Harmonic frequencies (IR) and basis set size Usually vibrations (harmonic), derived from harmonic approximations are bigger than for experimental values. The question is, how does this depend on the size of the basis set. Many papers give that as a statement in their introduction without any explanation.

My idea would be that, if I use variational methods like DFT, then my energy will always be bigger than the real energy, as long as I use a finite basis set. And also increasing the set gives better results in DFT. The harmonic frequencies I determine computationally are derived from the Hessian matrix (or better the force constants for the harmonic oscillator) and this matrix is the second derivative of the energy, which is much too big for small basis sets.

So can I say that, because a smaller basis set will give me a bigger energy, the derived frequency will have to be bigger as well?

No, you cannot say that larger energies will necessarily give larger frequencies.

Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry) $\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates to produce an energy, i.e. $$f(\mathbf{G})=E$$

This method, however, is a very special kind of method. It is one which gives an absolutely unreliable energy. It gives you a number that has absolutely no basis in reality and can't be trusted at all. But, one thing we know about this method is that the error, which is truly horrendous and unpredictable, is constant! From this fact, it follows that this method predicts the correct frequencies, the experimental ones. That is, at the minimum of the potential predicted by our hypothetical method and the actual minimum, the difference between our energy, $E_0$, and the true energy, $\mathcal{E_0}$, is,
$$E_0-\mathcal{E_0}=\epsilon$$
or,
$$f(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon$$
For all real methods, however, we would need to write the above equation as,
$$h(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon({\mathbf{G}})$$

The point of all this being that the hypothetical method I described was not variational or anything else you might desire out of a quantum chemical method. It only predicted energies for which the error term was constant. Whereas the real method, $h(\mathbf{G})$, predicts energies for which the error term is a function of the geometry $\mathbf{G}$.

The concise way of saying this is that frequencies depend only on the curvature of the potential well and not on the depth of the potential well.

With that established, we have answered the question that smaller basis sets, with their larger energies, do not necessarily give higher frequencies.

For instance, if you look at Sinha and Wilson's paper,[1] you will learn of something very empirical which computational chemists do sometimes that bothers me a lot but is illustrative for our purposes. In that paper, a big set of 42 organic molecules is taken, and the harmonic frequencies are calculated at a bunch of different methods and basis sets, then these frequencies are compared with the experimental values. A least squares fit is then performed to come up with scaling values to map the harmonic frequency onto the experimental frequency. As you say, these constants are numbers less than one, say 0.95. The reason these constants are less than one is what bothers me about doing this. Experimental values are anharmonic frequencies. We don't live in a harmonic world. Anharmonicity shifts the frequency down, by lowering one of the potential walls to accommodate bond-breaking, hence we need to scale our harmonic frequencies downwards. So we're basically comparing apples with oranges and there's no way to be sure that this will work in all cases.

More to your question, in this paper, the correction factors do not increase monotonically with basis set size as you suggested (this is the same as frequencies decreasing with basis set size). I won't reproduce their numbers here, because I don't know if that's frowned upon, but for both MP2 and B3LYP, the cc-pVnZ and aug-cc-pVnZ (n=D,T,Q) basis sets have T < D < Q for the scaling factor. So, the triple-zeta basis set invariably predicts a lower energy, but predicts a higher frequency.

Even more interestingly, if one only considers low-frequency modes, which are particularly poorly described by normal modes (these are torsions usually), the ordering becomes Q < T < D for the scaling factors. That's really a failure of the harmonic model more than anything, but it illustrates the point well that lowering the energy really doesn't guarantee you anything in terms of frequencies.

To address your comment, the reason there is any dependence on basis set size is because the error away from the equilibrium structure will change with basis set size, and hence the curvature of the potential changes which affects the frequency.

[1] Pankaj Sinha, Angela K. Wilson, J. Phys. Chem. A, 2004, 108 (42), 9213-9217. DOI: 10.1021/jp048233q


Clarification of Kinetic Energy during Phase Change This website states that:

During a change of phase the temperature does not change, but the internal energy does.  The internal energy is the sum of the kinetic energy of the molecules and the chemical potential energy of the molecules.  During a change of phase, the average kinetic energy of the molecules stays the same, but the average potential energy changes.

I'm confused as the two bolded statements seem to contradict each other.

My interpretation is that during a phase change, the temperature remains equal, but the kinetic energy of its particles increase/decrease.

Could this please be clarified and confirmed?

The temperature is not the average kinetic energy.

This is a bad habit which most chemists and physicists pick up because for an ideal gas, and for most systems at normal parts of the phase diagram, this is true. In thermodynamics, however, temperature is defined in terms of the thermodyanic beta as,
$$
\frac{1}{k_b}\frac{\partial S}{\partial E}=\beta
$$
where $\beta=1/(k_bT)$.

That is to say that during a phase change, because the temperature is constant, the ratio of the change in entropy to the change in total energy (kinetic plus potential) is constant. This means that when the phase change just begins and just ends, the ratio of entropy and energy must be identical, and unless this is quite an unusual phase change, this means both need to have changed from their initial values. It wouldn't be a phase change if neither entropy nor energy changed. Because temperature is defined in this way and not as $T\propto\langle \text{KE}\rangle$, then it is perfectly reasonable for the kinetic energy to either increase or decrease during a phase change.

For instance, in normal systems one would expect that the entropy of the liquid would almost certainly be larger than the entropy of the solid. This means that the total energy must decrease by an amount $T_m\Delta S$, which makes sense as one would expect the potential energy (and perhaps the kinetic energy) to decrease when going to the liquid phase. In other words, the entropic gain allows the system to energetically relax. With this picture, it is not too surprising that the average kinetic energy can decrease during a phase change.


Are sulfonyl-carboxylates stable in water? This time I am interested in a specific type of acid anhydride, sulfonyl-carboxylates, i.e. compounds with the following molecular structure:

where $\ce{R2}$ can be an organic substituents or hydrogen atoms. $\ce{R1}$ can be an organic substituent only.

In addition to the procedures described by Ian Fang, these mixed anhydrides (with $\ce{R_1, R_2 \ne H}$) have been used to prepare symmetric anhydrides of carboxylic acid in solid state reactions. If that's interesting to anyone, I can look up the reference.

As far as the effect of water is concerned, I do remember a procedure where the mixed anhydrides in question were actually prepared in a biphasic system of water and dichloromethane in the presence of a phase-transfer catalyst.


electron donation affecting carbonyl bond strength I have been told that electron donation decreases carbonyl bond strength and thus decreases the bond stretch frequency, but electron withdrawal increases it.

This does not seem to make any sense as I presume that increasing electron density between two atoms should increase the attraction of their nuclei to those electrons and thus to each other. Is this reasoning flawed? or does this just apply to carbonyls?

Let's take the above molecules, acetyl chloride and ethyl acetate, as examples of acyl compounds with electron-withdrawing and -donating groups, respectively.

We know that Cl has a higher electronegativity than O as a halogen, so the electron density from the carbonyl carbon should be pulled toward the chlorine in acetyl chloride and pushed toward the carbonyl oxygen in the case of the ester. This might not be immediately obvious, as the two oxygens should have the same electronegativity, so I like to relate this latter point to electric force laws:
$$ F=\frac{kq^2}{r^2} $$
Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O (this assumes electrons to be point charges, so it is only useful as a trend).

Therefore, we draw the following picture:

Meaning that there is more electron density on the carbonyl carbon for ethyl acetate than for acetyl chloride.

Now, that's somewhat of a problem for the carbonyl oxygen of ethyl acetate! It already has a lot of negative charge as an oxygen, so it will experience repulsive forces due to the increased electron density on its adjacent carbon. Remember that oxygen does not like having more than a charge of -2 formal charge, which is already satisfied by the double bond it has to the carbonyl carbon; a similar relation can be made for actual charge density.

These repulsive forces, well, repel the oxygen away from the carbon a bit, increasing the C=O bond length. And we also know from physics that increasing the bond length decreases the frequency of vibration. This should make intuitive sense: if the bond distance increases, then the atoms have more distance to travel to complete one cycle of vibration, thus decreasing the number of cycles that can be completed per unit time.

Now onto our example electron-withdrawing group:

Acyl chlorides pull electron density away from the carbonyl carbon, as mentioned earlier. Following the same logic, we can conclude that the C=O bond distance decreases(... to a point*). It follows that the C=O vibrational frequency should increase.

Well, that was a lot of words. But let's approach this from another, much simpler, way. In an acyl chloride, the chlorine acts as an electron-withdrawing group, destabilizing the already electrophilic carbonyl carbon. While this decreases the stability of the molecule, that does not mean that all bond strengths are diminished. In acyl chlorides, the C-Cl bond is the weak, labile bond, not the C=O bond. In fact, the C=O bond strength increases for the reasons outlined above. In our example ethyl acetate, the C-O donates electron density to the electrophilic carbon, thus stabilizing the compound. This does not mean that all bonds are strengthened. If anything, it might indicate a greater C-O bond strength than would otherwise be expected. The C=O bond however,  is a different bond that is being affected inductively, so it must be considered differently.

Personally, I am a fan of the first explanation, as that provides an actual understanding of what's going on, whereas the second explanation leads to a more memorization-focused approach to chemistry. I hope this helps!

*Extra information--Let's consult the following graph (ignore the specific numbers on it):

What this shows is the distance between two atoms and the potential energy of their "bond" (this is one of those areas where it should be noted that bonds are really just constructs to help us make sense of molecules... but this is not important for now). To the right of the minimum, attractive forces (nucleus --> electrons) dominate. To the left, repulsive forces (nucleus <--> nucleus) dominate. As two positive charges get further away from each other, their energy goes to zero (this can similarly be rationalized by simple physics II-type equations). As their distance approaches 0 (where a distance of 0 indicates superposition, or direct overlap), the energy between the two atoms goes to infinity. This is all just to say that there's a limit on how close the atoms can get to each other. You can see that the slope to the left of the minimum is much steeper and increases at a much faster rate than the slope the the right. It follows that there should be less difference in the C=O bond distance as compared with some fair reference molecule such as acetaldehyde than between the ester and that reference. Part of why I'm bothering with this at all is because of how useful this graph is in just about everything I have studied in some shape or form, so it is very useful to familiarize yourself with it.


Is this galvanic cell representation right? Question

13. In the galvanic cell:
$$\ce{Pt(s) | I2(g) | I-(aq) || Fe^3+(aq) | Fe^2+(aq) | Pt(s)}$$

Representation of anode is incorrect and cell will not work.
$[\ce{Fe^3+}] = [\ce{Fe^2+}] = [\ce{I-}] = \pu{1 M}$ is sufficient for $E_\mathrm{cell} = E_\mathrm{cell}^\circ.$
$\ce{I-}$ gets oxidized to $\ce{I2}$ and $\ce{Fe^3+}$ gets reduced to $\ce{Fe^2+}.$
None of these.

Given Answer

 3. $\ce{I-}$ gets oxidized to $\ce{I2}$ and $\ce{Fe^3+}$ gets reduced to $\ce{Fe^2+}.$

My Thoughts
The representation of anodic half confuses me. I am unsure whether it is right representation. According to me, it should be $\ce{I-} | \ce{I2}$ (anodic half). It seems to me that answer should be 1.

The chapter 1 of the reference 1 describes the Line Notation for Galvanic Cells as follows:

  A shorthand way of completely describing an electrochemical cell such as Galvanic cell is called line notation. It is a shorthand way of expressing the redox reaction progressing in the electrochemical cell. In this notation:

  A slash (/) or a vertical line (|) represents a phase boundary, and a comma separates two components in the same phase. A double slash (//) or a double vertical line (||) represents a phase boundary whose potential is regarded as a negligible component of the overall cell potential (e.g., salt bridge).
  When a gaseous phase is involved, it is written adjacent to its corresponding conducting element.
  The constituents of the cathode on the right and the constituents of the anode on the left.
  The phases of all reactive species are listed and their concentrations or pressures are given if those species are not in their standard states (i.e. $\pu{1 bar}$ for gasses and $\pu{1M}$ for solutions).

Thus, in the given cathode representation, both $\ce{Fe^3+ (aq)}$ and $\ce{Fe^2+ (aq)}$ are separated by a vertical line (|), which is an incorrect representation, according to the description (1) above. They should be actually separated by a comma (as $\ce{Fe^3+ (aq)}$, $\ce{Fe^2+ (aq)}$) because there is no phase separation between them (both are in same phase).

Also, according to the description (2) above, $\ce{I2 (g)}$ cannot be introduced into $\ce{Pt (s)}$ anode (as indicated in the representation) because $\ce{I2 (g)}$ is the product of the reaction, not a reactant.

Based on these, given answers (1) and (3) should be correct. Also, OP's representation of 'anode half' is also correct since there is phase separation.

References:

Allen J. Bard, Larry R. Faulkner, In Electrochemical Methods: Fundamentals and Applications; Second Edition, John Wiley & Sons, Inc.: New York, NY, 2001 (ISBN 0-471-04372-9).


What are "O-hydrocarbons"? A pharma translator's question (Russian to English, "О-углеводороды") A translator colleague is translating from Russian a plan for study of a number of leachable and extractable substances (pharmacology). This Russian text mentions one particular class of chemicals which theoretically could become leachables:

  "Hydrocarbons (alkanes, alkenes, alkynes, O-hydrocarbons)".

She can't understand what this could mean.

The Russian term is О-углеводороды.

My fellow translator sent a letter to the authors, and they replied that they used this term to stand for "oxygen-substituted hydrocarbons", just like andselisk suggested.


Looking for a non-toxic, environmentally safe highly colored or intense dye I need a dye that is "highly colored" or intense.  I want to put it in water and flow the water in the form of a liquid sheet that is around 100 microns thick.  A number of spray nozzles use this principle to produce droplets.  The liquid is first formed into a sheet that breaks down after a few milliseconds into droplets.
The issue is that at such thin thicknesses, colors often disappear even if they look very intense in a tank.  I'll be working with around 0.2 gallon per minute to 0.5 gallon per minute flowrates, so something that hopefully can be added in relatively small quantities.  But more important is that the dye is non-toxic and environmentally safe.
Thanks for any suggestions.

Many organic dyes which are fluorescent can be carcinogenic like the famous Rhodamine dyes. You can try a couple of tricks. One of best inks, at least for demonstrations, is the yellow highlighter's ink which fluoresces bright green in water (see my photo below). If you can shine a bright white LED or a bright blue LED, it will fluoresce very strongly. You can use an optical filter in front of the camera that can block blue light and let the green and longer wavelengths in. With that you can increase the sensitivity.
Note that fluorescence intensity is directly proportional to the incident light intensity in dilute solutions.


Is there a way to tell which one is the racemic or meso compound with 1H-NMR if all you had was the structure and spectra? I know that diastereomers will have different chemical shifts but can you predict which one is which via up/down field shifts in the spectra?

For instance, is it like alkenes vicinal H’s when cis and trans are up and down field respectively?

I ran a 1H-NMR of hydrobenzoin and when comparing it to the literature I noticed that the racemic mixture was further upfield than the meso.

I made the Newman projections below and assumed the most energetically favorable positions to show the cis and trans H’s in the diastereomers.

Is this the right way of thinking about this?

Hydrobenzoin has 3 stereoisomers: meso, R,R and S,S. You have not included the S,S. Also, your Newman projections might make sense based on steric grounds, but largely ignore electronic factors such as pi-stacking and hydrogen bonding. A very quick run through the Avogadro Bable geometry optimisation has these molecules optimising with the phenyl rings somewhat gauche.

While there are many methods in NMR that can assist in determining relative and absolute stereochemistry, there is no method that offers universal application. Generally it is not possible to identify stereoisomers from first principles in non-rigid skeletons. This is because conformers are rarely locked into a single position. Properties such as spin coupling, and nOe are useful, and the method by Murata is considered groundbreaking in this area. It relies on measurable  3J(HH) and 2,3J(CH).
Derivatisation, database comparison, measurement of residual dipolar couplings in aligned media (liquid crystals), and computational methods. Determination of absolute configuration usually involves complexation with a enantiomerically pure reagent.

And I'm not confident that your method for assigning cis/trans geometry based on chemical shifts is reliable.


Standard electrode potential of copper disproportionation I am having some difficulty with the below.
It says the standard potential for $\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
The relevant half equations are:
$\ce{Cu+ + e- -> Cu}$ (potential $=  \pu{0.52 V}$)
and
$\ce{Cu^2+ + e- -> Cu+}$ (potential $=  \pu{0.16 V}$)
To get the overall equation, I flipped the second half equation and so reversed the sign of the potential also to get $\pu{-0.16 V}$.
The first half equation is therefore reduction and the second is oxidation.
If standard electrode potential is $\mathrm{RHS (reduction)} -\mathrm{LHS (oxidation)}$, would the answer not be $\pu{0.52 V} - (\pu{-0.16 V}) = \pu{0.68 V}$? Why is it $\pu{0.36 V}$?

Alas, the confusions related to signs in electrochemistry will never vanish. You mention that
$\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
Let me start with a single equation, x-y = 10; There can be indefinite solutions if you can simultaneously change the value of x and y. However, the moment you fix the value of x, the value of y is fixed.
You stated that the overall cell potential is $\pu{+0.36 V}$. Electrochemically, this means that this reaction is spontaneous.
Now you also know that,
$$E_\text{cell} = E_\text{reduction} - E_\text{anode} \tag{1}$$
You are not supposed to change any sign of the half-cell from the electrode potential tables. People should stop teaching this nonsense to relatively innocent students. Suppose, I write
\begin{align}
&\ce{H2O (liquid) -> H2O (gas)} &T &= \pu{100 ^\circ C} \\
&\ce{H2O (gas) -> H2O (liquid)},    &T &=\pu{ -100 ^\circ C} ??
\end{align}
The equation (1), itself takes care of all sign flipping and all.
Your half-cell corresponding to the reduction is
$\ce{Cu+ + e- -> Cu}$ (potential $=  \pu{0.52 V}$)
And your half-cell potential for the oxidation is
$\ce{Cu^2+ + e- -> Cu+}$ (potential $=  \pu{0.16 V}$)
Using equation (1), what you do get (remember no sign flipping) = $\pu{+0.36 V}$


How to separate sulfur and carbon powders? How can someone separate two elemental solids, in powder form, that are mixed together?

If I have a powder, and its mass is 60% carbon and 40% sulfur, how do I separate one from another in a high school level laboratory?

Two possibilities:

Distillation: Sulfur has the comparatively low boiling point of 718 K (445 °C). Heating in apparatus not affected by sulfur (which leaves out iron) will leave behind the carbon. This is, after all, how sulfur winds up at volcanic vents, where it is harvested. You might even create a retort out of concrete, which would simulate volcanic rock.
Sulfur is somewhat soluble in toluene and in carbon disulfide, but both are toxic and flammable, with $\ce{CS2}$ being particularly easy to ignite. These are not safe to use without proper training and equipment.


Would adding acetic acid from household vinegar to washing machine make it too dilute to be effective? Many people advise adding a cup of acetic acid (vinegar) to one's laundry during the clean cycle. From my calculations, considering one cup of acetic acid and a top gallon machine (at 40 gallons), it would seem this would completely dilute the concentration next to nothing.

Is this correct or am I missing something? Someone had mentioned something about acetic acid being a weak acid, but I'm not sure what that means in this context.

As a weak acid, acetic acid will act as a buffer in the washing machine solution.

Following similar calculations from the first answer of the question,What's the pH of Vinegar when it generally contains 5% Acetic acid?, we can calculate the concentration and pH of acetic acid in the washing machine solution. 1 Cup is ~ .237 L and 40 gal is ~ 152 L. Using the concentration of acetic acid in vinegar from the post above (0.847 mol/L):

(0.847 mol/L)(0.237 L)=(X mol/L)(152 L), X=0.00136 mol/L

pH of the washing machine solution = 0.5[4.76-log(0.00136)] = 3.80

So, although the acetic acid is very dilute, there is still a significant pH drop from the addition of 1 cup of white vinegar. Changes in pH can strongly affect the solubility of many compounds, which would seem to be helpful in the cleaning process. Furthermore, the denaturing of proteins is promoted at lower pH, so lowering the pH could be particularly useful in the removal of protein-based stains (grass, bodily fluid, egg, etc.).


Is there an easy way to make PVC from PVA? Say, I took Elmer's glue (or a similar Polyvinyl Acetate) and I inserted some Chlorine Tabs and warm it up while stirring.

Would the electronegative Chlorine turn some of it it into Polyvinyl Chloride (PVC)?

Would I get a Dioxin Compound too?

Would I get any other toxic Side-Product?

Choring Tablets have positivly charge cholrine atoms you want negative, so chlorine tablets will not work.  And no you won't get a dioxin compound unless you burn it, then you may ge some.  If you want to make PVC from PVA you can use Hydrochloric acid but you may not get a good yield.


D6h symmetry clusters- orbital classification confusion! I was am simulating high symmetry clusters D6h in several programs such as molcas and gaussian. I understand that the highest symmetry used in these programs is D2h and therefore there would be mixing of the different symmetry representations upon correlating from D6h to D2h. However upon simulating Mn@Si12 cluster, I'm seeing that the silicon atoms in the silicon cage have for all representations: s, px, py and pz orbitals despite there being restrictions according to D2h e.g b3u representation has only px functions and b2u only having py functions. Could anyone please shed some light on this phenomenon?

The symmetry operations are usually defined with respect to the origin of the coordinate system. For single atoms one naturally chooses $(0,0,0)$, which coincides with the symmetry center.

For example the $\sigma_{xy}$ mirror plane in matrix representation (with $(0,0,0)$ being the symmetry center) would be:

$\sigma_{xy}=\pmatrix{1&0&0\\0&1&0\\0&0&-1}$

which means: invert every $z$ component. If applied to some object centered on $(0,0,0)$ (e.g. the AO of an atom on this position), the result would still be centered on $(0,0,0)$.

In case of molecules we can have at most one atom sitting on the symmetry center. Sometimes none, e.g. in $\ce{C2}$ where the symmetry center is in the middle between both atoms. If you apply the above matrix here (assuming the molecule axis is aligned with the $z$ axis), then it would map the AO from one to the other atom. Therefore it is not symmetric (with respect to the above defined symmetry operation) anymore.

Thus you cannot relate the IRREP of an atomic orbital in the single atom, with its IRREP in the molecule. The symmetry center changed! For example take an $s$ atomic orbital of any of your $\ce{Si}$ atoms. The position of the atom does not coincide with the center of the molecule. Therefore any symmetry operation (except for $E$) would map the $s$ orbital to some other $\ce{Si}$. The AO is no longer characterized by an irreducible representation.

What the code does instead, is first generating Symmetry Adapted Linear Combinations (SALC) of the atomic orbitals. Those are then again characterized by IRREPs (of the molecule) and used for the calculation. Now each obtained MO is indeed only constructed from SALC of the same IRREP. In other words: the Hamiltonian, and therefore the MO coefficient matrix as well, become block diagonal. However, in the output the MO coefficients are transformed back from the SALC basis set to the AO basis set, which is not block diagonal and may contain contributions from any AO for each IRREP of the MOs (an exception would be the AOs of an atom located in the center of the molecule).


Is iron the most stable element in the periodic table? According to the binding energy per nucleon vs mass number graph, it is observed that iron-56 has the maximum value of binding energy per nucleon ($\pu{8.75 MeV}$). It means that iron-56  is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. This is the approximate basic reason why iron and nickel are very common metals in planetary cores since they are produced profusely as end products in supernovae and in the final stages of silicon burning in the stars. So, in one word, iron is quite stable.

But, what about helium and other noble gases? They are considered the most stable elements in the whole periodic table. But their binding energy per nucleon value is less than iron-56. So, they are not stable as iron-56. Is that true? Is iron the most stable element in the periodic table both structurally and chemically?

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of this question).


Why is the buckminsterfullerene the purest form of carbon? Other websites say that $\ce{C60}$ doesn't have surface bonds that are attracted by other atoms as in graphite and diamond.

I understand that graphite may be attracted by other atoms because of its dangling electron. But why diamond? Each carbon in diamond is covalently bonded to $4$ other carbon atoms in a tetrahedral fashion.

Diamond has dangling bonds on the outer surface of the crystal for pretty much the same reason as graphite. If you understood graphite differently, then you understood it wrong.

See, a molecule of oxygen contains 2 atoms, a molecule of sulfur has 8; but how many atoms are there in a "molecule" of diamond or graphite? Try drawing one to the end, so as to count them. You won't be able to do that. There is no end. The thing is infinite. But the real-world objects are finite, which means that at some point you have to say "Enough" and crop your ideal structure, and in doing so, you leave dangling bonds which attract other atoms. Fullerene lacks those, and hence is "more pure".

There is an altogether different dimension to the problem. Our thought experiment implied that we are able to produce a huge crystal without defects except maybe some on the surface. This is not true. Real-world compounds always contain impurities, and once you have a wrong atom built into the crystal lattice of graphite or diamond, it is stuck there forever. You'll never remove it, short of destroying the entire crystal. Fullerenes, on the other hand, are molecular compounds. They can be dissolved. They can be put through chromatography, sublimation, and other purification techniques. We can always remove any impurity (not that we can remove all of them, because nothing is ideal).

Either way, fullerenes win.


What is more important, to have the parent chain be the cycle or the chain that has an hydroxyl group I know how to name alcohols, I just don't know what I should chose as the parent chain. Should I chose the cycle or the chain with the hydroxyl group. Here is an example:

My current algorithm tells me that this is 1-(propan-2-olyl)cyclohexane, because it always chooses the cycle with the most skeletal atoms for a parent chain no matter what else is present.

So my questions are:

In what situation an acyclic chain is chosen for a parent chain over a ring?
If this is one of those situations would the choice be affected if we add on hydroxyl group to both the ring and the chain?
If this is not one of those situations would the choice be affected if we increase the length of the acyclic chain and/or add more hydroxyl groups?
Something on a slightly different topic: is 1-(propan-2-olyl)cyclohexane a correct name or should it be 1-(propan-2-ol-1-yl)cyclohexane? I am not sure if the 1 is omitted or not.

P.S. Is there a place where I can find the Blue Book for free? I only have the 2004 draft and I am not sure how much of it is changed in the 2013 version?

The compound given in the question contains a characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the senior parent structure has the maximum number of substituents corresponding to the principal characteristic group.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

Therefore, the correct name for the compound given in the question is ‘1-cyclohexylpropan-2-ol’ (not: ‘(2-hydroxypropyl)cyclohexane’). (Note that in ‘(2-hydroxypropyl)cyclohexane’, the locant ‘1’ of ‘1-(2-hydroxypropyl)cyclohexane’ is omitted since, according to Subsection P-14.3.4.2, the locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings.)

If there is still a choice (for example, if the ring and the chain each have a substituent corresponding to the principal characteristic group), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).


The instability of hydrated carbon dioxide or "carbonic acid" Why is hydrated carbon dioxide - the predominant form of acid that one gets upon dissolution of carbon dioxide in solution - so unstable?

Is the below rationale valid?

Carbon in carbon dioxide has two (empty? no, but still vulnerable to attack) p-orbitals and bears a strong partial positive charge.
Oxygen's lone pair can attack an empty p-orbital and form a formal charge-separated complex with the carbon dioxide.
This is "hydrated" carbon dioxide or "carbonic acid." This form is extremely unstable and subject to disproportionation due to an unfavorable charge and the unfavorable nature of charge separation itself. The change in entropy also favors the products of disproportionation.
However, there exists a pathway to stability - that is - protonation of the oxygen with the negative formal charge by the oxygen bearing the positive formal charge.
This, however, is akin to a forbidden fruit; the ephemeral three-membered ring that would have to be formed exhibits "ring strain" (if you object to this term, can you please elaborate on your objection), and as a result, disproportionation is overwhelmingly favored, especially from an entropic standpoint (reconstitution of carbon dioxide gas is highly entropically favorable).

The situation is described by the following equilibrium
$$\ce{H2O + CO2 <=> HO-CO-OH}$$
In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.).  However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.

When all is said and done, water and carbon dioxide are two extremely stable molecules.  Since
$$\Delta G = -RT\ln K$$
Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.


Clarification on Photodissociation Can someone please explain what photodissociation is and if it is more efficient than electrolysis while splitting water into its components?

Thanks in advance.

Photodissociation is a unimolecular reaction triggered by absorption of light. You start with one molecule, a bond (or more) breaks and you end up with two (or more) products. In water you can get $\ce{OH}$ and $\ce{H}$ (etc), usually in excited electronic states. You can not get the products of the electrolysis.

Water electrolysis is a more complex reaction. The final balance is:
$\ce{2H2O -> 2H2 + O2}$,
so you break four bonds and make three bonds.

Both reactions are not spontaneous (in the electronic ground state of water). The net balance is not energetically favorable (you start with more energy in bonds than what you end up). However, the way you add this energy is also different. In photodissociation the energy is given by light; in electrolysis the energy is given by forcing an electron current.


How are reaction enthalpy and entropy affected by temperature The equation for Gibbs Free Energy demonstrates how the standard change in Gibbs Free Energy for a reaction is affected by temperature.

$$\Delta G^\circ =   \Delta H^\circ - T\,\Delta S^\circ$$

However, to my understanding, the equation makes the inherent assumption that the standard change and enthalpy and the change in entropy for the reaction is unaffected by temperature. I have heard that this is not true and that reaction enthalpy and entropy are indeed dependent on temperature. However, I do not know how exactly temperature affects the two and why.

Would a reaction occurring at a higher temperature be more or less exothermic? Or does it perhaps depend on the reaction itself?

I know that the entropy of a substance increases at higher temperatures but I do not know how this correlates to the situation of a reaction since I'd assume that both the reactants and the products would be at a higher entropy.

For each species you could use Kirchoff's equation to calculate the change in enthalpy with temperature, in its integrated form (at constant pressure) this is $\displaystyle \Delta H_2-\Delta H_1=\int_{T_1}^{T_2}\Delta C_p dT$. The heat capacity $C_p$ changes with temperature also, but for small temperature changes it may be considered to be effectively constant, then $\displaystyle \Delta H_2-\Delta H_1=\Delta C_p(T_2-T_1)$, if not, the heat capacity can be approximated by $C_p=a+bT+cT^2 \cdots$ and integrated but the constants should be determined experimentally.

The entropy is zero at zero Kelvin for a pure substance so must increase with temperature and as $\displaystyle S_T=\int_0^TC_p/T dT$ you can calculate the change in entropy between two temperature.

However, there are phase changes as the temperature increases (solid, liquid, gas) then there are discontinuities in the heat capacity and the entropy becomes $\displaystyle S_T=\int_0^TC_p/T dT+\Delta H_{fus}/T_{fus} + \Delta H_{vap}/T_{vap}$. The terms with phase changes cancel out in $\Delta S$ if your species are in one phase, i.e. all reactants are in the gas phase.

(If you are familiar with partition functions in Statistical Mechanics then thermodynamic properties of molecules can, in principle, be calculated directly.)


Second order reaction cases https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Second-Order_Reactions

When you will read this webpage ,then you will find that they have made cases for different types of second order reaction .

In  $situation 2A $ of case 2 in  the article (eq 1.13 and 1.14) ,they have taken change in concentration of A to be dx/dt instead of
d(a-x)/dt .What is the reason for this ?

Also what I am also finding difficulty is how did they combine the reactants to make it $[A]^2$.......  it would mean that individually the concentration of a single reactant A  is following 1st order reaction. So the rate constant in disappearence of A should be first order instead of second so what is the reason for this .I t would have make sense for 2A to follow second order instead of just A

If you are talking about the 1.13 and 1.14 equations, than whta he did was just a simple subistituction where $[A] = x,\ d[A] = dx$, if not, please identify the step.

And, yes, you are going to get diferent equations for $[A]_{(t)}$ depending on the order of reaction, $0^{th}$ order gives you a linear equation, $1^{st}$ order gives you an exponetial equation, and all the orders higher than 1 gives you a fraction diference like: $$\frac{1}{A_t^n}-\frac{1}{A_0^n} = n\ k\ t$$ where n is the order of reaction minus 2.

The step where they make $[A]^2$ is only because you have an elementary equation, it can not be assumed to be that if the equation was golbal or not elementary.


Would a carbamate be attacked twice by an alkyl lithium cuprate reagent? For example, say you react ethyl carbamate with methyl-$\ce{LiCu}$; would two methyl groups displace both the amide and the ester, creating an isopropyl alcohol?

I don't think so, carbonyls don't typically react with Gilman's reagent (dimethyllithium cuprate) unless the carbonyl happens to be $\ce{{\alpha}-{\beta} }$ unsaturated. So I would expect no reaction under normal reaction conditions.


What helps a sliced fruit or vegetables oxidize slower? What helps a sliced fruit or vegetables oxidize slower ?

A need of chemical compound to use for the slowing of oxidation... we have tried using a lemon juice it actually helped by 15 mins. However it change the quality of the vegetable after being dip for 10 mins and (the problem is how do we slow down the oxidation for 10 days without changing the quality of the fruit) since lemon juice changes the quality of its compound (2-3 days rather).

You can cover the cut pieces with aluminium foil which acts as sacrificial element and prevents oxidation of food.


Why is naphthalene less stable than benzene according to per benzene ring? Why is benzene more stable than naphthalene according to per benzene ring. Which results in a higher heat of hydrogenation (i.e. energy released on hydrogenation) of benzene than naphthalene according to per benzene ring, i.e. if we hydrogenate only one benzene ring in each. I think it should be opposite.

Both are aromatic in nature both have delocalised electrons but naphthalene has more resonance structures and more delocalisation so overall it must be more stable for a single ring.
May someone help?

A short answer is: you're forgetting that a double bond is shared between the two rings in naphthalene.
A long answer is given below.

In terms of heat of hydrogenation

Naphthalene can be hydrogenated to give tetralin.
Further hydrogenation gives decalin.

Scheme 1: hydrogenation of naphthalene.

On the other hand, the hydrogenation of benzene gives cyclohexane.

Scheme 2: hydrogenation of benzene.

In an old report it reads (Sherman, J. Oil Soap (1939) 16: 28):

  If benzene could be simply regarded as cyclohexatriene with no interaction between the double bonds, the heat of hydrogenation to form cyclohexane should be just three times the heat of hydrogenation of cyclohexene to form cyclohexane.
  From Table I, this is $3 \times -28.6 = -85.8$ kcal/mol.
  The experimental value is $-49.8$ kcal/mol.
  Hence benzene is more stable by an amount $-49.8 -(-85.8) = 36.0$ kcal/mol than it would be if it were completely unsaturated in character (with no interaction between the double bonds).
  From the theoretical viewpoint, this extra stability of benzene (and other aromatic compounds) is shown to be a consequence of the fact that the normal state of the molecule is not that corresponding to either Kekulé structure but is a sort of combination of the two.

This $36.0$ kcal/mol may be regarded as a direct measure of aromaticity and is cited in many places as resonance energy.

In particular, the resonance energy for naphthalene is $61$ kcal/mol.
This value is shifted by around $5 \times -28.6 = -143.0$ kcal/mol (five double bonds) from the actual heat of hydrogenation of naphthalene.
This means the heat of hydrogenation of naphthalene would be somewhere around $61 - (-143.0) = -82$ kcal/mol.
The actual value from NIST (to cis-decalin) is $-318$ kJ/mol, or $-76$ kcal/mol.

So it costs $-49.8$ kcal/mol to hydrogenate benzene to cyclohexane but only $-76$ kcal/mol to hydrogenate naphthalene to cis-decalin, less than twice a benzene.
Furthermore, part of this energy is due to the resonance energy, which is $36.0$ kcal/mol for benzene, but only $61$ kcal/mol for naphthalene, again less than twice a benzene.

What did we forget?

Naphthalene rings are fused, that is, a double bond is shared between two rings.
This makes the above comparisons unfair.

A better comparison would be the amounts of resonance energy per $\pi$ electron.
Benzene has 6 $\pi$ electrons, which makes its resonance energy equal to $36.0 / 6 = 6$ kcal/mol per $\pi$ electron.
Naphthalene (10 $\pi$ electrons) shows a remarkably similar value: $61 / 10 = 6.1$ kcal/mol.

Even comparison of heats of hydrogenation per double bond makes good numbers.
For benzene (hypothetically three double bonds) it costs $49.8 / 3 = 16.6$ kcal/mol to hydrogenate each double bond.
For naphthalene it would be $-76 / 5 = 15.2$ kcal/mol, again very similar.

What about tetralin?

Again NIST comes to our rescue.
The most recent value for the heat of hydrogenation from naphthalene to tetralin is $-125$ kJ/mol, or $-29.9$ kcal/mol (Scheme 1, first arrow).

As a per double bond value it gives us $-29.9 / 2 = -15.0$ kcal/mol, close to both benzene and naphthalene (all of them differ by less than $1.6$ kcal/mol).

In terms of electronic structure

  Both are aromatic in nature both have delocalised electrons but naphthalene has more number of $\pi$ bonds and hence more resonance structures and more delocalisation so overall it must be more stable.

I believe the highlighted sentence tells it all.
As you said, delocalisation is more significative in naphthalene.
A simple model for delocalisation is the particle in a box (in this case, more of a rectangle).

The $n$-th energy level for an electron confined in a single dimensional box (line segment) is

$$E_n = \frac{n^2 h^2} {8 m_e L^2}\text{,}$$

where $m_e$ is the mass of the electron, $h$ is the Planck's constant and $L$ is the length of the line segment.
So energy decreases with the square of the length of the confinement.
Results are analogous for other dimensions.

From this simple model, the more confined an electron, the higher will be its energy.
It is thus perfectly reasonable to rationalise that more delocalisation results in more stability.
Given its larger delocalisation, it seems rational that the energy levels of naphthalene have less energy in comparison to benzene.


Problems regarding Kc,Kp and rate of reaction In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_\mathrm f= k_1[A]^a[B]^b$

rate of backward reaction is $k_\mathrm b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get
$$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write
$$\ce{H2 + 0.5O2 -> H2O}$$ or
$$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

Well, there are two cases:

$a + b = c + d$; and
$a + b \neq c + d$

For case 2, it's fairly simple to tell how the equation was balanced because your $K_C$ will come in a different unit, but for case 1, because $K_C$ is dimensionless we depend on the convention that Chemical Reactions are balanced on the lowest whole number. So in the case of the example given, we would balance it as:

$$\ce{2H2 + O2 <=> H2O}$$

and $K_C$ would be defined on that basis.


The explanation on atomic size I always receive two type of explanation for the decreasement of atomic size for elements across the period.

Explanation according to proton number

This explanation is from my teacher.

The proton number of elements increases across the period.
The force of attraction between the proton in the nucleus and the electrons is stronger across the period as the proton will pull the electrons towards the nucleus.
So, the atomic size of the elements will decrease across the period

Explanation according to number of valance electron

This explanation is from my friends

The valance electron of elements increases across the period.
The force of attraction between the valance electrons and the proton in the nucleus is stronger across the period between the valance electrons is pulled towards to the nucleus
So, the atomic size of the elements will decrease across the period.

My question is which explanation is more suitable for the statement

  the atomic size of the elements decrease across the period.

Both of those explanations include part of the puzzle, but are not really complete on their own. To get a better understanding, you need to combine them, as well as understand something called Coulomb's law.

The basics of Coulomb's law are this:

The closer two oppositely charged things are, the stronger the attraction
The stronger the charge on either of the things, the stronger the attraction

With those points in mind, let's combine it all. I'll keep my explanation similar to the way you phrased yours:

Each step across a row on the periodic table adds both one nuclear proton and one valence electron. So, that is increasing both the total positive charge of the nucleus and the total negative charge of the valence shell, but not the distance between them (because electrons are added to the same shell).
The force of attraction between two charged objects depends on how strong each of them are (Coulomb's law). Since both the nucleus and the valence shell are stronger than in the previous step, and the distance between them has not increased (the electrons are added to the same shell) there is a stronger force of attraction between them, and the two are pulled closer together.
So, the atomic size of elements decreases across a period.


Synthesis of Benzaldehyde and Aniline Schiff Bases I have a problem while synthesize the schiff base. I synthesized using absolute ethanol as solvent and use reflux method for 4 hours generally. The problem occur on TLC where there is still two spot on the product which one of them are very clear (top one) (which I guess the ligand formed) and another one were very light, while on NMR spectra, whereby there is a singlet peak at approximately 10.0 ppm which I assumed that it is from the aldehyde itself which probably not completely reacted. I also tried several derivatives based on benzaldehyde and aniline but some of them have to be recrystallize which I havent found the suitable solvent system yet.

So with all due respect, I really need help especially those who have an experience on synthesizing schiff base and also share to me some ideas to recrystallize my ligands.

The reaction needs acid catalysis but ethanol is fine as solvent. There will always be some starting materials left, but since the product is a solid and the reagents are liquid, the best (and cheapest) is to purify it through recrystallization.

For this, dissolve the product in the smallest quantity of hot ethanol and then add water until precipitation occurs.


What is the nature of the oxide of boron? I think it should be acid, as I am aware that a compound by the name of Boric acid exists. But how can the oxide be acidic when it is electropositive? Also, other members of its group are well known metals. Why isn't the oxide amphoteric, when its group member Aluminium has an amphoteric oxide?

My teacher told us that the oxide of boron is basic in nature? So what exactly is its nature?

The most common oxide of boron is boron trioxide, $\ce{B2O3}$, which is the anhydride of boric acid:

$$\ce{H3BO3 -> HBO2 + H2O}$$
$$\ce{2HBO2 -> B2O3 + H2O}$$
$$\text{net} \ \ce{2H3BO3 -> B2O3 + 3H2O}$$

As the anhydride of an acid, boron trioxide is acidic. It reacts with water to regenerate the acid:

$$\ce{B2O3 + 3H2O -> 2H3BO3}$$

Boric acid's acidity does not derive from the dissociation of a proton (or abstraction of a proton by water), as does many acids, for example sulfuric acid:

$$\ce{H2SO4 <=> H+(aq) +HSO4- (aq)}$$
$$\ce{H2SO4 + H2P <=> H+(aq) + HSO4- (aq)}$$

Boric acid (written now as $\ce{B(OH)3}$, which is suggestive of its structure) instead reacts with water by forming a dative complex which is the active acid. A water molecule associates with the empty $p$ orbital on boron. This difference in behavior is enabled also by the electropositive character of boron. The acidic oxides of some metals behave similarly.

$$\ce{H2O + B(OH)3 <=>H2OB(OH)3 <=>H+(aq) + B(OH)4^- (aq)}$$


Why is the standard enthalpy of formation of black phosphorus not zero? Black phosphorus is the most stable allotrope of phosphorus. Then why is its standard enthalpy of formation not taken as zero?

Reproducibility is more important than stability. To obtain black phosphorus, you have to heat your sample under high pressure for quite a while, and even then it may still contain a significant number of random crystal defects, so its properties are not quite the same each time you make it. White phosphorus, on the other hand, is prepared by sublimation and hence considerably pure.

That's why white phosphorus, rather than black, is assumed standard and has zero enthalpy of formation.


How are these two molecules identical? I can see that these compounds are identical as we have basically just done a 180 degree rotation of the molecules.

But if we go by the concepts of chirality, the carbons 2 and 4 are chiral centers and since they have inverted configurations at both chiral centers, they should be enantiomers. Or is it just that I cannot assume the hydrogens not shown to be out of plane?

Sameer. The key to understanding this is that the two asymmetric carbons are indistinguishable because of the symmetry in the molecule. When you do your inversions on them, the molecule looks different, and should be an enantiomer, but after you do the 180 deg rotation, you have essentially switched them over (now the molecule looks the same as the first molecule, but only because you can’t differentiate the two asymmetric carbons).

Imagine that you switched one of the terminal methyl groups with an ethyl, then try the same trick. Because you have broken the symmetry of the molecule, the 180 deg rotation doesn’t get you back to the start. Hence, this meso isomerism relies on the mirror plane of symmetry (or a point of inversion) existing in addition to multiple asymmetric centers.


Half cells: difference between E(cell) = E(cathode}-E(anode) & E(cell) = E(oxidation) + E(reduction) When do I use each of the above? If I use one or the other they give me difference results.

For example, I'm trying to put the Volts into an equation to find standard free energy change.

$$\Delta  G^\circ= -nFE$$

If I use the cathode minus anode I get a different answer to the oxide + reduct.

Ah, I see what you did wrong.

The first equation is correct.

$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$

The second equation should be:

$$ E^\circ_\text{cell} = E^\circ_\text{reduction} - E^\circ_\text{oxidation} $$

Alternatively, you can write it as:

$$ E^\circ_\text{cell} = - E^\circ_\text{oxidation} + E^\circ_\text{reduction} $$

The cathode is actually the reduction step, and the anode is the oxidation step.

A useful mnemonic I learned from my chemistry professor is a red cat and an ox. Interestingly, cathodes are red and are positive. So I have found many uses for this along the way.


A basic solution of HCl So I had a test question the other day, and it essentially went as follows:

  A student measures the concentration of a $\ce{HCl}$ solution to be $10^{-9}$M by using a pH meter. Is the meter wrong?

This means that the $\ce{HCl}$ solution has a pH of 9, which seems unreasonable, yet that is not what I said. I said that the meter is not wrong. My reasoning was that you could have an $\ce{HCl}$ solution so incredibly dilute that the pH would in fact be 9. Since molarity is defined as moles per liter, if you hold the moles constant and increase the volume of solution, you will eventually reach a very tiny molarity.

In essence what I was wondering was: Is my reasoning correct?

Is this solution considered basic?

Can this be generalized such that any originally acidic solution is considered basic at a low enough concentration, or a high enough dilution?

Unfortunately, your reasoning is wrong, because you forgot to take into account the acidity of water. While in most cases we can ignore the acidity of water as the hydrogen ion concentration contribution by $\ce{HCl}$ dominates, at an extremely low concentration of $\ce{HCl},$ water becomes the main $\ce{H+}$ contributor.

It is well known that water has a $\pu{10^{-7} M}$ concentration of hydrogen ions at $\ce{25^\circ C}.$ Therefore, water's effect on the acidity of the solution is significant when looking at solutions of $\pu{10^{-6} M}$ or lower concentrations of strong acids. For example, if there is $\pu{10^{-9} M } \ce{HCl,}$ then the total $\ce{H+}$ concentration would still nearly be $\pu{10^{-7} M}.$ Therefore, no matter how much you dilute the acid, it can never turn into a basic solution.


Fluctuation in pH reading on pH meter I was conducting a titration of a saturated $\ce{Ca(OH)_2}$ solution against $\ce{HCl}$ using a pH meter (the kind with a glass electrode). At the beginning of the titration, the pH reading would drop a bit every time I added a drop of $\ce{HCl}$. However, near the equivalence point, when I added a drop of $\ce{HCl}$, the pH reading would drop rapidly, but then it would rise again at a steady pace for a few minutes. Why does the pH rise again? My teacher said that the pH meter's membrane is somehow becoming "saturated" with the solution, after being left in the solution for awhile. But what's actually going on in terms of the ions and the membrane?

Your teacher gave you a poor answer.

You are observing three different things.

(1) The solution isn't instantly homogeneous when HCl is added. It takes a couple of seconds of stirring to mix it thoroughly.

(2) The calcium hydroxide solution initially has considerable buffer capacity. As the solution gets less basic then a small amount of HCl can change the pH by a relatively large amount. So a drop of HCl in 50 ml of 0.1 molar calcium hydroxide won't change the pH much, but a drop of HCl in a  neutral solution at pH 7 would change the pH by quite a bit.

(3) The electrode and meter have a response time. So if you could move the pH electrode instantly from an acid solution to a basic solution it would take the electrode and meter themselves a few seconds to equilibrate. (The pH electrode has a hydrated silica layer that forms the boundary on the surface which takes time to equilibrate. The pH meter itself will have some RC time constant so that electrical noise is minimized.)


Two electron integrals between 1s STO I am trying to understand SCF cycle by trying to code up solved example from Quantum Chemistry by Levine (page 443, 5th edition). The problem is shown below:

One electron integrals are straight forward and I was able to get the correct answer, however I still can get correct values for two election integral, lets say (11|11). Given below is my attempt in octave

clear all;
N=2000;
zeta1 = 1.45;
zeta2 = 2.91;

r = linspace(0.000001,10,N)';
dr = r(2)-r(1);
chi = @(zetad,x) (2*zetad.^(3/2))*exp(-zetad*x).*x;
chichi = 0;
for i =1:N
    chichi = chichi + dr*(chi(zeta1,r(i))*chi(zeta1,r(i))*chi(zeta1,r')*(chi(zeta1,r)./((r(i)-r) + 0.000001)));
end
chichi*dr

However my values are way off in this case. Can anyone please shed a light on it? Value of (11|11) = 5/8 zeta1 = 0.9062.

Two electron integrals are defined in the book as:

@TAR86 was correct, I mistook the volume integral as simple 1d integral. Also integrating $1/|r_2 - r_1|$ would result in a singularity at $r_1 == r_2$ which would yield really wrong results.
As per @user1271772 suggestions, materials modelling stack exchange was quite useful (see question here). There as per Susi Lehota's suggestion I was able to get correct result by integrating Legendre Expansion.
My octave code for two electron integral (11|11) follows:
zeta1=1.45;
chi1111 = @(r1,r2) 16*zeta1^3*zeta1^3*exp(-2*zeta1.*(r1)).*exp(-2*zeta1.*(r2)).*r1.*r1.*r2.*r2./max(r1,r2);
dblquad(chi1111,0.000001,10,0.0000001,10,1E-3)
ans =  0.90623

Similarly other integrals can be encoded, and I get results exactly as Levine example.
Thank you everyone


Group theoretical condition for an integral to be zero In many aspects of chemistry it is of interest to find out whether a given integral $\int \psi \,\mathrm{d}\tau$ vanishes, i.e. becomes zero. Group theory provides us with a handy condition:

  If $\Gamma_\psi$ does not include the totally symmetric irreducible representation (TSIR) of the point group, then $\int \psi \, \mathrm{d}\tau$ must be equal to zero by symmetry.

The utility of this is that it can be easily expanded to different types of integrals, such as matrix elements $\langle i | A | j \rangle$, where the integrand transforms as $\Gamma_i \otimes \Gamma_A \otimes \Gamma_j$. If the direct product does not contain the TSIR, then the matrix element must be zero.

What is the theoretical basis of this statement, i.e. is there a proof of it?

In particular I'd like to see a proof that:

doesn't make assumptions about which symmetry operator(s) the integrand is not symmetric under; and
is able to deal with irreps with dimensionality larger than one.

But if such a proof doesn't exist, I'd be happy to compromise on one or both of these criteria.

I checked FA Cotton's book, but he skirts the issue by saying that "it is a generalisation of the case where $\int y(x)\,\mathrm{d}x = 0$ if $y$ is an odd function of $x$" (p 107, 3rd ed.). That does makes sense, and I have an idea of how it could be mathematically formalised, but it doesn't satisfy either of the two criteria above.

By Maschke's Theorem, every direct product of representations is decomposable into a direct sum of representations, that is, the function you are integrating can be rewritten as a sum of functions with symmetry properties equal to specific irreducible representations.

As you note, integrating an odd function (which has a certain symmetry, namely antisymmetry with respect to $\sigma_y$, the mirror "plane" of the y-axis) over all space is zero, while integrating an even function (which is symmetric with respect to $\sigma_y$) is generally not. Since integration is linear (that is, the integral of a sum is the sum of an integral), we can consider the integral of each irrep separately.

If the function being integrated is antisymmetric with respect to some symmetry, it will be zero when integrated over all space. The only way for the integral not to be zero is if it is symmetric with respect to all possible symmetries, i.e. it is the totally symmetric irrep.

Expressed in equation form:
$$\int{\prod_{i}\Gamma_i\,\mathrm d\tau}=\int{\sum_j\Gamma_j\,\mathrm d\tau}=\sum_j\int{\Gamma_j\,\mathrm d\tau}$$
and
$$\int{\Gamma}\,\mathrm d\tau=\left\{
  \begin{array}{@{}ll@{}}
    0, & \text{if}\ \Gamma \text{ contains any odd (antisymmetry)} \\
    \text{not necessarily } 0, & \text{otherwise}
  \end{array}\right.$$

Then, it is clear that the only time the integral is nonzero is when, amongst the irreducible representations of the direct product (regardless of its dimensionality), the totally symmetric one is present, all other ones integrate out to zero.

I should clarify that when I write the integral of $\Gamma$, I really mean the integral of some awful, complicated looking function that has the symmetry properties of $\Gamma$, which is why I used $\prod$ and $\sum$ instead of $\otimes$ and $\oplus$.

If your question was instead how do you know that the functions with a given antisymmetry really do integrate to zero, let's consider what it means to have a certain symmetry. It means that there is a particular line or plane or point about which the value of the function is redundant. That is, if you specify it on one side, then the value is exactly determined on the other. We are integrating over all space, which means we can choose bounds of integration that utilize this symmetry. This then separates your integral into a non-symmetric piece and a piece using the symmetry. You first integrate over the non-symmetric piece and get something, and then you integrate that over the symmetry-including part. If that symmetry is antisymmetry, as will be the case with at least one symmetry of any irrep other than the totally symmetric one, this will cancel out the non-symmetric part with an equal and opposite non-symmetric part and the whole integral will be zero. Regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric. Hence, only integrals of functions that are totally symmetric are non-zero.

As a basic example, consider your $\int{x\,\mathrm dx}$ integral that you started with. Since this function has a symmetry (namely antisymmetry with respect to $\sigma_y$), I can make the following choice of bounds:
$$\int_{-\infty}^\infty{x \,\mathrm dx} = \int_{-\infty}^0{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}  =-\int_{0}^\infty{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}=0$$

I can choose to integrate only over the nonsymmetric part and then use the symmetry for the rest of space. The presence of symmetry guarantees that I can do that. Remember, the irrep label is just telling you the symmetry properties of some complicated function. If $x$ in the above equation were some function $f$ that I knew nothing about except that is has this symmetry, I could still make this choice of bounds and use this property to get a 0 integral.

EDIT: To address orthocresol's question below:

How does this work for the $E$ set in $C_{3v}$? Let us first think of a function with $E$ symmetry. The $p_x+p_y$ orbitals are such a candidate. Let's consider a top-down view of a $C_{3v}$ molecule. Now, I will consider one of the mirror planes to lie in the $xz$-plane (see Figure 1). I can do this because I can rotate the $p_x+p_y$ ensemble anywhere in the $xy$-plane. That is, there is some linear combination of $p_x+p_y$ such that there is a new set $p_{x'}+p_{y'}$ where the mirror plane lies on the $x'y'$-plane.

Now we set about integrating the upper half plane. Clearly, the half of the negative lobe of the $p_x$ orbital will cancel the half of the positive lobe, leaving only the integral of the positive lobe of $p_y$. Now we consider the lower half plane, the two halves of the $p_x$ lobes cancel again and we have only the integral of the negative lobe of $p_y$. Since the integral of the upper-half plane is the negative of the lower-half plane, the integral over all space is 0.

Now, why was it important to consider $p_x+p_y$ together? This is what allowed us to orient the orbitals with respect to the mirror plane. Consider a $p_x$ orbital alone rotated away from the axis given. It will have no defined symmetries with respect to the plane (reflecting across it will not map the orbital to itself or its negative) and there is no way to rotate it without the $p_y$. This is why they need to transform together in $C_{3v}$. Indeed, the reason why $E$ functions have 0 character under this symmetry is because the $p_x$ got sent to $p_x$ and the $p_y$ to $-p_y$ and so the whole ensemble was sent to $1-1=0$.

Now, hold on you say, can't I use similar arguments to imply that if I integrated a $p_z$ orbital (which is $A_1$ in $C_{3v}$) in all space, I will also get zero. The answer is yes. The $p_z$ orbital is not zero by symmetry in $C_{3v}$. I alluded to this above, the integral of a totally symmetric function in a given point group can still be zero. However, by describing the point group as $C_{3v}$, we are indicating that there is something special about the $z$-axis that breaks the symmetry and therefore precludes coming to this conclusion based on symmetry alone. Consider the deformation of planar $\ce{NH_3}$ to pyramidal $\ce{NH_3}$. In the former ($D_{3h}$), nothing distinguishes the positive $z$ from negative $z$ axis and a $p_z$ orbital is not totally symmetric ($A_2''$) and so we could conclude that it is zero by symmetry. As we pyramidalize to $C_{3v}$ though, the negative $z$ axis becomes different (it's got hydrogens there while the postive $z$ does not) and we can no longer conclude by symmetry that the integral of the $p_z$ orbital will be 0 (indeed, if it mixes even slightly with those hydrogens, it won't be). If it didn't mix though, it's integral would still be 0 (it's still just a $p_z$ orbital), we just couldn't tell that assuming $C_{3v}$ symmetry.


By which mechanism would azide react with alcohols or alkyl halides? $\ce{NaN3}$ produces $\ce{N3-}$ ions. So the reaction with an alkyl halide will either be $\mathrm{S_N1}$ or $\mathrm{S_N2}$.  After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\ce{S_N2}$ mechanism is followed.
But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Is this correct?

After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\mathrm{S_N2}$ mechanism is followed.

Unfortunately, it is not as simple as that. Whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ is followed, depends on both partners but is more a function of the electrophile than one of the nucleophile. Typically, primary halides will undergo $\mathrm{S_N2}$ substitutions while tertiary ones will undergo $\mathrm{S_N1}$. Secondary ones can do both. However, note that secondary halides that are not 2-halopropanes will undergo Wagner-Meerwein rearrangements if a carbocation is formed.

In an $\mathrm{S_N1}$ reaction, the rate-determining step is the formation of the carbocation, hence weaker nucleophiles can react equally rapidly as stronger ones. In the $\mathrm{S_N2}$ reaction, the nucleophilicity of the nucleophile directly determines the rate of reaction, i.e. both partners are important. Hence these tend to prefer better nucleophiles.

  But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Again, primary alcohols will never react by an $\mathrm{S_N1}$ mechanism and secondary ones are very unlikely to do so. The $\mathrm{p}K_\mathrm{a}$ value of $\ce{HN3}$ is $4.72$ — which means that alcohols will practically not be protonated (you need strong acids for that, e.g. $\ce{HBr}$).

I would assume that $\ce{HN3}$ will not react with alcohols at all since it is too weak to protonate them. There will also be no elimination side reactions neither by $\mathrm{E1}$ nor by $\mathrm{E2}$ — a hydroxide ion will never be a leaving group in itself.


Predict the product of reaction of quinone and hydrochloric acid This is the question:

And this is my solution:

In the last step, I have tautomerised, and in the second last step, there is no double bond between carbon attached with Cl and OH (just some cutout work)

The answer given is B, but my answer comes out to be C.

The product of the addition of one chloride ion (C) is easily oxidized by atmospheric oxygen (radical mechanism). The obtained 2-chloro-1,4-benzoquinone is susceptible to conjugate addition of another chloride ion, this happens chemoselectively at the more electron deficient double bond which has already a chlorine attached to one carbon atom.


Are there limits to waters energy absorbtion capacity I have a system where I heat water in a flask.
The water vapor generated is condensed and returns to the original container (not a pressurized system, but condensation system is efficient enough that there is no loss of vapor).
My goal is to have maximum amount of vapor. (The real goal is the condensate droplets that filter through plant material to extract water soluble components - Kumagawa extraction)
Question:
Is there a limit or plateau to the amount of energy the water can absorb/turn into vapor?
or
I have a 10l flask and a 1500W heating mantle. If I run the heating mantle at full power is there a possibility that the energy generated couldn't be absorbed by the water in the flask and used in turning it into vapor and thus would heat the electrical control circuits and outer casing of the heating mantle instead.

The heat transfer is limited by the flask material heat transfer rate, which is proportional to the thermal gradient within the flask. The external flask temperature will follow.
On the water side, liquid can jump out of flask in form of the thick foam, powered by the vapour. This is the major limiting factor.
On external side, at high enough power, the eventual plastic holders/features may start to melt/decompose. At higher power, the flask may start shining red. Or start to melt if metallic.


Why are electrons added in the ionic counting method? Why do we add one electron to the ligands that have an odd electron number?

For example the cyclopentadienyl ligand has five electrons that can be donated to the metal but when we count the electrons by the ionic method we add one, so it become total of 6 electrons donated to metal.

In the ionic method of counting, you are not counting neutral compounds. This is reasonable on a molecular basis: the ligand is, in fact, not cyclopentadiene or the cyclopentadienyl radical (with four or five electrons to be donated) but the cyclopentadienyl anion $\ce{C5H5-}$. This species contains six electrons in its π system, meaning that six are donated.


How to determine the most reactive hydroxyl group of α-ᴅ-glucopyranose? In the following reaction, only one of the hydroxyl groups in D reacts under the given conditions:

I was wondering which OH group reacts and why? Also I have been trying to determine the structure of E based on which hydroxyl group reacts.
My thoughts are that OH in the side group can react via SN2 with the reactants, but the OH right adjacent to the ether O can react leaving a carbocation that can be stabilised by lone pair from O. This would mean SN1 and a completely different product.
Can someone please explain the correct mechanism?

$\ce{EtOH}$ is a rather bad nucleophile and we have the assistance of a lone pair of the neighboring oxygen atom (the positive charge can hence be delocalized over the carbon and oxygen) which greatly speeds up $\ce{S_{N}1}$ in comparison to $\ce{S_{N}2}$. So what will probably happen is protonation by $\ce{HCl}$ of the alcohol next to the ether oxygen followed by nucleophilic ($\ce{S_{N}1}$) attack of $\ce{EtOH}$ on the formed (delocalized) secondary carbocation. In comparison to the possible reactions at the other secondary centers (having the formulation of the question in mind) this primary reaction will be fast enough to justify the ignoring of them.


Is water acidic enough for an redox reaction involving H+? If you stored $\ce{Pb(NO3)2(aq)}$ in a $\ce{Cu(s)}$ container, would any reaction occur? Lead(II) ions will not react spontaneously with the copper solid as the $E^0_\mathrm{net}$ of the reaction would be negative. However, the $E^0_\mathrm{net}$ of the reaction between nitrate ions and copper solid is positive. The reduction reaction for nitrate ions is
$$\ce{2 NO3- (aq) + 4 H+(aq) + 2 e- -> N2O4(g) + 2 H2O(l)}\qquad{E^0=\mathrm{+0.80\ V}}$$

As is obvious, the reaction requires not only nitrate but hydrogen ions. However, water (the solvent) does have hydrogen/hydronium ions. Would this be enough to furnish the reaction? If not, why?

Yes the reaction would be spontaneous at standard conditions.  However standard conditions for electrochemistry are at pH 0, i.e. an $[\ce{H+}]$ concentration of 1 molar.

Solutions of lead nitrate by themselves are not going to give you pH 0.  Apparently very very concentrated lead nitrate solutions will reach pH 3 or 4, due to the weakly acidic nature of hydrated lead ions.

The electrochemical potential of the reaction you are interested in at pH 3 can be calculated from the standard (i.e. pH 0) potential via the Nernst equation.

$$E_\mathrm{cathode} = E_0\;– \frac{RT}{nF}\ln{Q}$$

Here, $Q$ is the "half"-reaction quotient, given by
$$ Q = \frac{\mathrm{p}_{\ce{N2O4}}}{[\ce{NO3-}]^2[\ce{H+}]^4}$$ Now assume that nitrate concentration and the partial pressure of dinitrogen tetroxide are in their standard states.

If you go through the math (see examples here or here), then you will see that there is a 118 millivolt-per-decade decrease in the potential, i.e. every increase in the pH by 1 unit decreases the potential by 118 mV.  So already at pH 3 the reaction will no longer be favorable.

Copper is unstable to concentrated nitric acid, but the acid needs to be concentrated.  Dilute acids cannot oxidize copper so easily.


Can halogens exist in their elemental state in nature? Question 7 from the 2015 Australian Chemistry Olympiad was:

Which list below gives non-metals that can be found in their elemental form in nature?
(a) neon, phosphorus, fluorine
(b) helium, hydrogen, iodine
(c) helium, carbon, fluorine
(d) nitrogen, chlorine, sulfur
(e) oxygen, chlorine, phosphorus

I was somewhat confused when all the options included reactive elements such as sulfur, phosphorus and the halogens. I know neon is a noble gas and is present in minute quantities in the atmosphere, and nitrogen and oxygen gases are present as diatomics.
However, where are hydrogen, phosphorus, sulfur, fluorine, chlorine and iodine found in their elemental forms in nature? (Answer is B)

I agree with Matt Black's comment on the question. The question is rather tricky and the answer can only be determined by the process of elimination.

Have to assume we're talking about STP.
"In nature" is ambiguous, so assume the question means "on earth."
It is also somewhat ambiguous what "elemental form" means. Assume that it means that element is only combined with itself since only the noble gases exist as monoatomic species.

The noble gases (helium and neon) and exist as monoatomic gasses at STP and can be found in the atmosphere.

Oxygen and nitrogen are diatomic gases and both exist in earth's atmosphere.

Sulfur is found in volcanic vents.

Carbon is found both as graphite and diamonds on the earth.

So, the list becomes:

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

Oxygen is ubiquitous on earth and phosphorus is so reactive with oxygen that elemental phosphorous just won't be found on earth. However, elemental phosphorous might be found on the moon for example.

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

So now we are left with the halogens -- fluorine, chlorine, and iodine. Iodine is the least reactive, so "elemental iodine" is the most probable to be found in nature.

Thus the answer must be #2. However this is really an "educated guess," not a "fact" since the question is so ambiguous.


1H (proton) NMR spectra for alkanes I do not fully understand the idea of coupling and why chemically equivalent hydrogens do not couple.

I wanted to work with the alkanes, so let's take propane first:

There would be a triplet signal due to the $\ce{CH3}$ groups in blue and green, due to the $n+1$ rule for neighbouring protons, so the neigbouring carbon for the two $\ce{CH3}$ groups would be the $\ce{CH2}$ group circle in red.

Then the $\ce{CH2}$ group in red would have a septet signal due to the two surrounding $\ce{CH3}$ groups is what I would predict.

Moving on to butane, this is where I get more confused:

For the $\ce{CH3}$ group in blue, there would be a triplet signal due to the neighbouring $\ce{CH2}$ in red, and the same can be said for the $\ce{CH3}$ in black.

Then for the $\ce{CH2}$ groups, I am not exactly sure what would happen but here is what I think:

The $\ce{CH2}$ group in red will have a quartet signal on the NMR spectrum because equivalent hydrogens do not couple and so the $\ce{CH2}$ will not couple with the green $\ce{CH2}$ group. This means that only the $\ce{CH3}$ group will couple. Using the $n+1$ rule, a quartet will be present and the same can be said for the $\ce{CH2}$ group in green. However, I think that the red $\ce{CH2}$ will have a triplet of quartets or a multiplet if that is correct, with the same being said for the green $\ce{CH2}$ group.

I would like to extend this further to pentane, which I am also confused about. Can this be clarified with the two examples above so that I can attempt to do pentane myself?

Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact magnetically non-equivalent. It serves as an example of how all too often NMR is poorly taught at the introductory level.

First, let's clarify a minor point of confusion that is widely taught at introductory levels - that equivalent protons do not couple to each other. They do; the reason that the coupling is not seen is that the transitions are degenerate and so sit directly on top of each other (ie not observable). Remember that the observed lines in a spectrum are the energy transitions between stable spin states - if the energy difference between two states is equivalent, then the lines in the spectrum are degenerate.

So, a better generalisation for interactions giving rise to splitting in a spectrum is that:

Coupling between magnetically equivalent nuclei cannot be detected
and does not affect the appearance of the spectrum
Coupling between
chemically equivalent, but magnetically non-equivalent nuclei will
usually affect the appearance of the NMR spectrum; commonly seen in
second order effects like that of AA'BB' systems.

Second, then, to see what a 1H spectrum of butane really looks like (well, a more realistic simulation at least):

Below is a more realistic simulated spectrum of butane (at 400MHz), which takes into account the fact that the two methylene groups couple to each other, with an average coupling of about 6.6Hz. Complicated isn't it? Have a read of Glenn Facey's page on his NMR blog, (an excellent NMR resource) where he discusses this very problem.

So why is the case? Well, we need to look at what gives rise to magnetic non-equivalence in molecules. There is an excellent on-line resource to explain about symmetry and non-equivalence at Hans Reich's NMR page. But for butane, let us consider the Newman projections:

At no time is the coupling between H2 and H3 the same as H2' and H3. Either it is psuedo-trans or pseudo-cis, but never the same. So, H2 and H2' are non-equivalent, and this will result in non-degenerate transitions, giving rise to complicated splitting patterns in the NMR spectrum. This is the same for other straight chain alkanes, with decreasing complexity as the chain length grows.

And just to show what a real spectrum of butane looks like: reproduced from Tynkkynen et al. (Magn. Reson. Chem. 2012, 50, 598–607), showing the real and calculated (at 500MHz):


Dead trees or charcoal is generally used as a reductant in the production of pure silicon for use in industry what other reductants could be used? What other known reductants for SiO₂ could potentially be used for producing silicon from silica in an arc furnace?

Read more: http://www.madehow.com/Volume-6/Silicon.html#ixzz6uKynirns

On a small scale, hydrogen gas is certainly an option. Here is a paper: Reduction Kinetics of Metal Oxides by Hydrogen. High temperatures induce the formation of monoatomic hydrogen as the active agent.
However, on a practical note, explosive mixtures of hydrogen with air exist over a wide range (Wikipedia has a good discussion on this topic), so this path is clearly not without danger.
[EDIT] To address a question on trials, not that am aware of, however, here is an interesting experiment one may conduct at a seemingly much lower temperature that is likely safe. For reducing, it is known that in situ creation of the hydrogen atom radical (cited as the major reducing species by Buxton, see Eq 3 on page 2 here). Suggestively, the chemisorption of the hydrogen radical on the surface of metals like Zn, Al and Mg (albeit, effective for a limited lifespan) may be an interesting trial path.
So, as an experiment, proceed to imbue the surface of Mg or Al metal with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH). Place the SiO2 on the imbued metal surface and heat.
As to the reaction mechanics, the •H radical functional behaves based on the equation:
$\ce{ e- + H+ <=> •H }$
where this seemingly reversible formation reaction apparently acts as (e-,H+) pair on ions. For an example, citing a work in Hydrometallurgy 2008: Proceedings of the Sixth International Symposium on p. 818, a suggested commercial reductive leaching equation, to quote:
$\ce{PbS + 2 •H = Pb + H2S (5) }$
Now, could a related reaction path exist with SiO2 at a much lower temperature? To answer, this suggested trial could be quite interesting.


What chemical properties make some grasses usable for producing textile fabrics? Textile fabrics are made out of cellulose fibres, that can be derived from cotton or bamboo. As far as I know all grasses contain cellulose, so why are only a few grasses used? What makes them so special? Why could you not use for example gamba grass/andropogon gayanus for that, which contains 30%  cellulose and 21% hemicellulose (http://www.lrrd.org/lrrd20/2/oded20027.htm)? Any insight on that would be very helpful

A large part is the agriculture behind the fabric and what the industry and customer demands. Cotton and bamboo are easy to farm because we have a lot of experience and techniques for raising and harvesting and processing the plants and fibers ( especially cotton, which is available as GMO round-up ready to simplify weed control, and centuries of breeding for better quality and quantity of fiber ). Additionally, the textile industry has machines that are calibrated for cotton, other plant fibers may not work as well and re-designing the machines would be expensive. Customer demand is important, and customers are used to cotton clothing.

As far as chemical reasons go I don't have a lot of specifics. However, different plants have different compositions of fibers. Though all plant fibers are mostly cellulose/hemicellulose, lignin is also important, and would make for woodier, stiffer, tougher fibers ( though bamboo gets pretty woody, I don't know much about bamboo textiles ). Other plant fibers may contain pigments that would have to be bleached out before it could dyed with other colors. Length of fibers is also important. Shorter fibers would be harder to work into threads and textiles.

And keep in mind that the textile industry is not limited to cotton and bamboo. For centuries, cotton was a luxury material because picking the seeds out of the cotton boll was very hard work, cotton was far more expensive than silk. The widespread adaption of the cotton gin ( and slavery ) made cotton cheap and easily available. Prior to this, fabric would have been made from flax to make linin, or from hemp. Flax and hemp are still commonly grown for fabric. ( and we haven't even mentioned animal sources of fiber such as silk or wool )


Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves? I'm not a chemist but I remember that the standard HF/nitric silicon etch used in micro-fabrication supposedly worked in two steps; the HF would dissolve the thin SiO2 oxide layer on the silicon surface, exposing silicon atoms, and the nitric acid would then re-oxidize the bare silicon, allowing the HF to then dissolve it.
Or at least that's how it was explained to me.
Despite being challenged here, Wikipedia explains that the aluminum bits in sodium hydroxide-based drain cleaner are included to produce heat via exothermic liberation of hydrogen gas so that the heat can loosen some of the clogging materials. The article provides the following two steps.

Breakdown of Aluminum Oxide: $\ce{ \ \ \ Al2O3 + 2NaOH + 3H2O → 2Na [Al(OH)4]}$

Oxidation of Aluminum metal: $\ce{ \ \ \ 2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2}$

Is what is happening similar to my understanding of the HF/nitric etch of silicon; it is the oxidized aluminum that is removed by the solution, and only because it happens to also oxidize the freshly exposed aluminum metal that the oxide is removed?

Snapshot of some drain cleaner with aluminum bits

If you want to understand the reaction of $\ce{NaOH}$ on Aluminium, you may describe it this way. First, the aluminium reacts spontaneously with water, like sodium or calcium, producing gaseous Hydrogen $\ce{H2}$ and an metallic oxide or hydroxide. With sodium or calcium, the hydroxide produced by this reaction quits the metal surface, and get dissolved, leaving a fresh surface of metal ready to react with more water. The reaction with sodium is:

$$\ce{2 Na + 2 H2O ->   2 NaOH  +  H2}$$

With aluminium, it is not the case. The reaction is:

$$\ce{4Al + 6 H2O -> 2 Al2O3 + 6 H2}$$

But the aluminium oxide is not soluble into water and it makes a waterproof layer on the surface of the metal, which prevents further contact between the metal and water. So the reaction stops, due to this protecting effect of $\ce{Al2O3}$.

There are two ways of allowing the reaction to continue:

Dissolve $\ce{Al2O3}$ as soon as it is produced by adding $\ce{NaOH}$, as @Uhoh tells it in its first equation.
Prevent the $\ce{Al2O3}$ oxide layer from adhering on the metal surface by adding some mercury salt in the water. In this case the mercury ions $\ce{Hg^{2+}}$ may go through the oxide layer, reacts with the metal and produces a mercury deposit which forms an alloy with Aluminium. The reaction is:

$$\ce{2Al + 3 Hg^{2+} -> 2 Al^{3+} + 3Hg}$$

The $\ce{Al2O3}$ layer cannot adhere on the surface of this alloy Al-Hg. So the
 reaction with water continues as long as there is some mercury atoms left on the
 metallic surface.

So, to go back to @Uhoh's question, it is not $\ce{NaOH}$ that acts on Aluminium. It is water. The action of $\ce{NaOH}$ is to remove the oxide layer.


Density of wooden materials Would all wooden materials have the same density? Why or why not? (Examples may help). I have the following solution to the question: No, not all wooden materials would have the same density. Due to the makeup and properties of the components found inside certain wooden materials.

You might find this table of Wood Densities helpful. For example, note the density of balsa and that of Lignum Vitae. In fact, Lignum Vitae is so dense that it has been used for bearings on ships and even nuclear sub!


Does the reaction become more or less favorable as temperature increases? Suppose a reaction has $\Delta S > 0$ and $\Delta H < 0$.

The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.

However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).

These two claims seem to be inconsistent, but I can't identify the error in my reasoning.

You should use the van't Hoff equation together with the standard enthalpy of reaction $\Delta H^\circ$ for the process to predict the response to a temperature change. This should provide you with a prediction consistent with Le Châtelier's principle that an endothermic reaction is pushed in the direction of products as $T$ increases.

The general equation for the Gibbs' free energy $$\Delta G = \Delta H - T \Delta S$$ should not be relied upon to determine the response of the reaction equilibrium constant to $T$. The expression is valid at constant $T$, but this is not the problem with using this equation, rather it is its generality: it is not limited to describing reactions at equilibrium. Note that $\Delta G = 0$ for a reaction at equilibrium.

On the other hand, you could consider using $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$to guide you, and the result should be consistent with analysis based on the van't Hoff equation. It is simpler however to work with the van't Hoff equation directly.  In fact, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are independent of $T$ you can use fairly straightforward mathematical manipulations to arrive at the usual integrated form of the van't Hoff equation.


Rearrangement in dehydration of alcohol I came across the following problem in my book:

Now my confusion is how will cyclohexene be obtained? I read that primary alcohols dehydrate by E2 mechanism because of formation of a primary carbocation. Further I read in Solomons organic chemistry that rearrangement in Primary alcohols occur due to the reversibility of the reaction:

So I tried to figure out the mechanism:

So I am unable to find how cyclohexene will be obtained. For ring expansion to take place there must be positive charge on the alpha carbon but in E2 mechanism there is no possibility of carbocation formation

Concentrated sulphuric acid is a highly acidic medium and thus it very exclusively prefers the $E_1$ mechanism over $E_2$ mechanism and thus a carbocation is surely formed in this conditions.
The example you provide proceeds through the following mechanism:

Now the mechanism you provide, other than following the wrong pathway(very minor pathway) also has a major mistake. During the protonation of 1-methylenecyclopentane the tertiary carbocation is not more stable than the primary carbocation. The tertiary carbocation is $\mathrm{sp^2}$ hybridized and thus prefers a bond angle of 120° which cannot be provided by the cyclopentane ring as it is not flexible enough.
It prefers the very less angle strain in case of $\mathrm{sp^3}$ hybridized orbitlas but with less positive inductive effect over higher angle strain but more positive inductive effect. That's the reason the below equilibrium is highly backward favoured.


How is the oxidation number related to the group number? If an element $\ce{X}$ forms the highest oxide of the formula $\ce{XO3}$, then it belongs to group:
  A) 14
  B) 15
  C) 16
  D) 17

How is an oxidation number of an element related to its group?

I tried comparing $\ce{X}$ with $\ce{N}$-, $\ce{S}$-, $\ce{As}$-compounds, but all 3 come in the form $\ce{XO3}$ ($\ce{NO3}$, $\ce{SO3}$, $\ce{AsO3}$), so I can't guess the group number.

For main group elements (groups 1, 2, 13-18), the highest possible oxidation state $O_\text{max}$ is given by the group number $G$ (for groups 1 and 2) or by the group number - 10 (for the rest):
$$ O_\text{max} = \begin{cases} G &\text{if } G \in \{1,2\} \\
G - 10 &\text{if } 13 \leq G \leq 18 \end{cases} $$

This means that for your compound $\ce{\stackrel{VI}{X}O3}$, so with the oxidation state VI, the corresponding group would be group 16.


Calculate the mass of Mg needed to produce 5 kg of vanadium I need to calculate the mass of magnesium needed to produce $\pu{5 kg}$ of vanadium from the following reaction:

$$\ce{3Mg + 2VCl3 -> 2V + 3MgCl2}$$

given relative atomic masses for $\ce{Mg}$ and $\ce{V}$ as $24$ and $51$, respectively.

I multiplied $24 × 3$ and $51 × 2$ to get

$72 : 102 $

I changed $\pu{5 kg}$ into grams, which is $5000$.

I cross-multiplied to find the answer and then divided the answer by $1000$ to turn it back to kilograms because the question stated it. My answer is $\pu{3.5 kg}$. Is this correct?

You answer is correct.

It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.

First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:

$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$

Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:

$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$

Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:

$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$