought it was small d and l. Would you be able to explain the difference?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why does tomato soup "explode" when boiled?
When you boil tomato soup it very easily "explodes" when reaching the boiling point. Why is that?
A related question has been answered here before: Why do some solutions eventually stop boiling and start "popping" as they get more concentrated?
However, the answer to that question does not fully answer the current question since it does not address what properties in a solution causes superheating/bumping. That answer describes what we see but does not describe WHY it happens (except for the words "heated too rapidly", but many solutions can be heated with the same speed as tomato soup without "exploding").
For example, to my knowledge tomato soup only suffers the risk of "exploding" THE FIRST TIME it reaches the boiling point. Therefore I have thought the reason for the explosion could be that it contains gases that leave the solution violently. (I have read this is the reason that milk also may "explode" at the boiling point.)
Friends if mine suggest that the reason for the explosive boiling is the viscosity of tomato soup. But this answer would not explain why tomato soup only risks "exploding" the FIRST time it boils. (Without extensive experience I would also say that NOT ALL viscous solutions boil explosively.)
Tha tomatoes are naturally peeled so their elastic peel/membrane does not come into account.
The previous question also relates to gradual concentration which is not the issue of the current question. The current question relates to reaching the boiling point.
So this question is not answered. To rephrase it slightly:
What properties in tomato soup causes it to boil "explosively"?

I think it's probably a combination of things you mention. First, tomato soup is a suspension of solids (tomato tissue) in water. So it's the water that's boiling. That water is turning into steam, which bubbles out of the water. The solid suspension likely traps some of that steam until the pressure builds and you get a big explosive bubble.
I happen to cook, and make a fair bit of tomato sauce, which is very similar to tomato soup. I disagree with your statement that tomato soup explodes only once upon boiling. I see bubbles popping quite continuously, and you have to stir often to prevent this. Which happens to also be the way you prevent bumping in laboratory conditions. Cooking is probably the most comparable chemistry application most people are exposed to.

Thanks. Sounds very reasonable and in line with other information I have found.


text:
NH2- and Primary Alkyl Halides: Sn2 vs E2
When we mix 1-bromopropane and $\ce{NaNH2}$, what happens? Which reactions wins out to make the major product?
My textbook says that primary alkyl halides give mainly SN2 products unless the base is hindered ($\ce{t-BuO-}$) (where E2 would win out).
But it contradicts itself when you have for example 1,2-dichlorobutane and you add three molar equivalents of $\ce{NaNH2}$ and then $\ce{NH4Cl}$ to have an alkyne. Why is there no SN2 going on in that case?

IIRC, Sodamide is generally insoluble, except in liquid ammonia.  And why do people like to prepare amines with ammonia as a nucleophile?  The basicity complicates things, as does the fact that the primary amine initially formed will react further to give secondary and tertiary amines, even quaternary ammonium compounds.  It's not a reaction useful in preparative chemistry.
That said, the balance of nucleophilicity and basicity is also influenced by the solvent.  It's well known that KOt-Bu acts mostly as a base in THF and mostly as a nucleophile in DMSO, there's plenty of examples in the literature.  Hydrogen-bonding solvents tip the balance towards elimination.

So, that means that the major product will be the alkyne because the immensely great basicity of NH2- will win out even though there could be some NH2- that could do Sn2?

My gut feeling is that from 1-bromopropane and NaNH2 in liquid ammonia you'd get mostly propene.


text:
Rate Expression for Reactions Involving 2° Benzylic Alkyl Halides

My teacher took off marks because he said the reaction was SN1. Before I go confront him, please tell me if my reasoning is correct.
So, basically, I thought: we have two possibilities:
 1. SN1: a 2°benzylic has the same stability of carbocation as a tertiary carbocation so, SN1 could work.
 2. SN2: we have a secondary carbocation and a weak base cyanide: SN2 could work.
So I just randomly put the rate law for SN2 since the problem did not specify SN1 or SN2 and there is no way I can know which mechanism is prominent.

The fact that a benzene ring is attached to the potential carbocation center pushes the mechanism to the SN1 side.

Ron, why? How can you qualitatively assess that? It is still a secondary carbocation and Sn2 will happen. How can you guess having Sn1 will be larger Sn2?

The phenyl group makes a big difference, it is  now a secondary benzylic carbocation.

When I consult Solomons Organic Chemistry textbook, it says that secondary benzylic carbocations can react by SN2 or SN1. It does not say which one prevails.

@ron any idea why?

Nope, time to ask your teacher I guess.

Ok. Thanks a lot. I had a similar line of thought. It is a secondary benzylic carbocation, but it's still an excellent candidate for SN2! So I was confused.

Maybe someone else could help clarify, although if you can't, other people may probably not too.

One more thought, Cl- is an excellent leaving group (see the link I supplied in your "Better Nucleophile" question), further tilting things towards SN1.

Even then, if Cl- is a good leaving group, it will also favour SN2 too no?

Read the link, it will explain.

My inclination is that you are correct and the reaction proceeds by an SN2 reaction. Secondary alkyl halides can react by either an SN1 or SN2 mechanism, so the other reaction conditions are important to make a decision of the predominant pathway. Since cyanide is a strong nucleophile (and often utilized as the prototypical SN2 nucleophile), that suggests that the reaction should go through an SN2 pathway. One other piece of information that we don't have is the solvent. If this is run in a protic solvent, that could tip the pathway toward SN1, while polar aprotic will favor SN2. I also would not be surprised if both pathways are operating.
It is true that the aromatic ring will stabilize a carbocation at the benzylic position, but it will also stabilize the transition state of the SN2 reaction.
A brief literature search has revealed less insight than I expected. The closest result I have been able to find is that an optically active benzylic bromide treated with trimethylsilyl cyanide in acetonitrile (polar aprotic solvent) does give an optically active benzylic cyanide, suggesting SN2.
DOI: 10.1055/s-0030-1260187


text:
Why do glucose, galactose and mannose differ in solubility in water?
I don't understand why they would be so different, when the only structural difference is the orientation of hydroxyl groups.

The thing is, they are different. They might be similar, but they are different. Anything that is different will behave differently.

The solubility of these sugars will depend on the relative energies of the crystalline form versus the dissolved form of the sugar. Without knowing the specifics, one could surmise that the crystal structures of these three sugars may differ substantially from each other - one may be more tightly or efficiently packed into a crystal lattice than another due to the specific shape of the molecule, resulting in a lower energy state than that of the other sugar's crystal form, and hence giving a lower solubility in water.


text:
What is the mechanism for demercuration?
The following image is from 'Organic Chemistry Second Edition' by Clayden, Greeves, and Warren.

The final step is the demercuration, but I'm unable to find a definitive mechanism.
The wikipedia page lists a mechanism sourced from the Journal of the American Chemical Society:
Wikipedia Page for Oxymercuration
But I've found videos on YouTube which outline an SN2 displacement of the acetoxymercuri group by hydride. e.g.: Mechanism of Oxymercuration Demercuration (new)
So I would like to know which mechanism is correct.

Apparently, the current assumption implies formation of $\ce{R-Hg-H}$ intermediate, followed by radical chain decomposition This, however, should be taken with a grain of salt, i.e. very skeptically. Proving mechanism is an incredible amount of work, and considering relative unimportance of organomercury chemistry in modern organic synthesis, it is unlikely any considerable resources will be allocated to the task.


text:
Assigning 1H NMR spectrum and predicting peak coupling patterns for 2H-1-benzopyran-2-one
I have a the proton NMR of 2H-1-benzopyran-2-one (2H-chromen-2-one):

There are 6 individual hydrogens and none of them are equivalent. In the spectrum I have been given there are only four peaks.
I have been able to assign two double peaks at chemical shifts 6.43 ppm and 7.73 ppm to the two hydrogens on the alkene in the ring on the left. This leaves the four hydrogens on the benzene ring which should all be non-equivalent, but this spectrum shows them as two 2H peaks. Does this mean they have overlapped, and if so please help me figure out how. Should the multiplicities have been two ddd for the two H's in the middle and then the two H's on the ends should be a dd or ddd?
This is the expansion of those two 2H peaks in my spectra shown below.

Don't you have an integral for this spectrum? This would show you clearly that you have two sets of overlapping signals, just as you have surmised.
Looking at the numbering for your molecule described:

You are correct with the assignment for the vinyl protons. Proton E will be strongly downfield shifted, and F only moderately shifted from the base double bond chemical shift of 5.25. These values could be determined using additive tables found in lots of NMR textbooks.
The proton shifts for the aromatic ring can also be roughly calculated using additive tables, but what you have are indeed overlapping signals; Protons A and C are at ~7.3 and protons B and D are at ~7.55 ppm.
To predict the splitting patterns you anticipate for the aromatic system, consider first the general couplings you'd expect:

ortho coupling ~7-8 Hz
meta coupling ~2-3 Hz
para coupling ~1Hz or less

Let's ignore the para couplings for simplicity - they are not often seen.
So,

Proton A will have one ortho coupling and one meta coupling, so a will be a dd
Proton B will have two ortho couplings and one meta coupling, so will be a ddd
Proton C will have two ortho couplings and one meta coupling, so will be a ddd
Proton D will have one ortho couplings and one meta coupling, so will be a dd

Let's now build in double bond :

Protons E and F have a cis coupling (~10 Hz).
Proton F will appear as a doublet, but proton E is likely to have a homoallylic type coupling to proton D (4J), splitting both of these signals to an additional doublet with a coupling of about 1-2 Hz. This means that proton D may be a ddd, and proton E a dd if this coupling is observable.

So that should be the expected splitting pattern from this molecule. Now to justify the appearance of the peaks. Overlapping signals can often make things difficult to interpret. Without knowing what frequency this spectrum was recorded at, you also need to consider whether there are second order effects starting to manifest, certainly with the meta couplings. There will definitely be tenting of the peaks giving rise to distorted peak intensities from what would normally be expected.
So, I wouldn't be devoting too much time to the interpretation of overlapping signals like these. One thing I will say is that if you label the peaks as (say) a ddd, then you should be able to report all couplings for each d. Otherwise, if you cannot report couplings (and this means couplings, not splittings) report them as a multiplet (or in this case overlapping multiplets).
Complex splitting patterns of second order and overlapping signals can be analysed and simulated, but it takes a lot of work.You can try drawing out a splitting diagram for each signal using the approximate couplings for ortho and meta couplings, and then start overlapping through constructive addition to see what you get. It gets very complicated very quickly.

How do you know that proton A couples with C and proton B couples with C. Why doesn't A and B couple and C and D couple?

You are confusing coupling with chemical shift. A and C have very similar chemical shifts, and their signals overlap. B and D have very similar chemical shifts, and their signals overlap. Chemical shifts of proton signals can be predicted using additive tables found in many NMR textbooks, among other places. The chemical shifts are a product of relative o/m/p substitution around the ring. It just so happens in this case that A and C have similar chemical shifts, as do B and D. I predict this using additive tables (and experience). I would confirm it using other NMR experiments.


text:
In a reaction if the equilibrium constant K was infinity, could you produce an infinite number of products at equilibrium?
It's impossible to produce an infinite amount products if the k constant was infinity?
no right?

Leaving aside the fact that you'd only have an infinite equilibrium constant if the denominator was exactly zero (i.e., not just very small, but zero), the amount of product you produce will be limited by the amount of starting material. At most, you could only produce 100% molar yield.

but at equilibrium, wouldnt the reaction constantly go forward and backwards and never end?

As stated in the answer, an infinite K would mean that the forward reaction will go to completion and the backwards reaction will never occur.

It's impossible, law of conservation of mass and energy immediately comes to mind. All an infinite K value means is that the forward reaction will go to completion.


text:
Does Combining Solid/Liquid Concentration with $K_c$ Matter when Comparing $K_c$ of Different Reactions?
I know it is convention to take liquid/solid concentrations as constants and lump it with the equilibrium constant. However, if we were trying to compare the equilibrium constants of two different reactions and if one of the reactions had a solid reactant, doesn't lumping the solid reactant concentration with the equilibrium constant make one reaction seem more product favoured than it actually is?

The equilibrium constant of a reaction containing solids or liquids doesn't contain hidden values for their concentrations.
The issue is that we use the concentration (or pressure) of a substance in the equilibrium expression because it is a shortcut.  What we are actually putting into the expression is the activity of that substance, a more complicated idea.  It very often reduces to the ratio of the concentration of a solution to the concentration at standard state.  Since the concentration of a solution at standard state is 1M, there's no point in dividing through by 1 for each term.
For solids and liquids, the concentration at standard state is its molar density. ($C = \frac{mol}{L}$)  This value doesn't depend on the quantity of substance available, so the ratio between the density at standard state and the density at any point in a reaction is 1.  The values ARE in the equilibrium expression; they're just all 1's.
As to your concern about comparing equilibrium constants, we can only do that directly for reactions with the same number of reactants and products.  Adding (or removing) any term to either side interferes with the comparison.
Compare:
$\ce{A(aq) +B(aq)->AB(aq)}$
$\ce{2A(aq) +B(aq)->A2B(aq)}$
We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.
It's just one of the limitations of the equilibrium constant.
Check out the wiki on Thermodynamic Activity for more information.


text:
Acidic nature of benzene and other alkenes
Is benzene acidic? It has $sp^2$ hybridization so doen it mean that it is acidic in nature? Are alkenes acidic?

A sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon, and you would expect it to be more acidic.
Experiments support this prediction.  Benzene has a pKa of ~ 43, compared to alkanes, which have a pKa of ~ 50.  You can prepare lithioaromatics from bromo/iodoaromatics and butyllithium via metal-halogen exchange.
The situation with alkenes is slightly more complicated; the pKa of vinylic protons is in the same ballpark as of aromatic protons, but allylic protons are slighty more acidic.
Consult any of the usual pKa tables on the web.


text:
What is the difference between configuration and conformation in stereochemistry?
I understand conformation to be any spacial arrangement of atoms that can be achieved through rotation of a single bond. I understand configuration to be the disposition of atoms/groups in space. However, would a conformational isomer also be a configurational isomer since the groups are arranged differently in space?

I recommend you this webpage: http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html.
Configuration and conformation are two concepts are often used interchangeably, but represent very different things. The conformation regards the relative spatial orientation of a portion of a molecule relative to another. Thus, it is an aspect that is not directly related to the covalent bonds that are established within the molecule, but with their possible rotation.

When we talk about rotation around covalent bonds, we are only referring to the single bonds, as they are the only ones that can suffer rotation. Basically this concept is easily understood if we think that the bounds work as an axis...
It should be noted that when we speak of different conformations, it does not necessarily involve all the covalent bonds of a molecule, it can account only for one or few of them.
 Taken together, it is possible to convert one conformation to another without cleaving or forming chemical bonds, simply by rotating some simple covalent bonds. Thermal energy at room temperature is sufficient to rotate some simple covalent bonds.
The configuration is a concept that is related to the order by which different substituents linked to the same central atom establish covalent bonds. That means, in this case it is clearly an aspect that is a direct consequence of the covalent backbone of molecules.

To change the configuration, you must always cleave and form new covalent bonds...This costs a lot of energy!!
In conclusion, the concept conformation encompasses portions of a molecule which are not directly linked to the same atom and do not involve the covalent backbone of the molecules, while the configuration comprehends parts of the molecule which are bound to the same atom, which means that there is a direct involvement of the covalent bounds of the molecule.

Configuration is arrangement of atom in three dimensional space around the molecule. Here there is no free rotation of atom usually we use configuration in optical isomerism and conformation is the one which allows free rotation of single bond with in  the molecule and we use conformation to represent the biomolecule structure.

Configuration differs from conformation by a fact that conformations are only due to the orientation of the atoms in molecule around the covalent bond considering the bond as an axis.  And changing the configuration causes the cleavage and formation of new chemical bonds.

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


text:
How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes
It's easy to apply CIP rules to chiral carbons but what about in chiral molecules without these carbon cneters like biphenyls, sulphoxides and allenes? Are they applied at all?

I think this Wikipedia article might answer your question.

Structures with axial chirality (e.g. an allene or a 2,2′,6,6′-tetrasubstituted biphenyl) with two pairs of substituents (ab and cd) are regarded as elongated tetrahedra and viewed along the chiral axis. The substituents (a, b, c, and d) are arranged in the elongated tetrahedral system.
In each pair (ab as well as cd), one higher-ranking substituent (a > b and c > d) is chosen using the sequence rule.
If the path going from a to b to c (while looking toward d) is clockwise, the chirality is described by the symbol Ra. The symbol is Sa if this path is anticlockwise.


text:
What makes carbon special and versatile?
My teacher told me that carbon's tetravalency and high catenation ability makes it special and is the reason why there are millions of compounds of carbon.
1) Silicon is tetravalent too but doesn't create as many compounds as carbon.Why?
2) The textbook says that no other element exhibits the property of catenation to the extent of carbon. Why is this so? Does it just happen magically?

There are two key factors that account for the ubiquity of carbon compounds.

Bond strengths: Look at the following table of bond strengths and notice how the strength of both carbon-carbon single and double bonds is much greater than the bond strengths found in other molecules.

\begin{array}\hline
Typical~Single ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C-C & 83    \\ \hline
N-N & 38    \\ \hline
O-O & 35    \\ \hline
Si-Si & 52    \\ \hline
P-P & 50   \\ \hline
S-S & 54    \\ \hline
\end{array}
\begin{array}\hline
Typical~Double ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C=C & 146    \\ \hline
N=N & 109    \\ \hline
O=O & 119    \\ \hline
P=P & 84   \\ \hline
\end{array}
link to reference
These numbers tell us that carbon-carbon bonds are significantly more stable than other bonds and that the driving force for their formation will be more exothermic.  Therefore, large collections of carbon atoms can be bonded together to produce very stable systems.

Hybridization: Saying that carbon is tetravalent doesn't fully convey the versatility carbon can display in forming bonds. Hybridization involves a re-mixing of s- and p-orbitals in response to bond formation and this leads to the formation of more stable bonds. Carbon is capable of extensive variation in the hybridization of the bonds it forms, much more so than most other atoms.  This wide range in hybridization that is possible for carbon allows it to form much more stable bonds in a wide variety of different structural situations.
\begin{array}\hline
Typical~Carbon-Carbon ~Bond~ Strengths~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
alkane & sp^3 & 84    \\ \hline
alkene & sp^2 & 150    \\ \hline
alkyne & sp & 200    \\ \hline
\end{array}


text:
Why does graphite conduct electricity?
The answer i know is delocalized electrons. But how?this site tells me that:
My textbook  tells me that one carbon atom is bonded to three others and two of these bonds are single while the third is a double bond.So all the electrons are bonded and there are no free moving electrons right? Where do these delocalized elctrons come from?

Conjugation delocalises pi electrons.

Carbon has four spare electrons. In graphite structure, two of them make two single (sigma) bonds while the other pair form a double bond (consists of one sigma and one pi bond). Electron that forms pi bond is freer to travel between atoms compared to the others and is theoretically associated with all adjacent atoms (imagine the 3d structure of graphite) rather than a single atom. Because of that, it is called delocalized electron. In other words, electrons that form pi bonds continously replace each other causing temporary, local crystallizations within the structure. According to Wikipedia "It is increasingly appreciated that electrons in sigma bonding levels are also delocalized". But it is not the subject we are discussing here.
Electric current is basically the flow of charged particles (ions and electrons, but electric currents in everyday life can simply be defined as flow of electrons). If graphite is exposed to an electric current in a circuit, "the pi electrons" which are already traveling between carbon atoms that forms the graphite structure begin to flow through the circuit and leave the structure from one point of connection between graphite and the circuit. At the same time, other electrons come via the other connection point to replace them as new delocalized electrons. So, the new delocalized electrons simply come from the current.

I am not familiar with the concepts of pi and sigma bonds. We haven't gone that far in school. I'm sorry but is there simpler way of understanding this?

@SmCcK  You might want to read (or study) a bit about orbitals too. Check these:  http://www.chemguide.co.uk/basicorg/bonding/benzene2.html#top http://www.chemguide.co.uk/basicorg/bonding/orbitals.html#top http://www.chemguide.co.uk/basicorg/bonding/ethene.html#top  http://en.wikipedia.org/wiki/Atomic_orbital


text:
Grignards do transmetallation?

Grignards don't add to alkyl halides to create linear carbon extensions because Grignards do transmetallation rather than SN2 displacement of halide.

Is this correct?
I thought transmetallation a) involved two metals and b) involved ligand exchange between the two metals … correct me if I'm wrong, but if I have a Grignard and an alkyl halide, I only have one metal.

I am not sure what the context of the question that your prof replied to was, but this (calling a metal/halogen exchange as transmetallation) is a pretty big mistake for en experienced chemist with one word "arghhh". I'm sure it was a slip of the moment (probably you bombarded him/her with questions -well done)

"Transmetallation" and "metal-halogen exchange" describe two different (ie not the same) and distinct (ie they don't partially overlap) chemical phenomena so you cannot use them interchangeably

Is Dr J happy for direct transcripts of your communications to be posted and scrutinized on a public forum? You don't want to burn any bridges.....

Prof told me Grignards don't add to alkyl halides

Grignards can react with alkyl halides to form the expected coupling product.  However yields are usually very low except under special conditions.  Improved yields (but yields that would still be considered poor for a synthetic route) result when the halide involved is a good leaving group (I ~ Br > Cl), or when the Grignard is made from a stabilized "R" group such as allyl or benzyl.
Here is an interesting example where the reaction of an aryl Grignard reagent with an alkyl halide was examined.
"When a THF solution of p-tolylmagnesium bromide 1a and an equimolar amount of 1-iodoheptane under nitrogen was stirred for 48 h at room temperature, only a very small amount (7% by GC) of the expected coupling product, 1-p-tolylheptane, was formed."
So in principle, Grignard reagents are capable of reactions that look like $\ce{S_{N}2}$ reactions (who knows what the mechanism is when a metal is involved), the yields are just very low.

How does stabilizing the R group affect the Grignard overall?

Allows the ionic form of the Grignard to be more significant.

You'd think that reaction of an alkyl grignard with a (primary) alkyl bromide/iodide gave all sorts of grignards via metal/halogen exchange.  I guess that's what your professor means by transmetallation.  You'd also get some elimination products.  Certainly that's what would happen with alkyllithiums.
That's why for carbon-carbon bond formation reactions you'd use cuprates.  Reduced basicity, increased nucleophilicity, all that.

Is metal/halogen exchange transmetallation? EDIT: okay, I saw your edit.

I wouldn't call it that.  Someone else may, though.  I would use the word transmetallation only for the reaction of a lithio/magnesio-organic compound with a more electronegative metal, such as zinc, copper, or boron.

Why would there metal halogen exchange?

As for metal-halogen exchange, you'd think where are -ate complexes.  Also, think Schlenk equilibrium.


text:
Most effective means of removing coffee stains from carpet
I have several old coffee stains on a carpet. Googling "coffee stain removal" reveals a variety of suggested cleaning substances-- vinegar, baking soda, laundry detergent, etc. I have used all of these on stains in the past to varying success, but always have a hard time totally eliminating the stain. Typically the instructions suggest iterative applications of the substance, and I usually end up with a huge pile of wet, wadded-up paper towels, each tinged slightly brown from the tiny amount of coffee absorbed.
What is the most effective way to remove old coffee stains from carpet using common household chemicals? Chemically, why is this the most effective means?

How long did you leave the chemicals on the stain? If the stains are quite old, the colour might be well in the fabric and only bleaching them will eventually remove them. Some cleaning agent which releases oxygen in the process should do it. However, usually after that procedure you have the problem that the part where your stain was is much cleaner than the rest of the carpet.

Coffee is brown either because it contains tannins or melanoidins (multiple sources site each...)  Both are large organic molecules, so if you can break them down into smaller organic molecules you should be able to decolorize them.  The easiest way to do that is as @Martin describes, via oxidation.  You might be able to do it with something like OTC hydrogen peroxide even (or OxyClean).  I'm not sure enough of the method to write this as an answer, but it's worth a try.

Oxidise it using a mild oxidizing agent like $\ce{H2O2}$ (hydrogen peroxide). When it undergoes oxidation coloured compounds get decolorized. Good luck on cleaning it.

Be sure the fabric is colorfast before you use the Hydrogen Peroxide on your carpet.
In trying to get out coffee stains I ended up discoloring my carpet.


text:
Concept of heat of reaction in a isolated system
One description of the concept is that heat of reaction (exothermic) is the quantity of heat that would given off to the surrounding by the product of a reaction if we allow the product to cool down.
If this description is valid for all exothermic reaction in a isolated system, would it also be valid for all endothermic reactions??

When chemical bonds are broken potential energy increases.  When they are made potential energy decreases. A loss in PE raises kinetic energy and a gain in PE lowers KE.  Since all reactions involve bond breaking and bond making the reaction will involve an increase or decrease in PE and a decrease or increase in KE.  When KE of the system increases, a system in contact with the surroundings will lose energy as heat (exothermic).  When KE of the system decreases a system in contact with the surroundings will gain energy as heat (endothermic).  In an isolated system "exothermic" reactions just raise temperature and "endothermic" reactions just lower temperature.

Wrong, potential energy would increase when bonds are formed. More heat capacity

Actually I kinda get it: if the reaction is endothermic, you can't really cool it off because the energy is used for bond forming, right?

No.  Bond formation is spontaneous compared to separate atoms.  Think of the reaction as the breaking of all the bonds in the reactants that aren't in the products.  The fragments then form the bonds that are in the products but not in the reactants.  It does NOT take energy to make a bond any more than it takes energy for an object to fall from a height to the earth.


text:
Electron affinity
Concerning the liberation of energy when an atom that is close to the configuration of a noble gas:
Where does the energy dissipated from an atom upon receiving an electron come from?

Well everything in nature wants to or tries to attain a configuration where it is at its lowest energy.
When an atom receives an electron and completes its octet it becomes more stable or in other words we can say that it has acquired a stable configuration.
The energy released was always present in the species before bonding took place.

In which form is that energy present?

in the form of potential energy.breaking a bond is endothermic.
to break a bond you need to provide energy and that energy is stored as potential energy.when that species reacts to form a bond then that energy is released that is we say that bond making is exothermic

The electron is attracted to the nucleus but repelled by other electrons.  In some atoms, adding an electron is an exothermic process because the potential energy of the lone electron and the atom apart are converted to kinetic energy.  In these cases the attraction of the nucleus overcomes the repulsion of the electrons.  Not all octet forming is exothermic.  The second electron affinity of oxygen is endothermic for example.


text:
Synthesis; effect of hydrogen gas and platinium

I'm pretty sure the last step in there is just something to throw us off. From what I understand, the first step will give us a Grignard, which will attack the carbonyl in formaldehyde to give us a primary alcohol, and the workup is just to neutralize any remaining Grignard and of course protonate the oxide ion that was just formed. I don't see any possibility of an intramolecular attack either ......
The H2/Pt step looks like a serious red herring from here; I know it doesn't touch the phenyl substituent. And there are no other pi bonds in this system for me to even consider. So is the last step just something to throw me off?

Is a beta-alkoxy grignard stable at all?

For the last step, your textbook should set you on the right track, just look up the table of functional groups reduced by catalytic hydrogenation.

It reacts with formaldehyde.

But there are no PI bonds to reduce.

debenzylation hydrogenolysis?

Dammit! I remember now!

The rate of reduction of benzyl ethers is faster than that of the double bond.

The selective debenzylation of benzyl ethers can be carried out under very mild conditions, leaving other functional groups intact. Commonly carried out under H2 using a heterogeneous catalyst such as Pd/C or Raney Ni, Pt can also be used. Solvent choice is critical, and reactions times are longer, but it is still effective.
I don't like red herrings.

Looks like the product is ethylene glycol!


text:
Sodium cyanoborohydride and iminium reductions

I understand that the LAH will reduce the nitrile into a primary amine. The addition of acid will likely destroy the acetal mask and leave us with ethylene glycol and a molecule with both a ketone and an amine functional group.
However, what does the sodium cyanoborohydride do here with the acid and heat? I'm again thinking this is a red herring; $\ce{NaBH3CN}$ only reduces iminium ions as far as we are concerned... and we have an amine. Amines don't exist in equilibrium with imines.

Still don't like red herrings. Cyanoborohydride is more useful than you think.

I just learned it reduces imines ...

I believe the reaction effected in the third step is a reductive amination. In this case, the ketone and amine moieties are both present within the same molecule, so the reaction proceeds intramolecularly. The $\ce{TsOH}$ is an acid, presumably serving to protonate the carbonyl oxygen and activate the carbon toward nucleophilic attack by the amine. As the iminium group forms (with concomitant loss of water), it is rapidly reduced to an amine by the $\ce{NaBH3CN}$. After workup, the final product is:

Cyanoborohydride has a number of functions as a selective reducing agent, which are very much pH dependant:

Reduction of aldehydes and ketones. At low pH (3-4), it effectively reduces aldehydes and ketones to the parent hydrocarbon.
Reductive amination of aldehydes and ketones. At pH 6-8, aldehydes and ketones undergo reductive amination
Reduction of oximes, at low pH

Have a close look at the pH of your system.


text:
How does H2O2 bond
I am asking this question on behalf of my seven year old daughter, so if possible please take that into consideration in the answer!
I am not a chemist, but I do try to teach my daughter the basics of diverse subjects. I explained to her that the right-most elements on Mendeleev's chart do not generally react with other elements as they have all the electrons that they need. Those to the left of them need a single electron to be satisfied, those another column to the left need two electrons, and so on. Additionally, those elements on the left of the chart have an extra electron that they want to be rid of, and those one column to the right of them have two electrons to get rid of.
On this basis she understands why water has a single oxygen atom, but two hydrogen atoms. Then she asks my why hydrogen peroxide has only two hydrogen atoms. What satisfies the second oxygen atom? I don't know myself, I'm not a chemist!
Does hydrogen peroxide use a different type of bond? Is the electron-swapping somehow satisfied for hydrogen peroxide in a different way?
Googling around I discovered hydrogen bonding but this doesn't seem to work with Oxygen. What else should I be reading? I've googled and found many "Chemistry for Kids" websites, but they seem to be either (fun) experiments, or explanations that are no more simple than those in wikipedia and obviously not aimed at children.

Tenuously. That's why it's such a good oxidizer (remember, the household stuff is very dilute but can still bleach your hair), and if you add more O-O bonds, it gets even nastier.

@PhilFrost: Thanks, that is an interesting link. Interestingly, the only other time that I had occasion to visit that site was for another material that also doubles as a rocket fuel!

You might also be interested in the Crash Course Chemistry (https://www.youtube.com/playlist?list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr) for educational purposes. It gets into high school/early college level stuff, but depending on your daughter it might be interesting anyway.

@MatthewScouten: Thank you, but those videos are terrible! I couldn't get through 5 minutes. I suppose that they are designed for someone with no appreciable attention span; thus the gimmicks, fast talking, and cut scenes every time the host takes a breath between sentences. I do appreciate the suggestion though, thanks.

Time's up for you to explain Covalent Bond to your daughter.
Getting rid of, or capturing an electron (ionic bond) is not the only way that atoms can get their outermost electron shells filled. If it were the only way, we would not have H2 or O2. Atoms can share an electron (or more) with another atom, if that fills the shells.
For example, H2 does not bond by giving one Hydrogen atom's electron to another. When two Hydrogen atoms get together, they share the two electrons between them (little dots denote electrons).
Before bonding                   After bonding
H ·      · H                          H : H

Now, each Hydrogen atom has two electrons to enjoy, and their shells are (sort of) filled.
Similarly, H2O isn't made by Oxygen capturing Hydrogen's electrons. The H--O bond is a covalent bond as well (hope you can see the electron pairs ¨).
Before bonding                  After bonding
         ¨                                ¨
H ·    · O ·      · H                 H : O : H
         ¨                                ¨

H2O2 has a covalent bond between the two Oxygen atoms
           ¨   ¨
       H : O : O : H
           ¨   ¨

See the structure of Hydrogen Peroxide from Wikipedia article:

source: http://en.wikipedia.org/wiki/File:Hydrogen-peroxide-3D-balls.png

Oh, thank you! Thank you for mentioning the names of both types of bonds, and explaining them exactly as I would have to explain it. One question: why does the unbounded O have electron pairs above and below? Are those just electrons that don't take part in the bonding?

@dotancohen: Yes. They are electrons that don't take part in the bonding. And they are pairs too, as you have observer. To know why, you have to understand about atomic orbitals and how electron shells fill.

@dotancohen: also note that in Oxygen, my diagrams only show the electrons in outermost shell.

@dotancohen: BTW, if your daughter is really clever, she might ask you about ozone or carbon monoxide, at which point you'll have even more fun explaining dative bonding and resonance structures.


text:
Is atom-clean aluminum active enough to react with alkali salts?
Thermally evaportated $\ce{Al}$ under high vacuum is free of oxidized species and highly active. But is it active enough to even substitute the alkali cationic metal in alkali salts, like $\ce{LiOH, NaOH, Cs2CO3}$.
Here $\ce{Al}$ is directly depositely onto the salt substrate.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Even normal metallic aluminum (with its spontaneously-formed corrosion-resistant oxide coating) reacts with concentrated sodium or potassium hydroxide to give the aluminate (e.g. NaAl(OH)4), plus hydrogen gas. Lithium hydroxide should also react (lithium is likely functioning as only a spectator ion) whereas alkali metal carbonates may not be sufficiently reactive.
This reaction is not what you refer to in your question as "substitution" reactions, where the alkali metal ion would be reduced to metallic form. Note that aluminum chloride is reduced by sodium or potassium metal to give aluminum metal, so the reverse reaction is unlikely to occur, but the energetics may be different for the hydroxides or carbonates in a non-aqueous environment.


text:
How can I estimate reaction rates in water with computational chemistry?
I want to estimate rates of substitution reaction of esters with nucleophiles in water.
How can I estimate them with a computational chemistry software (preferably GAMESS(US))?
Or is it practically impossible?
I know that this is possible in gas phase, but I don't know how to do it in solution.

Finding solvent  effect in computational chemistry is still a crucial part and subject to deep research effort. Now there are two types of models that are being used in computational chemistry.
One is explicit solvation model: Here you will use explicit solvent molecule with your reactant or products and then using first principles you will find the equilibrium configuration but the problem is, it is very expensive and the configuration space is so much high that it is practically impossible. For that you have to use molecular dynamics or monte carlo techniques. In our group we use 1000 to 5000 explicit water molecule but we can't do it for large reaction.
Another method is implicit solvation model: Here electrostatic potential is applied to account the solvent effect but this procedure is so expensive that now that potential is replaced by the solvent induced reaction field energy.
In our group we are using a hybrid approach that involves communication of several software at different level of theory. In my quantum chemistry class i used spartan and qchem and they have some implicit solvation but i didn't use GAMESS.
From GAMESS manual i found that they have solvation module. They are using QM/MM approach. That's actually a very good approach. Please see page 9 of this file:  http://www.sdsc.edu/~kimb/gamesstut.pdf
In this following file at page 3 & 4 you will also find the list of keywords that are needed to start a solvation model run
http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf


text:
Why gives NaBH4 more Felkin-Anh product than LiAlH4 gives?
In lecture we discussed the nature of the nucleophile for the attack of the carbonyl.
LiAlH4 acts as a better nucleophile (hydride) than NaBH4. I know there is no hydroboration! So I would have expected that LiAlH4 gives more Felkin Anh product as the NaBH4 gives.
However LiAlH4 gives a Felkin Anh : Anti Felkin Anh ratio of 3:1 and
NaBH4 gives a ratio of 5:1.
detailed reaction
Source: Carreira, E.M.; Classics in Stereoselective Synthesis
Is there any good rational?

Lithium aluminum hydride is such a reactive nucleophile that it reacts readily with any conformation of the ketone. The reduced nucleophilicity of sodium borohydride allows for some discrimination of the conformations.
Note that the difference in selectivity between aluminum hydride and borohydride small. A large boost in stereoselectivity can be achieved by increasing steric bulk of the reducing agent (L-selectride gives a 10-fold increase in selectivity).

Because NaBH4 is more sterically demanding than LiAlH4 the selectivity is higher for NaBH4?

I expect that NaBH4 and LiAlH4 to be similar sterically. L-Selectride (lithium triisobutylborohydride) is much more sterically demanding.

I mean there is though a small difference between 75% and 83% selectivity. Do you believe this two ratios (3:1 and 5:1) are random? I guess the experimentators have used the same measuring conditions so there must be a rational.

No, those are not random. I think LiAlH4 is less selective because it is more reactive.

yeah...this seems to be a good explanation.


text:
C=O and C=C antibonding orbitals
Why is the pi star orbital of the C=C bond higher in energy than the antibonding pi orbital of the C=O bond?
I thought it was the other way around; it's relatively easy to add something to a C=C alkene bond (i.e. halogenation) but halogenation with diatomic halogens doesn't occur with C=O bonds.

With halogenation it is not the $\pi^{*}$ but the $\pi$ orbitals that are involved in the reaction with e.g. $\ce{Br2}$ because the alkene is acting as a nucleophile here. And because $\ce{O}$ is more electronegative than $\ce{C}$ it should be logical that $\pi$ is lower in energy for $\ce{C=O}$ than for $\ce{C=C}$.

It's relatively easy to add an electrophilic species, such as X-X, to an alkene. As @Philipp points out, the alkene is operating through the (filled) pi orbital.
It's relatively easy to add a nucleophilic species to a carbonyl. In that case, the carbonyl operates through its (unfilled) pi* orbital.


text:
Metal vs. Styrofoam cup calorimeter?
For a metal cup as opposed to a styrofoam cup calorimeter, do you expect the enthalpy of the reaction to be higher, lower, or the same?
I think the enthalpy of the reaction would be the same in the sytrofoam cup as the metal cup because enthalpy is a state function? So if the same products are yielded at the end along with the same reactants being used in the beginning, they'll have the same enthalpy because it doesn't matter HOW you got that enthalpy, just the final result? That's my reasoning but I'm not sure.
So does it matter that metal cup will conduct heat better because if they're using the same reactants it'll give you the same change as a styrofoam cup?

What is your thinking on this question?  What qualities do metal and styrofoam have that differ that might alter your determination of the enthalpy of the reaction?  Please visit this page for advice on asking good homework questions.

You're right that IF the metal calorimeter and the styrofoam calorimeter were both perfect and behaved in the way that we usually pretend calorimeters do (they don't absorb or lose heat) then they should produce identical results, for the enthalpy reason you provide.
However, since we're asking about specific materials, metal vs styrofoam, we're considering real calorimeters instead of ideal.
There are a few things that calorimeters can do that will affect how large a temperature change you measure.

Heat loss - As you note, metal conducts heat much more effectively than styrofoam.  This means that the metal calorimeter will carry the heat of the reaction away from the reaction vessel to the outside world faster than styrofoam will.  If that is the case, which reaction vessel will get hotter, styrofoam or metal?
Heat absorption - The calorimeter (and thermometer) gets heated up along with the reaction mixture.  A calorimeter that absorbs large amounts of energy will have a relatively lower final temperature, since more of the reaction's heat went into heating up the calorimeter than went into heating up the reaction mixture.  This quantity is known as the calorimeter constant and varies depending on the materials and mass of the calorimeter.  Styrofoam absorbs more heat per gram than steel, but the steel calorimeter is probably heavier.  With a fast reaction there is the additional fact that the styrofoam calorimeter probably never comes up to the full temperature of the reaction before you quit recording temperatures.  This winds up leading to a small source of lost heat for the metal calorimeter as compared to the styrofoam.
The quality of the equipment -  Metal calorimeters are often made specifically for the purpose of calorimetry.  Styrofoam calorimeters are often a couple of coffee cups jammed together.  If the calorimeter allows a significant amount of heat loss to the atmosphere due to its construction, then it will result in a cooler reaction mixture and a lower calculated enthalpy for the reaction.

You've got to take all three of these into consideration in order to determine the particulars for your experiment.


text:
How can the pH value be unitless?
How can the definition of the pH value $\mathrm{pH}=-\log\left[\ce{H+}\right]$ yield a unitless result? Since $\left[\ce{H+}\right]$ is a concentration, wouldn't it have units $\mathrm{mol/dm^3}$ and therefore following the product rule you would get: $-\log\left(x\ \mathrm{mol/dm^3}\right)=-\log\left(x\right)-\log\left(\mathrm{mol/dm^3}\right)$.
How can these units in a mathematical sense suddenly disappear?

Good question. I didn't know the answer so here is a very enlightening IUPAC link from wikipedia. I guess chemists we are not very good at rigour. The pH definition could make a mathematician or a physicist cry. From the paper: "(pH) involving as it does a single ion activity, it is immeasurable". How you can introduce a definition of something that measures a common and useful quantity (acidity) that is itself "immeasurable" (what do they even mean by immeasurable?) is an interesting discussion topic.

I always thought that it was the number of H+ ions divided by the total number, so a relative ratio.  Note that the acidity should not depend on the density of the solution as it does in your viewpoint.

The reason there is a difference between the notional definition and the operational definition is that the activity of a single ion cannot actually be measured independently because it is not independent of its counter-ion. Even though it's not directly measurable, it's still useful conceptually, just as many quantum mechanical systems do not have exact solutions, numerical approximations are still of value.

As an aside, the log of a value never has a unit, even if the original value had a unit. I can't explain why, though.


text:
How are cathode rays produced?
Why cathode rays are produced if pressure of the gas discharge tube is lowered to about $10^{-4}$ atm?
How bombardment of glass of gas discharge tube from cathode ray result in a faint glow?

"Cathode rays" are simply beams of electrons. The name exists because cathode ray devices predate the discovery of the electron. The sharp shadows produced by Crookes tubes meant that something was travelling in a straight line from the cathode down the tube, so they were named cathode rays.
The vacuum doesn't have much to do with the production of electrons, per se, but without high vacuum, the electrons collide with gas particles before they travel very far. If there's enough gas, you get a plasma rather than a beam of electrons. The electron beam is produced by thermionic emission in hot cathodes—electrons are liberated from a hot filament when the thermal energy is enough for electrons to exceed the filament's work function. In cold cathodes, electrons are released by field emission, where a large applied electric field allows electrons to tunnel away from a metal electrode.
The glow of the glass envelope is luminescence—either fluorescence or phosphorescence (or both). Basically, the electrons striking the glass excite components of the glass to an elevated electronic state which releases light upon returning to the ground state. Normal glass isn't particularly luminescent so tubes designed to produce light (like a CRT) are coated with phosphors.

When a high voltage is applied to the tube the electric field accelerates the small number of electrically charged ions always present in the gas created by natural phenomena called photo-ionization and radioactivity.
These collide with other gas molecules knocking electrons off them and creating more positive ions in a chain reaction called Townsend discharge. All the positive ions are attracted to the cathode. When they strike it they knock large numbers of electrons out of the surface of the metal which in turn are repelled by the cathode and attracted to the anode. These are cathode rays.

When electric current passes through the gas inside the tube which is at low pressure, then the electric current manages to overcome the force between the electron and the nucleus and the the free electrons forms a ray which travels towards the anode in a straight line.

Cathode rays are so named because they are emitted by the negative electrode, or cathode, in a vacuum tube, also called as a cathode ray tube.
Cathode rays are just a beam of electron observed in vaccum tube. when the dicharge tube or evacuated glass tube is facilitated with 2 electrode and with high voltage, due to emmission of electron from and travelling in perpendicular to cathode,glass opposite of negative electrode(ie cathode) is observed to glow.
High electrical potential is applied between anode and cathode to ionise the residual gas in the tube. the ions are accelerated by electric field and collide with cathode which released electrons. As electrons are -vely charged they are repeled by cathode and are attracted bty anode.

Crookes tube is a rarefied tube evacuated to a pressure below 10−6 atm. It was used in discovery of cathode rays.
Source
Wiki
Link
Video

At low pressure ($10^{-2}$ atm) and higher voltage (10000 V) gases are partially ionised in the discharge tube. Positive ions of gases strike on the cathode. Due to the thermal effect, a beam of electrons releases from the surface of the cathode. This is called a cathode ray.

Cathode rays are originating from cathode as cathode is negatively charged. These rays then hit the gas sample inside the tube and ionize it. The electrons ejected from the ionisation of the gas travel towards anode. These rays are actually electrons produced from the ionisation of the gas inside the tube.

This addresses the second question, but not the first.

Cathode rays are the streams of fast moving negatively charged particles originated from the cathode.
At the pressure of 1 mm Hg in the tube, the positive column breaks into alternate and dark dics, then the cathode leaves the glow, this is cathode rays.


text:
Why is the standard enthalpy of formation of elements in their native forms zero?
Why is the standard enthalpy of formation of $\ce{O2}$ and other reference forms of elements equal to zero?
Well, from what I know, bonding releases energy.
Also, does anybody know a source for finding the reference form of an element at standard conditions?

It is purely a matter of definition. A standard enthalpy of formation describes the change in enthalpy during the formation of 1 mol of a target compound by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature. Now take the case of oxygen as an example: At standard conditions $\ce{O2(g)}$ is the most stable modification of oxygen. That means at standard conditions the reaction
$$\ce{2O <=>> O2(g)}$$
is defined in such a way that each of the elemental $\ce{O}$s are defined to be exactly the same state as an oxygen atom that is part of an $\ce{O2(g)}$, i.e. their enthalpy $2H^{\circ}(\ce{O})$ matches exactly that of $\ce{O2(g)}$: $H^{\circ}(\ce{O}) = \frac{1}{2}H^{\circ}(\ce{O2(g)})$. This gives you a reference point for $H^{\circ}(\ce{O})$ which is otherwise nearly impossible to measure much as the Standard Hydrogen Electrode is used to set up a reference system for the electrode potential.
A table showing some standard enthalpies of formation can be found here. There it is also indicated which phases are used as reference form.


text:
Which is the more stable enol form?
If I understand correctly, when we go from a keto to an enol form, we simply abstract an alpha proton and thereby create a double bond to a carbon-bearing the hydroxyl functional group.
When evaluating the stabilities of the two enol forms, it appears that the enol derivative of 2,4-pentadione is more stable because of intramolecular hydrogen bonding. Both have long-range conjugation, so that cannot be a factor. The only difference really is the hydrogen bonding ... and of course ring strain (a 5-membered ring with two double bonds can't be very stable).

I am sorry, but I fail to find the question. What are you asking here?

Which enol form is more stable @Martin?

Adding to all your points, second enol form has more number of alpha hydrogens (total 8) compared to first (total 3 alpha H). Thus more hyperconjugation is possible in second, hence second is more stable.

In acetylacetone the protons are not all equivalent.

Keto-enol equilibria reflect a delicate thermodynamic balance between the two forms.

The bonding differences between the keto and enol structures shown above are:
keto: C=O double bond, C-C single bond, C-H bond
enol: C=C double bond, C-O single bond, O-H bond
If we look up the bond energies for these bonds we find
keto: 745 + 347 + 413 = 1505 kJ/mol
enol: 614 + 358 + 467 = 1439 kJ/mol
This rough calculation tells us that, generally, the keto form will predominate.  However, the energy difference between the two forms is only 66 kJ/mol, a relatively small number; so it is likely that small differences can cause a dramatic shift in the relative concentrations of the two species.
Some general guidelines for predicting shifts in keto-enol equilibria

Aromaticity and conjugation - The keto-enol equilibrium for phenol (I) lies entirely on the enol side due to thermodynamic stabilization provided by aromaticity. Compound III would be expected to have a higher enol content than compound II due to the extended conjugation present in the enol of III
Substitution - Replacing hydrogen with alkyl groups on a double bond stabilizes the double bond. Therefore, we would expect compound V to have a higher enol content than compound IV
Dipolar repulsion - The carbonyl dipoles in butane-2,3-dione (VI) can reduce their electrostatic repulsion by adopting the geometry pictured with the carbonyl groups oriented away from one another.  Due to the constraints of the 5-membered ring, cyclopentane-1,2-dione (VII) cannot adopt a similar geometry.  In this case, the dipolar repulsion is lessened by increasing the enol content in the equilibrium to a point where the enol predominates.
Hydrogen Bonding - Hydrogen bonding can stabilize the enol form.  If the hydrogen bond is strong enough, and particularly if other factors also stabilize the enol form, the enol form can predominate.  Your molecule, 2,4-pentanedione (IX), is a good example.  There are two possible enol forms, VIII and X. X is by far the predominant enol as it has 1) extended conjugation and 2) substitution on the carbon-carbon double bond.  In addition, X can form a very stable hydrogen bond involving a 6-membered structure between the two oxygen atoms.
Solvent Effects - Especially in cases where hydrogen bonding is involved in stabilizing the enol, the solvent can have a dramatic effect. In benzene, where intramolecular hydrogen bonding predominates, the IX:X ratio is approximately 5:95. In water, where the intramolecular hydrogen bond is replaced by hydrogen bonding to the solvent, the ratio is roughly reversed.
Answer to your question
Cyclopent-2-enone can form two different enols, XI and XIII. Neither enol has any features that would be expected to provide significant enol stabilization.  There is no more conjugation than what is found in the ketone, there is no special hydrogen bonding, and no substitution of the carbon-carbon double bonds.  Indeed, the proton nmr of cyclopentenone is just as expected, no evidence for a significant amount of enol.  On the other hand, 2,4-pentanedione exists primarily as the enol form, at least in non-polar solvents, for the 3 reasons (conjugation, hydrogen bond, double bond substitution) discussed above.
As an aside, you mentioned that a 5-membered ring with two double bonds (a cyclopentadiene) might not be very stable; actually, they are quite stable and quite common, 2 double bonds in a 5-membered ring do not create a lot of strain.

I just have one question: Do we compare the stability of the carbanion intermediate formed(assuming basic medium) when acidic hydrogen is removed from carbon 2 (in IV and V) or the final enol product formed to estimate the enol content? Cuz on comparing the intermediate, I'm getting IV>V for enol content

@YUSUFHASAN The relative amount of enol present in an equilibrium is thermodynamically controlled (not kinetically controlled).  So it is the thermodynamic stability of the final product (enol) that matters.

Why is is thermodynamically controlled, exactly?

@YUSUFHASAN  All reactions that are equilibria are thermodynamically controlled.

Why don't we consider dipolar repulsion in compound VIII, IX, X?

@gauriagrawal VIII and IX can exist in extended conformations that will serve to minimize dipolar repulsions and in X the hydrogen bonding more than offsets any dipolar repulsion.

The 2,4-pentanedione enol form shown is the more stable enol, since the lone pair on the enol oxygen can delocalize across the five atoms to the electronegative carbonyl oxygen. Hydrogen bonding is also a minor factor in its favor. The enol form of cyclopent-2,3-en-1-one is indeed (also) a conjugated system, but the lone pair on the enol oxygen cannot delocalize to an electronegative atom, rather only to a carbon atom, and the delocalization occurs only over three atoms instead of five. In thinking about this problem, it may help to draw out and look at the various "resonance structures" of these enols, but from a molecular orbital standpoint, the governing principle is that a more delocalized orbital is lower in energy (when it contains electrons, that is).
The effect is more pronounced in comparing the enolate anions of the two species.

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


text:
In which direction does imidazole affect the pH?
I'm reading something about nickel affinity chromatography and it talks about using imidazole to affect pH. In which direction does it change the pH?

Imidazole is an amphoteric molecule and can therefore act as an acid and a base.
 (courtesy of wikipedia)
It's acidity is given with $\mathrm{p}K_a=14.5$ for the proton as $\ce{N~1}$. The basic site is the lone pair at $\ce{N~3}$. Its basicity is given through the following equilibrium: $$\ce{Im + H+ <=> IMH+}$$ with a $\mathrm{p}K_a=7.05$.
(source)
You can obtain more information about this nice little molecule on its wikipedia page.
So in conclusion, it very much depends on the solution with which you are working, which effect of imidazole takes over.


text:
Regarding common preservatives and their effect
If someone has an average of 1 litre of Coke or Pepsi a day, would this form too much benzene oxide in the person's body or stomach and eventually effect part of their genetics?

Benzene oxide isn't present in Coke or Pepsi.  Some drinks contain sodium benzoate though.  If that is what you're concerned about, there's not really any cause.  Sodium benzoate is not remotely harmful in the concentrations that humans ingest.  Here is the FDA's statement on benzoates and benzoic acid.
There were claims that a mixture of ascorbic acid and benzoates would react to form benzene.  That has been shown to not be an issue.  There were ~100 drinks containing benzoates tested by the FDA, and only 5 showed benzene levels higher than allowed in drinking water.  These were reformulated and now show less than 5ppb benzene.  I would suspect that their benzene came from another source rather than this unlikely reaction, but I don't have any evidence for that.
Finally, neither Coke nor Pepsi even contain benzoate, benzoic acid, or any other compound that is chemically close to benzene.

ssavec indicated benzene can form when one drinks Coke or Pepsi and one has some ascorbic acid already injested and then benzene oxide can form when the body tries to get rid of the benzene. Benzene oxide can accumulate and alter gene expression.

So if one drinks orange juice in the morning and the Coke or Pepsi during the day (many drink a lot of pop drinks during the day) then benzene may form , the body the forms benzene oxide to get rid of the benzene. The benzene oxide is harmful..

@user128932 Who or what is ssavec and what evidence does he, she, or it have that benzene is formed from Coke or Pepsi in the body?  Why make the same comment twice?  Your claim seems to indicate that if we tested regular cola drinkers we should find significantly higher quantities of benzene oxide in their systems.  Can you cite a study that shows that?

ssavec is a user of this very stack exchange. The idea was that benzene may be formed when ascorbic acid and sodium benzoate is injested by a person. If you tested many cola drinkers who drink a lot of cola in one day benzene MAY form in their body and benzene oxide would be used by the person's body to get rid of the benzene. I didn't know this , I'm just reporting what @ssavec indicated.

you asked who or what is ssavec and I replied and then no comment. The user with the designation ssavec should be concerned.

I previously wrote a question 'About all the pop drinks being sold' regarding sodium benzoate and a person called ssavec responded about the biochemistry effects of benzene oxide. Someone previously said 'who or what is ssavec..'; users should read more of the relevent references in these sites before criticising someone else.

@user128932 And I was notified about your comments exactly 1 time, today.  Why assume that the system works as intended, or that people check this place every day?  Regarding the actual point you tried to make, you seem to not understand that Coke and Pepsi do not contain sodium benzoate.  Therefore, the proposed mechanism, benzoate + ascorbic acid cannot possibly happen in these beverages.  Moreover, I posted actual links to scientific studies by the FDA indicating that your proposed benzoate + ascorbic acid reaction was false.  You never provided any information to support your claim.

I wrote the reply above on Nov.16 , then I got a response on Nov. 26 , I was wondering before if I would get any further response. According to the Food Production Daily.com ,there is an article called 'Pepsi and other companies settle benzene lawsuit..' Written in 2007. According to Live Strong.com , an article 'What soft drinks have E211 in them . Regular Coke doesn't have it now but it's other products like Fanta and Sprite do. Pepsi max has E211( sodium benzoate) as does Sunkist Orange , Mountain Dew and Nestea. This article was updated in 2013. Wikipedia also has articles about this.

I looked up the user ssavec who has a reputation of 1,449 and has answered 51 questions. ssavec wrote on one of my other question about pop drinks that there is an article at hpa.org.uk/webc/hpawcb_c/1194947391801 about the toxicology of sodium benzoate. He also mentioned about benzoic acid. I noticed you said 'who or what is ssavec and what evidence does he ,she,or it have..' I suggest you look up such things before suggesting I was distorting info. or somehow stupid..

@user128932 If you feel that you have evidence to support a better answer than the one I provided, please write an answer of your own and post it.  Select it as the answer to your own question.  It's a perfectly legitimate thing to do.  Arguing in comments is not the purpose of this site.


text:
Be and B ionization energies
Why does Be have a higher ionization energy than B? I know that it has a filled 2s orbital, but what's the more accurate explanation?

We essentially deal with two different trends here and I think the answer lies within analyzing these trends. First we have that the ionization energy obviously increases with increased nuclear charge/atomic number. Secondly we have that with a higher magnetic quantum number (the s-, p-, d- and f-orbitals) the energy of the orbital increases and therefore that the ionization energy decreases. The decrease in ionization energy that the higher magnetic quantum number introduces is larger than the increase that the single extra positive charge introduces.
We should also take Hund's rule in account, which is favorable for a lower ionization energy of $\ce{Be}$ compared to $\ce{B}$. This factor is however relatively small compared to the above ones.
To verify some of these trends yourself, you can take a look at this table of ionization energies.


text:
How does a redox reaction between ClO- and Br- take place?
How does a redox reaction between $\ce{NaClO}$ and $\ce{KBr}$ take place?
will bromide be oxidised to bromine?
$\ce{2 Br- + CIO- +  H2O  ->  Cl- + 2 OH- +  Br2}$
or will bromide be oxidised to $\ce{BrO-}$?
$\ce{Br- + CIO-   ->  BrO- + Cl-}$

In the pH range between 10 and 14, chloride and hypobromide are formed (DOI):
$$\ce{OCl- +Br- ->  Cl- + OBr-}$$


text:
Why is proton coupling through bonds and how does this affect identical protons?
I had the opportunity to work with a NMR spectrometer and decided to learn some of the theory behind it. After finishing Clayden's chapter on $^{1}\textrm{H}$ NMR (Ch. 11) I'm still left with some questions, mostly concerning coupling.
Proton coupling is described as the interaction between neighboring (2-4 bonds away) protons magnetic field. This can split the signals by a constant amount (the coupling constant), which depend on the environment of the protons but not on the strength of the magnetic field. Ideally creating a binomial distribution when the neighboring protons have the same shift.

How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?
Why do identical protons not influence each other? Or more generally: why is it that the more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their "outer" signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.

coupling constant does not depend so much on the environment but on the distance and connectivity

How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?

Through space coupling is known as dipolar coupling and is actually much greater than through bond coupling (J-coupling).  However, in low viscosity liquids the chemical shift effects of dipolar coupling are averaged to essentially a single value by molecular rotation.  Relaxation effects of dipolar coupling are still observed.

Why do identical protons not influence each other?

For two protons there will be four states $\alpha \alpha , \alpha \beta , \beta \alpha$, and $\beta \beta$.   In the general case, each state may have a different energy and there can be four possible one quantum transitions ($\alpha \alpha$ to $\alpha \beta$, $\alpha \alpha$ to $\beta \alpha$, $\alpha \beta$ to $\beta \beta$, and $\beta \alpha$ to $\beta \beta$), each transition having a unique energy and hence 4 peaks (two doublets).
However, when the two protons are equivalent, $\beta \alpha$ and $\alpha \beta$ will be the same energy, $E_2$ let's say.  So, letting the energy of  $\alpha \alpha$ be $E_1$ and $\beta \beta$ be $E_3$, and given a one quantum transition, the possible energies for transtions are only $E_3$ - $E_2$ and $E_2$- $E_1$.  But $E_3$ and $E_1$ are equally spaced above and below $E_2$, so there is only one energy of transition.
For a more-rigorous explanation see: http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html

How does a proton create a magnetic field that influences the neighboring protons (mainly) trough bonding and not trough space?

A proton doesn't interact with another proton by producing a magnetic field per se, but it interacts via the bonding electrons via the Fermi contact interaction. And because the Wikipedia article is kind of dry on this matter, here a figure that very well illustrates what is going on:

This interaction is not the only source for indirect coupling of spins. If we only consider protons, then we're fine. But if you'd like to consider fluorine-fluorine couplings, you also have to take into account the dipolar interaction between the nuclear and electron spins. This dipolar interaction also provides a source for anisotropic $J$ interactions.

Why do identical protons not influence each other? Or more general: why is it that how more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their "outer" signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.

Also here, a picture is worth more than a thousand words (I'll spare you the calculation of this spectrum):

The $k$ is defined as follows:
$$ k = \left| \frac{\Omega_2 - \Omega_1}{2\pi J} \right|$$
As you can clearly see, the outer peaks start getting smaller the smaller the difference in chemical shift ($\Omega$) of the two spins tends towards zero. This is sometimes called "roof effect".
The intensity of the left- and rightmost lines can be calculated (assuming $J>0$) with the following formula:
$$ I_{13} = I_{34} = \frac{1 - \sin(2\alpha)}{4} $$
where
$$ 2\alpha = \tan^{-1}\left(\frac{2\pi J}{\Omega_1 - \Omega_2} \right) $$


text:
is there a lead compound that crosses the blood-brain barrier when ingested?
for example mercury oxide has neurotoxic properties but isn't organic so, it isn't soluble in lipids and so doesn't cross the blood brain barrier when ingested, however, methylmercury chloride is organic and soluble in lipids thus it can cross the blood brain barrier

Do you mean "lead" as in the element $\ce {Pb}$ or lead as e.g. in "hit to lead compound"? What do you plan to do with it (as you asked about how easy it would be to obtain)

i mean the element Pb, i am building octane generators and need tetraethyllead, but i am also researching the role of lead in such generators, since i am obsessed with health i want to establish a list of how different lead compounds affect the body..

Tetraethyl lead (bp 180 °C) and tetramethyl lead (bp 110 °C) are easily resorbed through the lungs and the intact (!) skin. They pass the blood-brain barrier without any problems. The half-life for the elimination from the brain is about 500 days.
(Source: Günter Fred Fuhrmann, Toxikologie Für Naturwissenschaftler: Einführung in Die Theoretische und Spezielle Toxikologie, Springer-Verlag, 2006, p 209)
These compounds are neurotoxic and the literature is full of reported cases of intoxications and - sometimes - successful treatment, such as here.


text:
Are difluorine and hydrogen chloride considered isoelectronic?
Why are $\ce{F2}$ and $\ce{HCl}$ not considered isoelectronic? Both have 18 electrons.
$\begin{align}
\ce {H}\!:&\  \ce{1 e-}\\
\ce{Cl}\!:&\  \ce{17 e-}\\
\ce{F}\!:&\  \ce{9 e-}\\
\end{align}$
$\ce{F2} \implies 2 \times \ce{9 e- = 18\ e-}$
$\ce{HCl \implies 1 e- + 17 e- = 18 e-}$

If we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have the same number of electrons, they are actually isoelectronic. Please see "Ultraviolet Spectroscopy And Uv Lasers"
 edited by Prabhakar Misra, Mark A. Dubinskii, page 35 (available in google books)
But if we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have a similar electron configuration and the same structure, they are not.


text:
Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data
use these data to suggest the possible structure for the compound PF3(Ph)(NMe2):
19F Spectrum, two signals:
a doublet of doublets, J=56 and 819Hz, -39.5ppm, relative intensity 2
a doublet of triplets J=959 and 56Hz, -68.3ppm, relative intensity 1
31P Spectrum, one signal:
a doublet of triplets, J=959 and 819Hz, -53.0ppm
part b,
    suggest an experiment or measurement that might be helpful to discern possible isomers of the compound
My answer so far:
There are two types of structure that a Phosphorus with five bonds could be, trigonal bipyramidal and square based pyramid. I think that because there are two equivalent fluorines that this must be the trigonal bipyramidal structure as it had the two axial positions that would be equivalent.

What causes the 31P spectrum to give a double of triplets? I thought only adjacent phosphorus atoms with a different chemical shift would cause splitting. Also, the 19F suggests that you have 3 different environments for the fluorine atoms since you have a doublet of doublets instead of a single doublet.

I understood it as there is coupling between the P and F. So if you look at the coupling constants each peak for F is coupling with each other and the Phosphorous.

That's correct.

I agree that the trigonal bipyramid structure is what we are dealing with, so let me focus on that structure.
The nmr data shows that 2 of the fluorines are equivalent and one is different. This results in 3 possible isomers as shown below.

Both $\ce{^31P}$ and $\ce{^19F}$ are spin 1/2 nuclei, so each of these isomers would produce phosphorous and fluorine nmr spectra with splitting patterns and integration ratios consistent with the reported data. That is, the phosphorous will split each different fluorine into a doublet, the two equivalent fluorines will split the singular  fluorine's doublet lines further into triplets.  The one non-equivalent fluorine will split the other two equivalent fluorine doublet lines into doublets.  Similar analysis of the phosphorous signal leads to the expectation of a doublet (from the singular fluorine) of triplets (from the two equivalent fluorines).
Two additional pieces of information:

Fluorine prefers the axial position in pentacoordinate phosphorous
(see here)
Because the orbitals involved in axial and equatorial bonding in pentacoordinate phosphorous compounds are different, the axial and equatorial coupling constants will be different.  Generally, $\ce{J_{PF(eq)}}$ > $\ce{J_{PF(ax)}}$ because there is increased s-orbital density in the equatorial bond (see here, for example)

The larger reported P-F coupling constant (959 Hz) is associated with the intensity = 1 fluorine signal.  This suggests that structure I, with one equatorial fluorine (with the larger $\ce{P-F}$ coupling constant) and 2 axial fluorines (with the smaller coupling constant and as preferred electronically - point 1 above) is the structure being observed.
Part B Experiments:
The trigonal bipyramid structure could be confirmed by heating the sample while in the nmr spectrometer.  As Berry pseudorotation commences the 2 fluorine signals should coalesce and determination of a barrier height around 35-40 kcal/m would be expected.
Also, compounds II and III must have similar energies and both would be expected to be present at equilibrium.  If we heated our sample so that pseudorotation took place, and then upon cooling we saw an additional set of signals, that would be consistent with II or III being our starting compound.  No new signals upon cooling would support structure I.


text:
Balancing Chemical Equations + predicting products
How do I predict the products and balance the equation:
Li(s)+N2(g)---___

This question appears to be off-topic in its current state. Please see How do I ask homework question on Chemistry S.E..

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

If $\ce{Li_{(s)} + N2_{(g)}}$ react (I am not sure if they will or not), the product will be $\ce{LiN}$ and the balanced reaction will be:
$\ce{2 Li_{(s)} + N2_{(g)} -> 2 LiN}$
See this tutorial, it contains all the steps you should follow to predict the products when formula of reactants are given.


text:
Which ligands would be exchanged in this reaction?
I have this reaction and need to state what the products would be:
$\ce{(CF3)2PH + [W(CO)5(THF)] -> }$
THF stands for Tetrahydrofuran.
Would this just be a simple ligand exchange, and if so which ones would be exchanged? How would you know?

Have a look at this tutorial which shows you how math and chemical formulae can be nicely formatted on this site. Furthermore, what are your own thoughts on the matter? What do you think will happen and why?

I think that the two CF3 groups will replace either two of the CO or one CO and the THF. But i don't know which one or how to work it out.

Why do you think the CF3 groups will replace a ligand?

Complexes like $\ce{M(CO)5(solv)}$ are typically generated by photolysis ($\lambda$ = 350 nm) of $\ce{M(CO)6}$ in the presence of a non-inert solvent. This proceeds through a highly reactive "naked" $\ce{M(CO)5}$ intermediate.
$$\ce{W(CO)6 + solv ->[h\nu] W(CO)5(solv) + CO}$$

When offered another ligand $\mathrm{L}$, the weakest bond  ligand in the $\ce{W(CO)5(solv)}$ complex is replaced. This is THF.
Taking the lone pair on the phosphorous atom into account, is is conceivable that $\ce{(CF3)2PH}$ will do the job:

$$\ce{W(CO)5(thf) + (CF3)2PH -> W(CO)5((CF3)2PH) + THF}$$


text:
Bomb calorimeter and heat capacity of calorimeter
Can anyone describe the process through which we determine the heat capacity of calorimeter, for me?
The formula is
$q_{calorimeter} = C_{calorimeter} \times \Delta T$
The heat of bomb calorimeter is shown to be  the same as  negative heat of reaction. But without mass, their unit would be different, no?

The heat capacity of a calorimeter, $C_{calorimeter}$ is given in units of kJ/°C, with no units of mass included.  This is because the calorimeter's mass is a constant and each calorimeter's heat capacity (and mass) is different.
The general procedure that is used to determine the heat capacity of a calorimeter is as follows:

Place a carefully measured quantity of a compound whose combustion energy is well known and that is available in a pure form in the bomb, usually benzoic acid.
Seal the bomb and add oxygen so that the sample will burn completely.
Submerge the bomb and ignite the sample.
Measure the temperature change of the water bath in which the bomb is submerged.

Since you know the energy of the reaction and you know the heat energy in the water, you can find the heat energy of the calorimeter.
$\Delta E = q_{water} + q_{calorimeter}$
The heat of the calorimeter is the difference between the energy we know we put into the system through the combustion of our standard and the heat that was measured in the water bath.


text:
Real life chemistry :boiling carrot dices
This morning I was boiling carrot dices. And saw something quite interesting happening to the dices.
The water is not boiling. But before it boils, the dices started gathering at the center  of the pot.
Any one has any answer as to why this happened?

My guess would be natural convection. Hot water is less dense than cold water so when the heat source is at the bottom, the freshly heated water rises and the cooler water at the top sinks. The cooler water can't sink through the rising hot water, so the two flows run along side each other. Depending on the fluid properties, this can result in circular "rolls" of flowing water, known as convection cells. If you use a wide pot and small particles, you may be able to see many, but in a relatively narrow pot, you're more likely to observe a single recirculating flow around the centre, as shown in this simulation. The liquid loses more heat around the sides of the pot, so the descending flow rides down the sides and the ascending flow is in the middle where it can stay the hottest.
I'm guessing that your dice is too heavy to ride the ascending flow, but the flow from the edge to the centre pushes all the carrots towards the middle.

Would the water start boiling if there is a flow? Where does the flow from the edge to the centre came from?

If liquid is rising in the middle and descending at the edge, the two have to connect at the top and bottom. Look at the video, it's quite clear. Once it actually starts boiling, I expect this pattern is lost as the rising bubbles would disrupt it.


text:
Enthalpy (H) formula: what is the pressure volume product
$$H=U+pV$$
Above is the formula for enthalpy or the sum of the internal energy and pressure volume product.
$-p\Delta V$ is work done by or on the system. But $pV$, yet very similar to work, I think, calculates the whole “cylinder”. Instead of just calculating the height difference of the cylinder before and after the air is condensed, it calculate the height of whole cylinder.
But anyway this is just my guess, what is it really?

When you consider a change in enthalpy at constant pressure, the PV term becomes $P \Delta V$ and it and work (from the change in energy) cancel exactly.

Enthalpy $H$ is defined so that $\Delta H = q_P$, the heat at constant pressure. This is convenient for constant-pressure calorimetry---it makes it possible to track energy changes without considering volume changes in the system.
$$\begin{align*}\Delta U &= q + w & \text{(first law)}\\
\Delta U_P &= q_P + w_P & \text{(constant pressure)}\\
\Delta U_P &= \Delta H + w_P & \text{(definition of enthalpy)}\\
\end{align*}$$
...so the difference between $\Delta H$ and $\Delta U$ can be interpreted as minus the work for a constant pressure process.
But this doesn't answer your question for a general process. If the process is not constant pressure, we can still apply the definition of enthalpy, but the difference isn't minus the expansion work anymore. But what is it, then? How can we use something we've defined for a constant pressure process for all processes in general?
We'd like a new thermodynamic function $H$ that includes all of the information in $U$. We have $$dU = TdS - PdV$$ so the natural variables of $U$ are $S$ and $V$. We want this new function $H$ to have $P$ as a natural variable in place of $V$, so it will be convenient for constant-pressure calorimetry. Consider the following plot of $U$ vs. $V$:

The equation for the tangent line at some value of $V$ is
$$ U = \left(\frac{\partial U}{\partial V}\right)_S V + intercept$$
If we define the new function as the intercept of the tangent line shown in the graph, it will have the properties we need:
$$H \equiv U - \left(\frac{\partial U}{\partial V}\right)_S V$$
You can see that the derivative $\left(\dfrac{\partial U}{\partial V}\right)_S = -P$ from the equation for $dU$ above. Then
$$H \equiv U - (-P) V = U + PV$$
so
$$dH = dU + d(PV) = TdS - PdV + PdV + VdP = TdS + VdP$$
By adding $PV$ to $U$, we've transformed the information in $U$ (with natural variables $S$ and $V$) to a new function $H$ (with natural variables $S$ and $P$).
This trick is called a Legendre transformation, and it can be used to understand the definitions of other thermodynamic functions like $G$ and $A$ as well.

The best thing I found was on wikipedia, I couldn't explain it in a better way:

The U term can be interpreted as the energy required to create the system, and the pV term as the energy that would be required to "make room" for the system if the pressure of the environment remained constant. When a system, for example, n moles of a gas of volume V at pressure p and temperature T, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy U plus pV, where pV is the work done in pushing against the ambient (atmospheric) pressure.