Is there a field of chemistry that doesn't do experiments? In other words, a field of chemistry that is all theoretical? I have heard of theoretical chemistry, but my chemistry professor said you can't really major in theoretical chemistry. He said the closest thing is physical chemistry.

I like predicting in Chemistry, but I hate labs or experiments.

You are correct that it is possible to study theoretical or computational chemistry, however these typically only represent a handful of courses in a normal undergraduate chemistry curriculum. Additionally, most undergraduate chemistry curriculum are designed to instruct students broadly in the core disciplines of chemistry, hence it's usually not possible to specialize.

Non-chemisty degrees may be an alternative. However any science undergraduate program will require laboratory-based coursework. Hands-on experience teaches us how scientific concepts work in the real works, giving us an appreciation of how difficult experimental work can be. It also functions to reinforce content from books and lectures; to teach technical skills (which are usually necessary for future employment); and to develop lab safety habits. So there really is no means to escape it. Moreover, in order to take advanced chemistry electives, prior courses which include lab sessions are generally necessary.

That said, there are other routes to get into chemistry. I've know individuals whom have majored in mathematics, engineering, physics, and biology who either did research in chemistry or who did graduate degrees in chemistry subsequent to obtaining an undergraduate degree in those fields. However in order to obtain undergraduate degrees in those fields and to be able to transition to chemistry programs, lab experience and some senior-level chemistry courses are typically required.

However, once in graduate school, it is possible to avoid experiments altogether: theoretical and computational chemistry graduate degrees typically do not require that the candidate engage in any hands-on experiments. Many however do require that the graduate student work as a teaching assistant in undergraduate chemistry laboratories as a condition of funding. In such cases, laboratory experience is either necessary or a significant advantage.

So while it may be possible to minimize the amount of lab work that you engage in, some will always be required. Additionally I would not recommend avoiding lab work, as it greatly enriches the study of chemistry overall.


Is PCl5 really trigonal bipyramidal in shape? We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

In the case of $\ce{PCl5}$, why should the shape be trigonal bipyramidal? Obviously the top and bottom $\ce{Cl}$-atoms, above and below the plane subtend $90^\circ$ with each of the $3$ planar $\ce{Cl}$-atoms; whilst planar ones subtend $120^\circ$ in between each other. This means all bond angles aren't equal. Why so? Shouldn't there be regular distribution of $\ce{Cl}$-atoms in space?

To my opinion, the XY-plane distribution of $\ce{Cl}$-atoms should be such that each $\ce{Cl-P}$ is $72^\circ$ apart from the other $\ce{Cl-P}$ bond (atoms would not be in the same plane at all. But we should see that the orthogonal projection of each on XY-plane subtends equal angles, i.e. $72^\circ$). Same for YZ-plane. Hence there is regular distribution of atoms along all planes. Now it is to be calculated what the ultimate structure results in.

Sorry, I don't have the necessary tools/knowledge to represent my hypothesis. Else could've just modelled this figure and shown a representative model.

The trigonal bipyramidal shape of $\ce{PCl5}$ is indeed the lowest energy conformation.

I have had a very hard time figuring out what your description of the structure meant and I am still not convinced I have actually understood correctly. However I was trying to model it in my head I either arrived at a variety of different angles, or at least two of them were equal to $72^\circ$, or both of these cases.

The only structure $\ce{ML5}$, where all $\ce{L-M-L}$ bond angles are equal is a regular pentagon, i.e. all ligands are in one plane, with $\angle(\ce{L-M-L})=72^\circ$, e.g. $\ce{[XeF5]-}$.

Another possibility for (almost all) equal bond angles is a square pyramid, where $\angle(\ce{L-M-L})=90^\circ$, e.g. $\ce{ClF5}$. Note however, that there are also other angles, so not all bond angles are equal, $\angle(\ce{L-M-L})=180^\circ$.

There is no better solution to arrange 5 point 'uniformly' on a sphere than the trigonal bipyramid. This is also known as the Thompson Problem, which is addressed in the linked question on Mathematics.se, provided by orthocresol: 5 Points uniformly placed on a sphere.
Another point of view would be to ask the question whether there is a way to put 5 points on the surface of the sphere so that they are indistinguishable, which is also answered there. Spoiler warning: there isn't.

You can dive deeper into the matter. For example for $\mathbb{R}^n$ there can only be $n+1$ points equally equidistant to each other. In our perceived reality $n=3$, which makes the tetrahedron the most complex molecule with this property. See on mathematics.se: Proof maximum number of equidistant points for a given dimension (duplicate), Prove that $n+2$ points in $\mathbb{R}^n$ cannot all be at a unit distance from each other (duplicate), $n$ points can be equidistant from each other only in dimensions $\ge n-1$? (Which are all duplicates, but still have answers attached to them.)

And then there are only five Platonic solids, none of which has five vertices.

Furthermore, I would like to add to all of that another word of caution: Molecules are flexible. The trigonal bipyramidal arrangement of ligands just happens to be the low energy conformation for a variety of reasons (considering the necessary and useful approximations, e.g. Born-Oppenheimer approximation).
The molecule (and other analogues) are special though, as the undergo pseudorotation according to the Berry Mechanism (see Wikipedia).

Side note:

  We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

This is only half the truth. It is certainly true that protons repel each other, more importantly though is that electrons repel each other, too. The most important part of which is also built into the very simple VSEPR theory. While this theory has its uses, there is a lot of criticism attached to it and one should be aware of that. I shall stress again at this point that VSEPR theory is unrelated to orbital hybridisation; there are quite a few textbooks which do make this post-hoc rationalisation, which can be misleading and dangerous.


What's the most complete basis set for calculating chemical shifts with DFT? I'm trying to benchmark some NMR chemical shifts. The literature suggests that the pcS-4 basis set will be the most accurate. I was wondering if there were any basis sets that are more complete than the Jensen basis sets? Even for just general basis sets. I'm looking for something similar to the Partridge basis sets, except for DFT.

Using pcS-4 only tells you how much the error from your basis set is. This doesn't necessarily have any link to the total error, which comes from basis set, DFT functional, molecular geometries, treatment of solvation, and various other aspects of your chosen method. So, if all you want to do is to investigate the effect of basis set while keeping everything else the same, then sure, go ahead and use the pcS-n basis sets. They were designed with convergence to the complete basis set limit in mind, so pcS-4 is probably as close to the CBS limit as you can reasonably get. For a recent (and very thorough) example of benchmarking using pcS-4 (technically the segmented version pcSseg-4), see J. Chem. Theory Comput. 2018, 14 (2), 619–637.

If your aim is just to get the computed shifts as accurate as possible, then CCSD(T) with large pcS-n or cc-pVnZ (or even extrapolation to the CBS limit) is probably what you want. For benchmarking you're not likely to be using any excessively big molecules anyway, so for a one-off calculation I imagine the computer time needed is quite tolerable.

As far as DFT is concerned, as has already been pointed out, if you don't use a good functional then no matter what basis set you use, the shifts will not be great anyway. Hybrid functionals like PBE0 or mPW1PW91 are popular in the literature (and B3LYP of course), but there are some other interesting ones like M06-L (see J. Phys. Chem. A 2008, 112 (30), 6794–6799), and recently double-hybrid functionals have been suggested (J. Chem. Theory Comput. 2018, 14 (9), 4756–4771).

Aside from that, there are also empirical ways of increasing the accuracy of DFT-predicted shifts, which you may or may not want to take into account when benchmarking. A popular one is linear scaling of shifts (the slope and intercept being determined from a given test set). Tantillo has a nice collection of these scaling factors at http://cheshirenmr.info/index.htm. Although some of the work cited can be a bit outdated in such a rapidly-moving field, the methodology itself is still highly relevant.


Benzenesulfonyl chloride dehydration of an amide to give nitrile 1) I am unable to understand how benzenesulfonyl chloride acts as a dehydrating agent in dehydration of amides? What is the driving force for losing chloride and taking the oxygen from the amide?

2) Why I can't use other dehydrating agents like $\ce{P4O10}$, $\ce{H2SO4}$, or $\ce{CaCl2}$?

3) How can I estimate what is suitable dehydrating agent for a particular reactant?

4) Is it necessary to use only those dehydrating agents that undergo chemical change? Or can I use dehydrating agents that adsorb water like silica gel?

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

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


How would you determine the RDS from the rate and order of a reaction I know what a rate determining step is, and also have knowledge about zero order and first order reactions. But, in one of my tests, I was unable to solve the following question:

  A hypothetical reaction A2 + B2 --> 2AB has a rate law, Rate = k[A2]1 [B2]0 and mechanism of reaction is

  Step-I: B2 --> 2B

  Step-II: A2 --> 2A

  Step-III: 2A + 2B --> 2AB

  The correct statement is

  A) Step-I of mechanism is rate determining step of the reaction

  B) Step-II of mechanism is rate determining step of the reaction

  C) Step-III of mechanism is rate determining step of the reaction if step I and II are reversible

  D) Step-I and Step-III both are rate determining step of the reaction if step II are reversible

The correct answer is option B, but how would you approach the problem and solve it?

EDIT: What if rate was given as

1) Rate = k[A2]1 [B2]1

2) Rate = k[A2]2 [B2]1

The reaction is 1st-order in A2, so for starters we look for a step with A2 on the left, with a coefficient of 1. In fact, only step II has A2 on the left at all, so really we don't have to look any further.

Comments which may (or may not) make it clearer:

The rate equation says that greater the A2 concentration, the faster the reaction goes, in a linear relationship, and the rate doesn't depend on [B2] at all, since [B2] has an exponent of 0, and anything to the zeroth power is 1. (I suspect that was just put there to make it hard.)

If the whole reaction can be speeded up by adding A2 alone, [A2] must be rate-determining; i.e., [A2] determines the rate, and A2 reacts only in step II.


Why do some solvents bubble more than others when boiling? This question about boiling milk reminded me of my everyday lab work at the rotavap: Some solvents and solutes are feared for the intense bubbling they create while the solution is boiling — be it at ambient pressure like the milk or at reduced pressure in the rotavap. For most of the infamous solvents, the bubbles will even create a somewhat stable foam (that usually ends up on the wrong side of the rotavap).

Well known in my lab are dissolved triphenylphosphane and the solvent toluene while isohexane (‘mixture of hexane isomers including methylcyclopentane’) is known to hardly ever give bubbles.

Why is this so? Why are bubbles often so much more stable even for just a solvent (toluene)?

Boiling depends on certain properties of a solvent:

Tendency to form an emulsion of bubbles due to low surface tension and separation of solvent phases, as in soap bubbles. Try boiling some dish-washing detergent in water.
Tendency to superheat, i.e. to go well above the boiling point and then to suddenly form bubbles, which is more pronounced at lower pressure. This can cause violent bumping, in addition to foaming. The cleaner the apparatus, the more likely, so boiling stones (chips) are added to nucleate more gentle boiling. A glass mug of water heated in a microwave can bump violently enough to blow open the door (and blow the fuse).


how do i find how much energy needs to be removed? How much energy must be removed to convert $8.49\ \mathrm{g}$ of water at $24.5\ ^{\circ}\mathrm{C}$ to an ice cube at $-12.0\ ^{\circ}\mathrm{C}$?  Give your answer as a positive value in Joules since only positive energy can be removed.
$$c_{\mathrm{water}} = 4.184\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$
$$c_{\mathrm{ice}} = 2.03\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$
$$\Delta H_{\mathrm{fusion, water}} = 6.01\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$

O am just learning this material, and I'm struggling on what formula to use. Is it

$$q= n(H_{\mathrm{final}} - H_{\mathrm{initial}})$$

Firstly let us understand that heat removed would be directly proportional to the decrease in temperature and also the mass. $$\delta Q\,\,\alpha \,\, m\delta T$$  Here the specific heat capacity appears as the constant of proportionality.
Therefore we have, $$\delta Q\,\,=\,\,mc\delta T$$
Now all you have to do is to use this formula to convert 8.49g of water at 24.5°C to water at 0°C.
But after this you need to use the concept of latent heat i.e. heat you need to remove to convert water at 0°C to ice at 0°C.As this value is given as 6.01 kJ/mol you will have to remove that much amount of heat for every mole of water(so you need to convert grams into moles).
Now what you have is 8.49 grams of ice at 0°C. Now just use the formula again to convert ice at 0°C to ice at -12°C (using the value of c of ice this time).
At the end you need to add all those heat energies and you are done !!
The sum of heat energies is the amount of energy you need remove from 8.49g water at 24.5°C to convert it to ice at -12°C.
Hope this helps.


Mass fraction in a mixture of metal salts given the total mass of the metal I am trying to solve a question in my study and I don't get the answer required.

  A mixture of $\ce{NaNO3}$ and $\ce{Na2SO4}$ weight $5.37$ grams, the $\ce{Na}$ = $1.61$ grams. What is the percentage mass of the $\ce{NaNO3}$ in the mixture?

The correct answer according to the book is $45.1\%$, but I get $37\%$.

What is did is:

The weight of the Na is 1.61 total, which means the rest is 3.76 grams
total mass/Molar mass of the rest is 3.76/158 = 0.0238 mol
m(N) = 0.0238*14 = 0.33 grams
m(O3) = 0.0238*48 = 1.09 grams
m(Na) = 1.61/3 = 0.54 grams
total mass = 1.96 grams

1.96/5.37 = 37%

Where and what am I missing or mistaken?

You need to do a bit more math than that.
In order to solve the majority of problems in chemistry you need to switch to the amounts  of substances.

So, let's use the following notations: $\omega$ – mass fraction; $m$ – mass; $n$ – amount; $M$ – molecular mass.
I refer to $\ce{NaNO3}$ as compound $1$ and to $\ce{Na2SO4}$ as compound $2$.
Let's solve this algebraically first.
from the definition:

$$\omega_1 = \frac{m_1}{m} = \frac{n_1M_1}{m}$$

In order to find $n_1$, we use the sodium content in both salts

$$n(\ce{Na}) = \frac{m(\ce{Na})}{M(\ce{Na})} = n_1 + 2n_2 \implies n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - 2n_2$$

and total mass:

$$m = n_1M_1 + n_2M_2$$

Combining both equations (I let you do the math), you find $n_1$:

$$n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}$$

Now you can find the mass fraction of sodium nitrate:

$$
\begin{align}
\omega_1 &= \frac{\left(\frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}\right)M_1}{m} \\
  &= \frac{\left(\frac{\pu{1.61 g}}{\pu{23.00 g mol-1}} - \frac{2\left(\pu{5.37 g} - \frac{\pu{1.61 g}}{\pu{23.00 g mol-1}}\cdot\pu{85.00 g mol-1}\right)}{\pu{142.04 g mol-1} - 2\cdot\pu{85.00 g mol-1}}\right)\cdot\pu{85.00 g mol-1}}{\pu{5.37 g}} \\
  &= 0.4513~\text{or}~45.13\%
\end{align}
$$


Is probability of finding electron changing here? We all know what is a wave function. There is a technique to convert non normalized wave function to normalized wave function.

In case of non normalized wave function integration of square of wave function with respect to 'dt' in a length 'dx' will be any constant other than 1. In case of normalized wave equation the probability of finding will be maximum i.e. 1.

The process of converting a non-normalized wave function to a normalized wave function is known as Normalization technique. In this technique we multiply the wave function with some constant and makes it equal to 1 (condition for a normalized wave function). That means after conversion the probability of finding electron is becoming maximum.

If we multiply a wave function with any constant it's nature doesn't change. So, my question is after normalization of the wave function, the nature of the wave doesn't changes and how come the probability of finding electron becomes maximum i.e. 1???

I swear I read the question many times, but I still not quite sure what it is about.  So I will try to clarify some things hoping that they are relevant.

First off, any complex multiple of a wave function indeed describes exactly the same state of a system. This directly follows from the postulate of quantum mechanics which states that the expectation value of an observable $A$ represented by a self-adjoint operator $\hat{A}$ is given as follows,
$$
\langle A \rangle
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
where $\Psi$ is the wave function. It is indeed easy to show that given $\Psi$ we can multiply it by any complex number $c$ and this would not change the expectation value of any observable,
$$
\langle A \rangle
=
\frac{\langle c \Psi | \hat{A} | c \Psi \rangle}{\langle c \Psi | c \Psi \rangle}
=
\frac{c^{*} c \langle \Psi | \hat{A} | \Psi \rangle}{c^{*} c \langle \Psi | \Psi \rangle}
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
since $c^{*} c$ in numerator and denominator perfectly cancels each other. Also note that the following definition of the expectation value
$$
\langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle \, ,
$$
holds true only for a normalized wave function $\Psi$ and trivially follows from the above mentioned more general definition.

Secondly, for the simplest case of one particle in one spatial dimension, if one would like to interpret the wave function $\Psi(x,t)$ as a probability amplitude that the particle is at $x$, and consequently, it square modulus, $ \left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t)$ as the probability density that the particle is at $x$, he/she have to normalize it as follows,
$$
\int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1 \, .
$$
This is required because one of the axioms of the probability theory requires that

  the probability that some elementary event in the entire sample space
  will occur is 1.

The sample space is just a set of all possible outcomes in an experiment, which in our case is the set of events of finding a particle at any possible position, i.e anywhere between $-\infty$ and $+\infty$. Thus, in our case the probability of finding the particle anywhere between $-\infty$ and $+\infty$ have to be 1, and consequently, the wave function has to be normalized. Otherwise its interpretation as a probability amplitude would be inconsistent with the probability theory.

Hope it helps.


Nitrobenzene reduction with (tin) Sn catalyst: Why is C-H bond cleavage preferred over O-H bond cleavage? I am trying to understand this mechanism for nitrobenzene reduction with $\ce{Sn/HCl}$. I do not understand why it implies that you pull off the $\ce{C-H}$ hydrogen instead of the $\ce{O-H}$ hydrogen in the methanol.  Is it actually suggesting that the resulting $\ce{.CH2-OH}$ radical formed from the methanol eliminates to form formaldehyde or do you get a rearrangement where you form $\ce{(CH3-O)-}$? If the latter is the case, why not just pluck the proton from the alcohol group instead of breaking the $\ce{C-H}$ bond?

Also, it mentions that organic solvents are preferred for this reaction, why exactly is that over water?

I don't buy that mechanism.  Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available.

Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism.  Reductions of nitro compounds, be they catalytic hydrogenations or metal surface - acid catalyzed reductions, usually proceed through nitroso and hydroxylamine intermediates.  This link provides some discussion of the reaction and presents the following (more reasonable) mechanism where electrons are transferred from the metal surface to the various intermediates and the acid provides a ready source of protons.  These reactions are usually run in an organic solvent in order to bring the organic nitro compound into solution.


Do non-classical carbanions exist? There are some non-classical carbocations, such as the 2-norbornyl cation, in which the positive charge is heavily delocalised. Have non-classical carbanions also been discovered? I have not seen any non-classical carbanions in books. If there are any non-classical carbanions, please provide some examples.

Interesting question. It is much less studied and reported on than the case of non-classical carbocations, but I did find a few papers. Brown and Occolowitz (Ref.1) reported that deuterated bicyclo[3.2.1]octa-2,6-diene 1b, below, undergoes base-catalysed de-deuteration (potassium tert-butoxide in DMSO) to give 1a much faster (ca. $3 \times 10^4$) than the bicyclo[3.2.1]octa-2-ene counterpart, i.e. 2b $\rightarrow$ 2a. They suggested this is evidence for the intermediacy of a 6π-electron, delocalised (non-classical) carbanion 3.

Much more recently, Brown reviewed and provided additional DFT computational evidence for stabilisation and charge delocalisation in these species (Ref.2).

References:

J. M. Brown, J. L. Occolowitz, “A non-classical carbanion,” Chem. Commun. (London) 1965, (16), 376-377 (https://doi.org/10.1039/C19650000376).
John M. Brown, “Origins of Stabilization and Evidence for Charge Delocalization in the Bicyclo[3.2.1]octadienyl Anion and Related Species,” Australian Journal of Chemistry 2014, 67(9), 1296-1300 (https://doi.org/10.1071/CH14244).


Burns from boiling water and steam Why is a burn to the skin caused by steam more serious than a burn caused by the same amount of boiling water at the same temperature?

The temperature is the same, which implies that the kinetic energy of the particles in both the steam and the boiling water are the same; and yet the steam is more dangerous. Why so?

Let’s consider the following cases:

getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin

With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while cooling down to $40\,\mathrm{°C}$.

1st case

Assuming isobaric conditions (constant pressure), the heat energy liberated from $100\,\mathrm{°C}$ water that is cooled to $40\,\mathrm{°C}$ can be calculated as follows:

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 4.5\,\mathrm{kJ}$$

2nd case

A similar formula holds true. However, now we are considering a gas all the way, not a liquid. (And ‘air’ is technically not a single substance, but we can work with it.)

$$\Delta Q = \Delta T \cdot C_p(\mathrm{air}) \cdot n(\mathrm{air})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 29.19 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 1.7\,\mathrm{kJ}$$

3rd case

Unlike the previous cases, we start off with steam, which will condense to water. Therefore, we need to add the heat of vapourisation to the equation.

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O}) + \Delta H_{\mathrm{vap}} \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol} + 40.66 \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 45.2\,\mathrm{kJ}$$

You will note that the third case delivers a lot more heat energy than the first two cases.

(Part of the simplifications I used means that I grossly overestimated the heat transferred by hot air. Which is also why one can put one’s hands under a hair dryer and be fine with it.)


Why does a hydride shift take place in the cannizzaro reaction? Hydride is a poor leaving group so a hydride shift should not take place and instead the attacking hydroxyl should leave the carbonyl group. I don’t understand why this doesn’t happen?

You are correct that a hydride ion is a terrible leaving group and that it can under no circumstances be eliminated in e.g. an $\mathrm{S_N2}$ reaction. However, there are quite a few reactions where hydride does leave one atom and subsequently attaches to another; Wagner-Meerwein rearrangements jump to mind as one of the most prominent examples.

The feature of a Cannizzaro reaction is that it is not actually a leaving hydride. Rather, the reactivity can be explained in a cyclic transition state as shown in scheme 1 below.

Scheme 1: Transition state and products of a Cannizzaro reaction.

Instead of seeing the reaction as a displaced hydride that subsequently attacks another atom, we can see it as a type of pseudo-electrocyclic reaction. Breaking and forming the bonds will happen almost simultaneously. At no point in the mechanism do we actually have a ‘free’ hydride anywhere.

Note, by the way, that the Cannizzaro reaction is not the only one that features a hydride shift in a cyclic transition state; you may, for example, also be interested in the Evans-Tishchenko asymmetric reduction.


Can I melt Barium Titanate powder into a cylinder and preserve the high dielectric? I would like to make various solid forms of Barium Titanate e.g. cylinders, discs.

Barium Titanate powder seems widely available and has a melting point of 1,625'C. Can I melt it in an iron forge into a mould, and will this preserve the high dielectric constant?

Is there perhaps a low temp way to do this easier e.g. a solvent or additive to achieve the same? Others have tried embedding within epoxy, but with substandard K values (max 27 instead of 7,000 as per the raw powder)

As you know, barium titanate is a piezoelectric ceramic materials. It has its property as a piezoelectric material due to the specific spatial arrangements of its constituting atoms. Melting this material, means that it will lose its crystalline  structure upon heating (knowing also that its Curie Temperature is $T_C=120^o\mathrm{C}$, i.e. above this temperature it loses its ferroelectric property).
On the other hand, processing and shaping ceramic materials are often performed during the preparation of these materials, not after. Heating at temperatures around $T_C=1600^o\mathrm{C}$ for processing ceramics is not a common procedure and it is expensive.
If you want to prepare cylinders and discs from a powder of barium titanate, you can add it to a polymer matrix (wax, epoxy resin,..) in an appropriate mold using adequate processing techniques like injection molding (heating below Curie Temperature). Of course, the resulting dielectric constant of the composite material is less than the one of pure  barium titanate.


Tools like Litmus Paper Litmus papers are used to test acidity and alkalinity of a fluid. Are there other paper-based tools to test for other stuff like dissolved oxygen, presence of toxicity, electric potential etc?

It's easy to find a slew of test papers and strips online. For example, a quick search of Amazon for "test papers" turned up chlorine test strips for pools and sanitation, glucose test strips for diabetes, phenylthiourea genetic-linked taste tests, UV-C dose test cards, lead acetate paper test for $\ce{H2S}$, WBC and nitrite test strips for UTI... You could write a paper on all the test papers available.
Increasingly, medicine is making use of paper tests because of ease of distribution, storage and use, particularly important in rural location and during emergencies.
BTW, you can make a test paper for polarity testing, and for detection of peroxides, by dipping filter paper into a suspension of starch and potassium iodide, $\ce{KI}$, rather than buy it. Dampen and use it as you would a potato to determine the polarity of an electric cell.


Is pure water very corrosive? In other site, someone claimed that pure water is very corrosive.

Water distilled for some laboratory uses (like the Z-pinch reactor) is distilled to 0.00000009% purity, and is caustic enough to burn through most things.
Water, purified enough, cannot be distributed through stainless steel pipes. Instead, they use glass pipes to transport it within the plants where it's made.

For sure, caustic is not the correct word, since water is neutral.
But, is it that corrosive? Or is it even less than water with salt?

That quote had false statements all over it.

Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs.  They are not corrosive to stainless steel.

As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination.  Metal reacts not only with oxygen, but also many pollutants in the air.  Overtime on metal surface will build up a layer that has all kind of compounds in them.  When water flows through, ions get picked up.

However, the most major reason would be that glass is better in all respect.  The reason glasswares are commonly used in the lab have little to do with pure water.   Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced.   Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through.  You will really only need metal if extreme pressure or temperature is required.

As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well.  Purified water are actually almost always stored in plastic containers and distributed through plastic tubes.


Physical state of hydronium I am doing my chemistry homework and I am supposed to write the products when strong acids react with water, but I am unsure of the physical state for hydronium ions. Should it be liquid (l) or aqueous (aq)?

Example:

$\ce{HI(aq) + H2O(l) -> ?}$

Should it be

$\ce{H3O+(aq) + I^{-}(aq) }$ ?

Or should it be

$\ce{H3O+(l) + I^{-}(aq) }$?

This is a bit touchy.  The problem is that a bare proton, $\ce{H^+}$, won't exist in water.  It attracts electrons too strongly.  Once this was recognized, chemists started to write the bare proton as being bound to a water molecule, thus $\ce{H3O^+}$.  However, studies also showed that water molecules are typically linked together in clusters containing a fair number of water molecules held together by hydrogen bonds (Google for this if you are not familiar with it).

Thus today's version really is $\ce{H+_aq}$ but that looks awful, so it is as you note $\ce{H3O+_aq}$ is more often used.

The important point is that how you write it isn't really important.  It is the realization that the bare proton attaches itself to any nearby source of electrons.

How you should handle your homework depends on what your teacher does in class.  I'd follow that example.


What is the difference between free energy and standard free energy As I understood $\Delta G^0$ is just $\Delta G$ but at standard conditions i.e. $\pu{1 bar}$ and $\pu{298 K}$. But formulas like:
$\Delta G^0 = \Delta H^0 - T \Delta S^0$
$\Delta G^0 = - RT \ln K$
make me uncertain about that as these are dependent on $T$ and don't seem to be used at standard conditions only i.e. $T = \pu{298 K}$.
Could someone explain to me what the difference is between $\Delta G$ and $\Delta G^0$?

$\Delta G^0$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and pressure 1 bar, to corresponding stoichiometric quantities of  pure products, each at temperature T and pressure 1 bar.  So $\Delta G^0$, $\Delta H^0$, and $\Delta S^0$ are all functions of T, but not pressure (which is fixed in the initial and final states as 1 bar).  The values of these function changes tabulated in your text book are all for 298 K.  But values at other temperatures can be obtained from integration, for example, in the case of $\Delta G^0(T)$ using the van't Hoff equation.
$\Delta G$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and its own arbitrary pressure, to corresponding stoichiometric quantities of  pure products, each at temperature T and its own arbitrary pressure. If these arbitrary pressures happen to match the partial pressures of the same reactants and products in an equilibrium mixture, then $\Delta G$ will turn out equal to zero.  Otherwise, $$\Delta G=\Delta G^0+RT\ln(Q)=RT\ln{(Q/K)}$$and, for equilibrium, Q=K.


Determining electrode potential Can the anode always be considered as having a lower standard electrode potential than the cathode?
When I was studying galvanic cells, I found that for a number of reactions, it always appeared that the anode(i.e negative potential) mostly had a lower electrode potential than the cathode. Can we say that this is always true or are there some exceptions?

By definition, for a cell to be galvanic, you would need the difference $\ce{E^0_(cathode) - E^0_(anode)}$ to have a positive value. So yes, the standard potential for the anode to have a more negative potential than the cathode.

Lower and higher are terms that are best avoided when talking about electrochemistry. More positive and more negative are unambigious.


Salinic bridges in voltaic cells I'm doing a study a study on voltaic cells.
I've been working with them for a while and one of the questions that I've had is how does the salt bridge affect the cell. I mean, I know it allows ions to move and that it'a a basic component of the pill. But, does its' concentration affect the voltage of the cell itself?
Thank you.

A cell has an internal ohmic-equivalent resistance.  The salt bridge is a resistive path for the flow of ions.  Ions, plus their hydration shells and possible coordination shells, have low drift velocities through a medium.  The salt bridge can constrain the ion current of the cell and its material equilibrium rate.  A poor salt bridge by length, diameter, concentration, and permeability will increasingly adversely affect a cell's current and voltage as discharge rate increases.


Automated Identification of Isomeric Relationship from 3D Structure? Is there a software package that when given two sets of 3D structures can return the isomeric relationship between them?  Even if it was only able to state that they are conformational isomers that would be enough.

It appears that Jmol has the ability, but I am seeking a non-graphical solution that could be easily included in scripts that would compare many such pairs of structures - in addition, I'm not a java programmer.

You might become interested in OpenBabel.  Written in C++, it offers bindings to Python (ref).  It looks like stereochemistry is covered more in detail (here, here) than the obchiral command (here).

OpenEye's Chemistry Toolkit, on the other hand, equally capable to read numerous chemical file types (ref), offers several approaches to detect stereochemistry (examples).  Written in Python, it may offer an alternative starting point.  Perhaps you qualify for one of their academic licenses (ref).


Kinetic Barrier of ATP I do a fair bit of biochemistry, and one of the things I've taken on faith up until now is that ATP is stable because it has a high kinetic barrier for falling apart despite the spontaneity of its decomposition. I understand that it does, but what about the transition state of ATP hydrolysis in an aqueous solution is so high energy? For that matter, what is the transition state of spontaneous ATP hydrolysis?

This is not exactly an answer making strong use of references etc. but it does give a good hint. The most ‘energetic’ bonds in ATP are the phosphoric acid anhydride ones i.e. the $\ce{P-O-P}$, which, to hydrolyse ATP, must be attacked by water. The most logical path of attack is one according to the $\mathrm{S_N2}$ mechanism, where a lone pair of water attacks the rear side of a phosphorus-oxygen bond and liberates some phosphate species. ($\mathrm{S_N1}$ can be ruled out, since it would result in a phosphocation which is surrounded by three electron-withdrawing oxygens making the cation highly unstable.)

The problem in this mechanism is physiological pH. It is safe to say that the entire ATP molecule is at least a dianion, if not even a tri- or tetraanion. Now water’s lone pair, i.e. its partially most negatively charged area must approach this already highly negative molecule — but that would result in the repulsion of same charges leading to a high activation barrier.

This would be circumvented by protonation, and indeed even at equilibrium there is a certain probability that a higher number of protons attaches to ATP, making it less negative. This probably helps to account for the slow but steady decay of ATP.


Which element is undergoing oxidation in the combustion of methane? Chemical reaction:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}
$$

The example in our textbook for this type of problem involves the following reaction:

$$\ce{C + 2S -> CS2}
$$

Here, I can clearly see that $\ce{C}$ starts off with an oxidation state of zero and ends with an oxidation state of +4, meaning that it underwent oxidation.

In my given reaction, I see how the $\ce{C}$ in $\ce{CO2}$ has an oxidation state of +4, but I don't quite understand why it doesn't also have an oxidation state of +4 in $\ce{CH4}$. My assumption here is that, in the case of $\ce{CO_2}$, $\ce{C}$ has an oxidation state of +4 and $\ce{O2}$ has an oxidation state of -4; and in the case of $\ce{CH4}$, $\ce{C}$ has an oxidation state of +4 and $\ce{H4}$ has an oxidation state of -4. If that is incorrect, could someone explain why?

Edit: I just realized I overlooked a table that says that hydrogen has an oxidation state of +1, which would mean that in $\ce{CH_4}$, $\ce{C}$ would have an oxidation state of -4. I gather that this means that $\ce{C}$ loses 8 electrons in the reaction, which is why it is considered to undergo oxidation.

I believe I have answered my own question - I assumed that H had an oxidation state of -1, when it appears to have an oxidation state of +1.


Uncertainty Principle in the Ground State of a Rigid Rotor This is a fairly basic question which has bothered me for some time and I haven't been able to satisfactorily resolve.

In the ground state of the rigid rotor the energy is zero. That is, there is no zero point energy for this system. This means we definitely need to make sure we aren't violating the uncertainty principle for position and momentum.

So, the first place to look at is the wavefunction of the ground state. The wavefunction of the rigid rotor are the associated Legendre polynomials, and for the ground state (ignoring normalization) this is:$$P_0^0(x)=1.$$The point being, this is just a constant.

So, my first intuition is to say that we aren't violating position-momentum uncertainty because the wavefunction is only a constant so we know nothing about its location. This, however, I think is wrong because we solve this in spherical coordinates and the angle $\phi$ is constrained to be within $[0,2\pi]$. That is, our particle is somewhere on a sphere. Thus, this explanation doesn't work because the particle is somewhere on the sphere so we have some certainty of its position.

My next thought was that the uncertainty principle I'm thinking of is written as follows:$$\Delta p_x\Delta x \ge \frac{\hbar}{2}$$This, however, is wrong because it means that when the ground state energy is zero, the angular momentum is zero (that is, this is a closed spherically symmetric wave).

Thus, the next explanation I thought is that if I solved this in Cartesian coordinates rather than spherical coordinates, I would still get the same answer that the ground state energy is zero, but the x, y, and z position and x, y, and z momenta would all have positive, finite values which do not violate the uncertainty principle.

So, hopefully I've convinced you that this requires a slightly more nuanced answer than one might expect.

Is the last explanation correct in that when solving this in spherical coordinates, we should actually think about some other uncertainty principle than the one depending on Cartesian coordinates written above? Something like $$\Delta L \Delta r \ge \frac{\hbar}{2}$$

So, let me know if my reasoning has gone astray and where the solution lies. I feel a bit embarrassed that this isn't obvious, but I think my last point is where the answer lies.

Specifically, I'd like to see if anyone can address the question of the uncertainty principle in the ground state when this problem is solved in Cartesian Coordinates.

As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\partial\phi,\, \phi] = 1$, which leads to the problematic inequality $\Delta L_z \Delta\phi \ge \hbar/2$.

Remember that the variable $\phi$ is defined on a circle, i.e., $\phi = 0$ and $\phi=2\pi$ represent the same point. A function of the form $\phi\psi(\phi)$, where $\psi(\phi)$ is $2\pi$-periodic, has a discontinuous jump by the amount $-2\pi \psi(0)$ at $\phi = 0$ (or $2\pi$). Hence, when applying $\partial/\partial\phi$ to $\phi\psi(\phi)$, one should make sure to differentiate the discontinuity as well, which leads to an additional term: $-2\pi \psi(0)$ times a Dirac delta function. Then, the correct commutation relation should be $[\partial/\partial\phi,\, \phi]\,\psi(\phi) = \psi(\phi) - 2\pi\psi(0)\delta(\phi)$, and we have the modified uncertainty principle
$$
\Delta L_z \Delta\phi \ge \frac{1}{2}\big|\langle[L_{z}, \phi]\rangle\big| = \frac{\hbar}{2}\Big|\langle\psi|[\partial/\partial\phi,\, \phi]|\psi\rangle\Big| = \frac{\hbar}{2}[ 1- 2\pi|\psi(0)|^2].
$$
For a normalized angular momentum eigenfunction $\psi_{m}(\phi) = e^{im\phi}/\sqrt{2\pi}$, the right-hand side of the above yieids zero. This is exactly what should be the case when the uncertainties in the angular momentum and the angle are respectively zero and $2\pi$.


For molecules to be enantiomers, do the groups have to be non-superimposable mirror images even when rotated around C-C single bonds? For example, if I rotate the Carbon-Carbon bond in B around 120 degree anti-clockwise, I would get the molecule shown in image C. C looks to me like a non-superimposable mirror image of A, hence A and C are enantiomers.

However, as drawn, A and B do NOT seem to be mirror images.

Are A and B enantiomers considering what happens after a bond rotation?

Note: ChemDoodle shows those red triangles saying "questionable stereochemistry". I'm not sure what is wrong but the molecules I drew may have some issues, but I think the question still holds.

Real molecules surely do rotate around their single C-C bonds all the time(*). So yes, if two models can be superimposed after an internal rotation of such sort, this means they represent the same molecule, as is the case with your B and C.

This, BTW, is the reason why you should orient all substituents in a standard manner when drawing your Fischer projections. Otherwise any molecule would have just too many variants, and you'll get bogged down in them. Think of rotating the upper group the third way (the first two being B and C). Think of rotating the lower group too...

Then again, I've rarely seen Fischer projections used for anything other than sugars. Typically people just determine the absolute configuration (R or S) and go with that. It gives you a definitive answer, after all.

(*) That is, most of them do. When you have bulky substituents, rotation may pretty well get blocked, and then you may be able to separate the isomers that are (nominally) produced by this rotation - they are no longer equivalent. Look at BINOL or BINAP, for example.


Thought Experiment: Applying a high frequency high voltage to methane I am in the process of constructing a kitchen waste digester (vegetable/fruits) from which, hopefully, over time adequate methane may be collected to burn. The digester won't be complete for several months yet - a few minutes each week or so.

Anyway, if a high frequency high-voltage field were to be applied to the methane (taking due care to avoid sparks) would the methane decompose into carbon, and hydrogen? A google search seems to indicate that methane may decompose, or form a plasma; I understand there may not be a simple/straight answer for this - but a general direction would be nice to have.

To decompose methane molecule you have to apply energy about several electron-volts directly to molecule. Since microwave radiation wavelength measured in millimeters, to really decompose methane you'll need ridiculous voltage, much more them mere kilovolts needed to ignite electrical discharge.

And btw, electrical discharge is the easiest way to supply the energy needed for molecule to break.


Deciding more electrophilic centre between ester and thioester for SN type reaction My question: (though it won't alter the product)

  Which carbon is attacked first, the one adjacent to sulphur or the one adjacent to oxygen?

The solution says that the carbon adjacent to sulphur is attacked first. I don't understand why.

According to me, sulphur is less electron withdrawing than oxygen (electronegativity decreases down the group), so the carbon adjacent to oxygen must be more electrophilic.

Could someone please explain as to why attack should first occur at the carbon adjacent to the sulphur atom? (or is the solution incorrect?)

P.S. Here, due to syn stereochemistry (NGP is not favourable) choice of electrophilic centre undergoing attack first doesn't influence the product at all. However, in the case where the sulphur and oxygen atoms are anti with respect to the ring, and NGP is a possibility, and then the products would greatly differ (depending on whether attack first happens on carbon adjacent to oxygen or sulphur)

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q35 in Chapter Alcohols, Ether, and Epoxides

There is also another effect of delocalisation of the electron pair on $\ce{O}$ and $\ce{S}$ atom adjacent to the cabonyl carbon, which is dominant over the electron-withdrawing effect ($\ce{-I }$ effect).

In both the parts i.e. $\ce{OCOCH_3}$ and $\ce{SCOCH_3}$, there are two lone pairs on each of $\ce{O}$ and $\ce{S}$, which can delocalise with the carbonyl carbon. Thus, the bond between $\ce{O}$ and $\ce{C}$, and $\ce{S}$ and $\ce{C}$, will get a partial double bond character. Due to this delocalisation, the electrophilicity of the carbonyl carbon is decreased.

When the delocalisation happens in case of $\ce{O}$, the overlap between  $\ce{O}$ and $\ce{C}$ is $\ce{2p$\pi$ - 2p$\pi$}$ overlap, which is a strong overlap and thus the double bond character between $\ce{O}$ and $\ce{C}$ is more. Thus, electrophilicity is reduced to a great extent. But in case of $\ce{S}$, there is a relatively weaker $\ce{3p$\pi$ -2p$\pi$}$ overlap which introduces very feeble double bond character between them so, the electrophilicity is reduced to very less extent and thus the first electrophilic attack will happen at the $\ce{C}$ adjacent to $\ce{S}$. This is the reason to your question.


Why does steric hindrance cause a d8 complex to have a tetrahedral geometry rather than a square planar geometry? I'm currently working on a question based on two isomers of a nickel complex, where nickel exists in the oxidation state Ni(II) (i.e. has d-electron configuration d8). It has been deduced that the "normal" isomer is square planar and the "iso" isomer is tetrahedral.

The first of the two isomers (top) is the "normal" isomer and the second (bottom) is the "iso" isomer.

Apparently steric hindrance causes the "iso" isomer to be tetrahedral rather than square planar, but I am unsure how to explain this. Is it due to the bond angles causing less strain?

For a given number of ligands surrounding a central atom there is always an arrangement in which the distances between the ligands is maximised and which typically also results in the maximum possible angles between then. You can deduce them logically if you start at a single ligand (which has no steric clashes whatsoever).

A second ligand should approach the same centre from exactly the opposite side leading to a linear arrangement and a bond angle of $180^\circ$.

A third ligand should approach the same centre perpendicular to the existing bond axis. Upon its approach, the two ligands already present will ease away from the approaching one. The final arrangement is trigonal planar with bond angles of $120^\circ$.

A fourth ligand should approach the same centre in a perpendicular manner again; this time relative to the plane of the three existing ligands. Again, they will ease away to make more room for the incoming. The final arrangement is a tetrahedron with bond angles of $109.5^\circ$.

Going from a tetrahedron to a square-planar arrangement means squashing the ligands together i.e. reducing the space between them. Thus, if there are no electronic arguments favouring a different arrangement any tetracoordinated entity should assume a tetrahedral environment to minimise steric strain.


Ethanol discolored yellow when stored with a rubber stopper. Forming fine white precipitate when added to water I stored some Klean-Strip denatured alcohol in a test tube and capped it with a rubber stopper. I did this in preparation for a chemistry demonstration. When I opened my kit a few days later at the event, I was surprised to find that the ethanol in the tube had turned a definite urine yellow. It burned just fine but let off nearly imperceptible wisps of smoke. Even more interesting was when I added water into the tube, which caused an almost instant white precipitate to form. I had to improvise the demonstration and play it off as turning flammable urine into semen, which it closely represented (because it was meant to be a comedic demonstration for a mature audience). The next few days I ran an experiment of using the same denatured ethanol in a jar over the black rubber stopper. By the end of the experiment, the solution exhibited the same properties and characteristics as the accidental batch. Now, not even to save my own skin, I cannot come up with a reasonable explanation for why the ethanol turns yellow, what causes the discoloration, and how does it turn milky when water is added.

Why does this happen. Chemically, what is happening. And would this be a good indicator for a non-anhydrous chemical? For example, I can add this to methanol or NaOH, measure the "turned whitedness" and determine whether or not that chemical is truly anhydrous?

It’s not easy to know unless you decide to throw the full set of analytical chemistry at your yellowish ethanol. But we do have a few clues.

It is most likely that something from the rubber stopper got dissolved in the ethanol. This solution turned out to be yellowish so we can assume the compound itself to be yellow to brown-ish. Dissolution either came by the ethanol soaking the stopper while the flask was lying down or by evaporation/condensation.

We also cannot determine from the information you have given what the precipitate is that forms when you add water. But my first guess would be ions from the water being less soluble in an ethanol-water mixture and precipitating. If that is the case, it would be made up mainly of sodium chloride. It could also be that some component of the ethanol is insoluble in water and precipitates out. Considering your alcohol is denatured, it might also be the additive that makes it undrinkable. However, once again there are too many variables and you need the full set of analytical chemistry thrown at the precipitate to find out what it is.

Most certainly whatever is happening is not a good indicator of anything.


What aspects of a molecule's chemical makeup determine which portions of it will react? Is there a specific way to know how two or more arbitrary molecules will react?

An example that prompted this question: $\ce{(HOC6H4)2CMe2 + 2 NaOH → (NaOC6H4)2CMe2 + 2 H2O}$ (from here). How can you know how those molecules will react? Is it just a bunch of substitutions, with certain atoms in the reacting molecules reacting better than others? What factors would determine which atom in the molecule would react with the other molecule?

Let me answer with a question (or two):

How do you identify in an electric circuit which parts are just the wiring and which are functional parts?

How's that for a mechanical engine? Can you tell cogwheels and cranks from "just the basic structure"?

It's all about pattern recognition in the first to identify the functional groups.

Next step is to know how they react under certain conditions. As a rule of thumb, a group with a high electron density will probably react with something more positively charged (think plus and minus attract each other).

But it takes quite a while and learning to predict exactly what will go on exactly in the reaction. But that not different from any other scientific/technical discipline.

I tried to keep this as general and simple as possible ;)

Edit
Let me work you comment in:

  Generally molecules don't contain ions though

Says who? :D Molecules may contain ionic units, think in zwitterions of amino acids, think in all the mesoionic sydnones, münchnones and whatnot. Well, no. Forget about the special cases ;) But if it has more than one atom and is electrically neutral in total it is a molecule. At least in the less rigid sense of the definition.

But yes, the electronegativity is a good point.

Forgive me that I'm not going into detail why the example with fluorine and sodium is a not the best one, but in principle, you're right: Such a setup would allow to form a new $\ce{C-C}$ bond between a carbon that is negatively polarized (the one bearing the sodium) and another one that is positively polarized (the one bearing the halogene).


Why is chromate coloured but tungstate colourless? I have to try and work this out using the tetrahedral molecular orbital diagram for $\ce{[WO4]^2-}$ but I am confused as both are $\mathrm d^0$ right?

Both complexes (and many more such as permanganate, sulphate, …) share the same general MO scheme which I am going to shamelessly copy from my older answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.

The intense colour of permanganate and chromate is due to LMCT transitions: ligand-to-metal charge transfer. Essentially, an electron from the HOMO (a $\mathrm{t_1}$ orbital that can be thought of a ligand-centred p-type orbital) is excited to the LUMO (an $\mathrm{e}$-type orbital that can be thought of one of the metal’s empty d orbitals) by absorbing a photon from the visible range; we then see the complementary colour, meaning that chromate absorbs blue light and looks yellow to our eyes.

The energies of LUMO and HOMO – or rather, the energy difference between the two – determines the wavelength of the absorbed photon and thus the colour that we see. For both permanganate and chromate, the wavelength is well within the visible spectrum resulting in the strongly coloured complexes. In cases such as sulphate or tungstate, the absorption wavelength is in the ultraviolet range which cannot be perceived by human eyes; therefore, the solution appears uncoloured.

Without computer calculations, it is probably impossible to accurately predict the ultimate wavelength but note that such an absorption is practically always present if the corresponding orbitals are available although usually in the ultraviolet range. An obvious counterexample would be any zinc complex in which the metal has a $\mathrm d^{10}$ configuration and the LUMO is thus the much further removed $\mathrm a_1$ orbital; therefore, zinc complexes are typically colourless and do not display such an LMCT band.


What causes the blackening of light bulbs in 1878, Thomas Edison (or rather William Joseph Hammer) added an anode in a light bulb in order to catch the flux of electrons coming out from the carbon filament and reduce the phenomenon of blackening of the glass bulb.

However, am I right in saying that this was the wrong course of action? I understand the blackening is due to the accumulation of carbon oxides (such as CO and CO2) produced by the reaction between carbon (and, more recently, tungsten) vapour and the remainder of air present into the glass bulb. Thus, capturing electrons did not solve the problem of blackening (and that is why Edison did not find any use for the extra electrode).

A 14 page article (https://www.jstage.jst.go.jp/article/jieij1917/6/4/6_4_398/_article) describes the blackening of incandescent light bulbs as being totally due to evaporation of the filament. Platinum had been used as far back as 1840, and Edison's first patent was for a carbon filament (1879), but by this time (1922), tungsten was the metal used in almost all filaments.

Blackening of carbon-filament bulbs has not been a significant problem, due to the short commercial lifetime of carbon filaments. Any carbon deposition would likely be from sputtering of the filament, not evaporation. Blackening is a problem with tungsten filaments, and not so much from evaporation of metal, but from evaporation of tungsten oxides formed from adventitious water or oxygen, followed by deposition on the cool bulb surface. Inert gas is used instead of vaccuum to impede the mobility of tungsten atoms. But then, tiny amounts of oxygen or water vapor in the bulb can cause severe blackening; getters such as zirconium are used to combine with oxygen. Large bulbs used for theater lighting often contained some tungsten powder; when the bulb darkened, the bulb was shaken to abrade off the tungsten and increase the brightness. https://en.wikipedia.org/wiki/Incandescent_light_bulb#Bulb_blackening

Halogen bulbs use iodine to react with the deposited tungsten, which vaporizes as the iodide and is redeposited on the hot filament, where it decomposes. Halogen bulbs are smaller than the usual incandescent bulb and they run hotter filaments to keep the glass hotter, so that iodine can react there with the deposited tungsten and carry it back to the filament.

Your description of Hammer adding an anode to catch electrons emitted by the hot filament is correct, but I suspect that the original thinking went something like this: "We know that electrons are emitted by the hot filament. As a result, the filament must become positive; thus some metal atom at the surface might be repelled, carrying a positive charge (which would be neutralized as it passes thru the electron cloud). This metal atom might deposit on the cool bulb, producing a darkening. Perhaps we can prevent this repulsion/evaporation by making the filament negative so as to retain the positive metal ions." Apparently, this did not work, but only increased the electron emission, which could then be described as a bug or a feature.  OK, let's call it a feature and name it after Edison.


Reaction mechanism for reduction of nitrobenzene to aniline in catalytic and acidic media Can anyone explain the reaction mechanism of the two reactions given below?

They are the conversion of nitrobenzene to aniline in catalytic ($\ce{H2/Pd}$ in ethanol) and acidic ($\ce{Sn/HCl}$ or $\ce{Fe/HCl}$) mediums respectively.

$$\ce{Ar-NO2 ->[\ce{H2/Pd}][\ce{EtOH}] Ar-NH2}$$

$$\ce{Ar-NO2 ->[\ce{Sn + HCl} \text{ or } \ce{Fe + HCl}] Ar-NH2}$$

The first reaction is called a catalytic reduction.  Here is a picture of how it works with an olefin like ethylene.

You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form.  The same things happen with nitrobenzene.  As the following diagram shows, the reduction passes through the nitroso and hydroxylamine intermediates.

The mechanism for the Sn\HCl reduction involves reductive electron transfer of the nitro group while the tin is being oxidized. This diagram shows the electron transfers involved

and here is the mechanistic scheme.


Grignard formation and vicinal dihalides So let's say I have a symmetric 1,2-dihalide (the halides are identical) and I wanted to make a Grignard and I only add 1 eq. of magnesium metal ...

Will I get 1 eq. of asymmetric Grignards (i.e. only one side of the product has the C-MgBr bond) as my product or will:

a) Some of the product be "double" Grignards - i.e. both sides of what used to be the 1,2-dihalide now have C-MgBr bonds?

b) I get some E2 elimination product from Grignards robbing protons from unreacted 1,2-dihalides?

c) A combination of the above?

d) Something else entirely?

Let's say the above reaction works the way I want it to and we get the 1 eq. of asymmetric Grignard reagent.

We then react it with a carbonyl containing compound such as acetone ... do we get an epoxide? Are there even any routes to an epoxide? This suggested as a possible product ...

But I don't see the epoxide.

Taken from the Virtual Textbook of Organic Chemistry, which is maintained by William Reusch.

  If two halogen atoms are present in a given compound, reactions with
  reducing metals may take different paths depending on how close the
  carbon-halogen bonds are to each other. If they are separated by four
  or more carbons, as in the first example below, a bis-organometallic
  compound may be formed. However, if the halogens are bonded to
  adjacent (vicinal) carbons, an elimination takes place with formation
  of a double bond.

He does however not provide any explanation for this behavior. It could be that this result is largely empirically based, about some aspects of the Gringard reaction mechanism there is not much known about (or consensus among chemists). I looked at Clayden et al. and Vollhardt & Shore and neither give a satisfying answer. If I find any in an other perhaps more advanced textbook, I'll edit this answer.


Is Molarity constant? My friend and I were debating about if Molarity is constant in a disolution if this one changes volume due to evaporation.
There was this problem that stated that 125 [mL] of sodyum chloride 0.122 [M] evaporates over time and final volume becomes 113 [mL]
I say Molarity keeps being constant, he says no because volume changes.
What do you think?

As molarity is defined as the ratio of molar amount of the dissolved solute and solution volume:
$$c=\frac nV$$
molarity increases when solvent ( like water ) evaporates.
Molarity can be considered constant only in context of later refilling the evaporated solvent and mixing the solution. Then molarity returns to the original value. It assumes the solute ( like sodium chloride ) content is constant, not undergoing any reaction.


Reconciling multiple inorganic properties for the purposes of learning these elements How can I reconcile all of the chemical properties, physical properties and peculiar behaviour of metals in inorganic chemistry for the purposes of studying these systematically?

It is very hard to learn all of these for each group.

first of all take one group a day.
right whatever you have to learn and then try to identify some relationship. Then reread the material and write down whatever you remember . Now look what you miss.
everyday take a new group and do the same.
Also write whatever you remember of the previous groups you have learned.


Calculating concentration of dissolved ions A sample of $\mathrm{2.00~g}$ of iron(III) sulphate is dissolved in water to give $\mathrm{100~cm^3}$ of aqueous solution. What is the concentration of $\ce{SO4^2-}$ ions in $\mathrm{mol~dm^{-3}}$?

My attempt:

I found the number of moles which is $\mathrm{0.005~mol}$.

Then, 1 mol of iron(III) sulphate has 3 mol of $\ce{SO4^2-}$

So the concentration is $\mathrm{\frac{0.005}{0.1} \cdot 3 = 1.5 \cdot 10^{-1} ~mol~dm^{-3}}$

But the given answer is $\mathrm{1.5 \cdot 10^{-2}~mol~dm^{-3}}$. Why?

You're right and the given answer is wrong. Here's the math:

$$\left(1\;\mathrm{mol}\;\ce{Fe2(SO4)3}\over {399.88\;\mathrm{g}\;\ce{Fe2(SO4)3}}\right)\left({2.0\;\mathrm{g}\;\ce{Fe2(SO4)3}\over 100\;\mathrm{cm}^{3}}\right)\left({3\;\mathrm{mol}\;\ce{SO4^2-}\over 1\;\mathrm{mol}\;\ce{Fe2(SO4)3}}\right)\left({10\;\mathrm{cm}\over 1\;\mathrm{dm}}\right)^{3} =\ce{[SO4^2- ]} $$

$$\ce{[SO4^2- ]}= 1.5\cdot10^{-1}\;\mathrm{mol}\;\mathrm{dm}^{-3}$$


Method to know the number of isomers of metal complexes Whenever I try to find the number of geometrical isomers (including optical isomers) of coordination compound, I got confused, and mostly miss few isomers. Is there any standard method to know the number of isomers of compounds such as $\ce{Ma2b2c2}$ or $\ce{[M(A-A)a2b2]}$.

where $\ce{M}$ is metal and $\ce{a,b}$ are monodentate ligands while $\ce{A-A}$ is a bidentate ligand.

Do it many times over, so as not to get confused.

Alternatively, there is a thing called Polya's formula, but you won't be able to use it anyway. In trivial cases like this, the said formula is about 100 times more complicated than counting isomers by hand. It is not before polysubstituted fullerenes that its use in chemistry starts making any sense.


What is the problem with forming 2,3-dimethylcyclohex-2-en-1-one? I have suggested a mechanism for the formation of 3-ethyl-2-methylcyclohex-2-en-1-one A

but the question says problems would be encountered if a similar route was used to make 2,3-dimethylcyclohex-2-en-1-one B. I think perhaps a decarboxylation step is involved? Can someone please explain how to get to B and the difficulties in the mechanism?

Your starting product for A is symmetrical, but your starting product for B will be asymmetrical - one carbon less on one side.  This is going to mean you may get 2 or more different products; B may not be your major product, and you'll have to separate them afterwards.  I don't believe there's inherently a tricky difficultly in the mechanism for B.


Biological activity of the isotopes of the same element I have a question regarding biological activity of different isotopes of the same elements, notably $\ce{^206Pb}$ and $\ce{^208Pb}$. Lead is known to be toxic, such that it can replace calcium from bones and iron from hemoglobin for example. Would one isotope behave differently than another in such cases?
When it comes to hydrogen/deuterium chemical activities, some difference could arise as one isotope is twice as large as another. Would this be the case for heavier elements, such as lead?

I would not expect significant differences in the chemical properties (reactivity) of these lead isotopes, the differences in atomic weight being as you point out very small. There are reports  that claim an effect. Ref 1 for instance makes such a claim, but there is no plausible mechanism presented to explain the reported isotopic discrimination. Double check what peer reviewers think (see e.g. the response in Ref 2 to the results presented in Ref 1).
References

Wu J, Liu D, Xie Q, Wang J (2012) Biological Fractionation of Lead Isotopes in Sprague-Dawley Rats Lead Poisoned via the Respiratory Tract. PLOS ONE 7(12): e52462. https://doi.org/10.1371/journal.pone.0052462

Gulson, Brian and Kamenov, George D. and Manton, William and Rabinowitz, Michael (2018) Concerns about Quadrupole ICP-MS Lead Isotopic Data and Interpretations in the Environment and Health Fields. International Journal of Environmental Research and Public Health 15(4): 723. https://doi.org/10.3390/ijerph15040723


Why is there a sudden drop in the concentration of ammonia in this graph? So initially they are mixed with the mole ratio of $\ce{H2}$ and $\ce{N2}$ being 3:1 respectively. It seems as the reaction proceeds, both these quantities decrease and the quantity of ammonia increases until a point at which $\ce{NH3}$ stagnates. What happens at this point?

In the graph the following reaction will occur: $$\ce{N2 +3H2 <=> 2NH3}$$
This reaction is an equilibrium reaction and you can see that there is exponential decay for the reactants while the product is formed. After a while the concentrations do not change any more, so the system is in equilibrium. This is at the end of the time span "a".

At the beginning of time span "b" the system is disturbed, as the ammonia is completely removed. The system is changed and the equilibrium has to form again. Hence the concentration of the reactants is further reduced while new product is formed. Like santiago already pointed out in the comments, this is known as Le Châtelier's principle. At the end of time span "b", the system is in equilibrium again.


Bohr Model and Exciting Electrons So I've got a relatively simple question; I'm not currently taking any courses in chemistry but while I was helping someone review a quiz I came across a question we couldn't figure out. According to the Bohr Model, is the energy required to excite an electron from the n=3 to n=4 orbit generally greater or less than the energy required to go from n=4 to n=5? Also, why is this so? I can't seem to find a clear answer anywhere online, as just about every result involves directly computing energy required to bump the electron up a few levels in very specific cases.

The energy difference between the orbits gets smaller the higher "up" you go.

For the classical bohr model the energy of a given orbit $n$ can be approximated by:

$E_n=-\frac{Z^2R_E}{n^2}$

where $Z$ is the charge of the nucleus and $R_E$ is the Rydberg energy. Since there is a $n^2$ in the divider the difference $E_{n+1}-E_n$ gets smaller and smaller.

This is explained in more detail here.

As for why this is the case: here we have to look into quantum mechanics and I guess this is outside the scope here. But if you are interested look up explanations of a particle in a 1D box. The spacing of the energy levels there follows the same trend and you can find explanations there on why this is the case.


Phenol and Deuterated Sulphuric Acid What happens when phenol reacts with $\ce{D2SO4}$ in presence of $\ce{D2O}$?

How many hydrogens in total are replaced by deuterium?

Is the hydrogen of the hydroxyl group and three hydrogens ortho and para to the -OH group replaced by deuterium, and the remaining two unaffected?

OR are all hydrogens replaced by deuterium?

I'm not sure what really happens, but as far as I know - even though ortho and para positions are greatly activated, some small percentage of meta attack may also take place. In fact, if it does take place, the final compound will be totally deuterated, as bonds with deuterium are stronger than those with hydrogen, and backward reaction will not be thermodynamically favoured.

It'd be great if someone could explain what's going on, and if possible please quote sources supporting your answer. Thanks a lot.

The deuteration of phenol and aniline hydrochloride has been investigated by Werstiuk and Kadai by the High Temperature/Dilute Acid (HTDA) method. For phenol, ortho/para substitution is faster than meta (2.7 D/3.0 H vs. 1.3 D/2.0 H). OD/OH exchange was conducted at the end of the reaction. The presence of conc. HCl undoubtedly lowers the efficiency of incorporation of deuterium.
Incorporation of deuterium without the presence of conc. HCl allows for efficient deuteration of aniline hydrochloride. [RND3+ to RNH3+ exchange was conducted.] The authors note that conversion of aniline hydrochloride-d5 to phenol-d5 is efficient via diazotization.

Werstiuk, N. H.; Kadai, T., Can. J. Chem., 1974, 52, 2169.


Comparing the mesomeric effects of amino and mercapto groups I was quite surprised to find that $\ce{NH2}$ has much more $+M$ effect than $\ce{SH}$. But shouldn't we also consider the fact that $\ce{SH}$ has 2 lone pairs? Or is the greater electron density of nitrogen playing some role here?

How can I rationalise the stronger mesomeric effect of nitrogen?

For a mesomeric effect to occur, an orbital has to align with the π system it is supposed to resonate with. Since π systems have a defined $z$ axis — the orientation of the $\mathrm p_z$ orbitals that make up the π system — only orbitals that also feature such an alignment an interact. This limits any mesomeric interaction of any heteroatom with a π system to that heteroatom’s $\mathrm p_z$ orbital. Therefore, both nitrogen and sulphur can only have one orbital at their disposal for $+M$ effects.

The reason why the $+M$ effect of nitrogen is stronger despite its higher electronegativity is the period these elements are in. Nitrogen, being in the same period as carbon, has orbitals that fit together well size-wise with carbon orbitals. This is also reflected by the ease of forming $\ce{C=N}$ or $\ce{N=O}$ double bonds. Sulphur is one period down. Its orbitals are more diffuse than carbon’s or nitrogen’s; therefore, any π overlap is much weaker. This is also the reason why $\ce{S=O}$ double bonds are rare and $\ce{S=C}$ double bonds are typically more reactive than $\ce{O=C}$ double bonds.


Solving for gas law partial derivative I don't know where to begin with this question.

  A scientist discovered that the state of the unknown gas can be well described by the equation of state below:
  $$P = \frac{RT}{\overline{V}} + \left(\frac{a + bT}{\overline{V}^2}\right)$$
  Find the partial derivative $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ which is expressed in terms of $\overline{V}, R, b$, and $P$. (Hint: If $\overline{V}$ is a function of two variables $T$ and $P$ (i.e. $\bar{V} = f(T, P)$), then the total differential can be expressed as
  $$d\bar{V} = df(T, P) = \left(\frac{\partial \overline{V}}{\partial T}\right)_PdT + \left(\frac{\partial \overline{V}}{\partial P}\right)_TdP$$
  Use a particular condition to derive the expression $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ in terms of $\left(\frac{\partial \overline{V}}{\partial P}\right)_T$ and $\left(\frac{\partial P}{\partial T}\right)_{\overline{V}}$.)

This problem is related to the physical chemistry of the gas law. What is the correct approach?

Think of it this way.
$$dP = \left(\partial P\over\partial T\right)_VdT+ \left(\partial P\over\partial V\right)_TdV$$
Now, we want P to be constant (that is, $dP=0$). What does that entail? A certain relation between $dV$ and $dT$ (which, BTW, are now $\partial$ rather than $d$):
$$\left(\partial V\over\partial T\right)_P=-\left(\partial P\over\partial T\right)_V\left/\left(\partial P\over\partial V\right)_T\right.$$


Why gives NaBH4 more Felkin-Anh product than LiAlH4 gives? In lecture we discussed the nature of the nucleophile for the attack of the carbonyl.

LiAlH4 acts as a better nucleophile (hydride) than NaBH4. I know there is no hydroboration! So I would have expected that LiAlH4 gives more Felkin Anh product as the NaBH4 gives.

However LiAlH4 gives a Felkin Anh : Anti Felkin Anh ratio of 3:1 and
NaBH4 gives a ratio of 5:1.

detailed reaction
Source: Carreira, E.M.; Classics in Stereoselective Synthesis

Is there any good rational?

Lithium aluminum hydride is such a reactive nucleophile that it reacts readily with any conformation of the ketone. The reduced nucleophilicity of sodium borohydride allows for some discrimination of the conformations.

Note that the difference in selectivity between aluminum hydride and borohydride small. A large boost in stereoselectivity can be achieved by increasing steric bulk of the reducing agent (L-selectride gives a 10-fold increase in selectivity).


How to figure out which two solutions will emit largest heat when they are mixed in same volumes? How can I figure out which two solutions will emit the largest heat, when they are mixed in same volumes?
Then  let's assume that there are five pairs of solutions given:

\begin{align}
  \ce{2NaOH   + H2SO4   &-> Na2SO4    + 2H2O} \tag1\\
  \ce{Ba(OH)2 + H2SO4   &-> BaSO4     + 2H2O} \tag2\\
  \ce{2NH4OH  + H2SO4   &-> (NH4)2SO4 + 2H2O} \tag3\\
  \ce{NaOH    + H3CCOOH &-> H3CCOONa  + H2O}  \tag4\\
 \ce{2NaOH    + H2C2O4  &-> Na2C2O4   + 2H2O} \tag5
\end{align}

Which one emits the most heat, when we mix same volumes of given pairs?

These are the balanced reaction equation, that occur when the above solutions are mixed. All of these reactions are acid base reactions. And I know when there is an acid base reaction a salt is formed. So how can I proceed with it?

You should just think about a fact: all chemical processes will correspond to enthalpy changes.

The basic process of acid-base reaction is
$$\ce{H+ + OH- -> H2O}$$

This process will have an enthalpy change of $\pu{-55.8 kJ mol-1}$ at $\pu{298 K}$. (I don’t know why, but in different books or articles this value will be slightly different.) However,

Precipitate formation is usually exothermic, since there are some ion bonds formed.
Weak acid and base usually need some energy, for they are not completely ionized, during the reaction, they will to ionize more since $\ce{H+}$ or $\ce{OH-}$ is consumed. Ionization will be endothermic.

Therefore the answer would be the second reaction, since $\ce{BaSO4}$ is precipitated, which is exothermic, while weak acid and base reactions are endothermic.

(However, there seems to be always some exception, for example, ionization of $\ce{HF}$ will be exothermic. But to discuss this may be out of topic, and to be frankly, I’m not quite sure what its mechanism is.)


Numbering in IUPAC name of a trisubstituted cyclohexane This is the organic compound. CHO - aldehyde group. Br - Bromine

How do name this organic compound (IUPAC Standard)? Specifically, how do you number the carbon atoms? I thought the answer was 1-Bromo-3-Methyl-cyclohexane-2-carbaldehyde.
But that is not the right answer. I think I might have gone wrong in numbering the carbon atoms. Could someone explain why and what is the correct answer?

You have already correctly identified the principal characteristic group of the compound as carbaldehyde. You have also identified the senior parent structure as cyclohexane, which is correct since the compound has only one possible parent structure. The remaining substituents are correctly expressed as prefixes. The last problem is the correct numbering of locants.
The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The lowest locant is assigned first to the principal characteristic group according to rule (c), i.e. 1-carbaldehyde.
Next, low locants are assigned to the other substituents as a set accoding to Rule (f). However, this could be ‘2-bromo-6-methylcyclohexane-1-carbaldehyde’ as well as ‘6-bromo-2-methylcyclohexane-1-carbaldehyde’ since both names correspond to the locant set ‘2,6’.
According to Rule (g), the lower locant is assigned to the substituent that is cited first as a prefix in the name, i.e. ‘bromo’. Therefore, the correct name is 2-bromo-6-methylcyclohexane-1-carbaldehyde.


Value of the frequency factor in the Arrhenius equation Is the value of 'A' necessarily the same for two different reactions taking place at the same temperature?

It's actually a part of another question which is if the rate constant k1 of a reaction is found to be double that of rate constant k2 of another reaction the relationship between corresponding activation energies of the two reactions at the same temperature (e1 and e2) can be represented as e1__e2

The Arrhenius constant, A, does not have to necessarily be the same for two reactions taking place at the same temperature, due to the fact that it depends on the order of a reaction as well as the frequency with which successful collisions will occur during a reaction.

As for the second part, as k ∝ x-Ea, for 2k to be a larger value, Ea for 2k would have to be less than the one for k, i.e. e1 < e2 if the activation energy for 2k is e1 and that for k is e2.


Crystal meth in Breaking Bad Can good quality crystal meth really be made as in Breaking Bad?
Can a pharmaceutical drug really be reverse-engineered?

Can good quality crystal meth really be made and sold as in Breaking Bad?

Yes, most procedures are accurate, although the whole methylamine is so hard to get idea in Breaking Bad is nonsense. It is DEA List I chemical, but can be easily manufactured and no sane kingpin would go steal it from a cargo train (although it makes a good scene). The real problem is getting the P2P (phenyl-2-propanone, or phenylacetone) which the authors of breaking bad seem to ignore (probably on purpose, so that stupid kids won't try and make it at home).

  Can a pharmaceutical drug really be reverse engineered?

Reversed engineered? Assuming you are talking about the transition of pseudoephedrine into methamphetamine, this is just a chemical reduction, following this scheme (if this kid wants to make meth, he would have found this procedure anyway, I'm not giving out any special information that would help anyone manufacture methamphetamine).


How to find the second order perturbation to wave function? Today, I'm looking for how to find the 2nd perturbation to the base in Rayleigh Schrödinger Perturbation Theory (RSPT).

SETUP

Starting from the 2nd order perturbation in Dirac's notation:

\begin{equation}
\hat{H}^0 |n^2\rangle + \hat{V}  |n^1\rangle  = E_n^0  |n^2\rangle + E_n^1 |n^1\rangle  +  E_n^2  |n^0\rangle
\end{equation}

Multiply to the left with $\langle k^0 |$

\begin{equation}
\langle k^0| \hat{H}^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = \langle k^0|  E_n^0  |n^2\rangle + \langle k^0|  E_n^1 |n^1\rangle  + \langle k^0|  E_n^2  |n^0\rangle
\end{equation}

The hermicity of the Hamilton operator:

\begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle  + E_n^2 \langle k^0  |n^0\rangle
\end{equation}

Cancel the third term on the right

\begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle
\end{equation}

Gathering same inner products:

\begin{equation}
\langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle = ( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle
\end{equation}
\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle  = \langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle
\end{equation}
Replacing $| N^1 \rangle$ and $E_N^1$ with the first order perturbation to the base and to the energy:

    \begin{equation}
|n^1\rangle =\sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}
    \begin{equation}
E_N^{1} = \langle n^0 | \hat{V} | n^0 \rangle
\end{equation}

\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \langle k^0| \hat{V}  \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - E_n^1 \langle k^0 \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}

\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n}  \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}

At this point, I'm not sure if the bracket algebra was done properly not even forward.

Then ... are another some algebra steps that I cannot figure out the properly path to find the respective coefficient:

$$\langle k^0 | n^2 \rangle$$

In order to replace it in:

\begin{equation}
|n^2\rangle =\sum_{k \neq n} |k^0 \rangle \frac{\langle k^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_k^0 ???}
\end{equation}

QUESTION

Please, Would someone help me out to show me how to get the properly 2nd perturbation to the wave function?

Disclaimer This post is some kind of a legacy post. Find the notation used in the question in the other answer. I added this proof as I was not entirely certain I understood the notation correctly. As it turned out, I did not. As a result I posted the complete derivation of RSPT up to second order, trying to guide anyone through it using a different notation.$\require{cancel}$
Obviously, I will come back to the initial problem only at the end, if you are impatient, just skip ahead.
Please also note, that I removed the parts that became redundant when I posted the second answer. If you are interested in legacy versions, just look them up in the history.
Also, this page is heavily loaded with $\mathcal{M}^{ath}\mathrm{J_{a}X}$, give it time to load completely.

In Rayleigh Schrödinger perturbation theory, we set up the Hamiltonian as $$\mathbf{H}=\mathbf{H_0}+\lambda\mathbf{H'}.\tag1$$

The perturbed Schrödinger equation is therefore $$\mathbf{H}\Psi=W\Psi_\mathrm{ST}.\tag2$$
You can develop this into Taylor series:
\begin{align}
W                &=
     \lambda^0W_0    + \lambda^1W_1    + \lambda^2W_2    + \cdots
     &&= \sum_{i=0}\lambda^i W_i\tag3\\
\Psi_\mathrm{ST} &=
     \lambda^0\Psi_0 + \lambda^1\Psi_1 + \lambda^2\Psi_2 + \cdots
     &&= \sum_{i=0}\lambda^i \Psi_i\tag4\\
\end{align}

Your unperturbed reference system is a complete solution denoted as
$$\mathbf{H}_0\Phi_i = E_0\Phi_i,\tag5$$
where $i\in\mathbb{N}$.

You chose your system to be intermediate normalised.
$$\langle\Psi_\mathrm{ST}|\Phi_0\rangle = 1\tag6$$
Since all your solutions of the unperturbed system are orthogonal,
$\langle\Phi_i|\Phi_j\rangle=\delta_{ij},$
the same holds for your perturbed system, hence
$$\langle\Psi_{i\neq0}|\Phi_0\rangle = 0.\tag7$$

Now we can combine $(1)$ through $(4)$ and gather them according to their order.
\begin{align}
\lambda^0&:&
  \mathbf{H}_0\Psi_0
    &= W_0\Psi_0\tag{8a}\\
\lambda^1&:&
  \mathbf{H}_0\Psi_1 + \mathbf{H'}\Psi_0
    &= W_0\Psi_1 + W_1\Psi_0\tag{8b}\\
\lambda^2&:&
  \mathbf{H}_0\Psi_2 + \mathbf{H'}\Psi_1
    &= W_0\Psi_2 + W_1\Psi_1 + W_2\Psi_0\tag{8c}\\
\lambda^n&:&
  \mathbf{H}_0\Psi_n + \mathbf{H'}\Psi_{n-1}
    &= \sum_{i=0}^n W_i\Psi_{n-i}\tag{8d}\\
\end{align}

We do have a complete set of solutions, hence we can express perturbations as linear combinations of these.
\begin{aligned}
\Psi_0 &= \Phi_0 &
\Psi_1 &= \sum_i c_i \Phi_i &
\Psi_2 &=\sum_i d_i \Phi_i &
\dots
\end{aligned}

Now we project equations $8$ onto $\Psi_0$. For $\mathrm{8a}$ this is trivial and it leads back to the expectation value of the energy for the unperturbed system. (In the last transformation we use the linear combinations.)
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= \langle\Phi_0|W_0|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= W_0\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9a}\\
W_0 &= \langle\Phi_0|\mathbf{H}|\Phi_0\rangle = E_0\tag{9a'}
\end{align}
The next step is similar, and everything else will follow from there analogously. For $\mathrm{8b}$ we obtain:
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &=   \langle\Phi_0|W_0|\Psi_1\rangle
       + \langle\Phi_0|W_1|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cancelto{c_0}{\langle\Phi_0|\Psi_1\rangle} +
       W_1\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\\
\cancelto{c_o\cdot E_0}{\sum_i c_i \langle\Phi_0|\mathbf{H}_0|\Phi_i\rangle}
       + \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cdot c_0 + W_1\tag{9b}\\
W_1 &= \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle\tag{9a'}
\end{align}

Similarly we can obtain the coefficients by projecting on some other function than $\Phi_0$. I will cut this a little bit shorter.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \langle\Phi_j|W_0|\Psi_1\rangle +
       \langle\Phi_j|W_1|\Psi_0\rangle\\
\cancelto{c_j\cdot E_j}{\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle}
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \cancelto{c_j\cdot E_0}{\langle\Phi_j|W_0|\Psi_1\rangle}
     + \cancelto{0}{\langle\Phi_j|W_1|\Psi_0\rangle}\tag{10b}\\
c_j &= \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{E_0-E_j}\tag{10b'}
\end{align}

Now this is impossible to solve for $c_0$, but we still have the intermediate normalisation.
\begin{align}
0 &= \langle\Phi_0|\Psi_1\rangle
   = \sum_i c_i \langle\Phi_0|\Phi_i\rangle
   = c_0\cancelto{1}{\langle\Phi_0|\Phi_i\rangle} +
     \sum_{i\neq0} c_i \cancelto{0}{\langle\Phi_0|\Phi_i\rangle} = c_0\\
\end{align}

Similarly this applies to all zeroth coefficients, $c_0 = d_0 = \dots = 0$.

Let's get back to business and continue with $\mathrm{8c}$.
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle
  + \langle\Phi_0|\mathbf{H'}|\Psi_1\rangle
  &= \langle\Phi_0|W_0|\Psi_2\rangle
   + \langle\Phi_0|W_1|\Psi_1\rangle
   + \langle\Phi_0|W_2|\Psi_0\rangle\\
\scriptsize\cancelto{d_0\cdot E_0}{\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle}
   + \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
  &= \scriptsize W_0\cancelto{d_0=0}{\langle\Phi_0|\Psi_2\rangle}
   + W_1\cancelto{c_0=0}{\langle\Phi_0|\Psi_1\rangle}
   + W_2\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9c}\\
W_2 &= \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle\tag{9c'}
\end{align}

We do now insert $\mathrm{10b'}$ and get the second order correction:
\begin{align}
W_2 &= \sum_{i\neq0}\frac{
             \langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
             \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle
            }{E_0-E_i}\tag{9c''}
\end{align}

Now we have the expression for the second order energy, let's do the messy stuff for the second order contribution to the wave function.

Skip here for the short version, which is still very long, but that's how we roll for the second order correction to the wave function

We start by multiplying $\mathrm{8c}$ with $\Phi_j$ from the left and integrating (projection). I will include a few more steps for this problem, hopefully that will help transferring it to your notation.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
\end{align}

I will go through this term by term this time, as it is quite a long equation.
Let's start with the most simplest term, the third one on the right hand side. As you noted, it cancels.
\begin{align}
\langle\Phi_j|W_2|\Psi_0\rangle
   &= W_2 \cancelto{0}{\langle\Phi_j|\Psi_0\rangle }\\
   &= 0
\end{align}

Let's continue on the right hand side:
\begin{align}
\langle\Phi_j|W_1|\Psi_1\rangle
   &= W_1 \langle\Phi_j|\Psi_1\rangle\\
   &= W_1 \sum_i c_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_1 \left(c_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} +
         \sum_{i\neq j} c_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= c_j W_1
\end{align}

And the same for the last contribution:
\begin{align}
\langle\Phi_j|W_0|\Psi_2\rangle
   &= W_0 \langle\Phi_j|\Psi_2\rangle\\
   &= W_0 \sum_i d_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_0 \left(d_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
        + \sum_{i\neq j} d_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= d_j E_0 &\mathrm{9a':}&(W_0 = E_0)
\end{align}

Let's go to the left side:
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle
   &= \sum_i d_i \langle\Phi_j|\mathbf{H}_0|\Phi_i\rangle \\
   &= \sum_i d_i E_i \langle\Phi_j|\Phi_i\rangle\\
   &= d_j E_j \cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
      + \sum_{i\neq j} d_i E_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\\
   &= d_j E_j
\end{align}

And the last one can barely be simplified:
\begin{align}
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle
  &= \sum_i c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
  &= \cancelto{0}{c_0 \langle\Phi_0|\mathbf{H'}|\Phi_0\rangle}
   + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
\end{align}

Now, Avengers reassemble, rearrange, and insert the expression for $W_1$:

\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
d_j E_j + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle
 &= c_j W_1 + d_j E_0\\
d_j (E_j - E_0)
 &= c_j \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
   - \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
d_j &=
   c_j \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} c_i \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}
\end{align}

Now introduce the first order base, $c_i$ and $c_j$ from $\mathrm{10b'}$, there will be some minus shuffling as well.
\begin{align}
d_j
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)}
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}\\
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} \cdot
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(-1)(E_0 - E_j)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(-1)(E_0 - E_j)}\\
 d_j &=
   \sum_{i\neq 0}
   \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle%
         \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle%
        }{(E_0 - E_j)(E_0 - E_i)}
   -\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle%
          \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)^2} \tag{10c'}
\end{align}

Well, I am officially exhausted right now.

And lastly, a small remark on your notation (I keep this paragraph for nostalgic reasons). It took me quite some time understanding it, and as we saw (in some earlier version of this answer), I was mistaken.
While I do not mind the hat notation for operators, indicating order on top of the quantities is problematic in my opinion. To me this place is reserved for powers. I suggest you change it to one of the following: $E_{n,0},\ {}^{(0)}\!E_n,\ E^{(0)}_n$. When you have a look at the other answer, you will notice, I have decided to stick to the last option. But as I do not know which book your are using it is also not easy to recommend. I believe there are as many different notations as there are books out there. I personally preferred the book "Introduction to Computational Chemistry, 2nd Edition"by Frank Jensen.
As we settled in the comments, I appreciate that you stick to Dirac's original notation. And also, notation is like Marvel and DC, you pick one and stick with it, you can still appreciate the other, but just not as much. Or a football (Americans pronounce this as "soccer") club, or ...
As long as you understand it and can explain it to someone else, you are fine.
And now we all deserve some chocolate and some coffee and cake. Thank you for reading this post until the end.


Why does the phosphorus trifluoride molecule have dipole moment? My book says that from drawing the Lewis structure for the molecule below, you can conclude that it has dipole moment. Can someone please explain why?

I understand that the dipole moment expresses the polarity of the bond. And that the dipole moment of a molecule is the sum of the dipole moments for each bond in the molecule.

Fluorine is more electronegative than phosphorus.
The phosphorus is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex.
The F-P-F angles are 96 degrees.
Each F-P bond contributes to the net dipole moment, as a vector from the P to the F.  The net dipole moment is the vector sum of the vectors along the three P-F bonds.
If all four atoms were in a plane (a trigonal planar geometry), there would be no net dipole moment.  But because the geometry is pyramidal, the vector sum is not zero and there is a net dipole moment.
In some molecules lone pairs also contribute to the net dipole moment, in this particular case the lone pair of phosphorus is in a mostly s-like orbital, so in first approximation it can be neglected.