What is Meant by "Eluting with 1-2% EtOAc/pentane" in Terms of Flash Chromatography? I am reading the experimental section of a paper, and it states

  [compound of interest] Isolated as a pale yellow oil after flash chromatography on silica gel, eluting with 1-2% EtOAc/pentane...

This is all this paper states in regards to the work-up procedure. I fully understand everything in this section except what is meant by 1-2% EtOAc/pentane. Does this mean that in total only 1-2% by volume of ethyl acetate is used compared to the amount of pentane used in the flash chromatography? This could also mean that a solution of 1-2% ethyl acetate in pentane was used throughout the entire run (which I find most likely). Which of these interpretations is most likely? Would this mean the author of the paper did not change the polarity of the solvent during chromatography and only use a 1-2% EtOAc in pentane solution?

Here is the DOI to the paper: 10.1039/c6ob01026a

Based on OP's question, I tentatively assumed OP is a novice to the isolation of organic compounds using column chromatography. Thus I'd explain this question with as basis as possible.

This paper OP sighted is for the synthesis of 2$H$-chromenes (2$H$-benzopyrans). They can be non-polar to polar compounds based on the substitution on benzene nucleus. The paper describes a series of relatively non-polar chromenes. When you do organic reaction, you monitor the progress of reaction with TLC analysis. During that analysis, you can determine that what solvent should be the most suitable for the particular compound elution. That's how the authours have determined several solvents such as 1% ethyl ether/hexanes and 2% ethyl acetate/pentanes, etc. for different different chromenes.

Some of these chromenes have been isolated using just one solvent. For example, evidently, they have mentioned isolation of 7-Methoxy-4-methyl-2-propyl-2$H$-chromene, using flash chromatography on silica gel, eluting with 2% diethyl ether/hexanes solvent system (this means solvent made by 2 parts of diethyl ether and 98 parts of hexanes, by volume). Note that it would also mean that they might have used certain volumes of following systems: (1) pure hexanes; (2) 1% diethyl ether/hexanes; (3) 2% diethyl ether/hexanes; etc. However, expected compound has come out with only fractions eluted with 2% diethyl ether/hexanes (remember, you supposed to analyze all the fraction by TLC), which may have combined and further analyzed by spectroscopy to confirm the structure.

As in another data set, they have mentioned isolation of 5,7-Dimethoxy-2,2-dimethyl-2$H$-chromene, using flash silica gel column, eluting with 1-2% EtOAc/pentane solvent system.It would definitely mean that they should have used certain volumes of few solvent systems: (1) pure pentane; (2) 1% EtOAc/pentane; (3) 2% EtOAc/pentane; (4) 3% EtOAc/pentane; etc. However, their expected compound has come out with fractions eluted with 1% and 2% EtOAc/pentane (concluded by the fractions analysis by TLC).

Later Note: As commented by Waylander, these type of separation is called gradient elution. By the scale of reaction, total elution must not go over $\pu{100 mL}$ of solvent for each compound.


Chemical potential of mixtures of ideal gases and the Gibbs-Duhem equation As is well known for a constant composition system, $\mu$ (symbolizing chemical potential) is equal to the molar Gibbs energy. Using
$$
\mathrm{d}G = V\mathrm{d}p-S\mathrm{d}T,
$$
we can write
$$
\mu = \mu^{0} + RT\ln P.
$$
But for mixtures,
$$
\mathrm{d}G = V\mathrm{d}p - S\mathrm{d}t + \mu_{\ce{A}} \mathrm{d}n_{\ce{A}} + \mu_{\ce{B}} \mathrm{d}n_{\ce{B}}
$$
Can the original formula for chemical potential in the case of mixtures?

Moreover, using the Gibbs-Duhem equation,
$$
n_{\ce{A}} \mathrm{d}\mu_{\ce{A}} + n_{\ce{B}} \mathrm{d}\mu_{\ce{B}} = 0,
$$
which means that $\mu_{\ce{A}}$ changes upon adding B to the mixture, even when the partial pressure of A is constant. This contradicts the formula.

The chemical potential of a the two species in your ideal gas mixture are $$\mu_A=\mu^0_A(T)+RT\ln\left(P\frac{n_A}{(n_A+n_B)}\right)$$
$$\mu_B=\mu^0_B(T)+RT\ln\left(P\frac{n_B}{(n_A+n_B)}\right)$$
and the total free energy of the mixture is:
$$G=n_A\mu_A+n_B\mu_B$$So, the chemical potential of A depends on the number of moles of B and the chemical potential of B depends on the number of moles of A.  Do these equations satisfy the Gibbs-Duhem equation?  Try them out and see.


Why is use of calcium carbide as an artificial fruit ripening agent banned? Fruits like mangoes can be ripened using agents like ethylene, acetylene or calcium carbide. Of these, the latter is dangerous and is banned in most countries. Wikipedia says it is because calcium carbide has traces of arsenic and phosphorus. Is there any other reason?

Will it still be unsafe if pure calcium carbide is used?

According to Anwar et al. the use of $\ce{CaC2}$ is discouraged due to

  dangers of explosion and carryover of toxic materials like arsenic and phosphorus to consumers, thus making the healthy fruit poisonous. Since no technical knowledge is considered necessary for its anomalous use, higher quantity of calcium carbide needed to ripen immature fruit, makes them
  tasteless.

For references and more information, see: Anwar, R., A.U. Malik, M. Amin, A. Jabbar and B.A. Saleem, Int. J. Agri. Biol. 2008, 10, 35–41. Available at researchgate.net.


Specific conductance of 0.1 M acetic acid Find specific conductance of $\pu{0.1 M}$ acetic acid given that its molar conductivity is $\pu{4.6 S * cm2 * mol^{-1}}$.

I used the formula $\Lambda_m = \kappa/c$, where $\Lambda_m$ is the molar conductivity, $\kappa$ is the specific conductance and $c$ is the concentration of the electrolyte in $\pu{mol/L}$.

What's confusing me is the value of $c$. At first I thought it should be the given molarity, but according to my text book $c$ is the concentration of the electrolyte. Now the acetic acid molecule does not conduct electricity, rather its constituent ions do. So shouldn't the value of c be the concentration of its ions which turns out to be $\pu{\sqrt{1.8} \times 10^{-3} mol/L}$ (using the equation of the acidic dissociation of acetic acid)?

Also the answer given was $\pu{4.6 \times 10^{-4} S * cm^{-1}}$.

With the data provided with your problem I suspect you are only supposed to carefully use provided formula with the corresponding units of measurements. You need to convert the concentration from $[\pu{mol * L^{-1}}]$ to $[\pu{mol * cm^{-3}}]$ considering $\pu{1 L = 10^{3} cm3}$:

$$\kappa=\Lambda_m \times c = \pu{4.6 S * cm2 * mol^{-1}} \times \pu{0.1 * 10^{-3} mol * cm^{-3}} = \pu{4.6e-4 S*cm^{-1}}$$


"Hamiltonian operator has no effect on the spin function" what does it mean? I have read the Levine Quantum Chemistry book and it says "Hamiltonian operator has no effect on the spin function" in chapter 10 (Electron Spin and the Spin-Statistics Theorem)  and does the Hamiltonian operation like the following. Why the g(ms) state is taken out of the Hamiltonian?? Why is it constant with respect to the Hamiltonian?

$$\hat H[\psi(x,y,z)g(m_s)] = g(m_s)\hat H\psi(x,y,z) = E[\psi(x,y,z)g(m_s)]$$

Because the operator $\mathrm d/\mathrm dx$ only acts on $x$ and not on $y$, we can write

$$\frac{\mathrm d}{\mathrm dx}[f(x)g(y)] = g(y) \left[\frac{\mathrm d}{\mathrm dx}f(x)\right]$$

Likewise, in this context, the Hamiltonian operator $\hat{H}$ only acts on $(x,y,z)$ and not $m_s$. This is explained by Levine1 in the paragraph immediately preceding this

  To a very good approximation, the Hamiltonian operator for a system of electrons does not involve the spin variables but is a function only of spatial coordinates and derivatives with respect to spatial coordinates [...]

Is there something about this paragraph which you don't understand? This is pretty much all there is to be said on the topic.

Levine, I. N. Quantum Chemistry (7th ed.), p 268


Combustion that generates no gas? I'm learning on my own with Chemistry The Central Science, 13th ed. book. So far it's going well.

I've come across something I don't think I understand. I've googled and "stackechanged" with no avail.

How come a combustion generates no gas products? And in what circumstances?
This is the bit of the book that left me wondering:

For a noob like me, it's counterintuitive that something "burns brilliantly in air" and only generate solid products.

There are many reactions that do this. The simplest way is to substitute any other reactive metal for magnesium: aluminium, lithium, sodium, etc. All will burn in oxygen to give solid products / oxides.

This can be extended by using any of these metals and other reactive non metals, such as fluorine, chlorine, etc. in Group VII or sulphur, etc. in Group VI. Do not try it with $\ce{Se}$ or $\ce{Te}$ in combination with aluminium, however, as it is near-explosive and forms toxic gas on hydrolysis.

A simple experiment that does this is the reaction of sulphur with iron filings.


What is the chemical formula(s) for liquid potassium-based soap? Many places say that liquid soap is generally potassium-based, while the solid stuff is sodium-based.
But, while I have found several sites saying that solid soap is usually sodium stearate, I can't find any place that gives a specific name(s) to the chemical(s) in liquid potassium soap...

You'll get an answer quickly if you search for "formulation liquid soap". There are for instance DIY guides on making liquid soap starting from natural oils (sunflower, coconut) by saponifying the oil with KOH. The fatty acid composition of sunflower oil can vary significantly depending on breed (see e.g. the Wikipedia) but generally contains mono- and polyunsaturated FAs in large proportion. Unsaturation contributes to the fluidity of the product. It is also worthwhile to check common nomenclature for FA chain length (e.g. see the Wikipedia). Sunflower oil contains palmitic (C16:0) and stearic (C18:0) acids as principal saturated FAs. Coconut oil contains shorter (lauric, C12:0) and mainly saturated FAs. Mainly because of the extent of saturation, coconut oil has a higher melting point and is used as the minor component.
In addition there are plenty of commercial vendors willing to provide recipes you can follow to make your own liquid soap. Note not all require a potassium salt. For instance, the company Stepan provides a recipe which consists of mixing the following (in decreasing order of weight %) to form one phase (which is then mixed with another lacking detergents):

Deionized Water
BIO-TERGE® AS-40 (or BIO-TERGE® AS-40 HP)  ==  SODIUM C14-16 OLEFIN SULFONATE
STEPANOL® AM  == AMMONIUM LAURYL SULFATE
STEPANOL® WAT == TEA LAURYL SULFATE
NINOL® M10  == COCAMIDE MIPA

Admittedly ammonium counterion (and TEA) is like potassium in that they are larger and softer counterions than sodium, so similar in their effect on the detergent.


Why is AlF3 soluble in anhydrous HF only in the presence of KF? I am pondering over that what must be the reason that $\ce{AlF3}$ is insoluble or less soluble in anhydrous $\ce{HF}$? And pageants it is soluble to a good extent only in the presence of $\ce{KF}$? Is there any other reagent in presence of which $\ce{AlF3}$ is soluble in $\ce{HF}$?

I guess a probable answer for the cause of insolubility could be:
"the nonavailability of $\ce{F-}$ ions because of the presence of intermolecular hydrogen bonding in $\ce{HF}$, which makes $\ce{AlF3}$ insoluble in $\ce{HF}$."

  Can someone elaborate the cause and mention other such reagents like $\ce{KF}$ which can be used for the same purpose?

$\ce{AlF3}$ is insoluble in the anhydrous $\ce{HF}$ because the $\ce{F-}$ ions are not available in intermolecular hydrogen bonded $\ce{HF}$, but it becomes soluble in the presence of $\ce{KF}$ due to the formation of soluble complex, $\ce{K3[AlF6]}$.
$$\ce{AlF3 + 3KF -> K3[AlF6]}$$


How to calculate pH of solution made of H3PO4 and Na3PO4? I am unable to calculate the $\mathrm{pH}$ of a solution made of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05 M}$ $\ce{Na3PO4}$.

How to set the approximation and which ions are present at the equilibrium?

You have the the following six independent equations in order to specify the unknown concentrations of six species as present in an aqueous solution consisting of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05M}$ $\ce{Na3PO4}$:

\begin{align}
K_1 &= \frac{[\ce{H+}][\ce{H2PO4-}]}{[\ce{H3PO4}]}
     = 7.5 \times 10^{−3} \tag{1}\\
K_2 &= \frac{[\ce{H+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}
     = 6.2 \times 10^{−8} \tag{2}\\
K_3 &= \frac{[\ce{H+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]}
     = 3.6 \times 10^{−13} \tag{3}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{4}\\
0.15
    &=[\ce{PO4^3-}] + [\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{H3PO4}]\tag{5}\\
[\ce{H+}]
    &= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + 2[\ce{HPO4-}] + [\ce{OH-}] -0.15\tag{6}
\end{align}


Atom-condensed softness matrix In the ACKS2 polarizable force field paper, I found a thing called the atom-condensed softness matrix. In another paper, I found this expression for it:

  $$
\chi_{kl} = 2 \sum_{i}^{\text{occ MOs}} \sum_{j}^{\text{unocc MOs}} \frac{\langle\psi_{i}|g_{k}|\psi_{j}\rangle\langle\psi_{j}|g_{l}|\psi_{i}\rangle}{\epsilon_{i} - \epsilon_{j}} \delta_{\sigma_{i}\sigma_{j}},
$$
  where

  $\epsilon_{i}$: orbital energy of the $i$th KS orbital
  $\chi$: (non-interacting) response matrix
  $\psi_{i}$: spatial orbital
  $g_{i}$: potential basis function

What is the physical meaning of it, or at least what information we can get from this matrix?
If in KS-DFT we consider the system as non-interacting, why do we consider interaction between two species in this equation? (if I understand it right - between molecules $i$ and $j$)

Your first misconception is what i and j mean. These are indices of molecular orbitals.

A response matrix is something used in respone theory. It stems from Perturbation Theory.

Response Theory is a mathematical formalism to compute time-dependent molecular properties (in theoretical chemistry, that is).

This pdf (no guarantee for anything) describes it pretty well and concise.
http://www.lct.jussieu.fr/pagesperso/toulouse/enseignement/molecular_properties.pdf


Recycle unreacted ethene The question is this:

  You want to make oxirane from ethene and oxygen gas:
  $$\ce{2 C2H4 + O2 -> 2 C2H4O}$$
  The flow into the reactor is constant at $\pu{200kmol/h}$ of $\ce{C2H4}$ and $\pu{90kmol/h}$ of $\ce{O2}$. Only 6% of the ethene reacts in the reactor. 98% of the unreacted ethene flowing out of the reactor is filtrated and recycled (sent back into the reactor). Oxygen and oxirane passes through the filtrator.

  Calculate the flow of gas out of the filtrator:

  with no recycling, and
  with recycling.

There has to be something wrong with the question (or my translation of it, sorry) because given this information the net amount of ethene in this system is just going to increase.

As much as I understand, there is nothing wrong with the question, if your translation is correct. Let's see the reaction:

$$\ce{2 C2H4 (g) + O2 (g) -> 2 C2H4O (g)}$$

The flow reactor was fed with constant flow rates of $\ce{C2H4}$ at $\pu{200 kmol\:h-1}$ and $\ce{O2}$ at $\pu{90 kmol\:h-1}$. Only $6\%$ of the ethene reacts in the reactor. That means the reaction rate of the reactor is $\pu{12 kmol\:h-1}$ per $\ce{C2H4}$. Therefore, assuming a perfect world, $\pu{6 kmol}$ of $\ce{O2}$ would be consumed in each hour to produce $\pu{12 kmol}$ of $\ce{C2H4O}$ as well (see the chemical equation).

a) Thus, the gas outlet of the reactor would release 98% of unreacted $\ce{C2H4}$ ($0.98 \times \pu{188 kmol}$) and $\ce{O2}$ ($\pu{84 kmol}$), and $\ce{C2H4O}$ ($\pu{12 kmol}$) in each hour as a mixture. The sum ($188 \times 0.98+84+12 \approx \pu{280.2 kmol\:h-1}$) is the rate of gas flow out of filtrator of the flow reactor. Of course, this is assuming filtrator does not filtered off any of these materials (it wasn't clear on the translation). If filtrator separates the product and oxygen from the mixture (e.g., by gas partition by difference in melting points; melting points of oxirane and ethene are ~$\pu{161 K}$ and ~$\pu{104 K}$, respectively, and melting point of oxygen is ~$\pu{54.4 K}$), the outlet rate of $\ce{C2H4}$ out of the filtrator would be $\pu{188 \times 0.98 kmol\:h-1} \approx \pu{184.2 kmol\:h-1}$ (with no recycling of $\ce{C2H4}$). If  the flow of gas out of the filtrator means only the gases pass through the filtrator, that should be only oxygen and oxirane. Thus, the outlet rate of gases out of the filtrator would be $\pu{(84 + 12) kmol\:h-1} = \pu{96 kmol\:h-1}$.

b) I believe, the recycling of $\ce{C2H4}$ means unreacted $\ce{C2H4}$ is adding to original rate of $\pu{200 kmol\:h-1}$. Thus, rate of $\ce{C2H4}$ fed into reactor after first hour is: $\pu{(200+184.2) kmol\:h-1} = \pu{384.2 kmol\:h-1}$. Only 6% of it ($\pu{23.05 kmol\:h-1}$) reacts with $\pu{11.53 kmol\:h-1}$ of oxygen to give $\pu{23.05 kmol\:h-1}$ of oxirane. Therefore, outlet rate is:

$$\left((384.2 - 23.05) + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{462.67 kmol \: h-1}$$

This would be the outlet rate without filtration. If you have to consider only 98% of $\ce{C2H4}$, then the rate would be:

$$\left((384.2 - 23.05)\times 0.98 + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{455.45 kmol \: h-1}.$$


Can Ice Packs Freeze Water at 21 °C? I receive $120~\mathrm{mL}$ bags of medicine in sterile water. The medicine is refrigerated and then packed with a significant amount of ice packs in an insulated bag. Generally three times the volume of all the medicine. When I remove the ice packs, they are still solid two hours after being packed. So, they are very cold.

Is there enough energy to freeze the water directly in contact with the ice packs when packed and then melt immediately? The air temperature is around $21~\mathrm{^\circ C}$ at the time.

It really depends on the initial temperature of the ice packs and their composition. If they start at a temperature far below the freezing point of water, they could conceivably freeze the water as their container comes to thermal equilibrium. That said, they’re going to have to be quite cold.

To cool by $1°\mathrm{C}$, liquid water requires the removal of roughly double the amount of energy that ice would give up in moving up $1°\mathrm{C}$ (per gram.) If your bags of medicine started at $5°\mathrm{C}$, (stored in a refrigerator) then the ice will warm by about $3°\mathrm{C}$ as it brings the medicine down to its temperature because it has three times the mass and gives up half as much heat per degree gram.

Once the water has reached $0°\mathrm{C}$, it will begin freezing. For every gram of frozen water, about $160~\mathrm{g}$ of ice will warm by $1°\mathrm{C}$. The entire mass of ice will be enough to freeze ~$2.3~\mathrm{g}$ of water for each degree it is below freezing.

Assuming the ice packs came from a normal freezer, their starting temperature is about $−20°\mathrm{C}$. They would warm to about $−17°\mathrm{C}$ in bringing the water to its freezing point. From there the ice could then freeze about $40~\mathrm{g}$ of the medicine before it all came to equilibrium.

Note: There are significant assumptions (notably, that the heat capacity of the ice pack is equal to that of ice and that its melting point is equal to $0°\mathrm{C}$) and some healthy rounding here. If the ice packs melted at a temperature significantly lower than $0°\mathrm{C}$, their heat of fusion could be sufficient to freeze the water, given time.


Naming this organic compound What’s the name of the following compound?

$$\ce{CH3C(p-ClC6H4)2CH(Br)CH3}$$

My teacher said that she strongly thinks( with another teacher) that it should be 3-bromo-2,2-bis-(4-chlorophenyl)butane, though many other books and other students (similar to me) stated that it should be 2-bromo-3,3-bis-(4-chlorophenyl)butane.

She asked us try to find it out. Any organic chemistry experts out there?

I think your teacher is correct, numbering of carbons is done acording to the lowest locant rule.

The red numbering gives you : 2, 3, 3
The green numbering gives you : 2, 2, 3
We see that the first point of difference between the two numberings occurs at second number, so the green numbering is preferred.
Consequently, the correct name should be : 3-Bromo-2,2-bis(4-chlorophenyl)butane


Why does multipying a scalar in both sides of a balanced chemical equation, change related quantities? Stoichiometric coefficients appearing in a balanced chemical equation are unrelated to the actual concentration of reactant/products at any stage of a chemical reaction. They give information only about the ratio in which the reactants and products are consumed and produced. Multiplying by a scalar on both sides of the equation doesn't change that ratio.

Then, why does doing so result in changes in physical quantities related to the reaction, like equilibrium constant or Gibbs energy?

Maybe because the equilibrium constant, Gibbs energy, enthalpy, etc. are not the fundamental characteristics you think they are.

Really, what does enthalpy mean, for example? Say, a reaction $\ce{A + B -> C + D}$ has certain enthalpy. This means that running a reaction with 1 mole of A and 1 mole of B gives us certain amount of heat. Now let's multiply everything (including coefficients and enthalpy) by 2. That would be $\ce{2A + 2B -> 2C + 2D}$. In other terms, running a reaction with 2 moles of A and 2 moles of B gives us twice as much heat, just like we would expect.

Now think of the equilibrium constant. Say, we have a reaction $\ce{2NO2<->N2O4}$ and the constant just happens to be 4. This means ${\ce{[NO2]^2\over[N2O4]}}=4$. OK, and what if we write this reaction in a different way, like $\ce{4NO2<->2N2O4}$? Obviously, the equilibrium concentrations are still the same, but the expression for the equilibrium constant has changed to ${\ce{[NO2]^4\over[N2O4]^2}}$. That's a square of the previous expression, hence its numeric value would be $4^2=16$. So what? It's not like we can just up and measure this constant with some kind of "constantometer", anyway. It's the real concentrations that matter; and if we calculate the concentrations using either version of the constant, they will turn out the same.

This is also in accordance with $\ln K = {\Delta G\over RT}$. When you square the constant, its logarithm doubles, and so does $\Delta G$.


Finding Ka of a monoprotic weak acid from pH and concentration of added sodium salt A solution of $\pu{0.36 M}$ of the sodium salt of a generic monoprotic weak acid $\ce{HA}$ has a $\mathrm{pH}$ value of $8.782.$ $K_\mathrm{a} = ?$

Why doesn't the following formula work in this case:

$$K_\mathrm{a} = K_\mathrm{w} \frac{[\ce{A-}]}{[\ce{OH-}]^2}?$$

I don't know why shouldn't it work, this is my approach, leading to final result, what's the fallacy?

Consider the reaction
$$\ce{A-} + \ce{H2O} \ce{<=>} HA + \ce{OH-}$$
It can be broken into two steps
$$\ce{A-} + \ce{H+} \ce{<=>} HA$$

$K = \dfrac{1}{K_a}$
$$\ce{H2O} \ce{<=>} \ce{H+} + \ce{OH-}$$

$K = K_w = 10^{-14}$ at 25°C

If h is hydrolysis constant, approximating $h\ll 1$, then [$\ce{A-}$] = c(1 - h) $\approx$ c and [$\ce{OH-}$] = [$\ce{HA}$] = ch.

$\therefore$, $K_h = K_b = \dfrac{K_w}{K_a} = ch^2$.

$\implies K_a = \dfrac{K_w}{ch^2} = K_\mathrm{a} = K_\mathrm{w} \frac{[\ce{A-}]}{[\ce{OH-}]^2}$

$h = \sqrt{\dfrac{K_w}{K_{a}.c}}$

$[\ce{OH-}] = ch = \sqrt{\dfrac{K_{w}.c}{K_a}}$

pH = 7 + $\dfrac{1}{2} log {c} + \dfrac{1}{2} log {pK_a}$

Which is a well known result...


Boiling/melting points of cyclohexane and cycloheptane I’m looking at a chart in my book, and while cycloheptane has a higher boiling point than cyclohexane like I’d expect, cycloheptane has a melting point of −12 °C, while cyclohexane has a melting point of 6.6 °C. Why does it have a lower melting point only?

There are not many reputable sources on the matter (as I could find) since it seems that cycloheptane isn't the most popular chemical around (poor guy) but I will present you with the best explanation/argument/insight into the situation as I can.

Note that it would be solely my own interpretation, and a case of explaining results that we see, rather than formulating an explanation to encompass natural phenomena without observing the results prior.

The key ingredient that I noticed while looking through charts and whatnot was bonding which is connected to sterics. More on that later.

Melting and Boiling Points of Cycloalkanes

Above is a chart of the melting and boiling points of cycloalkanes. As expected, boiling point increases linearly. That is lovely! Thus, we can explain boiling point with natural intuition, similar to regular alkanes. More is more! If only life was so simple.

Now let's take a look at melting points. HOLY! Cyclobutane? Why you so high? (it is actually an error by the book. Nonetheless, it is higher by a little than cyclopentane and that is the point) And wow, is cyclohexane a monster, past the melting point of our dearest water. Something is wrong with this universe. It must be the martians.

But no. It is not the martians. At least I do not believe so. The answer, I believe, is sterics.

Sterics? you ask. Are you trying to say that cyclobutane bonds more structured than cyclopentane and cyclohexane bonds more structured than cyclohexane?

Precisely.

Although you may not expect much difference from cyclobutane and cyclopentane, cyclobutane is very close to a right angle (perfect square) and this would allow tight packing. Cyclopentane, on the other hand, is normally as an "envelope" which allows less close packing.

Similarly, cyclohexane has  very close cystalline packing which may account for its low melting point.

Then, after its crystalline structure melts away, our intuition gives way to the boiling point expectancies.

Again, this is not definite. If possible, I would like to juxtapose the solid structures of each cycloalkane, but unfortunately data seems scarce, especially for cycloheptane, which is of concern.


PEG (polyethylene glycol) and ITO (indium tin oxide) The binding of proteins to an ITO surface can be prevented by coating it with PEG groups (polyethylene glycol). PEG-silane is used to create the coating.

can somebody help with:

Why does PEG-silane bind to ITO (indium tin oxide)? and how? (which bond etc.)
Why does PEG repel proteins?
Are there better coatings to use to repel proteins?

PEG "silane" is $\ce{HOCH2CH2O-PEG-CH2CH2CH2-Si(OCH3)3}$.  (Pt-catalyzed $\ce{HSi(OCH3)3}$ addition to the terminal allyl monoether.)  A metal oxide surface can be hydrated, (bulk)M-OH.  The TMS end hydrolyzes and dehydrates to tetrahedrally covalently bonded $\ce{HOCH2CH20-PEG-0Si(OM#bulk3)}$.  The alcohol polyethoxy end hydrated waves in the water, and the protein ignores the PEGylated surface.

An especially fine biocompatibilizer rendering nearly any organic surface except silicone rubber invisible to protein deposition and denaturation is poly(N-vinyl pyrrolidone).  It can be linearly radiation grafted into organic surfaces (visualize with Congo Red).  Plexiglass gives a 19 degree average contact angle in saline and implants with neither cellular or humoral reaction.  NVP plasma deposition gives you sludge.  There is no obvious (inexpensive) chemical attachment of its low polymer or oligomer to a surface.  Pity.


How can I interpret this solubility data? I'm currently trying to make incense paper.  I've been instructed to mix ethanol 90% with potassium nitrate ($\ce{KNO3}$).  As ethanol 90% is hard to get where I'm located I'm currently using methylated spirits (95% alcohol) as an alternative.  The $\ce{KNO3}$ however is insoluble with the this.  If I dissolve the $\ce{KNO3}$ in water first then add to the methylated spirits the solubility improves but it is still very cloudy.
I went onto http://periodic-table-of-elements.org/SOLUBILITY to see if there were any suggestions.  The page had the below results for $\ce{KNO3}$ solubility with ethanol:

ethanol [$\ce{C2H5OH}$]: insoluble
ethanol [$\ce{C2H5OH}$] 50%: 3,7 (20°)
ethanol [$\ce{C2H5OH}$] 90%: 0,3 (20°)

I'm not sure what the 3,7 and 0,3 indicate though?  I'm not even sure what they refer to perhaps soluble parts per 1000 ?  If any one could help with what it means that would be great.  As you can guess I barely qualify as a hobbyist chemist.

Yes, the website is a little bit ambiguous. However, there is a clue in the colour legend. The solubilities are indicated in units of grams per 100 grams of water. (This is also a very common unit for expressing solubilities.) Therefore, a solubility of "$3.7$" means that $3.7~\mathrm{g}$ of $\ce{KNO3}$ dissolves in $100~\mathrm{g}$ of whatever solvent it is listed under. This interpretation is readily confirmed by checking the values against this Wikipedia page.

As for "$50\%$", my educated guess is that it refers to a mixture of $50\%$ ethanol and $50\%$ water.

And lastly "$20^\circ$" refers to $20~^\circ\mathrm{C}$.


Can Na2CO3 remove temporary hardness of water? This was asked in my test and I remembered that all compounds that can remove permanent hardness of water can remove temporary hardness of water also. But my teacher marked it wrong and I cannot find the answer anywhere.

Definition of hardness per a source:

The simple definition of water hardness is the amount of dissolved calcium and magnesium in the water. Hard water is high in dissolved minerals, largely calcium and magnesium. You may have felt the effects of hard water, literally, the last time you washed your hands.

Adding $\ce{Na2CO3}$ introduces the carbonate ions:
$$\ce{Ca++/Mg++ + CO3(2-) -> CaCO3/MgCO3 (s) }$$
which results in a precipitate of insoluble carbonates.
So, yes it is a temporary solution to address water hardness.
[EDIT] I recommend relatedly to hard water treatment with Washing Soda, a very interesting experiment of adding Na2CO3 (aq) to MgSO4(aq) as the entire solution, at times (it depends on respective concentrations),
can convert to a wet hydrate creating so-called Magnesium alba. Cited reaction per Wikipedia:

If magnesium chloride (or sulfate) is treated with aqueous sodium carbonate, a precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed:
$\ce{5 MgCl2(aq) + 5 Na2CO3(aq) + 5 H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10 NaCl(aq) }$

where the magnesium mix oxide/carbonate is known, but not uniquely in name or precisely in chemical composition, as Magnesium alba. Now, MgSO4 aqueous per a source:

Magnesium sulfate salts can form hydrated compounds with up to seven ... Magnesium sulfate hexahydrate is formed by coordinating 6 water molecules to Mg2+ ...

So, Magnesium sulfate forms $\ce{MgSO4(H2O)6.H2O}$ with six coordinating water molecules (which could be created upon concentration of the aqueous MgSO4 solution, further extracting water from the mix). Further, if Magnesium alba similarly also extracts water, with excess MgSO4, it may be understandable as to why the entire solution can apparently turn into a voluminous wet precipitate, somewhat surprisingly.


How much CO2 is produced in burning of a candle? I've been reading this book, page 149. In chapter 15 the following experiment is suggested,

  Fix a candle in a container. Fill the container with water. Light the candle and then cover it with an inverted glass.

It is said that the oxygen in the space of inverted glass will be used up and to fill the space that was previously occupied with oxygen the water level will rise.

My question is, when the wax burns since $\ce{CO2}$ is emitted, won't this $\ce{CO2}$ fill up the space of used up $\ce{O2}$? What amount of $\ce{O2}$ is used up and what amount of $\ce{CO2}$ is released in the burning of wax of candle?

Surely $\ce{CO2}$ will fill some of the space of used up $\ce{O2}$. Candles these days are commonly made out of paraffine, which is basically $\mathrm{C_{many}H_{many}}$. The equation may look as follows:
$$\ce{C31H64 + 47O2 -> 31CO2 + 32H2O}$$
You see that we get approximately 2 moles of $\ce{CO2}$ for 3 moles of spent $\ce{O2}$, hence the decrease in volume.

Another factor is that $\ce{CO2}$ is much better soluble in water than other components of air, so over time (though not immediately) it will probably vanish from the glass and thus free some more space.

Yet another factor is that air in the glass is quite hot while the candle is burning. As soon as it cools down, it shrinks.


Is the ideal gas constant not a constant if Celsius is used instead of Kelvin? For example, Charles' law does not work with Celsius. Why is this? I can only think that the gas constant is not a constant when Celsius is used rather than Kelvin.

Charles' Law certainly DOES work with temperature in degrees Centigrade!

If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.

More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.

The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X  = π^T$ could be a temperature scale.

The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)


Why can't element with paired electron have colour I know that elements with unpaired electrons are coloured especially the d-block element which can be explained by crystal field theory. But, why can't element with paired electrons give colour, is it because the energy gap is too large ?

Your assumption is wrong. There are elements with paired electrons which are coloured.

Colour is a product of electronic transitions in atoms or molecules. If those transitions match the energy of a photon of visible light, then the substance will be coloured (though there are some other rules about how intense the colour will actually be).

Some elements do show colour. Halogens, for example, exist as diatomic molecules in their native state and are all coloured. Chlorine is a greenish gas and bromine is an intensely red liquid. Sulphur is usually found as a yellow solid. All have paired electrons.


What can be used as substitute to dichloromethane for breaking optical cement bonds? An optics supply company recommends using a dichloromethane solution to break bonds between optical elements joined with its UV-cured optical cement.1 The page mentions that dichloromethane-based paint strippers also work. Use of dichloromethane in paint strippers has been phased out in the USA  as of last year. Is there an obvious chemical property that dichloromethane possesses which causes it to break UV cement bonds and if so are there any more readily available hardware store chemicals which could be used as substitute?

According to an email i got back from Norland, "It will work if the paint stripper has a chlorinated ketone in it"


Synthesis of methylenecyclohexane from cyclohexylmethanol I want to form methylenecyclohexane from cyclohexylmethanol

I'm thinking about an $\mathrm{E1cB}$ mechanism since the β-hydrogens will be quite acidic (similar electronegativity to fluorine). So I can use a strong and hindered base (to reduce $\mathrm{S_N2}$) to form methylenecyclohexane.

Another possibility is a weak hindered base for an $\mathrm{E1}$ Hofmann elimination (since the carbocation will rearrange). But not only will this lead to a some amount of Zaitsev product, $\mathrm{S_N1}$ will dominate because of better thermodynamics (net 1 σ formed vs net 1 π formed that is not very stabilized either).

I feel like $\mathrm{E1cB}$ would be a better idea.

Are there any better methods?

Whichever method of elimination you end up preferring, there is no way to directly eliminate the hydroxy group without derivatising it first in some way. The hydroxide ion is a very bad leaving group (but a rather good nucleophile) and transforming a hydroxy group into a better leaving group by protonation ($\ce{R-OH2+}$) requires acidic conditions that do not really allow the (basic) abstraction of a β hydrogen. For each of the simple options:

A hydroxide ion will never leave as part of an $\mathrm{S_N2}$ or $\mathrm{E2}$ reaction; it is too weak a leaving group.
An $\mathrm{E1}$ reaction would require strongly acidic conditions to protonate the hydroxy group; under these conditions the β hydrogen cannot be abstracted. $\mathrm{S_N1}$ would dominate.
An $\mathrm{E1cb}$ mechanism is absolutely impossible. The $\mathrm pK_\mathrm{a}$ value of the β hydrogen is around $50$ while that of the hydroxy group is around $15$. The only conjugate base being formed will be the alcoholate anion.

This means we have to resort to two-step procedures. A number of them have been published in literature in various different contexts. Some bear a name, others don’t. Some follow one of the paradigms mentioned above, others follow a completely different mechanism. Since there is only one elimination product possible, you need not worry about selectivity. I shall present a few possibilities grouped by the mechanistic type where possible but note that this list is not exhaustive and does not even begin to scratch the surface.

$\mathrm{E2}$ type mechanisms

The simplest is the one Orthocresol mentioned in the first comment: transform the alcohol into the corresponding tosylate, e.g. by adding $\ce{TsCl}$ and pyridine, then eliminate with additional potentially stronger base.

A variant of this would be to use mesyl chloride/DBU instead. Depending on how activated your proton is, a simple addition of more DBU to the reaction mixture may suffice.

Considering that your starting material is a primary alcohol, $\mathrm{E1}$-based reactions are very unlikely. Considering again the acidity of the β hydrogen, so are $\mathrm{E1cB}$; no matter which the reaction protocol.

syn-selective eliminations based on cyclic transition states

One example is the ester pyrolysis: use an acid chloride (e.g. acetyl chloride $\ce{AcCl}$) to synthesise an ester, then heat up to initiate thermal rearrangement to the corresponding acid and alkene. The transition state is cyclic, so this reaction is syn-selective.

A similar reaction is the Chugaev elimination. React the alcohol with a base and $\ce{CS2}$, then add an alkyl halide (typically iodomethane) to generate a xantogenate. This also fragmentates thermally to give an alkene and a dithiocarbonate while syn-selectively removing the β hydrogen.

And finally my favourite one: the selenoxide pyrolysis. First synthesise a selenoether with $\ce{NCSe-C6H4-NO2}$ (ortho-nitrophenyl selenocyanate) and $\ce{PMe3}$ in a Mitsunobu reaction, then oxidise the selenoether e.g. by using $\ce{H2O2}$ at room temperature to smoothly give the corresponding alkene — again via a cyclic transition state practically identical to the former two.


Error in calculation of rate of reaction through graphs I'm asked to find the Reaction Rate when t = 120 seconds (which involves differentiation) for an unknown Chemical Equation. The rate of a chemical reaction is defined as:

$$R =\frac{dC}{dt}\ \text{(Differentiation of concentration with respect to time)}$$

I'm given a set of 11 points $(x=t\ (\text{time}),\ y=C (\text{concentration}))$ which I need to plot in order to find a general trend for them.

After plotting all points and observing the resulting graph, I tried performing multiple regression analyses in order to find a function $C(t)$ that best fits the plotted points.

For each of the Functions I obtained from each individual Regression...

I derived each one
Evaluated each resulting derivative for $t = 120$, in order to obtain
the reaction rate for each case.

Finally, I compared each result (Reaction Rate) with the Literature Value, and calculated an error percentage for every case.

These were my results:

$$\begin{array}{|c|c|}
\hline
\text{regression type} & \text{error percentage} \\ \hline
\text{exponential} & 26.92\% \\
\text{quadratic} & 22.89\% \\
\text{cubic} & 19.1\% \\
\text{quartic} & 1.39\% \\ \hline
\end{array}$$

My question is:

Many sources (including my book) claim that a High-Degree Polynomial Regression (Cubed Regression and above) is not appropriate for describing a Physical Phenomenon, even when the Regression Fit is perfect (R=1).

However, based on the table above, the most accurate answer I got was obtained through a Quartic Regression, while the rest had considerably higher error percentages.

Am I missing something or is it just an exception? Any clarification would be much appreciated!

Am I missing something?

Yes. You are missing the rate law and everything that goes along with it. It is only easy to fit zeroth order raw rate data. To produce linear fits, you need to consider the possibility that your reaction has a first or second order rate.

In addition to rate being defined as the (microscopic) change in concentration respect to time, you can also derive a rate law representing this change since the rate of the reaction is going to be proportional to things like the number of collisions between particles (which increases as there are more particles, i.e. higher concentration). So, we get a rate law that is something like the following:

$$\mathrm{rate}=kC^x$$

where $k$ is the rate constant, which is dependent on things like temperature and the activation barrier for the reaction, and $x$ defines the order of reaction, which is dependent on the mechanism and can be loosely interpreted as being related to the number of particles the must collide in the fundamental step(s). The value of $x$ is determined empirically (one such way is the experiment you did), though common values for simple reactions are integers, with $x=0,1,2$ being most common.

With both equations for rate in hand, we can do some to obtain integrated rate laws, which take the form $C(t)$: just the sort of equations that you are looking for. Note the minus signs in my equations. Rate is usually expressed based on consumption of reactants.

  $x=0$, zeroth order

$$\begin{align}-\dfrac{dC}{dt}&=kC^0 = k\\
-dC &= kdt \\
-\int_{C_0}^{C} dC &= \int_0^t kdt \\
C - C_0 &= -kt \\
C &= -kt + C_0
\end{align}$$

A reaction with a zeroth order rate will produce a graph of $C(t)$ that is linear.

  $x=1$, first order

$$\begin{align}-\dfrac{dC}{dt}&=kC^1 = kC\\
-\dfrac{dC}{C} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C} dC &= \int_0^t kdt \\
\ln C - \ln C_0 &= -kt\\
\ln C &= -kt + ln C_0 \\
C &= C_0 e^{-kt}
\end{align}$$

A reaction with a first order rate will produce a graph of $C(t)$ that shows exponential decay. However, my derivation provides a linear equation: $\ln C = -kt + \ln C_0$. If you determine the natural log of each concentration and plot against $t$, you will get a line (if the reaction is first order).

  $x=2$, second order

$$\begin{align}-\dfrac{dC}{dt}&=kC^2\\
-\dfrac{dC}{C^2} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C^2}dC &= \int_0^t kdt \\
\dfrac{1}{C} - \dfrac{1}{C_0} &= -kt \\
\dfrac{1}{C} &= -kt + \dfrac{1}{C_0} \\
C &= -\dfrac{C_0}{1-C_0 kt}
\end{align}$$

A reaction with a second order rate will produce a graph of $C(t)$ that shows power decay. However, my derivation provides a linear equation: $\dfrac{1}{C} = -kt + \dfrac{1}{C_0}$. If you determine the inverse of each concentration and plot against $t$, you will get a line (if the reaction is second order).


How can concrete set underwater? I'm about to teach concrete to some high school children. I'm not a chemist but as I understand it the limestone cycle is this:
$$
\begin{align}
\ce{CaCO3 &-> CaO + CO2}\tag{R1}\\
\ce{CaO + H2O &-> Ca(OH)2}\tag{R2}\\
\ce{Ca(OH)2 + CO2 &-> CaCO3 + H2O}\tag{R3}
\end{align}
$$
Two questions:

The water has to go somewhere or the concrete will stay wet. Air drying I could understand but how can concrete set underwater? And, assuming concrete dries from the outside in, wouldn't osmosis predict more water would be drawn into the concrete?

Making kilograms of calcium carbonate will take kilograms of calcium hydroxide and kilograms of carbon dioxide. How can such a large amount of carbon dioxide reach the concrete? There isn't a lot of it in water, is there? I could imagine some reaction might produce carbon dioxide in the concrete, but then wouldn't it end up frothy/honeycombed?

The chemistry you have posted is for non-hydraulic cement, which requires air drying for precisely the reasons you enumerated.  The kind of cement that sets underwater uses silicates and various oxides (primarily calcium, aluminum and iron oxides). The basic chemical process that is occurring is hydration, whereby the mixture of oxides and silicates incorporates several water molecules directly into the crystal lattice.  The fact that the setting process both uses water, and also produces a structure that is insoluble and robust to attack by water, is what makes underwater setting of hydraulic cement possible.

With regard to teaching your high school class, it's probably better to stick to the lime cycle for non-hydraulic cement as far as detailed chemical reactions are concerned. One interesting point that you can tell them about hydraulic cement is that its chemistry is still not completely understood, and even the molecular crystal structure was only very recently sorted out by MIT scientists in 2009.


Electrophilic Substitution I need some help in a electrophilic substitution reaction.

What will be the major product formed in this reaction and why?

I know that
(1) Activator wins over deactivator.
(2) A powerful wins over a weak activator.
(3) When a substituent possesses a lone pair, it it usually extends the product determining influence.
(4) Steric factor always plays a key role.

$\ce{NO2}$ is a powerful deactivator, so I think the middle ring will not undergo substitution. I can't decide which of the other two rings will undergo substitution and at what position.

I no longer am a practicing chemist, so I have no access to a structure drawing program; I'll do my best with words:

Accept the initial premise that the NO2 is strongly deactivating so only the 'end' rings will be attacked.  As we know o,p- substituion have functionalyl equivalent resonance forms, so we only have to consider 4 possible products: p-, m-substitutuion on the 3-ring and similar on the 4-ring.

Drawing all four of these structures and the reasonanc forms indicates that the p-,4-ring susbtitution and the m-,3-ring susbtitution are poor choices because of high energy intermediates.

The final question to resolve is whether or not the m-,4-ring is better than the p-,3-ring.  Based on the intermediates, this is a even, but the nitro group is in conjugation with the 4-ring, whle only in cross-conjugation with the 3-ring  You can see this by drawing a rsonance form with a carbocation at the para position, and rearranging the bonds via resonance all the way to the nitro group.  This can't be done with the with the 3-ring, meaning the nitro group deactivates the 4-ring more than the 3-ring.

Thus the product is likely p-substitution on the 3-ring.

Advanced:
The resonance for arguments require all three structures be coplanar.  This is not the case; biphenyls have some torsion and twisting because o-hydrogens bump.  However, there is still enough 'flatness' and orbital overlap to make some use of the resonance arguments presented above, albeit the magnitude of better and worse are smaller than they would be for more classical EAS problems.


Delocalised electrons and stability of benzene In most high school chemistry books, it is said that the delocalised electrons in benzene makes it particularly stable. However, how does it contribute to stability? Is there any simple way of understanding this?

In physics (or you can google it) you will learn about the particle in a box experiment.  At the atomic level, if you confine a particle to a box, it will have some energy.  It turns out, if you make the box longer, the energy of the particle will be lowered, if you make the box smaller, the particle will have a higher energy.  The same concept applies to benzene.  Basically you start with 3 separate acetylene molecules in which the pi electrons are confined to a 2 carbon segment.  Once you trimerize the acetylene and form benzene, the pi electrons can now roam over 6 carbon atoms, a much longer "box" if you will.  So the pi electrons in benzene will be stabilized (lower energy, longer box) compared to the pi electrons in the starting acetylene (higher energy, shorter box).  Molecular Orbital theory tells us the same thing in a slightly different manner.  In each acetylene (let's just focus on the 2 pi electrons in the acetylene that will be the pi electrons in benzene) we have 2 pi electrons in the HOMO and none in the LUMO.  As we bring these 3 acetylenes closer together to form benzene, the orbitals interact.  The 3 initially degenerate HOMOs in the 3 acetylene molecules split (they are no longer degenerate) as they move closer to becoming benzene and a net energetic stabilization results.


Why is the water dissociation constant valid for acid/base solutions? The water dissociation constant was first defined with respect to dissociation of water into hydroxyl and hydronium ions. How then can one say that it is valid for an acidic solution, that is, even in an acidic solution the product of the concentrations of the hydronium and hydroxyl ions is $10^{-14}\ \mathrm{M}$?

If Kw is written in terms of concentration (as opposed to activity), as [H+][OH-], it is only $10^{-14} M^2$ in the limit of zero ionic strength, but it is approximately correct in general.

image source


Density of unknown solution? A glass vessel weighs $\pu{20.2367 g}$ when empty and $\pu{20.3102 g}$ when filled to the etched mark with water at $\pu{4 °C}$. The same vessel was then dried and filled to the same mark with a solution at $\pu{4 °C}$, the vessel was now found to weigh $\pu{20.3300 g}$. What is the density of the solution? (Assume the density of water is $\pu{1.00 g/ml}$)

Here’s what I did:

The mass of the water is just $\pu{0.0735 g}$ when you subtract the weight of (water + vessel) and (empty vessel). Then I found how many $\pu{ml}$ of water there was in the vessel by realizing that the density ($m/V$) must be equal to $1$, so $\pu{0.0735 g}$ of water must be $\pu{0.0735 ml}$ of water too. So with that, I figured out how many grams of the unknown solution were in the vessel ($\pu{0.0993 g}$), and knowing that it was filled to the same “etched mark,” we can say that the density of the unknown solution is $\pu{0.0933 g}/\pu{0.0735 ml} = \pu{1.27 g/ml}$. I went through all of that because I am not sure if my thoughts are on track with logic and I am pretty confused.

What you did in your explanation is correct. First find the difference in the weight:
$$20.3102-20.2367=0.0735 \ \mathrm{g}$$

We know that from your density of water:
$\mathrm{\frac{1\ g}{1\ mL}\ \text{that }\ 0.0735 \ g = 0.0735\ mL}$

Then the difference between the unknown and the vessel is calculated as:
$$20.300-20.2367 = 0.0933\ \mathrm{g}$$

and then density is expressed as mass over volume, so:
$$\mathrm{\frac{0.0933\ g}{0.0735\ mL}=1.2693_8\ g\ mL^{-1}}$$


Thermodynamics effect of change of temperature on equilibrium I'm studying thermodynamics and I don't understand, why, when we are asked which effect has a rise in temperature on the equilibrium constant of a reaction, we can't use the formula: $\Delta G= \Delta H-T \Delta S$, and say that if the reation is entropic, the change in free energy will more negative, therefore the reaction will be more spontaneous and the equilibrium shifts towards the products.

In my exercices books, it is said, that, according to le Chatelier's principle,the equilibrium will try to "fight" the new constraint and if it is exothermic,the equilibrium will shift towards the reactants, and if it is endothermic towards the products.

Le Chatelier's principle seems intuitive to me for the compression of a gas: when compressed, the molecules are now more narrow and they combine faster than they separate. But I don't see why it can be used in the raise of temperature.

Thank you in advance

First of all, spontaneity of a reaction depends on the kinetics, not thermodynamics.

Second, I always imagine LeChatelier's principle to work like this: When you raise the temperature for an endothermic reaction, it has more temperature to work with so the equilibrium shifts to the right. And if you lower the temperature for an exothermic reaction, that means there's more room for the energy created by the reaction.

But if you want to use $\Delta G = \Delta H - T\Delta S$ to check how the equilibrium behaves, you will only see the effect the temperature has on the free energy. It relates to the equilibrium via $$ \Delta G = -RT\ln K \Longleftrightarrow K = \exp\left(-\frac{\Delta G}{RT}\right)$$
This means that if you raise the temperature the expression in the exponent gets closer to 0, which in turn means that the overall $K$ becomes larger. So the equilibrium shifts towards the products.

However, I have neglected the influence of the entropy above. The formula should be, when completely written out:
$$ K = \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) $$

And with this you can always see whereto the equilibrium will shift.


Relative magnitudes of intermolecular forces from phase diagrams Is it possible to say whether or not a system A or B has stronger intermolecular forces from a phase diagram according to which:

The $T_{\text{b}}$ of A is lower than the $T_{\text{b}}$ of B.
The $T_{\text{melting}}$ of A is higher than the $T_{\text{m}}$ of B.
The triple points of A and B are the same.

Or is it not possible to compare A and B's intermolecular forces from a phase diagram?

I was sure that B has stronger intermolecular forces, as its boiling point is higher than for A at the same pressure.

This is an interesting situation. It would be easier to answer if the mp and bp of A were both either higher or lower than for B.

But your measure of the strength of intermolecular forces will be melting point - and boiling point. And they give different answers, so how do we make a decision?

Consider the boiling point first: the higher boiling point of B suggests higher intermolecular forces, as you have concluded. Heat of vaporization is needed to put a molecule into the vapor phase, and it seems that B needs more energy to go gaseous.

But A melts at a higher temperature, suggesting that its molecules are held together in a solid phase by stronger bonds. Solid A needs more energy to turn it into a liquid; perhaps A can pack into a tighter solid structure. It would then need more heat to melt it, so it has a larger heat of fusion.

Then your answer must have two parts: A is more tightly bound in a solid framework, but after you put enough heat into both A and B to melt them, B has more intermolecular bonding capacity as a liquid than A does. So B boils higher and is more strongly bound as a liquid.


Rapidly cooling Aluminum vapour to solid state? What sort of a microstructure would you get on bubbling Aluminium vapors through a large tank of Helium?Would it be a glassy microstructure and/or would you get nano-powders? and/or a new crystal structure formation because of the excess helium which would be present in the aluminium matrix? Ignore metal coolant reactions, if any. Aluminum vapour would consist of Aluminium atoms, unlike liquid aluminum which consists of clusters. So, on rapid cooling from gas to solid, would you still get an FCC crystal structure for aluminium? (There would be no time for diffusion)

There are a number of actual production processes that involve coating surfaces with metals via what are effectively vapour-phase metals.

Sputtering seems to be preferred to simple evaporation because it gives better results and allows for better control over the properties of the deposited metal coating (or non-metal, a very wide variety of compounds can be deposited this way including some created by vapour phase reactions). Sputtering creates aluminium (the same applies to many other elements or compounds) vapour by bombarding solid aluminium with inert atoms and allowing the atoms to hit the coated surface.

There don't appear to be any unusual crystal structures in the resulting coatings, though the grain size and other factors can be controlled by altering the conditions.


DNA dye: Are there any natural dye for DNA (instead of artificial ones)? In order to make easy and low-cost experiments, we are searching for natural dyes (or easy to make) that can color DNA, instead of for instance Ethidium Bromide that is hard to find and is expensive.

We are new to this field, but we learned that DNA dyes need to intercalate in between DNA bases. To do so, the molecule must be preferably poly-cyclic, aromatic and planar.

So the question is, is there any chemical that can be readily used for this purpose or if not, any lead about a molecule that can be easy to find by anyone.

Though not a naturally occurring dye, methylene blue has been used as a DNA stain. It is inexpensive and readily available. It is also comparatively nontoxic, and is used as a medication in pisciculture, so it can be found in pet stores.
If you choose to use methylene blue, save some for the blue bottle demonstration, a sensitive test for oxygen. Vitamin C or other reducing agents can be used instead of glucose.


Are decomposition and dissociation reactions the same? I have seen these terms being used almost interchangebly, but are decomposition and dissociation reactions the same ?

Decomposition and disassociation refer to similar processes, however they are different. The main difference between them is that decomposition reactions are usually irreversible while disassociation reactions are reversible and exist in equilibrium.

A decomposition reaction is where a compound breaks up into smaller elements or compounds. For example: $$\ce{2H2O -> 2H2 + O2}$$ is an example of a decomposition reaction. Note that in this case that this is actually a reversible reaction.

A disassociation reaction is the general process where molecules split into smaller particles which can include elements, compounds, ions and radicals. For example: $$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$ is a disassociation reaction.


Hybridization of sulfur in sulfur dioxide One of the canonical structures for sulfur dioxide - $\ce{SO2}$ - has sulfur (with a lone electron pair) double bonded to each oxygen atom to form a total of 4 bonds for sulfur - which can be achieved via valence expansion into empty d-orbitals.

What then is the hybridization of the valence-expanded sulfur?
It is described as sp². But how can that be? This seems unlikely
because d-orbitals are involved since the sulfur underwent valence
expansion.

On might imagine a pair of electrons from the 3s/3p oribital(s) being promoted to an empty d-orbital and then having the 3s and 3p orbitals hybridize in to sp².
If this is true it would mean that the lone electron pair of the
valence-expanded sulfur consists of 2 electrons occupying
and unhybridized d-orbital. But is this correct?

The structure of sulfur dioxide ($\ce{SO2}$) is quite complicated.
The image from socratic.org* explains it quite well, please find the $\ce{SO2}$ hybridization diagram via the Internet Archive.
As seen, all the atoms have $\ce{sp^2}$ hybridization.
I'll only focus on the central sulfur atom.

Two $\ce{sp^2}$ orbitals form $\ce{\sigma}$-bonds with the two oxygens.
The other $\ce{sp^2}$ orbital is where the lone pair lives in.

Now, we have dealt with 4 electrons, and only have 2 electrons to deal with.
The 2 remaining electrons actually live in the unhybridized $\ce{p}$ orbital.
At this point, the two oxygen atoms will also have 2 electrons left to pair with the sulfur atom.

The two oxygen atoms get to keep their own electron.
Sulfur shares two electrons among itself and the other two oxygen atoms.

Thus, no electron lives in the $\ce{d}$ orbital.

$\color{Red}{\mbox{red}}$ represents number of electrons (I am too lazy to draw the fish-hooks).
$\color{Green}{\mbox{green}}$ represents $\ce{sp^2}$ orbital.
$\color{blue}{\mbox{Blue}}$ represents $\ce{p}$ orbital.

Note that the two $\ce{sp^2}$ orbitals between sulfur and the two oxygen atoms are in $\ce{\sigma}$-bond.

* The website is no longer available; there is no version in the Internet Archive.


What is the difference between the mean radial distance and the most probable radial distance of an orbital? I was reading JD Lee's Concise Inorganic Chemistry and it has given some values of mean radial distance which is increasing from s to p to d orbital. I had earlier read that the most probable radial distance of an orbital has the order s>p>d>f. This makes me wonder what is the difference between the mean radial distance and most probable radial distance. Also, why is the order of mean distance reverse from that of most probable radial distance?
Also can you please explain why is the energy of an orbital proportional to mean radial distance as given in the book?

Image Source: Concise Inorganic Chemistry by JD Lee

In multi-electron atoms the energy of the s, p and d orbitals for a given $n$ is determined primarily by shielding by inner electrons and penetration of the orbital towards the nucleus. As s orbitals have some probability of being at the nucleus their energy is lowest, next is p then d orbitals in increasing energy. Also there is spin-orbit coupling which splits sub-levels of p and d orbitals or any orbital which has angular momentum so this does not effect s orbitals. This effect is much smaller in d than p orbitals and generally smaller than the separation of s to p to d.

The statements in the book extract you give seem contradictory. The mean radial value of an H type orbital is calculated using $\displaystyle \int_0^\infty r^3R_{n,l}^*R_{n,l} dr $ where $R_{n,l}$ is the (normalised) radial part of the wavefunction for an atom with quantum numbers $n,\;l$. The $r^3$ arises because the volume element in the integration is $r^2\sin(\theta)d\phi d\theta$ in spherical coordinates. The other $r$ is how the average is defined, i.e. $\int r\psi^*\psi d\tau$ for some normalised wavefunction $\psi$.

The radial parts of the wavefunction for the 3s,3p and 3d are define with $q=Zr/a_0$ ($a_0$ is the Bohr radius) as
$$\begin{align}
R_{3,0}&=N(27-18q+2q^2)exp^{-q/3}\\
R_{3,1}&=N(6q-q^2)e^{-q/3}\\
R_{3,2}&=Nq^2e^{-q/3}
\end{align}$$

and $N$ is the normalisation which is different for each wavefunction. The integrals are not hard to evaluate but very tedious as there are many terms. However, there is a general formula for the average of H atom type orbitals and is

$$\langle r\rangle =\frac{n^2a_0}{Z}\left( 1+\frac{1}{2} \left( 1-\frac{l(l+1)}{n^2}  \right) \right ) $$

The average values are

$$3s \;\langle r \rangle = 0.476 \text{ nm,}\quad 3p \;\langle r \rangle = 0.441\text{ nm,}\quad 3d\; \langle r \rangle = 0.370\text{ nm}$$

The s orbital value is the same as in your question but the others differ both in value and in order of size. Perhaps there are typos in the text you give or some other unstated factors producing these values.

The maximum values have to be calculated by making the derivative (of $q^2R_{n,l}^2$) zero and solving the equation. Alternatively the value can be taken with sufficient accuracy  from a plot of the function which is far easier. The values are

$$3s \;r_{max} = 0.46 \text{ nm,}\quad 3p \;r_{max} = 0.43\text{ nm,}\quad 3d\; r_{max} = 0.32\text{ nm}$$

the plot below shows the radial functions $q^2R_{n,l}^2$  each normalised to the same maximum height. The mean is greater than the maximum in each case as expected as the main part of the radial probability is 'lopsided' towards larger distances.

(In evaluating the integrals $\displaystyle \int_0^\infty r^ne^{-\beta r}= \frac{n!}{\beta^{n+1}}$ is useful.)


Can molecule have just h-bonds between its parts I'm just starting to study chemistry so sorry for the probably obvious question to you, guys.
According to the definition of molecule in Wikipedia:

A molecule is an electrically neutral group of two or more atoms held
together by chemical bonds.

According to that definition, I assume that parts of the molecule could be theoretically connected just with the h-bonds. As far as I understand the example of that case is DNA two strands of which are connected by only h-bonds (though many and many ones).
According to Khan Academy:

When atoms combine by forming covalent bonds, the resulting collection of atoms is called a molecule.

And now, as a newbie, I'm a little bit confused with all of that. So, could you help me with the following questions:

Is that true that two strands of DNA do not have covalent bonds to each other and connected only with h-bonds?
If the p. 1 is true could parts of molecule be connected only by h-bonds by definition?
If the p. 3 is true then why DNA is a molecule but two water compounds (H2-O***H-O-H) are not?

There are some grey areas in the normal usage of the word 'molecule'
The normal usage of the term molecule is accurately described by your quote from the Khan Academy. The majority of molecules chemists talk about are groups of atoms joined with covalent bonds. But, like many things in chemistry, there are some grey areas and the normal usage gets bent a little.
One of the major areas where the terminology gets fuzzy is in polymers. While on a strict interpretation of the normal definition a single polymer strand (be it DNA or a single strand of polyethylene) is a molecule the terminology is not particularly helpful as the key things that matter in bulk polymer properties are not easily reduced to the properties of individual 'molecules' so the very idea of molecules is less relevant. In polyethylene (a very simple polymer) there will not be a mix of identical molecules but, rather, a wide range of very similar molecules (even for linear polyethylene there will be a wide range of chain lengths (say from 1,000 to 10,000 carbons). So polymer chemists tend not to talk about molecules much.
DNA is a polymer but, by its nature, one where the structure of a given strand is constant (and can be precisely replicated). We do tend to say DNA molecule at least in casual conversation despite the fact that the two strands are connected only by hydrogen bonds. But there are good reasons why this is not an abuse of the normal definition of molecule. One is that many hydrogen bonds along a polymeric chain add up to a strong overall bond even though the individual bonds are merely hydrogen bonds. But the relative weakness of each bond is a key part of the process of replicating DNA as the two strands can be easily unzipped by the machinery in the cell nucleus. You could argue that a DNA strand consists of two molecules using the normal definition but this doesn't help much and, in normal discussion, we talk about both a DNA molecule with two hydrogen-bonded strands and a DNA strand as a molecule. This mild abuse of a strict definition of "molecule" doesn't matter much as it doesn't interfere with any understanding of how DNA works.
Mostly, though, chemists don't consider hydrogen-bonded units as "molecules" even when the bonds have significant effects on the bulk properties of the substance. The physical properties of, for example, acetic acid are heavily influenced by hydrogen bonding. In solution molecules often form pairs (joined by two hydrogen bonds) and this leads to a big elevation in their expected boiling point compared to a similar hypothetical molecule of the same size that couldn't do hydrogen bonds. But there is no benefit in describing this a "molecule" and nobody does this.
But the main point is that many properties in chemistry have an almost continuous range with no clear point where a black and white distinction can be made. The definition of what counts as a molecule is like this. Describing hydrogen bonded DNA as a molecule isn't a terrible use of the terminology; describing an acetic acid hydrogen-bonded pair as a molecule probably is. The term molecule can be useful in some circumstances but not others. The utility of the term is what matters not a simple black or white definition.


Why did the solution turn green? I mixed L Arginine with l Citrulline in Hibiscus and green tea water which was red in color. Once the red liquid hit the mix of powdered Aminoacids the solution changed green. Can someone tell me why? Thanks so much for any info!

Both citrulline and arginine are nitrogen-rich, basic amino acids. A basic acid sounds confusing, but that's the way it is.

The colour change indicates a change in the pH of the solution (the tea mix).

The pH indicator that changes the colour is in the hibiscus flower and belongs to the class of the anthocyanes.

Under acidic conditions, the dye exists as a deeply-red flavylium cation. At a slightly higher pH, the dye is reversibly converted to a colourless form. Here, the colour of the mixture is determined by the other ingredients. When more base (read: more amino acid) comes to action, further colour changes are conceivable and the double deprotonated form of some anthocyanes are known to be bluish-green.


Why does attractive forces of nucleus increase more than shielding across groups? When you study the electronegativity of the elements, the general trend is that it rises with increasing group number, and decreasing period.

Supposedly this is because the attractive forces of the nucleus increase quicker than the shielding effect of the electrons.

Is this shielding effect simply a repellant force, or is there another phenomenon at play?
Why does the shielding effect not increase as rapidly?
Why does the difference between the attractive forces and the shielding effect become less pronounced with increasing period?

(source: wikispaces.com)

Question (1)

To at least a very good approximation, the shielding effect is merely due to a repulsive force.  Like much of chemistry at the atomic level, electrons experience a balance of counteracting forces due to the nucleus and other electrons.

Question (2)

This is usually explained in the following way.  As one goes across a period from left to right, the charge on the nucleus increases at each step, increasing the attractive potential for electrons (which you already noted).  In addition, an electron is added to the valence shell for each step.  Since the added electrons are in the same shell as the preexisting valence electrons, they do not contribute as strongly to shielding as one might initially expect.  Electrons that are in lower shells contribute to shielding more effectively since they are "more likely" to be between a valence shell electron and the nucleus.

Question (3)

One way to justify the increasing effectiveness of shielding in later periods is an extension of the answer above to question (2).  There it was stated that valence shell electrons do not effectively shield other valence shell electrons since they are at the same level.  An natural extension of this would be that the further apart two electronic shells are, the more the lower shell screens the higher shell.  Compared to the valence electrons in Pb, the 1s electrons are essentially right up against the nucleus and, from the view of Pb's 6p electrons, should almost completely neutralize the charge of two protons in the nucleus (i.e. very effective shielding).

Another way of looking at this is that as the principle quantum number (n) increases, the orbitals become less penetrating.  This means they spend less time near the nucleus (where the shielding is weak) and more time further from the nucleus (where shielding is strong).


What is the maximum amount of medicine that could accumulate in the body if the compound has a half-life of 24 hours? If a patient is prescribed 25 mg per day of a compound that has a half-life of roughly 24 hours, what is the maximum accumulated amount of the medicine that would build in the patients body?

1st day: 25 mg
2nd day: (12.5 + 25) mg
3rd day: (6.25 + 12.5 + 25) mg
4th day: (3.125 + 6.25 + 12.5 + 25) mg
The amount get closer and closer to 50 mg but never exceeds it.


Saturation of multiple solutes Can a liquid which is saturated with a single solute, dissolve a different solute, or is saturation a universal thing? I ask because I’ve seen that different solutes have different points (amounts) at which they become saturated in water, for example, which caused me to think that maybe each solution was perhaps independent, even though that would produce some interesting extreme implications. Something like, take a water & sucrose mix that is saturated; can this mixture then dissolve salt?

The specific implication I am asking about is this: If I dissolve xylitol (a sugar alcohol) in water until it is saturated (200 g xylitol to 100 g water at 25 °C / 77 °F) can I then dissolve erythritol (a different sugar alcohol) into the solution?

Obviously, we could heat the liquid and enter super-saturation, but that’s not what I’m asking. I’m asking if this can be done without super-saturating (i.e. risking precipitation/crystallization) once cooled.

Different solutes in the same solution indeed can affect each others' solubility. One example where this effect can be used to advantage is the "salting out" of a protein from water solution. Adding a salt (ammonium sulfate is especially good for this purpose) reduces the solubility of the protein in water, causing it to precipitate or crystallize out. Salting out can also be used to more fully separate an organic solvent from water. I imagine that in your example with the sugar alcohols, adding one component to water will reduce the solubility of the second component by a similar degree. But there may be cases where one solute increases the solubility of another.


Preventing oxidation of copper(I) By aerial oxidation $\ce{Cu+}$ can easily oxidise to $\ce{Cu^2+}$. Which of the following should be added to $\ce{Cu+}$ to prevent it from oxidation: $\ce{F-}$, $\ce{Mg^2+}$, $\ce{Mg}$ or $\ce{F2}$?
$$
\begin{align}
E(\ce{Cu^2+/Cu+}) &= \pu{0.15 V}\\
E(\ce{F2/F-}) &= \pu{2.87 V}\\
E(\ce{Mg^2+/Mg}) &= \pu{-2.37 V}
\end{align}
$$

Why it isn't $\ce{Mg^2+}$?

We need a better reducing agent (that itself oxidises) than $\ce {Cu^+}$.
So choose the ones which have greater reduction potential than $\ce {Cu^+}$.
It isn't $\ce {Mg^2+}$ as it cannot oxidise any further hence it has a very low(negative) reduction potential, so it cannot stop $\ce {Cu^+}$ from oxidising.


Loading nickel metal hydride battery for the first time I just bought a new hair clipper and I was about to start shaving my head when I noticed that it was almost at 0%. It has a nickel metal hydride battery.
The manual says I have to load it for 24h hours before using it the first time or whenever it has been unused for a long time.

I started calculating. 24 hours are 1440 minutes. Dividing it by 100 makes it 14.4 minutes to gain 1%. To gain 20% it'll take 288 minutes (more than 4 hours).

Well I started to load it like an hour ago and it has almost 40% so far. That's not 4 hours. So it'll be at 100% in a few hours. I think in the evening it's done.

But the manual emphasized that it should be 24 hours ! So I wondered why do they want me to load it that long even when the percent-bar shows me it's fully loaded and ready to use? I also looked up some reviews where the customers wrote that the battery's lifetime is very short.

Would it cause a shorter battery-lifetime if I use it for like 10 up to 20 minutes in the evening (in about 6 hours)? Or more in general, is there something about the intrinsic chemistry of the battery which requires (in the most cases) so long before we use it the first time? Is that even necessary?

Battery on 0% usually still has some voltage, and battery on 100% usually can be overcharged. However, for many types of batteries it is not healthy, leading to premature degradation, the exact 0% and 100% charges as well as initial load procedure and storage requirements are marked at point that happens to prevent battery degradation and increase its capacity.

While I'm not aware of what exact process is intended to happen in Ni-hydride battery, I'm pretty aware that preparation procedure for battery MAY be required. For example, during initial cycles of lead accumulator, its capacity increases. The reason is in its chemistry, as it actually works on lead oxides transformations and the cycles increase layer of oxides on electrodes.

Since Ni-hydride cell includes electrode with half-reaction

$\ce{Ni(OH)2 + OH- = NiO(OH) + H2O + e-}$

I guess, that during initial charge the manufacturer expects that layer of nickel hydroxide is formed on positive electrode.  But this is just a guess, you'll have to search specific info on this particular type of battery to be sure. Ni-hydride batteries are resistant to overcharge at low currents, so if you use appropriate smart charger, say, shipped with the batteries, the overcharge is safe, so, since the manufacturer requires it, you should follow the manual.


What is the difference in mosM and mOsm units, if there is any at all? I'm reading Agre's paper on aquaporins [1] and he uses a unit called "mosM" throughout the paper, for example:

… ocytes were transferred from 200 mosM to 70 mosM modified Barth's buffer.

I just want to understand what this unit means, is it the same as mOsm? I know this might be an very stupid question but hey, I'm only a undergrad.
Reference

Preston, G. M.; Carroll, T. P.; Guggino, W. B.; Agre, P. Appearance of Water Channels in Xenopus Oocytes Expressing Red Cell CHIP28 Protein. Science 1992, 256 (5055), 385–387. DOI: 10.1126/science.256.5055.385.

I think the clinical chemistry notation has not been standardized but certainly it has been inspired by chemists.
The capital M in mosM emphasizes that it is milliosMolar. Chemists use capital M to denote molarity.
On the other hand, mosm indicates milliosmolal. Chemists use small m to symbolize molality.
This notation is not universal.
From this slightly dated article on Osmolality, which is behind pay-wall
Osmolality, Harry F. Weisberg Laboratory Medicine, Volume 12, Issue 2, 1 February 1981, Pages 81–85.

Terminology and Units
In 1934, Gamble, et al.,11 introduced the term ”osmillimole” which has been successively transformed into mosM (milliosmolar), mosm (milliosmolal), the noncommittal mOsm, and the present-day mosmol (milliosmole). Raoult’s law states that the lowering of the vapor pressure (one of the four ”colligative”
properties of dilute solutions) is proportional to the mole-fraction (more easily calculated from molality) of solute present in the mixture.


Identification of an unknown organic molecule from only 1H NMR and 13C NMR I am currently trying to determine the structure of my unknown compound with both 1H NMR and 13C NMR spectra.

My H NMR peaks:

1.16 ppm [singlet, 1H], 1.68 [singlet, 3H], 1.75 [singlet, 3H], 4.2 [doublet, 2H], 5.4 [multiplet, 1H], and 7.24 (solvent peak)

C NMR:

18, 26, 59, 77 (solvent peak), 124, and 136 ppm.

I have deduced that my molecular formula is $\ce{C5H10O}$. And I am fairly certain I should have a double bond in my structure. However, I do not know intuitively where this pi bond should be.

What should I be looking for to determine the placement of this pi bond?

I unfortunately do not have characteristic IR peaks to determine whether I have an alcohol, ether, or a ketone.

The doublet at $4.2\ \mathrm{ppm}$ is almost assuredly an alcohol group or ether group.  Since we only have one carbon that is bonded to an oxygen at $59\ \mathrm{ppm}$, we conclude this is a primary alcohol. The situation of being a doublet says smoething else is lurking around.

$5.4\ \mathrm{ppm}$ is far enough down field, we start thinking alkenes, but the multiplicity is mysterious for the moment.

$1.68\ \mathrm{ppm}$ and $1.75\ \mathrm{ppm}$ are in the range of allyl methyl groups, and the carbon signals are okay with this as well.  Additionally singlets for these groups indicate a geminal dimethyl compound with a third substituent on an alkene justifying different chemical shifts.

Finally the peak at $1.16\ \mathrm{ppm}$ is difficult to explain.  However, we know we have an alcohol, and they can have a fairly wide range of values.  If this peak is broadened at all, I would immediately assign it to the alcohol.  It being a singlet is also worrisome, but if exchange is occurring, this is not out of the question.

The multiplicity of the alkene proton indicates there are some long-range things going on, and these don't always resolve well.   My guess is you are picking up some long-range coupling and is not resolving nicely in the methyl groups, which appear to be singlets.


De-/protonation of R-OH group by acids and bases Let $\ce{R-OH}$ be a group with a $pK_a$ of 16. Are the following two statements correct?

In principle, the group can protonate another acid with $pK_a > 16$, and so (put differently), any "acid" with $pK_a > 16$ can be used to "deprotonate" the group
If I had a table with bases were only $pK_b$ values were given, in order to select these bases from the table which are able to deprotonate the group, I would need to choose the bases with $pK_b < -2$ because the $pK_a$ values of their conjugate acids (i.e. $pK_{a\text{H}}$) will be 16 or higher.

Your first statement is correct. The basicity increases with $pK_a$, so a base with $pK_a > 16$ will be able to deprotonate your alcohol.

Your second statement is also in principle correct; however, the value of the sum $pK_a+pK_b$ is solvent-dependent, and 14 is the value for water at standard conditions. $pK_a$ and $pK_b$ values of acids and bases depend on the solvent as well. For example, sodium amide is a strong base in liquid ammonia, but not so in water due to the leveling effect of the solvent. Dissolving sodium amide in water will give a solution of sodium hydroxide, which is not basic enough ($pK_a=13$) to deprotonate the alcohol.

Furthermore, the amount of deprotonated alcohol will depend on the difference between $pK_a$ values of alcohol and base, since the acid-base reaction is an equilibrium. For example, a base with $pK_a=33$ will deprotonate your alcohol to a greater extent than one with a $pK_a$ of 16.5 .


Why is the melting point of magnesium oxide higher than aluminium oxide? There's a graph of the melting points of period three oxides. The melting point of magnesium oxide is several hundred Kelvin higher than aluminiumoxide. I can't find any explanations for this on the Internet or in any of my A-level textbooks — perhaps this is something more advanced I haven't come across.

The reason I am questioning the value of its melting point is I'd expect aluminium oxide to be higher. Aluminium is a 3+ ion whereas magnesium is only 2+. Aluminium is also a smaller ion so the effective electrostatic attraction on the valence electrons should be greater. I'm also pretty sure aluminum oxide has greater covalent charterer than magnesium oxide. All of these would lead me to believe it has a higher melting point, so what is the explanation for it being less?

Two ideas to consider:

Alumina is not fully ionic.  Neither is magnesia, but magnesia has more ionic character and we may see the greatest amount of ionic attraction in magnesia versus alumina which is less ionic, or soda which has only singly charged cations.

When metal oxides are melted they do not necessarily produce free metal and oxide ions.  As described by Shi et al. [1], molten alumina retains mostly four-and five-coordination of oxygen to aluminum, rather than forming "free" aluminum and oxide ions and also contrasting with the six-coordination of aluminum in the solid conundrum phase.  Presumably magnesia would behave similarly to alumina when melted, but the greater ionic bonding character in magnesia makes reduced coordination less favorable and thus enhances retaining the fully octahedrally-coordinated solid phase.

Reference
1.
Caijuan Shi, Oliver L. G. Alderman, Diana Berman, Jincheng Du, Joerg Neuefeind, Anthony Tamalonis, J. K. Richard Weber, Jinglin You and Chris J. Benmore,  "The Structure of Amorphous and Deeply Supercooled Liquid Alumina",
Front. Mater. 6:38 (19 March 2019), https://doi.org/10.3389/fmats.2019.00038.


Why any other element except fluorine wouldn't take electrons from oxygen? My teacher said that fluorine is more electronegative than oxygen thus it will take electrons of oxygen.

But now, why won't other Group 17 elements, like chlorine, bromine, iodine, would take electrons from oxygen? They too have just 7 electrons in their valence shells.

Technically, they should but yet they have lesser electronegativity and this is something that I'm not able to digest.

You can consider thinking about charge to volume density. Fluorine is very small and has highest charge to volume ratio among other group 17 elements. For that reason only fluorine is able to snatch electrons from oxygen.


Ammonium nitrate in fuel-air bombs Are there any fuel-air bombs / thermobaric weapons that utilise ammonium nitrate as a fuel?

I know that BLU-82 is often mistakenly referred to as fuel-air bomb, so I became curious as to whether or not there were any actual fuel-air bombs that used ammonium nitrate.

Explosives that use ammonium nitrate and a fuel are mostly referred to as ANFO explosives, you might find more detailled information by using that term.

Most explosives used in military (especially bombs) are based on TNT, mostly in the form of Tritonal, a mixture of TNT and aluminium powder. Generally, only outdated bombs use ANFO explosives (due to their low price and ease-of-use). (See BLU-82, which does indeed use ammonium nitrate)

ANFO explosives are rather used in an industrial context, especially mining.

In the common mixture of ammonium nitrate with nitromethane (ANNM), no oxygen from the air is needed to sustain the explosion:

$\ce{3NH4NO3 + 2CH3NO2 -> 4N2 + 2CO2 + 9H2O}$

Since ANFO explosives carry their oxygen with them by definition (the ammonium nitrate molecule contains 3 oxygen atoms), they do not need oxygen from the surroundings and are therefore not thermobaric or fuel.air weapons.


Is home synthesis of cellulose acetate possible, and why? Is it remotely possible to synthesize detectable quantities of cellulose acetate from materials available in common department stores (such as cotton balls, vinegar, and battery acid) and using tools available in a home kitchen (such as ovens, stoves, freezers, glass jars, and metal pots)?  If so, how would the reagents need to be pre-processed, and what would the procedure be?  If not, where do the insurmountable difficulties lie?

On a webpage titled "From Cotton to Cellulose Acetate" it is indicated that, in general, cellulose acetate is manufactured from specially prepared cellulose in a solution of 100% acetic acid with acetic anhydride.  That same site goes on to specifically state that immersing cotton balls in vinegar will only result in "cotton balls that smell like pickles".  Other sources add concentrated sulfuric acid into the mix.  And an answer to another question on the Chemistry SE indicates that household vinegar could possibly be concentrated up to 58% using fractional freezing in a kitchen freezer.

It's difficult to tell from the sources available whether making homemade cellulose acetate is completely impossible, somewhat inconvenient, surprisingly hazardous, or merely commercially unviable.

The bottom line is that this would be a very bad idea to try due to safety reasons alone, and is not realistically feasible using only household materials regardless.

Glacial (100%) acetic acid is dangerous if not handled properly and requires an appropriate fume hood at a minimum. You also are not going to get to glacial acetic acid from vinegar, and water will kill this reaction.

Acetic anhydride is a reagent that you just won't be able to obtain or synthesize yourself. Even if you did, it would be hydrolyzed by the water in your impure acetic acid.

Sulfuric acid is commonly used as a catalyst, and while you could obtain this, it should absolutely not be handled by someone without appropriate equipment and training in it's safe usage.

So again, this is just not a feasible project and there would be serious dangers along the way if you tried. I like your creative thinking process and I hope you will pursue some other, safer, home science project.


Why do fluorescence intensities decline with replicate number? I am performing a determination of riboflavin with fluorescence spectrophotometry. I've dissolved my riboflavin in a dilute GAA solution, created a calibration curve and measured my unknown's fluorescence.

For these types of analyses I usually perform triplicate measurements, however I've noticed that spectrofluorometric measurements always decrease with replicate number. I assume this is due to 'photobleaching' of my samples. Since the phenomenon occurs with replicate rather than over time, and since my sample is in a dark environment, I would imagine this can only be the result of the excitation light emitted by the instrument itself. I find it surprising that this causes such a dramatic change in fluorescence from measurement to measurement. I also notice that with enough replicate measurements, eventually the intensity decline seems to bottom out. Can anyone explain the specifics of this phenomenon to me?

Am I right in assuming that this is photobleaching (deactivation of the chromophore) rather than photodegredation (breaking the molecule), or are these the same?
How exactly is the fluorophore deactivated?
Can the degree of fluorescence intensity decline be predicted quantitatively?
Would my calibration be more accurate if I used single measurements instead of taking the mean of three replicates?

That looks like a bleaching to me. The notion that bleaching only occurs with excesive light intesity on small populations of fluorophores is news to me. Might also that in my field we just use the bleaching term loosely. Best way to test is to take a long exposure of your sample and monitor the fluorescence intensity over time.
With our chemistry we usually encounter intensity loss that can be fitted with exponential decay function which would explain the bottoming out that you are describing. We often take multiple such "bleaching curves" and fit them to correct for intesity loss in our experiments that can take multiple hours.
In case you are only exposing your sample for short periods i would just make sure you measure each sample the same number of times that should rule out any error between samples. Thus reducing to only one measurment will not make it more correct.
Bleaching is poorly understood thus far. There is not a way to predict what excatly happens with a specific molecule.

Also how big is the effect for example (Start value)/(3rd measurment value) ?


Is Hoffman elimination anti or syn elimination? I read an amazing PDF from nptel on elimination reactions...but I was really confused when it used used $\ce{^+NMe3}$ as a leaving group in the reaction pathway via:

$\mathrm{Ei}$ (syn-elimination)
$\mathrm{E2}$ (anti-elimination)
And also, in an illustration for $\mathrm{E1_{cb}}$.

My organic chemistry teacher also did syn-elimination but in the problem solving book and on various online sites it took the $\mathrm{E2}$ elimination pathway.

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

(Taken from Ron's answer, in order to start an explanation)(1)

From Wikipedia:

  The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.

So on further thinking about the Hofmann elimination, you notice the following

$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))
Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage
In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^{2+}$, $\ce{NR^{3+}$ all have the same observation

What's the conclusion?

Of course, these undergo $E1cb$ when heated to a suitable temperature

There are sources which infer the same -

Wikipedia
Organic mechanisms by Xiaoping Sun (3)

  Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.

  The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion  character (e1cb) for the transition state.

Hence, $E_1cb$ is the correct answer


Nucleophilic acyl substitution leaving group From Wade's "Organic Chemistry" [1, p. 1001]:

  MECHANISM 21-3 Conversion of an Acid Chloride to an Ester

What I don't understand is the elimination of $\ce{Cl}$, even though a much better leaving group with a positive charge on it is present (which results in the backwards reaction).

Why doesn't the alcohol or acid chloride deprotonate the tetrahedral intermediate before step 2 and after step 2 $\ce{Cl-}$ can attack the protonated alcohol/acid chloride to turn into $\ce{HCl}$?

Reference

Wade, L. G. Organic Chemistry; Pearson Education, 2012,. ISBN 978-0-321-81139-4.

Indeed, there is no reason why the protonated alcohol should not be eliminated upon the lone pair rebound. It probably happens more often than not. However, it does not lead to an observable product: it just goes back to the reactants. Thus, we can write an equilibrium arrow on the first step.

$$\ce{R-C(=O)-Cl + R'-OH <=> R-C(-O^-)(-Cl)-OH+-R'}\tag{1}$$

Once the chloride has been kicked out, there is no way back as the ester is a much weaker electrophile than the acid chloride (and chloride a weaker nucleophile than the alcohol). Thus, the reason why we only care about the lone pair rebound eliminating chloride is because only that is productive.

$$\ce{R-C(-O^-)(-Cl)-OH+-R' -> R-C(=O)-OH+-R' + Cl-}\tag{2}$$

These arrows are also given in your source like that; if everything up to a single step is given as equilibria while that one step is given as non-reversible, that should be sufficient driving force to let a reaction happen.

At which point in time the proton is eliminated is somewhat up to debate, of little relevance for the overall reaction and dependent on what is in solution. If just a pure acid chloride and a pure alcohol are mixed the only potential bases that could accept the proton of the tetrahedral intermediate are other acid chloride molecules or other alcohols — or the tetrahedral intermediate’s $\ce{O-}$ atom. All of these are very weak bases and I see no preference for the proton to go anywhere. After we have eliminated chloride, that is added to the set of potential bases. It is also not a strong base; maybe even a weaker base than some of the other options. The big difference is that it is volatile and can diffuse out of the reaction mixture. Thus, with no other bases added, I would actually prefer this sequence of steps.

Once you add a base, I would also prefer deprotonation before chloride is eliminated.


What is the easiest way to concentrate a solution without causing it to hydrolyze? I’ve been working with iron acetate solutions in dilute acetic acid (vinegar) and I’ve been running into a problem. Every time I try to boil off the solvent the solution hydrolyzes to form iron oxides. I’ve given up trying to dry it completely because supposedly iron acetates can’t exist outside of solution, but is there a way of concentrating the solution?

Try adding Na2SO4 to the mix and freeze forming Na2SO4.7H2O. Slow, but at least, no oxygen-related chemistry, discussed below.

As a sidebar, you likely have formed basic ferric acetate from boiling, which adds oxygen, to your acidic (vinegar) and aqueous ferrous acetate mix.

Likely reaction mechanics are complex, see, for example, Table S1 here . Per the source my take of a possible reaction sequence:

$\ce{ O2 (aq) + Fe^{2+} → Fe^{3+} + .O2− }$ ( moderately fast) )

$\ce{ . O2− + Fe^{2+} + 2 H+ → Fe^{3+} + H2O2}$  (fast )

$\ce{ H2O2 + Fe^{2+} → Fe^{3+}  + .OH + OH− }$ ( slow)

$\ce{  .OH + Fe^{2+} → Fe^{3+} + + OH− }$ ( fast )

Net radical reaction:

$\ce{ O2 + 4 Fe^{2+} + 2 H+ → 4 Fe^{3+} + 2 OH-}$

The creation of FeOOH is also possible in less acidic conditions per his source: 'Air Oxidation of Ferrous Iron in Water'.


Theoretical Calculation of Lattice Parameter from Space Group If I am given a compound which belongs to, say the cubic $\text{Ia}\bar 3\text{d}$ space group, is there a known way to calculate the lattice parameter of the compound? In general, can lattice parameters be calculate from knowing what space group the compound belongs to?

No - the spacegroup only gives information about symmetry, not the lattice parameters.

The lattice parameters depend on what is in your unit cell, ie your compound - but even if you know your compound, this is not straightforward, as you don't know its arrangement. And if you do, you should already know your lattice parameters ;)

That being said: if you have additional information - e.g. the radius of a protein you crystallise - you might be able to get a rough estimate of what you are expecting as your minimal unit cell parameters. But this does not take into account any stoichiometry in the crystal, which is impossible to know in advance, so multiples thereof are just as likely.