Phase diagram of water using Gibbs free energy instead of pressure From my lecture course it was indicated that this is what a phase diagram looks like.

Is this diagram applicable to water? All the ones I can find online seem to use pressure on the y axis as oppose to G. If if it is applicable to water, where is the triple point on the graph? I understand it's where the 3 phase lines intersect but this doesn't appear to occur on this type of graph.

Phase diagrams are not limited to temperature vs pressure. They can be made with any set of thermodynamic quantities. The diagram in your question is for a particular pressure at which the liquid phase exists. If you changed the pressure, the lines would shift. A triple point will be represented by all three lines intersecting through the same point.


Why does branching decrease density and melting point? The general explanation given is that the chains are further apart, so the intermolecular forces are less, and as they are further apart, they occupy more volume, and the density is lesser. But the length of each chain (and thus its mass) also increases, so how do we know that the density overall decreases?

As for the melting point, longer chains = more electrons = more Van Der Waals forces, so the melting point should increase.

Edit: I'm referring to branching in the context of polymerisation.

I'm not sure that your statement about density is always correct. Looking at octane and its isomers the density changes very little, for example

octane           $703~ \pu{kgm^{-3}}$

2-methylheptane  $698~ \pu{kgm^{-3}}$

2,5-dimethylhexane  $694~ \pu{kgm^{-3}}$

2,2,4-trimethylpentane  $692~ \pu{kgm^{-3}}$

although the melting points do change considerably. In the same order as above they are $-57, -112,-93,-107$ so octane is the exception here; (data from Wikipedia pages).

This can be explained by the fact that melting can be considered to be governed by the repulsive part of the intermolecular potential this having a greater influence than the attractive part when the molecule has an odd shape. Rod like octane has to have deeper intermolecular potential than the odd shaped isomers as it has a higher melting point.

In the isomers the interaction should be dominated by the repulsive part of the potential as the odd shapes must try to push methyl groups into to one another. The potential itself has the same overall geometry, i.e. similar intermolecular minimum distance, as density is the same in all molecules above, but is shallower in nature due to the extra repulsion. Hence the isomers have significantly lower boiling points than octane.


Why are tertiary carbocations so stable? Today our lecturer provided this data to illustrate the relative stability of carbocations.

\begin{array}{|c|c|c|c|c|} \hline
\text{Substrate} & \ce{t-BuBr} & \ce{i-PrBr} & \ce{EtBr} & \text{MeBr}\\ \hline
\mathrm{Relative~rate~of~S_N1~reaction} & 1.2 \times 10^6 & 12 & 1 & 1\\ \hline
\end{array}

Additionally, in several places I have seen it said that tertiary carbocations are common and stable whereas secondary carbocations are only rarely stable and primary obviously not at all.

What is immediately obvious is that going from a primary to a secondary carbocation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change. The number of bonds available to participate in hyperconjugation increases linearly along the series but the rate clearly doesn't so there must be some other factor at work. The only idea I had was that there might be some similar effect to Y-aromaticity taking place in the trigonal tertiary cation (there are six electrons involved in hyperconjugation after all) but I have no idea whether this is a plausible suggestion.

What is immediately obvious is that going from a primary to a secondary cation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change.

Uh-Oh!

When we see a table reporting that the solvolysis rate for an ethyl and methyl compound are both equal (both 1 in your table), well that's a tip-off that something is grossly wrong.  No one would expect the methyl and ethyl carbocations to have anywhere near similar stabilities. Therefore we wouldn't expect the appropriate ethyl and methyl compounds to solvolyze (if they could) at anywhere near similar rates.

So if we accept that your table has bad data in it, then we really don't know that the change in going from primary to secondary is any different from the change on going from secondary to tertiary.

Carbocation Stability - Background

There are many factors that can affect carbocation stability (inductive effects, resonance effects, neighboring groups), but if we limit our discussion to the stability of saturated hydrocarbon carbocations then we need only consider two of these - inductive and resonance effects.

The carbocation center is roughly $sp^2$ hybridized with an empty p orbital; the alkyl substituents attached to this carbon are roughly $sp^3$ hybridized. There is a flow of electrons through the sigma bonds from the $sp^3$ carbon(s) to the more electronegative $sp^2$ carbon.  We call this an inductive effect and it will stabilize the carbocation center.

Carbocation stabilization can also result from the flow of electrons through p orbitals.  We call this a resonance effect, and in the particular case at hand it is a hyperconjugative resonance effect, as pictured in the diagram below.

(For more on hyperconjugation, particularly from an MO perspective, see the "Rotational barrier of ethane" section in this Wikipedia article.)

Prediction

The t-butyl carbocation has 3 methyl groups that can stabilize the carbocation center through inductive effects, there are also 9 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The 2-propyl carbocation has 2 methyl groups that can stabilize the carbocation center through inductive effects, there are also 6 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The ethyl carbocation has 1 methyl group that can stabilize the carbocation center through inductive effects, there are also 3 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The methyl carbocation has no inductive or hyperconjugative stabilization.

Consequently we might expect the relative carbocation stability to be

t-butyl > 2-propyl > ethyl > methyl

with the 2-propyl carbocation roughly midway between the t-butyl and ethyl carbocations.

Data

Chemistry is complex enough without adding extra unnecessary variables - like solvent. A relatively straight-forward method to compare carbocation stabilities is to measure the free energy ($\Delta G^o$) of the following equilibrium in the gas phase. The more negative the $\Delta G^o$ observed, the less stable the carbocation.

$$\ce{R^+ + H^- <=> R-H}$$

The following table shows the the results of this investigation and appears to  confirm the above predictions on carbocation stability.

\begin{array}{|c|c|} \hline
\text{carbocation} & \text{Delta} {~G^o} \\ &  \text{(kcal/m)} \\ \hline
\text{methyl} & -314 \\ \hline
\text{ethyl} & -270 \\ \hline
\text{2-propyl} & -252  \\ \hline
\text{t-butyl} & -237 \\ \hline
\end{array}

(data taken from Advanced Organic Chemistry, Part A: Structure and Mechanisms; fifth edition; Authors: Carey, Francis A., Sundberg, Richard J.; table 3.10, page 303)

The differences between the [t-butyl and 2-propyl] and the [2-propyl and ethyl] carbocations is 15 and 18 kcal/m respectively. That the 2-propyl cation is roughly in the middle of the t-butyl and ethyl carbocations is likely fortuitous since the ethyl carbocation is structurally quite different from the other two alkylated carbocations. The ethyl cation is non-classical in nature, involving a 3-center 2-electron bond with a bridging hydrogen as explained in this earlier answer.


Why do Alpha particles not collide with electrons during alpha decay? Alpha particles are positively charged, so my question is that during alpha decay, when these are released from the nucleus, why doesn't it ever occur that they collide and grab an electron or two since they are essentially a Helium nucleus?

Actually they do collide.  But, the emitted alpha particle carries much more energy than the binding energy of the electron(s) in a helium atom or $\ce{He^+}$ ion.  Therefore such a collision scatters the electrons (which carry away some of the original energy of the alpha particle) instead of forming an atom.


How does Borax Solution act as a buffer? What type of buffer is it? If it's a Weak Acid-Conjugate Base type of buffer, what is the weak acid and conjugate base?

Borax reacts in hydrochloric acid in the preparation of the buffer to give the boric acid through the following reaction $\ce{Na2B4O7.5H2O + 2HCl -> 4H3BO3 + 2NaCl + 5H2O}$

http://www.novedu.ru/anotes/refer/buffers.htm#bor1

The above link gives you the procedures to making the buffer. The reason why the borax solution is able to keep the system at around pH of 9 because the boric acid that is produced from this reaction has a pKa of 9.46. When the concentrations of the acid and its conjugate base, that is, the borate ions is exactly 1:1, the pH will be most difficult to change since any addition of hydrogen ions will only bond with the borate to form the undissociated boric acid.

So to answer you question, boric acid is the weak acid, and borate ions are the conjugate base.


How to convert P2O5 concentration to H3PO4 concentration? Why are some elements such as phosphorus reported as $\ce{P2O5}$ concentrations when the materials have no $\ce{P2O5}$ in them. I would like an explanation of how the conversion is done between real life and this idealized state.

For example. the Wikipedia page on phosphoric acid states that a $\ce{P2O5}$ concentration of 70% is equivalent to nearly 100% $\ce{H3PO4}$. Can someone explain this conversion?

Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $\ce{K2O}$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $\ce{P2O5}$ but the convention to report elements as oxides continues.

Second part: I will let you solve the rest of the problem: Start from here

1 mol $\ce{P2O5}$ = 2 mol $\ce{H3PO4}$

and do wt percentage conversions to moles. A 70% $\ce{P2O5}$ solution in water means 70 g $\ce{P2O5}$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.


Why does Vapor Pressure decrease on addition of solute? Consider a vessel with a volatile liquid in it. Let's say it has a vapor pressure $P^0$ when it is pure.
Now a non-volatile solute is added to it. Its vapor pressure decreases as $P^0\chi_{solvent}$.
Why does this happen at a molecular level?

The reason that I have come across says that less surface area is available for evaporation and hence this effect. But this does not make sense to me as we can have liquid in container of all shapes and sizes and will still get the same Vapor Pressure.

Some users have suggested me a link to a similar question. The answers in the link are remarkable but the explanation I was hoping for was more likely to involve forces (as in Physics) and maybe a sort of "free body diagram" explaining how these forces are acting(words rather than actual diagram will be fine too) rather than thermodynamic picture.

For solvent at temperature $T$, surface area $A$ and  having the molar fraction $x$, the rate of molecules leaving liquid is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$
Lower molar fraction means lower surface density and lower evaporation rate per surface area.
Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.
For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A  \cdot p \cdot  g(T) $$
The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.
For equilibrium, the net rate must be zero, therefore:
$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot  g(T)}$$
Therefore the equilibrium pressure is proportional to the molar fraction.
It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.


Harmless & pungent volatile liquid for use in a stage show If for some stage play/trick we would like to create some surprising effect in which something with a pungent odor has been released, but at the same time need to ensure that thing be absolutely safe for participants and audience. Is there any organic compound that can do this?

I assume that the compound should be volatile as possible to be effective which reminds me of benzene; however it is detrimental to health. Any recommendation for safer liquids?

I would be very, very very cautious in choosing the odorant here. And speed of odour spreading would be quite low unless specific adjustments are employed to fasten air circulation. And even then, it is quite easy to screw because of minor differences in air circulation.

In practice I would drop any unpleasant odours like mercaptans, sticking to powerful, but generally not unpleasant odours. An example would be benzaldehyde. It has quite strong odour, but autocatalitically oxidizes on air, so the odour should vanish reasonably fast. Some food odourant or parfume would be even better if possible.

Another possible option would be acetic aldehyde. It has strong apple-like odour, and can be easily produced by pouring ethyl alcohol into cromium (VI) oxide, immidiately producing strong odour.

Though I must say, most odoured compound are slightly toxic, so it is not impossible to overdo with troublesome consequences.

Still, I strongly recommend to move from odour to light and/or smoke effect, preferably using premanufactured and commercially available devices. It is too easy to screw up with it.


Openbabel is not converting Gaussian Z-matrix I ran into an issue when OpenBabel (a great and extremely useful problem) could not convert my custom-generated Gaussian Z-matrix to XYZ format. The problem is that probably major flaw of OpenBabel are very terse error messages (mostly just "0 molecules converted", and even with ---errorlevel 25 option I got only:

==============================
*** Open Babel Audit Log  in ReadChemObjectImpl
  OpenBabel::Read molecule Gaussian Z-Matrix Input
==============================
*** Open Babel Audit Log  in Clear
  Ran OpenBabel::Clear Molecule
==============================
*** Open Babel Audit Log  in ReadChemObjectImpl
  OpenBabel::Read molecule Gaussian Z-Matrix Input
==============================
*** Open Babel Audit Log  in Clear
  Ran OpenBabel::Clear Molecule
0 molecules converted

). Some other file was converting just fine, with no obvious differences between their contents! I spent a long time trying to understand what is wrong with the second file, but to no avail.

The file which converts without any problem:

# This is a dummy route line, required for OpenBabel for the conversion

   This molecule contains 26 atoms.

0 1
C
C   1    1.4141
C   2    1.3856  1    117.29
C   1    1.4077  2    121.06  3    -1.09
C   4    1.4088  1    121.27  2    2.37
C   3    1.4075  2    121.95  1    -0.93
C   3    1.5118  2    128.77  1    -176.79
C   7    1.4378  3    105.16  2    175.55
C   6    1.5156  3    108.80  2    -176.32
N   8    1.4317  7    122.53  3    -177.58
C   10   1.3629  8    120.12  7    -146.43
C   11   1.3942  10   120.44  8    -178.97
C   12   1.4024  11   119.82  10   -3.12
C   13   1.4110  12   118.94  11   1.63
C   10   1.3704  8    119.11  7    32.69
O   7    1.2445  3    125.27  2    -1.31
O   9    1.2448  6    124.96  3    -179.37
H   2    1.0882  3    121.72  6    -181.70
H   4    1.1010  1    119.57  2    -178.73
H   5    1.0951  6    121.12  3    179.30
H   11   1.0928  10   115.87  8    1.70
H   12   1.0908  11   118.95  10   178.60
H   13   1.0963  12   120.77  11   182.82
H   14   1.0959  13   121.94  12   -180.61
H   15   1.0855  10   115.78  11   178.99
H   1    1.0923  2    119.78  18   0.45

The file which fails to convert:

#T HF/3-21G Opt Integral(Grid=SG1)

  H2O2 H2O2 GeometryOptimization GeomOpt GeometryOptimization 1

0 1
O
H   1    1.0
O   1    1.4  2    105.0
H   3    1.0  1    105.0  2    176.0
O   1    5.0  2    90.0   3    90.0
H   5    1.0  1    105.0  2    52.5
H   5    1.0  6    105.0  1    178.0

This is OpenBabel version 2.4.1 on Debian 10.

P.S. I sincerely don't know whether this question belongs better to the Chemistry StackExchange or to other forum, but I presume the question might appear to be useful for someone someday.

Turns out that OpenBabel is sensitive to whether blank lines are really blank lines or lines with space characters only. In my case, this was the blank line between the route and the title line. One can see them by pressing the button with a paragraph sign (¶) on it in Your favourite text editor (Notepad++, or even MS Word or  LibreOffice Writer), or by entering :set list in vim or M-x whitespace-mode in Emacs.

What does OpenBabel require for conversion:

First line of the file must begin with a hash sign (#) – a substitute for the Route line
Next line must be absolutely empty
Next line can contain anything You want
Next line again must be absolutely empty
Then comes the charge and multiplicity
And then comes the Gaussian Z-matrix in either compact or full form.

What doesn't OpenBabel require for conversion:

the file to have the extension GZMAT, it can also be GJF
title line not to begin with the hash sign (#) or any other character
route line to contain actual keywords (if You are not trying to convert them to any other input file format, that is)


Do you know a simple fast front-end MD software for OS X I need a GUI front-end simple MD software for OS X, something similar to ascalaph maybe.

What I need to do is a rigid optimization of waters around a protein. I don't have many water molecules around, so a few tips about how to not let them fly away would be nice too.

I'm biased, but I recommend Avogadro which can do simple force field optimization using MMFF94 or GAFF. It's not MD, but it would certainly work. It runs on Mac.

The water molecules shouldn't "fly away" during minimization, since there should be hydrogen-bond and Van der Waals terms attracting them to each other and to the protein.

I'll also point out that VMD runs on Mac and are quite easy to use. It can set up MD calculations with NAMD, GROMACS, etc.


Which component acts as the electrophile in the synthesis of the Sudan-1 azo dye? In the two-pot synthesis, which acts as the nucleophile, 2-naphthol or phenyldiazonium?

Solomons et al. (2014) says that phenyldiazoium is weakly electrophilic. Yet in the questions to be answered in our laboratory manual, it asks for resonance structures on the attack of phenyldiazonium on carbon 1 and carbon 3 of beta-naphthol.

I'm confused. I can't seem to find a mechanism wherein phenyldiazonium acts as a nucleophile in the internet or in my books.

It was mentioned in the comments but I shall mention it again.

Attack does not have to mean nucleophilic attack. In your case, the attack is electrophilic, the electrophile phenyldiazonium attacking the naphtholate.


Melting point and boiling point in isomers We know, that cis isomers are less stable than trans isomers. What I don't get is why the boiling point is changing as opposed to the melting point. The change is melting point makes sense to me considering the stability.

Also the difference of mp and bp in iso and n-butene is not clear to me. How can I possibly explain the change in these physical properties?

Melting and boiling point in relation to cis and trans isomers

The factor that affects melting point is, for the lack of a better term- packability. Molecules that are symmetrical can fit into each other more snugly and are difficult to melt. Therefore trans isomers, being symmetrical, always have a higher melting melting point.

As for boiling, molecules that are held together by any sort of intermolecular force are harder to boil. Cis isomers being unsymmetrical will almost always have a net dipole moment, as opposed to trans isomers where the symmetric nature cancels out the dipole moment. Therefore they are slightly polar and attract each other. Therefore cis isomers have a higher boiling point

Edit: I don't know why isobutene has a higher melting point.


Mechanism of the formation of peracetic acid Wikipedia says that the equilibrium $$\ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.

The first possibility is that $\ce{CH3COOH}$ is protonated into $\ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $\ce{CH3C+O}$. Because the oxygen atom in $\ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $\ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $\ce{H2O2}$ attacks the carbon atom in $\ce{MeCOOH}$, then the $\ce{OH}$ in $\ce{COOH}$ and
one of the $\ce H$ in $\ce {H2O2}$ leave.

Is the mechanism above right or not? If it is not, what's the correct one?

P.S. (this question does not answer my question)

You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.


What is it with resonance structures? This is a concept I have never really understood. I mean to say is how can we include such a thing in a theory? How can we use them if we know that they don't actually exist? Are they some sort of calculation tool?

Resonance structures are a useful way of visualizing and representing the bonding in a molecule - they don't actually exist, but they are a useful tool to see what the bonding and structure of a compound is.

Bonding in a compound can sometimes be nicely represented using two-center, two-electron bonds - for example methane. The structure and reactivity of methane can be rationalised by representing it as a carbon with four hydrogens. Each "bond" can be represented by having two electrons, one from carbon, and one from hydrogen, in a bonding pair. You might draw this as a combination between an sp3 hybrid and a 1s orbital. In reality these hybrid orbitals don't exist either - they are merely a tool to simplify the bonding to localised bonding which is easier to understand. In reality there a multiple interactions between the carbons 2s and 2p orbitals, with the 1s orbital of the hydrogen. However, it is easier, and an accurate enough model to model methane with four localised bonds.

Resonanace is useful where this idea of two-center, two electron bonds breaks down - for example we know from experiment that the structure of benzene is a regular hexagon, with identical carbon-carbon bonds. However the idea of two-center two-electron bonds breaks down here - we can't assign a structure using the ideas we know - carbon makes four bonds, the shape of benzene is a regular hexagon with identical bonds. In this case we can represent benzene as an in-between structure between two structures that can be represented as nice two-center, two-electron bonds. When we draw benzene as a resonance structure of two rings with alternating double bonds, what we are trying to represent is that the real structure of benzene is somewhere between these two structures - it is neither of the structure, and it doesn't resonate - ie. flick really fast between the two structures - instead it is a mixture of the two structures. In benzene the structure we are trying to represent can be understood by considering each carbon to be sp2 hybridized - again these hybrids aren't real, but a tool to understand bonding more easily. Then we have the 6 p-orbitals in the plane of the molecule. In reality these combine two give 6 molecular orbitals. These give a final structure where the bonding is identical between all six carbon atoms, a kind of one and a half bond - but the shape and idea of this delocalised bonding is very important to the reactivity of benzene and other aromatics - which is where this idea of resonance starts to break down in some ways, but the structures can still be useful, but with a proper understanding of the bonding behind the structure.


Hydrolysis of Aspartame with a warmed alkali Aspartame has this structure below

Shouldn't the peptide group be oxidised along with the ester group?

Then why is the answer to the question below B?

EDIT : I'm sorry I misunderstood a friend's marking, the answer is A. I deeply apologise for my stupid mistake.

Thank you very much!

This looks like it's from an exam so there's a very high possibility that the person setting the question and answer intended the ester group on the right to be something else. I am largely of the opinion that the answer should be A.

However, there is one way I can imagine that you can try justify "B" being the answer (although it is a long shot):

  Under alkali conditions, we can expect the acid group from aspartic
  acid to be deprotonated to carboxylate which can act as an
  intramolecular nucleophile.

  We take note that intramolecular reactions are generally faster than
  intermolecular reactions. The carboxylate can attack either the ester
  group or the amide group. An attack on the amide results in a
  5-membered ring transition state, and an attack on the ester results
  in a 9-membered ring transition state. Since 5-member rings are more
  stable, the carboxylate attacks and breaks bond 1, leaving bond 2
  alone. Aqueous alkali then destroys the acid anhydride that is formed.

That said, this is highly unlikely given that esters are less stable to base-catalyzed hydrolysis than amides. So if Bond 1 Breaks, Bond 2 must have also broken.


Why does iodine solution turn transparent on adding base? Yesterday in biochemistry lab when I added sodium hydroxide to iodine solution. The iodine turned transparent. Can someone please explain why is this happening?

Iodine disproportionates, forming iodide and iodate ions, in the presence of base.  But it can be restored by neutralizing the base.  Neutralization will drop the hydroxide concentration low enough so that the disproportionation does reverse.

Let's punch in some numbers using the equilibrium constants given here.  We have

$$
\begin{array}{ccc}
\ce{I2 + 2OH- &<=>& I- + IO- + \ce{H_2O}}&K=30\\
\ce{3IO- &<=>& 2I- + IO3-}&K=10^{20}
\end{array}
$$

To superpose the reactions we take three units of the first one plus one unit of the second, canceling out the unstable hypoiodite ions.  So, rendering that three units of the first reaction mean three factors of that value of $K$, we end with:

$$\ce{3I2 + 6OH- <=> 5I- + IO3- + 3H2O}, K=2.7×10^{24}$$

If we consider a hypothetical case where all the iodine species are in their standard states, including iodine as the element, then the equilibrium hydroxide ion concentration is  $\pu{8.5×10^{-5} M}$.  Taking the base 10 logarithm and adding $\mathrm{p}K_\mathrm{w}=14$ then gives a $\mathrm{pH}$ just below $10$, so a neutral solution gets your elemental iodine back.


Does a halonium ion for astatine exist? I've just begun reading about halonium ions and I noted that only the first four halogens are acknowledged to be able to form halonium ions, but hydrogen astatide can be formed.

I can't seem to find any answers on the internet, but knowing that astatine is super reactive and radioactive, I suppose it may be due to this that forming an "Astatonium" molecule would be nearly impossible?

On the Wikipedia entry of hydrogen astatide, it states that there is an Astatonium — listed as a conjugate acid. However, a quick Google search of the term Astatonium only reveals 14 results.

No astatonium ion with hydrogen is known, but it appears that a cation $\ce{S7At^+}$ is formed by replacing a sulfur atom in $\ce{S8}$ with an isoelectronic $\ce{At^+}$ moiety (the actual oxidation state of astatine, with sulfur being more electronegative, is $+3$).  The ion is proposed to explain the incorporation of astatine into the sulfur precipitate when a sodium thiosulfate solution is decomposed by acid in the presence of astatide ion.

Primary reference:

G. M. W. Visser, E. L. Diemer, Radiochimica Acta 33 145 (1983).


Maximum Useful Work from a Process Negative of the $\Delta G$ for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way:  $\Delta H$ is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently $-\Delta H$ energy is obtained from the system after doing expansion work. Since $\Delta S$ is the entropy created in the process, at the very least $-\Delta S$ entropy must be created in the surroundings - that is, at least $-T\Delta S$ energy must be lost as heat. This comes from the $-\Delta H$, and thus leaves $-\Delta H + T\Delta S$ to do useful work. No more work can be done than this. $-\Delta H + T\Delta S$ is $-\Delta G$, so $-\Delta G$ is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why $-\Delta G$ is the maximum possible useful work.

Now if $|\Delta H|$ is more than $|T\Delta S|$, with both being negative, then one can think of the $W_{max}$ or $-\Delta G$ as a part of the $|\Delta H|$; since $|\Delta G|< |\Delta H|$. Some part of $|\Delta H|$ goes as $|T\Delta S|$ to increase the surroundings' entropy, and the other part in doing useful work. But if both are positive, with $|T\Delta S|$ being more than $|\Delta H|$, again $\Delta G$ is negative, allowing useful work to be extracted. But now it seems as though $|T\Delta S|$ heat will be extracted from the surroundings, $|\Delta H|$ used up in the process, while the rest can be converted to work - in other words, useful work is a part of $|T\Delta S|$ - with the other part used by the process as $|\Delta H|$. Is this correct? Is $|T\Delta S|$ extracted from the surroundings actually?

"Useful work" does not include expansion work. For a proof of why $w_{\text{by,add}} \leq \Delta G$, you may want to first refer to my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work? then come back here.

In that proof, I used equalities by assuming that the process is reversible. If we drop this assumption, then by the Second Law,

$$\begin{align}
\mathrm{d}S &\geq \frac{\mathrm{d}q}{T} \\
\mathrm{d}q &\leq T\mathrm{d}S \\
dU &\leq T\mathrm{d}S - p\mathrm{d}V + \mathrm{d}w_{\text{add}} \\
dG &\leq dw_{\text{add}} \\
\Delta G &\leq w_{\text{add}}
\end{align}
$$

Equality holds if the process is reversible, or more generally, if both the initial and final states are equilibrium states. Now this seems like an obvious contradiction of the equation in the very first paragraph. The key lies in the definition of the work - whether it is done on the system, or by the system. In my proof, I have been referring to the work done on the system; however, here we are interested in the work done by the system, since that is the work that can be "extracted". The work done by the system must be equal to the negative of the work done on the system: $w_{\text{by,add}} = -w_{\text{add}}$. This leads to the final result $w_{\text{by,add}} \leq \Delta G$.

Levine's Physical Chemistry 6th ed. writes:

  In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but $w_{\text{by,add}}$ is the useful work output. The quantity $-\Delta G$ equals the maximum possible nonexpansion work output $w_{\text{by,add}}$ done by a system in a constant-$T$-and-$p$ process. Hence the term "free energy". (Of course, for a system with P-V work only, $\mathrm{d}w_{\text{by,add}} = 0$ and $\mathrm{d}G = 0$ for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.

If you want a direct answer as to why your explanation isn't right, it's because you can't equate $\Delta H$ with the work done. Work done is given by $w = -\int\! p\,\mathrm{d}V$. You can only equate $\Delta H$ with the heat transferred at constant $p$. Apart from the First Law ($\Delta U = q + w$), heat and work are generally unrelated.


Modeling of photocatalytic and photoelectric devices on a macroscopic device level As a computational materials chemist studying photocatalysis, how close to realistic systems can your modeling get?

I have a sort of high-level question about the theoretical modeling of photocatalysis. Most of the computational work on photocatalysis involves DFT calculations looking at band gaps, free energy profiles, and reaction pathways. I have also seen a handful of papers that use reactive MD to probe how the surrounding water molecules get affected by adsorbed species.

As a computational chemist or materials engineer, and you want to take your work closer to a real system, what is the logical next step? Would it be device modeling, where you take inputs from DFT (such as overpotentials), and then try to calculate the I-V curves, for instance? Are there any other multi-scale modeling techniques or coarse-grain methods that can be used?

I guess this question can be extended more generally to different classes of applications, such as $\ce{CO2}$ sensors, other electrochemical reactions, and so on. I'm not really sure what the next step would be to go beyond DFT in order to model more macroscopic length and time scales.

Look,
I try to give you a broad answer.
If you want band gaps, you need electronic structure that means DFT!

If you do MD that means classical mechanics means no electronic structure no band gaps.

You want to have solvent effects for your, I presume large system, then DFT is costly, you would be left with QM/MM or sometimes called double resolution.

Reactive forcefields such as Reaxff use some valence bond theory to treat electronic structure which is low level of theory and can not be extended to inorganic compound unless it's parametrized.

Finally all these methods are highly approximate, but for engineering purposes everything is approximate.


Oscillator Strength and Absorption Intensity in TDDFT When I run a linear-response time-dependent DFT calculation in my favourite electronic structure program, for each excitation I'm given an oscillator strength, $f_i$, and transition dipole moment $$\boldsymbol{\mu}_i = \int\Psi_0\hat{\mu}\Psi_i \text{d} \bf{r}$$
What formulae relates the transition dipole moments, the oscillator strength and the observed absorption relative intensity?
And more specifically, how can I calculate the relative experimental intensity of TDDFT transitions from the oscillator strengths and transition dipole moments?

I'll address the last part of your question. I think the first part

What formulae relates the transition dipole moments, the oscillator
strength and the observed absorption relative intensity?

of your question could stand on its own.
The oscillator strength is connected to the transition dipole moment vector via
$$
f_{if} = \frac{2}{3}\omega_{if}|\vec \mu_{if}|^2
$$
in atomic units,i.e. $m_e=\hbar =1$, $\hbar \omega_{if}=(E_f-E_i)$. The equation in SI units can be found on Wikipedia. $i,f$ stand for the initial and final electronic state involved in the transition.
For spontaneous emission, you have to multiply $f_{if}$ by an additional factor of $\omega_{if}^2$ since spontaneous emission is proportional to $\omega_{if}^3|\mu_{if}|^2$.
The oscillator strength is already a dimensionless relative quantity and the ratio of different oscillator strengths should correspond to the ratio's in an absorption spectrum measurement. Just normalize everything by the highest value if you are going for a  plot that is comparable with a relative intensities plot where the peaks are also  normalized .
It is important to note that the calculation only describes a purely electronic transition and yields only a single line at the energy difference of the involved adiabatic electronic states. The result of a real measurement is not a single line and includes many effects that are not described by a simple excited electronic states calculation.
But the calculation gives you an estimate where the center of the peaks should be and how intense they should be. A simple TD-DFT calculation does not give you information about the detailed lineshape. The lineshape is typically assumed to have some form and then added by broadening a stick spectrum based on the oscillator strenghts with a spectral lineshape function. The width of the lineshape function is then adjusted to match the experimental measurement manually.  Common lineshape functions are here.


What kind of fire extinguisher for chemistry at home? The are various types of fire extinguishers. They can be based on

water
foam
dry powder
carbon dioxide
wet chemicals
halon
etc.

Suppose one is learning chemistry at home. What type of fire extinguisher one should ideally get? It seems the dry powder option is the cheapest ($\approx 30€$), foam mid-priced ($\approx~60€$) and the one using carbon dioxide the most expensive ($\approx~150€$). I did not find solutions based on water/wet chemicals/halon yet, though I know such exist. If there's no differences in achieved safety, I would probably pick the dry powder one.

You need to investigate the safety and handling of every chemical you use in your experiments individually. The handling and safety precautions can be very different depending on what substances exactly you use.

In terms of fire, the one thing you must know before each experiment is which methods won't work or even fail catastrophically. Trying to extinguish a sodium fire with water is a really, really bad idea, and that is something you must know in advance.

Metal fires are the most common kind of fire where you have to be careful about choosing the right method. Some metal fires can't even be extinguished by carbon dioxide. There are special fire extinguishers for this (class D), other methods might not work.

A bucket of sand is the most useful method for small fires in a lab. It works for almost everything and doesn't cause a big mess. But it is obviously useless if you cause a bigger fire and can't reliably cover it with sand anymore.

Powder extinguishers cause quite a mess, you'll spend a lot of time cleaning up the room afterwards. Obviously this is a lot better than burning down your home, but I wouldn't use them for small fires you could stop with some sand.

Foam and powder are probably the most useful general-purpose extinguishers. Carbon dioxide is a bit trickier to use from what I heard, I probably wouldn't recommend it in your case.

Thinking about a fire extinguisher is a good first step. Another thing I very strongly recommend is that you read up on general safety procedures in the lab. Simple things like not storing large amounts of solvent directly in your fume hood (or workplace) might prevent a small fire from becoming a dangerously large one. Knowing that you need to inactivate any reactive reagents before disposing them might save you from throwing away some oxidizing agent and accidentally setting your waste on fire. There is a lot more to chemical safety than just choosing the right fire extinguisher.


Nomenclature of a tertiary amine What is the name of the amine shown above? Is it N,N dimethyl (1-methyl-1-ethyl propane amine)?

You have correctly identified the given compound as amine. Amines can be named in various ways. According to Subsection P-62.2.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC names (PINs) are formed by using systematic substitutive nomenclature. Primary amines are systematically named by adding the suffix ‘amine’ to the name of the parent hydride; secondary and tertiary amines are named by the same method with further N-substitution (see Subsection P-62.2.2.1).

When there is a choice among acyclic parent hydrides, a principal chain must be chosen. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2).

In this case, the longest chain has five carbon atoms; i.e. the unmodified parent hydride is ‘pentane’.

Addition of the suffix ‘amine’ and the corresponding locant to the name of the parent hydride yields the name ‘pentan-3-amine’.

Finally, the substituent prefix ‘methyl’, the multiplicative prefix ‘tri’, and the corresponding locants are added to express substitution by three methyl groups, which yields the name ‘N,N,3-trimethylpentan-3-amine’ (note that ‘N’ is lower than ‘3’).


Temperature dependence of reaction enthalpy Estimate $\Delta_\mathrm{c}H^\circ(500\mathrm{~K})$ for methane by using the data given on the temperature dependence of heat capacities.

  $$\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2H2O(g)}$$

My working

First, heat capacities as functions of temperature.

$$\begin{align}
C_{p,m}(\ce{CO2}) &= 44.22 + 8.79\times10^{-3} T - \frac{8.62\times10^{5}}{T^2}\\
C_{p,m}(\ce{H2O}) &= 75.29 + 0 + 0 \\
C_{p,m}(\ce{O2}) &= 29.96 + 4.18\times10^{-3}T - \frac{1.67\times10^{5}}{T^2} \\
C_{p,m}(\ce{CH4}) &= 35.31
\end{align}$$

Note: a function of temperature dependence wasn't given in the data table for methane, this value is at $298 \mathrm{~K}$. I used this because I did not know what else to do.

Then, I used Kirchhoff's Law:

$$\Delta H(500\mathrm{~K}) = \Delta H(298 \mathrm{~K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T$$

I computed $$\Delta C_{p,m} = \sum_{\mathrm{products}}\nu C_{p,m}-\sum_{\mathrm{reactants}}\nu C_{p,m}$$ (here, I used the molar heat capacities values listed above, adding the functions after multiplying by the appropriate stoichiometric coefficients) and obtained:

$$\Delta C_{p,m} = 16.15 + 0.43\times10^{-3}T - \frac{5.28\times10^{5}}{T^2}$$

Therefore:

$$\begin{align}
&&\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T &= 2.581 \mathrm{~kJ~mol^{-1}} \\
&&\Delta H^\circ(298\mathrm{~K}) &= -890 \mathrm{~kJ~mol^{-1}} \\
&\Rightarrow & \Delta H^\circ(500\mathrm{~K}) &= -887.4 \mathrm{~kJ~mol^{-1}}
\end{align}$$

However, my result is incorrect. The correct result is $-803.07\mathrm{~kJ~mol^{-1}}$.

Could someone please help me out on this?

As a reference, the textbook is Physical Chemistry 10th ed by Atkins.

All enthalpies are taken to be at $298\mathrm{~K}$ if not specified.

The enthalpy of combustion that is shown in Table 2C.4, $\Delta_\mathrm{c} H^\circ = -890\mathrm{~kJ~mol^{-1}}$, corresponds to $\Delta H^\circ$ for the reaction:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

which is not the reaction you are interested in (since $\ce{H2O}$ is in the liquid state). Therefore, to calculate the enthalpy of the reaction, you have to use $\Delta_\mathrm{f} H^\circ$ values from Tables 2C.4 and 2C.5:

$$\begin{align}
\Delta H^\circ &= 2\Delta_\mathrm{f} H^\circ(\ce{H2O(g)}) + \Delta_\mathrm{f} H^\circ(\ce{CO2(g)}) - \Delta_\mathrm{f} H^\circ(\ce{CH4(g)}) \\
&= -802.34\mathrm{~kJ~mol^{-1}}
\end{align}$$

as I said in the comments.

Now, even though the question said to use the temperature dependence of heat capacities in Table 2C.1 (in the appendix), the solution manual irritatingly takes the values from Problem 2C.7, on page 106. Here, the heat capacity is expressed in the form $C_{p,\mathrm{m}} = \alpha + \beta T + \gamma T^2$:

$$\begin{array}{c|ccc}
\mathrm{Molecule} & \alpha / (\mathrm{J~K^{-1}~mol^{-1}}) & \beta / (\mathrm{mJ~K^{-2}~mol^{-1}}) & \gamma / (\mathrm{μJ~K^{-3}~mol^{-1}}) \\
\hline
\ce{CH4(g)} & 14.16 & 75.5 & -17.99 \\
\ce{CO2(g)} & 26.86 & 6.97 & -0.82 \\
\ce{O2(g)} & 25.72 & 12.98 & -3.862 \\
\ce{H2O(g)} & 30.36 & 9.61 & 1.184
\end{array}$$

Therefore, we have:

$$\begin{align}
\Delta C_{p,\mathrm{m}} &= 2C_{p,\mathrm{m}}(\ce{H2O(g)}) + C_{p,\mathrm{m}}(\ce{CO2(g)}) - 2C_{p,\mathrm{m}}(\ce{O2(g)}) - C_{p,\mathrm{m}}(\ce{CH4(g)}) \\
&= \left[21.98 - \left(75.27 \times 10^{-3}\right)(T/\mathrm{K}) + \left(27.262 \times 10^{-6}\right)(T/\mathrm{K})^2\right] \mathrm{~J~K^{-1}~mol^{-1}}
\end{align}$$

and integrating from $298\mathrm{~K}$ to $500\mathrm{~K}$ (I got lazy and used my graphing calculator):

$$\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T = -731 \mathrm{~J~mol^{-1}} $$

and:

$$\begin{align}
\Delta H^\circ (500~\mathrm{K}) &= \Delta H^\circ (298~\mathrm{K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T \\
&= (-802.34 - 0.731) \mathrm{~kJ~mol^{-1}} \\
&= -803.07 \mathrm{~kJ~mol^{-1}}
\end{align}$$

Now that's irritating, because if we use whatever data we can find from the appendix, we get the answer $-799.76\mathrm{~kJ~mol^{-1}}$, as mentioned in the comments. However, I suppose there's two things that can be learnt:

Do not assume that the combustion reaction gives $\ce{H2O(g)}$, especially if it is given for $T = 298\mathrm{~K}$.
Check the units in the heading of the table.

Your procedure itself is perfectly sound, which is a good thing.


Current with an ionic solution For the last assignment of the school year, my science teacher gave the class questions about chem, physics, etc. The third question had me confused and I'll be thankful for all the help given. The question is:

An ionic solution is created and tested to see how much current can go
through it. Two groups are told to create the following solution
consisting of $\pu{1.0 +/- 0.1 L}$ of water and $\pu{2.0 +/- 0.2 g}$ [of what?] . The battery used for the test is $\ce{8 +/- 1 V}$.
Group A measures a current of $\pu{2.1 +/-0.1 mA}$.  Group
B measures a current of $\pu{3.4 +/- 0.1 mA}$.
Describe 3 possible reasons why Group B measures a current greater
than group A.

No offence to you (certainly to the original author of this question), I call such home-work questions such as garbage in-garbage out type questions. No wonder a student will be confused if they see such open ended questions. The reason is that the author is not telling you or us of 2.0 grams of what?
Even if we assume it was 2 grams of salt, several things can affect the measured current
(1) What are the electrode areas?
(2) What are the electrodes made of?
(3) What is the distance between the electrodes?
Last but not the least, the identity of the salt matters a lot. Did both groups use the same salt? Monovalent, divalent, etc


Why does more heat transfer take place in a reversible process than in a irreversible process? We have defined $$\mathrm dS=\frac{\delta Q_\text{reversible}}{T}$$
and the second law of thermodynamics (Clausius inequality) says
$$\mathrm dS\geq\frac{\delta Q}{T}$$
(Here $\delta Q$ is just the differential of heat regardless of it being reversible or irreversible) and the equality holds for it being a reversible process (as defined).

Now,
$$\mathrm dS\geq\frac{\delta Q}{T}$$
If I put $\mathrm dS$ on the left as $\frac{\delta Q_\text{rev}}{T}$,
I will get $\frac{\delta Q_\text{rev}}{T}\geq\frac{\delta Q}{T}$,
or $\delta Q_\text{rev}\geq\delta Q$.

This implies that in a reversible process, more heat transfer takes place than in an irreversible one. Why is this so?

In the Clausius inequality, the temperature in the denominator is supposed to be the temperature at the boundary of the system where the heat transfer is taking place.  This important point is often omitted from many thermodynamics textbooks.  In a reversible process, the temperature at the boundary is essentially equal to the temperature of the system, because the system temperature is uniform throughout.  But, in an irreversible process, the temperature of the system is not uniform, and the temperature at the boundary can differ substantially from even the average temperature of the system.  So the Clausius inequality does not always require that $\delta Q_{rev}\geq \delta Q$.


Change in enthalpy equation If the formula for change in enthalpy is ΔH=ΔU+pΔV, and ΔU = q - pΔV then shouldn't the formula be simplified to ΔH = q regardless of whether the volume or pressure is constant? Is this only true when pressure and volume is constant?

The general formula is $\Delta H=\Delta U + \Delta (pV)$.

Since $\Delta U=q_v$ (heat at constant volume), then $\Delta H=q_V+V\Delta p$, so no, heat at constant volume does not equal enthalpy, unless p is constant, which just brings you back to the original relation $\Delta H = q_p$.


Cone voltage on mass spectrometers Why is the cone voltage on mass spectrometers positive for positive ions?
I would have thought that the positive ions would be attracted to the
cone by a negative potential. Is the voltage, perhaps, indicative of something else (and not of the degree of attraction between ions)?

Short answer: one should not consider the absolute value of a potential as long as the potential difference drives the ions in the appropriate direction.

Longer answer:
The potential difference is needed to accelerate ions in the appropriate direction of the mass spectrometer.

One should consider that in an atmospheric pressure or moderate vacuum ions need to be driven by the electric potential difference towards the MS high vacuum region by a constant decrease in potential (due to loss of kinetic energy through collisions with the gas) whereas in the high vacuum region, there is no need to keep this evolution true as long as the potential difference remains positive relative to the initial (entrance, near thermal kinetic energy) potential.

Just as an illustration, considering a simplified ESI atmospheric pressure introduction with only an ESI capillary, a cone to rough pumping region and a skimmer to the UHV region, one can have:

ESI Needle: + 4,000 V
Cone: + 100 V
Entrance skimmer: 0 V

Or

ESI Needle: + 4,200 V
Cone: -100 V
Entrance skimmer: -200V

The ions will see the same electric field in both cases, as it is the potential difference which creates the electric field and not the absolute value of the potential. There is just one issue: sometimes it is easier in instrument design to have one region of the instrument operated at ground potential (or close to ground potential), and it would be this part of the design which will decide on the absolute values of the potentials that will be applied.


Is lithium hydride a salt? Hydrogen is generally considered a non-metal, and a non-metal and a metal often makes a salt. However, lithium hydride seems like a special case. The wikipedia article does not clearly state if the compound is a salt or not, only describing it as "salt-like".

Structurally, it makes sense to classify it as a salt, due to the network structure, but are there any clear definition to settle where the compound belongs?

The Wikipedia article Salt (chemistry) starts off with "In chemistry, a salt is an ionic compound that results from the neutralization reaction of an acid and a base." This seems like a pretty good Chemistry 101 definition. So, in essence, this means having an anion and a cation in the salt, which further implies ionic bonding. This seems like a pretty reasonable place to start.

I'd think that the definition could be tweaked to say "a salt is a solid ionic compound..." That would indicate that one "molecule" isn't a salt. So one molecule of NaCl in the gas phase is a molecule, not a salt. I also have trouble thinking of molten NaCl as a salt. So I'd describe it as a "molten salt."

  Kind of like way I think German works. I'd expect that German would have some compound word for "molten salt" which better denotes that there is one new idea. So in English if a "salt" is a solid that has a 3D crystal lattice how can a salt be "molten"?

In general, there are a lot more organic salts than inorganic, so dissolving in water isn't a good restriction.

These last paragraphs point out the sorts of problem in trying to make such a definition absolutely unambiguous. You get a one sentence definition, then two pages of qualifications which render the definition useless.

I can't think of a solvent in which you could dissolve lithium hydride and I don't see how you could get lithium hydride as a neutralization reaction, so I wouldn't think of it as a salt. I don't know a better word than "salt-like" to describe such a crystal structure of anions and cations.


Why is gold sol used for intra-muscular injection? In my textbook it is written that gold sol is used for intramuscular injection. But, actually what is its use? Why would people use such a precious element for injection. Can other colloids be used like silver sol?

Note:I posted this question to get a deeper understanding of what is given in my text about the use of gold sol in intramuscular injection. Googling it didn't help.

Will this help?

Toxicity and cellular uptake of gold nanoparticles: what we have learned so far?

Gold has a long tradition as a disease-modifying antirheumatic drug in the treatment of rheumatoid arthritis and it seems to relieve pain and decreases swelling of the joints.


Why only ketone forms enolate in Claisen condensation between ester and ketone? Claisen condensation involves nucleophilic substitution in esters by an enolate (formed by removing an alpha-H from an ester or a ketone) giving a beta-keto ester or a diketone.

My textbook says that for the reaction between ketone and ester, the ketone always forms the enolate by losing it's alpha-H . But why can't the ester form the enolate here? What will happen if the ester has alpha-Hs and the ketone dosen't?

Attempts:

Does it the O on the alcohol side of the enolate ester help in delocalizing the negative charge via an inductive effect making it less effecive as a nucleophile?

(I can't find anything on other sites I've seen:

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch21/ch21-3-3.html

https://en.wikipedia.org/wiki/Claisen_condensation

http://www.chtf.stuba.sk/~szolcsanyi/education/files/Organicka%20chemia%20II/Prednaska%205_Aldehydy%20a%20ketony_Reakcie%20II/Doplnkove%20studijne%20materialy/Carbonyl%20compounds_full_Blackmond.pdf

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Claisen_Condensation )

The pKa of a ketone alpha hydrogen is of the order of 20, the pKa of an ester alpha hydrogen is of the order of 25 source. Bear in mind that this is a log scale, meaning the proportion of ketone enolate to ester enolate is 100000:1.

In answer to the comment by @orthocresol: the ketone does self-condense, but this is reversible. The condensation with the ester is irreversible due to formation of the anion of the product B-diketone.


Composition of a typical printed circuit board (PCB)? For the purposes of x-ray absorption, I'm trying to discover the average chemical composition of a typical PCB.

My research so far has indicated that the most common type of PCB is designated FR-4.  It is composed of fiberglass and epoxy resin.  The resin often contain Bromine to add fire resistance.  Let's ignore any printed copper for the mean time.  So we are looking at H, C, O, Si, Br, and probably some other bits and pieces.  But what are the percentage of these elements?

This article should be useful for you...

  Material Recovery and Characterization of PCB from Electronic Waste
  Jessica Hanafi, Eric Jobiliong, Agustina Christiani, Dhamma C. Soenarta, Juwan Kurniawan, Januar Irawan, (doi:10.1016/j.sbspro.2012.09.1194)
  Abstract
  Printed Circuit Boards are commonly exist in computers, laptops and mobile phones. Since it contains of a variety of metals and semiconductors, it is difficult to recover the materials in PCB. This paper aims to characterize the composition of these materials and a method to properly recycle and recover the materials in them. This paper utilized simple mechanical and chemical procedures to recover PCBs. Collected PCBs were disassembled, pulverized and separated by using density and magnetic separation method. By using the chemical recycling method, 98.82% purity of CuSO4 hydrate and Al2(SO4)3 hydrate were recovered from the PCBs. It is found that the milling methods, size of the sample and the density of the separating solution determine the effectiveness of the materials recovered.


Preference for tin or iron in the reduction of nitrobenzene My NCERT book(Class 12) said the following for reduction of nitrobenzene using $\ce{Sn/HCl}$ and $\ce{Fe/HCl}$ :

Now, I have been following this mechanism for this particular reduction, and it seems to me that the metal is getting oxidized to +2 state wherever it is shown that 2 electrons are being transferred. If that is the case, then:

•Why is $\ce{Fe + HCl}$ being preferred?

•Because I looked it up on Wikipedia and it says that $\ce{SnCl2}$ also gets hydrolysed to liberate $\ce{HCl}$ as follows:
$\ce{SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)}$

Or have I interpreted the mechanism incorrectly?

The book's reasoning is correct. Fe/HCl gives $\ce{FeCl2}$ on reaction and it gets hydrolyzed by steam vapors to produces more hydrochloric acid and hydrogen to push the reaction forward thus making the reaction self-sustaining to produce more aniline from nitrobenzene.

$$\ce{Fe + 2HCl ->[160°C,CH3OH] FeCl2 + H2}$$

$$\ce{3FeCl2 + 4H2O(steam) ->[Hydrolysis] Fe3O4 + 6HCl + H2 }$$

whereas $\ce{SnCl2}$ gives $\ce{Sn(OH)Cl}$ and 1 mole of hydrogen chloride which is insufficient and it gets exhausted after sometime.

The other thing is cost factor as @Waylander said. Iron scrap is way cheaper than tin.

Both Sn/HCl and Fe/HCl reduction(Bechamp reduction) is replaced by catalytic hydrogenation due to the latter being more efficient and more productive in industial production. Fe/HCl reduction is now an academic interest and is studied as a route to produce iron oxide based pigments.


How would I synthesize ethyl 2‐oxocyclohexane‐1‐carboxylate from cyclohexanone? I thought the 1st step would be deprotonation of a H atom to form the enolate. Then an aldol reaction, but the product I end up with is the alcohol, not the ketone.

As you will see if you look at the reference mentioned by @Mathew-Mahindaratne,   the standard conditions for this transformation is NaH/KH in THF at reflux with diethyl carbonate (EtOCOOEt) with an acetic acid workup.

The Organic Syntheses procedure here uses dimethyl carbonate, but will be equally applicable for diethyl carbonate.


Function of 'organic' vs regular pipe cleaner Recently, I needed to buy some pipe cleaner, as sometimes even with care hair and other detritus gets lodged so deep down the shower drain that it is not possible to get it out. I always try to avoid/reduce use of bleach and other hard chemicals or products that incorporate them, as in the end it goes down the drain is not good for the environment.

However, unfortunately the 'bio' product of the same brand of cleaner I had used before (didn't know it existed before) was not very satisfactory so far. Is there maybe someone here who could offer details of the chemical function of these types of cleaners, and either knows a method to improve the use of the bio/organic one or recommend another product or compound that is easy on the environment but cleans drains well? For the proper use, the instructions say that for the 'organic' one (yes sorry for that chemically improper use but ...) you should add hot or warm water to start the reaction, though my worry is also that this added water will just flush the product down the drain depending on the size of the drain.

I know some of these ingredients and how they likely function, but not all of them and I am not sure what the combined effects might be. To my understanding the 'non-organic' one relies on the reactive power of chlor(ide/ine) and resulting reaction products, whereas with the organic one I am not really sure.

Specifically, it is about these two products:
rorax and rorax bio (at least on the German market)
The ingredients of rorax (said to be chlorine based in marketing) are given as water, sodium hydroxide, sodium hypochlorite, myristamin oxide, sodium silicate, fatty acids, C6-12-, natrium salts as per http://detvo.werner-mertz.de/?lang=de&country=DE&prod_id=109655

The ingredients of rorax bio are given as water, sodium chloride, urea, sodim laureth sulfate, sodium thioglycolate, sodium hydroxide, Amide polyglycol ether, parfum, colorant
according to http://detvo.werner-mertz.de/?lang=de&country=DE&prod_id=114331

Congratulations on being environmentally sensitive!

The problem with sink drains is the device that lifts and lowers the stopper. It is a rod that protrudes into the drainpipe above the U-trap; it catches hair and provides a bacterial nest and eventually slows water drainage.

Older drain cleaners were based on alkali to dissolve hair; some used aluminum to foam and break up the hairy mass. Bleach plus alkali works well too, and kills bacteria. The newer, less aggressive formulas can use thioglycolate to dissolve hair (also used on human legs!).

In the overall scheme of life, with all the other bodily wastes going down the drain, a little drain cleaner may not be a big problem, and where it is important, the government usually has a statement or a regulation.

Even without regulations tho, it would be kind to the environment to limit our use of toxic chemicals. Using a lot of ineffective product may be worse than using a little of a more aggressive product. I would suggest two approaches:

1) Pour just a little of the effective product down the drain, but only when it drains slowly (you can measure the drain speed by filling the bowl to a certain line, then timing how long it takes to empty). This little bit (10-20 mL?) will saturate the clog because it is higher than the bottom of the U-trap. Give it some time (many minutes, even hours); see if it has opened up. If not, repeat. This may go against product recommendations to use a lot of product, but I suspect that the bottom of the U-trap is not blocked (any water flow usually cleans that out). This will reduce your environmental impact by using your brain.

2) Use a mechanical method. a) A stiff copper wire with a tiny hook at the end is useful for digging out hair, and often opens up the drain with just a lot of hot water. b) Cut a hollow rubber ball in half to make a tiny plunger. Place it over the drain (with the stopper lifted a little) and press it up and down to push the clog all the way down the drain.


Haloform reaction with limited bromine We were asked the reaction between 1 mole acetone and 1 mole $\ce{Br2}$ in basic medium. Since only a limited amount of bromine was given I got the final answer as 1-bromopropan-2-one (mono bromo acetone). However, in the given solution the entire haloform reaction was shown with $\ce{CHBr3}$ and acetate ion as the final product.

I am not sure how a single mole of $\ce{Br2}$ is sufficient for this(at least 3 moles are needed, I believe).

Let's take a look at the first product you propose: 1-bromopropan2-one.

The mechanism includes deprotonation by hydroxide to form an enolate. The resulting enolate is better stabilized with more electronegative atoms (Br in this case) and will form more readily than another molecule of acetone turning into an enolate ion. Thus, as the reaction proceeds, the product distribution will be determined by the most stable enolate anion. This will be the tribromoacetyl intermediate enolate ion and will drive the reaction forward. So a whole mol of bromine will be converted completely to the tribromoacetyl intermediate with a 1/3 mol of acetone. Notice how only a 1/3 will react to give the monosubtituted intermediate, than 1/3 mol of bromine will react to give disubstituted product, and lastly the last 1/3 mol of bromine will react to form the trisubstitued product. The last step is displacement by hydroxide via the nucleophilic acyl substitution mechanism (not SN2!) to form acetate and release bromoform. This reaction isn't really great if bromine isn't in excess.


How can a phase be thermodynamically stable at room temperature if it only appears at high temperature? How can a phase that appears at high temperature be thermodynamically stable at room temperature? For example, $\alpha$-alumina is a phase that is formed at temperatures above $1000^\circ C$. How is the same phase stable at room temperature?

$\alpha$-alumina is thermodynamically stable at room temperature, and also at high temperature (around $1,200-1,400 \space ^{\circ}C$) - however, alpha alumina is not what is typically formed when alumina is made from the various aluminum ores (ref). Typically, one of the several other crystal structures will form, with different particle sizes and amounts of water (hydration). These and all other phases that are found along the way to $\alpha$-alumina are called transition aluminas.

This link explains what happens upon heating. In general, first the crystal is dehydrated, which causes it to form a different, dehydrated phase. Then, as the temperature continues to rise, the alumina crystal structure rearranges into more compact structures, until eventually, near the sintering point, the atoms have sufficient kinetic energy that they can "find" the most thermodynamically stable configuration - $\alpha$-alumina.

In other words, the transition alumina phases are kinetically trapped in structures that are not the most thermodynamically stable, but heating allows them to get out of that trap and into the most favorable state.


Which alpha Hydrogens are more acidic? Why is the alpha hydrogen of a acetoacetic ester less acidic than the alpha hydrogen of 1,3 diketone ? I was under the impression that acetoacetic ester has more resonance structures so it more acidic. Please explain.

I was under the impression that acetoacetic ester has more resonance structures so it more acidic.

That's actually wrong. Deprotonated acetoacetic ester has structures where electrons are donated from ester oxygen into conjugated system. This, while stabilizing system as a whole, lessens effect of electron donation from deprotonated carbon. Think about it this way: carbonyl oxygen wants electrons. While in acetylacetone electrons may be harvested from deprotonated carbon only and in acetylacetic ester they may be harvested from both deprotonated carbon and ester oxygen, stabilization effect in the first case is higher.

It is well known, that stabilization effect on acidity of nearby $CH$ grows in row
$COO^- < -CO(NH_2) < -COOR < -COR < -CHO < -NO_2$

The more electron deficient the conjugated system is, the more stabilized carbanion becomes, thus the less resonance structures with transfer of the charge onto carbonyl oxygen the group has, the more it stabilize carbanion. Thus, $\alpha-CH$-acidity in carbonyl compounds correlates with electrophilic activity.

The reason is that negative charge from deprotonated carbon must be delocalized as much as possible, which is especially easy with strongly electron-accepting groups. In case the compound already has an electron pair conjugated with carbonyl group,  this is more problematic.


Finding membrane potential equation using the Nernst equation? I am very new to electrochemistry and ion selective electrodes. The membrane potential of an ion selective electrode is given by $$E=\frac{RT}{nF}\ln{\frac{c_2}{c_1}}$$
Where R is universal gas constant, T temperature in kelvin, n the number of electrons transfered, F Faraday of electricity, $c_2$ the concentration of internal standard solution and $c_1$ is the concentration of the ion to be calculated.

But according to the Nernst equation
$$E= E^\circ -\frac{RT}{nF}\ln k$$
Can someone explain how to come up with the membrane potential equation using the Nernst equation?

The Nernst Equation is $E = E^\circ - \frac{RT}{nF} \ln{Q_c}$. From that you can derive the approximation you gave. $Q_c$ is the product of the activities of the products over those of the reactants. This can be, for dilute solutions where activity approximates concentration, the typical [B]/[A] for the reaction A→B. So, the only thing you need to "explain" is why E° would be zero. Review the definition of E°.  hint:What is ln(1/1)? See the discussion on concentration cells in:  http://www.chem1.com/acad/webtext/elchem/ec4.html


Why does boron add to the less substituted carbon in the hydroboration of an alkene? In the hydroboration of a typical alkene, the boron fragment ends up attached to the less substituted carbon:

What are the electronic or steric factors that lead to this regioselectivity?

For the hydroboration of isobutene

there is a steric aspect:

And though sterics may play some part, it doesn't explain near 100% regioselectivity. More important in determining regiochemistry, I would say, are the electronics of the transition state:

Boron is electrophilic, and as such will draw the electron density of the alkene π-bond towards itself. This C–B bond formation occurs earlier than the C–H bond formation, which in turn imparts a partial positive charge to the other carbon of the double bond in the transition state of the reaction.

This somewhat resembles a carbocation, which you might remember are stabilized by additional substituents via hyperconjugation. Thus, hydroboration results in an anti-Markovnikov product for mostly the same reason that acid catalyzed hydration results in a Markovnikov product — stabilization of positive charge.


Lewis structure for NCO+ I proposed the following lewis structure for $\rm NCO^\oplus$

The problem is that this structure is linear, while my book tells that the structure is not linear. What will be correct canonical structure?

Also is it correct to call the following as resonance structures?

As always, writing a Lewis structure seems to be a trivial task, but sometimes it just isn't. The smaller the molecule becomes, more often than not the bonding situation becomes more complicated and complex. This can often be attributed to the high symmetry of such molecules. Prime examples are $\ce{O2}$, $\ce{CO}$, $\ce{CO2}$, $\ce{N_xO_y}$, etc..

I found no evidence anywhere that $\mathbf{\ce{NCO+}}$ is bent. I have tried calculations in the singlet and triplet states and all of them lead back to the molecule being linear. It would be nice to know which textbook you are using for general reference.

To be perfectly candid here, it is not even a trivial task to set up the right calculations. At the DF-BP86/def2-SVP level of theory, the triplet state is 151.8 kJ/mol lower in energy than the singlet state. A broken symmetry singlet (a biradicaloid electronic state) is still 111.3 kJ/mol lower in energy.
On the CASSCF(6,6)/def2-SVP level of theory the triplet wave function is 143.3 kJ/mol lower in energy than the singlet wave function.

Lewis structures were not meant to accommodate complicated bonding situations. Instead they are meant to be easily understood. Even the concept of resonance can be quite tricky in certain situations. Without knowing that the ground state of dioxygen is a triplet state, we would interpret its common Lewis structure wrong.

The same shall apply here. We need to treat the concept of Lewis structures within its limitations. Based on the triplet DF-U-BP86/def2-SVP calculation the best representation is probably a bit surprising: The positive charge would be located at the most electronegative atom, while the spin density is located at the nitrogen (1). The broken symmetry solution gives us instead two double bonds with the positive charge at the nitrogen (2).
In all cases we shall not forget, that the most electropositive element is carbon in this molecule, so the actual charge is most likely located there (3). Since the π-orbitals are stretched over the whole molecule, the spin density will not entirely be located at nitrogen alone (4). A meaningful interpretation in terms of resonance structures (NRT of NBO) is, however, almost impossible. I tried it based on the closed shell singlet DF-BP86/def2-SVP calculation and there are three contributing main structures (5).

For what it's worth, the structures you have proposed are valid resonance structures. They are not all of them, but they are a good start. As you can see, any of them capture only a tiny glimpse in the complicated nature of this molecule.


Assigning the IUPAC name of this compound Going by the rules that were given to me by my professor:

I first decide the seniormost group and its parent chain. In this case, cyclohexene carbonitrile.
Then, I decide direction to traverse the ring to assign locants by the first point of difference. Accordingly, I move towards bromine.

The final name I obtain is thus "2-bromo 3-hydroxy cyclohex-4-enecarbonitrile". The answer given in my workbook however is "6-bromo 5-hydroxy cyclohex-3-enecarbonitrile". Apparently, they decided to traverse the ring in reverse direction.

Obviously, this must be some special case. Please explain the exact rule involved here. Thanks!

This is because alkene has higher preference than alkyl halides. So when counting clockwise position of ene is less than that when counting anticlockwise. Hence the answer is as stated.

Check out Rule 6.2 of basic principles on Wikipedia.


Balancing of redox reaction I tried to balance the following redox reaction

$$\ce{S^0 + OH- -> S^2- + S2O3^2-}$$

by ion electron method, but I am unable to identify the oxidation and reduction half reactions. Please provide the correct method.

Genarally, sulfur doesn't disproportionate like that: in fact, it typically behaves very similarly to chlorine (see Balancing the redox disproportionation of chlorine by half-reactions), which upon heating with alkali disproportionates to chloride $\ce{Cl-}$ and chlorate $\ce{ClO3-}$:

$$\ce{3$\overset{0}{\ce{Cl2}}$ + 6 OH- ->[$\Delta$] 5 $\overset{\mathrm{-I}}{\ce{Cl}}${}^- + $\overset{\mathrm{+V}}{\ce{Cl}}$O3- + 3 H2O}$$

Without heating instead of chlorate $\ce{ClO3-}$ hypochlorite $\ce{ClO-}$ is formed, so this is an irrelevant process in this case. Back to sulfur: when reacting with alkali, sulphide $\ce{S^2-}$ and sulfite $\ce{SO3^2-}$ are formed as the primary products according to the scheme:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+IV}}{\ce{S}}$O3^2-} \tag{1}$$

Half-reactions and net equation for (1):

\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red1}\\
\ce{S + 6OH- &-> SO3^2- + 4e- + 3H2O} \tag{ox1}\\
\hline
\ce{3S + 6 OH- &-> 2 S^2- + SO3^2- + 3 H2O} \tag{redox1}
\end{align}

Note that (red1) has been multiplied by 2 to even the number the electrons in both (red1) and (ox1). The net equation (redox1) is obtained as a sum of both half-reactions.

Now, lets assume that the reaction you've written takes place, where where thiosulfate is formed instead of sulfite. In fact, it might as well happen when boiling solutions of sulfites with sulfur for a long time:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+II}}{\ce{S}}${}_2O3^2-} \tag{2}$$

Half-reactions and net equation for (2):

\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red2}\\
\ce{2S + 6OH- &-> S2O3^2- + 4e- + 3H2O} \tag{ox2}\\
\hline
\ce{4S + 6 OH- &-> 2 S^2- + S2O3^2- + 3 H2O} \tag{redox2}
\end{align}

Reaction (redox2) would be the net reaction for the process you asked about, even though I'd say it's more likely a hypothetical one. Keep in mind that for as long as the reaction takes place in aqueous solution and the redox process involves acid-base transformations (like formation of oxoanions in this case), you need to take water into account.


Stabilization of hydrogen peroxide It was given in a chemistry book that things like orthophosporic acid, acetanilide and sodium stannate stabilize hydrogen peroxide, while carbon, $\ce{MnO2}$ and $\ce{Pt}$ help decomposing it.
Why is this so?

Part of the explanation is that orthophosphate forms complexes -or even precipitate, traces of heavy metal ions present in the solutions. Many of these ions, like manganese and platin, will act as catalysts for the degradation of hydrogenperoxide. The phosphonic acid HEDP (Etidronic acid) is another example of a stabilizer.


Synthesis of the tetrachloroaluminate(III) anion Looking for random chemicals, I found a reference on the anion and thought that it could be synthesised from common laboratory reagents:

$$
\begin{align}
\ce{2 Al + 2 NaOH + 2 H2O &-> 2 NaAlO2 + 3 H2}\\
\ce{NaAlO2 + 4 HCl &-> NaAlCl4 + 2 H2O}
\end{align}
$$

The first reaction obviously occurs at high yield at STP. Does the second reaction occur at STP or not? If not, how can the $\ce{AlCl4^-}$ ion be synthesised?

Note that the first reaction would require substantial amount of heat to assure the designated product $\ce{NaAlO2}$ is formed (nowhere near STP):

$$\ce{2 Al(s) + 2 NaOH(l) + 2 H2O(g) ->[\pu{\approx 450 °C}] 2 NaAlO2(s) + 3 H2(g)}$$

In aqueous solution the product would be tetrahydroxoaluminate $\ce{[Al(OH)4]-}:$

$$\ce{2 Al(s) + 2 NaOH(aq, conc) + 6 H2O (l) ->[Δ] 2 Na[Al(OH)4](aq) + 3 H2(g)}$$

Synthetic procedure from "common laboratory reagents" is described in Brauer's Handbook of Preparative Inorganic Chemistry [1, p. 816]:

  Sodium Tetrachloroaluminate

  $\ce{NaAlCl4}$

  $$\ce{\underset{58.4}{NaCl} + \underset{133.4}{AlCl3} = \underset{191.8}{NaAlCl4}}$$

  The reaction is carried out in a Pyrex vessel (Fig. 245), which should be as compact as possible.
  The stoichiometric amounts of C.P. $\ce{NaCl}$ and freshly sublimed $\ce{AlCl3}$ are introduced into the pear-shaped part of the reaction vessel.

  Fig. 245. Preparation of sodium tetrachloroaluminate.

  The filling is done under nitrogen and the vessel should be very clean and dry.
  A melting point capillary is then affixed underneath the ground joint; the apparatus is evacuated to a high vacuum and torch-sealed.
  It is then immersed as deeply as possible in an oil bath at 200–240 °C.
  The $\ce{AlCl3},$ which tends to sublime onto the cool parts of the wall, is driven back by passing a luminous flame over the condensate spots.
  When the flask contents have become a nearly clear melt, they are poured hot into the side flask of the vessel.
  The product solidifies on cooling.

References

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1963; Vol. 1.


What is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid? Write an equation for the reaction you might expect to take place if trichloroethanoic acid, $\ce{Cl3CCO2H}$, $\mathrm{p}K_\mathrm{a} = 0.65$, were added to dimethylpropanoic acid, $\ce{(CH3)3CCO2H}$, $\mathrm{p}K_\mathrm{a} = 5.05$. Explain your answer in terms of the Brønsted–Lowry theory.

I know that $\ce{Cl3CCO2H}$ is the stronger acid as it has a lower $\mathrm{p}K_\mathrm{a}$, however, how can dimethylpropanoic acid be a base?

TL;DR
Contrary to what the answers/comments have suggested, I would say that no reaction happens here. The acid–base reaction between the two given species is thermodynamically unfeasible, with an equilibrium constant $K \sim 10^{-9}$.

If you want an acid-base reaction to occur between A and B, it's no use comparing the $\mathrm pK_\mathrm a$ values of A and B. That tells you how acidic A and B are, but you're not interested in both of them acting as acids. You're interested in one of them being an acid, and the other being a base. So the relevant quantities are the $\mathrm pK_\mathrm a$ of A and the $\mathrm pK_\mathrm b$ of B!
In this context, let's keep trichloroacetic acid $\ce{Cl3CCOOH}$ as the acid; we already know that it has $\mathrm pK_\mathrm a = 0.65$. The question tells us that for the "base" $\ce{Me3CCO2H}$, its $\mathrm pK_\mathrm a$ is $5.05$. But we're not interested in that value; we're more interested in its $\mathrm pK_\mathrm b$.
To find that out, we need to look for data beyond the question text itself. The veritable Evans tables states that the $\mathrm pK_\mathrm a$ of $\ce{PhC(OH)+OH}$ is $-7.8$. That species is the conjugate acid of $\ce{PhCOOH}$, so the $\mathrm pK_\mathrm b$ of $\ce{PhCOOH}$ is $14 - (-7.8) = 21.8$. Recall that a larger $\mathrm pK_\mathrm b$ means that a species is a weaker base. This gigantic number should already tip you off to the fact that $\ce{PhCOOH}$ is a terrible base. The species in the question, $\ce{Me3CCOOH}$, will likely have a value that's broadly similar.
How do we use these values to find whether an acid-base reaction is plausible or not? Returning to the reaction
$$\ce{AH + B <=> A- + BH+,}$$
we have the equilibrium constant
$$\begin{align}
K &= \frac{[\ce{A-}][\ce{BH+}]}{[\ce{AH}][\ce{B}]} \\
&= \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]} \cdot \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]} \cdot \frac{1}{[\ce{H3O+}][\ce{OH-}]} \\
&= \frac{K_\mathrm a(\ce{HA}) \cdot K_\mathrm b(\ce{B})}{K_\mathrm w}
\end{align}$$
For our molecules, we know that the relevant quantities are $\mathrm pK_\mathrm a(\ce{HA}) = 0.65$ and $\mathrm pK_\mathrm b(\ce{B}) \approx 21.8$. Putting in $K_\mathrm a = 10^{-0.65}$, $K_\mathrm b = 10^{-21.8}$, and $K_\mathrm w = 10^{-14}$, we get an equilibrium constant of
$$K = \frac{10^{-0.65} \cdot 10^{-21.8}}{10^{-14}} = 3.548 \times 10^{-9}$$
which is so small that it makes very little sense to claim that this reaction "happens".


H2O2 + NaCl + H(+) reaction? So I found these reduction potentials online:

$$\begin{alignat}{2}
\ce{H2O2 + 2H+ + 2e- \;&<=> 2H2O}\qquad &&E=1.78\ \mathrm V\\
\ce{2Cl- \;&<=> Cl2 + 2e-}\qquad &&E=-1.36\ \mathrm V
\end{alignat}$$

If the net reduction potential is positive will chlorine gas be spontaneously formed when salt and hydrogen peroxide are mixed in a acidic solution?

Not necessarily.  If you get the pH right and the stoichiometry absolutetly 1:1 then you might form $\ce{Cl2}$.  However your reaction will compete with other reaction and you will more likely oxidize the chlorine into chlorate ions ($\ce{ClO-}$, $\ce{ ClO2-}$, $\ce{ClO3-}$, or $\ce{ClO4-}$).


Structure diagrams indicate plausible structures? Structure diagrams have a hexagonal, "chicken wire" shape, which leads to more elegant representations of some molecules than others.

For example, the following diagram must deviate from the usual convention such as the horizontal connection above the nitrogen.

The following hydrocarbon would have the same problem.

  H H
  | |
H-C-C-H
  | |
H-C-C-H
  | |
  H H

Does this indicate anything about these molecules, for example they are rare or unstable, or is it just a weakness of this type of diagram?

Structural diagrammes illustrate connectivity in molecules but they are often very limited in their ability to represent the real 3D shape of complex structures.

A drawing on a 2D surface can never properly represent a 3D structure. This is as true for maps as it is for chemical structures (which is why all map projections contain distortion, e.g. the Mercator projection seriously distorts areas).

Chemical structure drawings also omit lots of detail from molecules like the relative size of atoms and the relative lengths of bonds. What they focus on is the connectivity of the molecule, showing us what connects to what. For many structures this gives a poor representation of the 3D shape, especially (as in the example in the question) where there are multiple ring systems. The flat diagram itself doesn't tell you much about the stability of the molecule. That is one reason why chemists need modelling kits that get the 3D geometry right. Even then, the kits are imperfect. Structural kits sometimes just do the connectivity and bonding (in effect a 3D version of the 2D connectivity picture). But, sometimes, you need to understand the spatial interactions that take into account the size of atoms. There are 3D space filling kits for that sort of modelling.

Having said all this, most modelling is done via computers not by creating hand-built models which are relatively expensive and limited. Computers can allow for extra calculations to represent the interactions of bonds and spatial factors to give plausible 3D geometries and can sometimes do detailed calculations of energy levels. Many of the software packages (e.g. Chemdoodle) that allow you to draw 2D structures will automatically attempt to generate plausible 3D structures based on those to give you better intuition about the geometry of the molecule.


What trends affect hydrophobicity in amino acids? I was wondering if the length of the carbon backbone affects hydrophobicity of amino acids (i.e. longer carbon backbone of side group = less polar = more hydrophobic).

Your assumption that hydrophobicity of amino acids increases with length of the carbon chain is correct. Also, a branched chain can have a lower hydrophobicity than a linear chain with the same number of carbons, because it has a decreased exposed surface area.

When polar acidic/basic functional groups are introduced, the side chain becomes more hydrophilic. For example, phenylalanine is classified as non-polar because of the hydrophobic benzyl side chain. Attaching a hydroxy group in para-position in the phenyl ring leads to tyrosine, which is more polar. This is due to its acidic phenol group which is deprotonated at physiological pH. Note that in both cases, the number and arrangement of carbon atoms in the side chain is the same.


Decomposition of Metal Nitrates An entrance JEE Main & Advanced AIEEE 2003 exam question:

Which one of the following nitrates will leave behind a metal on strong heating?
A. Copper nitrate
B. Manganese nitrate
C. Silver nitrate
D. Ferric nitrate

I looked up the answer and it was given to be silver nitrate, and I took it up as something to be remembered. However, when I checked online for the decomposition of other options such as ferric nitrate I found many research papers discussing the same which is not something one asks students who have just studied chemistry up to 12th grade.
Also, is there a way to decide which one will, because if it says strong heating, shouldn't most metals leave their salts and exist as free metals if we go on increasing temperature?

Most metal nitrates decompose on heating to give metal oxides. Decomposition of silver nitrate is different as it yields elemental silver.
$$\ce{2AgNO3 -> 2Ag + O2 + 2NO2}$$
Usually the metals which lie very low in the activity series decompose to give their elemental form.
I don't think it's possible to extract metals from just any nitrate by strong heating. They will yield their respective oxide which won't give the element rather they'll just fuse (melt) just like stable metal oxides like $\ce{Al2O3}$ does.


Concentrating acetic acid from vinegar Vinegar seems to kill poison ivy or at least severely stunt their leaves. I am trying to increase the concentration of acetic acid and I put a bottle of vinegar in the freezer. I have looked at it and it has turned cloudy. I'm thinking if I poured this through a coffee filter I should be filtering out solid acetic acid. I might do this in the freezer because it might take some time. Does anyone think this will work? I might be getting water ice instead. Either way I would be separating the two. I could repeat this step and concentrate it further. Would I not really be filtering out anything? On another blog someone suggested buying aluminum acetate which has more acetic in it instead, but this might not be as effective.

If I understand correctly, what you described can be seen as a form of fractional freezing.

This is done in order to make alcoholic beverages more alcoholic (Eisbock is an example): the beverage is left to partially freeze and the ice is removed.
Since ethanol has a lower melting point (-114 °C) than water (0 °C), this effectively removes more water than ethanol.

But in your case, since acetic acid has a higher melting point (16 °C) than water, the solids will have a higher concentration of acetic acid and you would want to keep that instead of the liquid.


Francium has isotopes so how can its atomic mass be a whole number? Why do francium, radon, radium, and actinium have whole numbers for average atomic mass even though they have isotopes? But then elements like beryllium, fluorine, and sodium are listed as having no isotopes but yet they have a decimal for their average atomic mass?

For elements with no stable isotope (i.e. Francium, Radium, and Actinium), the atomic mass is chosen to be that of the longest lived isotope.

More generally, the masses for stable elements are reflective of the natural abundance of each isotope in a sample of the element. Sodium has more than one isotope, so the statement is not really true, though only one of them is stable and included in the averaging. In this case, it's important to remember that the mass of protons and neutrons are only approximately 1 amu, so even if there is only one isotope there's no requirement that it have an integer atomic mass.


How many atoms are represented in this equation? $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$

The answer says 26, but shouldn't it be 13? Because actually, the number of atoms involved in the reaction are 13, and once the reaction is completed, again, number of atoms is 13.
Please explain.

The referenced atoms on both sides of the arrow are the exact same atoms? I like your thinking!

But, chemical equations do not refer to individual atoms.

The formula $\ce{Pb(NO3)2}$ can be read as "A molecule of this substance consists of six atoms of O, two atoms of N, ...". That is correct.

But the equation $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$, strictly speaking, does not talk about single molecules. The numbers (including the 1's that are nor written) are called "stoichiometric coefficients". The equation means: "Any number of $\ce{Pb(NO3)2}$ molecules react with the double amount of $\ce{KI}$ molecules to yield ...".

Wikipedia supports my view:

  In lay terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature) of any given component is the number of molecules that participate in the reaction as written.

  ...

  In more technically precise terms, ...

... the stoichiometric coefficient is the number of molecules divided by the progress variable or extent of reaction.

  ...

  The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules.


How do I show that a transition is magnetic dipole allowed with group theory/symmetry? I am taking an introductory course in symmetry and bonding and have so far learnt about SALCs and various point groups etc... one of my problems asks me to show a $\mathrm{^{2}T_{2g}\rightarrow\ ^{2}E_g}$ is allowed in a $\mathrm d^1$ metal. I am not sure how my knowledge of symmetry can help me answer this, all my searching flags up difficult quantum mechanics/group theory I don't understand! I think a relevant example I could use is $\ce{[Ti(H2O)6]^{3+}}$ as this is an octahedral, $\mathrm d^1$ system.

For a state-to-state transition coupled by an operator to be allowed, the direct product of the irreducible representations (irreps) of all three components must contain the completely symmetric irrep of the point group you're working in. That is, we are determining whether the following integral is forced to be zero by symmetry:

$$
\int \Psi_{i}(\tau) \times \hat{\mathbf{A}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \overset{?}{=} 0
$$

where $\Psi_{i}$ and $\Psi_{f}$ are the initial and final states, $\hat{A}$ is the operator (electric dipole, magnetic dipole, one-electron spin-orbit, there are many more), and $\tau$ is the integration coordinate, usually over electron position. Why bother with this if the integral can be evaluated exactly?

It's no more complicated than this pencil-and-paper calculation to gain qualitative understanding about whether or not a system with some symmetry will have a certain kind of spectroscopic response.
Why bother calculating the integral if it's going to be zero anyway? It's just extra work, though it usually isn't very expensive.

For $O_{\mathrm{h}}$, the completely symmetric irrep is $\mathrm{A_{1g}}$. It's always listed as the first irrep in a character table.

$$
\small\begin{array}{c|cccccccccc|cc}\hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & & x^2+y^2+z^2 \\
\mathrm{A_{2g}} & 1  & 1  & -1 & -1 & 1  & 1  & -1 & 1  & 1  & -1 & & \\
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0  & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T_{1g}} & 3  & 0  & -1 & 1  & -1 & 3  & 1  & 0  & -1 & -1 & (R_x,R_y,R_z) & \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  & & (xy,xz,yz) \\
\mathrm{A_{1u}} & 1  & 1  & 1  & 1  & 1  & -1 & -1 & -1 & -1 & -1  & & \\
\mathrm{A_{2u}} & 1  & 1  & -1 & -1 & 1  & -1 & 1  & -1 & -1 & 1  & & \\
\mathrm{E_u}    & 2  & -1 & 0  & 0  & 2  & -2 & 0  & 1  & -2 & 0  & & \\
\mathrm{T_{1u}} & 3  & 0  & -1 & 1  & -1 & -3 & -1 & 0  & 1  & 1  & (x,y,z) & \\
\mathrm{T_{2u}} & 3  & 0  & 1  & -1 & -1 & -3 & 1  & 0  & 1  & -1 & & \\ \hline
\end{array}
$$

The magnetic dipole operator is the angular momentum operator with some constants in front:

$$
\hat{\mathbf{m}} = \gamma_{\mathrm{e}} \hat{\mathbf{l}} = -\frac{e}{2m_{\mathrm{e}}} \hat{\mathbf{l}} \approx \left(\hat{l}_{x}, \hat{l}_{y}, \hat{l}_{z}\right)
$$

The angular momentum operator components have rotational symmetry, so $\hat{l}_{y}$ would use the irrep corresponding to $R_{y}$. Constant prefactors don't affect the symmetry of operators, so all three components of the magnetic dipole operator have $\mathrm{T_{1g}}$ symmetry.

Now we must show that

$$
\left< \Psi_{i} | \hat{\mathbf{m}} | \Psi_{f} \right> \equiv \int \Psi_{i}(\tau) \times \hat{\mathbf{m}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \rightarrow \mathrm{T_{2g}} \times \mathrm{T_{1g}} \times \mathrm{E_{g}} \overset{!}{\ni} \mathrm{A_{1g}},
$$

and to do this we need a product table.

$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$

Product tables are symmetric, so only the lower or upper triangle needs to be shown, and inversion symmetry is left out here as long as you consider the rule

$$
\begin{align}
\mathrm{g} \times \mathrm{g} &= \mathrm{g} \\
\mathrm{g} \times \mathrm{u} &= \mathrm{u} \\
\mathrm{u} \times \mathrm{u} &= \mathrm{g},
\end{align}
$$

just like multiplying even and odd integers. Some product tables show all of the redundant information.

Working from left to right, find the product of the symmetry of the initial state with the symmetry of the operator:

$$
\mathrm{T_{2g}} \times \mathrm{T_{1g}} = \mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}.
$$

The product of degenerate irreps with each other leads to multiple irreps, hence the sum. You might think that the next step is

$$
\left(\mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}\right) \times \mathrm{E_g} = \dots,
$$

however we take advantage of two nice rules:

For the transition to be allowed, the final product only needs to contain the completely symmetric irrep; irreps other than $\mathrm{A_{1g}}$ are allowed to be present as long as $\mathrm{A_{1g}}$ is there.
The product of any irrep with itself always gives the completely symmetric irrep; see the diagonal elements in any product table.

From the last equation, we see that $\mathrm{E_g}$ is in the sum, so the other terms can safely be ignored because $\mathrm{E_g} \times \mathrm{E_g} = \mathrm{A_{1g}}$. Additionally, because all three components of the magnetic dipole operator transform as the same irrep, this electronic transition will be allowed if you apply a magnetic field along any of the three Cartesian axes in the molecular frame.