Selectivity in aldol condensation of 7-oxo(6,6-²H₂)octanal I am supposed to find the product of the following reaction:

I know that six-membered rings are more stable than five-membered rings, but the C-D bonds are stronger, so it confuses me where will the attack happen.

There are two factors to take into consideration.

The acidity of the deprontonated $\alpha-\ce{H}$.
The electrophilicity of the carbonyl being attacked.

If 1 dominates, then the reaction is under kinetic control. While if 2 dominates then the reaction is under thermodynamic control. (Detailed arguments are omitted here.) Since addition to carbonyls is generally highly reversible, the reaction should generally be under theromodynamic control and yield 1-(2-hydroxycyclohexyl)ethan-1-one:

There are existing experimental results on aldol condensation of 7-oxo-octanal.
Ghobril et al. (Eur. J. Org. Chem., 2008: 4104-4108) used TBD as base and obtained the thermodynamic product, while Pidathala et al. (Angew. Chem. Int. Ed., 42: 2785-2788) used proline and obtained the kinetic product. Ghobril et al. also pointed out that

When unsymmetrical ketoaldehydes such as 3 are submitted to intramolecular aldol reactions, the ketone usually acts as the CH-acidic component, whereas the aldehyde plays the role of the carbonyl-active counterpart to afford regioisomer 4. This regiochemical outcome is also favoured when thermodynamic conditions are used.


What is the chemical redox equation between hydrogen peroxide and melanin? I am investigating hair bleach, and want to figure out the redox reaction between $\ce{H2O2}$ (hydrogen peroxide) and $\ce{C18H10N2O4}$ (melanin). $\ce{H2O2}$ is a strong oxidizing agent in this redox equation, but I don't know how to write that out. Someone please help?

Your formula for melanin is incorrect. Melanin does not have a clear and concise chemical formula; it is a class of natural polymers that differ in their exact composition. The only thing they have in common is that they derive from the amino acid tyrosine and that they at one point polymerised.

Therefore, unfortunately there cannot be an exact description of what happened between melanin and hydrogen peroxide. You can draw a half-equation for the hydrogen peroxide part which would be:

$$\ce{H2O2 + 2 e- + 2 H+ -> 2 H2O}$$

But what exactly happens in melanin and which groups are oxidised is hard to say. What probably happens is a series of radical reactions that in some way oxidises away the aromatic system and thereby moves the absorpion wavelength out of the visible spectrum. But that’s probably about as much as you can say.


Selective etchant for Al2O3 (Does not etch Al, Cu, Zn, Sn) I'm looking for a wet-etchant to remove aluminum oxide created by anodizing, but not etch the unoxidized metal (with greater than 10:1 selectivity)

I know about chromium(IV) oxide and fluorine-based etchants, but those are problematic for safety and disposal. Are there other etchants? It can be an acid or a base.

In Fabrication of Microstructures by Wet Etching of Anodic
Aluminum Oxide Substrates Chem. Mater. 2005, 17, 4049-4052,

5% phosphoric acid is used and is selective relative to aluminum, but you need to check how it would do with respect to the other metals you mention

  Finally, the AAO substrate was wet-etched in a 5 wt % H3PO4 solution
  at 20 °C for 8 h


How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent? I am doing a science fair project on OLEDs. I am testing whether polypyrrole can be used as an emissive layer in a 2-layer OLED. The polypyrrole I am synthesizing is water soluble, and will be spin-coated onto ITO-coated glass substrate. I also need to spin-coat on a hole-injection layer (PEDOT:PSS). I can only seem to find it in what I believe to be water-dispersed form. This will cause obvious problems since both substances are water-soluble. I read on some informal message board that I could drop droplets of water-dispersed PEDOT:PSS into a hot organic solvent with a boiling point higher than water, boil off the water, and be left with organic-solvent dispersed PEDOT:PSS. That makes sense, I am just confused about what organic solvent to use. I believe it needs to have the properties of not dissolving the PEDOT:PSS, ITO, or the polypyrrole, and allow the PEDOT:PSS to become dispersed.

Here is a link to the PEDOT:PSS I am looking at:

Even if you switch to an organic solvent, if the polypyrrole is in water, it will perturb the PEDOT:PSS layer.

Your best bet would be to use a poly-alkyl pyrrole or other conjugated polymer that's soluble in something that doesn't dissolve PEDOT:PSS.

Let's say you switch the solvent to DMF or DMSO and it dissolves the PEDOT:PSS. Then you spin-coat that new mixture onto ITO. Once you drop the aqueous solution of polypyrrole, it will dissolve the PEDOT:PSS.

I think you're looking at this the wrong way. Rather than trying to change the PEDOT:PSS (which will be tricky - PSS is anionic) you should consider changing your polymer. Alternatively, get rid of PEDOT:PSS and use another hole injection layer. (PEDOT:PSS is acidic and will etch the indium out of ITO.)


Does a change in partial pressure for a gas-solid reaction, changes the equilibrium? This is my reaction, $$\ce{Cd (g) + 1/2 Te2 (g) -> CdTe (s)}.$$
Considering the solid CdTe, $$K = P_{\ce{Cd}}^\mathrm{Eq}\cdot \left(P_{\ce{Te}}^\mathrm{Eq}\right)^{\frac{1}{2}}$$

I only have elementary knowledge on chemistry. I know that the equilibrium values of pressure are dependent on temperature. My question is - does increasing the partial pressure of one of the elements only changes the equilibrium pressure? If yes, for one or both?

The equilibrium constant will stay the same at the same temperature.  If the pressure of one component of the system in question is changed, equilibrium will be reestablished by either an change in the relative rates of the forward and reverse reactions.

Increasing the partial pressure of one reactant in this system will result in an increase in the rate of the forward reaction.  Pressure of both both reactants will be reduced until equilibrium is reestablished.


How to develop a retrosynthesis plan for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole? I need to retro-synthetically design a valid synthesis plan (in two steps, from cheap starting materials available over at Sigma Aldrich) for a target compound, pictured below.

How would I go about doing this?

Extempore, this should be available by the reaction of 3-bromocinnamaldehyde with 2-aminophenol. Samarium triflate as a catalyst might be a good idea.

In any case, I'd start the retrosynthesis at the benzoxazole moiety.


thermochemistry problem confusion I can understand upto where the ideal gas equation is used to calculate volume of steam. But on trying to find out $ \Delta V$ , i saw that there are no gases on the reactant side so then $P\Delta V=(\Delta n)\times R\times T$ where $\Delta n$ would be $1.39$. Then $\Delta V$ is simply the volume of the steam i.e., $42.51$. But when calculating the work a value of $42.8$ has been used for $\mathrm dV$. Why is that done? Secondly, there seems to be two different values for $\Delta$H. First they have written it to be $\pu{12470.6 cal}$. and then while calculating $\Delta U$ they have used the value of $\pu{13500 calories}$. (That value seems to have come from the given data: If we multiply $9.72\times 10^3$ (molar $\Delta H_{vap}$) by $1.39$ (number of moles) we get something around $13500$). How did those two different values appear? I also want to tell here that this is not a hw question for me, as you can see this is a solved example, so you can consider it as my attempt to understand thermochemistry better

The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid.  The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from.  I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically the way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal


The stability of the cupric and cuprous ions My teacher told me that $\ce{Cu^2+}$ is more stable than  $\ce{Cu+}$ in solution, but why? I think $\ce{Cu+}$ is $\ce{[Ar] 3d^10 }$ ,$\ce{Cu^2+}$ is $\ce{[Ar] 3d^9}$, why is $\ce{Cu^2+}$ more stable then?

Hund's rules, which are where we get the idea that half filled and completely filled orbitals are more stable than partially filled orbitals, are specifically about the ground states of atoms. Hund's rules do not apply to ions in solution which are complexed with ligands (like water).

What you fail to consider is that $\ce{Cu+}$ is larger than $\ce{Cu^{2+}}$ but with only half the charge, so its charge density is much lower. As a result, so its enthalpy of hydration for $\ce{Cu+}$ is much less, meaning it is much less stabilized by the formation of the metal-ligand bonds. As a result, in aqueous conditions $\ce{Cu+}$ tends to disproportionate to $\ce{Cu}$ and $\ce{Cu^{2+}}$.


Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%?

The answer is that for a simple dilution the following formula applies:

$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$

Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).


Calculating the theoretical Density of a Lithium oxide I am trying to calculate the theoretical density for lithium oxide using the following formula
$$\rho= \frac{n^{'}\sum M_c+\sum M_A}{V_c N_{AV}}$$
where

$n^{'}$ is the number of formula units in unit cell

$M_c$ is the sum of atomic weights of all cations in unit cell.

$M_A$ is the sum of atomic weights of all anions within unit cell.

$V_c$ is the unit cell volume

$N_{AV}$ Avogadro's number.

Here is what I have so far:
The structure for lithium oxide is similar to that of FCC so the number of oxygem atoms in this unit cell is $8/8+6/2=4$.

If the unit cell is to have an electrically neutral charge, it must have a total of $8$ lithium anions.
The radius of a oxygen ion with a coordination number of 8 is about $128$pm and for lithium with a coordination number of 4 is about $73$ pm

Since the crystal structure is cubic, the volume is $a^3$ but we need to write $a$ in terms of the radius of the lithium and oxygen ions.

From the structure we can deduce that $a=2\sqrt{2}(r+R)$ (since lithium and oxygen ions touch along the diagonal of a square).

The atomic weight for oxygen is $16$g/mol and the atomic mass for lithium is $6.491$g/mol

$$\rho = \frac{4\times 16+8\times 6.491}{16\sqrt{2}(73\times10^{-10}+128\times10^{-10})^3\ (6.022 \times 10^{23})}=1.05 \text{ g/cm}^3$$

The theoretical density however is closer to $2$ am I missing something?

Here's the unit cell of $\ce{Li2O}$, which adopts an antifluorite structure. The image is adapted from this webpage; technically it shows the fluorite structure but it doesn't matter.

The cation-anion contacts occur along the diagonal of the cube (black dotted line). By Pythagoras' theorem, the length of the black dotted line is $a\sqrt{3}/2$, as I indicated on the diagram. The sum of the ionic radii is half the length of this line, so

$$r + R = \frac{a\sqrt{3}}{4}$$

Using this I obtained a value of $\pu{1.929 g cm-3}$ for the density. By way of comparison, Wikipedia cites a density of $\pu{2.013 g cm-3}$. By the way, regarding your final equation, it's a good idea to leave the units in. At first I was fairly confused why you were using $10^{-10}$ instead of $10^{-12}$, until I realised it was meant to be in centimetres.


I'd like to model some compounds with DFTB, but some elements don't have parameter files in the official site. Is there a workaround? I was reading a bit about this technique, Density Functional based Tight binding (DFTB). It's claimed it can provide a speed improvement of several fold when compared to plain DFT, still giving acceptable results. I downloaded one of its free implementations, the DFTB+ package, and started to work on their tutorial. The problem is, like a semiempirical method, this technique demands parameters. In DFTB the parameters are downloaded separately, as a series of files for each bonded element pair. The problem is, in the page linked as the repository for such parameter files, the options listed don't cover every possible element pair in the periodic table. I my case, I'm interested in heavier members of the alkaline earth group, Sr and Ba. But the only elements in this group with parameter files available are Mg and Ca.

So, I wonder if there is a workaround, short of doing the parametrization on my own, what I imagine would probably be a whole research project on itself, and far beyond my league. Perhaps that repository page just doesn't list every Slater-Koster file in existence, ond others may be found somewhere.

I recommend you try out xtb by the Grimme group. It is a state-of-the-art Tight Binding program parameterized for elements up to Radon (already built-in). There is a small caveat: the only precompiled binary they distribute is for Linux (but the program is open source, so you can compile it yourself if you want.)

And yes, it is really fast.


Naming of halogen group in ethane Can we write "1-Chloro-ethane" when a single chloro group is present in the ethane? I think only one position is possible for the chloro group in ethane.

In general nomenclature, the practice of omitting locants when there is no ambiguity is widespread. For preferred IUPAC names, however, the current recommendations given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are prescriptive about when locants are omitted. In particular:

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(b) in monosubstituted homogeneous chains consisting of only two identical atoms;
(…)

Therefore, the preferred IUPAC name is indeed chloroethane and not the (correct, but not preferred) complete systematic name ‘1-chloroethane’.
(Nevertheless, note the correct capitalization and hyphenation of the names.)


Reaction stoichiometry for calculation of Kp NH$_2$CO$_2$NH$_4$(s)⇌$2$ NH$_3$(g)+ CO$_2$(g)

I want to calculate K$_p$ for the reaction above, given that the total pressure is $0.63$ bar.
My calculations:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $1/2$ p(CO$_2$) (For every NH$_3$ half a CO$_2$ is needed)
$0.63$ = p(CO$_2$) + ($1/2$ p(CO$_2$) = $3/2$p(CO$_2$) -> p(CO$_2$) = $0.42$ bar , p(NH$_3$) = $0.21$ bar
So, K$_p$ = ($0.21$)$^2$ * ($0.42$) = $0.0185$ bar$^3$
However, the answer key's way of calculating this is:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $2$ p(CO$_2$)
$0.63$ = $3$p(CO$_2$) , so that  p(CO$_2$) = $0.21$ bar and p(NH$_3$) = $0.42$ bar
How can p(NH$_3$) = $2$ p(CO$_2$)? Isn't that wrong? Or am I thinking wrong? I have contacted my professor but haven't received an answer yet and our exam is tomorrow so hopefully someone here can help me understand. Thank you!

Each time one molecule of salt is decomposed, it produced $\ce{2 NH3}$ for $\ce{1 CO2}$. The number of molecules $\ce{NH3}$ is always double from that of $\ce{CO2}$ in the gas phase. Now the partial pressure of a given compound is proportional to the number of molecules of this compound in the gas phase. As a consequence, the partial pressure of $\ce{NH3}$ is always twice as high as the partial pressure of $\ce{CO2}$


Write the combustion equation of an organic compound Write the combustion equation of the organic compound $\ce{C5H10O}$

I have no idea at all how to write the equation. Should I add $\ce{O2}$ to make it $$\ce{C5H10O + O2 → ... ?}$$

combustion is a reaction using O2 to burn the other reactant and will result in CO2 and H2O

$\ce {\_\_\ C5H10O + \_\_\ O2 -> \_\_\ CO2 + \_\_\ H2O}$

then just balance
so..

$\ce {C5H10O + 7O2 -> 5CO2 + 5H2O}$


Is [PdCl2(PMe3)2] diamagnetic or paramagnetic? Please tell me where I am wrong. The complex will have $\ce{Pd^{2+}}$ ion, which has a $\mathrm{d^8}$ configuration. So, it will have $2$ unpaired electrons. Unpaired electrons will mean that it is paramagnetic.

My reference book has this line:

  $\ce{[PdCl2(PMe3)2]}$ is a diamagnetic complex of Pd(II)

Shouldn't it say "paramagnetic"?

You have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\ce{Pd^2+}$ ion adopts the $\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic.

Image source: Wikimedia Commons

Image source:Wikimedia Commons


Hydrogen peroxide from water Is it possible that $\ce{2 OH-}$ from water could react to form $\ce{H2O2}$?

I mean most autoionization forms $\ce{H3O+}$ and $\ce{OH-}$.

However I think it is possible that $\ce{H-}$ and $\ce{OH+}$ forms, though the amount formed of these 2 is small. This is from water acting as an acid and H+ acting as a base.

The only "hydroxyl"-type species whose recombination has been observed to yield  hydrogen peroxide are hydroxyl radicals.

$$\ce{HO^. + ^.OH -> H2O2}$$

In order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\gamma$, electron beam).

The conceivable steps yielding $\ce{HO^.}$ are either

$$\ce{H2O  -> H^. + ^.OH}$$
or
$$\ce{H2O  -> H2O^{+.} + e-}$$

$$\ce{H2O^{+.} -> H+ + ^.OH}$$


Why doesn't the vapor pressure of water depend on the humidity? Correct me if I'm wrong, I'd like to think of vapor pressure of water as the number of water molecules escaping from the water surface per unit area. Greater the leaving number, greater the vapor pressure. If this is right, can I conclude that the rate of evaporation is directly proportional to the vapor pressure?

My textbook says that the vapor pressure of water depends only on the temperature of the water. It doesn't depend on external pressure or humidity. But we know that the rate of evaporation decreases with increase in humidity.(Wet clothes dry faster in less humid air.) Doesn't this imply a decrease in vapor pressure with increase in humidity?

The partial pressure of water vapor is equal to the equilibrium vapor pressure at the temperature of the liquid only at the surface of the water.  Above the surface of the water, the partial pressure of the water vapor decreases with distance, and far from the surface, approaches the partial pressure consistent with the relative humidity of the room air.  So there is a concentration gradient of water vapor in the air above the water surface.  Based on Fick's law of diffusion, this provides a driving force for diffusion of water away from the surface.  We call this diffusive flux of water vapor evaporation.  It is driven by the difference between the equilibrium vapor pressure of water at the surface and the partial pressure of water vapor in the bulk of the room air.


The reaction rate when the rate-determining step has a back reaction For some reason, I am reviewing general chemistry and I find that the elementary reaction for the rate-determining step (RDS) is assumed to be irreversible, that is, single arrow, in all questions that I see.

For example, think a reaction consisting of two elementary reactions:

$$\ce{A + B <=>[k_1][k_1^\prime] C} \tag{fast, reversible}$$

$$\ce{C ->[k_2] D} \tag{slow}$$

Then the overall reaction is $\ce{A + B -> D}$, and the rate is then

$$\text{rate} = k_{2}[\mathrm{C}].$$

Now, I'm curious about the rate when the equation becomes like

$$
\ce{A + B <=>[k_{1}][k_{1}^\prime] C \:(\mathrm{fast, reversible)} \\
C \color{red}{<=>[k_{2}][k_{2}^\prime]} D\:(\mathrm{slow)} }
$$

In this case, I think the rate expression should be changed.
From the overall chemical equation ($\ce{A + B -> D}$), the rate itself is defined as
$$
\mathrm{rate} = \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t},
$$
and checking the reactions related to $[\mathrm{D}]$ in the RDS,
$$
\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}].
$$
Thus, the rate is
$$
\mathrm{rate} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}].
$$

Though, I'm not sure if the reasoning is right. Is it correct?

Thus, the rate is
  $$
\mathrm{rate} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}].
$$
  Though, I'm not sure if the reasoning is right. Is it correct?

This is correct but does not give you much insight into how [D] changes over time. This system will reach equilibrium eventually. If you call the equilibrium concentrations $[\ce{C}]_{eq}$ and $[\ce{D}]_{eq}$, and $\epsilon$ the difference between current and equilibrium concentration for C, you get the following:

$$[\ce{C}] = [\ce{C}]_{eq} + \epsilon\tag{1}$$
$$[\ce{D}] = [\ce{D}]_{eq} - \epsilon\tag{2}$$

Also, the rates are equal at equilibrium:

$$ k_2 [\ce{C}]_{eq} - k{_2}'[\ce{D}]_{eq} = 0\tag{3}$$

Plugging all of this into the rate equation gives:

$$\frac{d[\ce{D}]}{dt} = -\frac{d\epsilon}{dt} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}]\tag{4}$$

$$= k_{2}([\ce{C}]_{eq} + \epsilon) - k{_2}'([\ce{D}]_{eq} - \epsilon)$$

$$= k_{2} \epsilon - k{_2}'(- \epsilon)\tag{*}$$

$$ = (k_{2} + k{_2}') \cdot \epsilon$$

The simplification yielding the expression labelled '*' is because the net rate at equilibrium is zero (substituting (3) into the previous expression). As a result, you get a first order behavior of approching equilibrium with respect to the "excess concentration" $\epsilon$. When you integrate this, $\epsilon$ decays exponentially, with the sum of the forward and reverse rate constant in the exponent. This is somewhat surprising because we started out with these rate constants appearing with opposite signs in the (correct) rate expression given by the OP. However, this result is correct and well-known from temperature jump kinetics.

What about A and B?

The derivation above is for a two-species system $\ce{C<=>D}$. If A and B are in fast equilibrium with C, equation (1) or (2) has to be modified to account for A and B supplying more C as it reacts to D. One way is to define $[\mathcal{C}] = [\ce{A}] + [\ce{C}]$ and write (instead of (1)):

$$[\mathcal{C}] = [\mathcal{C}]_{eq} + \epsilon\tag{1'}$$

Because the first equilibrium is fast, you can calculate $\ce{A, B and C}$ from $[\mathcal{C}]$ at any point in time, and you would have to adjust equation (4) accordingly. The details depend on whether $\ce{A and B}$ are present at stoichiometric amounts or one of them is in excess. If one of them is in large excess, the result would just be different by a constant $\beta$, i.e.

$$ \mathrm{rate} = (\beta k_{2} + {k_2}') \cdot \epsilon$$

where $\beta$ depends on the equilibrium constant of the first (fast) reaction and the concentration of the reactant in excess.


What are the assumption of Gibbs free energy equation referring to the system and the surrounding? The system for

$\mathrm dG=\mathrm dH-T\,\mathrm dS$

is a constant temperature and pressure throughout the process, only volume work is possible and it is in thermal equilibrium with the surrounding. Is that right? Are there any other constraints?

The only restriction on your equation is that the system is at constant temperature.  The system can be open or closed, the pressure does not have to be constant, and any type of work is allowed.

Let's start with the basic definition of G in terms of H, which is completely unrestricted:

$$G \equiv H-TS \Rightarrow dG = dH - d(TS) = dH -TdS -SdT$$

Then, to get your equation, the only restriction we need to add is constant $T$:

$$dG= dH -TdS$$

Why, then, might you be associating this equation with the more extensive restrictions of constant $T$ and $p$, and no non-$pV$ work?  I suspect it's because, to use $dG = dH -TdS = 0$ as a condition for equilibrium, these other restrictions are required. For more on this, see the end of my answer at What is wrong in this argument that dG must always be zero?


Can a compound with indirectly bonded carbon and silicon atoms within the same aromatic system be called organosilicon compound? Wikipedia says that organosilicon compounds are organic compounds with carbon-silicon bonds.

Does the class of organosilicon compounds include compounds in which the carbon and silicon are only indirectly bonded, for example the (hypothetical) molecule 1,3,5-trihydroxy-4,6-dioxa-2-lambda3-oxa-3,5-disilacyclohex-1-ene(3+), in which the carbon and silicon atoms are not formally directly bonded but are part of the same aromatic (six π-electrons) ring, sharing the same aromatic bonding orbitals?

Here's what the molecule looks like (direct uploadd won't work for some reason...)

Organosilicon compounds fall under the broader category of Organometallic compounds. It is a requirement that in organometallic compounds, atleast one $\ce{C-M}$ bond should exist. Since Si is only bonded to oxygen atoms in your compound, the given compound is not an organosilicon compound.

The following is also mentioned in the wikipedia page for Organosilicon compounds, under the image:

  A carbon–silicon bond present in all organosilicon compounds


Can we use prefixes like iso, neo, etc in IUPAC nomenclature of organic compounds? Shouldn't the name of the compound in the picture be 5-(1-ethyl-2-methylpropyl) nonane instead of the given name?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general nomenclature. However, many prefixes are no longer recommended.

  Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

For example, the prefixes “isobutyl”, “isopentyl”, “neopentyl” were contained in the 1993 recommendations but are no longer recommended as approved prefix. The prefix “isopropyl”, is still retained for use in general nomenclature; however, for the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”. (The prefix “1-methylethyl” may be used in general nomenclature).

Anyway, the name “5-(1-ethyl-1-isopropyl)nonane”, which is given in the picture in the question, is clearly wrong since no parent structure is given for the substituents 1-ethyl and 1-isopropyl:
“5-(1-ethyl-1-isopropyl what?)nonane”

If you really want to use such a name including the substituents 1-ethyl and 1-isopropyl, the corresponding systematic name would be “5-(1-ethyl-1-isopropylmethyl)nonane”. However, such a name would be highly unusual and not in accordance with the basic principles of IUPAC nomenclature.

The name “5-(1-ethyl-2-methylpropyl)nonane”, which is proposed in the question, describes the correct compound and may be used in general nomenclature. However, the PIN for this compound is “5-(2-methylpentan-3-yl)nonane” since prefixes of the “alkanyl type” are preferred if the free valence is not in position 1 of the substituent group (see this similar question).


What is the edge of a diamond like? The chemical structure of a diamond is defined as an endless lattice in which each carbon atom is covalently bonded to four other carbon atoms situated at the four ends of a tetrahedron.  But of course they aren't endless; diamonds have edges and faces.  So what happens when the lattice reaches the edge and there aren't any other carbon atoms to bond to?  What is the "surface layer" of a diamond composed of?

This is a nice question. Although I do not speak with authority on diamonds, most crystal surfaces have several imperfections which mitigate the hanging valencies that would otherwise appear for the atoms (constituents in general) at the surface.

Impurities. This is the primary method which helps satisfy the unfulfilled valencies at the surface. Many metallic, non-metallic or compound impurities are usually incorporated throughout the crystal, but are usually high at the surface. By bonding with these excess impurities, the surface carbon atoms can fulfill their valencies. An extremely pure crystal surfaces (lots of "hanging" valencies) will quickly adsorb most materials that it comes in contact with. Chemisorption or physisorption depends on the nature of the crystal, metals prefer physisorption while non-metals have at least unimolecular chemisorption usually.
Another way to mitigate surface valencies is distortion. The surface layer might not follow the exact space-filling Bravais lattice structure and may form short range distortions in structure to try to make the unfulfilled valencies satisfied. There might be kinks, depressions, steps etc which at least help in completing the valencies of a few atoms (at the boundary of distortion, say, a "step" structure).
If none of these operate, there is an inherent surface energy or surface tension between any two phases which quantifies the desire of the surface atoms to fulfill their hanging valencies to resemble the atoms in bulk which are stably attracted. In case of metallic and ionic crystals, this is comparatively low, but may be high in a covalent crystal like diamond.

In all cases, the surface layer of constituents (actually first few layers) are different from the atoms in bulk in energy, stability, forces acting (unbalanced on the surface and balanced in the bulk) and their structure/orientation.


Estimation of dipole moments in aromatic compounds If it asks to compare dipole moment in aromatic compounds then do we see inductive or resonance effects or do we see a mix of both?
E.g. we have to compare dipole moment in pyrrole and furan then in one we have opp direction of dipole moment in both why? Like it's a specific question but I wish to understand the mechanism behind every such tricky question so that I can do on my own.
Or suppose we have ortho hydroxyphenol then what would I say the direction of dipole moment would be? Would it be inwards towards benzene or outwards towards $-\ce{OH}$ group. It's not because of convention doubt, but like whether it's the resonance effect that dominates which would result in majorly + on $-\ce{OH}$ group and - on $-\ce{Ph}$ group or is it the inductive effect dominating which would result in opposite sign.

I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group.


What reaction is expectable from the atomized reactants? Let we have a reaction that starts from some reactants to produce some products. When the activation energy is provided to the reactants they will be atomized and produce a gas of individual atoms. The gas contain the whole atoms that can either form again the reactants, or form the products, or instead form the elemental molecules of themselves from which the reactants or products can be produced in other reactions. Now that the gas of all individual atoms is common for all these different reactions my question is what conditions decide which reaction to happen?

Can I say the reaction happens in the direction at which the Gibbs free enthalpy decreases more rapidly? Or do thermodynamic conditions like temperature, pressure and especially concentration of the species also play a role, independent of the variation in the Gibbs function?

The rate for a reaction is dependent upon the rate constant and, depending upon what type of reaction it is, the concentrations of the reactants.  The rate constant can be further broken down into an entropic factor (how likely are the reactants to collide in an orientation that will produce a reaction) and an energy factor (the activation energy).

Assuming for the moment that the reactant concentrations are such that they will not control the direction of reaction, and that the entropic factors don't differ dramatically for one pathway over another (that's usually the case for similar reactions), then the first reaction to happen will be the one involving the reaction with the lowest activation energy.  This will lead to the kinetically controlled product.

If the back (reverse) reaction from the kinetically controlled product also has a low enough activation energy such that it happens rapidly under our reaction conditions, then essentially we are setting up an equilibrium.  In this case, thermodynamics will eventually control the product distribution and we will eventually wind up with the product that has the lowest free energy, the thermodynamically controlled product.

Many reactions first produce the kinetically controlled product, but with time, the thermodynamic product becomes the major product.


When do lone pairs on nitrogen contribute to aromaticity? I've been told that 1,2,4-trimethyl-1H-imidazole is aromatic. As it has four obvious pi electrons (in the double bonds), I assumed that it was anti-aromatic as opposed to aromatic.

I've been operating under the assumption that a nitrogen does not contribute its pi electrons to the aromaticity count for Hückel's rule unless it is attached to a hydrogen. Does the nitrogen with a methyl group here donate its pi electrons to the count?

Both nitrogens in your drawing are roughly $\mathrm{sp^2}$ hybridized, therefore they each have a p-orbital and 3 $\mathrm{sp^2}$ hybdridized orbitals. All of the p-orbitals, including those on the 3 carbon atoms in the double bonds, are perpendicular to the plane of the screen (e.g. they project above and below the plane of the screen). So there are 5 p-orbitals and each contains one electron except for the one p-orbital located on your "bottom" nitrogen (in green). This p-orbital contains the nitrogen lone pair, so there are two electrons in this p-orbital.

If we count the p-electrons we have $1+1+1+1+2 = 6$ electrons in $5$ p-orbitals that form a continuous loop. Therefore, the molecule is aromatic ($4n+2, n=1$).

Let's take a closer look at the "top" nitrogen in your drawing (in purple). Where is its lone pair? It resides in an $\mathrm{sp^2}$ orbital on this nitrogen. The $\mathrm{sp^2}$ orbital lies in the plane of the screen. This orbital is orthogonal (perpendicular) to the 5 p-orbitals; therefore these 2 lone pair electrons are separated from the electrons that make the molecule aromatic. They are not involved in the molecule's aromaticity.

Here's how to approach this situation in the future:

Identify the nitrogen(s)
Determine if the lone pair is in a p-orbital or an $\ce{sp^2}$ orbital.

If the nitrogen is in a double bond, then the lone pair is in one $\ce{sp^2}$ orbital, 1 electron is in the p-orbital and 1 electron is in each of the 2 remaining $\ce{sp^2}$ orbitals.
If the nitrogen is not in a double bond, then the lone pair is in a p-orbital and 1 electron is in each of the 3 $\ce{sp^2}$ orbitals.

Count the electrons in the cyclic p-orbitals to determine aromaticity


How is it that the equilibrium constant does not depend on the mechanism? For a reaction of the form

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

the equilibrium constant is

$$K_c=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$

regardless of the mechanism of the reaction. Why is this the case?

I've seen derivations that use a reaction in one elementary step to demonstrate this, but this obviously doesn't work in general. The sample reaction used was $\ce{N2O4 <=> 2NO2}$.

You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see this answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction

\begin{equation}
  \ce{H2(g) + I2(g) <=>> 2HI(g)} \ .
\end{equation}

Its mechanism consists of three reaction steps:

$\ce{ I2 <<=> 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$
$\ce{I + H2 <<=> H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$
$\ce{H2I + I <=>> 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$

Now, take a look at the equilibrium constant of the overall reaction

\begin{equation}
  K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ .
\end{equation}

And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ .
\end{equation}

If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.


How can I clean my lab coat? Over the course of my studies, my lab coat has absorbed a non-negligible amount of different chemicals, some of them nasty organic polymers, some are more insoluble inorganic compounds like $\ce{Cr2O3}$.

Even after washing the lab coat some stains remain. Is there a way to get rid of those, or should I rather buy a new lab coat?

A lab coat will eventually get dirty, that's why it is a good idea to wear it. I had a lot of students complaining about the condition of their coat and they asked me if they could just wash it in the washing machine. My answer was always no.

As far as I remember - I am a theoretician now - a labcoat costs much less than a pair of trousers and a T-shirt. It should be renewed annually and disposed of according to the norms how you would dispose of an unknown chemical. You never know where the chemicals incorporated in your coat will end up, so dispose of them as safely as you can.

In some countries, it is required by the employer to provide equipment like that and taking care of either cleaning or disposing of it eventually. (In Germany that should be the case.) Wear it as a disposable layer of protection and please - for the safety of the environment and yourself - do never ever put it in a normal washing machine again.

As an additional note, if you wash it in the same washer like your normal clothes, how can you assure that it will not end up in your everyday clothes? I can imagine, that having chromium whatever in my underwear would not make me feel alright.


Is the structure of a crystalized protein the native one? When one finds the 3D structure of a protein by crystalizing it and then making a X-ray experiment, how does one know that the geometrical configuration of the crystal is the same (or even close) to the structure the protein adopt in aqueous solution?

I'll use quotes from B. Rupp, Biomolecular Crystallography (p. 7-8) to answer.

Generally the structure is similar...

  Comparison of many nuclear magnetic resonance (NMR) solution structure
  ensembles with crystallographic structure has shown that the core
  structure of protein molecules remains unchanged compared with the
  solution state during crystallization. In addition, enzymes packed in
  crystals even maintain biological activity.

with some parts not visible in the experiment...

  The maintenance of the core structure and of enzymatic function shows
  that crystal structures are a very good approximation of the native
  protein solution structure. Nonetheless, highly flexible or mobile
  regions, frequently the amino- or carboxyl-termini of the protein chain or
  flexible loops connecting secondary structure elements, can be
  poorly defined or even absent in the electron density and thus can be
  modeled only with limited confidence.

but beware...

  In certain situation flexible and dynamic regions of a protein
  molecule can be rigidly fixed in a specific conformation as a result
  of crystal packing interactions. In most cases this represents just a
  snapshot of one possible conformation out of many and it must be
  understood that such a specific conformation may not locally represent
  the protein structure in solution. A simple safeguard against
  misinterpretations -- which is usually assignment of certain
  biological relevance to regions where that is de facto not warranted --
  is to display all neighboring, symmetry-related molecules in the crystal
  structure and examine if any intermolecular interactions are
  present that are a result of crystal packing. Such packing induced
  artifacts can also hamper for example drug discovery by altering or
  blocking binding sites and thus preventing an otherwise active
  substance from binding.


Does mixing calcium dichloride and magnesium sulfate induce a reaction, and how does it affect the hygroscopy of the sample? I have a problem with a sample made from $\ce{CaCl2*2H2O}$ and $\ce{MgSO4*7H2O}$ in about equal amounts. I took the two salts, pressed them together, and the result was a weird brine that was very wet and almost seemed to turn into water before my eyes.

I am wondering what happened here, and was hoping that someone could walk me through it. As you can tell by the title, I believe that a reaction might be involved that I did not consider when making the sample.

As far as I know, the two salts are both very hygroscopic, so it could be that they simply adsorbed water and that is why the sample became wet. However, the effect seemed more severe after mixing them together, so I am thinking that a reaction might have been involved.

Could it be that the two salts reacted this way?
$$
\ce{CaCl2*2H2O + MgSO4*7H2O -> CaSO4*2H2O + MgCl2*6H2O + H2O}
$$
And if so, what happens with the leftover water that is not bound? Especially since $\ce{MgCl2*6H2O}$ is even more hygroscopic than any of the other salts, could it be that the water was attracted by the $\ce{MgCl2*6H2O}$, which then turned the sample into a brine?

I'm just trying to understand what happened, I did not add any heat, but I did add a lot of pressure.

You are correct.

In the system $\ce{Ca^2+*Mg^2+*Cl^-*SO4^2-*H2O}$, the least soluble phase is $\ce{CaSO4*2H2O}$.

Once you're mixing $\ce{CaCl2*2H2O}$ and $\ce{MgSO4*7H2O}$, your $\ce{CaCl2*2H2O}$ starts absorbing atmospheric moisture and then dissolving in it. Then you have a brine which dissolves the $\ce{MgSO4*7H2O}$, speeding up the kinetics of the thing. The solid that you precipitate by the recation has less $\ce{H2O}$ per mole of salt, so the excess $\ce{H2O}$ just ends up in the brine (the "water"). Pressing it speeds it up because there is more surface area of the two salts in contact, which is where the reaction happens.


Rotational-vibrational spectroscopy of a molecule I want to ask if the rotational-vibrational spectra of a molecule is related to any one mode of frequency or if it is related to the whole molecule?

Additionally, the vibrational levels that we say are composed of lower rotational energy levels. Are these vibrational energy levels of certain vibrational normal modes or of the whole molecule?

I'm not quite sure what you mean by 'vibrational levels composed of lower rotational energy levels' so I try to explain the whole thing, briefly.
In analysing the electronic, vibrational &  rotational spectra of molecules the Born-Oppenheimer approximation has to be used. This is applicable because the vibrational period (a few femtoseconds) is far faster than rotation period (a few picoseconds) thus these motions do to interact appreciably with one another to a good approximation. This allows the energy levels of, rotation and vibrational motions to be added together.
A normal mode vibration is the motion of all atoms in the molecule in a fixed phase relationship with one another. For N atoms there are $3N-6$ normal modes ($3N-5$ for a linear molecule). How how far and in what direction each atom moves has to be determined by calculation, but the overall symmetry species ($A_1, B_g, E_u$ etc) of the total motion of all atoms  is given by the Point Group of the molecule, $C_{3v},D_{2h}, O_h$ etc. The symmetry species determines the relative phase of each atoms motion with respect to all the others. By 'phase' is meant that as some bonds contract  others extend but they always do this together.
As energy levels add each vibrational normal mode has its own stack of rotational levels. (Because the B.O. approximation is not exact a more detailed analysis in needed to add the interaction between vibration and rotation.)
The figure shows a typical spectrum (calculated)  of a (perpendicular band) of a symmetric top molecule such as $\ce{PCl3 , CHCl3} \mathrm { ~or~ } \ce{CH3Cl}$. The total spectrum is shown on the bottom line and its separation into individual PQR rotational bands above this.  There will be similar complicated features for each vibrational normal mode. Clearly the analysis of spectra such as this is non-trivial :)
(In symmetric top molecules two of its three moments of inertia are the same but different from the third. Two quantum numbers are needed J for total angular momentum and K to fix the angular momentum about the symmetry axis and this is why there are so many rotational bands for a single vibration.)

(Spectrum is taken from G. Herzberg, ' Infra-red and Raman Spectra of Polyatomic Molecules')


Why are Erlenmeyer flasks shaped the way they are? The title says it all. We could conceive a good number of other possibilities. Why did Erlenmeyer arrive at this design?

Imagine a time when magnetic stirrers were unknown. You, or your lab assistant, would mix solutions in suitable glassware, shake manually, put it aside and then observe a change.

An Erlenmeyer flask is much more suitable than a simple beaker

the flat bottom gives it a stable stand on the bench
the narrrow neck allows to hold and shake a large flask with one hand
the conical shape prevents spills when the flask is shaken circularly


Why does Freundlich adsorption isotherm not have a term for surface area? Why does Freundlich adsorption isotherm not have a term for surface area?

The formula given by Freundlich adsorption isotherm:

  $$ \frac{x}{m}= k P ^{1/n}$$
  Where $x$ is the mass of gas adsorbed on mass $m$ of the adsorbent and pressure $P$. $k$ and $n$ are constants which depend on the nature of the absorbent and the gas at a particular temperature.

There is no term for the surface area involved for adsorption and I had read that adsorption increases with the increase in surface area so if I take a highly porous substance, it will adsorb more gas on it. What am I missing here?

Edit: Actually, what I mean is that, let's say I have a single piece of adsorbent, it has a certain value of $x$ for a given $P$. Then I break it into 10 pieces, The value given by the formula is still the same, while there will for sure be an increase in $x$, due to increased surface area.

The ratio

$$\frac{x}{m}= k P ^{1/n}$$

can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$:

$$\frac{x}{A}= k' P ^{1/n}$$

where

$$k'=k\rho/\lambda$$

The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) adsorbent to use in a particular application (see e.g. Table 4 in ref 1).

There are two types of adsorbent material you are usually interested in:

Porous material; in this case, breaking up the particles will not significantly increase $A$, and therefore not change k (since $\lambda$ will remain approx constant).
Small particles with high $A/V$ ratio and adsorption on an outer surface of the particle. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of $k$, since $\lambda$ increases and $k\propto\lambda$.

Parameter $k'$ is more fundamental than $k$, as it reflects how much surface area the adsorbed molecules require, keeping in mind the range of applicability of the equation, as it does not predict e.g. monolayer saturation. Note that in neither scenario would a reduction in the size of the particles be expected to change $k'$ by much - certainly not in the first case, and, in the second case, depending on the importance of changes in adsorbent curvature or edge/corner effects.

Reference

G. P. Jeppu, T. P. Clement, "A modified Langmuir-Freundlich isotherm model for simulating pH-dependent adsorption effects," Journal of Contaminant Hydrology 2012, 129–130, 46–53 (DOI: 10.1016/j.jconhyd.2011.12.001).


Does rearrangement occur in Friedel Crafts alkylation or not? In the reaction of alkylation of aromatic systems with the help of Friedel Crafts alkylation - the reagent used is alkyl halide in presence of appropriate Lewis acid.

When is rearrangement observed?

If the alkyl halide is primary, the carbocation is usually not stable - as a result, a free carbocation shouldn't be formed and the attack should occur from the acid base complex itself (this is what my instructor said). However, in cases where the alkyl halide is tertiary, or where the carbocation is stable enough to be formed - rearrangement (if possible) is observed and the reaction proceeds through a free carbocation intermediate which acts as electrophile.

So in the former case, where a full fledged carbocation isn't formed, is rearrangement possible?

For example, is rearrangement possible in alkylation of benzene with n-propyl chloride in presence of aluminium chloride? Clearly, the primary carbocation is not stable - and may arrange to secondary carbocation (only if it is formed in first place)

So what really happens, and when in general and in particular can we expect rearrangements in Friedel Crafts alkylation? Also, when not to expect rearrangements? (Of course, when the carbocation isn't formed, but when is that?)

It'd be great if you could quote some credible sources along with your answer, if you wish to add certain specific reactions etc.

P.S.

Which Lewis acids allow for rearrangement to occur? As far as I know, only in presence of stronger Lewis acids does rearrangement occur - but how strong?
Does temperature also play a role for rearrangement to occur?

When you conduct the alkylation of benzene with n-propyl chloride, the product you obtain is isopropylbenzene, so a rearrangement occurred by virtue of the fact that the primary carbocation that forms will undergo a 1,2-hydride shift to form a secondary one. Most alkylation ( aromatic ring and alkyl halides ) are conducted in the presence of aluminium chloride which is one of the strongest Lewis acids. The principle is that the alkyl halide reacts with the aluminium chloride to form a salt: a carbocation and a complex ion of tetrachloroaluminate. You can have methyl chloride react in this way because even if it forms an unstable carbocation, it will readily react with the aromatic ring ( which acts as a Lewis base ). You can have methyl carbocation in solution such as in alkylation reactions, but you can’t isolate a salt containing methyl carbocation in solid state. If you had methyl chloride and aluminium chloride in solid state ( in contact with each other, they would not react to form carbocations.


Titration with EDTA and Eriochrome black T I have a titration of Calcium Ions with EDTA. When I reach a volume of 9.6mL I the liquid turns slightly blue and after that the more I add the bluer it becomes. Now at what point is my titration finished? When I notice the first change or as soon as adding more EDTA doesnt change the color?

Assuming you've used Erichrome Black-T as an indicator, the second your whole solution changes from wine red to blue, as in the moment you realise there is a change in colour, however light it is, that's your end point and you should stop.

Make sure you use a white tile under your conical flask and remember to pause the titration and swirl the flask little bit every now and then as you near the end point. That should even out the colour and make the change noticeable before it's too late.


Why is the boiling point of ethyl fluoride lower than that of hydrogen fluoride? The book, Solomons' Organic Chemistry (for JEE Mains and Advance), contains the following question:

  Hydrogen fluoride has a dipole moment of $\pu{1.82 D}$; its boiling point is $\pu{19.34 ^{\circ} C}$. Ethyl fluoride ($\ce{CH3CH2F}$) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is $\pu{-37.7 ^{\circ} C}$. Explain.

Now while explaining the (very)  concept of boiling point and why it differs for different substances the book considers:

Dipole-dipole forces
London dispersion forces
Molecular weight

But when I tried to explain the phenomenon stated in the question, I found it hard to explain the lower melting point. Following is my approach on the explanation:

Since both have the same (similar) dipoles therefore there would be very negligible difference in their boiling points.
London dispersion forces increase in magnitude as the surface area of the molecule increases. So contrary to what was said $\ce{CH3CH2F}$ should have stronger dispersion force interaction (due to larger surface area)  and hence higher boiling point as compared to $\ce{HF}$ (given other properties were same).
Given that $\ce{CH3CH2F}$ has higher mass than $\ce{HF}$, therefore it should lead to higher boiling point (as is stated in the question itself).

Now it can be clearly seen that the above stated arguments are against the phenomenon, so I believe that I am missing something in my explanation.

So

What is (if any) wrong/missing in my explanation?
Why is the boiling point of $\ce{CH3CH2F}$ lower than that of $\ce{HF}$?

The three reasons that you present are not sufficient to explain the high boiling point of pure $\ce{HF}$. There is a fourth reason, that you have not mentioned : the Hydrogen bond. Molecules $\ce{HF}$ are interacting with one another through $\ce{H}$-bonds, exactly like $\ce{H_2O}$ molecules attract one another because of $\ce{H}$-bonds. In the liquid state, $\ce{HF}$ behaves as if it is made of heavier molecules, something like $\ce{(HF)_n}$, with $\ce{n}$ being between about $4$ and $7$. The boiling point of $\ce{HF}$ and of $\ce{H_2O}$ is extremely high when comparing to substances of nearly the same molar weight, like $\ce{CH_4}$ or $\ce{Ne}$. The boiling point of $\ce{H_2O}$ and $\ce{HF}$ does not depend significantly on the van der Wals forces. $\ce{H}$-bonds are much more important than van der Wals forces.

And there are practically no $\ce{H}$-bonds in ethyl fluoride, because $\ce{H}$-bonds happens only  when an $\ce{H}$ atom is attached to a strongly electronegative atom.


Is it possible to calculate the concentration percentage using chromatography paper? I'm trying to calculate the concentrations of specific compounds using paper chromatography. Would I be able to calculate a rough estimate by comparing the size of each concentrated spot? If not is there another way to calculate this?

Thanks!

Yes, it can be done.

I have done it on silica plates to compare the content of a natural product in different samples of plants.
It is , obviously, better to use HPLC, but in that case the sample preparation was very tedious so we decided to do it through TLC.
The content in the different samples was between 0 and 2%. Standards equivalent to a content of 0.5, 1, 1.5 and 2% we're prepared.

What you need to have succes is to be very accurate in the quantity that you apply on the plate and to have a good developer.
The quantities applied must be all the same, preferably not too much because then the spots are very big and it is more difficult to compare them.
Our product was developed with $\ce{KMnO4}$ yielding an intense yellow spot on a purple background.


Is Kirchoffs law valid at all pressures? $$\Large H_{T_\mathrm{f}, p}=H_{T_\mathrm{i},p}+\int_{T_\mathrm{f}}^{T_\mathrm{i}}c_p(T)dT$$
This can be derived by integrating: $$C_v =\left(\frac{\mathrm{d}U}{\mathrm{d}T}\right)_V$$
Applying this to both reactants and products in a reaction, and subtracting reactans from products, we can arrive to

$$\Delta H_{T_\mathrm{f},p}=\Delta H_{T_\mathrm{i},p}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_p(T)\mathrm{d}T \tag{1}\label{a}$$
Where $$c_p(T)=\sum_i\nu_\mathrm{i}c_{p,\mathrm{i}}(T)$$
This is essentially Kirchoffs law how enthalpy varies with temperature. However, whenever I see the the  equation $\ref{a}$, it always deals with standard enthalpy changes, i.e: $$\Delta H_{T_\mathrm{f}}^{⊖}=\Delta H_{T_\mathrm{i}}^{⊖}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_{p^{⊖}}(T)\mathrm{d}T $$
 Is this law not valid for all reactions? If so, what is the mistake in the above derivation?

It would say that you are making three separate mistakes:

The standard state pressure is any arbitrary pressure, not necessarily 1 bar, so there is no implication from the equation specifying standard state that it is only true at one particular pressure.
The important specification of being in the standard state is not the value of arbitrary pressure chosen, but that the substances are not mixed.  They are pure or in solvent as the single solute.  Also, for gases and solutions the states are fictitious.  In your derivation, you are not accounting for changes due to mixing and any changes due to being in real, rather than fictitious, states.
I don't see why you say to integrate Cv rather than Cp.


Why can temperature changes have units of either degrees Celsius or kelvin? One equation to calculate a heat change is $q = c_p \cdot m \cdot\Delta T$. I know that $\Delta T$ can be in either degrees Celsius ($^\circ{}\mathrm{C}$) or kelvin ($\mathrm{K}$). However, one thing that confuses me is how the units cancel if $\Delta T$ is in degrees Celsius, since $c_p$ has units of $\mathrm{J\ g^{-1}\ K^{-1}}$. How does this work?

The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.

So, $\Delta T$ will be the same whether the temperature was reported in $\mathrm{K}$ or $^\circ\mathrm{C}$.  As an example if $T_\mathrm{initial} = 0\ ^\circ\mathrm{C}\ (273\ \mathrm{K})$ and $T_\mathrm{final} = 100\ ^\circ\mathrm{C}\ (373\ \mathrm{K})$, $\Delta T$ is 100 degrees no matter whether you used $^\circ\mathrm{C}$ or $\mathrm{K}$ in your computation.


Geometry of buta-1,3-dien-1-ol Consider the molecular formula $\ce{CH2=CH-CH=CH-OH}$ $(\ce{C4H5OH})$. Tell the types of bond, hybridizations, bond angles and geometry of this compound.

I didn't have any trouble with the first 3 topics.
Carbon hybridizes in $\mathrm{sp^2}$ and oxygen hybridizes in $\mathrm{sp^3}$.
Each carbon makes 3 $\sigma$ bonds and 1 $\pi$ bond and the oxygen makes 2 $\sigma$ bonds. The angles of the bonds $\ce{H-C-H}$, $\ce{H-C-C}$, $\ce{C-C-C}$ are 120 degrees and the angle of bond $\ce{H-O-C}$ is 109 degrees.
My problem is with the geometry. How can I deduce it? It doesn't seem to be one of the standard geometries.

Let’s start off with the things you thought you got right: Oxygen will not be $\mathrm{sp^3}$ hybridised but somewhat closer to $\mathrm{sp^2}$. This is because that is how the oxygen lone pair in a p-type orbital can take part in π conjugation. I’m not sure what $\angle(\ce{C-O-H})$ I would expect but it’ll be somewhere between $109^\circ$ and $120^\circ$, that’s for sure.

To the entire molecule’s structure: Yes, at first glance it seems like a potato. However, there is still something that you can say: for proper conjugation of the double bonds (usually favourable), you need a planar structure. Thus, the entire molecule will be planar with all atoms in the same plane. That is about all that you can safely derive, since there can still be rotation around the $\ce{C-C}$ single bond.

Also, as Ivan pointed out, the molecule is likely to prefer the keto-tautomer rather than the enol one. The conjugation argument may apply, but it is rather weak.


Mechanism of Fritsch–Buttenberg–Wiechell rearrangement The following reaction was given, which uses potassium t-butoxide.

The solution to this problem was given as follows:

Step 1 is the attack of strong base, t-butoxide, on the vinyl proton.

Step 2 seems to have the chlorophenyl group shift to the neighbouring carbon, forcing the cleavage of the $\ce{C–Br}$ bond.

Here, the bromine is conjugated to the alkene, since the lone pair on bromine is in resonance with the alkene, leading to partial double bond character between $\ce{C}$ and $\ce{Br}$. In such a case, the $\ce{C–Br}$ bond usually cannot be cleaved.

My question is

Is this carbanion intermediate responsible for the product?
What is the mechanism to form 1-chloro-4-(phenylethynyl)benzene?

As your heading correctly states, this reaction take the path of Fritsch–Buttenberg–Wiechell rearrangement (Wikipedia).   The reaction mechanism of Fritsch–Buttenberg–Wiechell rearrangement (FBW) is described as:

  The strong base deprotonates the vinylic hydrogen, which after alpha-elimination forms a vinyl carbene. A 1,2-aryl migration forms the 1,2-diaryl-alkyne product. The mechanism of the FBW rearrangement was a subject of on-surface studies where the vinyl radical was visualized with sub-atomic resolution (Ref.1).

However, if you carefully read this reference, you'd realize they use special method to generate diradical from gem-dibromo-alkene stepwise. Thus, the proposed mechanism for every FBW-type reactions. Earlier, A. A. Bothner-By (Ref.2) from  Harvard reported that in a diaryl-substituted system of a bromoalkene (e.g., see structures 1 and 4 in the scheme 1), the major migrating group is the aryl moiety located anti to the halogen (the fact, which has been proven by $\ce{^{14}C}$-labeling experiments). Particularly, this migration of an aryl group between multiply-bonded carbons is one of the several unusual features of the rearrangement (Ref.2):

Yet, there are plethora of data to support the 1,2-phenyl group migration through single bonds (e.g., the benzilic acid rearrangement), and at least one example for migration of an aryl group between multiply-bonded carbons (e.g., the Beckmann rearrangement). Thus, based on these facts, I suggests following arrow-pushing mechanism, which would be very reasonable:

References:

N. Pavliček, P. Gawel, D. R. Kohn, Z. Majzik, Y. Xiong, G. Meyer, H. L. Anderson, L. Gross, “Polyyne formation via skeletal rearrangement induced by atomic manipulation,” Nature Chemistry 2018, 10, 853–858 (https://www.nature.com/articles/s41557-018-0067-y).
Aksel A. Bothner-By, “Stereochemistry of the rearrangement involving migration between multiply-bonded carbons,” J. Am. Chem. Soc. 1955, 77(12), 3293–3296 (https://doi.org/10.1021/ja01617a042).


In a mixture of NaOH and Na2CO3, why doesn't phenolpthalein change its color until Na2CO3 is half neutralised? Let us consider a mixture of $\ce{NaOH}$ and $\ce{Na2CO3}$. Based on what I have understood, double titration works in the below manner :

  Phenolpthalein turns pink when added to the mixture. As $\ce{HCl}$ is added, the pH of the mixture keeps reducing and phenolpthalein becomes colorless when the mixture becomes of a pH less than $8.3$ ie. at a stage when $\ce{NaOH}$ is completely neutralised and $\ce{Na2CO3}$ is half neutralized. Then, we add methyl orange which makes the solution yellow. As we add more $\ce{HCl}$, the pH reduces and hence, methyl orange turns red when the neutralization of $\ce{Na2CO3}$ is complete.

My questions are:

a) When phenolpthalein is added, I have read elsewhere that $\ce{NaOH}$ reacts with $\ce{HCl}$ first. In that case, phenolpthalein should make the solution colorless as $\ce{NaOH}$ neutralizes with $\ce{HCl}$ to form $\ce{NaCl}$ which has a pH of $7$ (not between $8.3$ and $10.5$). Why doesn't phenolpthalein change its color until $\ce{Na2CO3}$ is also half neutralized? Is it because the reactions take place simultaneously?

b) When methyl orange is added, $\ce{NaHCO3}$ is neutralized to $\ce{NaCl}$. But pH range of methyl orange is 3-4.5, but the solution obtained is neutral. How does methyl orange change its color, then?I think it is because of the $\ce{H2CO3}$ ($\ce{H2O + CO2}$) that is formed along with $\ce{NaCl}$, which makes the solution slightly acidic. Is that a correct explanation?

a) That's right, $\ce{NaOH}$ reacts first. But you are wrong in thinking that $\ce{NaCl}$ has a pH of 7. In a way, $\ce{NaCl}$ has "no pH at all". It just sits there and does nothing. It is other component(s) of the solution that determine its pH. At that point, the other component is $\ce{Na2CO3}$, which is alkaline due to hydrolysis. As the titration progresses, it turns to $\ce{NaHCO3}$, hence the change in pH.

b) Well, $\ce{H2CO3}$ would indeed make the solution slightly acidic, but not enough so to trigger the color change in methyl orange. It is $\ce{HCl}$ that does the job. After the neutralization is completed, one tiny extra drop is enough.


Why is energy released as electrons move? Why is it that energy is released from electrons when they move from one atom to a more electronegative one?

The previous answer is correct. I just wanted to add that atoms, electrons and generally matter, tries to ''organise'' itself in order to adopt a state, where it would have the lowest energy possible.
An electronegative atom tends to attract electrons and has the ability of handling their charge better. Thus when an electron moves from a less electronegative atom to a more electronegative one, it is like moving from a state of high energy to a state of low energy. This is also described as realease of energy.


Can cis-trans isomers have same InChI? From all we know only tautomers should get the same InChI key, or, in case of the second block, if there are lots of stereo centers. However these cases are just collisions, they still have different InChI. I have found a class of molecule pairs, with four real-world examples in ChEBI and PubChem, where both web input pages of ChEBI and PubChem generate identical InChI and keys, despite there being a different E-Z configuration on a double bond. And these are also entries in the said databases. My question:
Is this normal/expected? Is it a bug in both software, which part of it?
The pairs:

5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)-1-benzopyran-1-ium-3-yl 6-O-[(2Z)-3-(3,4-dihydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)-1-benzopyran-1-ium-3-yl 6-O-[(2E)-3-(3,4-dihydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

2-(3,4-dihydroxy-5-methoxyphenyl)-5,7-dihydroxychromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 2-(3,4-dihydroxy-5-methoxyphenyl)-5,7-dihydroxychromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)chromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]--β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)chromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

5,7-dihydroxy-2-(4-hydroxy-3-methoxyphenyl)chromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3-methoxyphenyl)chromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucoside-glucopyranoside

Addendum: I also confirmed with two different SDF files that the current inchi-1 Linux binary generates the same InChi.
P.S. This is a pair of minimal molecules:

Both have the InChi InChI=1S/C11H10O3/c1-14-8-11(13)7-4-9-2-5-10(12)6-3-9/h2-8H,1H3/p+1. If the methyl is removed, leaving a =OH+ group at the end, both isomers get different InChi.

Professor Jonathan M Goodman gave an explanation on the InChi mailing list:
https://sourceforge.net/p/inchi/mailman/inchi-discuss/thread/B7D99356-8B55-43BE-8C2A-B8CC3EB37799%40cam.ac.uk/#msg37289766
In summary, the given class of molecules represents a special case where the current InChi algorithm (1.06) produces less than useful results in order to being able to handle all other cases satifyingly. Prof Goodman hints at work being done in organometallics which may affect this behaviour. So, it's not a bug in the implementation of the algorithm, it is the algorithm itself that needs improvement.
The linked message also contains a smaller minimal case, confirming that it is the hetero-ion together with the keto-enol configuration causing the phenomenon.


Order of reactions in kinetics In kinetics, zero order reactions are those reactions who are independent from the concentration of the reactant or the product. So their rate is equal to the constant k. But still we build the diagram where we find the correlation between concentration and time, where the slope is negative.
Why do we even write a diagram to find the correlation between concentration and time, since these reactions are completely independent from concentration ?

So their rate is equal to the constant k.

With other words, the change of concentration over time is linear:

\[ \frac{d[\ce{A}]}{dt} = k\]


Physical chemical property that determines redox potential? Can someone explain (on an advanced level) what phys-chem property of compounds/elements determines what their half-cell redox potential will be? For example, Zn and Cu are structurally very similar and their ionization energies, electronegativities, etc. won't be all that different. So, why is the 2-electron redox potential for each so different? ie Zn is -0.76 V (SHE) and Cu is +0.34 V (SHE), which is a significant 1.1 V different.

Isn't electric potential the amount of "stored" energy? So, wouldn't removing or adding 2 electrons to Zn or Cu be very similar since they are next to each other on the periodic table?

Edit: I've added some sources and clarity to the discussion in the answer below.

As I understand it, the potential is related to the difference in energy between the oxidized and reduced state (via the Gibb's energy as Mithoron noted). While their individual electron orbital energies are probably similar given the similar nuclear charges, the oxidized and reduced states are differently occupied and so have different energies.

In zinc for example, you have the reduced state of 4s2, 3d10 and oxidized state of 3d10. Copper on the other hand goes from 4s1, 3d10 to 3d9. $\ce{Zn^2+}$ has a fully occupied d-shell while $\ce{Cu^2+}$ does not, meaning that the reduced state for zinc is lower energy and more stable. As a result, zinc has a stronger tendency to oxidize and consequently a lower reduction potential.


Ionisation energy of Nickel Look for the following data for Nickel
IE1+IE2  2490 kj/mol
IE3+IE4.  8800kj/mol
Total.       11290kj/mol

Hence Nickel can show +2 oxidation state but not +4.
But according to me Ni in +4 oxidation state can acquire 3d6 and 4s0 which has t2g3 configuration.
Why do it has high ionisation energy?

Metals with a $d^8$ oxidation state are stabilized by a square-planar molecular orbital configuration. In this configuration, the $\ce{d_{yz},d_{xz},d_{z^2}}$ orbitals are stabilized, while the $\ce{d_{xy}}$ orbital is slightly destabilized and the $\ce{d_{x^2-y^2}}$ is severely destabilized (however, in $d^8$ species this orbital is unoccupied). The amount of stability provided by the square planar configuration is far greater than the amount of stability provided by the low-spin $d^6$ configuration.

This is a good online powerpoint that shows the stabilization.


How will liquid nitrogen affect a steel instrument's ability to function? I want to retrieve cryotubes from the bottom of a liquid nitrogen freezer without taking all the freezer racks out and draining the liquid nitrogen. I think I can accomplish this using a tool like this but I am not sure how well the tool will function at temperatures as low as -160 Celsius. Also, this particular tool is probably only about half as long as it needs to be.

Assuming I can find a longer version of such a tool, how can I determine whether or not I need to be concerned about using to retrieve cryotubes from a liquid nitrogen freezer? My main worry is that parts of the tool may become brittle and break or that the claw's ability grab may somehow become compromised. I am not concerned about the metal becoming excessively cold because I will be wearing gloves intended for such work.

Is there a chart or other resource that I can use to get a sense of metal's shear strength at cold temperatures?

Metals become brittle as low temperature (and that's why you cool your Parr bomb in a dry ice bath, never in liquid nitrogen when you condense gases in).  But the worst thing that can happen is that a claw might break off, which shouldn't cause any damage to equipment.  Why not try in a small dewar first.  Liquid nitrogen is cheap.

A good alternative to racks of storage boxes are storage canes, such as these.


Dry adiabatic lifting If an unsaturated air parcel ascends adiabatically in the atmosphere with an initial temperature, would the final temperature of the air parcel, when it reaches it's destination, be the same at the initial temperature since it does not exchange heat with it's surroundings?

No. When the gas - let's assume it is an ideal gas - rises up it will expand because the pressure $p$ sinks the higher it comes. If it is allowed to expand quasi-statically under adiabatic conditions ($\delta Q = 0$) then it does work $W$ (pressure-volume work $W = - p \mathrm{d}V$ to be more precise) on its environment, and, hence, its internal energy $U$ is reduced, and its temperature $T$ changes because according to the first law of thermodynamics

\begin{align}
  \mathrm{d} U &= \delta Q + \delta W \\
  &\overset{\delta Q = 0}{=} \mathrm{d} W = - p \mathrm{d}V
\end{align}

and also for an ideal gas you have

\begin{align}
  \mathrm{d} U &= C_{V} \mathrm{d}T
\end{align}

where $C_{V}$ is the heat capacity. Hence the temperature changes according to

\begin{align}
  \mathrm{d}T = - \frac{p}{C_{V}} \mathrm{d}V \ .
\end{align}


What is the major product of the reaction between 4-chlorobut-1-ene? When acetic acid looses hydrogen, an acetate ion is formed which attacks 4-chlorobut-1-ene, forming methyl butenoate. How does the cyclic structure form?
Does the lone pair of oxygen attack the butene moiety internally?

In solvolysis of simple primary cyclopropylmethyl systems the rate is
  enhanced because of participation by the $\sigma $ bonds of the ring. The ion
  that forms initially is an unrearranged cyclopropylmethyl cation that is
  symmetrically stabilized, that is, both the 2,3 and 2,4 $\sigma $ bonds help
  stabilize the positive charge(page 464 ,JerryMarch ,Advanced Organic Chemistry
  Reactions, Mechanisms,and Structure,Sixth Edition).

The reaction of various cyclopropylmethyl systems are given below.

In the question 4-chlorobut-1-ene  (1) undergoes rearrangement as shown in the scheme 3 and scheme 4 to give 1a, 1b and 1c.Assuming solvolysis with acetic acid ,1a, 1b and 1c would give 2, 3 and 4.


How to use a modified Peng-Robinson equation to calculate a density? I'm trying to use the Peng-Robinson-Stryjek-Vera 2 (PRSV2) Equation of State to calculate the density of a compound, in this example: methanol. At ambient temperature and pressure, shouldn't I be able to solve the cubic equation of state for the molar volume and calculate a density corresponding reasonably well to the density of the liquid given in the literature at those conditions of temperature and pressure? The values are considerably off, at least farther off than I would expect from such a well tested equation of state. For instance, for methanol, I obtain a density of 0.672 g/mL rather than the literature value of 0.791 g/mL. For acetonitrile, it's even worse: 0.480 g/mL rather than 0.786 g/mL. Does anyone have experience with such cubic equations of state? Are they simply not that accurate? Am I doing something wrong? I've checked my implementation numerous times for typos, but cannot find any.

This implementation in MATLAB comes from the following two papers:

Stryjek, R. and J. H. Vera, (1986). "PRSV: An Improved Peng-Robinson Equation of State for Pure Compounds and Mixtures", Can. J. Chem. Eng., 64, 323-333.

Stryjek, R. and J. H. Vera, (1986). "PRSV2: A Cubic Equation of State for Accurate Vapor-Liquid Equilibria Calculations", Can. J. Chem. Eng., 64, 820-826.

T = 298; % Temperature, in K
P = 1; % Pressure, in atm

% Values for methanol
M = 32.04; % Molar mass, in g mol^-1
T_c = 512.58; % Critical temperature, in K
P_c = 8095.79; % Critical pressure, in kPa
omega = 0.56533; % Acentric factor, from PRSV2
kappa1 = -0.16816; % From PRSV
kappa2 = -1.3400; % From PRSV2
kappa3 = 0.588; % From PRSV2

R = 8.3144598; % Ideal gas constant, in Pa m^3 mol^-1 K^-1
R = R*1000; % Ideal gas constant, in kPa cm^3 mol^-1 K^-1
T_R = T/T_c; % Reduced Temperature

kappa0 = 0.378893 + 1.4897153*omega - 0.17131848*omega^2 + ...
    0.0196544*omega^3;

kappa = kappa0 + (kappa1 + kappa2*(kappa3 - T_R)*(1 - (T_R)^0.5))*...
    (1 + (T_R)^0.5)*(0.7 - T_R);

alpha = (1 + kappa*(1 - (T_R)^0.5))^2;

a = (0.457235 * R^2 * (T_c)^2 / P_c)*alpha;
b = 0.077796*R*(T_c)/(P_c);

P = P*101.325; % Pressure, in kPa

% Using the PRSV EOS cubic expansion for molar volume
C3 = 1;
C2 = b - R*T/P;
C1 = a/P - 2*R*T*b/P - 3*b^2;
C0 = (b^2 + R*T*b/P - a/P)*b;

cubic  = [C3 C2 C1 C0];
v = roots(cubic); % Molar volumes, in cm^3 mol^-1

rho1 = M/v(1) % Density, in g cm^-3
rho2 = M/v(2) % Density, in g cm^-3
rho3 = M/v(3) % Density, in g cm^-3

According to this source:

  The popular EoSs such as Soave-Redlich-Kwong (SRK) [1] and Peng-Robinson (PR) [2] predict liquid density with an average absolute error of about 8%, much higher than several good density correlations. This large magnitude of error is not acceptable by industry; therefore, they are not used for this purpose.

Which is what I first thought when I thought of your question. Liquid density is a hard issue, specially in compounds with hydrogen bonding like ethanol. I wouldn't predict a gas-oriented (as in its primary use) EoS to perform just as well for liquids.

The article further suggests

  In order to overcome this deficiency, volume translated methods have been developed.

So that's what I'd recommend you to look into.


In regard to collision theory, why do molecules have to have the correct orientation in order to react? Hypothetically speaking, if a reagent in a bimolecular reaction has enough energy to break the bond of the other reagent but is orientated in the wrong direction, why does the bond not break? Does the reagent's orientation affect the activation energy of the reaction somehow? The point I'm trying to make is if you have enough energy to break a bond, why would molecular orientation matter when the same amount of energy is being used to distort bonds regardless of orientation?

Suppose we have a simple reaction of a diatomic and a single atom $\ce{AB + C ->A +BC}$. So the reaction is breaking the bond between $\ce{A}$ and $\ce{B}$, which then bonds to $\ce{C}$. So we might expect the transition state to look like $\ce{A--B--C}$. This suggests the reaction is highly orientation dependent.

Looking at the process as a function of the angle formed by ABC we would expect that a 0 degree angle would vary likely lead to a reaction, since everything is lined up as we would expect, while a 180 degree angle would have almost no chance of forming the desired product as it would aligned more like this $\ce{C--A--B}$. While its not a perfectly linear relationship, we should expect that the reactivity will decrease monotonically as it goes from 0 to 180 degrees.

For more complex structures, the orientation becomes even more important. A large molecule may only have a very small reactive site, and if the other reactant collides with a completely different section of the molecule, there is virtually no chance the desired reaction will occur.


Under what conditions does MnO2 acts as an oxidizing agent? It's very strong in acidic medium. Does it have low or nil oxidizing ability in neutral and basic media?

Does heating solid $\ce{MnO2}$ give it better oxidizing properties?

It certainly is  powerful oxidizer when heated! Thermite can be made with powdered aluminum and $\ce{MnO2}$, and is at least as spectacular as with $\ce{Fe2O3}$ as the oxidant.


Electrochemistry conditions for voltage to increase in a galvanic cell I understand that increasing the concentration of reactants in a galvanic cell generally causes the voltage of the cell to increase. But is that really the case always?

Consider a standard cell of zinc and copper, 1 M and 1 L each solution of copper nitrate and zinc nitrate in each half cell. Another cell, increased the concentration of copper nitrate to 4 M, but added 0.25 L of it and kept the same zinc nitrate 1 M and 1 L.

This way I increased the concentration, but the amount of substance is the same. Does the voltage increase in the new cell? I would really appreciate someone to explain to me the specifics of what causes the voltage of a cell to increase based on the reactants

Yes, your voltage will change in accordance with the Nernst equation

$$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$

So, in your copper-zinc battery, where

$$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$

with a standard $E^\circ_\text{cell} \approx \pu{1.10 V},$ setting the $\left[\ce{Cu^2+}\right] = \pu{4 M}$ should raise your cell potential by

$$\Delta E_\text{cell} = -\frac{RT}{2F}\ln\frac{1}{4} = \pu{+0.0178 V}$$

for a new cell potential (at $\pu{25 °C})$ of

$$E_{\text{cell}} = \pu{1.10 V} + \pu{0.0178 V} \approx \pu{1.12 V}$$


How to distinguish between a coordinate bond and a covalent bond? I’m confused as I am unable to find a completely trusted way of finding coordinately bonded compounds from a bunch of other compounds.

Coordinate bonds can be distinguished from (other or ‘normal’) covalent ones experimentally.

Let’s look at the bond creation first as Newbie suggested. Typically, one would use a Lewis base and a Lewis acid to create a coordinate bond between them.

$$\ce{NH3 + BH3 -> H3N\bond{->}BH3}\tag{1}$$

In contrast, one would usually consider a normal covalent bond to be created from two radicals:

$$\ce{H3C^{.} + {}^{.}CH3 -> H3C-CH3}\tag{2}$$

But we can also create covalent $\ce{C-C}$ bonds by using a carbanion source and a carbocation source (slightly different example, because both ions are actually known and react with each other):

$$\ce{(H3C)3C+ + {}^{-}CN -> (H3C)3C-CN}\tag{3}$$

Thus, we can’t really tell a difference here. The difference in fact lies in the way a bond dissociates if the bond dissociation enthalpy $D = \Delta_\mathrm{D} H^0$ is provided. Standard covalent bonds will dissociate homolytically by the following equation:

$$\ce{H3C-CH3 -> H3C^. + ^.CH3}\tag{4}$$

A coordinate bond will, however, dissociate heterolytically thus liberation the original Lewis acid and base:

$$\ce{H3N\bond{->}BH3 -> NH3 + BH3}\tag{5}$$

In cases such as the ammonium ion ($\ce{NH4+}$) or say $\ce{[AlCl4]-}$ this means we have a certain number of normal bonds and another certain number of coordinate bonds (three and one, respectively in both examples). Although they are all between the same atoms, we can distinguish if we remember that neither ion can exist by itself, there will always be a corresponding counterion. Thus, if we consider $\ce{NH4Cl}$ or $\ce{Na[AlCl4]}$, we realise how the first bond is broken differently:

$$\begin{align}\ce{NH4Cl &-> NH3 + HCl}\tag{6}\\[0.6em]
\ce{Na[AlCl4] &-> NaCl + AlCl3}\tag{7}\end{align}$$

And all further bonds are cleaved homolytically. We gain two different bond dissociation energies, one corresponding to the coordinate bond and one corresponding to the remaining bonds. (Note that salts such as $\ce{AlCl3}$ or $\ce{NaCl}$ are cleaved homolytically by heating in the gas phase — flame-colouring would not happen if we had $\ce{Na+}$ in the flame.)


If temperature and volume are same, the pressure of gas obeying van der Waal's equation is smaller than that of an ideal gas" Can you state the reasons for it. I think " It is smaller because mutual forces of attraction would resist the collision of particle and container".

Your statement is false. There are cases where the pressure from the Van der Waals pressure is higher than the pressure from the Ideal Gas Law.

The Van der Waals equation introduces two corrections to the Ideal Gas Law:

It accounts for attractive forces between gas particles. This means that the Van der Waals pressure can be lower than the Ideal Gas pressure.
It accounts for the small but nonzero volume of the particles. This means that the Van der Waals pressure can be higher than the Ideal Gas pressure.

Depending on the amount and nature of your gas, the volume and the temperature, either one of these effects can dominate.

Qualitatively:

In a situation where gas molecules are squished together very closely, the Van der Waals equation yields:

Higher pressures for molecules that are big but not sticky, like isobutane, and
Lower pressures for molecules that are small but sticky, like hydrogen chloride.

Quantitatively:

I'm attaching a spreadsheet with a bunch of examples; pardon the German. For a number of situations described by:

substance, column A
volume (liters), column B
amount (moles), column C
temperature (°C), column D

the pressure is given according to the Ideal Gas Law (column I) and the Van der Waals equation (column J).

There are examples that follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 6.

There are examples that do not follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 11 or 22.

Keep in mind that both equations only apply to gases; look up phase diagrams and make sure the gases in question don't condense under the conditions given. Rows 12, 13, 23 are supercritical fluids.

You'll find that the Ideal Gas Law is quite precise in most cases, and the cases where it's not, are those where the gas is close to not being a gas anymore.


How will metallic lithium react with atomic oxygen? Very high in Earth's atmosphere, there's a lot more atomic oxygen than anything else. Although I'm sure the energies involved in keeping oxygen in this state are an important variable, I'm interested (for now) in what compounds would form.

Mistake here: should be targeting $\ce{Li2O}$, not $\ce{LiO2}$. Would you get $\ce{LiO}$, then $\ce{LiO2}$? If so, will $\ce{LiO}$ be stable enough to bond with another $\ce{O}$ atom later? I can easily believe that lithium metal will simply erode away faster than a film of oxide could develop, given temperatures in the thermosphere and above. I just haven't seen much chemistry on this, however.

As explained in Long-term ozone decline and its effect on night airglow
intensity of Li 6708 ˚A at Varanasi (25◦N, 83◦E)
and Halley Bay (76◦S, 27◦W) Journal of Earth System Science 120, No. 2, April 2011, pp. 291–300:

$\ce{Li + O -> LiO}$

$\ce{Li + O2 -> LiO2}$

$\ce{Li + O3 -> LiO + O2}$

$\ce{LiO2 + O -> LiO + O2}$

The article gives rate coefficients for each of these reactions.

The article also gives $\ce{Li}$, $\ce{O}$ and $\ce{O3}$ concentrations as a function of altitude for the range 80-99km (85km being considered the lower boundary of the thermosphere)


Is solubility coefficient affected if ion data is given in Ksp? Recently I've been looking at this website, which says:

The $K_{\mathrm{sp}}$ of $\ce{Cr(OH)3}$ is $6.70 \times 10^{-31}$
Problem 1: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{\mathrm{sp}}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$[\ce{Cr^{3+}}]=s$$
$$[\ce{OH^{-}}]=3s$$
Problem 2: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate if the solution has $[\ce{Cr^{3+}}] = 0.0670~\mathrm{M}$?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{sp}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$6.70 \times 10^{-31}=[0.0670~\mathrm{M}][s^{3}]$$

I don't understand why the site wrote this:

Note that $3s$ is not necessary. In example one, $s$ was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is not expressed in terms of some other unknown value (as it was in Problem 1).

Can anyone tell me why?
Also is also used to other equilibrium constant equation? What did I miss?

Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.

So, the site says that writing the concentration as $3t$ is not necessary.

What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?

Now, for example we had not done that and calculated the value of $t$ by using the equation

$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$

You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.

From which you will find the $\mathrm{pH}$. So the site says:

"Why not calculate the value of $3t$ directly by using a variable $s$?"

Therefore: $s=3t$

Therefore: $\mathrm{pOH}=-\log{s}$

What you are thinking is also correct, and the answer with that will be:
$\mathrm{pOH}=-\log{3t}$

So you both are doing the exact same thing here.


Leaching of copper using zinc Why can we not use zinc for extracting copper from a solution containing $\ce{Cu^2+}$ ions even though it is a better reducing agent than $\ce{Fe}$?
I am studying metallurgy and I cannot understand why zinc cannot be used.

We can.

But I see few reasons why it is not used:

Iron is much cheaper than zinc.
There can be remaining residue of iron/zinc, coated by copper, or just being excessive. While copper can be melted away and iron stays, zinc would melt together with copper, causing unwanted impurity (unless wanted for making brass alloys)
If we remove copper for environmental concerns, then by using zinc, we would just replace one evil by another. While both metals are essential for life in small amounts, both are toxic in larger amounts. ( So does iron, but at much higher level)
Removing dissolved iron from waste is easier than removing zinc.


Does adding NH3 to an HCl solution facilitate the dissolution of aluminum compounds? If we compare the two solutions hereafter, which one would better leach/dissolve aluminum metal/compounds present at/near the inner surface of the glass:
Solution A: $10\,\%\, \ce{HCl + H2O}$
Solution B: $10\,\%\,\ce{HCl + 1\% NH4OH + H2O}$
I'm asking because I'm looking for a solution to leach away aluminum (and other lesser trace metal ions) from the surface of borosilicate glass such that it won't contaminate puriss solutions in the glassware, and one suggestion was to use "salmi" (which I believe is $\ce{NH4Cl}$), so I'm now wondering if $\ce{NH3}$ in solution somehow facilitates the "grabbing" / solubility of aluminum in an $\ce{HCl}$ solution.

Aluminum metal is attacked by $\ce{HCl}$ solutions. But adding $\ce{NH3}$ or $\ce{NH4Cl}$ has no effect on aluminum reactions.
The same thing can be said for aluminum compounds. Adding ammonia or ammonium compounds is not recommended, because it does not help. Aluminum does not make complexes with ammonia or ammonium salts.
But I suspect this addition of ammonia or of ammonium compounds may have another effect. It may help dissolution of other metals if they are deposited on the glass together with aluminum, making complexes containing the metal and ammonia.


Why does solubility of solids sometimes decrease with temperature increase? Recently I was researching $\ce{Ca(OH)_2}$, and I found this on Wikipedia:

\begin{array}{|c|l|}\hline
{\rm \color{blue}{Solubility~in~water}}
 & {\mathrm{0.189\,g/100\,mL}\   (0\,^\circ\mathrm{C})} \\
 & {\mathrm{0.173\,g/100\,mL}\  (20\,^\circ\mathrm{C})} \\
 & {\mathrm{0.066\,g/100\,mL} \ (100\,^\circ\mathrm{C})} \\\hline
\end{array}

Can you explain why solubility decreases when temperature increases?

In short, for this group 2 hydroxide crystal, the dissolution and solvation process is an exothermic equilibrium. Increasing the temperature drives the reaction in the reverse, favouring crystallisation!

However, it would be nice to rationalise some of these terms, so I will lay out a very brief overview of general solubility factors so you can understand the process in a little more detail.

Solubility is a complex thermodynamic process that takes many factors into account. It can be understood as the process of dissolving a crystal in a solvent, and then forming a solvent cage around the solute (solvation). It is a function of three interactions at heart;

solute-solute
solute-solvent
solvent-solvent

Large solute-solute interactions hinder solubility and are associated with high lattice enthalpies (of crystalline solutes), these must be overcome in order to break up the solid and dissolve it. Solvent-solvent interactions must also be overcome in order to solvate the dissolved solute, therefore for a highly interacting solvent the solubility is reduced.

The formation of ordered solvent structures such as clathrin cages etc orders the solvent, and therefore reduces solvent entropy significantly. Therefore solvents with high degrees of freedom hinder solubility.

Solute entropy favours solvation, this reflects the higher solubility of flexible molecules, this is because in the solid they are all locked up, while in solution their rotational and vibrational modes are populated.

In addition to these if the solute has multiple ionic forms then the solubility will be a function of pH too, so must be held constant in practice. The surface profile of the solid too, as well as the extent of cavitation and porosity are also key factors (since they affect the number of solvent-solute interactions possible).

Importantly the type of solvent into which you are trying to dissolve your crystal in is also of fundamental importance! The rule "like dissolves like" is best understood by the (semi-empirical) general solubility equation,
\begin{equation}
\log S =\frac 12-0.05 (T_m-T)-\log P
\end{equation}
A large partition coefficient (the ratio of how soluble the crystal is in organic solvent to aqueous solvent) reduces the solubility, (and mathematically since $\log x$ follows $x$ in limits).

To further complicate the situation there are several definitions of solubility that we can take. In practice we choose intrinsic aqueous solubility, mainly used for high throughput and speed in drug development.

Solubility is really an experimentalists tool for parametrising the extent of Gibbs energy for the dissolving-solvation process, which in turn is usually modelled by complex thermodynamic cycles via gas phase species or super cooled liquids. Modern research into solubility focuses on finding either molecular descriptors through random forest/QSPR or support vector models etc (which are all just number crunching techniques to find correlations between large sets of data) or through fundamental physics and chemistry of the process.
\begin{equation}
\log S=\frac{(-\Delta G_{sol}/RT)}{\ln 10}
\end{equation}
So we understand solubility as an equilibrium constant for the Gibbs change of the solvation process (whatever particular cycle you use to get that data).

Now that we know something about solubility we can ask ourselves, under what conditions does increasing the temperature reduce solubility?

In practice the temperature can effect the entropic term, but also the enthalpies of the crystal lattices, the heats given out by the solute-solvent interactions etc. It is the balance of these process that determines the solubility of something.

In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.

Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :)


How does critical temperature affect adsorption? I don't really understand how does a high critical temperature favour adsorption?

What I think is that the lower the temperature, more easy is it for the gas to liquefy and therefore adsorption is favoured

Higher critical temperature = more adsorbed.

So, what is critical temperature? When gas is heated above it, it can't become liquid, even if you'll increase pressure. Why? Because it's large kinetic energy makes intermolecular forces kinda "irrelevant". And adsorption is all about those forces.

High critical temperature means that gas particles are good in intermolecular forces - "tenacious". This allows them to adsorb (cling to surfaces) more easily than those that won't hold on (low critical temperature).


Dipole moments of pyrrole and furan Why do pyrrole and furan have dipoles oriented in different directions?

Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system.

Where pyrrole and furan significantly differ is that,

in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring
whereas in furan, there is a full lone pair of electrons in roughly the same position.

The localized lone pair of electrons pointing away from the ring has a very significant effect on the dipole vector and is enough to cause the observed reversal in dipole moment direction between furan and pyrrole.