Dipole moment of cis-2-butene I need someone to back me up on this before I go confront my teacher:
I was doing some analysis of the dipole moment of cis-2-butene.
Let's say that the alkyl groups are both on top.
Would the dipole moment be a vector that has its tail on the top and its head on the bottom?

How would one calculate that?
I made some calculations assuming all C-H bonds have the same dipole moment vector and came to the conclusion that the three H's attached to the alkyl groups actually end up overcoming the dipole moments of the C-H bonds of the two unsubstituted H's on the double bond.

Am I right?

Mini Research Project Time

Updates

Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete).

Preamble

This response is in no way meant to be contrary to what Geoff has already posted.  I happen to enjoy these types of questions and I like to tackle them as little research projects for fun.  Therefore, my answer will be brutally detailed and in-depth.  That said, this should be a good introduction to how I would approach this problem if I were going to do this at the 'production-level' of research but this definitely goes far beyond the purpose of any reasonable response for an SE answer.

Its All About the Dipoles Baby

Okay so I wouldn't use THAT heading in a paper but maybe a talk depending on who my audience was...

We can easily determine the dipole moment of cis-2-butene via electronic structure theory.  I have modeled two conformers of cis-2-butene as shown below.  I will refer to the geometry on the left as StructA and the geometry on the right as StructB.

A couple of things to note since I don't include any captions to tables.  Energies are always reported in $\mathrm{kcal\ mol}^{-1}$ and dipole moments ($\mu$) are given in Debye.

Computational Methods

Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with second-order Moller-Plesset perturbation (MP2) theory and a variety of density functional theory (DFT) methods using the Gaussian 09 software package for the conformers of cis-2-butene.  Both conformers were characterized in $C_{2v}$ symmetry.  The DFT methods implemented include B3LYP, B3LYP-GD3(BJ), M06-2X, MN12SX, N12SX, and APFD.  The B3LYP-GD3(BJ) method employs Grimme's 3rd generation dispersion correction as well as the Becke-Johnson damping function. All DFT computations employed a pruned numerical integration grid having 90 radial shells and 590 angular points per shell.  The heavy-aug-cc-pVTZ basis set was employed for these computations where the heavy (non-hydrogen) atoms were augmented with diffuse functions (e.g. cc-pVTZ for H and aug-cc-pVTZ for carbon).  This basis set is abbreviated as haTZ.  The CCSD(T) (i.e. coupled-cluster method that includes all single and double substitutions as well as a perturbative treatment of the connected triple excitations) was similarly employed using the CFOUR software package.  The magnitudes of the components of the residual Cartesian gradients of the optimized geometries were less than $6.8\times 10^{-6} E_h\ a_0^{-1}$.

Single point energies were computed with the explicitly correlated MP2-F12 [specifically MP2-F12 3C(FIX)] and CCSD(T)-F12 [specifically CCSD(T)-F12b with unscaled triples contributions] methods in conjunction with the haTZ.  These computations were performed with the Molpro 2010.1 software package using the default density fitting (DF) and resolution of the identity (RI) basis sets.

Natural bond orbital (NBO) analyses were performed for the MP2 optimized structures using the haTZ basis set and the SCF density.

All computations employed the frozen core approximation (i.e. 1s$^2$ electrons frozen in carbon).

Computational Methods in English

Two different conformations of cis-2-butene were characterized with a variety of cheap (but usually okay) approximations (i.e. DFT methods) as well as reliable (but generally more expensive) wave function methods (i.e. MP2 and CCSD(T)).  The wave function methods are necessary to validate the DFT results.  CCSD(T) is the gold-standard and gives very good results for single-reference closed-shell well-behaved systems so we will use this as our 'best estimate'.  We use a variety of methods in order to look for agreement in the results.  If we see good agreement across the board, we can be confident in our results.  If we see massive discrepancies then we will have to be careful when we analyze the results.  Geometries have been converged to a tight threshold (i.e. we have good molecules) and our DFT computations use a relatively dense integration grid (which leads to more accurate results).

Notice that I employ a heavy-aug-cc-pVTZ (haTZ) basis set.  Why leave the diffuse functions off hydrogen?  The purpose of diffuse functions is to describe electron density far away from the nucleus of an atom.  Therefore we slap these functions onto carbon which are relatively large atoms compared to hydrogen.  Hydrogen, on the other hand, has only one electron and therefore has a small electron density when isolated.  In cis-2-butene, hydrogen is bonded to carbon via a rather small bond distance.  The electron density around the hydrogen is even more reduced than an isolated hydrogen atom in the gas phase.  Therefore, it would be impractical to include diffuse functions on hydrogen.  Doing so may even lead to erroneous results since we will be trying to describe electron density far away from the hydrogen nucleus when in reality there is virtually none to be found.

Finally, we perform single point energies using explicitly correlated methods.  Because the CCSD(T) opts and freqs will likely not be done in time for this posting, we can gauge how the resulting geometry from each optimization procedure will vary from another geometry given by a different method.  If all of the energies are similar (within a few tenths of a $\mathrm{kcal\ mol}^{-1}$), then we can be confident that our geometries are not only very similar but that small deviations in the geometry will have little effect on the corresponding energies at least in this region of the potential energy surface (PES). Large deviations will usually mean that the method which produced the 'outlier' is not a good approximation for the system (I do not expect cis-2-butene to be problematic at all).  Explicitly correlated methods accelerate convergence to the CBS limit.  These methods have been shown to give results that a large basis set and a canonical method would provide but with a much smaller basis set.  For example,  the result that I get with a regular CCSD(T)/aug-cc-pV5Z basis set could be obtained using CCSD(T)-F12/aug-cc-pVTZ.  This makes the computations less intensive and much more feasible.

Results

The number of imaginary frequencies ($n_i$), relative MP2-F12 and CCSD(T)-F12 energetic  ($\Delta E^{\mathrm{MP2-F12}}$ and $\Delta E^{\mathrm{CC-F12}}$, respectively, in $\mathrm{kcal\ mol^{-1}}$) and dipole moment ($\mu_z$ in Debye) are given in the following table for both conformers of cis-2-butene for a variety of a methods.
The relative energies were determined by taking the difference of the respective geometry and the reference CCSD(T) geometry [e.g. E(CCSD(T))-E(MP2)].

StructB is a second-order saddle point ($n_i = 2$) on every single PES considered and therefore is not a minimum energy structure.  StructA is, however, a minimum ($n_i = 0$) on every PES considered.  The characterization of the nature of the stationary point is consistent between CCSD(T), our best estimate, MP2, and DFT methods.

Single point energies reveal negligible differences in the optimized geometries.  The energies associated with the MP2 optimized structures is the reference point for all other relative energies.  Deviations grow no larger than 0.27 $\mathrm{kcal\ mol}^{-1}$ for the MP2-F12 and CCSD(T)-F12 relative energies.  In addition, there is good agreement between the MP2-F12 and CCSD(T)-F12 relative energies, suggesting that higher-order correlation effects are small.

The dipoles for StructA and StructB are very similar with very small magnitudes, on the order of a couple tenths of a Debye.  To put these quantities into perspective, the dipole moment of water is 1.85 D.  We all know that water has a pretty large dipole moment so by comparison, cis-2-butene has a very WEAK dipole moment.  You can compare to the dipole moments of other molecules by referring to this NIST reference.  Clearly, the rotation of the methyl groups have a very small effect on the dipole moments of each conformer.  The MP2 and DFT dipole moments deviate from the best estimate by no more than 0.03 D, but remains in qualitative agreement for both StructA and StructB.

The figure below shows the directionality of the dipole.  The head of the (unscaled) arrow points toward the negative pole while the tail of the (unscaled) arrow is oriented to the positive pole.  The numbers on the atoms represent 'natural charges' from an Natural Bond Orbital (NBO) analysis of the MP2 optimized Struct A.  Clearly the carbon atoms have a small negative charge to them as these atoms are sucking (great scientific term here) electron density away from the neighboring hydrogen atoms.  This is because the nuclear charge (i.e. the number of protons) on carbon is much greater than that of hydrogen (6 vs. 1).

It was suggested that I add some data highlighting the energy difference between the two different conformers.  The following table presents the energy difference (where $\Delta E_{\mathrm{A-B}}$ is equivalent to E(A)-E(B)) of each optimized geometry using the MP2-F12 and CCSD(T)-F12 single point energies (in $\mathrm{kcal\ mol}^{-1}$).  We can see that StructA is about 1.5 $\mathrm{kcal\ mol}^{-1}$ lower in energy than StructB.  This makes sense because StructB is a higher order saddle point and StructA is a minimum on the PES. (I'm actually quite surprised that the energy difference is this large for a couple of methyl rotations...)

Conclusions

The dipole moment for two conformers of cis-2-butene has been examined using seven different computational approaches and a triple-$\zeta$ quality basis set.  The performance of these methods have been tested by evaluating the energies of each geometry with the 'gold standard' CCSD(T) method (the explicitly correlated variant).  Good agreement is seen across the board with respect to the Hessian indices, energies, and dipole moments. Only StructA was a minimum on each PES.  Both conformers of cis-2-butene have a very weak dipole moment on the order of 0.2 -- 0.3 D.  The positive pole is in the vicinity of the methyl groups whereas the negative pole is centered around the sp2 hybridized carbons.

FAQ

So you may be asking yourself (or rather, should be asking yourself) questions such as these listed below.  I will tackle them one at a time.

1.) Why did we look at two conformers of cis-2-butene and why does it matter?

StructA is a minimum which means that if you were to characterize this guy in the gas phase, you'd find StructA rather than StructB.  This is important because if we were to report these results to other scientists, they will want to know what they can expect to find without wondering.  Therefore, StructA is going to be our conformer of interest rather than StructB.  The latter still will provide insightful results but that is about it for the purposes of this examination.

2.)  Why did we use a variety of methods to characterize these molecular systems?

Computational methods are simply approximations.  They are not guaranteed to give the 'correct' answer.  So we try to address this by using a variety of methods (i.e. approximations) and we analyze the results accordingly.  If the results are in agreement, then we can feel confident that the results are correct since we tested them against a list of methods.  You can never rely on just one method unless it is rigorous, well-tested and has been shown repeatedly to perform well in the literature.  When we say 'perform well', this usually means that the computational results are in reasonable agreement with experimental results.  This is important because experimental results is the LAW (for all intents and purposes).  If the computations disagree with experiment, 99.9% of the time this means that your computational approach sucked and that your approximation was either flawed or misapplied.  By using a host of methods, we can put a little more faith into the results we observe because the chances of massive disagreement between the results of the methods is very unlikely in a normal situation.

3.) What is the point of doing a bunch of energy points?

Again, because we used a slew of methods to characterize the geometries of cis-2-butene, we end up with non-identical geometries each time we use a new approximation.  For instance, the methyl C-H bond lengths from B3LYP are going to be a little bit different from those obtained with MP2.  So then that begs the question, "How do these small differences effect the property of the system that we are interested in?"  Generally, minute differences will have little effect on the resulting energies of each molecule under consideration.  Energy is a very important property that chemists love to look at.  So, if we take each geometry (and each one is unique from the other) and we evaluate the energy of the molecule at the same level of theory (in this case, MP2-F12 and CCSD(T)-F12), then we can quickly see how 'resolved' each geometry was.  There should be very good agreement between the relative energies of each geometry (probably to within a few tenths of a $\mathrm{kcal\ mol}^{-1}$).

4.) Okay, so why MP2-F12 AND CCSD(T)-F12?

We use MP2-F12 AND CCSD(T)-F12 to test for 'higher-order correlation effects'.  MP2 methods are much cheaper than CCSD(T) methods but MP2 is not as rigorous and can be error prone in a host of molecular systems.  Therefore, we test the performance of MP2 by busting out our 'gold-standard' which is the CCSD(T) method.  If MP2 agrees well with CCSD(T), then we can feel confident in our MP2 results and never have to revisit the difficult, time-consuming CCSD(T) computations ever again.  ALso, CCSD(T) will also tell us how DFT performed as well.  DFT methods must always be calibrated against something more rigorous since DFT is known for 'getting the right answer for the wrong reasons' and it isn't always right.

The 'F12' bit just means that these are 'explicitly-correlated' methods.  Rather than give an introduction to what this means, you should understand why we use it instead.  You may have noticed that whenever we do a computational job, we specify a method AND a basis set (e.g. heavy-aug-cc-pVTZ).  These basis sets can be measured by how many atomic orbitals (or functions) are given in the set.  The more that are given, the better the 'basis-set approximation'.  Think of this in terms of Riemann sums where you try to approximate the area under a curve using a set of rectangles.  Each rectangle is a basis function and the number of rectangles you use forms a basis set.  The more rectangles you use, the better your approximation of the area under that curve will be.  Basis sets in computational chemistry behave the same way.  When you approach an infinite set of rectangles, you approach the exact answer.  When you approach an infinite number of basis functions, you approach what is called the CBS (complete basis set) limit.  At the CBS limit, you have an exact answer.  We cannot implement an infinite basis set in chemistry (for obvious reasons), and very large basis sets are cost prohibitive.  Therefore, people have devised these F12 approximations that are constructed in such a way to give results that are comparable to those that you'd get with a large basis set, but you can get them by using a relatively small basis set instead!  This is a powerful approach to convergent quantum chemistry that saves you a lot of time while maintaining a set of very good results.

5.) Why didn't you provide more pretty pictures?

That's just the nature of the beast.  Computational chemistry is usually short on graphics but very dense on spreadsheets.  Plus... I'm no artist.  I actually spent a good couple hours trying to get some electrostatic potentials posted but the new version of G09 hates the molecular viewer programs I currently use so I ditched that idea.


Temperature change when volume halved For this multiple question:

The correct answer is B. However, I don't understand.
You are halving the volume
so for the original solution
the number of moles of HCl is 0.1 moles
and in the new solution, the number of moles of HCl is 0.05 if we are using the same concentration.
Therefore as there are less number of moles, won't the temperature increase (ie. energy release) get halved?

The correct answer is B in both the case the final concentrations are not changing. Therefore same will be the heat of rise. But heat released will be less but volume of the final solution also got halved. So temp. raise will be same.


Is there a "Comprehensive Solid State" book? We have Comprehensive Coordination Chemistry, Comprehensive Organometallics, Comprehensive Organic Synthesis and probably few other similar works that aim to do a wide review of recent works at the time of writing.

Is there a similar review book, that reviews modeling, characterization, preparation, structures and properties of crystalline, non-molecular solids, maybe as a part of a wider topic or a series on smaller subjects?

I think the closest comprehensive analogical literature source for solid state materials to the aforementioned series like Elsevier's Comprehensive Coordination Chemistry II, Elsevier's Comprehensive Organic Synthesis or Wiliey's Inorganic Syntheses would be Springer's Springer Series in Solid-State Sciences currently consisting of 223 titles:

  The Springer Series in Solid-State Sciences consists of fundamental scientific books prepared by leading researchers in the field. They strive to communicate, in a systematic and comprehensive way, the basic principles as well as new developments in theoretical and experimental solid-sate physics.

As for the universally recommended textbooks on solid-state chemistry in general, there are the following titles issued for the past 15 years:

West, A. R. Solid State Chemistry and Its Applications, 2nd ed.; John Wiley & Sons, Inc: Chichester, West Sussex, 2014.
Simon, S. H. The Oxford Solid State Basics, 1st ed.; Oxford University Press: Oxford, 2013.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012.
Tilley, R. J. D. Understanding Solids: The Science of Materials; J. Wiley: Chichester, West Sussex, England; Hoboken, NJ, USA, 2004.
Müller, U. Anorganische Strukturchemie, 6., aktualisierte Aufl., unveränd. Nachdr.; Studium; Vieweg + Teubner: Wiesbaden, 2009. (in German)
Dronskowski, R. Computational Chemistry of Solid State Materials: A Guide for Materials Scientists, Chemists, Physicists and Others; Wiley-VCH: Weinheim, 2005.


Which causes of quenching increase specifically in electrochemical experiments? If quenching is seen as a result of running electrochemical experiments on a solution, what kind of processes may be responsible? I'm thinking about UV/Vis spectroscopy, if it makes any difference.

The redox couple would be iodide ions, $\ce{I-/I^3-}$. I understand iodide ions to be quenching agents, so my suspicion would be that their varying concentrations would cause a variation in quenching over the course of the experiment.

I wonder if, when you say UV/Vis spectroscopy, you are also including fluorescence in that. If you are, then the soft iodide / triiodide ions are pretty good reductive quenching agents and they may be responsible for the reduction in fluorescence emission you observe.

Typically, the $\text{I}^-$ will react with the excited state of your dye $\text{D*}$ and quench it by photo-induced electron transfer:

$$\ce{I^{-} + D}^*\ce{\quad ->[\text{photo-induced electron transfer}]\quad I^{\ .} + D^{\ .\ -}\quad ->[\text{back electron transfer}]}\quad \ce{I- + D}$$

As you can see, this process provides a non-radiative pathway for the excited state of the dye to decay back to its ground state. The process is essentially catalyzed by the iodide anion, which is not consumed in the process. The more iodide you have, the more prevalent this process, the more likely it is for the excited state of your dye to go through this process rather than fluorescing, so ultimately the more quenched your dye.

Such a process can be modeled by the Stern - Volmer equation:

$$ \frac{I_0}{I}=1+k_q\tau_0[\text{Q}]$$

Where $I_0$ is the fluorescence intensity in the absence of quencher Q; $I$ the intensity in the presence of a certain concentration [Q] of quencher; $k_q$ is the kinetic constant of the quenching process; $\tau_0$ is the intrinsic lifetime of the fluorophore, i.e. the lifetime you would measure in the same conditions, but in the absence of quencher Q.


What is the hidden meaning of electrochemical stability window? I don't really get the real meaning of the electrochemical stability window. I googled a lot, and the only thing I understand about the electrochemical stability window is that it's the voltage range that the substance is not oxidized and reduced. However, what does this term imply? I mean, I wonder that how we use this concept in practice.

The electrochemical stability window is most important when considering components of an electrochemical system that you do not want to be oxidized or reduced. This refers most often to the electrolyte or protective coatings.

For example, in lithium ion batteries it is highly desirable that the electrolyte does not react/change/degrade in any way as a result of battery charging/discharging or even in storage. A lithium ion battery typically has a voltage of 3.6 V. The electrolytes used are either lithium salts in organic solvents (ethylene carbonate, etc), solid ceramics, or polymerics electrolytes. If the electrolyte was water based, the electrochemical stability window would only be 1.2 V.

Why 1.2 V?

The standard reduction potential of water is 0.0 V (vs SHE)

The standard oxidation potential of water is 1.2 V (vs SHE)

You can not operate a 3.6 V lithium ion battery with an aqueous electrolyte because the electrolyte itself would react, creating dangerous hydrogen and oxygen gas. So you need an electrolyte with a larger stability window.


Preparation of Grignard reagent. Free radical rearrangement Mechanism of formation of Grignard Reagent
$$\ce{R-X + ^.Mg^. ->[\text{r.d.s.}] R^. + ^.Mg+X-} $$
$$\ce{R^. + ^.Mg+X- -> R-Mg+X-} $$

As the mechanism of preparation of Grignard reagent suggests, it involves the formation of free radical. And so, it is observed that racemic mixture is formed, in case the carbon on which the free radical is formed is chiral. But, why there occurs no rearrangement of this free radical ($\ce{R^.}$) ?

Grignard reagent is an organomagnesium halide in which magnesium and halide bond is ionic. It is not like magnesium is always inserted in between carbon and halogen bond. Once formed, magnesium-carbon bond should be configurationally stable under the conditions of the reaction (Ref.1), and is polar covalent (at least partially) so that $\ce{Mg}$ has to be connected to its bonding partner $\ce{C}$ in the first place. In this polar covalent bond, carbon ends up being the negative end of the dipole, giving its nucleophilicity. The reaction undergoes in free radical mechanism (yet it does not involve a radical chain mechanism) to form Grignard reagent:
$$\ce{R-X + ^.Mg^. ->[\text{r.d.s.}] R^. + ^.Mg+X-} \tag1$$
$$\ce{R^. + ^.Mg+X- -> R-Mg+X-} \tag2$$
Magnesium metal has two valance electrons. The first step is the rate-determining step in this two-step mechanism, which happens on the surface of the metal. In this first step, $\ce{Mg}$ metal transfers one of its two valance electrons to the halogen radical, cleaving carbon-halogen bond homolytically. The process in step by step mode to explain the outcome:
$$\ce{R-X -> [R^. + X^.] ->[^.Mg^.] R^. + ^.Mg+X-} \tag3$$
This first step forms two radicals: $\ce{^.Mg+}$ and $\ce{R^.}$. These two radicals then get terminated by coupling each other to give the Grignard reagent in second step. The rearrangement of $\ce{R^.}$ (if there any) during the second step depends on the rate of rearrangement compared to the rate of the radical termination. However, according to the Ref.1, both chiral cyclopropyl-magnesium bond and vinyl-magnesium bond do in fact retain their configurations from parent alkyl halides. The authors carefully chose non-ionizable alkyl halide in solvent for this study so that any rearrangement should happen during the reaction (e.g., cyclopropyl tosylate solvolyzes at a rate of $\pu{5.7 \times 10^{-13} s-1}$ in actic acid at $\pu{25 ^\circ C}$, Ref.2). The excistence of $\ce{R^.}$ intermediate is proven by trapping it using $\ce{TEMPO^.}$ radical (Ref.3).
References:

H. M. Walborsky, "Mechanism of Grignard reagent formation. The surface nature of the reaction," Acc. Chem. Res. 1990, 23(9), 286–293 (https://doi.org/10.1021/ar00177a004).
P. von R. Schleyer, W. F. Sliwinski, G. W. Van Dine, U. Schöellkopf, J. Paust, K. Fellenberger, "Cyclopropyl solvolyses. III. Parent cyclopropyl derivatives and methyl-substituted cyclopropyl tosylates," J. Am. Chem. Soc. 1972, 94(1), 125–133 (https://doi.org/10.1021/ja00756a024).
Karen S. Root, Craig L. Hill, Lynette M. Lawrence, George M. Whitesides, "The mechanism of formation of Grignard reagents: trapping of free alkyl radical intermediates by reaction with tetramethylpiperidine-N-oxyl," J. Am. Chem. Soc. 1989, 111(14), 5405–5412 (https://doi.org/10.1021/ja00196a053).


Oxidation of tertiary alcohols Why is it that tertiary alcohols cannot be oxidized? I was told that it's because there is no hydrogen to remove, but I don't understand why another OH could not be formed where there is a C-H bond.

The meaning of the answer you've gotten, "it is because there is no hydrogen to remove", is with respect to oxidizing the carbon of the tertiary alcohol itself. You could oxidize the rest of the molecule into any number of things, but there is no way for the tertiary carbon to be further oxidized than the alcohol.


How does dimethyltryptamine cross the blood-brain barrier? Is there a specific chemical reaction that enables dimethyltryptamine to cross the capillary edothelial cells?

Although serotonin and dimethyltryptamine (DMT) have similar chemical structures (see Jori's answer), serotonin is not transported across the blood brain barrier (BBB) ("serotonin does not cross the blood–brain barrier").  On the other hand, not only does DMT cross the BBB, it is one of the few compounds that is actively transported across the BBB.  While most compounds that cross the BBB do so by being lipid soluble (relatively small non-polar compounds) some molecules are assisted across the barrier through a variety of mechanisms.  Some of the known assistive paths include

glucose carrier: transports glucose and several other sugars across the BBB
phenylalanine carrier: transports a number of neutral amino acids across the BBB
arginine carrier: transports several cationic amino acids across the BBB
lactate carrier: transports several monocarboxylic acids across the BBB
adenosine transporter: transports purine nucleosides as well as uridine
organic cation transporter: transports a variety of cationic compounds, including monoamine neurotransmitters

There are more classes and sub-classes within each.  The following two references provide nice, readable overviews

Small Molecular Drug Transfer across the Blood-Brain Barrier via Carrier-Mediated Transport Systems
Drug transport across the blood–brain barrier

I don't know which of the above transporters is active in the case of DMT, but I strongly suspect it is the "organic cation transporter".  Here are some further references that relate specifically to DMT active transport across the BBB

Takahashi T, Takahashi K, Ido T, Yanai K, Iwata R, Ishiwata K,
Nozoe S (1985) 11C-labeling of indolealkylamine alkaloids and
the comparative study of their tissue distributions. Int J Appl
Radiat Isot 36:965–969
Yanai K, Ido T, Ishiwata K, Hatazawa J, Takahashi T, Iwata R,
Matsuzawa T (1986) In vivo kinetics and displacement study of
a carbon-11-labeled hallucinogen, N, N-[11C]dimethyltryptamine.
Eur J Nucl Med 12:141–146

They were found in this article, "Bufotenine: Toward an Understanding
of Possible Psychoactive Mechanisms"

P.S.  We've all read fairy tales where kissing a frog turns the frog into a prince.  When the skin of the Bufo toad (Colorado river toad, sonoran desert toad and other members) is licked, a psychedelic experience results.  It is not illegal in the US to own the toad, although the Bufotenine is a controlled substance!  Anyhow, just wondering if this, or something similar, is what the fairy tales are based on.


Why is there a greater degree of separation in a dipole-dipole interaction? My textbook says this:

  Electrostatic forces between two ions decrease by the factor $\frac{1}{d^2}$ as their separation distance, $d$, increases. Dipole–dipole forces, however, vary as  $\frac{1}{d^4}$. Because of the higher power of $d$ in the denominator, the attractive forces diminish with increasing $d$ much more rapidly than does  $\frac{1}{d^2}$. As a result, dipole forces are effective only over very short distances.

Now mathematically I understand why this is true, but I don't understand why this is the case in reality. Can someone shed some light on this?

Like I answered my other question, this is in some ways very similar. Since electrostatic attractions are so strong, it is decreases less at a longer distances. The $+$ & $-$ are attracted together so strongly.

Dipole-dipole attractions aren't strong attractions, electrostatic attractions are stronger, since they are created by ionic bonding. Dipole-dipole occur in covalent bonding. Thus dipole-dipole attractions decrease with increasing distance faster.

This is just a piece of the whole picture.

Check the refs: 1, 2, 3. They provide more in depth detail to the entire picture.


Relation between rate of adsorption and the critical temperature of a gas How is the rate of adsorption related to the critical temperature of a gas? My textbook says more the critical temperature, better are they adsorbed. But I cannot really understand why. They threw in another confusing statement that says that easily liquefiable gases are readily adsorbed. I am thoroughly confused.

I did read the question How does critical temperature affect adsorption? before posting this question, and while that question was answered, it did not answer my question fully.

Edit: To clarify my area of doubt, let me say what I find confusing. Gases with lower Ct will 'prefer' the gaseous phase over the liquid phase. Which is what we want. We want gases to be adsorbed.

You can argue that the effect of the solid is to "seed" formation of the condensed phase, as a set of interactions that lower the chemical potential below that of the gas, leading to "condensation" (well, adsorption). This is all very approximate as the details matter of how (as in, through what interactions) the adsorbent interacts with the gas molecule.

In addition, the magnitude of the critical temperature may be considered as a proxy for the strength of intermolecular interactions. For instance, if you assume the gas follows the van der Waals equation of state, then the critical temperature $$T_\mathrm{c} = \frac{8a}{27bR}$$ where $a$ is a measure of the strength of attractive interactions. You can find other ad hoc justifications for associating higher $T_\mathrm{c}$ with stronger strength of adsorption. For instance, larger, more polarizable molecules will tend to have higher boiling points and critical points. Such molecules will also adsorb more strongly. However the argument that the rate of adsorption increases should be considered carefully, since steric effects may be associated with larger molecules.

Regarding the point that "we want gases to be adsorbed", well, the point is that you want to predict behavior. The book merely arms you with guidelines that allow you to make predictions. So that, for instance, if you have a number of compounds in a gaseous mixture, you can estimate which adsorb more. Or perhaps you can use that ability to select new substances with desirable properties based on the adsorption behavior of other compounds.


Why is carbonic acid a weaker acid than acetic acid? Is it because the $\ce{OH}$ group next to $\ce{COOH}$ in carbonic acid donates electron density to the $\ce{COO-}$ by resonance, making it a stronger base, compared to a weaker electron donating inductive effect of $\ce{CH3}$ and thus its conjugate acid is weaker?

What does the resonance contribution look like in this case?

What makes resonance a stronger factor than induction?

Before you ask "why", ask "if".

Carbonic acid is not weaker than acetic. In fact, it is stronger, just as one might expect by looking at that extra electron-withdrawing substituent. The trouble is, carbonic acid is also pretty unstable. At any given moment, most of it exists as $\ce{CO2}$. This leads to much lower apparent acidity.

Think of it this way: you add some acetic acid to a carbonate salt. An equilibrium sets in, much closer to the starting compounds than to the products. In other words, you'll have just a very tiny amount of $\ce{H2CO3}$. But as soon as you have it, it starts decomposing, thus shifting the equilibrium more and more to the right.

Look again at that Wikipedia page, it says:

  3.6 ($\rm pK_{a1}$ for $\ce{H2CO3}$ only), 6.3 ($\rm pK_{a1}$ including $\ce{CO2(aq)}$)...


Why can copper have both four and six bonds? I'm trying to understand the Biuret test by first understanding how copper dissolves in the aqueous solution. We first have added the $\ce{CuSO_{4}}$ to a substance than the $\ce{OH^{-}}$.

I've read the following page on the reaction, but I still confused: http://www.docbrown.info/page07/transition09Cu.htm

How is it possible to have tetraaquacopper(II) and hexaaquacopper(II) ions in the solution? How could this configuration be explained using electron configuration? $[\ce{Ar}]\mathrm{3d^{10}4s^{1}}$

$\ce{Copper(II)}$ complexes are $\ce{d^9}$ type complexes. This means that there are 9 d-electrons for the metal centre. When a complex forms, a number of ligands coordinate to the metal centre and the species adopts certain geometry. The five d-orbitals are no longer the same energy. They split into sets of different energies depending on the geometry and how ligands interact with them.

For an octahedral complex, orbitals split into two sets of degenerate orbitals - a lower $\ce{t_{2g}}$ ($\ce{d_{xy}},\ce{d_{yz}},\ce{d_{xz}}$) which is bonding to non-bonding and a higher $\ce{e_g}$ ($\ce{d_{z^2}}, \ce{d_{x^2-y^2}}$) which is slightly anti-bonding.

Copper is a bit funny as it experiences what is called a Jahn-Teller distortion where there is an elongation of the metal-ligand bonds along the z-axis. This is due to uneven occupancy of the $\ce{e_g}$ orbitals set. Then, the orbitals further split to remove the degeneracy and gain more stability. For the square planar complex of $\ce{tetraaquacopper(II)}$ to form you would need to go even further and remove those two ligands along the z-axis. The stabilization energy gained overall by going from octahedral to square planar configuration is what allows this switch to occur; it wouldn't happen if it were unfavorable. The gain in stability is borderline and that is why I assume that both configurations are possible.

Also, square planar fields tend to prefer stronger field ligands and water is not such ligand and this might also be considered as a reason why you get a mixture of the two types of geometry.


Can I perform electrolysis without electrolyte? (with distilled water) When performing electrolysis at home, I usually used salt water as electrolyte. The problem is, that this method also produces chlorine and chlorine is aggressive to both humans and electrical equipment.

I was thinking of creating pure hydrogen and oxygen by using distilled water. I was taught that water is not conductive.

But since water has "positive" and "negative" part (which allows for it to dissolve polar bonds, doesn't it?). Couldn't I just tear the molecules apart if I use electrical field strong enough? If I can, how high the voltage must be?

If I can, how high the voltage must be?

Not possible at home.

For obtaining pure hydrogen and oxygen it is still required to use an electrolyte.  However, there is some amount of them that are not electrolyzed, allowing to electrolyze only water. I'd try sodium/potassium hydroxide or sulfate.


How may the water from rain be tested to determine it is natural, or from cloud seeding? This article lists several mechanisms that may be used to control the weather and/or induce rainfall. Apparently many states, other than the one listed in the headline, have experimented with such techniques. The article also lists concerns over contamination from such artificial rainfall.

Say I, a layman without access to a laboratory, collect the rain in a glass. What test/tests may I apply to determine whether it is natural rainfall, or the outcome of cloud seeding?

The AgI would act as a cloud condensation nuclei (CCN) on which water vapour would condense allowing cloud droplet formation. Presumably, you could use scanning electron microscopy to determine the composition of the CCN and determine whether cloud seeding has taken place.

Apparently, his has already been done with cloud chamber (so, artificial scenario) measurements (Zimmerman et al., 2007 - "Environmental scanning electron microscopy (ESEM) as a new technique to determine the ice nucleation capability of individual atmospheric aerosol particles").

However, considering that there has been ESEM measurements in field situations before (Pöschl et al., 2010 - "Rainforest Aerosols as Biogenic Nuclei of Clouds and Precipitation in the Amazon") to determine CCN composition, I suppose a similar set up could be performed.


Expansion under isobaric adiabatic condition A 3 mole sample of a triatomic gas at $\pu{300 K}$ is allowed to expand under isobaric adiabatic condition from $\pu{5L}$ to $\pu{40L}$. The value of change in enthalpy is:

  $\pu{12.46 KJ}$
  $\pu{-14.965 KJ}$
  $\pu{-24.62 KJ}$
  $\pu{-10.24 KJ}$

In case of adiabatic process, there is no transfer of heat so $Q=0$. Again for an isobaric process, pressure is constant that is $\Delta P=0$.
So, change in enthalpy:

$$\Delta H = \Delta U + \Delta(PV) = Q + W + P\Delta V + V\Delta P = W + P\Delta V = -P\Delta V + P\Delta V = 0$$

Thus, we see that change in enthalpy is zero. But the answer matches with none of the options. What is incorrect in my solution?

If, at time = 0, we suddenly drop the external force per unit area to $P_{\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:

$$nC_\mathrm v(T_2-T_1)=-P_{\text{ext}}(V_2-V_1)=-\frac{nRT_2}{V_2}(V_2-V_1)=-nRT_2\left[1-\frac{1}{8}\right]$$

Solving this equation for $T_2$ yields: $$T_2=\frac{T_1}{\left[1+\frac{7}{8}\frac{R}{C_\mathrm v}\right]}$$

But, for $C_v=3R$, this gives:$$T_2=0.774T_1=\pu{232 K}$$

So, $$\Delta H=3(4R)(232-300)=\pu{-6.78 kJ}$$

So the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.


Alcohol protection with ethers Various ethers are used for hydroxyl protection. My question is about their removal. I know that there are various methods but I want to specify on the use of acids.

Certain groups are removed easily like Trityl, by using aqueous acetic acid. If I'm not mistaken, the mechanism is Sn1, as the trityl carbocation is stabilised through delocalisation on the phenyl groups.

However, other similar groups like benzene or PMB are not removed that easily. Can't they follow the same logic, as the postive charge could again be delocalised in the aromatic ring? Is it because the delocalisation is not good enough, as in the case of trityl?

Also, I've noticed that in order to remove smaller groups like methyl, haloacids are mainly used. Why can't a more weak acid do the job?

You are right.In case of benzene and PMB, the delocalization is not good enough. So a strong acid is required.

As for the case of OMe, a strong acid as well as strong nucleophile will be needed. A weaker acid can't do the job. Let me explain why.
Note that carbon oxygen bond is quite strong.
So what we need is a strong nucleophile. But in case of OMe as a protecting group, even a stronger nucleophile will not work if the oxygen is not heavily protonated due to two reasons:-
1. Strength of C-O bond.
2. SN1 can't occur as methyl carbocation not stable.

Here acid is a stoichiometric reagent, not simply a catalyst. If a weaker acid is used, it will just establish a protonation-deprotonation equilibrium. However, a stronger acid will shift the equilibrium towards protonation.
And haloacids are best suited.

Conclusion being that a weaker acid will render nucleophile incapable to cleave ether in case of OMe.


What is the effect of conjugation of a carbonyl group in IR spectroscopy? Doing an exchange term so I've missed some classes that the others have attended, and unfortunately we're getting contradictionary explanations, so I thought I'd try and ask you.

Would be extremely thankful if someone could explain the effect of conjugation of a carbonyl group on the resonance frequency in IR spectroscopy.

Our 'assignment' is basically to figure out if it's supposed to be lower or higher than an unconjugated carbonyl.

Thanks in advance!

It would shift the IR peak to lower wavenumbers. You can rationalise this by understanding the bonding in the carbonyl group; in essence, by conjugating the carbonyl, the $\ce{C=O}$ bond is weakened, which shifts the IR peak down. You can think of it in terms of, if there is little or no conjugation, the electrons are firmly based in the $\ce{C=O}$ bond. As soon as we introduce some resonance structure, or conjugation, now those electrons can delocalise over 3-4 atoms or even more. This weakens the $\ce{C=O}$ bond, moving the IR peak down to lower wavenumbers.


How can I synthesize phenylpiracetam from styrene or benzaldehyde? I am trying to design a proper synthesis route for phenylpiracetam.

I am thinking a good starting material would be styrene or benzaldehyde. Here is my proposed route.

Is there a better method? Please identify any mistakes I may have made in my synthesis and suggest improvements, even if you have a different synthesis route in mind.

This is a reasonably sound route. I have some observations:
The prep of nitrostyrene from benzaldehyde is well known, it is best to use freshly distilled benzaldehyde. Here are the conditions from org syn prep.
The sidechain can be added by treatment of the phenylpyrrolidine with $\ce{NaOMe/MeOH}$/Methyl chloroacetate then heating with $\ce{NH3/MeOH}$
(patent).
You are better off reducing the nitro group at the diester stage as it will spontaneously cyclise to the pyrrolidine. On hydrolysis it will then decarboxylate. In my experience, hydrogenation of alkyl nitro compounds does not give good yields. $\ce{Al/Hg}$ amalgam is better. Sodium hypophosphite /phosphinic acid with cat. 5% Pd/C in water/THF and ultrasound was recently reported here as an efficient method for reduction of alkyl nitro, and the paper contains an example of spontaneous cyclisation of the product amine onto a methyl ester.
A synthesis of phenylpiracetam is described here in this patent.


Why is the rate of nitration greater for thiophene than toluene? Toluene has a $\ce{CH3}$ group attached to the benzene ring and as a result due to the +I effect of $\ce{CH3}$, the electron density in the ring increases and it undergoes nitration faster than a normal benzene ring.
But what causes thiophene to have an even greater rate of nitration ?

I would suggest two reasons:

Thiophene is an electron-rich compound because of presence of two lone pairs on the sulphur atom. Hence, it attracts electrophiles readily.
Though its aromaticity is lesser than that of benzene, the delocalisation is among five atoms in thiophene whereas in benzene it is among six atoms. Hence electron density in thiophene is more.


Anions produce flame colour? I know that flame tests can be used to distinguish between some metal ions, and that the colours come from excited electrons returning to the ground state.  My question is, why don't non-metal anions also produce colours?  Are their emission spectra in the non-visible range?  Is there a simple reason why?

There are two reasons, depending on the anion of interest:

1) For the flame emission of any species, anion or cation, metal or not, to be observable, it must emit light in the visible wavelength range, and most non-metal anions do not do this very strongly.

2) Even if there is some emission from the non-metal anion, in many cases their flame excitation efficiency is very low.

The answer then is the combination of these two things, and the fact that many metals have the opposite tendencies (e.g. high excitation efficiency, strong emission in the visible region of the spectrum), which overrides any weak visible emission from the non-metal anions.

As noted in a comment, this is not always the case, but it is the explanation for this strong trend.


Does glycerin promote bacterial growth the same as water? A sodium chloride + water solution draws the water out from bacteria through the process of osmosis.

Would a sodium chloride + glycerin solution have the same antibacterial effect?

If I understand correctly if the concentration of water is greater on the outside of a bacterial cell the water will go inside the bacteria allowing it to multiply or grow faster. Does glycerin have the same effect?

If I understand correctly if the concentration of water is greater on the outside of a bacterial cell the water will go inside the bacteria allowing it to multiply or grow faster. Does glycerin have the same effect?

Usually the point of adding salt to water is to lower its activity, in order to make the water concentration outside the cell lower than on the inside. This leads to water wanting to escape the cell, causing osmotic pressure to swell the cell.  With strong enough salt, the osmotic pressure causes water to flow out of the cell, so much that the cells dry up and aren't able to function.

Water is miscible with glycerol, which makes glycerol an osmolyte.  That means that glycerol would have the same effect as salt.  Water mixed with glycerol is lower in concentration than pure water.  Putting a happy aqueous bacterium into 100% glycerol will cause immense osmotic shock, probably resulting in death for most bacteria, due to too much of their intracellular water trying to escape to the extracellular, water-free (but glycerol rich) environment.

The downside of using glycerol is that at low concentrations (say 10 g/L or even 100 g/L) it is a great food for many types of microorganisms.  Sure at >1000 g/L it is a deadly osmolyte, but at low concentrations, its a yummy food.  Salt would merely lose its effectiveness as it gets diluted, but glycerol on dilution becomes a potent agent that would enhance microbial growth.  For this reasons, and others (costs, viscosity) it is not often used (outside of laboratory settings)


Double alkylation of benzene to form tetralin I have been asked to suggest a synthesis for the following compound (tetralin) from benzene:

I am unsure how to do this. It surely can't be Friedel–Crafts alkylation because of carbocation rearrangement and polyalkylation, right? Possibly a double acylation followed by a double reduction? Is there an issue with this; maybe a competing aldol type reaction in the double acyl chloride reagent that would be required.

You've got it right, double acylation and double reduction - but the first reduction occurs after the first acylation.  FWIW, the name of the reaction is the Haworth reaction and is commonly used to synthesize substituted tetralins, naphthalenes and higher aromatic polycyclics.

In your particular case, here is a diagram of the first 3 steps in the reaction

Friedel-Crafts acylation of benzene with succinic anhydride
reduction of the ketone using a Wolff-Kishner or Clemmensen reduction
a second Friedel-Crafts acylation using strong acid; it is an internal cyclization and forms the needed ring
and finally (not shown in the above diagram) a second Wolff-Kishner or Clemmensen reduction of the resultant tetralone to produce the desired tetralin


Create Sodium Hydroxide from Sodium Silicate During a reaction I use sodium hydroxide and create sodium silicate as a by-product.

I am curious if there is a way to recycle that sodium silicate into sodium hydroxide?

I am not a chemist, just have to use chemistry and I love it.

My best guess is to use Hydrogen Peroxide and Heat.

or Hydrogen gas.

or maybe even Aqua Ammonia...

For any reaction I have abundant electricity available, to create heat.

I have thought about this for months and stuck in the do-nothing phase until I can get some help from experts and see the reaction on paper.

Any help would be greatly appreciated.

The reactions you proposed are pure fantasy. They will never happen. Na2SiO3 does not react with either H2O2 or Hydrogen gas or CH2 (which does not exist, by the way).

The only possibility would be to have silica precipitate by bubbling CO2 in the solution. This will produce a precipitate of Si(OH)4. But this precipitate is so gelatinous that it cannot be easily filtrated. If you are able to filtrate (under pressure), the remaining clear solution contains Na2CO3.
$$\ce{Na2SiO3 +CO2 + 2 H2O  ->  Na2CO3 + Si(OH)4}$$

which could be transformed into NaOH by adding calcium hydroxide (or chalk) Ca(OH)2. This addition creates a new precipitate, CaCO3, which can be filtrated, and the residual solution contains NaOH.
$$\ce{Na2CO3 + Ca(OH)2 -> CaCO3 + 2NaOH}$$


Rutherford's Alpha Ray Scattering Experiment I understood the result of this experiment that the nucleus is nearly empty and things like that.
 But what I have on mind is that when an alpha particle goes nearer to the thin gold foil why couldn't it capture electrons from gold and convert to very stable helium gas? It could also happen right?
All the textbooks only mention the deviation of alpha particles. Why couldn't the above happen or is there anything wrong about my question?

Your assumption is correct. For alpha particles, the main contribution to the total stopping power can be attributed to the electronic stopping power, i.e. inelastic collisions with electrons. Only a small contribution comes from the nuclear stopping power, i.e. elastic Coulomb collisions in which recoil energy is imparted to atoms.

The stopping power of a material is defined as the average energy loss per path length that the alpha particle suffers when travelling through the material.

According to the International Commission on Radiation Units and Measurements (ICRU) Report 49 Stopping Powers and Ranges for Protons and Alpha Particles (1993), the contributions to the total stopping power for alpha particles in gold are as follows.

Typical low-energy alpha particles with $E=1\ \mathrm{MeV}$:

Electronic stopping power: $3.887\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $8.394\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Typical high-energy alpha particles with $E=10\ \mathrm{MeV}$:

Electronic stopping power: $1.650\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $1.315\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Since the energy that is required for excitation or ionization is only a few $\mathrm{eV}$, an alpha particle with an initial energy of a few $\mathrm{MeV}$ can liberate many electrons on its path. When the alpha particle has sufficiently slowed down due to the stopping interactions, it can finally catch two electrons to form a neutral helium atom. Hence, you can release and detect helium when you dissolve old minerals of uranium or thorium.

In the Rutherford experiment, however, the used gold foil was very thin ($<1\ \mathrm{\mu m}$; note that gold is very malleable and can be beaten thin enough to become semi-transparent).
Even for low-energy alpha particles of only $E=1\ \mathrm{MeV}$, the range in gold is still $r/\rho=3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}$, which can be calculated using the continuous-slowing-down approximation (CSDA), i.e. by integrating the reciprocal of the total stopping power with respect to energy. Considering a density for gold of $\rho=19.3\ \mathrm{g\ cm^{-3}}$, the range $r$ can be calculated as

$$\begin{align}
r&=\frac{3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}}{19.3\ \mathrm{g\ cm^{-3}}}\\[6pt]
&=2.06\times10^{-4}\ \mathrm{cm}\\[6pt]
&=2.06\ \mathrm{\mu m}
\end{align}$$

I.e., the experiment was deliberately designed so that the alpha particles were not completely stopped and converted to neutral helium atoms within the gold foil.


A question about acidity Since everyone knows an acid would dissociate(partially or fully) in its aqueous solution and liberate $\ce{H+}$. We also calculate the $\pu{pH}$ of the solution using $[\ce{H+}]$ and this$(\ce{H+})$ is what gives the solution acidity. Now consider we have just got a strong acid i.e pure acid without having any water in it, for example, $\ce{H2SO4}$ and no water in it, would the sulphuric acid dissociate in this case i.e without the aid of water? If it does not dissociate would this substance be acidic, since there would be no $\ce{H+}$ in that case?

Not really. Like lots of things with polar bonds, sulfuric acid can undergo ion transfer reactions in the liquid and make ionic species in solution.  And it is indeed protons that get transferred just like water.  But as with water, at least under ordinary conditions, these ionic species are only a minor portion of the liquid which is mostly $\ce{H2SO4}$ molecules.

Strong acids such as sulfuric acid are indeed acidic, capable of acting as acids.  But they still need a base with which they can react, to actually act as acids.  In the case of diluting sulfuric acid, water serves as such a base.


Deriving the equation for reaction spontaneity I'm trying to make sense of the derivation of the equation $$\Delta G = \Delta G^0_r + RT\ln(Q)$$

While searching for a derivation of this equation that made sense to me I came across a publication in the Journal of Chemical Education that suggests it is incorrectly applied if not completely wrong:

"Following this development, the paper investigates the source of the mathematical
and logical errors contained in textbooks ca. 1950 to the present that led to the
fallacious statements still present in most introductory chemistry textbooks and some
more advanced texts that the conditions for spontaneity are ΔG < 0 at constant T and
p and ΔA < 0 at constant T and V, whereas the corresponding conditions for
equilibrium are ΔG = 0 or ΔA = 0. This investigation shows the principal errors to be
(i) incorrect evaluation of definite integrals; (ii) failure to determine whether the
results of such integrations produce criteria for spontaneity and equilibrium that are
necessary conditions, sufficient conditions, both, or neither; and (iii) incorrect logical arguments related to equilibrium. (Spontaneity and Equilibrium III: A History of Misinformation)"

https://pubs.acs.org/doi/full/10.1021/ed500253e

Also, by the same author: https://pubs.acs.org/doi/10.1021/ed400453s

Here is the standard derivation:

For a given chemical reaction, let $v_i$ be the signed stoichiometric coefficient of reagent $i$ (That is, $v_i$ is negative for reactants and positive for products, with $|v_i| = $ the stoichiometric coefficient of reagent $i$).

Define the extent of the reaction, $\xi,$ as: $$\xi := \frac{n_i - n_{i,o}}{v_i} =  \frac{\Delta n_i}{v_i}$$

where $n_i$ and $n_{i,o}$ are the instantaneous number of moles and initial number of moles, (of reagent $i$) respectively. This quantity will have the same value regardless of which reagent is chosen to compute it and is therefore well-defined. It is effectively a measure of the reaction coordinate.

Note that: $$n_i = n_{i,o} + v_i\xi \to dn_i = v_i d \xi $$

$ \\ $
 $$dG = Vdp - SdT + \sum_i \mu_i dn_i \to \left(dG\right)_{P,T} = \sum_i \mu_i dn_i = \sum_i \mu_i v_i d \xi$$

In particular, $$\left( dG/d \xi \right)_{P,T} = \sum_i \mu_iv_i = \sum_i \left(\mu^o_i + RT\ln(a_i)\right)v_i = \sum_i \mu^o_iv_i + RT \sum_i \ln(a_i)v_i = \Delta G^o_r + RT \sum_i \ln(a_i^{v_i}) = \Delta G^o_r + RT \ln\left(\prod_i a_i^{v_i}\right) = \Delta G^o_r + RT \ln(Q)$$

In particular, $$\left( dG/d \xi \right)_{P,T} = \Delta G^o_r + RT \ln(Q)$$

I assert this is where it should stop. It tells you everything you need to know about the direction of spontaneity and the condition for equilibrium. For instance, the condition for equilibrium is that the free energy of the system is at a (local) minimum. To find the minimum of a differentiable function, you set its derivative equal to zero. Thus at equilibrium you have $Q = K$ and $dG/d \xi = 0$. Plugging this in gives $$K = e^{- \Delta G^o_r / RT}$$What I don't understand is how the left side gets equated with $\Delta G$. The left side is $\sum_i \mu_i v_i$. This is not the "instantaneous difference in free energy between the reactants and products". That would be $\sum_i \mu_i n_i$ (assuming such a notion even makes physical sense). I suppose you could call it $\Delta G_r$, but that perspective is confusing and would have no physical significance in the context of this equation (as you are viewing the reagents' concentrations as the variable that is moving as opposed to pressure or temperature).

I'm wondering if I am conceptually missing something?

The method you described for deriving the desired relationship is one of two that I am aware of.  In my judgment, it is important to mention that this approach looks at a reaction in progress,  determining the free energy of the mixture of reactants and products as a function of the reaction conversion.  As such, it makes the tacit assumption that it is valid for this irreversible process of a reaction occurring spontaneously to determine the free energy at each conversion as if one is considering a thermodynamic equilibrium state of an unreacting mixture.  In my judgment, this is an excellent approximation, but some purists might argue that such an approach is not valid.

The other approach I am aware of is illustrated for the case of a reaction involving an ideal gas reaction.  It evaluates the change in Gibbs free energy between the following pair of initial and final thermodynamic equilibrium states:

Stoichiometric proportions of pure reactants in separate  containers at temperature T and 1 bar pressure
Corresponding stoichiometric proportions of pure products in separate containers at temperature T and 1 bar pressure

The process for evaluating the change from state 1 to state 2 is reversible, and involves successive steps involving isothermal gas expansions and compressions within cylinders in series with a van't Hopf equilibrium box.  The net result is the standard free energy change of the reaction $\Delta G^0$.

If one then modifies the process slightly, by starting with pure gases in separate containers at pressures different from 1 bar, this merely changes the compression and expansion steps in the reversible process, and results in the value of the Gibbs free energy change altering from $\Delta G^0$ to $$\Delta G=\Delta G^0+RT\ln{Q}$$Moreover, if the initial and final pressures in the separate containers are chosen to exactly match the equilibrium partial pressure of the various species in the equilibrium box, then the compression and expansion steps are not necessary, and the $\Delta G$ for this process is zero.  Under these circumstances the reaction parameter Q is equal to the equilibrium constant:  $$0=\Delta G^0+RT\ln{K}$$


Rates of solvolysis of p-dimethylamino benzyl halides What will be the order of $\mathrm{S_N1}$ solvolysis rates for these?

I'm confused between the second and third options. $\ce{I-}$ is a better leaving group than $\ce{Br-}$, but the carbocation formed in 2 is more stable than 3. So how are we supposed to decide between them?

In the second compound, steric repulsions (denoted by the curly lines) between the N-methyl groups and the methyl groups on the ring occur:

This forces the $\ce{-NMe2}$ group to rotate such that the nitrogen lone pair is no longer parallel to the ring's $\pi$ system. As a result, the nitrogen lone pair can no longer contribute to the mesomeric stabilisation of the second carbocation.

The rate of solvolysis therefore increases in the order: $2 < 1 < 3$. The effect described above is called "steric inhibition of resonance".

A similar effect is observed in azo coupling:

In the second reaction, the nitrogen lone pair is not delocalised into the ring and the ring is therefore not nucleophilic enough to attack the diazonium ion.


Which chemical should I use to wash soot precipitate from cast iron? The soot deposition is quite strong (almost like graphite) and my iron object has a complicated shape with many grooves.
Which chemical would dissolve the carbon without oxidizing the iron?

Depends on your resources and how delicate the object is (is it an antique?).

You can put it in a furnace or self-cleaning oven to burn-off the soot. Determining the correct temperature for this will be an art.
You could put it in a furnace under a hydrogen atmosphere.  the hydrogen will react with the carbon to form hydrocarbon gases and it definitely won't oxidize the iron.  You can also use carbon dioxide which will for carbon monoxide with the soot.  You may not have this sort of equipment though.
You can clean the object in an ultrasonicating bath. These are widely available and the best part is that you will not require chemicals to remove the soot.  This method may take a while but it is the most gentle.


what keeps an identical atoms of an element together? I have learned that the atoms of different elements are held together by covalent bonding and ionic bonding but in the case same element which force keeps the atoms together i.e in  sodium which is the force that keep sodium atoms in position?which is the force acting between the atoms of sodium?

well atoms of the same element will typically either, forma metallic lattice, a giant covalent structure ( a macro-molecule), or from a simple molecular structure. but all the bonds will either be metallic or covalent bonds. if it forms a simple covalent structure, there will be van der Waal's forces acting between the molecules to keep it as a solid or liquid.


How can silver work as a disinfectant in a bathtub if precipitates in the presence of chloride? I use a silver based disinfectant in my hot tub. It is in the form of silver nitrate beads. There is chlorine in the tub that I presume becomes chloride ion. Why doesn't the silver form a precipitate of silver chloride in the water?

If you're referring to chlorination of public water supplies, keep in mind that this is likely not the main source of chloride ion. Elemental chlorine disproportionates via the following reaction:

$\ce{Cl2 + H2O <=> HCl + HOCl}$

Most chloride ion probably derives from various natural and industrial sources, as well as miscellaneous human activity. One source (somewhat outdated) I found indicates an average chloride ion concentration in U.S. public water of $11.5$ mg/L, which equates to a molarity of $3.24 \cdot 10^{-4}$. Using the $K_{sp}$ of $AgCl$ you can calculate the concentration of silver ion necessary to produce a precipitate, but given that it has a value of $1.77 \cdot 10^{-10}$ under standard conditions, you can assume that a fairly minimal amount of silver would be required, even with the large volume of a tub and increasing solubility in hot water. It's likely the case that any precipitate is just finely dispersed in solution and not clearly visible.


How can the concentration of a solid be constant during a reaction? The molar concentration is given by the ratio between density and molar mass. I understand the proof.
But how does this remain true during a reaction?

For example consider the reaction,

$$\ce{4Al(s) + 3O2 (g) -> 2Al2O3(s)}$$

If the amount of Al and O2 moles are equal, according to stoichiometry Al ought to be the reactant to end first. But, being a solid, Al’s concentration must be constant.
How can this be? What is wrong in my reasoning?

**Please note that I mistakenly assumed that the amount of Al and O2 are equal, without stating it, leading to the confusion that Al ends first at all times, I have edited the question appropriately

In physical chemistry one does not use the concentration of a chemical but its activity (see https://en.m.wikipedia.org/wiki/Thermodynamic_activity)

In a solid the standard condition is the pure chemical. In case of the solid block of aluminum it is therefore in its standard condition which leads to the fact that the activity is 1 (the chemical potential is equal to the chemical potential in the standard condition as the block of aluminium is in its standard condition - see the link I provided for the definition of the activity).

Therfore it doesn't need to be considered in calculations like the law of mass action as multiplying or diving by 1 does not affect the result.

However in "real chemistry" one does not calculate with activities too often and uses concentrations instead and in order to apply the same trick one simply says that the concentration is constant.

So in the end it's just a result of sloppiness because one doesn't want to work with activities.


Why are they called ‘dispersion forces’ if they are an attractice force? From the name, I thought dispersion forces would be repelling or dispersive forces between atoms, and not attractive forces. What is the dispersion part of the name actually referring to? Is it the electrons moving around?

Although Danny Rodriguez has already excellently exposed what the dispersion force is in simple terms, the word dispersion still demands a better explanation in my opinion.

According to Wikipedia:

  The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase "dispersion effect."
  In physics, the term "dispersion" describes the variation of a quantity with frequency, which is the fluctuation of the electrons in the case of the London dispersion.

So the word "dispersion" here is a mere analogy.
The original work is this one (in German).


Which textbook is this image on protein tertiary structure from? I am trying to find the original textbook image that is the inspiration for many schematic figures showing tertiary interactions (hydrogen bonding, ionic interactions, hydrophobic interactions, disulfide bridges). The earliest I found was in a blog, without source attribution, and is reproduced here:

It must be from Chapter 8 of some textbook. The same blog shows other figures and tables from the same textbook, I think, for example:

The quantitative nature of the table points at a upper-level text (rather than the last chapter of organic chemistry, or a biology introductory text). The black and white figure (intended as such as judged by the rastering of the gray shade) suggests 1960s or early 1970s. There is a picture of the structure of hemoglobin in another figure (a schematic of a wood slab layer model, and a low resolution model next to it), so it cannot be written earlier than 1959.
I don't have access to a library with old textbooks at the moment, so I am asking here.
I did find a newer colorized version (with at least one mistake) here:

The illustration is likely taken from a later edition of Baum's Introduction to organic and biological chemistry.
The first edition [1, p. 292] presents nearly identical illustration under a different number alongside with a note that the image was adapted from Bennett's Graphic Biochemistry [2]:

Unfortunately, at the moment (2020-10-16) [1] is available on The Internet Archive only for a limited amount of time for borrowing, and I was unable to find a full-text source for [2].
References

Stuart J. Baum. Introduction to Organic and Biological Chemistry, 1st ed.; Macmillan: London, 1970.
Bennett T. P. Graphic Biochemistry. The Chemistry of Biological Molecules; Macmillan, 1968; Vol. 1.


Are these metamers, if not why? Please tell me if these are metamers ie Metamerism: In Chemistry the type of isomerism in which chemical compounds have identical proportions of the same elements and the same molecular weight, but have radicals differing in type or position, with resulting differences in chemical properties.

The term metamer isn't terminology that is widely used today, and indeed doesn't appear in the IUPAC Gold Book (the International Union of Pure and Applied Chemistry publishes guidelines for standardised nomenclature in chemistry and is widely adopted).

The description you're looking for is structural isomerism, with the compounds you show being examples of positional isomers (isomers which differ by the position of the functional groups, for instance in this case the relative positions of the methyl and methoxy moieties).


Are phosphomonoesterases solid? What state of matter are phosphomonoesterases?  I am not a biochemist; while I've found various lists of such enzymes, and plenty of (to me) barely comprehensible articles describing their production, it's not clear to me whether the final product in isolation is a solid, liquid, or gas.

More generally, is there some reference work in which one can look up the state of matter (at standard temperature and pressure) for a given chemical?

Enzymes are very, very large in comparison to the typical molecules you encounter in chemistry (excluding any polymers). A 10 kD protein is considered to be small, and there are proteins of several hundred kD.

If you remove all the water from a typical protein you end up with a solid. But for most enzymes this would be an irreversible step, and they would lose their activity.

Usually enzymes are kept in aqueous solution, often with some glycerol or other cryoprotectants, and frozen.

Strictly speaking, they would probably be solid, but likely inactive in that form so I'm not sure if it counts. The relevant form of an enzyme is dissolved in water.


Are the total number of electrons present in the central atom of triiodide equal to 10? My book states that

  the total number of electrons present in the central atom of $\ce{I_3^-}$ is equal to 10.

What if I put a negative charge on the central iodine atom such that it gains a noble configuration and then I say that it is coordinatively bonded with other two iodine atoms? Then the number of electrons which the central atom will have would be 8. Wouldn't it?

Am I wrong somewhere in my analogy?

Looking at the wikipedia article here, your explanation seems to be exactly right:

  The ion is linear and symmetrical. According to VSEPR theory, the central iodine atom has three equatorial lone pairs, and the terminal iodines are bonded axially in a linear fashion, due to the three lone pairs bonding to the central iodine-atom. In the molecular orbital model, a common explanation for the hypervalent bonding on the central iodine involves a three-center four-electron bond.

Thus, the central iodine atom has $3\cdot 2$ valence electrons from the lone pairs and 3 three-center four-electron bonds, so that I speculate you end up with $3\cdot 2$ VE + $4/3 \cdot 3$ VE = 6 VE + 4 VE = 10 VE. Which also fits to the reference given by you.


Compound having the highest enol content will be? The answer written in book is 3 but no solution is given.
After referring to some other question I think that hydrogen bonding can take place in compounds like 3 but am not sure about it. Am I right that it is the only reason why it has the highest enol content?

I would like to ask one more thing that, while comparing the keto and enol content, if in a compound more than one ketone is present, than do we enolize any one of them only to check?

If you can form an enol where the carbon-carbon double bond is conjugated to one or more other double bonds, the enol is more stable and the equilubrium mixture will therefore have more.  With that in mind, try seeing what happens to (3) if you tautomerize with the proton on the bridging carbon between the two carbonyl groups.

For a further exercise: render the structure for cyclohexa-2,4-dien-1-one where the carbonyl group is conjugated to two carbon-carbon double bonds in the ring.  Now tautomerize to the enol and see what "enol" you get.  When the conjugation of the enolic double bond forms an aromatic ring, the stabilization effect is so strong that the enol takes over completely.


Which of the following has higher enol content? Which of the following has higher enol content?

The enol form of (A) and (B) is:

Both the enols have hydrogen bonding.

If the oxygen were to accept the pi electrns from carbon forming a carbocation like this (a resonance structure):

the compound becomes stabilized due to resonance as well. So isn't (B) the answer?

The two main factors affecting the stability of the enol forms of (A) and (B) are

Hydrogen bonding
Aromaticity

In the enol form of both compounds, hydrogen bonding is present:

But the resonance structure of enol form of (B) is aromatic. It contains $6$ $\pi$-electrons:

The enol form of (A) lacks this stability. Hence, the enol content of (B) is more than that of (A)


How's the IUPAC name of basketane determined? A book of mine mentions a certain molecule, basketane, and provides its corresponding structure:

Now, I appreciate the book went on to mention this particular compound (I didn't know that organic molecules could be shaped like...well...a basket, so I find this interesting) but the problem is, that it didn't mention basketane's IUPAC nomenclature, much less how to go about naming it.

A quick internet search reveals the following IUPAC name for basketane:

  Pentacyclo[$4.4.0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane

But I'm still having trouble finding out how people even arrived at this name. Heck, I don't even know if the above mentioned name is even correct. I consulted all my Org. Chem books, and predictably, there was no mention of how to even begin naming basketane. I'm still in high school, so this particular kind of nomenclature is still alien to me (I haven't dealt with any super-scripts or carbons assigned the locant: zero yet).

Would anyone be able to:

1) Validate that name I've come across

2) Explain (citing a decent reference will do as well) how I ought to go about even naming this thing.

EDIT- @DHMO has been kind enough to point out two possible duplicates in the comment section below; but I'd still beg to differ.

Validation
Let us build this alien step by step:

We will build bicyclo[$4$.$4$.$0$]decane.
We will build tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane.
We will build tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane.
We will build pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane.

Bicyclo[$4$.$4$.$0$]decane looks like this:

The part "decane" means that there are $10$ carbons and the molecule is saturated, just as indicated.
The part "bicyclo" means that there are two rings. One ring is $1$-$2$-$3$-$4$-$5$-$6$ and the other is $1$-$6$-$7$-$8$-$9$-$10$.
The part [$4$.$4$.$0$] is based on $\ce{C}1$ and $\ce{C}6$:

The first $4$ refers to the first link between $\ce{C}1$ and $\ce{C}6$, namely $2$-$3$-$4$-$5$ which contains $4$ carbons.
The second $4$ refers to the second link between $\ce{C}1$ and $\ce{C}6$, namely $7$-$8$-$9$-$10$ which contains $4$ carbons
The $0$ refers to the third link between $\ce{C}1$ and $\ce{C}6$ which is just a line with $0$ carbons between.

Tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane looks like this:

It is now "tricyclo" because there are three rings: $1$-$2$-$5$-$6$, $2$-$3$-$4$-$5$, $1$-$6$-$7$-$8$-$9$-$10$.
The extra part $0^{2,5}$ means that $\ce{C}2$ and $\ce{C}5$ are linked by $0$ extra carbons.

Tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane looks like this:

(from now on, a strong intuition on topology is required)
The new part $\mathrm{0^{3,8}}$ refers to the new linkage between $\ce{C}3$ and $\ce{C}8$ with $0$ carbons between.

Finally, pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane:

Now you might ask: how on earth is that basketane? Let me show you how:

Not convinced? Here is the proof:

$1$ is connected to $2$, $6$, $10$.
$2$ is connected to $1$, $3$, $5$.
$3$ is connected to $2$, $4$, $8$.
$4$ is connected to $3$, $5$, $7$.
$5$ is connected to $2$, $4$, $6$.
$6$ is connected to $1$, $5$, $7$.
$7$ is connected to $4$, $6$, $8$.
$8$ is connected to $3$, $7$, $9$.
$9$ is connected to $8$, $10$.
$10$ is connected to $1$, $9$.

In the above two pictures, you can verify the connectivities, and they match exactly.

Derivation
Firstly, we try to find the largest ring that can contain the most carbon atoms. In this case, it is already indicated above (trace the ring $1$-$2$-$3$-$4$-$5$-$6$-$7$-$8$-$9$-$10$).
Then, you might ask, "How do we judge which numbering is correct?"
Curiously, the 1993 IUPAC Recommendations do not contain the criteria, but the 1979 IUPAC Recommendations do. Rule A-$32.31$, Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H. Pergamon Press, Oxford, 1979. Copyright 1979 IUPAC.:

$32.31$ - When there is a choice, the following criteria are considered in turn until a decision is made:
(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.
(b) The main bridge shall be as large as possible.
(c) The main ring shall be divided as symmetrically as possible by the main bridge.
(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

Let us look at the first rule:

(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.

This means that we should maximize the sum of the first two numbers and minimize the sum of the other numbers.
Also, this means that we need to choose two carbons which are each connected to three carbon atoms.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{2}$.$\color{green}{2}$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]
This name is incorrect because the first two numbers, $\color{red}{4}$ and $\color{orange}{2}$, add up to $6$, which is smaller than $8$ as derived from the correct name.

Let us look at the second rule:

(b) The main bridge shall be as large as possible.

The "main bridge" refers to the third number, which is to be maximized.
This is not applicable here, as every number beyond the third is zero.

Let us look at the third rule:

(c) The main ring shall be divided as symmetrically as possible by the main bridge.

This means, that although the first number is to be larger than the second, the difference between the first two numbers is to be minimized.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{6}$.$\color{orange}{2}$.$\color{green}{0}$.$\mathrm{0^{2,7}}$.$\mathrm{0^{3,10}}$.$\mathrm{0^{6,9}}$]
This name is incorrect because the first two numbers, $\color{red}{6}$ and $\color{orange}{2}$, have a difference of $4$, while the difference between the first two numbers in the correct name is $0$.

Let us look at the fourth rule:

(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

This is what A-$2.2$ says:

[...] When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. [...]

For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\mathrm{0^{2,9}}$.$\mathrm{0^{5,8}}$.$\mathrm{0^{7,10}}$]
Let us compare the locants in the incorrect version and the locants in the correct version:

Incorrect: $2,9,5,8,7,10$
Correct: $2,5,3,8,4,7$

The first difference is $9$ vs $5$, in which the correct version is smaller.

Conclusion
Therefore, we have arrived at the correct systemic IUPAC name for basketane:

Pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\color{purple}{\mathrm{0^{2,5}}}$.$\color{blue}{\mathrm{0^{3,8}}}$.$\color{pink}{\mathrm{0^{4,7}}}$]decane.


Why neglecting OH Ions from charge balance? I have a problem, in which the concentration of $\ce{Ca^{2+}}$ ions in water should be related to the partial pressure of $\ce{CO2}$ (assuming the water is an open system and in equilibrium with atmosphere).

All the approaches I saw so far, started by looking at the charge ballance of this system:

$$\ce{2 [Ca^{2+}] + [H+]} =\ce{2[CO3^{2-}] + [HCO3-] + [HO-]}$$

This is totally logical. But they continue with the assumption that the terms $\ce{[CO3^{2-}]}$,$\ce{[HO-]}$, $\ce{[H+]}$ are insignificantly small so that you can work with:

$$\ce{2 [Ca^{2+}]} =\ce{[HCO3-]}$$

That's the point I don't get.

Assuming at the beginning there is only $\ce{CaCO3}$ and Water. Both dissociate and let's say $\ce{CO3^{2-}}$ immediatly combines with $\ce{H+}$ to $\ce{HCO3-}$, then it is reasonable that the contribution of $\ce{H+}$ and $\ce{CO3^{2-}}$ can be ignored. But what about the remaining $\ce{OH-}$? Stupidly written, I would get something like:

$$\ce{H2O + CaCO3 } =\ce{ HCO3- + Ca^{2+} + OH-}$$

So why can I also neglect the charge caused by $\ce{OH-}$ Ions? For each bicarbonate/calcium-ion produced I would also get a hydronium ion. why can this still be ignored?

This is not totally unreasonable.

Calcium carbonate as solubility product of $K_{\mathrm{sp}} = 3.3\times 10^{-9}$ (source).

This means that if you're dissolving the solid, you'd expect a maximum calcium ion concentration of $\ce{[Ca^{2+}]} = 5.7\times 10^{-5}\ \mathrm{M}$. This is a 2.5 orders of magnitude larger than either hydronium or hydroxide concentration in neutral solution. You'd expect the carbonic acid to be mostly dissociated (at least 50%), so the bicarbonate concentration will be similar to the calcium ion concentration (maybe factor of 2 smaller).

The moment you step away from a neutral solution though, this analysis totally falls apart. You'd need more details on the assumptions of the model in that case.


Do electronic balance weighings still need to be corrected for buoyancy error? Buoyancy error is a weighing error due to the difference in the buoyant force exerted by a medium (commonly air) on the object and on the standard masses. The difference in buoyant force occurs when the object and the standard masses have standard masses (Skoog et al., 2014).

Knowing this, I am confused as to whether electronic macroanalytical balances need to be corrected for buoyancy error because of the following factors:

Electronic macroanalytical balances still get calibrated by weighing standard masses.
Closing the doors of the balance does not make the system inside of it evacuated. It still has air which can exert buoyant force on the object and on the standard masses.

Given the factors I mentioned, do electronic balance weighings stil need to be corrected for buoyancy error?

References:

Fundamentals of Analytical Chemistry, Skoog, et al., 9e

A balance cannot determine the density of the weighted material and air automatically. For this reason a buoyancy error correction must be applied for high precision measurements.

Suppose you weigh $\pu{100.0000 g}$ water and have $\pu{100.0000 g}$ reference weights with a density of $\pu{8.0000 g cm^{-3}}$. Then the static buoyancy in air is equivalent to about $\pu{120 mg}$ and $\pu{15 mg}$ respectively, accounting for a difference of about $\pu{105 mg}$ in the reading of the balance.

Besides the static buoyancy there may also be dynamic buoyancy due to air flow caused by temperature differences between the weighted material and the surrounding air.

Dynamic buoyancy can be avoided be weighing in thermal equilibrium. Static buoyancy can be corrected by formulae involving the density of air, the density of the weighted material and the density of the reference weights.


Why are partial molar Gibbs energy and Gibbs energy per mol equal? I'm not a chemist, but I need some insight into chemical thermodynamics.

In the book that I'm reading (Nevers, 2012), equation (4.28) relates the chemical potential of a pure species to it's Gibbs energy per mol:

$\qquad\mu_a^{(1)} = \bigg( \dfrac{\partial G}{\partial n_a} \bigg)_{T,P,n_b...} = g_a^{(1)}$

where the subscript stands for the species $a$, the superscript stands for it's phase, $\mu$ is the chemical potential and $g$ is the Gibbs energy per mol.

I understand the first equal sign. Though, I don't understand the second equal sign. What am I missing?

The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as
$$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$
where N is the number of particles and g is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then G will also double.

From the above expression you can see that the chemical potential is just the Gibb's free energy per particle:
$$\mu\left(T,P\right) = g\left(T,P\right)$$


Aspartic acid conjugate base acidity Question

Which hydrogen among the four in the conjugate base of aspartic acid is most acidic?

Answer

 No answer given

My attempt
I am confused between Hb and Hc
I know Hd has $\mathrm pK_\mathrm a \approx  9$, so it will not be the most acidic.
$\ce{NH3+}$ has very high -I effect but no mesomeric effect, so Hc appears to be quite acidic.
But at the same time it is at the alpha position of the carboxylic acid and may not be effectively conjugated in the $\ce{C=O}$ bond.
Here can the electron pair if formed by loss of Hc be conjugated on the $\ce{C=O}$ bond? This will prevent the lone pair on oxygen from being delocalized, however, something that is not characteristic of carboxylic acids.
Hb may be acidic but no special effect may be attributed to it, similar to Ha.
I think the order will be Hb > Hc > Ha > Hd
Is my assumption correct?

A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion.
A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point.

Let's consider this molecule to have reached the isoelectronic point. We know that the value would be less than $7$ since there are two acidic groups (meaning that this is an acidic amino acid). When we reach that point, we see that the both the carboxyl groups are de-protonated and the amine group still exists as $\ce{-NH3+}$.
This means that the $\ce{-NH3+}$ group is less acidic than the the acid groups. Therefore, Hd is is less acidic than Ha or Hb.
As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before Hc is removed. This means that Hd is more acidic than Hc.
Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to Hb than Ha, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of Hb would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of Ha.
Therefore the order of stability would be:

Hb > Ha > Hd > Hc


Does mercury have an NFPA 704 classification? I've searched on a lot of websites if elemental mercury (Hg) has an NFPA classification but I can't find any.

Does mercury have an NFPA classification?

According to Sigma-Aldrich

Some symbols:

GHS06: Acute toxicity (oral, dermal, inhalation), categories 1,2,3
GHS08: Respiratory sensitization, category 1; Germ cell mutagenicity, categories 1A,1B,2; Carcinogenicity, categories 1A,1B,2; Reproductive toxicity, categories 1A,1B,2; Specific Target Organ Toxicity – Single exposure, categories 1,2; Specific Target Organ Toxicity – Repeated exposure, categories 1,2; Aspiration Hazard, category 1
GHS09: Hazardous to the aquatic environment; Acute hazard, category1; Chronic hazard, categories 1,2

And according to ScienceLab:

Health: 3
Fire: 0
Reactivity: 0

I believe the ScienceLab information is the 'NFPA 704 classification'.

EDIT****

An email exchange between NIST and myself asking for clarifiation:

My Email (Pre-filled form on the website, I don't have the original):

Something along the lines of "Hello, I see NIST and a few other manufactures
of have different NFPA classifications (I provided links to the SDS's),
could you explain the difference between your ratings and theirs?"

Their response:

Good afternoon,
Our NFPA rating Health = 2 Fire = 0 Reactivity = 0 matches other manufacturers.
You can check Aldrich website for 99.99 Mercury
http://www.sigmaaldrich.com/catalog/product/sigald/215457?lang=en&region=US

Thank you,
Office of Reference Materials

Other manufacturers (Note, plural) apparently now equals 1 example


How do you figure out the positional and chain isomer possibilities of a compound? I need to figure out the chain and positional isomers of C4H7Cl. I am not looking for someone to directly give me the answer, I just want to know what steps are needed to be able to figure them out for a given compound.

This is different to my prior question. Here I am asking for the technique, not the direct answer. Thanks.

Is the parent hydrocarbon unbranched? That would be n-butane, $\ce{C4H10}$. Replacing one hydrogen atom by chlorine would lead to $\ce{C4H9Cl}$. Apparently, you're not having that!
Move a methyl group in n-butane, yielding 2-methylpropane, $\ce{C4H9}$. Here, replacing one hydrogen atom by chlorine gives $\ce{C4H8Cl}$. You're not having that either!
Does your target molecule have $\ce{C=C} $double bonds? Calculate the number of double bond equivalents in $\ce{C4H7Cl}$! I count one.

Work your way down from here.


Claisen ester condensation Why is it a requirement that the ester should have 2 alpha hydrogens in this reaction although we actually need just one for the reaction to occur?

Great question!

The Claisen reaction happens under equilibrium conditions, and the addition of the ester enolate is highly reversible. Your mechanism omits the very important end of this reaction, which is that the β-keto ester product is highly acidic and easily deprotonated to yield the corresponding enolate.

The enolate formation drives the reaction forward according to Le Châtelier's principle. Very importantly, your question about the α-hydrogen. The deprotonation of the β-keto ester is not possible without two hydrogens on the original ester that will serve as the nucleophile.


Freezing trend of water In this experiment, I used a digital temperature probe to measure the freezing point of water and some other salt solutions.

I was just wondering if there was any explanations for 1. The first plateau and 2. The dip in temperature right before the freezing point. (2.) can be observed for other salt solutions.

I think that the dip before the freezing point is caused by the formation of ice needing a 'nucleation point', which requires the temperature to be much lower such that a regular structure can be formed. I have no good explanation for 1. but it could be due to uneven cooling or some sort of issue with the probe.

Please give some suggestions that explains 1. and 2. if possible, even if you are unsure. Thank you for the help!

First, it looks like you supercooled water. This at least explains the dips in temperature below zero degrees Celsius. It is not difficult to supercool water up to a few degrees below the normal freezing point. It remains metastable provided you avoid perturbing it, say by shaking. See for instance: http://www1.lsbu.ac.uk/water/supercooled_water.html. And yes, availability of nucleation points (small crystals) accelerates freezing.

The plateau observed around $\pu{2^\circ C}$ may be due to density differentials across the sample. At about $\pu{4^\circ C}$ water reaches its maximum density. Above this temperature, the colder water at the surface sinks, creating convection currents and accelerating the cooling process. As you dip below about $\pu{4^\circ C}$ the convective currents cease (around 3000 s), since the density gradient is stable: colder, less dense water is at the top. The water at the top of the sample continues to cool and decrease in density without sinking. If the thermometer is placed away from the air interface, it should at this point measure a slowly changing temperature as the temperature gradient toward the interface builds up. The temperature behavior will be a response to heat diffusion instead of heat convection.

You may gain additional information if you add a little dye near the thermometer and attempt to follow its movement, either before or after the convective behavior is presumed to have stopped. Convection should result in mass flow, mixing the dye more rapidly. In the diffusion regime dye should disperse more slowly and locally.

This discussion makes various assumptions of course with regard to the purity of the water and the geometry and insulation of the container it's in. It also assumes that the water has not frozen at all before about 4800 s. It also makes assumptions about the location of the thermometer.


Inhibitors and poisons in a reaction What are the basic differences between inhibitors and poisons which are used in a reaction
Actually both the names confuse me so please help me out.

My understanding of catalyst poisons and inhibitors is that:

Inhibitors curb the catalyzing powers of a catalyst but do not react with any of the reactants and the reaction is reversible.
Poisons react with the catalyst and/or the reactants and the reaction is irreversible.


Homemade diode working mechanism Hello,
I will list the questions directly, if you want to know details please read below.
As in the figure, we have water+baking soda in a plastic cup, there are 2 versions in figure that affects LED's brightness and current that flows in circuit.

In 1st version of figure, LED is bright and more current flows
(measured 9.76 mA with multimeter)
In 2nd version of figure, LED is
dull and less current flows (measured 0.22 mA with multimeter)

So my question is, why changing aluminium's place from negative to positive affects current flow of circuit? is it related with aluminium's chemical properties?
Details that explain how i got to this question:
My main aim was to make a home made diode by following instructions in this wonderful site, i did one and it worked.To understand the working mechanism better, i decided to setup things as in the figure upper and faced with this question that looks like main working logic of this home-made diode.In site there was an explanation but i couldn't understand it.

When aluminum is the anode (connected to the positive battery terminal), a thin layer of insulating aluminum oxide is produced via the “anodization”. It serves as a dielectric, so basically you have a leaky capacitor: current is low. But when the aluminum is connected to the negative terminal, it is the cathode and then you get reduction of hydrogen ions to hydrogen gas. So considerably more current can flow. This is essentially how tantalum and aluminum electrolytic capacitors have rectifying properties: their oxide films are the cause. Fun link, by the way.


Would it be correct to say that ozone is an element? My level of chemistry knowledge is restricted to that which is given at high school level.

I was tought that an element is a substance that consists of only one type of atoms, meaning the number of protons is the same for every atom. Is this definition sound?

In case it is sound:

Since ozone gas only consists of oxygen atoms, would it be correct to say that ozone gas, as an entity, is an element?

Would it be more correct to say that ozone is an allotrope of the element oxygen?

Why is it called the table of elements, when it seems that we are actually listing the atoms and not the elements themselves?

It would be more correct to describe ozone as an allotrope of oxygen. Dioxygen (about 1 in five of the molecules making up air) is another allotrope.

The periodic table describes the atoms that make up elements as this is the most fundamental property. Many other elements, though, exist in multiple allotropes. Carbon, for example, has several: graphite, diamond, buckminsterfullerene and quite a few others.

All allotropes consist of only one type of atom but the bulk physical properties depend on how the atoms are joined up (flat plates of carbon hexagons with multiple bonding in graphite, a three-dimensional network of singly-bonded tetrahedral carbons in diamond).

We don't normally add too much about allotropes in a periodic table as it would make the table a little big, although some include limited information about how many there are. And, usually, only one is thermodynamically stable under normal conditions (so the physical properties will be about that one).