Dehydration of secondary alcohol via E1 mechanism Write the mechanism for this reaction:

I am trying to do a dehydration of an alcohol using the reagents sulfuric acid and acetic acid. I don't know how I am supposed to use the acetic acid, instead I've been using water. This is my proposed mechanism:

But it doesn't include the acetic acid, and it doesn't say in the question whether it is done in solution with water or not but I only know how to do it with $\ce{H2O}$. My main issue is that I need to put the acetic acid into the mechanism.

Your proposed mechanism looks reasonable. I think that acetic acid primarily serves as an acidic solvent for the reaction. However, in the presence of concentrated sulfuric acid, $\ce{CH3CO2H}$ can be protonated by $\ce{H2SO4}$, like in the first step of this esterification. The resulting resonance-stabilized cation is strongly acidic ($\mathrm{p}K_\mathrm{a} = -6.1$ according to this source), and should readily protonate the alcohol. This would be a possibility to include acetic acid into the mechanism.

$$\ce{CH3CO2H + H2SO4 <=> CH3C(OH)2^{+} + HSO4-}$$

(Image source: ChemGuide)

So in the first step, the alcohol can either be protonated by $\ce{H2SO4}$ or $\ce{CH3C(OH)2^{+}}$, and in the end, the eliminated proton can either be picked up by $\ce{HSO4-}$ or $\ce{H2O}$.


Regarding change in temperature in Bomb Caloriemeter In a Bomb Calorimeter, we measure the change in temperature of water $\Delta T$. But when it comes to calculating the change in internal energy we use $$\Delta U = C_{cal}\Delta T$$ Where $C_{cal}$ is the specific heat capacity of the whole calorimeter. Why is that? What I mean is that we're measuring the change in temperature of the water and assuming that heat from the reaction transferred completely to water, so shouldn't we apply $$\Delta U = C_{water}\Delta T$$ since we're calculating the change in temperature of the water?
Or are we assuming that the change in temperature of the water is the change in temperature of every element in the calorimeter?

Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings), and ignition circuit connected to the bomb. (*)

The whole device known as calorimeter consists of a bomb and a water bath, they are not two separate entities. So, $ C_{cal}$ is actually referring to the heat capacity required of the steel vessel and also the water. The construction of the calorimeter is such that the thermal mass of the calorimeter system completely absorbs the heat of reaction (**).

*: This Wikipedia
**: Refer 29:27 of this video


Estimating for atomic number for an "unknown element" A questions tells me there's this element with 3 isotopes (270.51 amu, 34.07% abundance; 271.23 amu, 55.12%; and 269.14 amu, 10.81%), and to solve for average atomic mass. I have done so and got 271 (270.758) for its atomic mass. Then it asks me to estimate its atomic number, and its location on the periodic table. How do I exactly "estimate"?

ps. I dont think this element actually exists...

Well I found it out. It is seaborgium atomic number 106, mass number 271.

To be found in period B6, group 7.


Mechanism of hydrolysis of polycyclic oxetan The reaction is this:

The only idea I have thus far is if we have a chloride acid that can form an ester, then we might be able to find a suitable mechanism. This would require a chloride reagent.

I have drawn a mechanism but I haven't checked for any references as I am at home now, so it's more like paper chemistry. The chlorine atoms should play some role and possibly disfavour formation of certain carbocations that you would expect in a standard Wagner - Meerwein. It is possible to draw a concerted mechanism (not shown here) but seems very unlikely

**EDIT: I have changed my initial mechanism so it explains better the methyl ester formation (see the comments below). The stereochemistry at C* has also been corrected.

As a general comment you cannot propose a non trivial mechanism without drawing one, it is handwaving. So I did.


Point group from mono to bulk MoS2 Many articles say

even numbers of layers or bulk crystal MoS2 have D6h point group
and odd numbers of layers (including monolayer) MoS2 have D3h point group

But whenever looking at the figure of MoS2, I can't recognize there is a symmetry representing the number '6' which means 6-fold(?). With the point group flow chart,

The final answer is C3h for the bulk MoS2.

If I included the inverse symmetry with C6 symmetry, then it would give D6h? Is the inversion symmetry necessary, then why? Especially, if you can point out the six ford axis in bulk case or two layers MoS2, please let us know.

I got the hind from the reference document that Mo has D3h and S has C3v. And these two are factor group of D6h. But I don't much know about group theory... So definitely it is not D6h?

Here the unit cell of bulk 2H-MoS2

Here the top and side view of 1H-MoS2

This is not a proper answer but can give you a clue. I arrived here looking for the same answer.

I didn't arrive to find the C6 rotation axis in bulk 2H-MoS2. On the other hand, if you look for the symmetry operations of the space group of 2H-MoS2 (P63/mmc, No. 194, D46h):

http://img.chem.ucl.ac.uk/sgp/large/194az1.htm

You can see that there are not any C6 rotation axis for this space group. Nevertheless, there are 6-fold screw axes (http://img.chem.ucl.ac.uk/sgp/misc/sixscrew.htm) and 6-fold inversion axes (http://img.chem.ucl.ac.uk/sgp/misc/barsix.htm).

You are right that the point group corresponding to this space group is D6h (https://en.wikipedia.org/wiki/Space_group#Table_of_space_groups_in_3_dimensions). And according to character tables a D6h point group should have C6 (http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=606&option=4).

Both things are contradictory. My background is Chemistry, and I am used to group theory applied to molecules rather than crystals. My guess is that the 6-fold screw axis, which is a symmetry element that is only possible in crystals (as it requires translational symmetry) might be related with labelling as D6h. But this is just my speculation.

In this article, they also talk about the symmetry operations of single-layer, multi-layer (odd and even layers) and bulk MoS2, and could be usefull for you too:

http://www1.spms.ntu.edu.sg/~qihuagroup/data/Xiong/Papers/nano%202013%20yanyuan.pdf


Deducing the shell when transition occurs such that two photons are discharged Find $n$ (quantum number) corresponding to the excited state of $\ce{He+}$ ion if on transition to ground state the ion emits two photons in succession with wavelengths $\pu{108.5 nm}$ and $\pu{30.4 nm}$.

Solution:

Using Rydberg's equation where $n_1$ is the shell from which the electron transitions to shell $n = 1$ (ground state). $(1)$. $n_2$ is the excited state $n$ from which the electron transitions to $n_1$. $(2)$

When finding $n_1$ the wavelength $\pu{30.4 nm}$ is used, while the wavelength $\pu{108.5 nm}$ is used for the other transition.

My Question: How are the wavelengths coupled with those transitions specifically?

My Guess: Since the absolute value of energy in a transition is inversely proportional to the wavelength of the photon and since the energy to remove electrons from outer shells is lower, the smaller wavelength corresponds to the transition closer to the nucleus (i.e : to the ground state). Is this the reason for this correlation?

You've got yourself all twisted around with 1's and 2's.

The problem states that "the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm." So the first photon emitted has a wavelength of 108.5 nm and the second photon has a wavelength of 30.4 nm.

Furthermore the problem stated that "on transition to ground state." So after the second photon was emitted the helium ion is in its ground state which means that the single electron is in the 1s subshell.

The gist here is that you have to work backwards. Since you know for the 30.4 nm photon was from a transition that ends up in the 1s subshell, then did the electron start in the 2s subshell, or the 3s subshell and so on. So once you figure out the two subshells for the 30.4 nm photon, then you repeat the procedure for the 108.5 nm photon.

The relative wavelengths of the two photons isn't the important clue here. In other words you can't always say that the shortest wavelength will always be to the ground state for some electronic transition.


Meaning of electrochemical cell notation without a salt bridge An electrochemical cell is shown here.

$$\ce{Ag(s)~|~AgCl(s)~|~KCl(1M)~|~Hg2Cl2(s)~|~Hg(l)~|~Pt}$$

I am used to seeing cell representations with a $||$ for a salt bridge between the two electrode solutions (generally with the anodic solution written first followed by the cathodic solution).

I am seeing the above type of representation for the first time. Can someone tell me what it means? (including which are the anodic and cathodic solutions). Also, how do we write the reactions at anode and cathode for the above cell?

This is an example of an electrolytic concentration cell.

Both the cathode and the anode have the same electrolyte present and both are in same solution. Thus, there is no need to separate the electrolyte in two different solution and connect by a salt bridge.

The anode and cathode both are metal-sparingly soluble salt-salt anion electrode. The $\ce {KCl}$ solution here is the common electrolyte.

The anode reaction is:$$\ce  {Ag(\text{metal})(s) + Cl^-(\text{salt anion})(aq) ->  AgCl(\text{sparingly soluble salt})(s) + e^-}$$

The cathode reaction is also similar: $$\ce {Hg2Cl2(s) + 2e^- -> 2\ce{Hg}(l) + 2Cl^-(aq)}$$

Observe that the $\ce {Cl^-} $ is present in both the reactions as a common electrolyte's ion.


Determining solubility of a gas in a molten salt If pressure and temperature are known constants, how can I determine the solubility of a known amount of gas in a mixture of molten salts?

Assume the mixture of salts is homogeneous and the composition (as well as amount of said compounds) is known.

If you know the phase diagram of the gas, and can calculate the exact pressure the gas should have given the volume outside of the molten salt, you can use the measured pressure/temperature to determine any difference in pressure. Then you can calculate the number of moles of gas that must be absorbed to generate the pressure difference.
If the gas is relatively inert (which it probably needs to be if it wont react with molten salt), you can probably use the ideal gas equation:
$$P_0V = NRT$$
If you compare to the actual pressure, you can determine the number of moles absorbed by the salt:
$$\delta N = (P_0-P_\mathrm{exp})V/(RT)$$


SN2 vs SN1 in Acidic Environments for Secondary Alkyl Halides I have researched three books: Brown's, Solomons' and Wade's. I could not find enough information about this.

My prof. says that secondary alkly halides undergo SN1 only under acidic environments.

We had this quiz. I got a terrible mark on one question.

I think these are the missing structures.

Now, I am wondering, did Sn2 occur here, if yes, at what rate? Was it more important than Sn1? What was the major pathway?

What is the major pathway for all secondary alkyl halides with acidic environment and with NONacidic environments?

Could anyone also just check my missing structures?

Note that the starting material is drawn as a chiral molecule.  Solvolysis of this starting material will generate a symmetric, achiral, allylic carbocation.  Since a stabilized allylic carbocation is generated, the mechanism likely proceeds through an $\ce{S_{N}1}$ mechanism.

Both ends (there is positive charge delocalized at both ends of an allylic carbocation - draw the resonance structures and convince yourself)  and both faces (top and bottom) of the allylic carbocation can capture the nucleophilic solvent molecule to restore neutrality as shown below. This results in the formation of two diastereomers of the 3,5-disubstituted cyclohexene - the cis and tran isomers. For each diastereomer, both enantiomers are also produced, as they must be, having passed through an achiral intermediate.


How To Obtain A Reducing Sugar At Home? I'm looking for a reducing sugar such as glucose or maltose (any reducing sugar will do) for some home chemistry experiments. Is there a readily available reducing sugar that I might source without contacting a chemical company? A simple "synthesis" of glucose from sucrose would work as well.

As permeakra mentions, glucose can be purchased pretty easily. It might be labeled as "dextrose".

Another commonly available is maltose (or malt sugar).

Sucrose (or table sugar) is not a reducing sugar, but one of its constituents (glucose) is. Glucose can be made from sucrose by heating with a little bit of acid to make invert sugar syrup. Invert sugar is approximately a 1:1 ratio of glucose to fructose, and only the glucose in this mixture is reducing. An example procedure can be found here:How does adding lemon juice to sugar make better caramel?. Honey is a naturally occurring invert sugar.


Why is the peak-to-peak separation of cyclic voltammetry is 57 mV? I'm studying CV, and I wonder if I can calculate the peak-to-peak separation of CV for a reversible redox system.

I found that the separation is 57 mV on my textbook, and my teacher taught me that is can be calculated theoretically using Nernst equation,
but I think it is more complicated.

Is it easy enough (for sophomore) to calculate the value?
And if it is, I want to know the way.

(Sorry for my poor English)

The derivation of equation for cyclic voltammetry requires a lot of advanced calculus and differential equations based on the Fick's law of diffusion. This is not to discourage you but to show that the teacher is oversimplifying it. The derivation are provided in Allan J. Bard's Electroanalytical Chemistry. Why are they teaching cyclic voltammetry in sophomore classes?


Where to store Sodium Azide? I have about 100g of Sodium azide and would like to store it properly. Any suggestions as to where? I though of storing it in a locked vault but there are chemicals such as Arsenic sulfide, aniline, and Potassium cyanide, but the presence of metals in the cabinet makes me a bit queezy.

Many thanks in advance.

Have a look at the MSDS of the hazardous materials, which you are planning to store together. At least in Germany, these materials are assigned to storage classes (in German: Lagerklasse (LGK)).

The German regulations are to be found in the document TRGS 510, TRGS stands for Technische Regeln für Gefahrstoffe (Technical Regulations for Hazardous Materials). Document 510 of this series specifies Lagerung von Gefahrstoffen in ortsbeweglichen Behältern (Storage of hazardous materials in non-stationary containers). Within TRGS 510, you'll find a "compatibility table" that specifies, which storage classes may (not) be stored together.

Sodium azide belongs to the LGK 6.1 A (flammable , acute toxic hazardous materials of category 1 and 2).

Materials from LGK 6.1 A may be stored together with other materials of the same LGK, but not, for example, with explosives (LGK 1), strong oxidants (*LGK 5.1 A), organic peroxides (*LGK 5.2), pyrophororic materials (*LGK 4.2), etc.


Derivation of the isothermal Joule-Thomson coefficient The definition of the Joule-Thomson effect is:

$$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$

And this is defined in an isenthalpic process, i.e. $dH=0$.

My book shows the derivation of the isothermal Joule-Thomson coefficient ($\varphi$) using the cyclic rule:

$$\left(\frac{\partial T}{\partial P}\right)_H\times\left(\frac{\partial P}{\partial H}\right)_T\times\left(\frac{\partial H}{\partial T}\right)_P=-1$$

Using this cyclic rule,

$$\mu=\left(\frac{\partial T}{\partial P}\right)_H = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_P} = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{C_P}$$

And therefore

$$\left(\frac{\partial H}{\partial P}\right)_T = -\mu C_P = \varphi$$

I understand the mathematics of this derivation, but I don't understand it on a conceptual level.

$\mu$ was derived in an isenthalpic process, with $dH=0$. It owes it existence to the fact that there is no enthalpy change. So how can we now use this variable when $dH\neq 0$?

This question may be a more fundamental misunderstanding on my behalf of how thermodynamics equations are derived. Often I see an equation derived under the assumption that some variable is held constant, but then the equation applied when that variable is not constant any more. The above example is just one where I think this discrepancy is obvious.

When you write a total differential such as
$$dH = \left(\frac{\partial H}{\partial T}\right)_p dT + \left(\frac{\partial H}{\partial p}\right)_T dp \tag{1}$$
you are applying the methods of differential geometry, so in reality the answer to your question lies in the applicability of these methods in thermodynamics (the rest being "math"), something which statements such as "so-and-so is a state function" implicitly justify. One can write a total or exact differential of a state function, as for the enthalpy in the equation above. This equation can be interpreted as follows: small (differential) changes in p and T, which are orthogonal dimensions (in the sense that they can be varied independently), additively cause a linearly proportional differential change in the function H. In the differential limit, the surface of H looks like a plane. The partial derivatives describe the slope of the plane in the orthogonal dimensions.

The cyclic rule can be derived from the above equation by taking the partial derivative wrt one of the independent variables while holding H constant.

$$ 0 = \left(\frac{\partial H}{\partial T}\right)_p \left(\frac{\partial T}{\partial p}\right)_H + \left(\frac{\partial H}{\partial p}\right)_T \tag{2}$$

What does it mean here to hold H constant? It means we are looking for an isenthalpic path on the enthalpy surface, from the initial point at which we computed the partial differentials of the surface wrt T and p, in direction $\left(\left(\frac{\partial T}{\partial p}\right)_H dp, dp\right)$, where the partial differential $\left(\frac{\partial T}{\partial p}\right)_H$ also happens to be given (thanks to the geometry of the problem) by

$$ \left(\frac{\partial T}{\partial p}\right)_H = -\frac{\left(\frac{\partial H}{\partial p}\right)_T }{\left(\frac{\partial H}{\partial T}\right)_p} \tag{3}$$

Alternately, consider the line resulting from intersection of a horizontal isenthalpic plane $c(T,p)=H_0=H(T_0,p_0)$ and the plane $s(T,p)$ tangential to the surface $H$ at the point $(T_0,p_0,H_0)$, the tangential plane given by $$s(T,p) = H_0 + C_p \Delta T + \varphi \Delta p$$

where

$$C_p=\left[\left(\frac{\partial H}{\partial T}\right)_p\right]_{(T_0,p_0)}$$
$$\varphi=\left[\left(\frac{\partial H}{\partial p}\right)_T\right]_{(T_0,p_0)}$$

are the partial derivatives of $H$ evaluated at $(T_0,p_0)$, and $\Delta T = T-T_0,~ \Delta p = p-p_0$. Solving for the intersection line by setting $s(T,p)=c(T,p)$ gives

$$T = -\frac{\varphi}{C_p} p + T_0 + \frac{\varphi}{C_p} p_0 +\frac{c-H_0}{C_p}$$

The slope of the intersection line can be recognized to be the same as $\left(\frac{\partial T}{\partial p}\right)_H$ given by Eq. (3). The geometric nature of the problem and relation between the different derivatives should then be clear.

  So how can we now use this variable when dH≠0?

A Joule-Thompson coefficient $\mu$, like any other thermodynamic (or state) properties, is strictly valid at the conditions under which it is determined (it may have a broader useful range depending on how much it varies with T and p and the tolerated error). This is different from the question of the mathematical accuracy of the relationships used to derive the properties.    $\mu$ is derived at a specific state defined for a pure substance by a specific point (T,p) and as such is a fixed property of the substance at that point.


In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack? In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack ?

My approach:I considered hydroxy group as the stronger +M group.So my guess was para position w.r.t hydroxy group.But that's not the correct answer.

The answer is actually given as ortho w.r.t hydroxy and para w.r.t -CHO group!!

This is known as the ortho effect. According to it if an ortho/para and meta directing group are meta to each other, then the incoming electrophile prefers to come on ortho of meta directing group.

See the reference.


Why 1-ethenyl-3-ethyl-5-ethynylbenzene has that name? I would priorized these funtional groups based on two options (following the ciclohexane rules):

Molecular weight
Representing each pi bond as a alkane branch. e.g. Ethenil as a isopropil backbone.

But, none of these option works. Why this molecule has that name?

Ordering substituents by complexity has been abandoned long ago (before 1979) in favor of the much simpler alphabetic order. Here, ethenyl < ethyl < ethynyl.

Note, however, that simple substituents are ordered alphabetically before adding a multiplicative prefix, which means that "dimethyl" is ordered as "methyl".

Complex substituents are ordered by their first letter, e.g. isopropyl (I), dimethylphenyl (D)


Why it is not yet possible to produce ethanol selectively in the Fischer Tropsch reaction? The Fischer Tropsch reaction between $\ce{CO}$ and $\ce{H2}$ has the potential to give a wide range of products. However by judicious choice of catalytic metal it is possible to produce $\ce{CH4}$, $\ce{CH3OH}$, and other alkanes selectively. $\ce{Fe}$ and $\ce{Co}$ are most commonly used catalysts where $\ce{Fe}$ produces less $\ce{CH4}$, more oxygenated products and more alkenes and $\ce{Co}$ produces more $\ce{CH4}$, less oxygenated products and more alkanes. There are a number of other catalysts that can be used, for example $\ce{Rh}$ which can produce $\ce{C2}$ and higher oxygenates and hydrocarbons. This includes the production of $\ce{CH3OH}$. However, it is not selectively produced but is produced in a mixture of other products and in small amounts. My question is why can $\ce{CH3OH}$ not be produced selectively?

in Fischer-Tropsch synthesis-like reaction there are (roughly) following processes:

$\ce{CO \rightarrow  Surf-CO}$
$\ce{H2 \rightarrow  2 Surf-H}$
$\ce{CO + Surf-(CH2)_{n}-H \rightarrow  Surf-CO-(CH2)_{n}-H}$

3'. $\ce{CO + Surf-H \rightarrow  Surf-CO-H}$
$\ce{Surf-CO-(CH2)_{n}-H + 4 Surf-H \rightarrow
    Surf-(CH2)_{n\,+1}-H + H2O }$
$\ce{Surf-CO-(CH2)_{n}-H + 3 Surf-H \rightarrow
    HO-(CH2)_{n\,+1}-H }$
$\ce{Surf-CH2-CH2-(CH2)_{n}-H \rightarrow Surf-H  +
    CH2=CH-(CH2)_{n}-H}$
$\ce{Surf-(CH2)_{n}-H + Surf-H \rightarrow H-(CH2)_{n}-H}$

All processes happen simultaneously, and tuning catalyst and reaction conditions it is possible to hinder some processes. However, as you can see, as long as you have chain growth at all (that is necessary to have $\ce{C2}$ products, like ethanol), you cannot stop $\ce{C2}$ surface particles to participate in further chain growth, producing $\ce{C_{3+}}$ products. So, if chain growth happens at all, a mixture of products with different chain length is always obtained, and there is no way around it. On the other hand, on proper catalyst in proper conditions it is possible to prevent chain growth, but still allow reduction of carbon monoxide, gaining either methane or methanol depending on catalyst. So, in practice it is possible to obtain one of the following

Methane
Mixture of terminal alkenes and alkanes
Methanol
Mixture of linear terminal alcohols (OK, this case is a bit more complicated, as other oxygenates may be produced as well, giving rather complex mixture, but that's beyond the point)

It is still possible to obtain pure ethanol indirectly (using hydration of isolated ethylene), or using other processes. It is also possible to obtain product mixture where some products, say C2-C4 alcohols, will be dominating.


Reaction of ethene with ammonia and sodium Is the following sequence of reactions correct?

$$\ce{CH2CH2 ->[][\ce{NH3},\,\ce{Na}] NaHC=CHNa}$$

$$\ce{NaCH=CHNa + 2CH3CH2Br-> CH3CH2CH=CHCH2CH3}$$

No, because for ethene, $\mathrm{p}K_\text{a} = 44$. You won't be able to abstract even one proton.

Maybe, you were thinking in ethyne ($\ce{HC#CH})$? In that case, you are right! The bis-alkylation would work with sodium amide as a base!

But is the simultaneous abstraction of both protons the most plausible reaction then?


What does the "N" in N-nitrosoamine mean/stand for? What does the "N" in N-nitrosoamine mean/stand for?  Also, if the functional group is connected to a benzene ring what do I call it?

The "N" means that the nitroso group ($\ce{-NO}$) is attached to the amine nitrogen.

So, for example, in N-nitrosomorpholine the nitroso group is attached to the nitrogen atom of the saturated, heterocyclic morpholine ring:

You could never have an "N-nitroso" group attached to a plain benzene ring, since there is no nitrogen atom in a benzene ring for the "N-" prefix to refer to.

In general, though, you would name aromatic nitroso compounds in the same way as you would those with any other substituent like 'chloro-' or 'nitro-'. For example, the following compound would is named nitrosobenzene:


Are gem-diols on alkenes stable? Geminal diols are unstable, except for a few exceptions like chloral hydrate.

I wish to ask if a gem-diol alkene is also stable. Their structure:

suggests to me that the intramolecular hydrogen bond should stabilize the gem-diol.

What you call a "gem-diol alkene" might also be called an "ene-diol" or "endiol" by analogy to the name enol.

Enols are a tautomeric form of aldehydes or ketones.  They have one double bond and one C-O single bond.  Since "ene-diols" have one double bond and two C-O bonds, the oxidation number of the O-bonded carbon is 2 higher than with enols.  This corresponds to the oxidation number of carboxylic acids.  Thus, "ene-diols" are (theoretically) tautomers of carboxylic acids.

Ene-diols are not stable because they would rapidly tautomerize the more stable carboxylic acid form.

Hydrogen bonds like the one you have drawn are unlikely to stabilize enols very much, for reasons that Jan noted in a related question here on chem.SE.


Relation between speed of sound c and compressibility We know that
$c^2=\frac{\partial p}{\partial ρ}$

The adiabatic compressibility is defined as: $\beta_S=-\frac{1}{V}\frac{\partial V}{\partial p}$ such that the subscript "S" stands for "adiabatic"

How can I show that $c^2=\frac{1}{\rho \beta_S}$ ?

I tried replacing $V$ by $\frac{m}{\rho}$ but I get for $\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}$

With your result

\begin{equation}
\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}
\end{equation}

you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:

\begin{equation}
  \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ .
\end{equation}

Using this and the definition of $c^2$ you get to the desired result:

\begin{equation}
\beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} =
\underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ .
\end{equation}


Does sodium ethoxide (sodium ethanolate) have a boiling point, or does it decompose? According to Wikipedia, $\ce{C2H5ONa}$ melts at 260°C, but no boiling point is given. What will happen if $\ce{C2H5ONa}$ is heated further?

Yes, sodium ethoxide starts to decompose on heating and thus its boiling point cannot be determined. According to [1]:

  The decomposition of sodium ethoxide is observed to start above 573
  K agreeing well with that reported by Blanchard et al. The gaseous
  products formed on decomposition were mainly methane and ethylene with
  minor quantities of ethane, propylene and butylene. Solid residue was
  found to be a mixture of sodium carbonate, sodium hydroxide and
  amorphous carbon. However, in an earlier study on decomposition of
  sodium ethoxide, by collecting and condensing the evolved gases in
  liquid nitrogen and analysis by gas chromatography, the authors have
  reported the gaseous products to be mainly of ethylene and minor
  quantities of propane, heptane, ethanol and water. The authors have
  also stated that the non-condensable gas at liquid nitrogen
  temperature could be hydrogen or methane. The TG trace for
  decomposition of sodium ethoxide in their study indicated a slow and
  steady decomposition right from 273 K with a sharp weight change above
  573 K which implied that the sodium ethoxide could contain some
  amount of ethanol and moisture that evaporated on heating.[...]

Reference:

Chandran, K.; Kamruddin, M.; Ajikumar, P.; Gopalan, A.; Ganesan, V. Kinetics of thermal decomposition of sodium methoxide and ethoxide. J. Nucl. Mater. 2006, 358 (2-3), 111–128. DOI: 10.1016/j.jnucmat.2006.07.003.


Role of solvents in ozonolysis and oz0nolysis of alkynes with water My text book says that when an ozonide (formed from an alkene) is treated with $\ce{Zn/CH3COOH}$, it cleaves to give carbonyl Compounds.

1) Does $\ce{ Zn/CH3COOH}$ mean Zinc in acetic acid (or) zinc or acetic acid. I at first thought that it should be zinc in acetic acid, but then I found a separate mechanism for acetic acid alone, (given below) which made me confused. So, which one is it?

2) If the above mechanism is correct, then why doesn't the oxygen-oxygen bond (being unstable and weak) cleave in the step I've encircled?

Ozonolysis:

Ozone, an allotrope of oxygen, is highly reactive molecule that can undergo a 1,3-dipolar addition across alkenes, forming a 1,2,3-trioxolanes. This compound is unstable and undergoes a cycloreversion to form a carbonyl compound and a carbonyl oxide. These resultant compounds (the carbonyl compound and the carbonyl oxide) then proceed through a second 1,3-dipolar addition to form a 1,2,4-trioxolane, commonly known as an ozonide. To complete the cleavage of the alkene, the ozonolysis (implies that ozone causes to break (-lysis)) is followed by a work-up designed to break apart the ozonide.  The work-up, which can be oxidative or reductive, determines the oxidation state of the products. For example, cis- or trans-alkenes gives two corresponding aldehydes with reductive work-up while same alkene gives two corresponding carboxylic with oxidative work-up.

The most common reductive work up procedure uses dimethyl sulphide (DMS) in the
presence of methanol. Methanol reacts with the carbonyl oxide to form a hydroperoxy acetal, and this is in turn reduced by the dimethyl sulphide to give relevant carbonyl compound and dimethyl sulfoxide. It is also possible for dimethyl sulphide to reduce the ozonide without the presence of methanol as well, although reation would proceed slowly (A new and convenient method for converting olefins to aldehydes: Tetrahedron Letters, 1966, 7(36), 4273–4278). Other reductive methods: $\ce {Zn/AcOH}$, triphenylphosphine ($\ce {PPh3}$), or thiourea ($\ce {NH2C(=S)NH2}$).

Note: Recently, there is a  publications claiming an efficient and convenient method for reductive workup of the ozonolysis reaction using sodium hydrosulfite. However, the final product could be either aldehyde or acid, thus the method is not selective.

Oxidation: The direct conversion of terminal and 1,2-disubstituted alkenes to carboxylic acids is described as oxidative ozonolysis. This can be easily achieved through treatment of initial ozonolysis products with hydrogen peroxide ($\ce {H2O2}$) or other oxidants.

You may also find answers to your Q3 and Q4 from the reference given above.


Why is the reduction by sugars more efficient in basic solutions than in acidic ones? Why is it said that reduction by sugars is much more efficient in basic media than in acidic media as in the following quote taken from Satyanaryana’s Biochemistry, 4th edition page 16?

  In the laboratory, many tests are employed to identify the reducing action of sugars. These include Benedict’s test, Fehling’s test, Barfoed’s test etc. The reduction is much more efficient in the alkaline medium than in the acidic medium. (sic emphasis)

Is it something to do with the reagents used because many of them are just basic solutions, but if the reagent happens to be an acidic solution then will be that statement valid?

This is based in the underlying redox rection. If we take e.g. mannose and attempt to oxidise that, the (unbalanced) half-reaction we need is the following:

$$\ce{C6H12O6 -> C6H10O6 + 2e-}\tag{Ox1}$$

To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about alkaline solutions, let’s do the latter:

$$\ce{C6H12O6 + 2 OH- -> C6H10O6 + 2 e- + 2 H2O}\tag{Ox2}$$

As you can see, the oxidation of sugars consumes hydroxide ions. If you increase the concentration of hydroxide, that will typically increase the reaction rate.

Of course, one should always check what the reduced compound is. In your case, it is always a complexated copper(II) ion and is reduced by the following equation:

$$\ce{2Cu^2+ + 2 e- + 2 OH- -> Cu2O v + H2O}\tag{Red}$$

Even here, if we set up the half-reaction, we notice that we are consuming hydroxide. Thus, the overall redox reaction is:

$$\ce{C6H12O6 + 2 Cu^2+ + 4 OH- -> C6H10O6 + Cu2O v + 3 H2O}\tag{Redox}$$

If a reaction consumes hydroxide ions, it is enhanced by basic conditions. Likewise, if a redox reaction consumes protons, it is enhanced by acidic conditions.


Bromination of a benzene based molecule Using two equivalents of $\ce{Br2}$ and catalytic $\ce{FeBr3}$, compound A can be converted into B. In the same reaction vessel but in multiple steps, the conversion of B into C ($\ce{C11H12Br3I}$) can be obtained using $\ce{H2SO4, NaNO2}$ and $\ce{NaI}$.
The molecule A is below:

My Attempt
What I am struggling with is determining what double bonds the $\ce{Br2}$ is added on. I am pretty sure that the double that is not on the benzene ring will get brominated, but I am not sure what other double bond will also get brominated.
I know the mechanism for the conversion for B into C, which is basically the sandmeyer reaction followed by substitution of iodine. However what further confuses me is that the end product only has 3 bromine atoms, not 4 bromine which is what I expected (since bromination occurs twice).

You can deduce the end product from considering the change in the molecular formula from start to finish. Aromatic bromination results in the loss of one hydrogen and gain of one bromine, bromination of the double bond results in no loss of hydrogen but gain of two bromines and the Sandmeyer reaction results in loss of two hydrogens. Therefore the intermediate must have fourteen hydrogens which means only one aromatic bromination has taken place.

The aromatic bromination likely produces a major product with the bromine at the ortho position shown. The amine group strongly activates the positions ortho and para to it so substitution will occur at one of these three positions. I suspect the one shown is most likely due to the bulky nature of the other attatched group, which stericically hinders attack at the other two positions.


4-Bromophenol vs 4-aminophenol pKa in water comparison confusion Recently the other week, I was asked the following question:

  Would you expect 4-aminophenol to have a higher or lower pKa than 4-
  bromophenol when measured in water?

And I thought about this on the way home, and decided to try and justify this using resonance.

My feeling was that Nitrogen couldn't handle 4 bonds and an extra lone pair on its atom, so the negative charge won't leave the benzene ring, whereas the Bromine could handle the double bond (I might have been mistaken with the negative charge on the Bromine - if so, do let me know please.)

Using this, I thought that this is the reason why 4-aminophenol would have a higher pKa than 4-bromophenol as its resonance form is not as extended as Bromine's, which is at a lower energy level to be able to handle the extra negative charge, plus all the carbons in the ring have 4 bonds as normal.

Is this a correct answer and justification?

Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).

Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while  p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol  and  p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).

Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.


Finding electron configuration by following the Aufbau principle In the question, "Electron Configuration of Tellurium", there is mention of the 'follow yellow brick road' method of finding electron configuration.

What I'd learned in the past was to find the nearest noble gas (in the row above the element of interest) and count from there.

I searched in Google for this method and see that it appears to also called the Aufbau principle & diagonal method. However I'm not getting a good grasp on how it works.

Could someone explain this method to me?

Also, would this method work for something with a weird electron configuration, like copper?

The so-called rule is to fill the subshells in increasing order of $n+l$. For subshells with the same value of $n+l$, the subshell with lowest $n$ is filled first.

$$\begin{array}{cc} \hline
n+l & \text{Orbitals} \\ \hline
1 & \mathrm{1s} \\
2 & \mathrm{2s} \\
3 & \mathrm{2p,3s} \\
4 & \mathrm{3p,4s} \\
5 & \mathrm{3d,4p,5s} \\
\vdots & \vdots \\ \hline
\end{array}$$

For all intents and purposes, this is exactly equivalent to your heuristic of "go from the nearest noble gas", since to get the configuration of the noble gas itself you have to go through this same ordering of subshells. You can easily verify that the table above reproduces the order of filling of the Periodic Table.

Chromium and copper are exceptions to the rule (obviously there are many more).


Why does potassium react the most vigorously with water? Link to exam

Which metal reacts most vigorously with water?
(A) Ca (B) K (C) Mg (D) Na

The given answer is B and the reasoning is that is is the most electropositive since it the furthest left and down according to periodic trends.
However, isn't $\ce{K}$ the conjugate acid from a strong base -- $\ce{KOH}$? So isn't the conjugate base of a strong acid quite weak? So why would $\ce{K}$ react strongly?

I interpret "reacts most vigorously" to mean which metal reacts the fastest.  This is a question of kinetics, not thermodynamics (e.g. the answer is not based on the enthalpy of the reaction).  Kinetic rates are determined by the height of the energy barrier that needs to be surmounted, we need to determine and compare the activation energies in this process for the various metals.  The activation energy in this process is determined by the following two steps: $$\ce{M(s) -> M(g)}$$ $$\ce{M(g) -> M^+(g) + e^-}$$ $$\ce{or}$$ $$\ce{M(g) -> M^{+2}(g) + 2 e^-}$$ depending on whether we are analyzing a group I or group II metal.  The following link provides a nice analysis and tabulation of the energies required for these two steps with the group I metals

Group I link

and this link does the same for the Group II metals

Group II link

I'll summarize their findings

Metal ...   Activation Energy (kJ/mol)

K   ...         +508

Na  ...        +603

Mg  ...       +2200

Ca  ...       +1950

CONCLUSION: Potassium has the lowest activation energy and should react the fastest.


What happens if a buffer is made up with only half as much NaH2PO4 compared to Na2HPO4? A solution of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$, when provided in equal amounts at the same concentration, form an excellent buffer. What happens is that the two sodium-phosphate compounds each dissociate independently:
\begin{align}
\ce{Na2HPO4 &<=> HPO4^2- + 2Na+}\tag1\\
\ce{NaH2PO4 &<=> H2PO4^- + Na+}\tag2
\end{align}

What would happen if you made up a solution that had only half of the reactant in (1)?

Am I right in saying that Le Chatelier's principle is not involved, and that shifts in equilibrium don't occur?

Will the solution just have a lower buffering capacity for bases relative to acids?

You are not right, Le Châtelier's principle plays a very important role.

Let's have a look at the reactions involving disodium monohydrogen phosphate, as it is an amphoteric substance and therefore a buffer on its own:
\begin{align}\ce{
Na2HPO4 + H2O &<=> 2Na+ + H3+O + PO4^3- \\
Na2HPO4 + H2O &<=> 2Na+ + {}^{-}OH + H2PO4^- \\
}\end{align}
To a small extent, there will also be the formation of phosphoric acid, but we'll ignore that for the moment.

Let's ignore the counter ion and formulate the equilibrium constants:
\begin{align}
\ce{HPO4^2- + H2O &<=> H3+O + PO4^3-} &
K_1 &=\frac{c(\ce{PO4^3-})\,c(\ce{H3+O})}{c(\ce{HPO4^2-})\, c(\ce{H2O})} &
= K_\mathrm{a}\cdot c(\ce{H2O})\\
\ce{HPO4^2- + H2O &<=> {}^{-}OH + H2PO4^-} &
K_2 &=\frac{c(\ce{H2PO4^-})\,c(\ce{{}^{-}OH})}{c(\ce{HPO4^2-})\, c(\ce{H2O})}&
= K_\mathrm{b}\cdot c(\ce{H2O})\\
\end{align}

The whole reaction can be described with a coupled equilibrium constant:
$$K = K_\mathrm{a}\cdot K_\mathrm{b}\cdot c^2(\ce{H2O})$$

At all times in a pure solution there will be phosphate, monohydrogen phosphate, dihydrogen phosphate and maybe even phosphoric acid in solution. Changing any of the concentrations of these species will change the equilibrium.
The same equations can be applied to the sodium dihydrogen phosphate system. The initial concentrations of course will be different.
Depending on the ratio of the two solutions, the pH will also change.

The buffer capacity is of course not only dependent on the overall concentration, but also if it is near the ideal buffer point.

Overall, polyprotic acids are very complicated systems and it solely depends on your understanding of the word "excellent" how the buffer behaves. There are many fine tuned buffer systems for various purposes, this website has quite a few neat tables.


Benzyne formation with Grignard reagents Nucleophilic substitution through an elimination-addition mechanism (involving benzyne formation) occurs with a strong base. $\ce{NH2-}$ can show this reaction. Can Grignard reagents $\ce{RMgX}$, which form $\ce{R-}$ - an even stronger base - show this reaction?

NAS reactions normally require strong electron withdrawing groups to facilitate the addition.  Normally these groups, nitro, carbonyl, cyanide, are preferentially good electrophiles for Grignard reagents (nucleophiles).

So no, organomagnesium reagents will not react via NAS.

You also mention the Benzyne mechanism (which is somewhat different).  In this case, where nucleophilic attack on other groups is not possible, I expect metal-halogen exchange due to aryl anions being slightly weaker bases than alkyl anions.  After said exchange benzyne formation is possible.


What makes cola efficient as a rust remover? I actually tested submerging a bunch of old rusty knives blades in a cup of plain cola for a few days not expecting much. To my amazement the myth held true, coke is a really good rust remover.

So what is the chemistry behind it?

Rust consists of insoluble iron oxides and hydroxides, $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Acidifying these compounds protonates the oxide and hydroxide anions to for water:

$$\ce{O^{2-} + 2H^+ \rightarrow H_2O}$$
$$\ce{OH^{-} + H^+ \rightarrow H_2O}$$

This leaves the soluble iron $\ce{Fe(III)}$, which can be washed away.

Many colas contain phosphoric acid, and are quite acidic (Coke has a pH of ~2.5 according to this science fair project), thus they can dissolve the oxides.


Reason behind more Reactivity of Unsaturated Hydrocarbons My textbook says that unsaturated hydrocarbons are more reactive than saturated ones. But double and triple bonds are stronger than single bonds.
What is the explanation for this?

The unsaturated regions are electron-rich and typically allow for electrophilic attack, such as by halogens or protons. Saturated hydrocarbons usually react only with radicals (radical halogenation). C-H activation is often hard in saturated hydrocarbons.

While double and triple bonds are stronger than a single bond when looking at complete cleavage of a bond, one notices that the bond strength of a double bond is less than twice the bond strength of a single bond (with other factors being comparable, there are surely some exceptions). When a double bond reacts, it typically is changed into a single bond, not fully cleaved. Similar arguments hold for triple bonds.


Equilibrium constant for heterogeneous equilibria having aqueous as well as gaseous reactants Suppose we have a heterogenous equilibrium :
$$\ce{A(aq) +B(aq) <=>  C(g) +D(aq)}$$
Which equilibrium constant is used here?
Both pressure and concentration terms are there. So, according to me, both $K_\mathrm{p}$ and $K_\mathrm{c}$ can be written separately. If that is the case, can we relate both of them by $K_\mathrm{p} = K_\mathrm{c} (RT)^{\Delta n}$?

Your conception of the equilibrium constant is flawed. $K_\mathrm{p}$ is preferably used when the reaction only contains gaseous and solid components whereas $K_\mathrm{c}$ is used primarily for  a reaction taking place in water.
In reality, however, both of these equilibrium constants are special cases of equilibrium. This is because in the actual scenario, what we use to calculate the equilibrium constant is a concept known as activity.
$K_\mathrm{p}$ and $K_\mathrm{c}$ are said to be special cases of $K$ since, in $K_\mathrm{p}$, the activity is expressed in terms of the partial pressure of the components and in $K_\mathrm{c}$, the activity is expressed in terms of concentrations of the products and reactants.
In case of reaction as stated above:
$$\ce{A(aq) + B(aq) <=> C(g) + D(aq)}$$
The equilibrium constant would be expressed as follows:
$$K = \frac{a_\mathrm{C}\cdot a_\mathrm{D}}{a_\mathrm{A}\cdot a_\mathrm{B}}$$
Now, for the aqueous components, the activity is expressed in terms of their concentration in the solvent. For the gaseous component, we express it in terms of its partial pressure. Substituting these conditions, we get:
$$K = \frac{P_\mathrm{C}[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}$$
An example of where this is used is in electrochemistry.
A galvanic cell may be made which has a heterogenous reaction in which solid, aqueous and gaseous components may be given.
The Nernst equation used to calculate the EMF of such a  cell is given as:
$$E_\mathrm{cell} = E^\circ_\mathrm{cell}-\frac{2.303RT}{nF}\log_{10}Q$$
Here, $Q$ is the reaction quotient. Using this, we calculate the  EMF taking all the activities into consideration.
Note: The activity of a solid or the solvent is taken to be $1$ in such cases.


What happens to the chlorine in epichlorohydrin when a two-part epoxy is air-cured? Where does the chlorine go? Does it evaporate as a by-product or does it remain in the epoxy? I'm using a two-part resin and hardener and curing them at room temp. The resin is bisphenol-A-epichlorohydrin, and I believe the final cured epoxy does not have chlorine as a part of its main structure. I need to know the whereabouts of Cl because I'm trying to study neutron interaction with materials in my detector, and I suspect there's some chlorine in there but not sure entirely.

There is little if any chlorine in a epichlorohydrin / bis phenol A expoxy resin, it is normally lost during production as sodium chloride into the aqueous phase while the resin forms an organic phase.

The normal way to make the resin is to react bisphenol A with sodium hydroxide and the epichlorohydrin. What happens is that a mixture of the three reagents and water is heated up together. The phenol groups on the bisphenol A are deprotonated and then they attack the epichlorohydrin to then form 2,2'-(((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(methylene))bis(oxirane). The epoxide (oxirane) groups in this then react with phenolate anions from new bisphenol A molecules to form 4,4'-((((((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(2-hydroxypropane-3,1-diyl))bis(oxy))bis(4,1-phenylene))bis(propane-2,2-diyl))diphenol. The unreacted phenol groups on this then are reacted with more epichlorohydrin to form new reactive groups.

This reaction continues on and on to form a polymer, it is normal to use a small excess of epichlorohydrin to give a epoxy terminiated polymer (resin). This is then reacted with a cross linking agent such as an amine to form the final epoxy. As a result the chlorine content of the epoxy should be very very low.


What is the purpose of trimming the bottom corners of TLC plates? During my chemistry lab course, I was told by my instructor to make 2 small cuts on the bottom corners of the TLC (Thin-Layer Chromatography) plate. Does this make the solvent rise uniformly on the plate? Is this a standard practice in labs? I could not find any information online. In this question the OP has made the cuts as well, but this does not seem to make the solvent rise uniformly.

Yes, your question contains the answer. Think about capillarity.

Without the little cuts the eluant would be "sucked" and flow from both the bottom and the left ( and right) edges of the thin plate, resulting in a tilted flow that

tends to move the elute faster along the vertical edges while clumping the spots to the center;
decrease the resolution at the given conditions by deforming and enlarging the spots.

Think of it as a weird bad executed two dimensionally-developed plate, if it helps.

Note that beside resolution, the cuts should be there to guarantee a levelled horizontal and straight flow front, as requisite to obtain reliable and useful ratio front values.

(Good practice: draw the real appearance of the developed plate or even post it to the lab journal, so that a future user get a feeling of the separation whatever it happened. The latter not recommended if a product spots out to be irritant/allergenic).

Left: correct - Right: not horizontal flow frontline due to capillarity at vertical edges.

As OP mentions an example with cutted corners but not levelled flow, obviously cutting the corners but plunging the plate in too much eluant is useless. The idea is to have eluant sucked up just by the bottom edge.


Why is NH3 a stronger base than PH3? I know that the orbitals of nitrogen overlap better with the 1s orbitals of hydrogen, but I don't know how to use this information to say why ammonia is stronger than $\ce{PH3}$.

From the Wikipedia article for phosphine:

  The low dipole moment and almost orthogonal bond angles lead to the
  conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H)
  and phosphorus 3s orbital contributes little to the bonding between
  phosphorus and hydrogen in this molecule. For this reason, the lone
  pair on phosphorus may be regarded as predominantly formed by the 3s
  orbital of phosphorus. ... This electronic structure leads to a lack
  of nucleophilicity and an ability to form only weak hydrogen bonds.

And, one might assume, likely also explains the lowered basicity of phosphine relative to ammonia.

Ammonia is classically $sp^3$-hybridized, which entails that the lone pair resides in an an $sp^3$-orbital. Based on empirical evidence (some of which is mentioned above), phosphine is less $sp^3$-hybridized, with the lone pair residing in an orbital that is more like an $s$-orbital than an $sp^3$-orbital. Thus phosphine's orbital has more s-character, which means the electron pair will be held closer to the nucleus and be less inclined to bond. This is an argument rationalizing phosphine's lower basicity than ammonia.

Countering this argument, one might argue that phosphine's lone pair orbital is a $3s$, which is inherently bigger than the $2s$-orbital ammonia uses for hybridization. So "after"* hybridization, ammonia's $sp^3$-orbital is certainly bigger than its original $2s$ orbital ($2s$ has more s-character than $sp^3$), however, it may not actually be bigger than phosphine's $3s$. Thus, perhaps ammonia's hybridized $sp^3$-orbital is not so different in size than phosphine's $3s$-orbital, I suppose we would need a quantum computation to know for sure.

At the end of the day, we are only trying to rationalize an observed phenomenon after the fact. In spite all of our human ideas of orbitals, hybridization, etc, nature does as it pleases. Perhaps our models can offer a correct explanation and someone can interpret it better than I did (which shouldn't be too difficult), but also for some phenomena we must realize that our models don't fit perfectly and there are exceptions to every trend.

*Quotes used because ammonia is inherently hybridized, the act of nitrogen's $\{s, p_x, p_y, p_z\}$ atomic orbitals hybridizing when the molecule forms is a human idea that helps explain empirical observation, but may not necessarily take place physically. Indeed, the very $\{s, p_x, p_y, p_z\}$ orbitals themselves are only exact solutions for the hydrogen atom (as treated by quantum mechanics), and are only approximations of the orbitals of all other elements. But let's not get bogged down in all that.


How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide? Bismuth(III), if I am not mistaken, has only two valence electrons in the 6s orbital. However it is able to form 3 bonds with hydroxide. This goes against my intuition. How is it able to form 3 bonds with 3 hydroxide molecules with only 2 valence electrons?

my best guess is that i am mistaken in thinking that the bismuth(III) is forming 3 covalent bonds when in actually it is forming ionic bonds. As a result, bismuth does not need to offer/share 3 electrons to each hydroxide, but simply take 3. Is this correct?

There isn't a unique sturcture for bismuth hydroxide, but ions such as

$$\ce{[Bi6O4(OH)4]^{6+}}$$

are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.

See Solvation of the Bismuth(III) Ion by Water... Inorg. Chem. 2000, 39, 4012-4021 as a starting point.

But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.


When I heat up a balloon, does the air inside increase in pressure as well as volume? When I heat up a balloon, does the air inside the balloon increase in pressure as well as volume?  I thought pressure and volume were inversely proportional?  Or does pressure and volume increase as temperature increases?

If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first.

If the gas were completely free to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force per area (gas pressure) as the force per area acting on it (ambient pressure), so that the forces cancel out and the piston stops moving. Here, a temperature increase in the gas would translate solely to a volume increase.

If the gas were confined in a perfectly rigid box, then an increase in gas temperature would cause the molecules inside to bump harder against the inner surfaces, but to no avail, as the walls do not budge and the box stays exactly the same size. Here, a temperature increase in the gas would translate solely to a pressure increase.

In a balloon, the gas is free to expand, but not completely free. In other words, it's a situation somewhere between the two described above. The balloon skin is an elastic which pulls itself, creating a force vector pointing towards the interior of the balloon, and the more the skin is stretched, the stronger the force becomes. Now the gas inside the balloon has to create enough pressure to compensate for both the ambient pressure, and the elastic force trying to pull the balloon skin inwards. This means that after heating, the gas inside a balloon will expand since the balloon is not perfectly rigid, but the equilibrium pressure of the gas inside the balloon will be higher than before because the balloon is pressing against a more tightly stretched balloon.

This should be provable experimentally without much difficulty. Take a rubber balloon, preferably one with as thin of a membrane as you can get, and open its mouth to the atmosphere. Then, clamp it shut without blowing any air inside (in reality, you might have to blow some air in to unstick the rubber walls, but then let all the excess air out). The air inside will have a pressure exactly equal to ambient pressure, because the elastic is not being stretched. Now cover the balloon completely in warm water for a few minutes, and it should inflate slightly. Finally, remove the balloon from the water and quickly perforate it with a sharp object. You might be able to hear a small pop, and feel a rush of air. Both are an indication that pressure inside the heated balloon was higher than ambient pressure.

Notes:

Here is a very related Physics.SE question - Why is the pressure inside a soap bubble higher than outside? which deals with bubbles rather than balloons. The diagram and equations are applicable in both cases.
Evidently, taking into account the elasticity of the balloon requires a more subtle treatment of the problem. Take a look at Floris' interesting input in the comments. It seems that even an idealized balloon starts off acting as a quasi-rigid container. After reaching a maximum pressure, the balloon starts expanding, and from there the walls get weaker as expansion continues. This means that for a range of temperatures higher than some critical value, the pressure of the gas actually decreases as the temperature increases. The interior pressure of the balloon will still always be greater than ambient pressure, though.


Is Blue Silica Gel more harmful than the Orange ones? I am currently searching for a portable dehumidifier but notice that most of them come with a indicator that actually using the Blue Silica Gel, which I believe that it may contain Cobalt Chloride.

So, I would like to ask,
(1) if that is the case, would the portable dehumidifier become a harmful product?
(2) would an Orange Silica Gel contain other harmful chemical that is currently not found?

I had a poke around in an effort to determine the dye used in orange silica gel and was able to find two material safety datasheets (MSDSes) for orange silica gels which both gave different empirical formulae for the dye ($\ce{C25H30ClN3}$ and $\ce{C20H14O4}$), which suggests that there are multiple formulations of orange/yellow silica gel. Both of these dyes are organic molecules (as opposed to transition metal salts), and now their possible identities are given at the bottom of this response. Suffice to say that orange silica gels were designed specifically to avoid the use of cobalt chloride, which is fairly toxic.

As a document1 from the US National Parks Service indicates, the chief danger of cobalt chloride is its environmental toxicity if it enters soil or water through improper disposal. Cobalt salts are often rather evil, and this is actually reflected in the name, which is ostensibly an alteration of 'kobold'. However as long as the beads are kept in a suitable enclosure such as a filtered dessicator plug, they should not present a hazard. Inhalation of dust from the silica gel beads, direct handling of beads or cobalt chloride-impregnated materials or complete inundation of the dessicant (which could dissolve some of the cobalt chloride) are three possible ways the cobalt chloride could present a direct hazard, but the disposal of blue silica gel-containing products in landfill and subsequent leaching of cobalt chloride is probably the chief concern.

As far as the orange beads go, as I was not able to determine the specific indicator molecule(s) I can't ascertain their toxicity, suffice to say that they are by design less harmful than cobalt chloride and appear to be generally considered safe under normal use.

P.S. (finally, after chrome crashing multiple times)

F'x points out that $\ce{[C25H30N3]^{+}Cl^{-}}$ is crystal violet, which is not super toxic but for which there is evidence of carcinogenicity2 and which is considered an environmental pollutant.

$\ce{C20H14O4}$ may very well be phenolphthalein, which is considered to be a carcinogen by some authorities, most notably appearing on the California Prop. 65 listing3.

I guess the idea is that cobalt chloride is less desirable than either of these dyes. Don't eat the beads, and wash your hands!

(1) U.S. National Park Service; Cobalt Indicating Silica Gel Health and Safety Update; 2005, pp. 1-3

(2) Littlefield, N.A.; Blackwell, B.-N.; Hewitt, C.C.; Gaylor, D.w.; Chronic toxicity and carcinogenicity studies of gentian violet in mice; Fundamental and Applied Toxicology, 2004, pp. 902-912

(3) Chemicals Known to the state to cause cancer or reproductive toxicity; State of California EPA; March 16 2012, p. 17


How much of Ag has formed on the Cu tile? $\pu{9.547 g}$ $\ce{Cu}$ tile is added to an $\ce{AgNO3}$ solution. After some time, $\ce{Cu}$ was taken out of the solution, washed, dried and weighed. The mass appeared to be $\pu{9.983 g}$. How much of $\ce{Ag}$ has formed on the $\ce{Cu}$ tile?

The right answer to be found is supposed to be: $\pu{1.08 g}$.

My imaginary equations (with unlikely scenarios):

\begin{align}
\ce{Cu + 2AgNO3 &-> Cu(NO3)2 + 2Ag}\\
\ce{Cu + AgNO3 &-> CuNO3 + 2Ag}
\end{align}

I cannot solve this problem. There are a few others, but this is bugging me the most. How does one solve it? I believe this is something I wasn't taught before or there is again a mistake in the references given.
I've tried calculating in ways I thought was right and anyhow, I cannot find the answer.

After discussions, I have come to the conclusion that I am taking $\pu{0.618 g}$ as the solution to this problem and will look forward into it tomorrow with the teacher as well.
This is the only closest answer I seem to arrive at by myself and others, as we get $\pu{0.618 g}$ when using the $\Delta m_\mathrm{p}=\pu{0.436 g}$ and $\Delta m_\mathrm{t}=\pu{22.853 g}$ differences. However, the problem regarding the other exercises still persists, which are of similar nature.
I am starting to question my teacher's logic and abilities.

There is a close answer to the referenced answer, which assumes the $\ce{Cu}$ having a charge of $+1$, thus $\ce{Cu + Ag+ -> Cu+ + Ag}$ and $\Delta m_\mathrm{p}=\pu{0.436 g}$, $\Delta m_\mathrm{t}=\pu{16.2 g}$.
However, I assume that to be a very unlikely scenario considering the current circumstances. This scenario lets you acquire \pu{1.0617 g} as the answer.

In this experiment the tile lose mass due to oxidation and dissolution as Cu dication (Cu++) and acquire mass due to the reduction and deposition of Ag.

This is what is measured by weighting before and after.

This difference in mass correspond to 2n moles of Ag minus n moles of Cu so is just matter of solving a first degree equation.

Quickly, without being consistent in digits and truncation.,
solving it for the current data gave about 0.6 g of Ag.

All others outcomes seems to me to assume an unlike stoichiometry or refuses perpetuating around.

A way to get into an "effective stoichiometry" of 1:1 is to consider formation of Cu+ followed by disproportionation AND assuming that all metal copper finely precipitates without depositing back to the tile from which originally came from. While, partially occurring, this scenario would explain discrepancies found performing a real experiment, it seems unlikely that is intended as to give a fixed result as for it  requires the assumption in italic above. Doing that the amount of silver is now about 1.06 g.


Number of atomic and molecular orbitals in different Hartree-Fock references For a given molecule and electronic configuration,

Are the number of contracted atomic orbitals (AOs) different in a restricted open-shell Hartree-Fock (ROHF) calculation from an RHF calculation? What about unrestricted Hartree-Fock (UHF)?
Are the number of molecular orbitals (MOs) different in a ROHF calculation from an RHF calculation? What about UHF?

For example, if we are using the STO-3G basis set for a water molecule, we get 1 + 1 + 5 = 7 contracted functions for AOs in an RHF calculation, leading to 7 MOs. Are any of these numbers different in an ROHF calculation? What about UHF?

Essentially, what are the core differences between AOs and MOs within each approach?

For water molecule using sto-3g basis I obtained the orbitals in the following order For RHF: Occupied (A1) (A1) (B2) (A1) (B1) Virtual (A1) (B2) For ROHF: Occupied (A1) (A1) (B2) (A1) (B1) Virtual (A1) (B2) But if we take alcohols for an e.g., the number of such orbitals is not the same. I want to know how they are calculated, so that atleast I could do them by hand. For the RHF one, it is nicely explained in "Modern Quantum chemistry" by Szabo-Ostlund. I want some similar information on ROHF and UHF (if possible).

Straight answer to your question: whatever method you use, RHF, ROHF or UHF, your calculations will use the same number of basis functions (7 contracted functions for your water example). How will they be used and combined, however, is different. In your comment on the question you talk about molecular orbitals - it's important to keep these conceptually separate from the basis set you are using to construct the orbitals.

How many functions do you use for an orbital in ROHF?
In general, UHF/ROHF approaches use a separate molecular orbital for each electron, so you would use basis functions twice to describe an unrestricted pair compared to a restricted pair.

In other words, basis functions aren't "split" between the electrons when treated as an unrestricted pair; rather, two orbitals are calculated using the same basis functions, one corresponding to the alpha electron and another for the beta electron.

So, if you compare UHF with RHF, you'll be using the same basis functions to construct twice the number of molecular orbitals.

Let's take an example to illustrate it: $\mathrm{O_2}$.

For singlet $\mathrm{O_2}$ ($\mathrm{S=0}$), there are 8 alpha and 8 beta electrons. Using STO-3G, we'll have 10 contracted functions, those corresponding to $\mathrm{1s}$, $\mathrm{2s}$ and 3 x $\mathrm{2p}$ for each oxygen atom, each of them composed of 3 primitive gaussians.

In RHF, we can combine these 10 basis functions in a single Slater determinant, producing 10 molecular orbitals. The first 8 of these will be doubly occupied, the last 2 will be vacant.
In ROHF, we have exactly the same as for RHF, because ROHF is exactly the same as RHF for molecules with $\mathrm{S=0}$.
In UHF, we use the same 10 basis functions, but we use them to build two sets of 10 molecular orbitals - 10 for the alpha electrons and 10 for the beta electrons - in a single Slater determinant. We occupy them with 8 alpha electrons and 8 beta electrons, 1 electron per orbital, with 2 alpha and 2 beta orbitals left vacant.

For triplet $\mathrm{O_2}$ ($\mathrm{S=1}$), there are 9 alpha and 7 beta electrons. We have the same basis functions as before.

In UHF, we do the same as before, and we also get 10 alpha and 10 beta molecular orbitals in a single Slater determinant. Occupation changes, though, so now we have 9 occupied and 1 vacant alpha orbitals and 7 occupied and 3 vacant beta orbitals.
In ROHF, and since now we have an unequal number of alpha and beta electrons, we have a more complex construction of the Fock operator - one that is not unique. Essentially, we'll calculate alpha and beta electrons separately under the hood, and then combine them; and we'll be treating full, partially full and empty orbitals as three coupled sets. We'll produce 10 orbitals, of which we'll doubly fill the first 7 and partially fill the following 2, leaving 1 unoccupied.

In conclusion: the number of basis functions in all cases is the same: 10 contracted functions comprising 30 primitive gaussians. However, these basis functions are used a different number of times for RHF, UHF and ROHF; RHF uses them once in a simple Slater determinant, UHF uses them twice to form a Slater determinant with a two sets of orbitals, one for alpha and the other for beta electrons, and ROHF uses them twice but ends up producing a single set of full, partially or non-occupied orbitals that share their spatial part.


difference in boiling point of 2-iodopentane and 3-iodopentane There is no difference in boiling point of 2-fluoropentance and 3-fluoropentane, so is chloro- and bromo-. But why is there a difference in the boiling point of 2-iodopentane and 3-iodopentane, up to 10 degree Celsius?

I have spent a quite a time to figure out "so-called" irrational behavior of the boiling points of iodoalkanes compared to relevent othe halpalkane, but I have seen noting really radical, what OP claimed there are. The following chart listed boiling points of all possible long chain halides of pentane (all have been measured at $\pu{760 mmHg}$):
$$
\begin{array}{ccc}
\text{locant} &\text{fluoropentane} & \text{chlororopentane} & \text{bromoropentane} & \text{iodopentane}\\\hline
1 & \pu{62.8 ^\circ C} & \pu{108 ^\circ C} & \pu{130 ^\circ C} & \pu{155 ^\circ C}\\
2 & \pu{58.4 ^\circ C} & \pu{99 ^\circ C} & \pu{117 ^\circ C} & \pu{150 ^\circ C}\\
3 & \pu{59.5 ^\circ C} & \pu{97.5 ^\circ C} & \pu{119 ^\circ C} & \pu{145 ^\circ C}\\
\end{array}
$$

Please be aware that, all of these values are extracted from EPA(USA) website. Also note that values of 2-fluoropentane and 3-fluoropentane are predicted on EPA, but ChemSpider gave experimental values of $\pu{56 ^\circ C}$ and $\pu{56.7 ^\circ C}$, respectively. Also, the boiling point of 3-iodopentane ($\pu{145 ^\circ C}$) has also been predicted, but it is available commercially at Alfa Aesar, which listed the boiling point of it as $\pu{146.6 ^\circ C}$. Thus, these listed values are very reliable. Therefore, I'd say that the values OP has must be not so reliable. I admit that I have to dig deeper to find a reliable source of those predicted values. All three bromo compounds have no problem because all of them are available in Sigma-Aldrich.


Can a product be alcohol free while containing isopropyl myristate or isopropyl palmitate? I always find isopropyl myristate and isopropyl palmitate in cosmetics, and I want to know if they have alcohol in them or not. I read that they are composed of alcohol and acid, but they are considered to be conditioning agents, so does that mean that there is no alcohol left in them? So can a finished product be alcohol free while containing isopropyl myristate or isopropyl palmitate?

Isopropyl myristate and palmitate are esters formed from isopropyl alcohol and myristic and palmitic acids. These products will have been purified before being used in the cosmetics to remove any residual isopropyl alcohol. If the labelling does not mention isopropyl alcohol as a component then it will not be present.

Note. I don't know your reason for wanting to know about the presence of isopropyl alcohol but if you are concerned for ethical reasons about the presence of alcohol then you should be aware that what is generally referred to as 'alcohol' is ethanol. Isopropyl alcohol is a different substance.


What is the balanced equation of BH3/THF with an amide to form an amine? I am doing a green chemistry project as an undergraduate and need the balanced equation to complete the calculations. I cannot find the right information online so any help would be greatly appreciated.

The reduction of secondary amides with BH3 requires 3 equivalents, mechanism shown below image sourced from here. The borane becomes HB=O which may transform further depending on work up.


Is it possible to make polyethylene in a high school laboratory? We learned how to make small amounts of ethene from ethanol using sulfuric acid.

Is there a way to make a visible amount of polyethylene from that using materials I could get for a school lab? We can't do pressures of 1000+ bars obviously, but we don't need high efficiency either -- just enough of a solid for the students to see that something happened.

Not polyethylene, but I recall doing an experiment in school where hexamethylenediamine and adipic acid were poured carefully so one was layered on top of the other.  They did not mix in bulk, but reacted at the contact surface to form a Nylon-type polymer barrier.  We poked a rod in to pull the polymer out (enabling more reaction) and wind it around the rod!


Which reaction between Lugol and Hydrogen peroxide? The Lugol solution is a mixture of $\ce{(K+, I- )}$ and $\ce{I2}$ such that $\ce{I3-}$ are formed by:

$$\ce{I- + I2 -> I3-}$$

I know that the brown color comes from $\ce{I2}$, but in this solution one have $\ce{I3-}$ not $\ce{I2}$, why is it brown?

When I mix Lugol with $\ce{H2O2}$ without the presence of $\ce{H3O+}$ there is a variation in the conductivity of the mixture but the brown color doesn't disappear. If I add $\ce{H3O+}$ the solution becomes transparent after few minutes.

My guess is that either $\ce{I2}$ is reduced by $\ce{H2O2}$ (hence the disappearing of the brown color) or a redox reaction between $\ce{H2O2}$ and $\ce{I-}$, but in the later case I don't understant why the brown color disapear instead of appearing.

Can you enlighten me please, thanks.

I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:

$\ce{I+ , IO3-}$ don't absorb in the visible region
$\ce{I3-}$  I think should be red
$\ce{I2}$ is blue if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see Hildebrand (1909). In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.

Because there is an equilibrium:

$$ \ce{I_2 + I^- <=> I_3^-}$$

So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water).
Adding $H_2O_2$ should oxidize the iodide in this way:
$$ \ce{2I^- + 2H^+ + H_2O_2 -> I_2 + 2H_2O}$$
It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter,  you can make the solution transparent reducing the colored compounds. If you have a look to the iodine clock reactions iodine can be reduced by bisulfite:
$$ \ce{I_2 + HSO_3^- + H2O -> 2I^- + HSO4^- +2H^+}$$

But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:

FIRST: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$),  see Piranha solution (I think this could be very dangerous please take all the precautions!). But now I don't think that the equation that if I wrote before (see J. Am. Chem. Soc., 1931, 53 (1), pp 38–44 and
J. Am. Chem. Soc., 1978, 100 (1), pp 87–91):

$$ \ce{I_2 + 5H2O2 -> 2H^+ + 2IO3^- + 4H2O}$$

could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!

SECOND: simply you are  modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the  light absorption to take place.

I'm sorry this is now quite not an answer, but I hope it helps!

Iodine color reaction are wonderful (but very complex)!


Why we make pyrogallol alkaline to absorb oxygen? Pyrogallol is activated by -OH groups but what is the use of making it alkaline?

Because that makes them $\ce{\dots-O-}$, which is even better for oxidation. See, this thing is just ready to give away two electrons and become the corresponding quinone without any structural rearrangement.


What is the name origin of ester? Ester is quite a random nomenclature for a compound derived from a parent acid and parent alcohol.
Is there any reasoning behind using the word 'ester' to name such compound (for memorising purpose)?
One explanation for such nomenclature I found is quoted below from this website

The word "ester" was coined in 1848 by German chemist Leopold Gmelin,
probably as a contraction of the German Essigäther, meaning acetic
ether.

But even then, how does 'acetic ether' has to do with anything about ester being a nomenclature for a compound derived from a parent acid and parent alcohol?

The term "Essigäther" is actually the German name for ethyl acetate i.e Essig = "vinegar" + Äther = "ether". "vinegar" becomes acetic, hence ethyl acetate becomes "acetic ether". (etymoline.com)

But, ethyl acetate is an ester. So, why is it named "acetic ether"?

According to Leopold Gmelin, ester was called "oxy-acids ether" or "ether of third type". The original German text is:

b. Ester oder sauerstoffsäure Aetherarten. Ethers du troisième genre.
Viele mineralische und organische Sauerstoffsäuren treten mit einer
Alkohol-Art unter Ausscheidung von Wasser zu neutralen flüchtigen
ätherischen Verbindungen zusammen, welche man als gepaarte
Verbindungen von Alkohol und Säuren-Wasser oder, nach der
Radicaltheorie, als Salze betrachten kann, in welchen eine Säure mit
einem Aether verbunden ist.

which translates to:

b. Ester or oxy-acid ethers. Ethers of the third type.
Many mineral and organic acids containing oxygen combine with an
alcohol upon elimination of water to [form] neutral, volatile ether
compounds, which one can view as coupled compounds of alcohol and
acid-water, or, according to the theory of radicals, as salts in which
an acid is bonded with an ether. (italics mine)

You see the name "ether" was used to define an ester but this usage has become obsolete to avoid confusion.
Other compound classes originally named ethers
Gmelin reports in his 1848 Handbook of Chemistry that French chemists use the terms ether of the first, second and third type to refer to ethers, thio- and halo-alkanes, and esters, respectively. While the modern term ether refers to compounds with a specific functional group, the term ether was historically used to refer to a variety of volatile substances.

The names ether and ester are currently used, while the name "afer" for the thio- and halo-alkanes has not caught on.
Source: https://www.google.com/books/edition/Hand_book_of_Chemistry/U4sMAAAAYAAJ?hl=en&gbpv=1&bsq=Ethers%20du%20troisi%C3%A8me%20genre
Reference

Leopold Gmelin, Handbuch der Chemie, vol. 4: Handbuch der organischen Chemie (vol. 1) (Heidelberg, Baden (Germany): Karl Winter, 1848)


Where should I publish my formula related to Aufbau Principle? I have made a formula for determining the sequence of orbitals which is currently obtained using the Aufbau diagram. So, I wanted to know where should I publish it (if it is new). I searched the same on internet but did not get any such formula.

Have you searched the Journal of Chemical Education? There are so many articles on this topic you are interested in. If you don't have access to formal databases, start from Google Scholar and read abstracts.

Journal of Chemical Education is the right platform for such "teaching" tricks. However, you have to justify with some theory as to why your idea works. What is the logic behind it? As far as I can recall (>10 years ago), there was an article by a high-school student in that journal. So, students can also publish but you have to be quite good at that subject - similar to a professional.

However, sometimes I receive such articles for review but the tricks or analogies are so horrible that there is no option but to reject them. Similarly, make sure you know where your approach fails. Can it predict the electron configuration of all the elements? If you are a student share the idea with a chemistry teacher. See their reaction (no pun intended). Do they find it useful or have they seen it before? If yes, certainly consider writing a short note.


Can sugars dissolve in liquid ammonia? Can monosaccharides and disaccharides dissolve in liquid ammonia due to hydrogen bond formation?
My rationale is that these sugars may be able to form hydrogen bond with ammonia ($\ce{NH3}$).

Can sugars dissolve in liquid ammonia?

Yes, according to Ref.1, liquid ammonia is used to extract sugars in sugar-beet chips:

5.88 kilograms of sugar-beet chips having a moisture content of 5.4 percent and a saccharose content of 68 percent (calculated on the dried substance) are treated in an autoclave of 50-litre capacity with 18 kilograms of liquid ammonia. [...] The extracted pulp, after driving off the ammonia, contains only 1.2 percent of saccharose, which constitutes 0.4 percent of the total sugar.

Accordingly, 99.6% of sugar in sugar-beet chips are isolated in one extraction using liquid ammonia as the solvent. However, it was noted that the saccharose is not present as such in the solution, but is in the form of a compound of saccharose with ammonia. These ammonia-saccharose compounds can be obtained in a well crystallized form, which are composed of $\pu{1 mol}$ of saccharose and $\pu{2 mol}$ of ammonia and even those of other compositions Ref.1). I assume this is analogous to hydration by water molecules in aqueous solutions.
The high solubility of sugar in liquid ammonia seemingly due to H-bonding capabilities similar to water. This is evident by sucrose solubility in nonaqueous solvents such as morpholine (with a $\ce{N-H}$ bond) and N-methylmorpholine (without a $\ce{N-H}$ bond), which are $39.8$ and $\pu{0.37 g}$ in $\pu{100 g}$ of solution, respectively at $\pu{85 ^\circ C}$ (Ref.2). This difference in solubility is analogous to sucrose solubility in water and methanol, which are $210$ and $\pu{1 g}$ in $\pu{100 mL}$ of solution, respectively at $\pu{25 ^\circ C}$ (Wiki, and Ref.3). Further, sucrose is insoluble in diethyl ether.
References:

Georg Hingst, Werner Emte, "Process for recovering sugar from natural products containing it," US Patent 1958, 2,829,985 (Patent Application, 1955, US2829985A) (PDF).
Oleg K. Kononenko, Karl M. Herstein, "Nonaqueous solvents for sucrose," Ind. Eng. Chem. Chem. Eng. Data Series 1956, 1(1), 87–92 (DOI: https://doi.org/10.1021/i460001a017) (PDF).
Z. Bubník, P. Kadlec, "Chapter 5: Sucrose solubility," In Sucrose: Properties and Applications; M. Mathlouthi, P. Reiser, Eds.;  Springer: Boston, MA, 1995, pp. 101-125 (ISBN: 978-1-4613-6150-3).


Determining the oxidation state of a bridging platinum complex I want to ask a question about determining the oxidation state of a Platinum complex.

I was presented today with the following complex and asked to comment on the oxidation state and hence the d-orbital configuration of the central metal ion:

My working was as follows:

The $\ce{Me}$ groups do not contribute to any charge.

The $\ce{SMe2}$ ligands are neutral similar to $\ce{H2O}$ and don't contribute to any charge either.

However, if I take this into account, this would therefore mean that each Platinum central metal ion has the same magnitude of charge, but in opposite magnitudes i.e. $-2$ or $+2$ and this is not valid chemistry.

My other thought was whether the electronegativity of the $\ce{SMe2}$ ligand central S atom compared to the $\ce{Pt}$ meant that the bonding pair of electrons is electronegatively attracted to the S atom, meaning that for each $\ce{Pt}$ atom, the $\ce{Pt}$ atom would lose two of its own electrons, one in each opposite direction giving a +2 oxidation state.

The answer given is +2 for both metal centres but I was unsure using the approaches above whether this was valid.

What other steps am I missing to reach the final oxidation states of the $\ce{Pt}$ centres?

For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$:

$\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{H2O}$, alkenes (via the pi bond), ... and $\ce{SMe2}$ falls comfortably in this category.
$\ce{X}$ refers to anionic ligands: for example halides, hydroxide, azide, alkyl anions such as $\ce{Me}$, ...

Other ligands can be constructed as sums of these: so for example η4-butadiene would be $\ce{L2}$ (two pi bonds coordinating to the metal), an allyl group would be $\ce{LX}$ (one pi bond plus one anionic centre - ignore resonance), and so on.

The oxidation state of the metal, then, is simply the charge on the metal plus the number of X ligands. In this case each Pt centre is neutral, and has two L and two X ligands. That makes for a total oxidation state of +2.

As another example, the platinum in Zeise's salt has one L ligand (ethylene), three X ligands (chloride), and a negative charge: so its oxidation state is $-1 + 3 = +2$.

Of course, this method will fall apart very quickly when there are non-innocent ligands: $\ce{NO}$ and $\ce{O2}$ are the examples that come to mind, although there are many more. But for relatively simple complexes they suffice.


Can diastereomers form a racemic mixture? In my book, it says racemic mixture is the mixture of + and - optical isomers and it is optically inactive. This means enantiomers (optical isomers which are mirror images to each other) can form racemic mixture. But can diastereomers (optical isomers which are not mirror image to each other) form a racemic mixture?

The short answer is no.

A racemic mixture is defined as an equimolar mixture of enantiomers. Since each enantiomer rotates plane polarized light equally, but in opposite directions, the overall optical activity is zero.

Describing diastereomers as optical isomers is incorrect. Diastereomers are stereoisomers (that is, compounds that have the same connectivity of atoms but differ by the orientation of atoms in space) but are not enantiomers. In many cases, diastereomers are not even chiral (have no optical activity). If there's a mixture of diastereomers that are chiral, this is not described as racemic. Most likely that mixture would be optically active, but even if it was optically inactive, that would just be coincidental and not considered racemic.

The situation where a mixture is made up of two enantiomers in equal ratios is so common (and important) that it gets the special name of racemic.


How to use Google’s Alpha-fold to predict the structure of a two-protein complex? Alpha-fold won the CASP13 and CASP14 competitions last year and this year. It used deep learning to predict the secondary structure of a protein given the primary amino acid sequence.
Google has released its code a few weeks ago, and I was wondering how to use Alpha-fold to predict the complex of two proteins.
Naively, I'd concatenate the amino acid sequences of both proteins as if they were a single protein. Then I’d input the neural network with the concatenated primary sequence.
The network will predict the lowest energy conformation  of the concatenated proteins. I am afraid that the artificial backbone connection between the two proteins might lead to a completely wrong complex.
I could do some docking of the two individual proteins, but I want to find the denovo prediction of the two-protein complex.
Edit: here is an Pytroch implementation of Alpha-fold2 on github.

Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the former, I would recommend contacting one of the code developers to ask them for guidance.  It's possible that they have already developed a prototype option to address this.
If you're asking the latter, that approach is problematic, for two reasons:

Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first amino acid in the second protein.  That's not a constraint that would make sense a priori for a two-protein complex.

You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constraint mentioned in point #1, above).  If, on the other hand, you would like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that.


NaHCO3 decomposition dynamics and extent vs Temperature I'm wondering approximately how fast and to which extent $\ce{NaHCO3}$ will decompose (to $\ce{CO2}$ and $\ce{H2O}$) at $50$, $60$, $70$, $80$, $90$, $100$, $110$, $120$, and $\pu{130 ^\circ C}$.
It seems as if J. Environ. Sci. 2019, 78, 74–80 would answer my question but unfortunately I can't access the content.
Does anyone know ?

The paper doesn't tell how fast will it be.
As for the rest, have this:


Does an onion completely remove "fresh" paint VOCs and other chemicals emitted? I have read that paint contains VOCs(Volatile Organic Compounds), and solvents generally, and I could wait for them to dissipate, but i've read there are ways to remove the smell of paint such as half an onion in a bucket of water, such as slices of lemon on a plate, such as small bowls of distilled white vinegar.

Do these techniques actually remove all the nasty elements from the paint fumes, or just some of the nasty elements from them?  Do they remove all the paint fumes?

I wouldn't want to remove the smell but leave the VOC or other bad elements there.  If that's the case i'd rather just leave a window open in the room, closing it periodically, seeing if there's finally a time when the paint smell doesn't build up when the window is closed, then i'll know the paint isn't emitting any more fumes.

I've never heard of onions, lemons or vinegar removing a paint smell, but I would guess that they'd just cover up the unpleasant odor with something nicer.

As to your later comment:

  I wouldn't want to remove the smell but leave the VOC or other bad elements there.

VOC stands for "volatile organic compound." (See the Wikipedia page on VOCs) Volatile compounds are ones that evaporate easily. Technically, the smell of that onion (or lemon, or vinegar) is a VOC, just not a particularly harmful one. (Not knowing the particular chemicals that make up a onion's scent, I can't say for sure that it's a volatile organic compound, but it's a pretty good bet.)

That fresh paint smell comes from VOCs (and potentially from other volatile compounds), so by the time the smell is gone, any VOCs you couldn't smell are likely gone too.


Why does hydrolysis of ADP to AMP yields the same amount of energy as ATP to ADP, if ADP is more stable? I learned about the resonance stability of ADP and the fact, that ATP is less stable due to intramolecular instability. I was surprised to see that the energy net of conversion of ADP to AMP is the same as ATP to ADP. Shouldn't the energy net of a reaction ADP to AMP+P be lower, because it's not so readily hydrolysed?

ATP has two high energy bonds which are phosphoanhydride bonds. Both bonds release the same amount of energy. Therefore, when ATP converts to ADP, breaks one phosphoanhydride bond, and releases -30.5 kJ/mol. When the 2nd bond is broken in ADP releases -30.5 kJ/mol because it is the same phosphoanhydride bond. However, AMP has only a phosphate ester bond which releases -61 kJ/mol.
The "less stability" of ATP doesn't affect energy yield.


calculating Kp with partial pressures This is a question I solved in class:

  At 35 degrees, $\ce{N_2O_4}$ is 15% dissociated in a flask at a pressure of 1.2 atm. Calculate
  $K_p$  Equation:
  $$\ce{N_2O_4(g)<=>2NO_2 (g)}$$

This was how the answer went...

Partial pressure of $\ce{N_2O_4} = 0.85\times 1.2$

Partial pressure of $\ce{NO_2} = 0.1\times 1.2$

then $K_p$ was calculated using the standard formula

My question:

The part I cannot agree to with my teacher is that how is the partial pressure of $NO_2$ equal to 0.15x1.2. If I begin with 1 mole of $\ce{N_2O_4}$, if 15% dissociated at 1.2 atm, I would end up with 0.3 moles of $\ce{NO_2}$... the the partial pressure would be ${0.3\over 1.15}$x$1.2$... this does not go with the general answer stated above

I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:

$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$

Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:

$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

and:

$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0}  -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.

Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:

$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$

You will obtain by substitution, molar ratio values:

$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15},  \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$

Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.

Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!