Balancing a reaction with unknown products I'm given this equation in word form:

$$\ce{sodium (s) + water (aq) -> unknown~products (s/g/aq)}$$

and I have to convert this to a chemical equation. I understand how to covert this normally, that equation can hence be represented by:

$$\ce{Na + H2O -> unknown~products\:(s/g/aq)}$$

This is indeed a single displacement reaction so it should become (according to what I know so far) $\ce{Na2O + H2}$ but after doing some research, I got that it should be:

$$\ce{Na + H2O -> NaOH + \frac{1}{2}H2}$$ or $$\ce{2Na + 2H2O -> 2NaOH + H2}$$

I'm lost now, where did $\ce{NaOH}$ come from? Especially the H since there is no reason for hydrogen to be present.

Please explain why and how to balance this reaction or convert this and what factors to look at when dealing with these types of equations.

Water is normally a mix of hydroxide and hydrogen, at very dilute levels.  For example, neutral water (pH of 7) is 10-7 of OH- and H3O+ ions.  Add sodium and it combines with the OH-.  The loss of electrons by sodium are available to the hydrogen ions, which combine to make hydrogen gas.  Sodium in metal form is so reactive, it will sustain this reaction even at the low hydroxide levels in pure water, and even in slightly acidic seawater.


Naming Ionic Compounds with Transition Metals Look at $\ce{FeO2}$.  We know that $\ce{O2}$
is $-2$ then we must have $-4$ for $\ce{Fe}$

So $\ce{Fe}$ must have $+4$ charge so then is it $\ce{Fe(IV)}$ ?

As stated in the Wikipedia article for iron:

Like other group 8 elements, iron exists in a wide range of oxidation states, −2 to +6, although +2 and +3 are the most common.

Thus, assigning $\ce{FeO2}$ as an iron(IV) compound is not something that is impossible, but it is unlikely. There are only a few compounds with that are iron(IV).
A more likely scenario is that $\ce{FeO2}$ may contain other anions of oxygen:

Oxide $\ce{O^2-}$
Superoxide $\ce{O2^-}$
Peroxide $\ce{O2^2-}$
Ozonide $\ce{O3^-}$

While oxide is the most common anion, there is another on the list above that is reasonably common. Which is it? If you have that anion, what is your oxidation number for iron?


Which of the following species has an enthalpy change of formation = 0 kJ/mol? Below was a question in the 2012 Australian Chemistry Olympiads:

  The enthalpy change of formation ($\Delta_\mathrm{f}H^\circ$) for a species at
  $\pu{298 K}$ is defined as the enthalpy change that accompanies the
  formation of one of the following of one mole of a substance from its
  constituent elements in their standard states. Which of the following
  species has $\Delta_\mathrm{f}H^\circ = \pu{0 kJ/mol}$?

  a. $\ce{H2O(l)}$  b. $\ce{Na(g)}$  c. $\ce{Na(s)}$  d.
  $\ce{CO2(g)}$  e. $\ce{O3(g)}$

Original

Solutions stated the answer is b, but isn't the standard state of $\ce{Na}$ solid?

The correct answer is indeed c. $\ce{Na(s)}$

Since for the heat of formation to be zero, the molecule should be the must abundant chemical formula of the element.

$\ce{Na}$ is a metal and is predominantly found in solid state (I dont think it exists as gas anywhere in the atmosphere of earth).

$\Delta_\mathrm{f}H^\circ$ is a quantity which is taken relative to the heat of formation of the most abundant naturally occuring form of an element.


Obtain chemical insight from neural network results I have recently read many papers where neural networks (NNs) are trained to predict chemical properties (starting from the structure of small chemical compounds) for compounds rather close to the ones with which they were trained. In some cases the predictions are rather good. But I see that apart from the predictive capability, people use NNs as "black boxes" in the sense that they do not tell you or give you chemical insight about why that compound has that value for that predicted property. At least that is my impression now. So my question here is, do you know cases where useful chemical information can be extracted from NN predictions?

Since you are interested in Artificial Neural Networks that also explain their decisions/predictions you may be interested in transparent neural networks, although they have not been used to my knowledge on chemical information.

Other interesting approaches include inductive logic programming which have among other things been applied to predicting binding of proteins to hexose


Which of these carbocations are more stable? The line of text that follows this figure in my work-book reads:

  Cation (A) is more stable than cation (B), since the former is resonance stabilized but the latter isn't (due to steric effects).

Don't both carbocations show resonance stabilization? More importantly, isn't cation (B) more stable than (A)?

Based on what I've learnt, the following should occur:

1) Cation (A)

The sp2 carbon (in the isopropyl group) acts as a -M group and causes the shift of a pair of π electrons from the benzene ring towards the $\ce{C-C^{+}}$ bond, thereby creating a double bond in its place. So the resulting structure should look like,

The mesomeric effect would also ensure that the positive charge "travels" across the ring, and decreases the electron density at ortho- and para- positions.
The delocalization of the positive charge across the benzene ring should confer some degree of stability to the carbocation.

2) Cation (B)

Once again, a pair of π electrons from the benzene ring would shift towards the $\mathrm{C-C^{+}}$ bond, leaving behind this:

And again, the mesomeric effect would delocalize the positive charge over the benzene ring and create electron deficient centers at the ortho- and para- positions.

However, the presence of two isopropyl groups at ortho- positions (with respect to the orignal charge-carrying isopropyl group) will result in the "supression" of the positive charge via inductive and esomeric effects, hence, conferring additional stability to this molecule.

Hence, cation (B) should be more stable than (A).

Yet, my conclusion is different from that provided by the book. Well, I guess that the presence of three adjacent isopropyl groups (in cation-B) does seem sterically unfavorable, but how would that prevent resonance effects in the molecule (as the book seems to imply)?

So, is my conclusion correct? Have I arrived at it correctly? If the book's right, then why is it right?

As we know (to some extent), resonance in a molecule occurs due to sideways overlapping of p-π orbitals.

Now in the molecule B, there is high steric hinderance near the carbocation due to which the plane of the "carbon with carbocation" changes, so that it can feel minimum repulsion from the other groups. This forces the carbocation to remain in the mutually perpendicular plane, which inhibits the resonance.

Thus, this makes A (the less hindered carbocation) more stable than B.


Is there a difference between imidazolidinone and imidazolidone? I've had my nose in the early literature on GFP (Green Fluorescent Protein), which contains a 4-(p-hydroxybenzylidene)imidazolidin-5-one fluorophore, and in one paper the author refers to the "imidazolidone ring".

I assumed that this change of spelling was equivalent, but I can't find anything that looks reliable on Google (there are also pages referring to imidazolinone, which looks like it refers to an aromatic ring with Nitrogens at positions 2 and 4...)

I don't think I'd know quite where to look on the IUPAC website for the info so trying here...

Can anyone help me understand the variations?

According to the NIST ChemBook, 2-imidazolidinone and 2-imidazolidone are synonymous. The official IUPAC name is imidazolidin-2-one. 1,3-ethyleneurea is also another name for this same molecule:

But, as you note, tiny changes in spelling can sometimes be much more tricky than this. To take an example from this family: imidazolidinone and imidazolinone (note that the later ends in –inone, not –idone) are ketones formed from imidazolidine and imidazoline, respectively. Imidazolidine and imidazoline are partly saturated and fully saturated imidazole derivatives:

Imidazolidine is a proper IUPAC name, but the official name of imidazoline is 4,5-dihydro-1H-imidazole.


Will KOH work in place of NaOH as the electolyte in the electrolysis of water? I am making a hydrogen and oxygen generator by electrolysis water. I found out that using salt as an electrolyte creates a compound of chlorine and oxygen with my electrodes clouding up the water and using all of the oxygen from the reaction. Bubbles are coming off of only one electrode. In my research I found that sodium hydroxide is a better electrolyte because it doesn't create chlorine gas and shouldn't react with my electrodes. Unfortunately I don't have any, but I am making lye ($\ce{KOH}$) from some hardwood ashes, will this work in place of $\ce{NaOH}$? I'm not sure if the potassium ions will conduct electricity in the same way. Thanks!

Using $\ce{KOH}$ rather than $\ce{NaOH}$ will work just fine.

Solutions of $\ce{KOH}$ are actually about $\pu{5\%}$ more conductive than solutions of $\ce{NaOH}$ at the same molar concentration. In terms of mass concentration, a given mass of $\ce{KOH}$ will have about $\pu{30\%}$ fewer equivalents of reactant than the same mass of $\ce{NaOH}$ because of the difference in molecular weights.

The combination of the difference in molar conductivity and molecular weights between $\ce{NaOH}$ and $\ce{KOH}$ just means that a given concentration of $\ce{KOH}$ should be about $\pu{35\%}$ less conductive than the same mass of $\ce{NaOH}$. Of course you can increase the concentration of $\ce{KOH}$ relative to the amount of $\ce{NaOH}$ you had planned to use if you like.

Please don't hesitate to ask for any clarifications in the comments below, and best of luck!


Heat generation in exothermic reactions Im thinking specifically of N2 + 3 H2 ⇌ 2 NH3. Its an exothermic reaction, therefore heat is generated. Im trying to understand how Chateliers principle interacts with this reaction. I understand that if you remove heat, Chateliers principle dictates that the equilibrium shifts to the product-side. I cant quite understand why though, do the reactants kind of know how warm it is and then decide, if its too warm "No, its too warm, lets not react and make it too warm"?

A followup on that question too, these reactions happen in enclosed places where temperature and pressure can be manipulated.Is it just efficiency of these containers that prevent the containers to not get as hot as the sun? If temperature wasnt manipulated, would the temperature keep rising?

Unlike the other factors affecting chemical equilibrium, (namely concentration, pressure), temperature follows the Le Chatelier principle via a different mechanism.

Temperature alters the equilibrium constant $K$, and this change in $K$ also changes the relative concentrations of the reactants and the products. The way in which the equilibrium constant depends on temperature is given by Van't Hoff equation:
$$\frac{d \ln K}{d T}=\frac{\Delta H}{RT^2}$$
Therefore, for exothermic reactions, the change in $K$ acts as if heat is one of the products, since increasing the temperature causes the $K$ to decrease, making the equilibrium at lower temperatures more favorable for Haber's process for making ammonia. (While this only talks of the equilibrium concentrations, lower temperatures in general result in slower rates, hence a balance is always sought)

While for your other question, as the temperature increases, the equilibrium is reached faster by both increasing ammonia (the cause of the increasing temperatures) and the decreasing $K$, until at a certain temperature, the reaction reaches equilibrium. I am not sure of the practical constraints employed in temperature regulation in the industry.


Application of the electronegativity effect on bond angles There's a rule in determining molecular geometry that says that if the electronegativity of the central atom increases, the bond angle of the molecule increases.

Is this is only applicable in the cases where the molecular geometry of the said compounds are similar? For example, considering $\ce{ClF3, BF3,}$ and $\ce{ PF3}$, they're primarily differentiated on the basis of them having different geometries.

(I will simply be converting my comments above into an answer here)

Yes, generally, you can apply this in cases with similar geometry. You can see why from the following rationale:

Imagine that the bond pairs look like a ring-shaped cloud, and one is worn on your index finger and another on your middle finger, both towards the fingertip. For example, to represent the two  $\ce{O-H}$ bonds of $\ce{H2O}$.
More electronegative central atoms will pull the ring towards the knuckles and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger

Here's an illustrative image for $\ce{H2O}$, with blue rings representing bond pair density :

Hence, you can see why the bond angle of $\ce{H2O}$ would be bigger than, say, $\ce{OF2}$ and $\ce{H2S}$.

In case of the former, the fluorines won't led the ring move towards the depression easily, hence the index and middle fingers can come closer to each other in contrast to $\ce{H2O}$

For the latter, $\ce{S}$ is less electronegative than $\ce{O}$, so again, the rings will show less movement towards the knuckles.

This will also help you to understand that you can't do this in cases of radically different molecular geometries, as this visualization will not be very useful there


Is the Arrhenius Equation a solution to a differential equation? I just learned about the Arrhenius Equation,

$$k(T)=A\exp{\left(\frac{-E_\mathrm a}{RT}\right)}$$

and noticed that it appears to be in the same form as the solution to a linear homogeneous differential equation, so I attempted to find the associated differential equation.

Since $E_\mathrm a$ and $R$ are both constant parameters, you can treat it as a single constant. Defining $$\beta = -E_\mathrm a/R$$

we have $$\ln(k(T)) = \ln[\exp(\beta/T + \ln A)]$$

Take the derivative of each side and solve for $\mathrm dk/\mathrm dT$ ($A$ is treated as a constant of integration and goes away), and we get:

$$\frac{\mathrm dk}{\mathrm dT} = -\frac{\beta k}{T^2}$$

Here is the slope field for the differential equation when $β=-1$

and here is the slope field for $β=1$

However, since $k$ is a rate constant, I'm not sure if this method is actually applicable. Is this, or a similar differential equation actually used on this topic?

To answer the question of the title. In Adam-Gibbs theory the activation enegy is defined as [1]:

$$E_\mathrm a=-R\left[ \frac{\partial \ln k}{\partial (1/T)} \right]$$
Now clearly this is a first order linear differential equation, as you were looking for. This can easily be solved by seperation of variable:
$$\int E_\mathrm a\, \mathrm d(1/T) = -R\int \mathrm d\ln k $$
$$\frac{E_\mathrm a}{T}=-R\ln k+A'$$
$$k(T)=A\exp\left( \frac{-E_\mathrm a}{RT} \right)$$
Here $A=\exp(-A'/R)$, now it can be seen that the Arrhenius equation has been regained.

[1] S. Petrucci; Ionic Interactions: From Dilute Solution to Fused Salts; equation 125


Source of the no. of isomers in alkanes, and other organic molecules I would like to ask if there is a source where I can find all the isomers for the alkanes, or other organic molecules. With other words, a list where each molecule of that group is stated in ascending order, and beside there is a value for the no. of isomers it can form.

Not counting stereoisomers and enantiomers, the number of isomers of the first ten alkanes is as follows:

Methane: 1
Ethane: 1
Propane: 1
Butane: 2
Pentane: 3
Hexane: 5
Heptane: 9
Octane: 18
Nonane: 35
Decane: 75

Source: http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf and Wikipedia articles for these alkanes. The PDF has the number of isomers for the first 30 n-alkanes.

The number of isomers possible with an n-carbon alkane can be calculated by a recursive combinatorial function; however, as the linked paper states, this math isn't for the weak. The number of structural isomers of a 30-carbon alkane is over 4 billion, and a 50-carbon alkane has over 1031 possible isomers. Even organic chemists and petrochemists usually don't bother to identify any one isomer beyond about the pentane mark. Some of the more symmetrical isomers are interesting on their own, since a few of the more compact ones (like the n-methyl types) have markedly different behaviors such as boiling point from longer isomers (for example, tetramethylbutane, an isomer of octane notable for its relatively narrow liquid window). The 2,2-dimethylpropane isomer of pentane is visually interesting because of its isohedral geometry, and likewise the tetraethylmethane/diethylpentane isomer might find interest among a few supremacist groups but is otherwise just another 9-carbon alkane.


Why does treatment of a geminal dihalide with sodium amide lead to an alkyne and not an allene? I've come across the following problem in Klein's Organic Chemistry, 2nd edition:

However, I'm a little bit confused about why the terminal alkyne is produced. I believe that an allene should be formed via stepwise elimination of HBr, using the strong base NaNH2. The first elimination should give the more substituted, more thermodynamically stable alkene (Zaitsev's rule):

Why does the former occur instead of the latter?

Sodamide is bloody dangerous.  If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion.

Even with good stuff, generally the reagent is not heated to avoid similar complications.  This means normally this reagent works at low temperatures (often times dry ice/acetone).  Under these conditions kinetic products are favored over thermodynamic ones.

The exomethyl group is much more accessible for deprotonation, and so kinetically, I would expect the amide ion to abstract the terminal hydrogen.

After that, the resulting bromoalkene still has the possibility of forming the allene product, but the transition state resulting in elimination to the alkyne is favoured as the vinylic hydrogen and the bromine are always locked in an anti conformation due to alkene geometry, as indicated in the above diagram. On the other hand, the allylic hydrogen may not necessarily be anti to the bromine due to C–C bond rotation.


Why is molality per mass solvent rather than per mass of solution? Knowing the moles of substance per mass of solution would be useful. Pour some solution and you'd know how much substance you have. Why is molality defined not this way, but as moles of substance per mass solute instead? It seems like this unit was engineered to have a value similar to molarity rather than to be maximally useful.

Why was molality defined this way? Is there an advantage I'm missing? This measurement seems easy to prepare but more cumbersome to use, and I can't see why that would be an advantage.

Molality $b$ is defined its way because being proportional for the given solvent to the solute/solvent molecular ratio.

Molarity or molar/mass fractions or molar_amount/solution_mass are not proportional to it.

The boiling and melting point shifts are proportional via ebullioscopic/kryoscopic solvent constants to molality.

$$\Delta T_\mathrm{b}=K_\mathrm{e} \cdot b$$

$$\Delta T_\mathrm{m}=K_\mathrm{k} \cdot b$$

The number of solvent (s) molecules per 1 solute molecules

$$\frac{n_\mathrm{s}}{n}=\frac{m_\mathrm{s}}{M_\mathrm{s}\cdot n}=\frac 1{ M_\mathrm{s}\cdot b}$$

If we provide as the input the solute mass and solvent volume and density, the above molecule ratio is

$$\frac{n_\mathrm{s}}{n}= \frac{m_\mathrm{s}\cdot M}{M_\mathrm{s}\cdot V \cdot \rho}$$


How to qualitatively compare bond strength? Without any thermodynamic data, how would one compare the bond strength of one molecule to another?

I understand a major factor is bond order: triple bond strength > double bond strength > single bond strength. But what if the molecules have the same bond order.

Also, generally a higher electronegativity difference gives a stronger bond (Chemistry LibreTexts article). But what if the difference is insignificant?

For example, in an exam, I was asked to compare $\ce{O-O}$ vs $\ce{S-S}$, and $\ce{O=O}$ vs $\ce{S=S}$ bond strengths. How would I compare them?

You can compare the bond strengths of $\ce{O-O}$,  $\ce{S-S}$,  $\ce{S=S}$ and   $\ce{O=O}$ by considering all possible effects that can alter the bond strengths. The effects which needs to be considered here are extent of π-overlap between the orbitals and inter electronic repulsions between the bonding atoms.

Firstly, it is obvious that double bonds will be relatively stronger than the single bonds. So, it is for sure that  $\ce{O=O} $ and  $\ce{S=S}$ will have higher bond strengths than  $\ce{O-O}$ and  $\ce{S-S}$. Now, we have to compare between two double bonds and two single bonds.

In  $\ce{O-O}$, two similar $\ce{2p}$ orbitals of two oxygen atoms form head-on overlap (σ-bonding) with each other, whereas, in  $\ce{S-S}$, two similar $\ce{3p}$ orbitals overlap. As it is a head on overlap, the extent of overlapping doesn't differ significantly, but for $\ce{2p}$ orbitals being smaller in size, two highly electronically dense $\ce{O}$ atoms comes much close to each other which hugely increase the inter electronic repulsions between the atoms. But in $\ce{S}$, the two bonding atoms remains much far from each other, and the electronic charge density is not also too much on each $\ce{S}$ as the $\ce{3p}$ orbitals are large in size, where the charge can easily diffuse. For this reason, $\ce{S-S}$ bonds become much stronger than  $\ce{O-O}$ bonds. (The reason is similar for the bond dissociation energy order  to decrease as in $\ce{Cl-Cl} >  \ce{Br-Br} > \ce{F-F}$)

But if we compare $\ce{O=O}$ and $\ce{S=S}$, the charge density on oxygen is reduced as lone pairs on oxygen decreases, therefore that effect becomes negligible. In this case, the extent of π-bonding starts to dominate. Being small sized,  $\ce{2p\pi-2p\pi}$ overlap in oxygen becomes much more strong and effective rather than  $\ce{3p\pi-3p\pi}$ overlap (where extent of overlap is not so much) in $\ce{S=S}$. That's why the introduction of double bond becomes much more thermodynamically favourable for oxygen and therefore, $\ce{O=O}$ bond is stronger than  $\ce{S=S}$ bond.

Thus the overall order of bond strength becomes, $$\ce{O=O > S=S > S-S > O-O}$$


Boiling-point elevation paradox This is the boiling-point elevation equation:

$$
\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}.
$$

If we do the limit of the above, for $b\rightarrow\infty$ we find that the boiling-point elevation is infinite. How this could be explained under a physical view?

$$
\lim_{b~\rightarrow~\infty}ib\mathrm{K}_{eb}=\infty,~\text{if}~i\wedge b > 0.
$$

For example, if we use a solute molality that is so big that the solution would be something like the pure solute, the final boiling-point of the solution, in the case of water as solvent for example:

$$
\mathrm{T}_{b}=ib\mathrm{K}_{eb} + 100\mathrm{°C}.
$$

wouldn't be like the boiling-point of the pure solute? I mean, something like this:

$$
\lim_{b~\rightarrow~\infty}\mathrm{T}_{solution}=\mathrm{T}_{solute}.
$$

Now, please forget what I've said above about $b\rightarrow\infty$. And take a look at this:

$$
T(b) = 0.86b + 64.7.
$$

As you can see is, the boiling-point of a solution of $\mathrm{CH_3OH}$ with $\mathrm{H_2O}$ as solute would be $322.7\mathrm{°C}$ for $b = 300$. This it's weird to me  'cause that temperature would be about three times higher then the boiling-point of the pure water!

So the following:

$$
\mathrm{T}_{b}=ib\mathrm{K}_{eb} + \mathrm{T_{solvent}}.
$$

holds true if

$$
b = \frac{x~\mathrm{g}_{solute}}{1000~\mathrm{g}_{solvent}} \le 1
$$

right? If yes why (in the above case):

$$
0.86\frac{1000~\mathrm{g}}{18.01528~\mathrm{g}/\mathrm{mol}}\frac{1~\mathrm{mol}^2}{1~\mathrm{kg}} + 64.7~\mathrm{°C} = 112.4~\mathrm{°C}
$$

?

In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.

This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.

A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.

So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.

When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.

It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the liquid solvent molecules at the liquid-gas-interface will enter the gas phase. The gaseous solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it.

The amount of solvent molecules going from gas->liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid->gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.

Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -> gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower...  So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -> liquid.

Why boiling point goes infinite: At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.


Minimal basis CI of H2 dissociation In Modern Quantum Chemistry by Szabo and Ostlund, they give an example of a minimal basis, full CI calculation of the dissociation of $\ce{H2}$ (on p.$241$ if you have the book). In explaining why this correctly describes the dissociation, they say:

  To see this, recall that as $\mathrm{R}\rightarrow\infty$, ... that all molecular orbital two-electron integrals tend to $\frac{1}{2}(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$, where $\phi_{1}$ is a hydrogenic orbital.

I could not find where they justified this statement. The closest I could find was in a previous section describing the dissociation using RHF (p.$166$) they state:

  [Besides $H^{\text{core}}_{11}$] All other integrals go to zero as $\mathrm{R}\rightarrow\infty$, except the one-center electron-electron repulsion integral $(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$.

What justifies this first statement? Is this unique to this case of a diatomic at infinite separation? These two comments seem to contradict rather than support each other.

Using Szabo and Ostlund's notation (page 85):

\begin{align}
J_{ij} &= (\psi_i\psi_i|\psi_j\psi_j)&
\text{and}&&
K_{ij} &= (\psi_i\psi_j|\psi_i\psi_j)\tag1\label{integrals}
\end{align}

For minimal basis set $\ce{H2}$ (page 162):

\begin{align}
\psi_1 &= [2(1+S_{12})]^{-1/2}(\phi_1+\phi_2)&
\text{and}&&
\psi_2 &= [2(1-S_{12})]^{-1/2}(\phi_1-\phi_2)\tag2\label{minimal}
\end{align}

Substitute equation $\eqref{minimal}$ into $\eqref{integrals}$, set $S_{12}$ and any integral involving both $\phi_1$ and $\phi_2$ to zero (large $R$) and you see that

$$J_{11} = K_{12} = \frac{1}{2}(\phi_1\phi_1|\phi_1\phi_1)$$

Remember that $(\phi_1\phi_1|\phi_1\phi_1)=(\phi_2\phi_2|\phi_2\phi_2)$ since you are dealing with a homonuclear diatomic.


proton NMR deshielding The protons alpha to the oxygen in methyl propionate are more shielded than their corresponding protons in ethyl propionate.

.

We can see from the NMR data that this is the case as well.

My TA says this is due to the inductively withdrawing effect of the extra carbon in ethyl propionate.

However, this strikes me as blatantly wrong. The carbon should be inductively donating ... I don't see how it pulls electron density away from the specified hydrogens.

In your example we are comparing the electron releasing/withdrawing properties of a hydrogen substituent and a methyl substituent, and the effect this has on proton ($\ce{H_{a}}$) chemical shifts.

Carbon (2.55) is more electronegative than hydrogen (2.2).  Therefore, the methyl substituent will be more electron withdrawing compared to the hydrogen substituent.  This means that the methyl group will remove electron density from the carbon to which it is attached, compared to the hydrogen substituent.  The fact that this carbon has lower electron density in the methyl case, is further transmitted to the protons ($\ce{H_{a}}$) attached to this carbon. Consequently the $\ce{H_{a}}$ protons will also experience lower electron density in the R=methyl case.

Lower electron density at an nmr-active nucleus results in deshielding (downfield shift). Consequently, the $\ce{H_{a}}$ protons are deshielded in the ethyl ester (R=methyl) relative to the methyl ester (R=hydrogen), in agreement with the spectral data you've provided.


Balancing Redox reaction : with 2 species being oxidized I have a redox reaction I found from a chemistry book from the 1970's.
Two chemical species are oxidized, the Chromium and Iodine.

$\ce{CrI3 + KOH + Cl2 -> K2CrO4 + KIO4 +KCl + H2O}$

Hope someone can help on how to balance this.

Your best bet is to consider the entire reactant $\ce{CrI3}$ as the reducing agent.

The chromium (III) iodide starts out as that compound and gets oxidized to a combination of chromium (VI) (in the chromate) and iodine (VII) (in the periodate) with a ratio of chromium to iodine matching the original compound:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII}}$

Clearly the originally neutral reactant has been oxidized to atoms having a combined oxidation state of $+27$, so we balance this reaction with $27$ electrons:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII} + 27 e^-}$

Each chlorine molecule takes up two of the electrons so the redox ratio must be $\ce{2 CrI3 : 27 Cl2}$.  Put those in and balance the remaining compounds as you normally would.


How to determine the empirical formula for terephthalic acid from its combustion products? Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?

We have:
$$\ce{C_{$x$}H_{$y$}O_{$z$} + O2 -> CO2 + H2O ->[\ce{CaCl2 + NaOH}] CaCl2 + NaOH}$$

The anhydrous $\ce{CaCl2}$ absorbs the $\ce{H2O}$ produced, and the $\ce{CO2}$ is absorbed from $\ce{NaOH}$? So the products are $\ce{CaCl2. 2H2O}$ and $\ce{Na2CO3}$

I may have just figured the empirical formula out, and was wondering what the final answer is so I can compare and see where I went right/wrong. I realised that instead of $\ce{Na2CO3}$, $\ce{NaHCO3}$ would be produced.

Working out:
Via conservation of mass, the $\ce{CO2}$ and $\ce{H2O}$ must also be $0.4953~\mathrm{g}$.

Therefore: conservation of mass (mass of $\ce{CO2 + H2O}$) - increase in mass of $\ce{CaCl2 + NaOH}$

$0.4953 - (0.1613 + 1.0495) = 0.7155~\mathrm{g}$ <--- mass of $\ce{CaCl2}$ and $\ce{NaOH}$

Mass of $\ce{CaCl2.2H2O} = 0.7155 + 0.1613 = 0.8768~\mathrm{g}$

Mass of $\ce{NaHCO3} = 0.7155 + 1.0495 = 1.765~\mathrm{g}$

     C:               H:              O
 1.765x(12/84):     0.8768x(4/147.1):

 (m) 0.252142857:     0.023842284:   0.4953-(0.252142857+0.023842284)= 0.219314859

 (n) m/12       :     m/1         :    m/16

 0.021011905 : 0.023842284     :  0.013707179

       (Divide by lowest value)

( 1.53      :     1.74        :    1 ) x4

  6         :       7         : 4

Therefore $\ce{C6H7O4}$

You are on the right track, but if you google terephthalic acid, you will know that your formula is not correct.

One mistake you are making is using the final mass of the $\ce{CaCl2}$ and $\ce{NaOH}$ and not the change in mass. For example, let's take the $\ce{CaCl2}$:

First, we don't know what the initial mass of $\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass.

However, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.

Similarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.

The second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products.
$$\ce{C}_x\ce{ H}_y\ce{ O}_z\ce{ +}\left(\frac{4x+2y-z}{4}\right)\ce{ O2 -> }x\ce{CO2 +}\frac{y}{2}\ce{ H2O}$$

We need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:

Mass of water produced $\ce{->}$ moles of water produced $\ce{->}$ moles of hydrogen in the water produced (and in the original sample!) $\ce{->}$ grams of hydrogen (in sample)
Mass of carbon dioxide produced $\ce{->}$ mass of carbon (in sample)
Mass of oxygen in sample (by subtraction) $\ce{->}$ moles of oxygen in sample

You are doing everything correctly after this.


How can one say pressure is constant in thermodynamics? When a gas expands by  b charles law pressure increases
$$\frac{p}{t}=k$$
thermodynamics says:
$$dQ=dU+PdV$$
so herfe one is considering constant pressure but how b charles law says it changes?So shouldn't it be:
$$dQ=dU+PdV+VdP$$
can't we apply a variable external pressure?

Sometimes in thermodynamics we wish to constrain pressure to be constant.  [Or we may wish to do a thermodynamic analysis for a system that is at constant pressure, which is a common laboratory condition.]   However, that is a choice.  Pressure is not generally constrained to be constant in thermodynamics.
For a closed system:
$$d\text{U} = \text{đ}q + \text{đ}w$$
[Note that I'm using a different sign convention from you for work, but the results come out the same.]
If the only type of work is $pV$- work, then  $\text{đ}w = -p_{ext}dV$, where $p_{ext}$ is the external pressure applied to the system by the surroundings.  Thus we have:
$$d\text{U} = \text{đ}q - p_{ext}dV,$$
where the pressure can change in any way you please during the process.
If the process is reversible, then (in that idealized case) the system is in continous equilibrium with its surroundings. As as a consequence, $p_{sys} =p* = p_{ext}$, and we can write:
$$d\text{U} = \text{đ}q - pdV$$
[*The convention is that if a subscript is not present, the variable is assumed to apply to the system.]


Single line equilibria notation I'm reading a paper with a reaction scheme where an enzyme binds loosely to inhibitor reversibly and then the Enzyme-Inhibitor complex undergoes another reversible conformational change to a tightly bound complex. My question is about the notation used (which I cannot type because I can't find the appropriate symbols). The second reaction uses the regular equilibria symbol which is fine - no problem. However, the first reaction uses a modified form of the equilibria symbol which is one line rather than two with 'half arrows' at either end. What does this symbol mean?

...equilibria symbol which is one line rather than two with 'half arrows' at either end

I'm pretty confident you mean "⥋" arrow within the reaction scheme for enzyme $\ce{E}$ and inhibitor $\ce{I}$ which I tried to recreate from your description:

$$\ce{E + I ⥋ \underset{\rm loose\\bond}{EI} <=> \underset{\rm strong\\bond}{EI'}}$$

I've seen this arrow in literature before, and in general it's been also utilized to denote equilibrium. This notation is not officially supported by IUPAC standards. However, there is a couple of references tangentially mentioning it.

From the article "The Chemical Equation Part I: Simple Reactions" [1, p. 189] (emphasis mine):

  Sometimes reactants interact quickly, but the reaction never goes to completion because the reverse reaction is competing with the forward reaction, and the system ends up in a state of chemical equilibrium. A pair of arrows pointing in both directions indicates an equilibrium system, but the second arrow need not be there. Even when two arrows are used, the equation still cannot identify the point of equilibrium.

There is also a recent outstandingly well-written review with the title that speaks for itself: "Chemistry: A Panoply of Arrows" [2, p. 592], which pretty much says the same (emphasis mine):

  Van't Hoff's double arrow was modified in 1902 by Hugh Marshall,... who removed the inner barb of each arrow. This created the half-headed double arrows, ⇌, which have since been further simplified to ⥋. In justifying his proposal, he noted that single arrows were being used in organic chemistry
  “to indicate merely the stages and methods by which a substance can be produced from some other substance as a starting point. In such cases, as no attention is paid by the bye products (sic), there is nothing of the nature of an equation
  involved.”

My humble opinion is that authors' intent was to visually separate those two processes, e.g. due to exceptional difference in rates, or maybe they wanted to denote conformational change aside from the chemical equilibrium.

Notes

"⥋" arrow belongs to the Supplemental Arrows-B Unicode block under the name "Left barb down right barb up harpoon" and is invoked via U+294B.
This and many other symbols you are having troubles identifying can be recognized by services like Shapecatcher or Detexify.
I also asked a question on Mathematics.SX hoping that someone could provide more definitive answer regarding the proper usage of this arrow (no answer yet).

References

Kolb, D. Journal of Chemical Education 1978, 55 (3), 184. DOI: 10.1021/ed055p184.
Alvarez, S. Angewandte Chemie International Edition 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.


Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants? I understand Le Chatelier's principle and how every change to an equilibrium system causes an opposing reaction from the system. I also understand how, when pressure is increased, the equilibrium shifts to the side with the lowest number of moles of gas and vice versa.

So, why does this happen? How is it more energetically favourable - or is it something else?

Also, are these explanations of why the ration changes for the other factors right:

Temperature - equilibrium moves to the endothermic side of the reaction as temperature increases because the activation enthalpy of the exothermic reaction increases as the overall energy of the system is less
Concentration - ratio of products/reactants changes immediately after one part of the reaction's concentration increases, it then approaches the same ratio (assuming one whole side of the reaction is added)
Catalysts do nothing more than decrease the time it takes to reach equilibrium as they catalyse both sides of a reversible reaction and reduce the activation enthalpy

With gasses, what you're doing by changing the pressure is you change the partial pressures or the reactants. As long as there's the same moles of gas on either side, the equilibrium is unaffected, but if there's an un unequal number, the reaction quotient is changed. The same would happen if you added water to an aqueous reaction. You can play with the numbers yourself, I'll give you an example to use:

$$\ce{N2(g) + 3H2(g) <->2NH3(g)}$$

We can use the reaction quotient with partial pressures, but it's more clear if we use the one with concentrations:

$$Q_c\ce{ = \frac{[NH3]^2}{[N2][H2]^3}}$$

Using $c=\frac{n}{V}$:

$$Q_c\ce{ = \frac{\frac{n(NH3)^2}{V^2}}{\frac{n(N2)}{V}\times \frac{n(H2)^3}{V^3}}}$$

Take notice of how this fraction depends on volume! So it's really just the system reacting to attempt to reach equilibrium again (making it so that K = Q).

As for temperature. My understanding is that it's not to do with activation energy. It IS related to the enthalpy of the reaction though, and your understanding of what a temperature change means for a particular reaction is correct.

Concentration is the same deal as I've talked about in the beginning of this post, and incidentally it matches your own understanding of it.

You've got catalysts right too. Have a look at the following diagram:


Which resonance structure would be favorably contributed to the actual structure? Total number of pairs of resonating structure in which second is more favorable than first:

I know  (d) and (e) have second structure more stable.

(f) has both same.

According me to, (a) and (c) should have second structure unstable as oxygen has positive charge.
I don't have a clue about (b) and (g).
Please help me out.

(a) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Even though $\ce{O}$ is more electronegative than $\ce{C}$, its octet is fulfilled while $\ce{C}$'s is not (only 6 $e^-$s on positively charged $\ce{C}$). Therefore, second structure is more favored over the first one.
(b) Negative charge on $\ce{O}$ versus negative charge on $\ce{C}$: Here, both negatively charged $\ce{O}$ and $\ce{C}$ have their octet fulfilled. Yet, it is always stable when the negative charge is on more electronegative atom ($\ce{O}$) than on comparatively less negative atom ($\ce{C}$). Therefore, first structure is more favored over the second one.
(c) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Here, same argument of (a) applied here. Therefore, second structure is more favored over the first one.
(d) Charge separation versus neutral: it is always stable when the compound has no charge (neutral) than on has charge separation (with net charge is zero). Therefore, second structure is more favored over the first one.
(e) and (f) Kekulé structures of naphthalene and benzene, respectively: Of each pair, first and second structures have the same energy. Therefore, both structures of each pair have same stability equally favored.
(g) Positive charge on $2^\circ$-$\ce{C}$ in both structures: Since positive charge is on a $2^\circ$-$\ce{C}$ of each compound, the relative energy due to carbocation is same on both structures.  Yet, the first structure has its double bond conjugated to the carbonyl carbon while the second structure has isolated double bond to the carbonyl carbon (cross-conjugated). Therefore, first structure is more favored over the second one.


Why is nitrous acid prepared in situ? Why is nitrous acid prepared in situ?

I am not sure about the above question. First of all, I don't exactly understand what is meant by "in situ". Secondly, how would preparing nitrous acid in situ, in the diazotisation of amines with nitrous acid, make it less toxic or unstable?

Well, if you don't make it in situ, you would have to add the nitrous acid you have already kept in your laboratory shelf, and surprise! It's not there! Because, it has already decomposed.

Nitrous acid ($\ce{HNO2}$) has an intermediate oxidation state of +3, and hence, it tends to disproportionate into more stable oxidation states:

$$\ce{2HNO2 -> NO2 + NO + H2O}$$

$\ce{NO2}$ itself dissolves in water to give $\ce{HNO3}$ and $\ce{HNO2}$. This gives an overall reaction in aqueous solution as:

$$\ce{3HNO2 -> HNO3 + 2NO + H2O}$$

This should give you a good idea on why we make nitrous acid within the reaction mixture. You can't store it beforehand. It must be made when it's time to use it.


Apparent contradiction in the name of group 2 elements From my understanding (I might be very wrong), the "earth" in alkaline earth metals means non-metallic, insolube in water and resistant to heating, therefore the oxides and hydroxides of group 2 metals should be insoluble in water by definition of "earth".

However, doesn't the term "alkaline" by definition mean bases that are soluble in water? How can the two definitions be co-applicable?

"Insoluble (in water)" is a relative term, especially for ionic compounds, and so is "alkaline".  That relativeness gives us enough wiggle room for two of our most common bases to slip through.

Roughly speaking, it takes a thousand volumes of water to react with and then dissolve one volume of calcium oxide, so the solubility of lime in water would not be evident from casual observation.  Yet if you put an indicator into the water it reacts in a way that indicates the water has turned alkaline.  The solubility of lime in water is hard to see with the eye, but it overwhelms the tiny intrinsic autoionization of water, so lime gives evidence of being both an "earth" and "alkaline".  Ditto (despite the still lower solubility) for magnesia.

Lime and magnesia, then, are the quintessential alkaline earth's, far exceeding anything else with similar properties in their common occurrence.  They have become the namesake of the all the "alkaline earth metals" in Group 2.


Is it possible to find fire outside the Earth? Excuse me if the question is silly, I have very little knowledge of chemistry.
We can find oxygen in many places in the universe, but all the combustibles we use to light fire seems to be organic: carbon, oil, wood, alcohol... so if that is true, we could only find fire on Earth, as long as we don´t find organic materials in other planets.

Another way of asking the same question could be: can we light a fire without organic material?

Note that I mean fire as a result of a combustion. The Sun would be a non-valid example, as it works by fusion.

Many non-organic substances burn. For example:

Reactive metals such as potassium burn when exposed to oxygen. (Thanks to @Jon Custer in his comment)
Non metal elements such as carbon, sulphur and phosphorous also burn in oxygen when heated.

So yes fire is definitely possible outside earth.

Additionally simple organic compounds such as methane are known to exist in large quantities on other bodies such as Titan, a moon of Saturn, which has a whole 'hydrological' cycle based around methane, and to a lesser extent ethane and heavier hydrocarbons, with a methane atmosphere and lakes of methane on the surface. (Thanks to @ron in his comment)


Reaction of hydrogen peroxide and catalase All Google searches have simply returned more info on catalase. I'm looking for a catalyst that isn't found inside living organisms which can break down $\ce{H2O2}$. Looking for any resources to look at, or just examples of other catalysts.

Side question: Can catalase be used for anything other than the decomposition of  $\ce{H2O2}$?

Manganese (IV) oxide, iron (III) chloride, and lead (IV) oxide are common examples for a catalyst that isn't found inside living organisms and which can break down $\ce{H2O2}$. Potassium iodide can also be used as a catalyst, along with other transition metal compounds, of metals such as silver or platinum.

This didn't take much effort to find so I'm sure you can find more examples easily.

As for catalase, it seems to just decompose $\ce{H2O2}$, but this is useful in many different areas. For example, decomposing $\ce{H2O2}$ in contact lens cleaning solutions, preventing foods from being oxidised in wrappers (in conjunction with glucose oxidase, another enzyme), and removing $\ce{H2O2}$ from milk before cheese is made. The decomposition reaction is also used as a test for catalase in bacteria.

Sources: General use - National Center of Biotechnology Education - and more uses (including the cheese example) - Worthington Enzyme Manual


Find volume of ammonia gas required to prepare a solution of given pH Calculate the volume of gaseous ammonia $(K_\mathrm{b} = \pu{1.8E-5})$ to dissolve (at NTP) in $\pu{100 mL}$ of water in order to obtain a solution with $\mathrm{pH} = 11.0.$ Consider volume variation negligible.

I started by evaluating molar concentration of $\ce{NH4+}$ and $\ce{OH-}$ knowing the $\mathrm{pH},$ then I used the $K_\mathrm{b}$ inverse formula to evaluate the molar concentration of ammonia. These are the results:
$$c(\ce{NH3}) = \pu{0.056 mol L-1}$$
$$c(\ce{NH4+}) = c(\ce{OH-}) = \pu{1E-3 mol L-1}$$
Then my idea was to work with water. Since $\ce{NH3}$ is a weak base it does not dissociate completely, the amount of water is
$$n(\ce{H2O}) = \frac{\pu{100 g}}{\pu{18.02 g mol-1}} = \pu{5.55 mol}$$
Then I run out of ideas. I tried to find the mamount of ammonia (but I think the result was not correct) by evaluating the degree of dissociation, but then I got lost.
The final result of the problem is $\pu{0.13 L}.$ How do I pursue the problem further?

The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$
Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations of $\ce{NH4+}$ and $\ce{NH4+}$ are $\alpha$. Thus:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{\alpha \cdot \alpha}{c - \alpha} = 1.8 \times 10^{-5}$$
$$\alpha^2 + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5}c =0 \tag1$$
Since, $\mathrm{pOH}$ of the solution is $3.0$ ($14 - \mathrm{pH}$), $[\ce{OH-}] = \alpha = 1.00 \times 10^{-3}$, from equation $(1)$:
$$(1.00 \times 10^{-3})^2 + 1.8 \times 10^{-5}\times 1.00 \times 10^{-3} = 1.8 \times 10^{-5}c $$
$$1.8 \times 10^{-5}c = 1.00 \times 10^{-6} + 1.8 \times 10^{-8} = 1.018 \times 10^{-6} $$
$$\therefore \ \ c = \frac{1.018 \times 10^{-6}}{1.8 \times 10^{-5}} = \pu{5.66 \times 10^{-2} molL-1} $$
This is exactly what you got without showing your work. However, you made only $\pu{100 mL}$ of solution. Thus, amount of ammonia dissolved in $\pu{100 mL}$ of water is:
$$$\pu{0.100 L} \times \pu{5.66 \times 10^{-2} molL-1} = \pu{5.66 \times 10^{-3} mol}$
Assuming gaseous ammonia is behaving like ideal gas, you can use $PV = nRT$ here to find the volume of $\pu{5.66 \times 10^{-3} mol}$ of ammonia at STP:
$$V= \frac{nRT}{P} = \frac{(\pu{5.66 \times 10^{-3} mol})(\pu{0.082 LatmK-1mol-1})(\pu{298 K})}{\pu{1 atm}} = \pu{0.138 L}$$


Skin adhesive that can be used daily without damaging skin What is the branch of chemistry that deals with skin adhesives? Is there an adhesive that is not water soluble that is safe to use every day?

what is the branch of chemistry that deals with skin adhesives?

Likely a combination of biomedical engineering and organic chemistry, with some materials science in there as well.

  Is there an adhesive that is not water soluble that is safe to use every day?

Yes. Mastisol fits your criteria.


Are nucleophilic aromatic substitutions stepwise or concerted? Recently, there has been a lot of discussion about nucleophilic aromatic substitution (SNAr) reactions apparently being concerted instead of stepwise (e.g. Krämer, K. Textbook aromatic substitution mechanism overthrown. Chemistry World, 23 July 2018):

This is in contrast to the usual mechanism taught at an undergraduate level, where the anionic Meisenheimer complex is considered to be an intermediate:

What is the current consensus on the mechanism, and what evidence is there for it?

TL;DR Most SNAr reactions proceed via concerted mechanisms without a discrete anionic intermediate. This is typically the case when there is a good leaving group (e.g. chloride or bromide) and if the ring is not extremely electron-poor.

Over the years, there has been increasing evidence of concerted nucleophilic substitutions at sp2-hybridised carbons. One notable example was published by Neumann et al. in 2016, who studied the mechanism by which PhenoFluor (2) converted phenols to aryl fluorides:[1]

For the conversion of [1,1′-biphenyl]-4-ol to 4-fluoro-1,1′-biphenyl (i.e. the above reaction with a phenyl substituent at the para position), a 16O/18O kinetic isotope effect of 1.08 ± 0.02 was observed. This implies that Ar–O bond breaking takes place in the rate-determining step, which is inconsistent with the traditional stepwise mechanism. The concerted mechanism shown above was further supported by DFT studies which revealed a transition state 4 with partial C–O and C–F bonds.

Building on this, Kwan et al. went one step further to show that many SNAr reactions were indeed concerted.[2] In particular, for three examples, the authors used 13C/12C kinetic isotope effects to evaluate the mechanism. These KIEs were determined by integrating the 13C satellites in 19F NMR spectra (an adaptation of Singleton's method[3] which improves sensitivity).

The presence of a (primary) 13C/12C KIE suggests that the C–X bond is being broken in the transition state. However, the KIE itself does not a priori differentiate between stepwise and concerted mechanisms, since in both pathways the respective rate-determining steps involve weakening of bonds in the transition state. In order to identify the mechanism, the authors compared the experimental KIE to a calculated "maximum" KIE. Compounds with stronger bonds (such as C–F in 6) have more vibrational zero-point energy to lose in the TS, and hence have larger maximum KIEs (1.070 for reaction 1).

For a concerted mechanism which "alters both bonds simultaneously", the ratio $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is expected to be large; and for a stepwise mechanism which "alters one bond at a time", $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is small. [The paper does not go into detail about exactly why this is so; I suggest reading pp 225 onwards of an article by Westaway[4] which explains this in the context of SN2 reactions.]

This allows the three reactions above to be classified into three different categories:

a stepwise reaction (6 to 7) with a discrete anionic intermediate. This intermediate is heavily stabilised by the electron-withdrawing nitro groups, and also because fluoride ion is a poor leaving group;
a concerted reaction (8 to 9) with the Meisenheimer complex as the transition state. Here, the presence of a good leaving group (chloride) means that the Meisenheimer complex is not sufficiently stable to be an intermediate;
and a borderline case (10 to 11) where there is formally no intermediate but a shallow region in the potential energy surface near the TS.

For the concerted pathway, DFT calculations revealed that the transition state was essentially a Meisenheimer complex: the ring bears a significant amount of negative charge, is no longer aromatic (as indicated by NICS values), and NBO analysis shows that the electron density in the TS is mostly described by the Lewis structure of the Meisenheimer complex. Consequently, even though many "textbook" SNAr reactions are in fact concerted and not stepwise, the same factors that stabilise the previously purported intermediate (electron-withdrawing groups at 2- and 4-positions, and heteroatoms in the ring) also stabilise the transition state and facilitate the reaction.

Finally, a range of SNAr reactions on different carbocyclic and heterocyclic rings were then surveyed by DFT. It turns out that, at least within the scope of the reactions investigated:

All SNAr reactions on nitrogen-containing heterocycles (2-halopyridines, 2-halopyrazines, and 2-halopyrimidines) proceed in a concerted fashion, regardless of leaving group or nucleophile.
Likewise, the theoretical SNAr reactions on unsubstituted benzene are concerted.
For electron-deficient benzene rings (with 4-nitro, 2,4-dinitro, 2,4,6-trinitro, or 4-acetyl groups), SNAr reactions with fluoride as the leaving group are stepwise, but chlorides and bromides mostly react in a concerted manner.

References

Neumann, C. N.; Hooker, J. M.; Ritter, T. Concerted nucleophilic aromatic substitution with 19F− and 18F−. Nature 2016, 534 (7607), 369–373. DOI: 10.1038/nature17667.
Kwan, E. E.; Zeng, Y.; Besser, H. A.; Jacobsen, E. N. Concerted nucleophilic aromatic substitutions. Nat. Chem. 2018, 10 (9), 917–923. DOI: 10.1038/s41557-018-0079-7.
Singleton, D. A.; Thomas, A. A. High-Precision Simultaneous Determination of Multiple Small Kinetic Isotope Effects at Natural Abundance. J. Am. Chem. Soc. 1995, 117 (36), 9357–9358. DOI: 10.1021/ja00141a030.
Westaway, K. C. Using Kinetic Isotope Effects to Determine the Structure of the Transition States Of SN2 Reactions. Adv. Phys. Org. Chem. 2006, 41, 217–273. DOI: 10.1016/S0065-3160(06)41004-2.


Pathway from Sucrose to Cholesterol I was reading Linus Pauling's book How to Live Longer and Feel Better, and in various places in the book he discusses a correlation between increased sucrose intake and increased blood plasma cholesterol levels, as reported in a study by Milton Winitz.

Pauling writes:

  ... the metabolism of sucrose yields in it's first step equal amounts of glucose and fructose.  ... Fructose metabolism goes in part by a different route such that it produces acetate, which is a precursor of the cholesterol that we synthesize in our liver cells.

I was wondering specifically what the metabolic pathway from sucrose to cholesterol is and given say 100 grams of sucrose, how much will be turned by the body into cholesterol.

As you have known that metabolism of sucrose yields glucose and fructose.So formed glucose undergoes glycolysis producing pyruvate or lactate(under anaerobic condition). In case of fructose, it is converted into fructose-1-phosphate and which on action of aldolase enzyme produces dihydroxyacetone phosphate and glyceraldehyde-3-phosphate which are the intermediate products of glycolysis and finally converted to pyruvate.

The end product of glycolysis i.e. pyruvate is converted into acetyl coa in presence of coenzyme A , which is the precursor for the synthesis of cholesterol.

For the synthesis of cholesterol , acetyl coa undergoes different enzymatic reactions finally forming cholesterol at the end.


Find the half equivalence point How to find the half equivalence point knowing the pH, molarity, titrant added at equivalence point?

This a fairly straightforward and simple question, however I have found many different answers to this question. I originally thought that the half equivalence point was obtained by taking half the pH at the equivalence point. However, I have encountered some sources saying that it is obtained by halving the volume of the titrant added at equivalence point.

Half equivalence point is exactly what it sounds like. It is the point where the volume added is half of what it will be at the equivalence point.

There are 3 cases

Strong Acid vs Strong Base: Here one can simply apply law of equivalence and find amount of $\ce{H+}$ in the solution. Similar method for Strong base vs Strong Acid.
Weak Acid vs Strong Base: Here one can apply Henderson-Hasselbalch equation as a general result one gets pH = pKa.
Weak Base vs Strong Acid: This case is similar to WA vs SB and one can easily use Henderson-Hasselbalch for bases again one gets pOH = pKb


Can methanol act as a ligand? As methanol is both a lewis acid and a lewis base, the lewis base part suggests that it can act as a ligand as many bases are. So, is methanol a ligand? And what is the reason behind it?

There are several complexes where methanol acts as ligand. Some examples are:

methanol coordinated vanadium(V) chalcogenido complexes: $\ce{V(Chal)Cl2(OMe)(HOMe)2}$ where $\ce{Chal = O, S, Se}$
bis(μ-methoxo)dichromium(III) complexes, $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)2]}$ and $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)(CH3O)]}$, where $\ce{L^{Se}}$ represents the dianion of 2,2‘-selenobis(4,6-di-tert-butylphenol)
di-µ-chlorido-bis[dichloridobis(methanol-κO)iridium(III)] dihydrate  $\ce{[Ir2Cl6(CH3OH)4]2H2O}$
$\ce{M(CH3OH)3SO4}$ where $\ce{M = Ni, Co, Zn}$

Reference

Google: Methanol ligand complex. First 4 search results.


Is it possible for Cu to reduce Cu2+? I tried a slightly different set up of a Galvanic cell by inserting the $\ce{Zn}$ electrode into the $\ce{CuSO4}$ solution and the $\ce{Cu}$ electrode into the $\ce{ZnSO4}$ solution. This gave a voltage reading of 0.85 V. I also observed that the $\ce{Zn}$ electrode was getting coated with copper as the reaction progressed.

I had two hypotheses in mind -

The reaction is limited to only one half cell: the $\ce{Zn}$ electrode is oxidised and the $\ce{Cu^2+}$ in the solution is reduced. This explains the coating on the $\ce{Zn}$ electrode, but it does not explain the voltage. Doesn't the voltage mean that there is some interaction between the two cells?
This is my main question: Is it possible that the $\ce{Cu}$ atoms on the electrode are losing electrons that then travel through the circuit and reduce the $\ce{Cu^2+}$ ions in the second half cell? A test for $\ce{Cu^2+}$ ions in the $\ce{ZnSO4}$ solution after the reaction had progressed was inconclusive.

What exactly is happening in this set up?

Zinc is higher in the reactivity series than copper, so simply by placing a piece of zinc metal into a copper sulphate solution, you will get a displacement reaction where the zinc metal loses electrons and becomes $\ce{Zn^2+}$ ions, and $\ce{Cu^2+}$ ions gain those two electrons to become copper metal.  This will happen until the zinc electrode is coated with copper and no more zinc ions can get into the solution.

The standard half-cell potentials are measured using standard 1M solutions, and deviations from this concentration cause the equilibria to shift according to Le Chatelier's principle:

$\ce{ Cu^{2+}_{(aq)} + 2e^{-}<=> Cu_{(s)}} \quad  E^{o}=+0.337V$

Since the $\ce{[Cu^{2+}_{(aq)}]}$ is zero for the copper electrode, this system is far from equilibrium and the half-cell potential will be closer to zero.  In the other half of the system, the $\ce{[Cu^{2+}_{(aq)}]}$ is less than 1M due to the reaction with the zinc electrode, but it is still significantly greater than zero.  The result is a slight potential for copper metal to become ions and for the electrons to travel through the circuit to reduce some of of the copper(II) ions in the other half cell.  Using the Nernst equation for this isn't valid since concentrations close to 1M are too high for accurate results on one hand (concentrations greater than $10^{-3}$M deviate from experimental values), and the use of the equation is on the assumption that the ion is in equilibrium.  If we use it naively we do get something:

$E_{cell} = E^{o}_{cell} - (\frac{0.059}{2})log(Q) \approx 0-0.0295log(\frac{\ce{[0.0000000001]}}{\ce{[0.5]}}) = 0.37V$

As the approximation of the copper(II) ion concentration approaches zero the potential difference for the cell approaches infinity, which goes some way to demonstrate this is only valid within limits since you don't get a 100,000 Volts electric shock when you put a piece of metal into pure water.  Nevertheless it describes a trend and supports the Le Chatelier interpretation.

However, if the circuit is connected using a voltmeter this has a sufficiently high resistance to prevent current from flowing, so although there is potential for the copper to reduce itself through the cell, this can only happen if electrons can flow.  Even without the high resistance the reaction of the copper metal forming ions may be so slow, that very little current flows and few ions are formed which may explain the inconclusive test result.


How to calculate the standard Gibbs energy at equilibrium? Given that $K_c = 1.7 \times 10^{-13}$, calculate $\Delta G^{\circ}$ for this equilibrium mixture at $\pu{298 K}$.
$$\ce{N2O (g) + \frac{1}{2}O2 (g) <=> 2 NO (g)}$$

I've calculated:
$$
\begin{align}
\Delta G^{\circ} &= -RT \ln K \\
  &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln (1.7 \times 10^{-13}) \\
  &= \pu{72.8 kJ mol-1}
\end{align}
$$
The result I'm expecting is $\pu{68.9 kJ mol-1}$. Where have I gone astray?

I think the author of this problem forgot to add units to $K_c$ since it's not a dimensionless entity:

$$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$

and should be $K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$. In general

$$[K_c] = \mathrm{dim}(c)^{Δn}$$

where square brackets denote the dimensions of the quantity $K_c$, $c$ is concentration and $Δn$ is the difference in the amounts between gaseous products and reactants, e.g. here

$$Δn = 2 - (1 + 0.5) = 0.5$$

On the other hand, the equilibrium constant $K$ you use for determining the standard Gibbs energy, must be dimensionless. Since we are dealing with gases only, the easiest way is to use $K_p$, which is dimensionless as required (when normalized to the standard state of pressure $p^\circ = \pu{1 bar}$):

$$K_p = \frac{\left(\frac{p(\ce{NO})}{p^\circ}\right)^2}{\left(\frac{p(\ce{N2O})}{p^\circ}\right) \left(\frac{p(\ce{O2})}{p^\circ}\right)^{0.5}}$$

so that in general

$$[K_p] = \mathrm{dim}(p)^{Δn}\cdot \mathrm{dim}(p^\circ)^{-Δn}$$

$K_p$ and $K_c$ are related (via the ideal gas law):

$$K_p = K_c (RT)^{Δn}$$

So, the equilibrium constant is

$$
\begin{align}
K &= K_p(p^\circ)^{-Δn}\\
  &= K_c(RT)^{Δn}(p^\circ)^{-Δn} \\
  &= \pu{1.7e-13 mol^{0.5} L^{-0.5}}\cdot(\pu{8.314e-2 L bar K−1 mol−1}\cdot\pu{298 K})^{0.5}(\pu{1 bar})^{-0.5} \\
  &= \pu{8.5e-13}
\end{align}
$$

Note that here I used gas constant  expressed as $\pu{8.314e-2 L bar K−1 mol−1}$ since in this case all dimensions are cancelled out and $K$ is left dimensionless. Now we can finally find the standard Gibbs energy:

$$
\begin{align}
Δ G^\circ &= -RT\ln K\\
  &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln\left(\pu{8.5e-13}\right)\\
  &= \pu{68.9 kJ mol-1}
\end{align}
$$

Here the product before the logarithm includes $R = \pu{8.314 J mol-1 K-1}$ to get answer in $\pu{kJ mol-1}$ straight away.


Aren't hydrogens on less substituted groups more electrophilic? I am having some trouble understanding the behaviors of some Alkene reactions that I have been studying. I took two semesters of O-chem a couple of years ago and am trying to refresh before an advanced class, but I have found two reactions in my studies that aren't making sense.

At the top of page 13 of the e-book found here: http://crab.rutgers.edu/~alroche/Ch08.pdf ... I see that the water molecule is attacking the carbon with two methyl groups on it... but wouldn't the carbon with only one methyl group be more electrophyllic? Because of induction, I would think that the carbon with two methyl groups would be more stable than the other, and so the carbon with less substituents would be more electron-poor.... So why does the water attack the more substituted carbon?

Secondly, in the image below (sorry it's turned), where the second proton transfer is taking place, you see the water molecule pulling the hydrogen off of the more substituted carbon. But wouldn't the hydrogens on the methyl group adjacent to the positive charge be way more electrophyllic, thus creating a double bond on the end of the chain?

I appreciate any help on this. Thanks for your time and effort!

Water can attack both carbocation and hydrogen atom. Sometimes it attacks C, sometimes it attacks H.

Whenever it attacks C alcohol forms so the reaction goes back to reactants. When it attacks H the product forms. We are interested in product, so we show only this path.


Prevention of rusting I'm currently learning about corrosion prevention, and have a question regarding galvanising.

Say that you have an iron nail. You galvanise it (coat it in zinc), so that is doesn't rust by preventing water and Oxygen from reaching its surface. But then wouldn't the Zinc coating rust? What would be the point of galvanising an object when its coating is going to rust anyway?

Galvanizing isn't just a protective coating in the sense that it avoids contact between the iron and the oxidant. Zinc oxidizes more easily than iron, and since both conduct electricity, you can have zinc in contact with the iron anywhere, and it will oxidize first. The term for zinc in this case is "sacrificial anode."

Full coating does help additionally though, by forming a protective layer of zinc oxide. Iron deteriorates when it rusts, so new iron metal is always being exposed as the process continues. However, some metals form oxides that protect the metal underneath from further oxidation. Zinc and aluminum are both examples of these metals.


Why does a wood fire create benzene? According to this article called
The Fireplace Delusion
a wood-fire will create benzene. This surprised me and I can't find an article online that explains the chemical reaction behind that.

This is actually just one part of a much more general phenomena, which is the release of large, polycyclic aromatic compounds due to pyrolysis or combustion of organic matter. It is not specific to wood. So, to understand why these compounds form in a wood fire, or any other kind of fire, we must first understand what wood is, and then we must understand how wood burns.

Wood is biomass - that is, matter composed primarily of carbons, hydrogens, and oxygens. We can generalize here, and talk about any biomass - wood, dead or dry grass, leaves, animal carcasses, fibers, sugar, fat, and even things like coal (which is fossilized biomass with special properties). All of these things contain cellulose or similar molecules, which consists of long chains of aromatic rings (the wikipedia article on cellulose is a good place to start, and any textbook on biochemistry will give much more detail). In the context of combustion, this is called "volatile organic matter." Volatile just means that if you heat it up, it burns, gives off heat, and turns into gas.

Living things also have non-volatile and non-organic matter. As an example, people require some trace amounts of metals for biochemical functions, so humans contain some percentage of mineral matter. In the context of combustion, this material is often called "ash." If you have a fuel like wood, this ash is what's left behind after the wood burns, and composes much of what's released into the air in the form of smoke. Some fuels can have large percentages of ash (as an example, when organic matter fossilizes and forms coal, a large amount of rock or dirt interspersed with the organic material can lead to coals with very high ash content).

When you heat up organic matter, what you're doing is dumping a large amount of thermal energy into the molecules composing that matter, and after a certain point (depends on the fuel, but we're talking about 400 or 500 C) the large chains that compose organic matter begin to break apart. If this happens in the presence of oxygen, the material will also begin to oxidize, and the process is called combustion. If it happens in the absence of oxygen, the material will gasify. As the temperature gets hotter and hotter, a soup of big organic macromolecules is formed. (When I say big, I mean, BIG - hundreds of carbon, hydrogen, and oxygen atoms, and thousands of side chains and functional groups). It is in this macromolecular soup that you can find benzene and its many, many, many cousins. (These molecules are too complex to have names, and are so large that they can't be characterized exactly - instead, chemists can only statistically describe the molecules, using techniques like nuclear magnetic resonance (NMR), mass spectrometers (MBMS), or infrared spectrometers (FTIR).) This process of solid matter turning into volatile gases is called pyrolysis.

Big polycyclic molecules form when there is not enough oxygen (if there were enough oxygen, all the carbon and hydrogen would form carbon dioxide and water). They are huge and hydrophobic, and agglomerate to form soot and tar. When you see sticky brown films on cigarettes or smokers' hands or in pipes, this is caused by these macromolecules.

If you looked very closely at a wood fire, and were able to see the actual flame-wood interface, what you would see is jets of hot, flammable gases shooting out of the pores of the wood. It is these gases that ignite - not the wood itself. (Of course, as the wood heats up, and the organic matter is burned, the mineral matter is left behind, and that becomes extremely hot but does not vaporize, so you still have radiant, red-hot material.)

You can still find benzene, even in a hot fire, because flames are not perfectly mixed, and you can get pockets of cold gas interspersed in a cloud of hot gas. If it is cold enough, benzene (which can be very stable) will not be decomposed.

So the next time you watch a campfire, or smoke a cigarette, or set your dinner on fire in the oven, you should be able to better understand what you're seeing - or know where to look.


How can I produce hydrogen gas efficiently and cheaply? I have been intrigued by one of the principles of chemistry lately- electrolysis. In my knowledge, electrolysis is a quite dangerous operation as the decomposition of water produces hydrogen and oxygen gas, both which are flammable. However, its potent to be volatile isn't what I am concerned with; rather, it is the production of these gases sufficiently and cheaply that I have been wondering about.

Thus, what is the cheapest way available right now to build an efficient electrolysis apparatus that could be used for a relatively long amounts of time? In addition to this, what are the side effects with electrolysis of water? In addition to this, could solar panels be used in the process of electrolysis? I want this electrolysis apparatus to be as independent from the grid as possible!

Electrolysis of water involves the splitting (lysis) of water molecules using electricity.

$\ce{2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)}$

From the above equation, we see that the production of hydrogen will occur twice as fast as oxygen (on a volume basis). The electrode materials must be inert (non-reactive with water or its constituents). Platinum is good but very expensive, so titanium and/or nickel alloy catalysts are sometimes used. The current flows into the anode where oxygen gas is produced, whereas hydrogen gas is produced at the cathode. Generally, the higher the current, the higher the rate of production of hydrogen gas.

Since this is not a spontaneous reaction, energy is required from an external source in order to drive it. This is provided by a direct current (DC) electrical power supply, such as a battery or solar cell. The amount of energy required can be determined from the Gibbs energy of formation of water, which is -237.2 kJ/mol (at 298.15 K or 25 °C).

The theoretical minimum voltage needed for the reaction to occur is the standard electrode potential of oxygen, which is +1.23V. In practice, a slightly higher voltage (about 1.48V) is needed because of an over-potential of about 0.25V which is used up as heat.

Pure (demineralised) water can be expensive and has very low electrical conductivity, so some impurities in the water are desirable. Salt (sodium chloride) dissolved in the water will increase its conductivity dramatically, thereby reducing the current necessary to produce a produce hydrogen at a given rate. However, the chlorine gas can be produced at the anode (instead of oxygen) which is highly corrosive and hazardous. Minerals such as magnesium or calcium (present in sea water) will form insoluble precipitants on the cathode which could build up a 'passive' layer, reducing the rate of hydrogen production. Some acids or bases such as sulfuric acid or sodium hydroxide are therefore sometimes preferred.


Mechanism of dimedone preparation From the Wikipedia article on dimedone,

  Dimedone is prepared from mesityl oxide and diethyl malonate.

I can't really figure out the mechanism for the reaction. Could someone please help me out?

You will need to start in basic conditions. Dimethyl malonate is a good nucleophile on account of the two ester groups on either side of the central carbon atom. It has a $\mathrm{p}K_\mathrm{a}$ of 12.9, which is extremely acidic for a $\ce{C-H}$ bond. The first step is a Michael addition.

The resultant enolate will equilibrate, enabling it to form a much more stable 6 membered ring. We then have a Claisen condensation.

Lastly, decarboxylation will yield dimedone.


What form of zinc's found naturally in vegetables & meat? What kind of zinc naturally occurs in plants? Zinc oxide, zinc sulphide, or something else?

The most likely form is $\ce {Zn^{2+}}$ bound inside a protein, or maybe chelated to some amino acids. It is not very likely that there is zinc oxide, or zinc sulfide, or zinc sulfate in organic matter.


Magnetic non-equivalence of acetal protons I have been using bromoacetaldehyde diethyl acetal recently and noticed something interesting about the 1H NMR. The methylene protons of the ethyl acetal are magnetically non equivalent. You see two quartets integrating for two protons with different shifts.

You can see this in the spectra here.

I have found a short description, however it is a little unclear:

  Diethyl acetals such as (1 6) and triglycerides such as triacetin (1 7) have a molecular symmetry plane and consequently are optically
  inactive. However, this plane does not bisect the connecting line of
  the geminal protons, which are magnetically non equivalent. This is
  manifested by the relative complexity of the AA'BB'X type proton
  spectrum of the glyceryl moiety of triacetin (Figure). Similar
  magnetic non-equivalence can be induced across an ester bond; e.g.,
  the two methyl groups of the isopropyl ester (18) area non-equivalent.

I don't really understand as there is no restriction of rotation. And does this mean that this would disappear using high temperature NMR?

This link on SE explains why the methylene hydrogens of the ethyl groups are different or diastereotopic. For your specific case, if a deuterium is substituted for Ha on the left side of the structure, it leads to an R-configuration while substitution of deuterium at Hb leads to an S-configuration. Both of these hydrogen sites "see" the same handedness at the methine carbon. Thus, the two hydrogens, Ha and Hb, are in different environments, they have different chemical shifts and they are coupled to each other and the vicinal methyl group. The same argument applies to the ethyl group on the right side of the structure. This difference in chemical shift is more apparent at high field (the QE300 MHz spectrum) rather than at low field (60 MHz).
There is no restricted rotation and temperature will not make a difference. Don't feel bad, these arguments were made for a monoalkyl derivative of diethyl malonate by a chemistry Nobelist!
ADDENDUM: Here is an additional link to this answer. Here is a related spectrum: Intermediate Level #12 at this link.


Conservation equation in an enzyme-catalysed reaction Consider the reaction scheme:

$$\ce{S + E ->[k_1] C1} \qquad \ce{C1 ->[k_2] E + P} \qquad \ce{S + C1 <=>[k_3][k_4] C2}$$

where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme-substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have used the law of mass reaction to convert this reaction into a system of differential equations:

\begin{align}
\frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\\
\frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2\\
\frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1
\end{align}

I am asked to find a conservation equation and use it to simplify this system however I haven't been taught as to what a conservation equation is. Can anyone push me in the correct direction? (and check the above differential equations are correct).

The reaction scheme is:

$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$

Let's define a bit of nomenclature.

S = compound S in chemical reaction equation
s = $[S]_x$, the concentration of S at t=x.
$S^*$ = ds/dt

The whole set of differential equations has 5 equations with 5 unknowns.

\begin{align} \\
S^* = \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2  \tag{4} \\ E^* = \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\tag{5} \\
C^*_1 = \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2 \tag{6}
\\ C^*_2 = \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2 \tag{7}\\ P^* = \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \tag{8}\end{align}

What is a conservation equation

In mathematical terms a conservation equation is some linear combination of the time derivatives which add up to zero on the right hand side.

The gist is that the various conservation equations allow some simplification in the various interactions in the system.

Rather than just search blindly, there are a couple of relationships that can be exploited to find time derivatives which add up to zero on the right hand side. First is a sort of mass balance. The initial concentration of the reactants has to be spread as the various intermediate species and the final product. A second line of attack is to exploit and equilibrium conditions such as equation (3).

First conservation equation from "mass balance"

The first conservation equation is found noting that at t=0 then all is [S], a constant. Noting that it takes 2 $\ce{S}$ to make one $\ce{C_2}$, at some later time t=x, thus:

$$[\ce{S}]_{t=0} = [\ce{S}]_{t=x} +  [\ce{C_1}]_{t=x} + 2[\ce{C_2}]_{t=x} + [\ce{P}]_{t=x}\tag{9} $$.

Now for some intermediate time we can take derivative of the mass balance equation with respect to time. All the dt's get cumbersome so lets use just $S^*$ from now on to mean ds/dt. From the mass equation then since the initial concentration is a constant we get

$0 = S^* + C^*_1 + 2C^*_2 + P^*\tag{10} $

Most likely solving for either $S^*$, or $P^*$ would give an interesting result.

A second conservation equation from mass balance

The second conservation equation comes from noting the enzyme participation in the reaction.

$$[\ce{E}]_{t=0} = [\ce{E}]_{t=x} +  [\ce{C_1}]_{t=x} + [\ce{C_2}]_{t=x}\tag{11}$$.

Taking the derivative we get:

$ 0 = E^* + C^*_1 + C^*_2 \tag{12}$

which can be rearranged to

$E^* = - C^*_1 - C^*_2\tag{13}$

Substituting equation 6 for $C^*_1$ and equation 7 for $C^*_2$ this gives us equation 5 again which isn't helpful.

A third conservation equation from the equilibrium condition

From the third reaction we can derive the equation.

$ \frac{k_3}{k_4}= \frac{[S]_x[C_1]_x}{[C_4]_x} = \frac{sc_1}{c_4}\tag{14}$

Taking the derivative:
$ 0 = \frac{c_1}{c_4}S^* + \frac{s}{c_4}C^*_1 - \frac{sc_1}{c_4^2}C^*_4\tag{15}$

rearranging

$  S^* = -\frac{s}{c_1}C^*_1 + \frac{s}{c_4}C^*_4\tag{16}$

Infinite time as a boundary condition

Obviously at $t=\infty$, all the S has been converted to all P which is a constant. This factoid isn't very useful though in predicting how long that an "effective infinite time would be. Rather it is just a boundary condition.

$$[\ce{P}]_{t=\infty} = [\ce{S}]_{t=0}\tag{17}$$


Meaning of hydrogen bonds for life Why are hydrogen bonds so important for life? Or more generally asked: Why is water so important for life / biological systems? Or maybe again in other words: Why does especially water provide this special property? And not a single other substitute(?) ?
I can read what the hydrogen bonds cause to happen but I would rather like to know: Why is that in detail or what makes so water so much different from any other (similar?) liquid?

I know this touches heavily biology (or also physics) but I think I will find the reason in an chemical explanation?

Water is (almost) everywhere on earth in copious amounts. It is liquid at earth conditions (H-bridges!). It's rather non-toxic (or organisms were able to evolve to tolerate water well). It is a good solvent for the molecules involved in energy metabolism and biostructure ana- and catabolism. Water has neutral pH, it happily donates and accepts protons which makes it ideal for participation in metabolism.

I'd broaden the question, asking for other molecules that could support life as we know it in a similar way like water does it and also check other liquid molecules which are available in large quantities in the universe (e.g. ammonia, nitrogen, hydrogen) if they could support other forms of life that depend on analogous chemical processes, or even go beyond that (silicon- and aluminum oxides, etc.)


Why is cyclooctateraene shaped like a "tub" rather than a "chair"? I understand that for cyclooctatetraene (COT) to "escape" the horrible prospect of becoming an anti-aromatic molecule, it must adopt a non-planar conformation.

The widely touted conformation of COT is said to be "tub" shaped. My sources (random links Google threw up) also concur on this particular "tub" conformation.

But why isn't it "chair" shaped instead?

Going along the same line of cyclohexane (non-aromatic), wouldn't a "chair" conformation reduce (even if it's a little bit) the steric repulsions due to crowding of hydrogens atoms on one side/face of the molecule?

Or even if "avoiding steric issues" isn't really the big idea here, then I could re-phrase my question as:

What stops COT from assuming a "chair" shaped structure?

The "chair" structure that I have in mind for COT is something like this:

(I don't have access to a decent enough organic chemistry sketcher to have this drawn any better, sorry.)

It's all about the 3D structure of double bonds.

If we look at the tub form we see that all dihedral C-C=C-C angles are 0°. the C=C-C angles are 125°, also pretty close to the optimal value. Everything works out fine, there's basically no strain on the whole molecule.

I tried to build the chair form of all cis, but every optimization ended up in either the all cis tube form or the cis-trans-cis-trans chair form. The best structure that looks like the all cis chair form I ended up with is this:

First of all, it's pretty flat, so antiaromaticity would again play a role and the dihedral angles are also ~33° for the double bonds in front and in the back, which is energetically quite bad and unstable.

The cis-trans-cis-trans compound is definitely no monstrosity. I'm working a lot with trans-cyclooctenes and I don't see why this shouldn't be possible if a trans,trans-cycloocta-1,5-diene is possible.

Here we've got 0° for the dihedral angles for the cis bonds and 125° for the trans. Yes, this is far away from the optimal value of 180° but quite similar to what you expect from trans-cyclooctenes (around 135°). I feel like it would be quite reactive and unstable but it could definitely exist. There's just no good way to produce it I guess.

So overall: all angles in the tub form are much closer to the optimal values than in either chair form and the all cis form is quite planar.


Hydrogen's confusing purple glow While reading about hydrogen on Wikipedia, I read that hydrogen has a purple glow in its plasma state. This confused me because although hydrogen is present in its plasma state on the sun, the sun doesn't appear purple to us (rather, it is yellow-orange). Why is this the case?

The radiation of hydrogen in plasma with low density comes from discrete energy levels, resulting in spectral lines. Those spectral lines can be observed in sunlight, too.

But the sun consists of a plasma with a much higher density. Most photons are the result of a long series of energy exchange inside the sun following the emission of high energy photons and kinetic energy of the nuclear processes. They are part of a continous spectrum as there is not a discrete number of energy levels at the interactions in this high density plasma.

In the photosphere, some of them are absorbed again and result in negative spectral lines (Fraunhofer lines), but the main portion of light remains a continuous spectrum.


Electrophilic attack of X+ on double bond Is it the double bond attacking the $\ce{X+}$ ion or the other way around?

Also, does it form a cyclic transition state if it isn't bromine or chlorine?

Eg. cyclohexene $+ \,\ce{Cl+ ->}$ cyclic transition state

But would the same thing happen if I put any other electrophile?

These kind of reactions are called Electrophilic addition reactions. These are usually given by alkenes and alkynes. In your example, the electrophile is $\ce{Cl+}$. The alkene will act as a source of π electrons and will donate it to the electrophile which is in need of electrons. So, alkene is called as nucleophile in this reaction.

  Is it the double bond attacking the X+ ion or the other way around?

It depends on how you look at it. We can say that alkene polarizes the $\ce{Cl-Cl}$ bond by pushing electrons into the anti-bonding molecular orbital. This weakens the bond and hence electrophile is getting generated. In the next step,the electrophile attacks the double bond [or double bond attacks the electrophile] and hence electrophile gets added across across the double bond

Why does the 3-membered cyclic transition state form?

The cyclic transition state is just a temporary relief for the electrophile that's just been added. The π molecular orbital donates electrons to the electrophile.

While the p-orbitals in the π-bond overlap with the electrophile, each p-orbital forms a σ-bond.

  But would the same thing happen if I put any other electrophile?

Generally,electrophilic addition reactions take place through carbocation intermediate.
But this reaction can take place through cyclic 3-membered intermediate,if the electrophile has a lone pair.

Apart from $\ce{Cl+,Br+}$ even $\ce{Hg^2+}$ (when the reagent is $\ce{Hg(OAc)2/H3O+}$) and $\ce{NO+}$ (when the reagent is $\ce{NOCl}$) can form the 3-membered cyclic intermediate.


Hellmann-Feynman Forces with Hartree-Fock The energy in the Hartree-Fock approximation is given as:

$$E_{HF}=\left<\psi_{HF} \left| \hat{H} \right| \psi_{HF} \right>=\sum_{i,j}P_{i,j}H_{i,j}^{core}+\frac{1}{2}\sum_{i,j,k,l}P_{i,j}P_{l,k}(ij||kl)+V_{NN} \tag{1}$$

The geometrical derivative of the Hartree-Fock energy can be shown to be [1]:

$$\frac{\partial E_{HF}}{\partial X_A}=\sum_{i,j}P_{i,j}\frac{\partial H_{i,j}^{core}}{\partial X_A}+\frac{1}{2}\sum_{i,j,k,l}P_{i,j}P_{l,k}\frac{\partial (ij||kl)}{\partial X_A}-\sum_{i,j}Q_{i,j}\frac{\partial S_{i,j}}{\partial X_A}+\frac{\partial V_{NN}}{\partial X_A} \tag{2}$$

The Hellmann-Feynman Theorem states:

$$\frac{dE}{d X_A}=\left<\psi \left| \frac{d\hat{H}}{d X_A} \right| \psi \right>$$

The Hellmann-Feynman Theorem implies that the energy derivative only depends on the parts of the Hamiltonian that have a dependency on the derivative. This leads to the geometrical derivative being equal to:

$$\frac{dE}{d X_A}=\left<\psi \left| \frac{d\hat{V}_{NN}}{d X_A} + \frac{d\hat{V}_{Ne}}{d X_A} \right| \psi \right> = \left<\psi \left| \frac{d\hat{V}_{NN}}{d X_A} \right| \psi \right> + \left<\psi \left| \frac{d\hat{V}_{Ne}}{d X_A} \right| \psi \right>=\left<\psi \left| Z_A\sum_B\frac{Z_B(X_B-X_A)}{R^3_{AB}} \right| \psi \right> + \left<\psi \left| -Z_A\sum_i\frac{X_i-X_A}{r^3_{iA}} \right| \psi \right>$$

Now the Hellmann-Feynman theorem can be applied to find the derivative geometrical of the Hartree-Fock energy:

$$\frac{\partial E_{HF,\ce{Hellmann-Feynman}}}{\partial X_A}=\left<\psi_{HF} \left| \frac{\partial\hat{H}}{\partial X_A} \right| \psi_{HF} \right>=\sum_{i,j}P_{i,j}\frac{\partial V_{Ne,ij}}{\partial X_A}+\frac{\partial V_{NN}}{\partial X_A} \tag{3}$$

I tried to implement equation $(2)$ and equation $(3)$, and used them to calculate, the force between H and Li for different distances, with cc-pVDZ and cc-pVTZ. and got the following:

Hartree-Fock refers to equation $(2)$ and Hellmann-Feynman refers to equation $(3)$. I have read that the Hellmann-Feynman theorem is only valid for exact solutions, but should become a better and better approximation, when going towards the Hartree-Fock limit. As can be seen in the picture, equation $(3)$ performs worse with the larger basisset cc-pVTZ. This now leads me to my question, were am I wrong in my understanding of the application of the Hellmann-Feynman theorem to the Hartree-Fock approximation?

[1] Attila Szabo, Neil S. Ostlund; Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; equation C. 12

It is not at all surprising that the forces you get from application of the Hellmann-Feynman theorem do not match the true force. The reason is the appearance of Pulay forces: the atomic orbital basis set depends on the nuclear coordinates, and changing the position of the nuclei also changes the basis set. This yields the contribution you're missing in the plots.

cc-pVDZ and cc-pVTZ are still pretty small basis sets. I would expect that if you add diffuse functions, the result improves quite a bit, since the basis set covers more degrees of freedom. Furthermore, to get something meaningful you'd have to compare e.g. cc-pV5Z vs cc-pV6Z, which are closer to the complete basis set limit.

However, already in your plots you can see that the agreement is better from cc-pVDZ to cc-pVTZ when you look at the far-range behavior. A better way to plot these would be to plot the difference, i.e. the Pulay force.

I would also imagine that decontracting the basis sets helps, because if you imagine just moving the electron density but not the basis functions, then you will see that the description of the core orbitals is impoverished quite a bit as the contraction coefficients are wrong.

Actually, the result in cc-pV6Z is probably still not very close to the Hellman-Feynman limit. When you move the nucleus, 1s core orbitals start to gather p, f, etc character. Thus, when you move the nuclei but not the basis functions, your original basis set should have e.g. very tight p functions that are able to describe this effect.

If you want to toy around with this, you should look at calculations in larger and larger primitive basis sets. Even-tempered exponents should work fine. Start with a small set of s functions, and study how the result changes when you add steeper or more diffuse functions. Next, look at the p shell: which exponent changes the result the most? And then again start adding steeper and more diffuse functions. Repeat ad nauseam. (This is called completeness-optimization, and I've published an automatic procedure for this.)