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Bimolecular and Isomerization reactions
Is it possible to describe a reaction of the
$$\ce{A + B <=> AB}$$
exactly using the solution of an isomerization reaction?
$$\ce{A <=> B}$$
I just can not figure it out the equations seem so similar but the solution in my script is so convoluted that I can not prove they are equal.
Equations for top reaction:
$${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$
and for the bottom:
\begin{align}
{\partial c_\ce{A} \over \partial t} &=
  -k_{\ce{A -> B}} c_\ce{A} + k_{\ce{B -> A}} c_\ce{B}\\
{\partial c_\ce{B} \over \partial t} &=
   k_{\ce{A -> B}} c_\ce{A} - k_{\ce{B -> A}} c_\ce{B}\\
\end{align}
Could we not just represent $\ce{A + B <=> AB}$ as a new reaction with $\ce{C $=$ A + B}$ and $\ce{D $=$ AB}$ and thus use the isomerization results on $\ce{C <=> D}$?
I could write a set of equations for the $ A+B \rightleftharpoons AB $ reaction :
$${\partial c_{AB} \over \partial t} = -k_{off} c_{AB} + k_{on} c_A c_B$$
$${\partial c_{A}c_{B} \over \partial t} = +k_{off} c_{AB} - k_{on} c_A c_B$$
which looks suspiciously like the $$A \rightleftharpoons B$$  reaction.
A starting point might to prove they can be used to express each other would be to use the equilibrium relation
$$k_d = { c_\ce{A} c_\ce{B} \over c_\ce{AB}}$$
with the relation for the total number of species A $c_\mathrm{total} = c_\ce{A} + c_\ce{AB}$ to get :
$$c_\ce{A} c_\ce{B} = k_d (c_\mathrm{total} -c_\ce{A})$$

This is a bit like "can't we describe a kitten as a dolphin"? Sure, both are cute, but other than that, they are hardly similar. The same applies to your reactions.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.  A bimolecular reaction will never be a unimolecular reaction. They occur with a completely different mechanism, e.g. UC Davis. However, you can tweak the experiment to run a bimolecular reaction in a pseudo-first order fashion.

If you are only concerned about the fact that they are mammals that this might be a good example. It might depend.. But is there a mathematical reason why these can't be equivalent?

not shure what happened to my edit just now sorry

Thanks for the link martin but in the article the isomerisation is not talked about. maybe that is the difference?

TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.
Elaboration:
Your idea $\ce{C = A + B}$ changes the order of reaction, and has no
physical basis.
The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation:
$${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$
has the physical interpretation that $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.
Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.
To help clarify, consider the following reaction:
$$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$
In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}] = 2 [Br^{\cdot}]}$.  The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$  as opposed to second order.
Other comments:

You also forgot to include the other reactions for the bimolecular case:
$${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.
Little k is a rate constant.  Equilibria are described with big K.  It is unclear what $\ce{k_{d}}$ means in your question.


text:
Can we find 100% probability region in an orbital if the position of an electron is limited by speed of light?
The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.
The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.
My question is, provided that we knew position of an electron at a give time, could we now draw a boundary surface diagram that encloses a region in which probability of finding the electron is 100% for some point in time in future?
This would be possible because the electron cannot travel faster than the speed of light, so we could draw a sphere of radius $3*10^8$ meters centered at the current known position and say that probability of finding the electron after one second will not be beyond the region enclosed by the sphere.
Or is this question invalid because we can't know the position of an electron ever, not even within a finite uncertainty that would allow us to draw such a sphere?

Your own objection to your idea is right: we can never know the position of an electron.

@IvanNeretin is it because of Hiesenberg's uncertainty principle? (shouldn't that still allow for knowledge of position within a finite uncertainty)?

Well, yes, we might put it this way. Finite uncertainty of position does not imply finite distance. Look again at that Gaussian bell curve (or just about any other probability distribution, for that matter): it surely has finite uncertainty, but it continues to infinity and never quite reaches 0.

@IvanNeretin I think I got it now. I always thought that uncertainty principle related $\Delta x$ and $\Delta p$ where these give a range of values that velocity or position may take (i.e. x will lie with 100% probability between $x - \Delta x$ and $x + \Delta x$ ) but actually it relates $\sigma_x$ and $\sigma_p$ where these are the standard deviations in the probability distribution. Am I correct?

That's right, these are the standard deviations. There is no 100% probability interval.

The (Wave)functional forms for the surface which contains the 100% probability for finding the electron extends to the literal edge of the universe.  In the context of 'wave' it is nonsensical to measure an electron's position, and hence its speed may not be determined.
In the context of particle, an electron, having nonzero mass, may not be accelerated to the speed of light.

An electron can be prepared in any (proper wave function) state where its confined to an arbitrarily small volume. In such a state its momentum wave function is in general necessarily diffuse; this diffuseness is regulated via the uncertainty principle.  When its energy is then really measured, the originally prepared wave function collapses into a certain eigenfunction of the Hamiltonian of the system with a probability proportional to the square abs value of the complex coefficient of the proportion that eigenfunction contributed to the originally prepared state. Also such eigenfunctions can be very strongly localized (e.g. particle in a box). Moreover in quantum mechanics the electron does not "travel" with some speed between some points in space, it has no trajectory. The wave functions much more show some time evolution which neither in the non-relativistic (Schrödinger eq.) nor the relativistic case (Dirac eq.) would inhibit it to "appear" (get probability amplitude) suddenly at places which it could not reach if it would be a particle obeying classic equations of motions. I think this can be seen in analogy to the famous tunneling where the electron can appear at classical "forbidden" regions.
Comment: I have edited the answer, the position measurement results in a complete localization of the electron; so I have replaced it with energy measurement to get the argument working.

The probability density  has some value howsoever small it may be, at any finite distance from the nucleus. It is therefore not possible to draw a boundary surface diagram of rigid size in which the probability of finding the electron is 100 percent.

As OP points out it's not taking into consideration speed of light - orbital functions don't do this because they're only approximate descriptions.

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Oleum is used to prepare sulfuric acid. Why do H2O+SO3 not give H2SO4?
I encountered the fact that $\ce{SO_3}$'s solubility in water is low but even when $\ce{H2S2O7}$ (oleum) is added to water it is $\ce{SO3}$ only that react with water . Please explain the process.

What process? The Contact process?

Solubility of $\ce{SO3}$ in water is not low at all; in fact, they react quite vigorously.

Related: What is the difference between these equations for the Contact process?

Also see this and this

The reaction $\ce{SO_3}$ and $\ce{H_2O}$ is exothermic, which means it releases a lot of heat.The heat produced will boil $\ce{H_2SO_4}$ and become $\ce{H_2SO_4}$ gas. The $\ce{H_2SO_4}$ gas is hard to condense causing this method to be unefficient.

It would boil water before reaction would end

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Why are the given compounds in a homologous series?
Why are the given compounds in one homologous series?
$$\ce{CH3-OH}$$
  $$\ce{CH3-CH2-OH}$$
  $$\ce{CH3-CH2-CH2-OH}$$
  $$\ce{CH3-CH2-CH2-CH2-OH}$$
I know that homologous series can be represented by a common general formula and have the same chain length. But I could not answer this question. Can anyone help me?

You got it all wrong: compounds in a homologous series have different chain length.

@Ivan Neretin, I am sorry for that.

Because they can all be described by the general formula $\ce{CH3-(CH2)_n-OH}$. The only varying parameter in this case is the number of middle $-\ce{CH2}- $ groups.
So while the the number of middle $-\ce{CH2}- $ groups may vary $\left(n = 0,1,2,\ldots\right)$ they all fit in this general formula.
The formula $\ce{CH3-(CH2)_n-OH}$ tells you that:

The chain starts with a $\ce{CH3}-$ group
The chain ends with an $-\ce{OH}$ group
Contains a variable number, $n$, of $-\ce{CH2}-$ groups in between.

Could you explain a bit more? @ttdijkstra

For the series in the question, another possibility is n=0

Thanks for the note on formatting @IͶΔ I just noticed myself.

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chem.se""="), this page and this one on how to format your future posts better with MathJax and Markdown. Welcome to chem.SE!

I cleaned up the formatting and the wording a bit, but @ttdijkstra nailed the correct idea.


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H NMR of 4-chlorochalcone
In an aldol condensation reaction, I synthesized a molecule out of two unknown reactants. These reactants I found to be 4-chlorobenzaldehyde and acetophenone which made 4-chlorochalcone. I was given an H NMR spectrum to identify and label the hydrogens for the molecule but am having trouble reading it!
There is a quintuplet with an integration of 6 but no where in the molecule am I seeing 6 hydrogens that share a chemical environment (I counted 7 environments). Is this possible or could I have come up with the wrong product structure?
Any help deciphering this would be great!

The region corresponding to 6 hydrogens doesn't have to be 6 equivalent hydrogens. It could be 6 hydrogens that are in a similar chemical environment. So that could be two different hydrogen types in a 1:5 ratio, or a 2:4 ratio, or a 3:3 ratio. Or it could be three hydrogen types in a 1:1:4, or a 1:2:3 ratio, or a 2:2:2 ratio. And so forth up to six hydrogen types in a 1:1:1:1:1:1 ratio.

Based on the structure, you have already determined correctly that there are 7 hydrogen types (environments). How many of each type do you expect? What multiplicity do you expect for each? What chemical shift makes the most sense for each environment? Assign the peaks as best you can for your prediction, and whatever is left is likely in that multiplet with integration 6.

Interesting... But I don't  see 6 similar hydrogens. I see 5 hydrogens on the phenyl on the left that could be similar. But where is the sixth?
I'll try assigning them and see where it gets me! Thank you!

The integrals in a quintuplett should be 1:4:6:4:1. Pascals triangle. Not here. The dubletts at 7.4 and 7.77 both look suspiciously like they are part of an AB coupling each.

Could it possibly be an overlapping doublet triplet doublet? What would that mean for the structure?

As with all NMR analysis problems like this, it is good to have a strategy to interpret data. Perhaps rather than looking at your spectrum and trying to rationalise your structure, look at your structure and predict what you might expect to see in your spectrum. You'll find it much easier to rationalise the appearance of your peaks, especially in the case of overlapping peaks as you have here.
So, in your molecule, what splitting would you expect?
You have 11 proton environments, so your integrals look fine. Your chloroform doublet is  a bit of a worry though. Perhaps your sample needs to be mixed a little better? You've got 3 distinct spin systems that will give characteristic splittings:

trans double bond. These two peaks will appear as distinct doublets, with couplings of around 12-17Hz. I can see one very clearly. The other, which we know to be upfield of the one at 7.77 because of the tenting is probably under the mess of stuff between 7.5-7.6. Being next to the carbonyl will have strong influence on the chemical shifts of these, and through a bit of electron-arrow pushing we can show that the proton closest to the carbonyl is shielded, and the other deshielded. Hence, the signal at 7.77 is likely to be your proton β to the carbonyl group (and closest to the para-substituted benzene ring)
para-substituted benzene ring. Again a distinct splitting pattern; these will look like a pair of doublets (actually a second order pattern) with coupling about 7.5Hz. I can see one probably at 7.4. Its partner will also be under the mess between 7.5 and 7.6
a mono-substituted benzene ring, which will have three sets of signal, with ratio 2:2:1. A carbonyl group is strongly electron withdrawing, deshielding the ortho protons, which is probably the signal up at about 8. The others.... under the mess between 7.5-7.6.

This is clearly a homework/lab problem, so this should be enough to keep you going until the next question.


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Free radical substitution - bromination of ethane
I am new to this service so please overlook any mistakes.
My question is with regards to free radical substitution. I understand the three step process of the bromination of ethane and what would occur if chlorine was in excess.
I do not however understand what would happen if ethane was in excess. What would occur and what would be the major product produced?

You'd definitely get the monobrominated product, but if your ethane was in large excess you'd get practically no yield.

So bromine or chlorine? That's a typo saying chlorine in the middle there, isn't it?

If there is an excess of ethane, you would be getting the monosubstituted bromoethane, $\ce{CH3CH2Br}$, as the main product.
When ethane, $\ce{CH3CH3}$, is in excess, for any bromine radical formed, most of the reacting particles around it would be an unreacted ethane molecule (high concentration of $\ce{CH3CH3}$), this means that there is a greater chance of a bromine radical colliding with and reacting with $\ce{CH3CH3}$ to form the monosubstituted bromoethane.
This is in contrast with when bromine is in excess, where the excess of highly reactive bromine radicals mean that most of them react fully with the available ethane molecules to form fully substituted $\ce{CBr3CBr3}$.


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Hybridization in Oxygen and Nitrogen
What is the proper orbital hybridization O and N in $\ce{H_2O}$ and $\ce{NH_3}$? The rules say that they are sp and sp2 hybridized, respectively, but some sources say both are sp3. I have problems with both, and I'll try to explain them.
1) In the first case: According to elementary rules, the total number of hybridized orbitals should be equal to the number of sigma bonds. Thus, oxygen should be sp hybridization and then nitrogen should have sp2 hybridization (in the above molecules). However, this possibility does not relate well to the VSEPR model: in water, the hydrogen atoms would be striaght and perpendicular to the two perpendicular unhybridized p-orbitals.
2) If they have sp3 hybridization, (and why would they?), then there are some contradictions with electron delocalization/conjugation. In phenol, the delocalization in the benzene group is supposed to combine with an nonbonding orbital in oxygen. However, there's a 109.5 degree angle around the O atom, so then both nonbonding domains would be in the wrong orientation for conjugation. (I reference this page: http://chemguide.co.uk/analysis/uvvisible/bonding.html).
As a side question, does the conjugation/delocalization in phenol change oxygen's formal charge?

see http://chemistry.stackexchange.com/questions/5789/why-is-h%e2%82%82o-v-shaped http://chemistry.stackexchange.com/questions/33818/bond-angles-for-the-hydrides also you should use proper tags

see the "Note on Reality" section of this answer for a discussion on the hybridization in water.

You're right. Welcome to the real world where an model (analogy) can only be pushed so far...
So you build this:

Isn't it "real"? I can see it, I can hold it in my hand. I can weight it. If it was a variety of kit that could fly would it then be a "real"?
So let's look at an isolated gaseous ammonia molecule. As $\ce{NH3}$ we could say that it is $\mathrm{sp}^2$ hybridized with another $\mathrm{p}$ orbital. However we would expect 3 equivalent $\mathrm{sp}^2$ orbitals to be planar. But the H-N-H is 107.8 degrees and the molecule is a trigonal pyramid. In methane there are 4 equivalent $\mathrm{sp}^3$ orbitals and the H-C-H angle is 109.5 degrees. So an ammonia molecule isn't purely $\mathrm{sp}^3$, but it is more $\mathrm{sp}^3$ than $\mathrm{sp}^2$. If we consider an isolated gaseous $\ce{NH4^+}$ cation then it is $\mathrm{sp}^3$ hybridized since all the N-H bonds are equivalent.
But what happens when we put these isolated species into an aqueous solution. Because of the hydrogen bonding we don't have a "pure" ammonia molecule, nor a "pure" ammonium cation. So we're left in a situation where we are trying to model a unicorn as some sort of hybrid between a horse and a rhinoceros.
So the gist is that you have to pick the model depending on what you're trying to model, and be careful not to push that model too far.
As a simpler example consider an ionic and a covalent bond. A pure ionic bond and a pure covalent bond are two extremes that don't really exist. So a bond can be much more ionic than covalent, but that isn't the same thing as being "purely" ionic.
Ah... but what about diatomic molecules like $\ce{O2}$ asks the young padawan. Since there is no difference in electronegativity isn't that bond purely covalent? Well what time scale are we considering? The $\ce{O2}$ molecule has the resonant forms $\ce{O+O^{-}}$ and $\ce{O^{-}O+}$.


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H2 Triplet State
How does $\ce{H2}$ triplet state exist if there are electrons in both bonding and antibonding ($1\sigma$ and $1\sigma^*$) orbitals?
Or am I being taught the hypothetical triplet states of the hydrogen molecule?

Are you sure you are talking about the electronic and not nuclear spin states?

Assuming that you're talking about electronically excited states and not spin isomers of the $\ce{H2}$ molecule (as anticipated by Ivan Neretin), note first that there are many triplet states, of which the lowest lying one corresponds to $1 \sigma 1 \sigma^∗$ configuration. Under ordinary conditions this particular triplet state is indeed unstable towards the dissociation to the hydrogen atoms. But it is not hypothetical, as you said, it is just unstable. Besides, in some extraordinary conditions, say, in an ultrahigh magnetic field, the triplet state of the hydrogen molecule is the ground state.


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Calibration of a spectrophtometer
In my lab, I had to use spectroscopy to determine the concentration of protein in milk whey from buttermilk and understand how absorbance of light is related to concentration (Beer's Law).
This was my data: The standards were dilutions.
Concentration (mg/mL) --> Absorbance
Standard 1: 0.4  --> 0.19
Standard 2: 0.2  --> 0.16
Standard 3: 0.1  --> 0.07
Buttermilk: 0.6  --> 0.34
One of my post-lab questions was:
Draw a picture of what you think your calibration graph (Beer's law plot) would have looked like if you had blanked the spectrometer with a tube that only contained water?

Possible duplicate of Bioanalytical Spectroscopy

Do not repost your questions! I can even see how you used copy paste

@Mithoron Not a duplicate - just a different question about the same prac problem, I suspect.

@Mithoron Yes, this is NOT a duplicate. I copy and pasted my data and what I had to do in the lab in order for people to be able to understand my question better.

In an ideal situatiuon, you might expect that at zero protein concentration there would be zero absorbance. This means your calibration plot should go through the zero-intercept. However, there may be a number of reasons why this is not quite so: your cuvette will absorb or scatter a small amount of light, your solvent may absorb light, there may be stray light, or there may be some other systematic error being introduced into the system. Running a blank sample of just water allows you to compensate for these systematic errors. The blank contains no analyte, but will still incorporate any systematic error, and so adjusts your calibration graph to compensate for these.
A couple of comments:

it is unusual for you to have your calibration curve that contains concentrations that are outside the expected concentration of your real sample. This is not generally very accurate, as it relies on extrapolation of your calibration curve. Normally, one would run a series of standards that extend beyond the expected concentration range of your sample.
A calibration curve of just three points is also not very accurate. Your plot is not linear (would you expect it to be?), but with only three points, it is not possible to know whether you have a random error, or your graph is in fact non-linear resulting from a possible concentration-dependent equilibrium shift or other chemical phenomenon.


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What does Molecular Orbital Theory get wrong?
I am learning about MO theory in my advanced inorganic chemistry course and am starting to realize that it is truly the most accurate representation of how molecular orbitals look like, where they are located in the molecule, and their relative energies to each other and the original atomic orbitals from which they are composed of. We are using Symmetry Adapted Linear Combinations as the approximation method and so far this method has successfully explained all chemical/magnetic/electronic properties for any molecule investigated thus far.
My question is: Is MO theory perfect? Or does it have a flaw somewhere like every other bonding theory (Lewis, VB, hybridization) I have learned about so far in my chemistry career?

Everything is flawed because we can't solve the Schrodinger eqn. By MO theory I'd assume you are referring to a Hartree-Fock wavefunction, which gets electron correlation pretty wrong.

Thank you for orthocresol for the great answer. No thank you to the other two for making me feel stupid. I struggle enough with chemistry, no need to rub it in. Just trying to learn.

Nova, I don't think that anyone wanted to make you feel stupid. I believe the key problem is - and I came across that many times - that you are learning quantum chemistry from an inorganics point of view. The theory you learn there is usually a short version of the whole deal. It is good that you question the concept, stay curious. I think (hope) that @Karl wanted to express that to gain deeper insight into the matter it would have been better if you were introduced properly to quantum chemistry first.

I do not think that this question is too broad. It is basically asking for the pitfalls and shortcomings of MO theory. While there are many, I think that a concise not too long answer is certainly possible. @orthocresol even provided the first stepping stone.

Yea, sorry, I didn't mean to be snippish. It's just that studying advanced concepts without the complete basic training in physical chemistry is imo bound to be very hard for both you and your teacher, who then has the choice of loosing half his audience by overcharging them or the other half by boring them with endless repeating of introductory stuff. I know a lot of universities allow or even advocate that. :-/

Some good additional context, as well as the beginnings of an answer to your question, can be found here and here.

The problem with MO theory are... the orbitals themselves! Orbitals are, by definition, single particle wavefunctions and they have no real physical meaning since you are dealing with a many-body system. For this reason MO theory is far from perfect!

MO is a way to approach FCI by using orthogonal one-electron basis. Its limitation is quantum mechanics itself. Since FCI can reach the exact solution, even it typically takes infinity long expansion.

The trouble with MO theory is that, in practice, it is largely descriptive, not predictive. When we see a molecular structure, MO theory often gives good explanations of why its properties exist. But the math is too complex to make high quality predictions before observation of some key properties like the symmetry.

@matt_black Not sure what your point is. The trouble with any quantum-chemical procedure is that it is generally too complex to make robust, accurate predictions without doing the actual computational work.

@Antimon My point was, partly, that the computational power needed for good calculations is too high for calculation to be a really useful tool for anything other than simple situations. You have to be prepared to do a humongous amount of work and that isn't worth it for most situations.

I still fail to see how that is an exclusive feature of MO theory.

MO theory fails pretty fantastically in a wide-variety of cases.
While it works well near the equilibrium geometry of a molecule, it cannot dissociate even the simplest bonds correctly.

It is well-known by computational chemists that any quantitative predictions of MO theory (ionization energies, HOMO-LUMO gaps, etc.) cannot be trusted, even for the simplest molecules.
In more complex molecules, such as those with transition metal centers, calculations can even fail to qualitatively describe the electronic structure correctly, even giving ground states with incorrect symmetries.
In my opinion, inorganic chemists like MO theory so much because it's is the simplest theory of molecular bonding and geometrically its results are quite intuitive. It is an excellent conceptual tool for understanding quantum chemistry, but a horrible one for actual predictions of chemical properties.

Depends on what exactly you call "MO theory". If you're talking about only Hartree-Fock, then you're right, but it's also a bit of a moot point because noone in their right mind uses HF as a standalone method. If you extend it to include post-HF methods that are based on those MOs, then your answer is not accurate.

Post-HF or Multi-Reference methods, I should have said. - Also, what happened to UHF? That one describes the dissociation of e.g. H2 just fine.

UHF has its own set of problems with spin-contamination. Also, UHF theory emphasizes spin-orbitals, most inorganic classes focus entirely on spatial orbitals. Of course post-HF methods can be used to improve HF results, but post-HF methods are no longer really orbital theories. For example it no longer makes sense to talk about a HOMO/LUMO, because even in the ground-state virtual orbitals can be partially occupied.

More concretely, you give a graph showing how far off things are, but for what molecule? (I'm guessing H2.) And what reference?

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Does the order of protonation/deprotonation in the formation of hemiacetals matter?

In the mechanism depicted above, deprotonation occurs first and then protonation.
Can this order be changed? If not, why?

Yes you can change it and it's no problem. Just don't deprotonate the alcohol starting material and you'll be fine

The first (productive) step of this mechanism in acidic medium MUST be protonation of the carbonyl.  Alkoxides and hemiacetal-alkoxides have no appreciable concentration in acidic solutions, and the carbonyl is not a strong enough electrophile to be attacked by alcohol without protonation.

I realize now that teacher mentioned this rxn favors left side. but the rxn to the right side does occur, right? If a answer is yes, can you answer my original question?

Reaction rates are dependent on concentrations of reactants.  The concentration of a deprotonated alkoxide in aqueous solutions is dismally small, and hence its contribution to reaction rate negligible.

why should ROH be deprotonated to react? can't it just attack carbonyl carbon as nucleophile though this reaction is not preferred much?


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Do you say "reduce something to polystyrene"?
this question may not fit in this forum because it's about English, but biochemistry knowledge is required, so please allow me to make this thread.
When you measure the molecular weight of a polymer, you may use polystyrene standards and say "polystyrene equivalent Mw of --- is..." or "the molecular weight relative to polystyrene (standards) is..."
I have seen these usages in various articles and documents. The other day, however, I ran into an expression, "the Mw was reduced to polystyrene." I understand that it means the same thing, but my question is, do native speakers really say like this?
I googled "reduced to polystyrene" and got many hits, but it seemed many of them were written by Japanese. And when I did the same thing on Google Books, there was only one hit and the "reduced" meant "converted/changed."
So my question is, do you say "reduce to polystyrene" in this particular situation, meaning you use polystyrene standards?
Thank you very much!

As a native English (American version anyway...) I can't say that "the Mw was reduced to polystyrene" is absolutely wrong but it seems very awkward. I'd use "the Mw was compared to polystyrene" or "the Mw was determined relative to polystyrene."

Thank you very much! Yes, that's what I wanted to know, if the expression sounds natural or awkward. Thank you :)

Just write what you actually did: You used the calibration curve of your SEC column for polystyrene in solvent xx. Those phrases you cite are meant to sound impressive, but are actually just scientifically wrong, bad English, and therefore stupid.
Nothing gets reduced here like a reduced temperature or a chemical reduction. Also there is no "PS equivalent Mw", because it depends strongly on the solvent. "Relative" is as wrong, because that implies that there is simply a factor between the two scales, while in fact there are two calibration curves, one of which is unknown.

Thank you for your comment, but are those phrases really scientifically wrong?
For example, "PS-equivalent Mw of Polymer A" means the Mw of Polymer A as determined by using the calibration curve of your SEC column for polystyrene, no? There are two calibration curves elicited under the same conditions ( which means, the same solvent for both) except the object is polystyrene or the unknown analyte. Equivalent (= corresponding to PS standards) and relative (= in relation to PS/ not absolute) sound right to me...?

Those adjectives both imply there is a direct, linear relation. That you cannot know.

It is likely, as i see it, if your sample elutes in the range of retention time/volume where your calibration curve is linear. The slope of the log (Mw)/v_ret should always be similar there. IF your sample coils to an isotropic ball, described by one radius of gyration.


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Measuring reaction rates when reactants and products both have (different) molar extinctions at the chosen wavelenghth
In my Biochem labs we would always measure the consumption of NADH, and they always noted that it was useful that NAD+ had essentially no absorbance at 340 nm, which allowed us to directly relate absorbance to concentration by NADH's known molar extinction coefficient (i.e. all of the absorbance at 340 came from NADH).
I'm imagining that it would be possible to measure a reaction rate where the reactants and products both absorb at a given wavelength, but there is a difference between the reactants and the products. Is this as trivial as saying that for every M reacted, the absorbance will change by the difference in molar extinction coefficients?
I'm just wondering about it because they stressed the value of the lack of A340. Was this just a matter of high responsivity/good signal, and maybe convenience? There is no inherent reason why this wouldn't work, right?

Seems right to me. Of course, the closer their molar absorptivity, the bigger of a pain it would be to determine significance of a change. So using the absorbance on the left there would require more precise measurements.


text:
Why does fluorine act as a leaving group in organophosphates?
C-F bond is the strongest single bond in organic chemistry with a bond energy of  $\ce{453 kJ\,mol^{-1}}$. And it is very difficult to break this bond and F does not act as a leaving group.
However, in organophosphates, F acts as a leaving group. For example, take an organophospate such as sarin(below). When this binds with acetylcholinesterase (enzyme), F is leaving as a fluoride ion. How can this be explained w.r.t. the P-F bond energy of $\ce{490 kJ\,mol^{-1}}$, which is even higher than that of C-F. Is this possible because of the catalytic activity of the enzyme?

According to David O'Hagana, Chem. Soc. Rev. 2008, *37 , 308-319, C-F bond is very strong due to the difference of electronegativities of C(2.55) and F(4.0), and the resultant ionic character of the bond. However, the electronegativity of P is 2.19 and according to this, P-F bond must be much difficult to be broken. But, this is not the reality. Sarin can be hydrolyzed (Andrew F.   Kingery  and  Herbert E.   Allen, Toxicological \& Environmental Chemistry 1995, 47, 155-184.) at pH 6 ($\ce{25\,^{\circ}C}$) with a half life of 312 hours in aqueous solution.

Well, did you at least check proper values? Bond energies vary... Not that this approach is terribly useful, as you can see.

So what approach should I use then? I tried looking for the actual bond energies, but couldn't find.

Enzymes can make miracles.

I am reluctant to post this as an answer because it is not really backed up by much research, but since nobody has said anything yet, here's my theory:

It is probably to do with the fact that the covalently bound sarin (which necessarily involves kicking out either $\ce{RO-}$ or $\ce{F-}$) mimics the transition state for acetylcholine hydrolysis too well, such that the expulsion of one of those two leaving groups is much more favoured than in aqueous solution.

Acetylcholinesterase (AChE) is very similar to a typical serine protease, and serine proteases are very well-studied. I am going to postulate that the mechanism of acetylcholine (ACh) hydrolysis in AChE is exactly the same as it is in a typical serine protease. So first, let's look at what the enzyme does normally. It's nothing special, just a nucleophilic acyl substitution:

(There's an additional glutamate residue, analogous to the aspartate in a serine protease, that fixes the conformation of His440, but it's not directly relevant to the mechanism. Also, there's an anionic site which binds to the positively charged quaternary nitrogen, but that's also irrelevant.)

The acyl-enzyme intermediate then goes on to be hydrolysed by water (not relevant here).
The reason why the enzyme can catalyse the reaction is because it stabilises the transition state for the ester hydrolysis, which means that the activation energy for hydrolysis is decreased and the rate is increased. (Strictly, it provides an alternative hydrolysis pathway in which the activation energy is lower - but the point is that since the catalysed hydrolysis is mechanistically similar to the uncatalysed hydrolysis, the enzyme must play a role in stabilising a tetrahedral intermediate or transition state.)
The transition state looks something like this (partial charges omitted):

Sarin is a potent AChE inhibitor because it covalently binds to the active serine residue as in the left-hand side of this diagram:

You can see that structurally, this is very similar to the transition state shown above. If the enzyme stabilises the TS above, this covalent binding of sarin to the enzyme must also be extremely stable. I do not pretend to know exactly how it is stabilised, but it must be related to the specific 3-D conformation of the enzyme. If you do not kick out the fluoride ion (right-hand side), then it does not quite resemble the transition state so well.
What this means is that there is a strong driving force for the fluoride ion to be expelled, in order for it to reach that stable, 4-coordinate phosphorus form.
That leaves one question that I feel is worth addressing, which is why fluoride is expelled instead of isopropoxide. Honestly, I do not know. As pointed out by jerepierre, fluoride is certainly a better leaving group than isopropoxide ($\mathrm{p}K_\mathrm{a}(\ce{HF}) = 3.17$, $\mathrm{p}K_\mathrm{a}(\ce{^{i}PrOH}) = 16.5$ - quite a large difference), and it might very well just be that. On the other hand, as with most biological systems, it is difficult to give a full explanation of why things happen the way they do, and applying (relatively) simple chemistry heuristics might not be the full picture. As to what other factors might influence this, I can only guess that is that there might be some hydrogen bonding arrangement that requires the specific size of the oxygen in the $\ce{-OR}$ group over the size of the fluorine (in the catalysis TS, there is an oxygen there as well). Essentially, I am saying that the LH diagram resembles the TS more than the RH diagram (don't quote me on it though...):

Note that, in case of tabun, the reaction also works with $\ce{CN-}$ instead of $\ce{F-}$.

"... why fluoride is expelled instead of isopropoxide." Fluoride is a better leaving group than an alkoxide.

@jerepierre Yes, I think it's not unlikely that isopropoxide picks up a proton from somewhere else when it's leaving though.

Wow It's so cool that this came up. I've been wondering the exact same thing for the last week! I asked a couple of my professors about it and this is the first time I've seen a reasonably satisfying answer.

I would point out (as did the author of the question) that sarin hydrolyzes in water in days to weeks without the need of an enzyme, and also that carboxylic acid fluorides (e.g. acetyl fluoride) also rapidly hydrolyze, despite the strength of the C-F bond. The ability of these acyl fluorides to hydrolyze is due to the relative ease of addition of water or hydroxide with formation of a tetrahedral intermediate (or pentacovalent intermediate in the case of phosphorus) - likely helped by the electron withdrawing characteristic of the fluorine atom - followed by expulsion of fluoride, which is a relatively stable anion and is a pretty good leaving group. Keep in mind that the term "bond strength" typically refers to the energy required to cleave a bond homolytically to form radical species. That is quite unfavorable in the case of P-F and C-F bonds.


text:
The time scale of calculations in computational chemistry
Is it possible to determine or get a range of the time required for a specific calculation with knowledge of the type (MM or QM), class (semi-empirical, ab initio or DFT), the degree of theory (MP, CI, etc.) and the basis set (STO-3G, 3-21G, 6-31G(d), etc.)
 used in the calculation and the computer ability (processor, RAM, etc.)?

A more accurate reference for ab initio might be Ab initio quantum chemistry methods.

http://scicomp.stackexchange.com/

No, that is not possible. What is possible is to estimate how much more time you need to calculate the electronic energy if you increase the system size by some factor. For any method which uses LCAO-MO expansion, it is the number of basis functions $m$ which primarily determines the computational cost, so that it is usually used as a measure of the system size. For instance, HF (without any tricks like density fitting, or taking the spatial symmetry into account) scales approximately as $m^4$, i.e. if you double the number of basis functions by using a basis which is twice as large, the computational cost will be approximately $2^4=16$ times larger. But that is it: estimating the time for a calculation using the well-known scaling behavior is possible only if you already know some timings on the same setup. And besides, this is all about just the electronic energy, which is quite often not the end of the story.

But, why is it not possible ?

@M.Ibrahim, because there are too many parameters that can vary significantly: dozens of job types and methods, plenty of important implementation details, settings & speed-up tricks that can be used in a particular program, thousands of basis sets, myriads of computer architectures - all this makes it is absolutely impossible to estimate the time needed for a calculation.

Trial and error is basically everything we have at hands: try to submit your job for a time which you think would be more than enough for a successful completion and wait for the result possibly periodically monitoring the submitted job. Start from something simple to get a feel of the scaling behavior of a particular code you run on a particular machine for a particular type of calculation you do. The more you work with a particular code on a particular machine the better your initial estimations of time needed will be.

You mean to develop a "sense" calibration, don't  you ?,,but actually I'm working on enzyme inhibitors so I wanted a method or source of estimation for the computational cost which will help me in decision making; whether to do Quantum calculations or just use the cheap Molecular Mechanical approach.

As I said, there is no general rule to answer such requests. To get an advice on a level of theory you need to specify few things. 1) What is the size of a system? How many light and heavy atoms are there? 2) What kinds of calculations you want to do? Just the single-point energy + properties (if so which properties?), or geometry optimisation and/or something else? 3) Which hardware and software you have access to? How many cores, memory, disk space and how much wall time are at your disposal?

For some methods it would be theoretically possible to estimate the total time required to perform them, because they requires some specific steps (system dependents or not) for which the number of operations involved can be estimated. Knowing that and a measure of speed of the computer used for that mater (see for example: FLOPS) you can in principle to estimate the total calculation time.
In practice we can't. There is not too detailed info easily available. Even more, as it would be computer dependent (in the sense that any two random computers can take different relative times for any too parts of the calculation, due to specific hardware) there is not too interest in this kind of prediction. But, due to personal experience, computational chemist can have a vague idea about the time required for a calculation, more or less accurate depending of their experience.
There is other "kind" of methods (enclosed by quotes because no one classify methods in this way) for which it is in essence impossible. Those methods require iterative steps to get closer to the solution, and finish when some closeness criteria is reached. As is not possible to know how many steps they require for a particular system ( even more, there are many many tweaks that can change it a lot) it is not possible to know how many operations will take to perform the method.
Resume
For some method it is theoretically possible to estimate the time needed, but as there is so many variables it is not done. In practice the estimate are done through personal experience, depending the method, basis set (if have sense), system and calculation type.
Advice
If you have an specific problem, ask about the specific case here, including in the question the result (with accuracy required) that you are in need of, and the system in question.


text:
vertically growing icicle
I made some "super" soap bubbles with my kids using the following ingredients:

liquid dish soap
tap water
corn syrup

We left the solution out over night last night here in Maine and the temperature got down to −3 °C. This morning we found that the solution left in the lid had grown a vertical icicle.
There is no trickery here. I stepped outside and found these items as you see them below.
How could this icicle have formed?

That is a fairly unusual but previously documented formation called an ice spike. Curiously, it seems that details on how they form are still somewhat elusive.

It pretty much has to be a supercooling phenomenon.  This page from the University of Toronto Physics department presents an explanation that I think fits the scenario in the question nicely.  The description and photos used suggest that this is likely the case for this question:

The short explanation is this: as the ice freezes fast under
  supercooled conditions, the surface can get covered except for a small
  hole. Water expands when it freezes. As freezing continues, the
  expanding ice under the surface forces the remaining water up through
  the hole and it freezes around the edge forming a hollow spike.
  Eventually, the whole thing freezes and the spike is left.
A slightly longer explanation: the form of the ice crystals depends on
  the cooling rate and hence on the degree of supercooling. Large
  supercooling favors sheets which rapidly cover the surface, with some
  sheets hanging down into the water like curtains. These crystallites
  tend to join at 60 degrees and leave triangular holes in the surface.
  Hence, spikes often have a triangular base. The sides of the spike are
  sometimes a continuation of pre-existing subsurface crystallites, and
  can extend from the surface at steep angles.

Notice the triangular shape at the base of the ice spike in the photos in the question, as is also mentioned in the description above.  The spike also appears to possibly have a hollow region through the center.  Additionally, the flexible plastic lid would allow for some pressure buildup before the release through some weak spot in the ice sheet.


text:
Isotropic Fermi Contact Couplings vs. Dipolar Coupling Constant
I need to calculate a Dipolar Coupling Constants for some fancy triplet molecule that has been measured by EPR spectroscopy.
My "normal" output from Gaussian gives me the following:

Isotropic Fermi Contact Couplings
Spin Dipole Couplings
Anisotropic Spin Dipole Couplings in Principal Axis System

Am I right that none of these are Dipolar Coupling Constants? And If so ... how can I compute them either directly with the program or by hand through some formula?

Can you share your actual main line of keywords?

@user1420303 currently nothing special, as those data are automatically printed when an open shell system is calculated.

I am not sure, but I would try adding:   NMR Prop(EPR)

This doesn't change the output as far as I can see.

I can not assure that it will give you the info. you need, but it must print a lot more of information.  I've tried the NMR command by myself and it really worked for me.

The output sections you mention are the major components of the electron-nuclear hyperfine interaction. There is also an orbital contribution, but this tends to be small compared to the Fermi contact and spin dipole terms, unless you go down the periodic table.
These are the terms in the effective spin Hamiltonian that contribute to the hyperfine interaction (from Wikipedia):
$$
\begin{align}
\hat{H}_{\textrm{hyperfine}} = & 2 g_{I} \mu_{N} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{L_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{l}_{zi}}{r_{i}^{3}} \mathbf{I} \cdot \mathbf{L} \\
&+ g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{s}_{zi}}{r_{i}^{3}} \left\{ 3(\mathbf{I}\cdot\hat{\mathbf{r}})(\mathbf{S}\cdot\hat{\mathbf{r}}) - \mathbf{I} \cdot \mathbf{S} \right\} \\
&+ \frac{2}{3} g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \mu_{0} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \hat{s}_{zi} \delta^{3}(\hat{\mathbf{r}}_{i}) \mathbf{I} \cdot \mathbf{S}
\end{align}
$$
The first term is the orbital contribution due to spin-orbit coupling, the second term is the spin dipole term due to the electrons and nuclei behaving as dipoles in the presence of each other's magnetic fields, and the third term is the Fermi contact term arising from the non-zero probability of the unpaired electron(s) being inside the nucleus.
Isotropic Fermi Contact Couplings are the eigenvalues of the Fermi contact part of the effective spin Hamiltonian. Because of the presence of $\delta^{3}(\hat{\mathbf{r}}_{i})$, there is only one value, hence the "isotropic" descriptor.
Spin Dipole Couplings and Anisotropic Spin Dipole Couplings in Principal Axis System are two different presentations of the spin dipole term. Spin Dipole Couplings shows the eigenvalues of the spin dipole part of the effective spin Hamiltonian (in a.u.). Because this is a symmetric 3x3 tensor, only the 6 unique elements are shown. Diagonalizing the tensor gives the 3 principal values Baa, Bbb, Bcc as the eigenvalues and the vectors from the origin for each principal value as the eigenvectors. These are the values you're interested in, most commonly reported in MHz by the computational/theoretical chemistry community and in Gauss by the spectroscopy community.


text:
Isotopes and their percentage existence in atmosphere
When we talk about percentages of isotopes of an element in the atmosphere, is it percent by mass or volume or weight or something else?

Short answer: it's the percent by number of particles.
Longer answer: It's the proportion of particles of a particular mass for that element.  To chemists, a per particle basis is also commonly referred to as a per mol basis or a molar basis.
A way that it can be measured is by mass spectrometry - you pass the atoms through the apparatus which shows the mass to charge ratio. E.g. if you could pass $\ce{Cl}$ molecules, you'd see 2 peaks in the resulting graph - one is at 35 and the other at 37 marks which shows the mass/charge. First peak would be about 3 times larger (thus there were 3 times more particles). Thus chlorine would be about 75% isotope 35 and 25% isotope 37. So while this approach measures mass of each isotope, it also determines the relative molar amounts of each isotope.


text:
Benzyl (3,4-difluorophenyl)carbamate reaction with epoxide in basic conditions
I attempted this mechanism, however I am uncertain if I am correct.

The product you've drawn on the bottom line has one more carbon than the product you've drawn on the top line.  Carefully count the number of carbons in the various reaction intermediates.

What is the uncertainty? Is there a particular step that seems unreasonable to you?

Thank you I see were I went wrong with that, I'll edit it accordingly. Bar that slip would you say I'm on the right track with the mechanism? @ron

I'm just asking for confirmation about my method, organic chemistry isn't my strong suit and I was just looking for a second opinion. @jerepierre

Your mechanism looks correct to me.  Thanks for taking the time to draw clear structures.


text:
How to mix small amount of alcohol into coconut oil
I want to mix some alcohol (say rum) into a coconut oil base.
At the minute when I try this the alcohol seperates and sits on the bottom as a liquid. Is there any way I can mix these together so the alcohol remains suspended into the coconut oil.

Do you really want to add rum (= water + ethanol + frangrances) or do you just want to give the coconut oil a rum flavour? In the latter case, ethyl formiate might be the component that you want.

You need an emulsifier. I recommend mustard.

When you consider that rum and most other hard liquors contain at least 50% water, it is not surprising that the result is a two-phase system. If you began with absolute or 95% ethanol, or perhaps even grain neutral spirits, which vary but often come as 180 proof (90% ethanol) you might get the homogeneous solution you want. Another approach would be to add an edible emulsifier such as lecithin, which will help the ingredients to form a stable emulsion.
The earlier suggestion to hydrogenate the oil is not likely to work, as hydrogenation does not make oil able to mix with water. Coconut oil already has a high saturated fat content, as can be seen by the fact that it is often a solid at room temperature.

Coconut oil is solid at room temperature, but melts at temperatures slightly higher. It is physically similar to butter, which is typically 16% water (plus emulsifiers).
Butter may be a readily available additive to investigate. The first question is whether butter and coconut oil are compatible, then whether you could add some rum to this mix, since there is already some water in it, and the third thing to assess is whether you could mix the combination warm (liquid), then quickly cool it (say on a cold plate) to make it solidify.
Another readily available additive is dry milk solids. These could be blended into the rum - mmm - milky rum! - and then mixed well - really well - with some melted coconut oil. It would be like reconstituting an alcoholic superfatted milk.


text:
Does crystalline glass exist?
As far as I know, glass is a non-crystalline amorphous solid. Does glass exist in crystalline forms? Or is it the definition of glass that it's non-crystalline, in which case my question is based off of a false assumption?

That's right, the definition of glass is that it's amorphous, so the question as it is does not make sense. Some compounds just tend to be amorphous in certain conditions; when treated differently, they may become crystalline as well. When crystalline, however, they do not qualify as glass.

@IvanNeretin Thank you for explaining, I'm not well versed in chemistry.

Some of the confusion may be that glass with a high lead oxide content is typically referred to a "crystal" because of its sparkle and the fact that the glass is usually cut to highlight the high refractive index. But it too is amorphous. https://en.wikipedia.org/wiki/Lead_glass

Many compounds that usually form "glasses" can be cooled from the melt slow enough so they crystallise. In some cases this would take years, in others (e.g. atactic polystyrene) it is impossible.

As Ivan explained, a glass is by definition amorphous. However, you can take a quartz crystal (silicon dioxide, which is the main ingredient of glass) if it is of high purity and optical quality, and machine it into a transparent object such as a window pane.

There are very very few quartz crystals in the world big enough to get a regular window pane. Even then it would almost certainly be flawed.

Sapphire (i.e. corundum) is regularly grown into huge single crystals and sawn and polished into "not-glass" windows, mostly for expensive watches.

Huge crystals of quartz are also routinely produced.

"Lead crystal" usually refers to glass cut decoratively to partly resemble the facets of a cleaved crystal, so it's not really misleading, considering that most people recognise crystals by their faceted shape rather than their internal structure.
I looked into "crystalline glass" after being astonished at this term in a promotion involving what looked like ordinary glassware, with no flat surfaces. In European rules, "lead crystal" implies more than 24% lead oxide in the formula. There have been concerns about lead leaching from lead glass, so that alternatives such as barium have been replacing lead. Manufacturers are permitted to call low-lead or "lead-free" glass "crystal glass" or "crystalline glass". That seems a gross deception in the case that drew my attention, but in glass with similar designs to cut lead glass, perhaps not, although "crystalline" still seems more of a misnomer than "crystal".

Glass is by definition amorphous. If you were to crystallize it then it would be considered a ceramic, even though compositionally you have not changed your material. Often times the crystallization may occur as a result of slow cooling rates which then allows the formation of ordered arrangements of atoms (crystals) which are usually disallowed by more rapid cooling.
Sometimes partial crystallization occurs and is even desired. This produces a "glassy-ceramic" material which can give the best of both materials. Corelleware is famously strong and durable because its structure consists of ceramic particles in a glass matrix. This also gives it it's famous million needle-like shards when broken.


text:
How to read solubility graph

The following table is showing the solubility of CO2 at different pressures.
However, I cannot figure out how to read this graph. The y-axis is molar fraction so that should mean that its value = mol CO$_2$/(mol CO$_2$ + mol H$_2$O). The journal that I got this graph from does not specify how many moles/what mass of water is used. Can I assume that it's 1 and that the y-axis is actually just the moles of CO$_2$

Remember that molar fraction is an intensive property http://www.wonderwhizkids.com/conceptmaps/thermodynamic_properties.html

Given some amount of dissolved CO2, how would you calculate the molar fraction?

would't it be the way i have mentioned in my question?


text:
Mechanism for ring contraction from 2-methylenecyclobutanol to 1-methylcyclopropanecarbaldehyde

I am stuck between two paths. I was going to protonate the alcohol to give a better leaving group, leaving a 2 degree carbocation and then go on with a hydride shift. But, I'm not sure what that may turn out to look like, and when exactly the ring is supposed to contract. Or, I was going to have the alkene attack the acid, but after that I wouldn't know what to do with the alcohol.

How might acid react with this compound? Perhaps start there...

Mechanistically, both of those protonation options (at the alcohol or the alkene) are reasonable. Without any other information, I would probably try both and then see if either gets closer to the product.
However, two things suggest to me protonation of the alkene:
1) As suggested, protonating the alcohol generates a good leaving group (water). However, the product has a C-O double bond, so breaking the C-O single bond appears to be unproductive. That's not to say it's impossible, but it leads me to think about trying something else first.
2) The product has a methyl group, but the starting material does not. Somewhere in the mechanism, that methyl group has to be formed. If protonation of the alkene is the first step, then a methyl is formed, which I see immediately as productive. Again, that's not to say that this protonation is definitely correct, but at least something productive has happened.
One last tip before I show my suggested mechanism. For any mechanism that is difficult to draw, but especially rearrangements, it can be very helpful to draw in the hydrogens. In this case, drawing in the hydrogens just about gives away an exact mapping of the carbons from the starting material to product. With a mapping, it becomes easier to determine all of the bonds broken and formed, which is essentially a checklist of all of the goals of the mechanism.
I'll leave it to you to add the curved arrows, but the intermediates are shown below. Protonation of the alkene in 1 gives carbocation 2, which contains the methyl group seen in the product 4. Next is the ring contraction, sliding a C-C bond over the the carbocation and generating new carbocation 3. Compound 3 might appear to be less stable than 2, but it turns out to be quite stable due to resonance structure 3', which is fully octet. Deprotonation gives 4.

Contraction only takes place when the oh group is axial right? If the oh group is eqitorial then then hydride shift can take place to give a planar carbonation through pinacol-pinacolone rearrangement to give more stable 4 membered ring structure.

Hi Dhruv, you should consider supporting your answer with more sources/information. Leading with a request for confirmation makes it seem like you aren't sure of your own answer.


text:
Why do acetylene and chlorine react to yield carbon and HCl, but acetylene and bromine react less intensely?
In the famous "underwater fireworks" reaction we get
$$\ce{C2H2 + Cl2 -> 2C + 2HCl}$$
On the other hand, when we react acetylene with bromine, we get
$$\ce{CH#CH + Br2 -> CHBr=CHBr}$$
$$\ce{CHBr#CHBr + Br2 -> CHBr2-CHBr2}$$
Why don't we get carbon and HBr? Or do we get it?

Bromine is not as strong of an oxidizing agent as chlorine.  In essence the reaction of acetylene with chlorine is the same as burning, just with elemental halogen instead of elemental oxygen.
The higher electronegativity of chlorine explains why it is a good oxidant as compared to bromine.
Other notes:

Describing something as 'radically' (this phrasing removed  from original question edit) when you mean 'intensely' will
cause much confusion around here (Stack) since 'radically' refers to reactions involving single electron mechanisms.


text:
Minerals in water solidify when heated?
Is it possible for minerals in drinking water to solidify when high temperatures are applied? The minerals in mind are: iron, copper, aluminum, magnesium.
Background: I have small, brittle, darkish flakes appearing in my hot water system. The manufacturer of my tankless hot water heater said it's possible for these elements to solidify in this manner over time.
Looking at their website, I found this brief explanation:
http://support.noritz.com/article/faq-what-is-water-hardness-46.html

"Water hardness is the measure of dissolved calcium, magnesium, and other minerals in your drinking water. It is typically measured in grains of hardness which equates to 17.1 mg/L of dissolved calcium carbonate. If not treated hard water can cause mineral build-up in your heater. Mineral build-up reduces flow and produces flakes at your fixtures."

We got out water hardness measured, and it came back as 68 mg/L.
The temperature of the water heater is set to 130F. I believe that the heater has a higher internal temperature.
I'm not sure why they are dark/black in color, all the pictures I'm seeing for calcium carbonate are white.
Thank you

How often do you drain your hot water heater? If you have ever done it, you will see all kinds of gunk coming out. Exactly what depends on the details of the water coming in to your house and the condition of the tank and any sacrificial electrodes in it.

Hi, it's a tankless hot water heater, not sure if there is a place for them to collect in the same manner.

Well, my tankless states to flush it every year (or sooner if you have hard water). You get gunk out...

https://en.wikipedia.org/wiki/Hard_water

@JonCuster thanks I updated my question with a test result of the water's hardness. It is 68 mg/L.

Assuming that those $68~\mathrm{mg/l}$ are truly calculated as hypothetic $\ce{CaCO3}$ all I can say is: wow, your water is damn soft!

Limescale (calcium and magnesium carbonate) precipitates in the heater, the brown colour is most likely some iron compound that co-precipitates. A fresh surface of limescale likes to adsorb other ions, which then get trapped in it as the scale grows. If you had a water kettle, you'd see the same stuff in there after using it a few times. (Not exactly, because carbonate precipitation depends on the pressure, but iron (hydroxide?) not. I wouldn't want to predict.)
The limescale first appears on the surface of the heater cell, and then breaks off during temperature changes, i.e. when you open the faucet. As you have no tank, the flow rate of the water is high in the heater and carries the scale out of there.
Very convenient, that. In a tank, the scale builds up and can fill the whole thing after a few years, depending on your water source.

Thanks, do you know what could be causing the dark color? They flakes are generally black / dark grey in color (a few have brown or green smatterings). All the pictures I'm finding for calcium carbonate appear to be white.

also I updated the question with some test results we got from city water tests, the water hardness was measured as 68 mg/L fwiw.

It's most likely iron (or manganese). The stuff in the water heater in my kitchen has a light brown colour. The colour might vary with the pressure at which the stuff precipitates. Larger pressure will let less carbonate precipitate, so the amount of adsorbed ions (like Fe,Mn) could be higher. Thats speculation from my side now.

Ok thanks - sorry last question - if  the flakes are calcium carbonate, will they dissolve in vinegar overnight? Just want a household way to test if that's what it is.

The part that is calcium carbonate in them, definitely.

Your heater is set way to hot.
Set it at 108° then turn your shower as hot as it goes.
You can't handle any hotter.
Your adding cold water to cool it down to 108.
Calcium starts dropping out of solution around 125°
Two birds with one stone!


text:
Identify products of Propiophenone using nmr
I've been analyzing a compound "propiophenone", using NMR and IR, and
I can't tell which of the nmr spectrums relates to which isomer of C9H10O2 is which.

At first I thought the two isomers were ethyl benzoate and phenyl propanoate, which would make the 1.5 peak from the CH3 group and the 2.7 and 4.2 peaks from the CH2, and the CO stretch in ethyl benzoate is about 1730. But then I didn't know why the stretch in the other isomer would be 1770 and I thought that the ethyl benzoatate would be the major isomer, but I was told that the isomer D (with a stretch at 1730) is the minor isomer.

Someone can freely expand this hint into an answer - conjugation lowers the frequency of IR absorption for carbonyl compounds.

As this is clearly a homework type question, this answer will just give you some helpful pointers to head you in the right direction, and leave you to piece together the answer.
My usual approach for these types of problems is to have a look at the NMR data first, and ignore all the IR data. We might use that later for double checking.
From both spectra, it is possible to identify the following spin systems:

-CH2CH3
-C6H5 (monosubstituted benzene).

We also know we must have a -OC(O)- group in there also. That satisfies our molecular formula.
Looking at Isomer C first:

CH2 group at 2.7, indicative of it being adjacent to the carbonyl of an ester
Ortho aromatics at 7.0  and upfield of the meta, indicative of substitution with an electron donating (or aromatic activating) group, Ar-O-
I'll let you piece together the structure of C.

Looking at Isomer D:

CH2 group at 4.4, indicative of being adjacent to a heteroatom, so -O-CH2
Ortho protons at 8.0  and downfield of the meta , indicative of substitution with an electron withdrawing (or aromatic deactivating) group; Ar-C(O)-
I'll let you put this structure together also.

And what about the IR data? We know the following, which is consistent with what we propose from NMR:

Al-C(O)-O-Ar 1740-1770cm-1
Ar-C(O)-O-Al 1715-1730cm-1

The general rule with IR absorptions is that conjugation lowers the frequency - I'll leave you to rationalise this with some fancy arrow-pushing from the aromatic ring to the carbonyl group.

It is very obvious on NMR that C is the benzoate since the 2.7 2H is very typical for protons on the carbon next to the carbonyl. D is the propanoate since the 4.4 2H is for protons on the carbon with C-O bond on it.


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How to explain the different regioselectivity of ketones/imines reacting with LDA?
When 1 equivalent of LDA is reacted with, for example, 2-pentanone, you get the thermodynamic (more substituted) product. But when you react butan-2-imine with 1.0 LDA, you receive the kinetic (less substituted) product. Why?
I found out that the 2-pentanone reaction conditions included a temperature of around 0 degrees celsius, explaining the thermo product. The reaction conditions needed for the kinetic product would need a temperature of about -78 degrees celsius with more than 1.0 equivalent of LDA base. The reaction condition for the butan-2-imine included a temperature of -78 degrees celsius.

Well, I found out that the 2-pentanone reaction conditions included a temperature of around 0 degrees celsius, explaining the thermo pdt. The reaction conditions needed for the kinetic product would need a temperature of about -78 degrees celsius with more than 1.0 equivalent of LDA base. The reaction condition for the butan-2-imine included a temperature of -78 degrees celsius.

Think you answred your own question with your edit.


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Allotropes of Phosphorus
Something I noticed whilst revising about group 15 is that the darker the allotrope of phosphorus, the more stable it is.

white/yellow – very volatile, soluble, pyrophoric
red – air stable, polymer, more stable, one cross link;
black – most stable, polymer, maximum amount of cross link.

What's the reason for this?
Or rather, why would a more stable allotrope have a darker colour?

Random chance for the ordering of three items seems likeliest.

In phosphorus’ case, the connection between stability of the allotrope and its colour is only accidental.
White phosphorus, i.e. $\ce{P4}$ molecules, has very strained bonds because of the very small tetrahedral angles the atoms need to accomodate for. Going by the edges of a regular tetrahedron, the bond angles are $60^\circ$ which is very strained. This is the reason for its reactivity. Its colour is white, because there are no electron transitions excited by visible light, much like similar compounds of other elements.
Black phosphorus, on the other hand, forms large extensive networks which are well on the way to being metallic. Metallic bonding is typically such a widespread bonding network that every atom contributes to every molecular orbital with more or less the same contribution. You can compare this to large organic π systems. This allotrope is more stable because each phosphorus atom has bond angles that well suit the corresponding angles between atomic orbitals (somewhere just above $90^\circ$) and because that large network is hard to break. And since we have a lot of molecular orbitals with rather small differences in energy, light of the entire visible spectrum can be absorbed. There will always be a corresponding transition. Because of this, black phosphorus appears black.
Red phosphorus is somewhere in-between. The network is not as extensive, it is not as ordered and thus there are not as many molecular orbitals with good energy differences. Therefore, not all visible light is absorbed but green light is preferentially absorbed (giving a red colour). And also therefore, the reactivity is somewhere in-between white and black phosphorus.
(There is also violet phosphorus, which is somewhere between red and black.)


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Is a strontium–fluorine battery the highest voltage battery using pure elements?
Strontium has a very low standard electrode potential and fluorine has a very high one.
\begin{align}
\ce{F2 + 2e^- &<=> 2F^-} &\quad E^\circ &= \pu{+2.87 V} \tag{R1} \\
\ce{Sr &<=> Sr^+ + e^-} &\quad E^\circ &= \pu{-4.10 V} \tag{R2}
\end{align}
In theory, a strontium–fluorine battery would have a voltage of $\pu{6.97 V},$ although there are many practical reasons such as danger and rarity of materials for such batteries not to be made.
Is a strontium–fluorine battery the theoretically highest voltage chemical battery using pure elements, or is it possible to obtain a higher one?

You could always change the concentrations to change the EMF. But, if fluorine and strontium are indeed at the extreme opposites, you are right.

The half-potential you've given for strontium is only for the first ionisation. The half-potential you'd actually get is $\pu{-2.899 V}$ for the stable dication to give a cell potential of $\pu{5.769 V}$. From the CRC Handbook [1], lithium has the lowest element to stable ion potential of $\pu{-3.0401 V},$ which is why it is common in batteries.
Reference

Lide, David R., ed. CRC Handbook of Chemistry and Physics, 87th ed. Boca Raton, FL: CRC Press. 2006. ISBN 0-8493-0487-3.

Can more exotic stuff such as electrides, halides and alkalides achieve higher potentials? The only standard electrode potential I can find for one of these strange anions is for the hydride anion which is 2.23V which is still not better than flourine.

gsurfer04 is right that it wouldn't be strontium. If you're going really outlandish and willing to use explosive compounds, then the highest possible cell voltage would be obtained with the following redox reactions, using nitrogen to azide as the anode and krypton difluoride as the cathode:
\begin{align}
&\text{Anode:} &\quad \ce{3 N2 + 2 H+ + 2 e- &<=> 2 HN3} &\quad &\pu{-3.09 V} \tag{R1}\\
&\text{Cathode:} &\quad \ce{KrF2 + 2e- &<=> Kr + 2 F-} &\quad &\pu{+3.50 V} \tag{R2}
\end{align}
This would give $\pu{6.59 V}.$
More practical is to use lithium for the anode which is nearly as good at $\pu{-3.04 V}$ for $\pu{6.54 V}$ total. Hirashige et al. [1] tried using a $\ce{Li}-\ce{XeF2}$ cell (theoretically $\ce{6.49 V}),$ however they only managed $\pu{4.2 V}.$
Reference

Hirashige, T.; Hagiwara, R.; Ito, Y. Chemical Stability and Electrochemical Activity of Xenon Difluoride in Propylene Carbonate. Journal of Fluorine Chemistry 2000, 106 (2), 205–209. DOI: 10.1016/S0022-1139(00)00332-8.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Note that DOI links and complete references (ACS style) are preferred over publisher URLs which rot over time. Also, you might want to correct anode reaction and provide reputable reference(s) for the given reduction potentials.

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


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Basic Intro to Quantum Chemistry. Two Slit Experiment and Bohr Model
We just learned about the 2 slit experiment in Quantum Chemistry today, where electrons behave as waves when nobody is looking and behave as particles when they are being observed.
So, what would happen If I were to observe an atom (such that all it's electrons now behave as particles instead of waves) for a long period of time?
I know that real atoms have wave electrons, but when you observe them, don't they "turn" into particle electrons and obey Newtonian law like the 2 slit experiment? So what keeps the electrons from falling into the nucleus when you observe them? I was considering how the earth doesn't "fall" into the sun because of a momentum perpendicular to the centripetal force, but the mass of an electron is so much smaller than the nucelus, I'm not sure it will work.
If anyone can give a simple explanation that would be great, I asked the professor and he only said the the Bohr model was inaccurate and it's difficult to observe electrons.
Thanks in advance.

Firstly, electrons don't switch between particles and waves; that's a major misconception. Electron is neither wave nor particle. Waves are not real but are of mathematical construct; they impart probability amplitude of finding the electron at a certain coordinate at a specified time. When you measure, you are collapsing the wavefunction into Dirac-Delta function; even after a minute time-interval, the wavefunction again starts to evolve according to the Schroedinger's equation. Electrons don't move in orbits like solar system; we don't need  centripetal force.

And for why electrons don't fall to nucleus; this can be traced to the uncertainty principle; electrons can't get too localised for then the uncertainty of momentum would be very high- that means there might be chance of the electron having very high kinetic energy and getting ejected.

We cannot observe the electron without photons to (to see it by human eye) or radiation ( to at least detect it ) and both of these probes interfere with the electron in it's state of the electron gun but not in the ground state in the atom.

Electrons do fall on the nucleus. But they cannot do anything with the proton. They cannot form a particle like the neutron. The neutron has a mass which is significantly higher than the mass of a proton plus the mass of en electron. So the electron stays as near as possible to the proton, but it cannot do anything with it.

As some commenters have stated, this is a complicated topic, and if you really want to understand it you would probably be best to study quantum mechanics itself. Although it's detailed it's also interesting!
However, as to your question, the first thing to ask is what does it mean for an electron to be "observed"? It can't mean just looking at them with your eyes, because then any matter we looked at would suddenly go crazy when all of its electrons change their behaviour. It also doesn't mean just looking in a normal light microscope, because electrons are too small to see in a light microscope. In fact to really properly "see" an electron, we would need to use very powerful high energy equipment. So high energy in fact, it would give the electron enough energy to fly out of the atom it was attached to. After that it perhaps could well fly off like a "normal particle" electron with a Newtonian path.
Now the obvious question everybody then asks is, is there a way to "look at" the electron bound to an atom without giving it so much energy? And the answer according to physics is no, it is actually impossible, to really "look at" a bound electron enough to make it act like a particle, it always takes a big amount of energy. You could "look at" it just a little bit, just enough to find what energy level it is in for example - but as you probably know this isn't enough to stop it behaving like a spread out wave (atomic orbital).
Now as to your question about why it doesn't fall in to the nucleus, you have actually already answered it. Why would it? The earth doesn't fall in to the sun. You mentioned that electrons are very light, but that doesn't really matter. If we hollowed out the earth to make it much lighter that wouldn't affect its orbit. Now the real reason is a kind of "quantum mechanical" version of this argument that I could explain if you like, but it is complicated and really this is the key reason.


text:
Why are metals conductive in terms of orbitals?
Back in high school I was taught a definition of metals:

Metals are a lattice of positive nuclei embedded within a sea of delocalised electrons

Metals are conductive because the electrons aren't bound to any particular atom, and hence can flow freely throughout the entire metal.
However, I was recently taught about electron orbitals, and I'm not sure how to reconcile the two pictures. If the metal atoms were sharing pi or sigma bonds, I'm not sure how any of the electrons could be truly delocalised, and how conductivity arises. In terms of the atomic orbital theory, how does conductivity arise in metals?

AO theory explains atoms and ions. sigma and pi bonds are MO theory. That is a model for covalent bonds, molecules. Both are not very useful for metals.

@Karl Actually the bonding in solids can be discussed in a way that shows very strong links to MO theory for molecules and pretty much uses the same concepts. This has been done in a masterfully comprehensible way by Roald Hoffmann.

@Philipp So? All those models are based on  the same concepts.

@Karl Your first comment seemed to me to be implicating that there is little similarity between the bonding (and the concepts used to describe that) in molecules and solids. There I disagree, because you can find things like $\sigma$ and $\pi$ bonds (with slight differences compared to molecules) in solids and the LCAO-MO method to analyse the electronic structure of a molecule can be used (again with only slight alterations) for solids as well. To some extent you really can think of a solid as just a very big molecule...

... The problem is just that the vocabulary for the description of the electronic structur of molecules and solids differs due to being developped by different communities: for example a Bloch function is essentially just a symmetry-adapted linear-combination of atomic orbitals, i.e. an MO of the solid; a density of states is not much different from a MO diagram; the valence band edge is very much like the "HOMO" of a solid etc. Of course, one has to be aware that there are some differences and that additional effects come into play for solids but the basics can be understood with the...

... chemist's tools and intuition coming from a molecular perspective. I am only stressing this point so much because I feel that the chemists at my university (and maybe at a lot of other universities too) actually don't know this and the electronic structure of solids is not part of the general curiculum here in much detail. So, many chemists get deterred from this subject because they think it is a completely new world that is much more math-heavy than the way they think about molecules.

Hm. Are we talking about solids or specifically metals? There can be a lot of directed bonding in metals too, but then there are the electrons which make up the conductivity. How do you apply regular LCAO-MO to them? It's not two or three atoms as in regular bonds, but the whole solid! Of course the principles behind it are the same quantum mechanics.

It's not that the electrons are completely unbound from the atoms - they still attract each other and form the bonds with other atoms. But the attraction between the atom and its electron is pretty weak so it's easy to detach the electron from the atom.
The properties of the metals are better to be described via ionization energy - the energy needed to remove an electron from an atom. Elements to the right of the periodic table have high ionization energy - they are not willing to lose electrons. Therefore these elements are bad conductors (and are not metals).
Metals have lower ionization energy - their valence shell just started to be filling and the strength of the proton is not enough to keep them tight. Therefore these electrons are free to flow with minimum efforts applied - which would form electrical current.
The intention of the "sea" model is to describe (in simplified manner) the fact that electrons are floating around from one atom to another. And while they participate in bonding, this is a dynamic structure that breaks and forms all the time.
Here is a nice video from Khan Academy on the metallic nature.


text:
Safe endothermic reaction that gets below freezing?
I have a project/idea I am working on but I am only 14 years old and I need some help.
I'm looking for an endothermic reaction that gets below freezing, but the chemicals before the reaction and after the reaction is over are safe to touch and safe for the environment. The reaction can be harmful to plant life, but only for a short time before it decays or whatever needs to happen for it to be safe. I do not want it to have long term effects.
Additionally, two important parts are that these substances need to be easy to obtain and need to not to be a fertilizer.

A classic one is ammonium nitrate.

@YunfeiMa Not especially safe to handle (oxidizer, can explode if mixed with the wrong thing), potentially difficult to obtain (restricted e.g. in USA b/c of explosive uses), and it's a fertilizer.

The reaction of baking soda and citric acid (both are easy to come by and safe) in water does cool the water noticeably. I have tried it multiple times, and while it does not reach freezing, it will reliably reach $6\ ^\circ \mathrm{C}\ \left(43\ ^\circ \mathrm{F}\right)$, which is cold to the touch.

Add 4 teaspoons each of sodium bicaronate (baking soda) and citric acid to a cup. Then add the smallest amount of water you can and start stirring!

You can tell a chemical reaction is occurring because a gas is produced:
$$\ce{3NaHCO3 + H3C6H5O7 -> Na3C6H5O7 + 3CO2(g) + 3H2O}$$
But, wait! We want a temperature below freezing! It might seem a little like cheating, but the reaction still works. Carbon dioxide bubbles are still produced, and the temprature gets down to around $-6\ ^\circ \mathrm{C}\ \left(21\ ^\circ \mathrm{F}\right)$.
If you do not have a thermometer, you can prove the second version is colder than the freezing point of water by doing the following:

Get a small, flat piece of wood and dampen the top surface of it.
Do the reaction in a glass, using 4 teaspoons each baking soda and citric acid and replacing the water with crushed ice.
As the temperature drops, the water between the wood and the glass will freeze, freezing the glass to the wood. If you lift the glass, the wood will stick to it.

The evaporation of hexane is endothermic. n-Hexane has a low enthalpy of vaporization at $31.56\ \mathrm{kJ\ mol^{-1}}$ at $25^\circ \mathrm{C}^{[1]}$, compared to water, which is $43.99\ \mathrm{kJ\ mol^{-1}}$ at $25^\circ \mathrm{C}^{[2]}$. This means that hexane quickly and easily absorbs heat from its surroundings, and can get well bellow $0^\circ \mathrm{C}$. Indeed, I recall performing an experiment in high school whereby paper towels dipped in different solvents were wrapped around the end of thermometers, and the lowest temperatures were recorded. By far hexane was the lowest at something below $-10^\circ \mathrm{C}$.
n-Hexane is definitely not a fertilizer, and doesn't do any serious damage to the environment like chlorofluorocarbons do. Your main concern would be storing it in a cool well ventilated space and keeping it away from open flames, as it is very volatile and flammable. Additionally I would recommend you work with nitrile gloves, as they are insoluble in hexane. Hexane is a very non-polar molecule and can dissolve your epithelial cell membranes. It's not immediately dangerous, however, so just take caution not to get too much on your skin.
If you find hexane too hard to obtain, acetone would have similar qualities, and is easily obtained at the grocery store as nail polish remover.

[1] Majer, V.; Svoboda, V., Enthalpies of Vaporization of Organic Compounds: A Critical Review and Data Compilation, Blackwell Scientific Publications, Oxford, 1985, 300.
[2]
Marsh, K. N., Ed., Recommended Reference Materials for the Realization
of Physicochemical Properties, Blackwell, Oxford, 1987.

Came across this and wanted to add something. n-hexane has been classified as causing nerve damage and polyneuropathy at high levels of exposure. Neurotoxicity has been observed in rats. Not one to use. Always check all reactants and products for a reaction with CLEAPPS, the EPA or another hazcard provider. https://www.epa.gov/sites/production/files/2016-09/documents/hexane.pdf


text:
Why is 2H2O2 -> 2H2O + O2 a first order reaction?
My book says that the reaction $\ce{2H_2O_2 -> 2H_2O + O_2}$ is a first order reaction and that it's rate equation is $\ce{k[H_2O_2][I^{-}]}$.
However i don't see why that would be first order? Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

I- is catalyst not a reactant

Since the concentration of iodide ion remains constant its concentration can be taken under equilibrium constant( just as we do in pseudo 1st order reactions) so the order will be 1st

It is better to say "a first order reaction in $\ce{H2O2}$". That would be right and unambiguous.

@IvanNeretin      Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- " .  So, what is meant by term with respect to in this context?

It simply means that the reaction rate is proportional to the first power of $\ce{[H2O2]}$.

@IvanNeretin       Can't we deduce from the  statement that the reaction is proportional to 1st power of I-

Yeah, that too.

Catalyst decrease  $Ea$ ,so $k$ increases

Summarizing from the excellent comments:

Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

Yes, that is how you get the overall order of reaction. If you compare the rate with a given set of concentrations to the rate with all concentrations doubled, the rate goes up by two to the power of the overall order.

I don't see why that would be first order?

It is not, but it is pseudo-first order. If you start the reaction using certain initial concentrations, you will find an exponential decay of the reaction rate over time (i.e. there is a constant half-life).
As JM97 comments, the iodide is a catalyst, so its concentration does not change over time. So while the concentration of hydrogen peroxide drops over time and causes the reaction to slow down, the iodide concentration in the rate law does not contribute to a slowing down (no matter what order the reaction is in iodide).

[OP responding to Ivan...] Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- "

As Ivan comments, that means the reaction rate is proportional to the first power of the hydrogen peroxide concentration (if the iodide concentration is unchanged). It is also proportional to the first power of the iodide concentration (if the hydrogen peroxide concentration is unchanged). If both are changed at the same time, you use the rate law to see what the effect is.


text:
Volatile nature and odour
Will it be correct to make a general statement that all compounds having odour are volatile?
My reasoning: if a compound has odour then to smell that it must evaporate ( ie. some molecules must diffuse in the surrounding) . So I conclude that for smelling the compound must be volatile.
Am I right?

It's true that all odors come from compounds with at least some volatility.
However, some of the smells we associate with certain materials (like some metals) aren't the metal themselves but products of reactions caused by those materials, so you have to be careful about that.


text:
Why is this stereocenter by the Br an R? I thought it was S?
Why is this stereocenter by the Br an R? I thought it was S? For problem b in the following image:

It is S. That image is confused. Image, go home. You're drunk, image.

The descriptors for the first structure are also wrong. The name for a. should read (2S,3S,4S)-4-amino-3-(1-ethyl-1H-pyrrol-2-yl)pentan-2-ol.

Both structure (a) and structure (b) have misassigned chirality centers. There should be only one $R$ assignment over all four structures. The correct assignments are as follows:

Structure (a)

The misassigned chirality center is the one in the center. The carbon atom with the OH group has the highest priority (one C-O bond beats any number of C-C or C-N bonds). The second highest priority is the pyrrole ring (one C-N bond and one C=C bond beats one C-N, one C-C, and one C-H). The carbon atom with the primary amine is the third priority. The groups are arranged counter-clockwise and the lowest priority (H) is in the back.

Structure (b)

The misassigned chirality center is the one with the bromo group. The bromine atom is the highest priority. The carbon atom bonded to chlorine is second, and the carbon atom bonded to fluorine is third. The groups are arranged clockwise, but the lowest priority (H) is forward.


text:
Reference request - Nomenclature of organophosphorus compounds?
Can someone please provide me a good reference for the nomenclature of organophosphorus compounds such as pesticides and chemical warfare agents? Something from the IUPAC would be great.
Thanks.

The usual substitutive nomenclature for organic compounds was extended to various other elements including phosphorus. The current rules for name construction of organic compounds containing phosphorus are included in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). These recommendations cover typical organophosphorus compounds that are used as pesticides and chemical warfare agents. However, they do not include the nomenclature for organometallic compounds, coordination compounds, polymers, and some natural products.


text:
How to calculate the abundances of certain gases from the peak areas, GC-MS
In an experiment we had a series of air samples which were analysed through GC-MS, before using TargetLynx software to integrate the mass spectrum obtained.
So I now have the specific peak areas for a number of compounds (for example CFC-12 has an area of 28335299.35m^2) and a calibration standard for each specific gas investigated. For example CFC-12 has a mixing ratio (ppt) of 539.894.
With this information how could I determine the actual abundance of CFC-12 from the air sample?
Apologies for any confusion!
Thanks!

What is the peak area for the CFC-12 at 540 ppt?  That is MUCH smaller that your sample so you really need to use a concentration of CFC-12 standard that gives a similar or somewhat greater peak area.  Running the samples at a few concentrations comparable to that of your sample (similar peak areas), ranging over a factor of 10 or so (not a factor of 50k like in your case!)

If you only have a single calibration standard concentration for each compound as your question implies, then you will simply need to calculate a calibration factor for each compound then multiply your measured peak area by the calibration factor.
This is also know as a single point calibration and it is important that the peak area of the standard is greater than that of your sample or you might be out of the linear response range of the instrument.  Also, a single point calibration assumes that the instrument response is linear from zero up to the concentration of the standard, or at least that the concentration of your sample is very close to that of your standard.
In your example you would calculate the calibration factor as:

$\mathrm{539.894\ ppt / 28335299\ m^2 = 1.905\ x\ 10^{-5}\ ppt/m^2}$

Then if you had a sample with a peak area of 18335299.12, the concentration of CFC - 12 would be:

$\mathrm{18335299.12\ m^2\ x\  1.905\ x\ 10^{-5}\ ppt/m^2\ =349\ ppt}$

Note: The calibration factor is often calculated in the inverse manner of my example above.  If calculated in that way, then of course you would divide your peak area by the calibration factor.


text:
Is ethylene diamine tetra acetate a misnomer?
Since there is no double bond, I thought EDTA should be ethane... instead. Am I missing something?
I understand this is not the complete IUPAC name, but why name it wrong on purpose?

Related: Why doesn't EDTA have a double bond? and Why are PTFE and FEP named “ethylene” when they don't have double bonds?


text:
Fraction of cube Volume empty
The unit cell in a body centered cubic lattice is given. Each sphere has a radius r and the cube has a side a. What fraction of the total cube volume is empty?

The fraction is known as the "packing fraction". https://en.wikipedia.org/wiki/Atomic_packing_factor

Assuming the $a$ is large enough to accommodate a BCC lattice with spheres of radius $r$, we may define
$V_s = \frac{4}{3}\pi r^3$ is the volume of the sphere
$V_c = a^3$ is the volume of the cube.
in a BCC lattice, there are 8 corner points, and one central. Because the lattice points on the corners are defined with respect to the center of each of the spheres, we know that each of the corner spheres will contribute $\frac{1}{8}$ of their total volume to the occupied volume of the unit cell. In addition, the sphere at the center will contribute its entire volume to the occupied volume. We may then write the total volume occupied in the unit cell as
$V_{occ} = 8\left(\frac{1}{8}\right)V_s + V_s =2V_s$
We know that the total volume of the unit cell is that of the cube, meaning that the unoccupied (empty) volume of the unit cell may be written
$V_{unocc} = V_c - V_{occ} = a^3 - \frac{8}{3}\pi r^3$
Wherefore we may write the empty fraction as
$\frac{V_{unocc}}{V_c} = \frac{a^3 - \frac{8}{3}\pi r^3}{a^3} = 1 - \frac{8\pi r^3}{3a^3}$