How to determine if a base is bulky? I remember my teacher saying that during E2 (elimination bimolecular) reactions of alkanes, if the strong base that is used as a reagent is a bulky one, it has higher probability of giving a Hofmann (less stable) product as the major product even if there are possibilities for a Saytzeff (or Zaitsev) product to form.
How should I find if a base is bulky or not?
Some bases like t-butoxide ion are pretty obvious. I usually just guess it is based on how much steric-clouded it seems when I draw a representation on a paper. Is there any better way to figure this out?

This question about the effect of the bulkiness of bases on the mode of E2 elimination has been studied and was published in 1956 by Brown, et al.[1] The elimination of 2-bromo-2-methylbutane was examined using solutions of potassium ethoxide, potassium t-butoxide, potassium t-amyloxide and potassium t-heptoxide in their respective alcohols. The clear result was that increased branching of the alkoxide increased the yield of the Hofmann product over the Zaitsev product. The bulkier the base the less selective for attacking methylene hydrogens over methyl hydrogens. The numbers in parentheses are the selectivities corrected for the number of hydrogens available in each elimination.

H. C. Brown, I. Moritani and Y. Okamoto, J. Am. Chem. Soc., 1956, 78, 2193.


Pyrolysis of calcium chloride dihydrate Does the pyrolysis of $\ce{CaCl2 · 2 H2O}$ give $\ce{CaCl2 + 2 H2O}$ or $\ce{Ca(OH)2 + 2 HCl}?$

All calcium chloride hydrates give away water of crystallization upon heating without forming a hydroxide, and the anhydrous $\ce{CaCl2}$ melts without decomposition [1, p. 162]:

  anhydrous salts melts at 772 °C, while the mono-, di-, tetra- and hexahydrates decompose at 260 °C, 175 °C, 45.5 °C and 30 °C, respectively; the anhydrous salt vaporizes at 1935 °C.

$$\ce{CaCl2 · x H2O ->[\pu{260 °C}] CaCl2 + x H2O}$$

Calcium chloride dihydrate would remain intact until its melting/decomposition point at 175 °C.
When heated quickly in the absence of desiccant, the powdered salt is dissolved in its own water of crystallization forming glass-alike solid upon cooling.

References

Patnaik, P. Handbook of Inorganic Chemicals; McGraw-Hill handbooks; McGraw-Hill: New York, 2003. ISBN 978-0-07-049439-8.


R,S Configuration of asymmetric carbon atom when group priority 3 and 4 are not on wedge bond? I am not blessed with the gift of spatial awareness. I understand how to determine the configuration when a group with priority 4 is on the hashed wedge or when the group with priority 4 is on the bold wedge and priority 3 is on the hashed wedge. However are there any rules I can follow for situations as shown in the image? The numbers show the priority.

To do this, Put the lowest priority group as far away from you as possible.

In the 1 example, if I look at 4 from between 1 and 2 it would be farthest away from me. And i will see that the 3 group is on top, 2 on my left and 1 on my right.

And now if i move from highest priority group to lowest (1-2-3) it would be in clockwise sense. So the configuration is R.

2nd and 3rd example has S and R configuration respectively.


Inversion of Configuration - Opening of Epoxide This doesn't look like a traditional inversion of configuration:

At first I dismissed the textbook author as making a typo. But then I realized that this "typo" was repeated throughout the solutions manual.

So I tried drawing a Sawhorse projection of the textbook's answer and my answer, and I found that both are equivalent.

So would my answer also be acceptable? In my answer I change the methyl from a wedge to dashes and the hydrogen from dashes to wedge and I place the SH group appropriately.

It's slightly difficult for me to see clearly, but your drawing appears to be correct, and it corresponds to a $180^{\circ}$ rotation about the $\ce{C_1-C_2}$ bond axis relative to the textbook rendering ($\ce{C_1}$ being the carbon bearing the oxygen atom and $\ce{C_2}$ the sulfur atom, respectively). An unambiguous method of determining whether the two representations are equivalent is to assign the absolute configuration for the chiral carbon(s) in question. It may also help to use other projections (Fischer or Newman) if visualization of bond rotation and stereochemistry in the skeletal formula becomes difficult, as you did with the Sawhorse projection.


Allred-Rochow vs Pauling Electronegativity scale While observing the trends in electronegativity of the elements, I found a discrepancy between Allred-Rochow and Pauling scales. In Pauling's scale, chlorine is the third most electronegative element ($\ce{F > O > Cl > N}$). In Allred-Rochow's, it is the fourth ($\ce{F > O > N > Cl}$).

While of course the numeric values of the two scales are different, I thought that the rank of the elements would be the same, being electronegativity related to the properties of the atoms (for instance, the partial charge of molecules such as nitrogen trichloride relies on the electronegativity).

So, I was wondering: which scale is more accurate? And, if one is wrong, why are both scales still in use?

Pauling's electronegativity scale is determined by considering the strengths of A-A, A-B, B-B bonds.

Allred-Rochow's electronegativity is determined by considering the charge experienced by an electron on the "surface" of the atom.

While the correlation between the two scales is not perfect, it is quite good, since the scales are calculated with different methods:

  The correlation between Allred–Rochow electronegativities (x-axis, in Å−2) and Pauling electronegativities (y-axis).
  By Physchim62 GFDL, CC-BY-SA-3.0, from Wikimedia Commons

Also the partial charge is a model, useful in order to make rapid evaluations on real things. Sometimes, when molecules with almost covalent bonds (such as nitrogen trichloride) are modeled with different electronegativity scales, the partial charge may be different, but the actual electron cloud may not agree with the model.

The real electron density can be then calculated with quantomechanical methods, which are much more accurate and yet complicated than the simple model of electronegativity.


Proper typography of "NOx" I am a journal typesetter, and recently I came to a problem that I'm not sure I know to solve. The question is how to properly typeset the "NOx" abbreviation for nitrogen oxides. Which is typographically correct:

$\mathrm{NO}_x$ (because $x$ is a numeric variable); or
$\mathrm{NO}_{\mathrm{x}}$ (because all names of chemical compounds are strictly upright)?

I quote the Green Book by IUPAC, 2nd printing (2008), section 1.6, enumeration item 2:

  The overall rule is that symbols representing physical quantities or variables are italic, but symbols representing units, mathematical constants, or labels, are roman. [...]

As such, the correct way to write it is $\ce{NO}_x$, because $x$ is a variable. Please note that it is only the sum formula for the compound and not its name. Since the element names still are in roman (upright) font, everything is fine.

The same goes for a formula like $\ce{C_{$n$}H_{$2n+2$}}$, which is typeset exactly like that in the Gold Book (hat tip to @Martin).

Here at StackExchange, you can type it as $\ce{NO_$x$}$ ($\ce{NO_$x$}$) or $\ce{NO_x}$ ($\ce{NO_x}$).


Restrictions on transitions in Spectroscopy Just for the sake of an example, let's assume a transition from one energy state to a higher energy state in atomic absorption spectrum

If the energy difference between the ground state and first excited state is E and that between ground state and second excited state is 3 E (just for say...). Now if an incident photon has an energy of 2 E so;

(Assuming the electron is initially in ground state)

Will the electron show transition to the first excited state and emit a photon of energy E

OR

Will the electron not show any transition and not accept the photon at all, since the energy of the photon does not exactly match with the energy difference between any two specific states?

The example of atomic absorption spectrum is taken up just for the sake of explaining the question, the question is rather general and pertains to any kind of spectroscopy experiment in which their occurs transition from one state to another by absorption of EM waves.

If the energy is to be transferred from the photon to your atom one energy gap must match the energy of the photon. Thus absorption may occur from the first to second excited state in your example as the energy in both is 2E. However, there has to be some population in the first excited state for this to happen, zero population zero absorption. In fact it is a bit more subtle in that there has to be a difference in population between the two levels involved. It is possible for the photon to stimulate a transition from the upper level to the lower one, just as from the lower one to the upper one. The reason why this seems odd is that in most cases the population in the upper level is vastly smaller than in lower ones (via Boltzmann distribution) and the population in the upper level can often be set to zero without serious error. This is particularly true in electronic spectroscopy, as in the case of atoms, but far less so for pure rotational transitions in molecules which occur in the microwave region of the spectrum. You will find details in any good spectroscopy book.

Thus neither of your cases occurs but something different; transitions between the first to second excited state. How strong this transition is depends on the population in the two levels and on selection rules involving angular momentum and electron spin changes.


Bond length increase when replacing nitrogen with phosphorus for NH3 and NF3 There is an increase in bond length of about $40\rm~pm$ from $\ce{NH3}$ to $\ce{PH3}$, but only by about $20\rm~pm$ from $\ce{NF3}$ to $\ce{PF3}$. I understand that the bond length will increase with the larger $\ce{P}$ atom, but I cannot see why the changes are so different when the atom is changing in the same way.

This answer of mine basically leads the way to answering your question but it is focussed on bond angles, not lengths. So let me rewrite it to focus on lengths.

All four compounds would ideally want a set of $90^\circ$ angles which nitrogen cannot achieve due to steric stress; instead, the angles are expanded to $102^\circ$ ($\ce{NF3}$) and $107^\circ$ ($\ce{NH3}$). This seems couner-intuitive, because the smaller atoms require a larger bond angle. But looking at it closely, we realise that the $\ce{N-H}$ bond is simply shorter so the hydrogens are closer together hence more strained and hence further apart. Hydrogen being a very small atom the bond lengths cannot be extended well.

The fluorine atom on the other hand is much larger, allowing not only for a greater standard bond length but also for greater bond length variations. It can extend the bonds when compared to $\ce{NH3}$ to allow for a smaller bonding angle and in turn allow for a larger s-contribution to nitrogen’s lone pair. Lengthening (that is, weakening) the bonds slightly allows for a greater stabilisation on the central atom and thus a more stable overall molecule.

Another important issue is the fact that a smaller bond length in $\ce{NF3}$ would bring the fluorine atoms closer together causing electrostatic repulsion of their lone pairs (although that is what is commonly meant with steric stress). Hydrogen, having no lone pairs, does not have this issue.

In phosphorus’ case, both bond angles are almost perfectly $90^\circ$, so we can expect an almost unstrained molecule. The smaller hydrogen still calls for an equally short bond but the larger fluorine can now come closer making the bonds stronger, because the angles are still near-perfect and phosphorus’ lone pair has a very high s-contribution. In fact, the $\ce{P-F}$ bonds are rather strong; much stronger than their $\ce{N-F}$ counterparts.

Rationalising the entire thing we come up with:

The difference between $\ce{NH3}$ and $\ce{PH3}$ ($40~\mathrm{pm}$) is approximately equal to the larger phosphorus atom’s size. (The difference between the empirical atomic radii is $35~\mathrm{pm}$.)
In $\ce{NF3}$’s case, the bonds are lengthened by some $15$ or $20~\mathrm{pm}$ to accomodate for steric stress both. This weakens the bonds but stabilises the atoms, especially nitrogen. Also, steric stress between the fluorine atoms is reduced.
$\ce{PF3}$ has the bond lengths one would expect.


Why does alpha cleavage occur? I am studying fragmentation patterns in mass spectrometry and I am confused about why alpha cleavage occurs. Why does the carbon-carbon bond adjacent to a functional group break instead of the bond between the alpha carbon and the fucntionalized moiety?

After ionization occurs, the carbonyl donates its free radical into its own double bond.  This causes the bond adjacent to the carbonyl to split into two electron s, one with spin antiparallel to that of the carbonyl radical and one of parallel spin.  The one of antiparallel spin combines with the free radical to form another pi bond.  This creates a carbon oxygen triple bond and carbon atom with a free radical (alkyl radical).  The carbon atom with the radical is the atom alpha to the carbonyl, hence alpha cleavage.  The oxygen bears a positive formal charge.  Alpha cleavage basically makes substituted carbon monoxides, and side chain radicals.  You will see the "carbon monoxide derivative" on the spectrum.

To do this quickly, just take the original molecule, draw lines showing fragmentation on both sides of the carbonyl, and then draw the carbon oxygen triple bond structure.


Truly Understanding the Second Law of Thermodynamics and Entropy There are some seemingly contradictory things I’d like cleared up:

Second Law states: $\mathrm dS_\text{univ} = \mathrm dS_\text{sys} + \mathrm dS_\text{surr}$, and $\mathrm dS_\text{univ}$ is always positive in a spontaneous process.

First off, what does “spontaneous” truly mean? “It means the reaction occurs without any external energy input” – I’ve heard this one before, but it doesn’t make sense. In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs, though at a slower rate, than the spontaneous reaction. What is the most fundamental definition, intuitive and universally applicable definition of spontaneous?

Moving on, according to ChemWiki the Second Law states that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes. Who is right and why (explain in quantum terms, since I understand that in the long run of course the entropy of the universe increases by the laws of probability)?

An example to work through:

In a living cell, large molecules are assembled from simple ones. Is this process consistent with the second law of thermodynamics?

Book’s solution: To reconcile the operation of an order-producing cell with the second law of thermodynamics, we must remember that $\mathrm dS_\text{univ}$, not $\mathrm dS_\text{sys}$, must be positive for a process to be spontaneous. A process for which $\mathrm dS_\text{sys}$ is negative can be spontaneous if the associated $\mathrm dS_\text{surr}$ is both larger and positive. The operation of a cell is such a process.

First off, are we to consider the operation of a cell in this problem spontaneous? As I recall from biology, life processes require ATP, the carrier of energy, so these would technically “require external energy input,” would they not? However, the book treats it as a spontaneous process. Again, this confuses me because the definition of spontaneity is inadequate as I now semi-understand it.

Sorry for such a long post, but these are very important and fundamental ideas in spontaneity, entropy, and free energy, and I don’t want to read on until these ideas are well ingrained, and correctly so.

Be careful in citing equilibrium as an example, as it fully integrates entropy in its definition. See this article for the details, but in a spontaneous reaction the entropy is increasing, and part of reaching equilibrium is that the energy gain and loss among products and reactants is balanced. Also, most experimental reactions are (nearly) closed systems for short periods of time, and if no energy is gained or lost (again, for short periods), you can have an equilibrium where there is no change in entropy. If you let your experiment sit and the heat of the system changes (for example), the equilibrium will change as well. And eventually you end your experiment, clean out your flask, and thus increase the entropy of the universe again!

@MARamezani's comment pointing to the article on Biology looks like a good answer to your biology questions. Earth is most definitely not a close system with that giant energy source 93 million miles away.


What is/are the product when this reacts with Lucas reagent? What would be the product(s) if this alcohol reacts with Lucas reagent?
Would Methyl shift happen ? I heard that in Groove's process, only primary and secondary alkyl chlorides are formed.

A shift of the methyl group is likely, because this would yield a more stable tertiary carbocation. In concentrated hydrochloric acid, the double bond can also be protonated, producing a secondary carbocation which is captured by $\ce{Cl-}$. If both happens, the product would be 2,4-dichloro-4,5-dimethylhexane.


Formation of radicals from C-H and C-C bonds Why do radical mechanisms favor the homolytic dissociation of a $\ce{C-H}$ bond over a $\ce{C-C}$ bond?

The $\ce{C-H}$ bond is stronger in magnitude (and it's also slightly polarized), so why would it break more easily as compared to the $\ce{C-C}$ bond?

My instructor explains this by claiming that the products of $\ce{C-H}$ bond dissociation are stabler than$\ce{C-C}$ bond dissociation.

I know that radicals are unstable and highly reactive, so why does it make any difference as to which radical is formed? Wouldn't the radical easiest to form be the one formed the most?

Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens.

  Homolysis

$\ce{C-C}$

$$\ce{H3C-CH3 -> 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{H3CCH2-H -> H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$

However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.

$$\ce{Cl2 -> 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$

The chlorine radicals then react with the alkane by abstraction, meaning that bonds are being formed as well as being broken.

  Abstraction

$\ce{C-C}$

$$\ce{Cl. + H3C-CH3 -> Cl-CH3 +H3C.}$$

$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3C-CH3}  & \ce{Cl-CH3} \\ \hline
\Delta H^\circ & 377\ \text{kJ/mol} & -350 \ \text{kJ/mol} \\ \hline
\end{array}$$

$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{Cl. + H3CCH2-H -> Cl-H +H3CCH2.}$$

$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3CCH2-H}  & \ce{Cl-H} \\ \hline
\Delta H^\circ & 423\ \text{kJ/mol} & -432 \ \text{kJ/mol} \\ \hline
\end{array}$$
$$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$

Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.


Do molecules with bridges through rings exist? Do molecules with bridges through rings (in a manner illustrated by this) exist?

I sometimes get results like this when doing Energy Minimization on molview.org. For example:

Is this actually a thing?

EDIT: As a slightly more realistic example, consider this:

I'm not sure about the existence of molecules with bridges through rings. However, there are several publications of synthesis of molecules mimicking wheels and axles ([2]rotaxanes; The “[2]” refers to the number of interlocked components) as one shown below (Ref. 1):

(The diagram is from Reference 1)

This specific molecule (8; an “impossible” [2]rotaxane) represents a macro-cycle with a straight-chain molecule with bulky end groups going through its center. The inclusion of two bulky end groups prevents the straight-chain molecule leaving the macro-cycle (mechanically interlocked) as depicted in the diagram (See Ref. 2 for the total synthesis of the molecule).

Note that Ref. 1 also cited articles for the synthesis of [2]catenanes, which contain two interlocked rings (instead of one axle and one macrocycle). Keep in mind that there are some advanced catenanes and rotaxanes that exist (e.g., [3]catenanes and [3]rotaxanes).

(The structures are from Reference 1)

References:

Edward A. Neal, Stephen M. Goldup, "Chemical consequences of mechanical bonding in catenanes and rotaxanes: isomerism, modification, catalysis and molecular machines for synthesis,"  Chem. Commun. 2014, 50(40), 5128-5142 (https://doi.org/10.1039/C3CC47842D).
Jeffrey S. Hannam, Stephen M. Lacy, David A. Leigh, Carlos G. Saiz, Alexandra M. Z. Slawin, Sheila G. Stitchell, "Controlled Submolecular Translational Motion in Synthesis: A Mechanically Interlocking Auxiliary,"  Angew. Chem., Intl. Fd. 2004, 43(25), 3260-3264 (https://doi.org/10.1002/anie.200353606).


Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that? I know the history of the discovery of the rare earth elements, that they were discovered near the village Ytterby in Sweden in the 19th century. Four elements are directly named after that village: erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb).

Why are they named like this? It makes sense to name one element after the locality, but why name the rest as misspellings of that village? It's not like we have ermanium, germium and ermium in addition to germanium (although we do have holmium).

This site mentions this:

  In 1842, Swedish chemist Carl Gustaf Mosander separated the "yttria" component found in the mineral gadolinite into three segments that he called yttria, erbia and terbia.

Is the any record as to why Mosander chose these names? Because...

  As might be expected considering the similarities between their names and properties, scientists soon confused erbia and terbia.

The links provided below are to the website "Elementymology & Elements
  Multidict".  It is a fascinating site with information on the
  discovery, naming and various other facts for each element. As the author (Peter van der Krogt) says, "I am not a chemist, but a (map) historian much interested in the origin of names". Enjoy!

A nice piece on the complex etymology for Erbium, Terbium and Yttrium can be found here.  In brief the article says, "All three names were derived from the Ytterby quarry where the gadolinite was originally found in 1787. It is also said that since the original earth was divided into three new earths, Mosander split the name of Ytterby in three parts: ytt, terb, and erb."

As to ytterbium's etymology see here.   Again the history is complex with the name changing as recently as 1925. Another chemist isolated this element and again, wanted to pay homage to the geographic point of discovery, the Ytterby quarry.


Biopolymers as alternative for PVC and PU? I'm working on a project making a product line more sustainable, environmentally friendly and ready for recycling.

Now I'm searching for good and suitable biopolymers for a part of the product that is made from PVC and one for PU (Polyurethane). For PES I already have chosen PLA. Is there someone who can help me select the right biopolymers with similar characteristics of PVC and PU?

Please let me know, thank you!

Ethylene vinyl acetate (also known as EVA) is the copolymer of ethylene and vinyl acetate and is a replacement for PVC.

Biomass > (fermentation) > Bioethanol > Ethene

Ethylene + acetic acid + 1/2 O2 → Vinyl acetate + H2O

Tung oil can be used to create a plastic like water proofing film as a replacement for PU


Finding the concentration of ferrous thiocyanate in calibration solutions Concentration of $\ce{Fe(SCN)^{2+}}$ in calibration solutions

Recall that the concentration of $\ce{Fe(SCN)^{2+}}$ in the calibration solutions is essentially the concentration of the $\ce{SCN-}$ initially in the solution. So, determine the concentration of $\ce{SCN-}$ initially in each of the $\pu{10.0 mL}$ calibration solutions. This is a simple dilution!

The lab checker is saying that this is wrong. Am I even going about doing this problem correctly?

So, to find the concentration of $Fe(SCN)^{2+}$, the problem says "the calibration solutions is essentially the concentration of the $[SCN^−]$ in the solution." So, you simply take the volume of $K(SCN)$ solution divided by the total volume, then multiply by $[K(SCN)]$:

$[Fe(SCN)^{2+}]=[SCN^−]=[K(SCN)]=0.002M*\frac{0.5mL}{10.0mL}=0.0001M$ (repeat for 2, 3, 4)

The reason we do not involve $K_c$ is because the large amount of $[Fe^{3+}]$ drives the reaction to the right. The absorbance measurements are going to be useful for determine the amount of $[Fe(SCN)^{2+}]$ in a system where $[Fe^{3+}]$ is not high. In these cases, the concentrations $[Fe^{3+}]$ and $[SCN^-]$ are known, and you use the absorbances you collected to draw a line relating concentration $[Fe(SCN)^{2+}]$ to absorbance. When you know the concentration of your three parts of $K_c$, you can simply use arithmetic to find $K_c$.


Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields We're learning about the Crystal Field Theory (CFT, under coordination chemistry) at school now, and my teacher mentioned that the 5 d-orbitals split into orbital sets, namely the $\mathrm{t_{2g}}$ (triply degenerate-gerade) and the $\mathrm{e_{g}}$ (evenly degenerate-gerade) for an octahedral ligand field; whereas for the tetrahedral ligand field, the notation is the same except that the subscript 'g' is removed, indicating it is ungerade (That bit was my interpretation … I think I'm wrong though)

Now a lot of the books and sites I've looked up mention the same gerade notation for both octahedral and tetrahedral ligand fields.

So naturally my question would be "Which one's right and which one's wrong?" and a reason if possible.

The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes belong to the $T_\mathrm{d}$ point group with a considerably lower symmetry (i.e. a smaller number of symmetry operations).

Octahedral complexes are probably more present in the average non-coordination chemist’s brain. Since $O_\mathrm{h}$ includes a centre of symmetry, all irreducible representations have a subscripted g or u to indicate whether equal phases of an orbital are mapped onto each other upon inversion (g for gerade) or whether phases are mapped onto their opposites upon inversion (u for ungerade). Due to the d-orbitals having two nodal planes, they are automatically g, which is why they as a whole transform as $\mathrm{e_g + t_{2g}}$. Since, as I said, this is the more prevalent case in general, this is what a lot of chemists have in mind.

The tetrahedral point group $T_\mathrm{d}$ does not contain a centre of symmetry. This is easily seen when looking at a tetrahedron, which is not inversion-symmetric. Thus, the irreducible representations do not gain a subscripted g and u. The d-orbitals thus transform as $\mathrm{e + t_2}$. Adding a g to these is wrong, but common (see above).

Thus, you were correct in assuming: ‘just because everyone does it does not mean it is correct.’ The same thing can be said for $\ce{S=O}$ double bonds in sulfate.


Slater Matrix of Methane Molecule I am going through Hartree-Fock calculations in detail, but the textbooks I am referring all considered homonuclear molecules as examples. So I am trying to come up with my own the Slater determinant for Methane. Kindly let me know if my application is correct!
For Methane, we have in total 10 electrons (Carbon 6e, and four Hydrogen with 1e each). We also know that valence shell electrons of C and H engage in bonding. The $1s$ shell of Carbon does not engage in bonding. I am also going to denote $+$ for spin-up and nothing for spin-down. Please note that $1s$ for first two columns in the following matrix describes the $1s$ orbital of carbon which does not involve in bonding.
$$\begin{pmatrix}1s(1)^+ & 1s(1) & S_1(1) & S_1^+(1) & S_2^+(1) & S_2(1) & S_3^+(1) & S_3(1) & S_4(1) & S_4^+(1) &\\\ 1s(2)^+ & 1s(2) & S_1(2) & S_1^+(2) & S_2^+(2) & S_2(2) & S_3^+(2) & S_3(2) & S_4(2) & S_4^+(2) \end{pmatrix}$$
For brevity, I am not writing out the rest of the eight rows in matrix. Now  $S_n$ (spatial part of the wavefunction $S_n$), above in the matrix, is defined as the following:
$S_n=c_{n,1}1s + c_{n,2}2s + c_{n,3}2p_x + c_{n,4}2p_y + c_{n,5}2p_z$, where $n$ runs from $1$ to $5$. Why? because we know that these electrons engage in bonding, hence we have to take a superposition of orbitals (mixing). So in total we have 5 equations e.g. $S_1, S_2,...,S_5$.
Now each such superimposed MO (the $S_n$) can hold 2e each for a total of 10 electrons. Now the carbon $1s$ took away 2 electrons, so 8 electrons remained. And I have hit redundancy, because theoretically each $S_n$ can take 2e, so I have 10 electrons which can be accommodated, but I have only 8. Moreover in the Slater matrix above, I can't find a way to fit in $S_5$.
I am missing something. I would be glad if someone can throw some light on this.

Whenever some says they are just starting out with electronic structure theory, I always recommend that they read Szabo and Ostlund's Modern Quantum Chemistry. The book is great introduction to HF and post-HF methods (some of the discussion of coupled cluster is a bit outdated) and its very affordable. It is nice to have around as a reference.
Getting to your actual question, I think you are confusing yourself by trying to hybridize the orbitals. During the HF procedure, you will solve for a set of MOs that is are linear combinations of the AOs, so there isn't a need to form these $S_n$ functions beforehand.
There is also the issue of actually forming the Slater determinant; while the solution to HF will be a Slater determinant, you generally don't explicitly write it out this way. Generally, you assume that each MO in the Slater determinant is a linear combination of the AOs and you solve for the coefficients of each AO, such that the energy is minimized. So the Slater determinant for methane will have 10 (initially unknown) MOs in it to account for the 10 electrons and you will then need to build these MOs using the Roothaan-Hall form of the Hartree-Fock equations.


Can metal oxides react with water to form a hydroxide, then that with HCl to make a salt? Lets say I have a metal oxide and I put it in water, and it reacts. As far as I know, only metal oxides from group 1 and 2 react in water, making hydroxides.

Now here is my question: If I add HCl to the hydroxide, can I be sure that it will make a salt and water? Or are my previous assumptions wrong?

An example that works would be Sodium Oxide which reacted with water and then with HCl produces NaCl and water. Does it work the same way with all metals from this group though?

tl;dr: Yes, depending on extent of reaction

The group 1 and (group 2) hydroxides for example: $\ce{NaOH,  KOH, RbOH, CsOH, Ba(OH)2, Ca(OH)2}$ are strong bases, being essentially fully ionized in aqueous solution; (although some are weaker $\ce{LiOH}$, $\ce{Mg(OH)2}$) . These should react with $\ce{HCl}$ - a strong acid forming water and a neutralisation product a typical corresponding salt $\ce{MCl}$ or $\ce{MCl2}$ .

Being soluble in aqueous envronment implies the salt is also obtained in this phase, so the question will be on extration method to use, in order to get this salt product in solid form.

Why I said depending on extent of reaction, is beacuse some metal oxides e.g $\ce{MgO}$ hardly form the hydroxide in the first place, and also insolubility with render the second reaction effectively out of question.


How different are molecule with same connectivity Let us represent molecules by graphs (atoms are vertices and bonds are edges).
Let $A$ be the adjacency matrix of the molecular graph. $A_{ij} = b$ where $b$ is either binary: 0 when atom i and j are not bonded else 1 (Repr. 1). Or, further we may take b from the set $\{0,1,2,3\}$ depending on whether the no bond (0) or single(1) double(2) or triple (3) bond is present (Rper. 2). Here were are not distinguishing the atoms - for our purose they represent a node in a graph.

Let us consider only non-isomers and molecules with at least 10 atoms. How different will be two molecules with same graphs (with respect to Repr. 1 or Repr. 2) in terms of their properties or particularly about drug-like properties. My interest is to know how faithfully the topological indices characterizes the molecular activity. Specific or general examples of pairs of  non-isomer molecules with same graph but very different property could be useful. It would be helpful if I can find enough examples to convince myself about the usefulness or otherwise of topological indices which are calculated from the grpahical representation outlined above. Topological indices are popular in drug discovery research, but I am not very much convinced about their utility since most the results published are not statistically significant (use of very small data size - from 15 to some hundreds molecules, among other weakness) nor rigorously validated.

Thank you.

In my experience, calculated parameters like topological indices are not viewed as significant if considered alone. It's a strategy for finding trends or correlations and, subsequently, candidates from  large volumes of chemical structures. Many studies will publish trends they find in a target chemical groups which are not meant to apply to every chemical group.

An example of how all these seemly useless data is useful - see the Merck Kaggle Challenge.
https://www.kaggle.com/c/MerckActivity

http://blog.kaggle.com/2012/11/01/deep-learning-how-i-did-it-merck-1st-place-interview/

Statistical significance is harder to comment on. I work with between 8 and 20 million data points and the idea of how to define statistical significance is evolving in cheminformatics as it evolves in big data analytics. It gets much more complicated because standard methods, such as the t-test, fail.

NOTE
I just realized there might be some fundamental confusion. Mol files include atomic information as do most graph based chemical representations. They're stored as properties of the nodes. An algorithm can choose to use or ignore this information. Not all topological parameters ignore atomic information.


Will a Galileo's Thermometer containing liquid paraffin explode in my car if the temperature is above 35 degrees Celsius? I'm driving from Sydney to Melbourne (approximately 12 hours trip) and I'm wondering if it's safe to have this Galileo's Thermometer in my car during the trip if it's going to be a scorcher of a day. It could easily reach 35 degrees Celsius or over during the trip. I'm a little worried that it might explode. Any help much appreciated!

Do not confuse the flash point with the autoignition temperature.

I don't know what atmosphere you have inside the thermometer, but either way I think it's hardly the case that you reach the autoignition temperature of paraffin (gasoline undergoes autoignition at 246-280 °C).


What type of chemical reaction is sulfuric acid with calcium carbonate? $$\ce{H2SO4 + CaCO3 -> H2CO3 + CaSO4}$$

I thought it would be a double replacement and also a decomposition reaction because $\ce{H2CO3}$ immediately breaks down to $\ce{H2O}$ and $\ce{CO2}$.

While double replacement is correct, you could also classify it as a neutralization reaction (what I think would be the most common answer if one had to pick a single description).  I agree with the comment explaining that you can't consider it a decomposition reaction as written.  You would need to explicitly show that as part of the reaction for that to be an acceptable description.


What is the accuracy of medical syringes? I am interested in both low and high volume syringes.

The requirements for ordinary sterile single-use syringes made of plastic materials are standardized in the international standard ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use.

According to this standard, the tolerance on the graduated capacity depends on the nominal capacity of the syringe $V$ and the expelled volume $V_\text{ex}$.

For syringes with a nominal capacity of $V\lt5\ \mathrm{ml}$, the tolerance amounts to

$\pm5\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +2\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

For syringes with a nominal capacity of $V\geqslant5\ \mathrm{ml}$, the tolerance amounts to

$\pm4\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +1\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

The values apply to the volume of water at a temperature of $T=\left(20\pm5\right)\ \mathrm{^\circ C}$ expelled from the syringe when the line circumscribing the end of the piston traverses a given scale interval.


Find the Ionisation Potential and Electron affinity of X $N_0/2$ atoms of $X_{(g)}$ are converted into $\ce{X^+_{(g)}}$ by energy $E_1$. $N_0/2$ atoms of $X_{(g)}$ are converted into $\ce{X^-_{(g)}}$ by energy $E_2$.
Hence ionisation potential and electron affinity of $\ce{X_{(g)}}$ per atom are:
I tried solving this by unitary method, which gave ionisation potential and electron affinity as, ${2E_1/N_0}$, ${2E_2/N_0}$ respectively.
But the answer given is  ${2E_1/N_0}$, ${2(E_2-E_1)/N_0}$
Can anyone please tell me where I went wrong and help me with the solution. Thanks in advance!

The Ionisation Energy of an atom is defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. In this case, since $E_1$ amount of energy is used to ionise $N_0/2$ atoms, the ionisation energy of each atom is $\frac{E_1}{N_0/2} = \frac{2E_1}{N_0}$.
The Electron Affinity is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion. In this case, $E_2$ amount of energy is released when $N_0/2$ atoms accept electrons to form negative ions, hence electron affinity of each atom is $\frac{E_2}{N_0/2} = \frac{2E_2}{N_0}$.
Your answers are correct. The textbook must have made a typo in the second question. For the textbook's answer to be correct, the second question should read

$N_0/2$  atoms of $\ce{X+_{(g)}}$ are converted into $\ce{X−_{(g)}}$ by energy $E_2$

This gives the answer as $\frac{E_2 - E_1}{N_0/2} = \frac{2(E_2 - E_1)}{N_0}$ for electron affinity.


Decline in current during electrolysis I observed a sharp decrease in the current when I carried out the electrolysis of water. Just to set the context, I used a voltage of $\pu{12V}$ and the concentration of my electrolytes were $\pu{0.15 mol dm^{-3}}$ for $\ce{Ca(NO3)2}$ and I used graphite rods which were flaking but I am not sure whether that might cause such a significant dip in the current. I did also notice some white material forming on the cathode. So I want to know why there is such a dip?

Given the whitening observed at the cathode, one potential explanation is that calcium salts were precipitating there, forming a layer blocking passage of part of the current.
I usually write the cathodic hydrogen-evolution reaction in neutral solution as:
$$
\ce{2 H2O + 2 e- -> H2 + 2OH-}
$$
Thus, the pH in the vicinity of the cathode will be considerably greater than 7 during the electrolysis process, due to the $\ce{OH-}$ formed. Since you're using a calcium-based electrolyte, it seems likely that some combination of $\ce{CaO}$/$\ce{Ca(OH)2}$ is precipitating in that alkaline surface layer. Once formed, in my experience such solids will dissolve only slowly in neutral aqueous solution.
If possible, I'd recommend trying the experiment again but in a different electrolyte, such as $\ce{KNO3}$ or $\ce{NaNO3}$; those alkali cations shouldn't exhibit any precipitation.


What are common procedures for lab absorption spectroscopy of gases? I'm doing research for applications of absorption spectroscopy. I'm trying to find common practices such as how the gasses are mixed, measured, contained, and discarded.

The gases involved will be those commonly found in the atmosphere or the human body. Top among these will be $\ce{NO2}$, $\ce{CO2}$, and $\ce{CO}$.

Non-dispersive infrared absorption spectroscopy is commonly used to measure $\ce{CO}$ and $\ce{CO2}$ (and other gases, like methane and $\ce{NO2}$). The absorption overlap from water interferes with $\ce{CO2}$ measurements in particular but various algorithms can be used to mitigate this.

$\ce{NO2}$ is more commonly measured by chemiluminescence because of its high sensitivity, but FTIR and other techniques are used for higher levels (i.e. exhaust pipe and stack emissions).


When diagonalising the overlap matrix, what happens if the square root of the diagonal matrix does not exist? I was looking at the code required for a simple RHF computation when I came across the section for the diagonalisation of the Fock Matrix.

Based on what I know, the diagonal matrix $\mathbf{s}$ is evaluated as

$$\mathbf{s} = \mathbf{U}^\dagger\mathbf{S}\mathbf{U},$$
where $\mathbf{S}$ is the overlap matrix.

The matrix $\mathbf{X}$ is then evaluated as

$$\mathbf{X} = \mathbf{U}\mathbf{s}^{-1/2}\mathbf{U}^\dagger$$

and the diagonalised Fock operator $\mathbf{F'}$ as

$$\mathbf{F'} = \mathbf{X}^\dagger\mathbf{F}\mathbf{X}.$$

But what happens if the matrix $\mathbf{s}$ has terms that are negative? What happens then? Do we simply take the magnitude and carry on?

Thanks!

The matrix $\mathbf{s}$ in your question are the eigenvalues of the overlap matrix $\mathbf{S}$, further the matrix $\mathbf{U}$ are a matrix of the eigenvector of the overlap matrix.
Now looking at the equation that gives you concern:

$$\mathbf{X} = \mathbf{U}\mathbf{s}^{-1/2}\mathbf{U}^\dagger$$

Since $\mathbf{s}$ is a diagonal matrix then taking the power of $\mathbf{s}$ is just taking the power of every element:

$$\mathbf{s}^{-1/2}=\begin{bmatrix}s_1&0&...&0\\0&s_2&...&...\\...&...&...&0\\0&...&0&s_n\end{bmatrix}^{-1/2}=\begin{bmatrix}s_1^{-1/2}&0&...&0\\0&s_2^{-1/2}&...&...\\...&...&...&0\\0&...&0&s_n^{-1/2}\end{bmatrix}$$

The only way as you noted this can cause trouble is if any of the elements of $\mathbf{s}$ are negative.
Now here we have to remember that $\mathbf{s}$ are the eigenvalues of the overlap matrix, it is therefore guareenteed that all of the elements will be equal to or greater than zero!
For the proof of this see my answer to one of your other questions, Basis sets - negative contraction coefficients?


exchange integral at large distances I want to ask a question about the exchange integral $\ce{H_{AB}}$ -

If I consider an exchange integral $\ce{H_{AB}}$ the following exchange integral can be evaluated:

so $\ce{S_{AB}}$ is $0$ at $\ce{R = \infty}$

but in this presentation I was shown today, at large values of R, $\ce{H_{AB}}$ is preportional to the overap integral $\ce{S_{AB}}$.

Surely at large distances and towards infinity, as $\ce{S_{AB}}$ goes to zero, $\ce{H_{AB}}$ also goes to zero too.

Where does the preportionality

$$H_{AB} \propto S_{AB}$$

come from then using these approimations when R is large?

The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations:
$$
\frac{e^2}{4 \pi \epsilon R} = k/R
$$
when $R \rightarrow \infty$: slowly to zero

$$
S_{AB}
$$
when $R \rightarrow \infty$: to zero

$$
\left< 1s_A\left| k/r_A \right| 1s_B\right>
$$

when $R \rightarrow \infty$: rapidly to zero

So there is some regime somewhere, in which the change of the entire expression is dominated by the change in $S_{AB}$. The verification of the correctness of the different notations is apparently left to the reader ...


Crossed Cannizzaro products I have understood the basic mechanism of cannizzaro reaction. But I am unable to understand the products formed in case of a crossed Cannizzaro reaction.

Suppose Benzaldehyde and Formaldehyde are heated in presence of $\ce{NaOH}$ to carry out Cannizzaro transformation. I have read that it is Benzyl Alcohol and Formic Acid formed in the greatest proportion and not Benzoic acid and Methyl alcohol.

Why is this so? How can we understand this in terms of the stability
 of Cannnizzaro intermediates?

The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.

$$\begin{aligned}
&\ \ \ \ \  \ce{O} && \ \ \ \ \ \ce{O-} \\
&\ \ \ \ \  \parallel && \ \ \ \ \ \mid \\
&\ce{R-C-H +OH- ->}  && \ce{R-C-H} \\
& \ && \ \ \ \ \ \mid \\
& \ && \ \ \ \ \ \ce{OH}
\end{aligned}$$

Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:

Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde
Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.

If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.


Is a 2-D periodic structure isomorphic with the surface of a torus, a sphere, neither or both? When I was reading through the ADF-BAND tutorials, one of the toy systems presented was a 1-D periodic structure involving 3 collinear hydrogen atoms. The tutorial pointed out that, topologically speaking, this is cylindrically symmetrical (more specifically, it's ring symmetrical).

In the instance of a 2-D structure, can the calculation be considered a model for the surface of a torus (this seems logical), a sphere (I doubt this because if you fit a rectilinear grid to a sphere you end up with two poles and dissimilar meridians and parallels), or other?

Bonus questions: Has anyone used periodic calculations to model electronic structure/chemistry on the surface of a sphere or torus? Can you introduce a curvature term to account for these structures having a finite size?

Yes, a 2D-periodic space can be mapped to a torus, but that's more a question for the math.SE…

Regarding your bonus question, why would there be? What would you do with it? Molecular structures are intrinsically 3D, so I don't see what you would do in a 2D (periodic or not) space? Even when we talk about planar or pseudo-2D structures (buckyball, nanotube, etc.) they are 3D objects with 3D electronic densities and wavefunctions.

Edit: 3D structures that are periodic in two dimensions and finite in the other one can be studied by many computational chemistry codes. They are often referred to as slab calculations or surface calculations. The most common issue is that of Coulombic interaction (or Poisson equation solver), which typically requires special treatment in the 2D case.


Possible methods for copper (II) reduction to copper (I) How can I make $\ce{Cu2O}$ from copper?

I am seeing if electrolysis is a good route (in the context of kitchen style chemistry) to cuprous oxide, instead of the traditional heating copper and removing the cupric oxide, etc.

With Epsom salt electrolyte, it is very easy to make copper II hydroxide (cloudy ammonia soluble, according to the wiki). This can be heated to $\ce{CuO}$ easily enough, but I have no idea if this can be reduced.

Can $\ce{CuO}$ be reduced easily? Or is there a simple way to produce (from copper) a copper (I) salt, from which a precipitate of $\ce{Cu2O}$ may be formed? What reducing agents could be used?

There are several ways to make copper(I) oxide from copper:-

Heat it

$$\ce{4Cu + O2 ->[\Delta] 2Cu2O}$$

Copper react with oxygen to produce copper(I) oxide. This reaction takes place at a temperature of over 200°C.

Heat it with nitrous or nitric  oxide

$$\ce{2Cu + N2O->[\Delta] Cu2O + N2}$$
$$\ce{4Cu + 2NO->[\Delta] 2Cu2O + N2}$$

Copper react with nitrous and nitric oxide to produce copper(I) oxide and nitrogen. Both reaction takes place at a temperature of 500-600°C.

Heat it with copper(II) oxide

$$\ce{Cu + CuO->[\Delta] Cu2O}$$

Copper react with copper oxide to produce copper(I) oxide. This reaction takes place at a temperature of 1000-1200°C.

If you made copper (II) oxide, just heat it

$$\ce{4CuO ->[\Delta] 2Cu2O + O2}$$

The thermal decomposition of copper oxide to produce copper(I) oxide and oxygen. This reaction takes place at a temperature of 1026-1100°C.

Since you are doing this at home, method 2,3,4 cannot be used as it requires laboratory grade heater but method 1 can be used since it require 200°C which is achievable.

But the best method(according to the comment) I would recommend is the Fehling's test. All you need is Fehling's solution and glucose which can be purchased from sigma-aldrich at affordable price.

$$\ce{RCHO + (2Cu^2+ + 5OH^{-})->Cu2O + RCOO^- + 3H2O}$$


How do bonds form without violating the second law of thermodynamics It might be obvious that entropy related to reactions always increases and I understand (somewhat) why this happens. My question is why if you have two hydrogen atoms in a vacuum without being bonded to a molecule they would form a bond. Does that not show a movement from chaos to order and thus a decrease in entropy?

Note that 2 isolated hydrogen atoms will not form a stable hydrogen molecule, unless they are able to dissipate the energy, released by the bond formation. It can be done e.g.  passing mechanical energy to neighbour molecules, or emitting a photon.

In a macro scale, a system can decrease its entropy, if its Gibbs energy decreases, as it means the dissipated heat causes the bigger entropy increase of the surrounding, than is the entropy decrease of the system, so the total entropy increases.

$$\Delta G = \Delta H - T \cdot \Delta S$$

If $\Delta G_\mathrm{sys} \lt 0$ then $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} \gt 0 $

as $\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}=\Delta S_\mathrm{sys} - \frac{(\Delta G_\mathrm{sys} + T \cdot \Delta S_\mathrm{sys})}{T}=-\frac {\Delta G_\mathrm{sys}}{T}$


Esterification of Glycine I came upon the following statement in a true/false question:

  Glycine can be esterified by an alcohol in the presence of inorganic acid

Is this statement correct?

Glycine contains the basic $\ce{-NH2}$ group, which would get protonated in the presence of inorganic acid. As a consequence, would the esterification process still be carried out on the $\ce{-COOH}$ group? This paper says that the reaction needs trimethylchlorosilane as a catalyst and not an inorganic acid.

The statement is correct.

The most common method for the esterification of amino acids is with anhydrous acid in alcohol. The most convenient way of doing this is with thionyl chloride example procedure or chlorotrimethylsilane example procedure as a source of anhydrous HCl.

The original procedures with HCl gas or conc. sulfuric acid are slower and lower yielding recent example


How do you know that Sodium Hydrogencarbonate is amphiprotic? Do you just have to remember that it is?

For example, I know that substances with CHs are hydrocarbons. Does such a 'code', if you will, exist for amphiprotic substances ?

You have to consider whether or not the species is capable of both losing and gaining a hydrogen.

When sodium bicarbonate dissociates in water, it becomes a sodium ion ($\ce{Na+}$) and a bicarbonate ion ($\ce{HCO3-}$). The bicarbonate ion itself is what is amphiprotic, and this is because it can lose the hydrogen it has to become a carbonate ion ($\ce{CO3^{2-}}$), or it can gain a hydrogen to become carbonic acid ($\ce{H2CO3}$).

Many species that contain a hydrogen and an unbonded electron pair are amphiprotic, like $\ce{NH3}$ and $\ce{H2O}$, for example.


Oxidation State of Amino Group What is the oxidation state of nitrogen the following amino acid:

I figured it must be -III:

Hydrogen is always +I
Both electrons of the N-C bond are near the nitrogen (higher electronegativity).

However, I've also heard that the N-C bond sometimes is seen as neutral (+0), which would make the oxidation state of Nitrogen -II. Which one is correct?

Do the zwitterions play a key role here (as the molecule could also be H3N-C-COO)?

I think -III is the correct 'formal' oxidation state of N in your molecule.

The method I was taught is the following. Write down the molecule without any bonds. Sum the number of 'valence' electrons from all atoms. For each covalent bond (counted as single regardless whether it is single, double or triple), put two electrons on the most electronegative atom of the pair (if same type of atom, put one electron on each instead). Distribute all remaining valence electrons in decreasing order of electronegativity (of course without going beyond 8 electrons on any atom). When you have multiple atoms of the same type, put one electron on each alternately. At the end of this process, the formal oxidation state of each atom is its number of valence electrons minus the number of electrons that ended up on it.

For an organic molecule with 'R' moieties, if you can assume the atoms have equal electronegativity (e.g. the C-R bond in your molecule would be a C-C bond), you can just put an electron on C from the beginning and ignore R.

If you try that in your example, you have in total 4*1 (from 4 H) + 1*5 (from 1 N) + 2*4 (from 2 C) + 2*6 (from 2 O) = 4 + 5 + 8 + 12 = 29 valence electrons. Put 1 of them on the C-R carbon. You are left with 28 more electrons to place. Take care of the bonds first. You put 6 electrons on N (3 bonds to less electronegative atoms), 2 electrons on the C=O oxygen (1 bond to a less electronegative atom), 4 electrons on the COH oxygen (2 bonds to less electronegative atoms), 1+2 electrons on the CR carbon (1 bond to another carbon, 1 bond to a less electronegative atom), 1 electron on the carboxyl carbon (1 bond to another carbon). You are left with 12 more electrons to place. 10 of them go on the two oxygen atoms. 2 of them go on the nitrogen. You end up with: 8 electrons on each oxygen (6-8 = -2); 8 electrons on the nitrogen (5-8 = -3); no electrons on any hydrogen (1-0 = +1); 3 electrons on the carbon bound to R (4-4 = 0); 1 electron on the carboxyl carbon (4-1 = +3).

You can also verify that 2*(-2)+1*(-3)+4*(+1)+1*(0)+1*(+3) = - 4 - 3 + 4 + 0 + 3 = 0.

The nice feature of this method is that when you have atoms of the same type but with different oxidation state (e.g. a nitro and an amino group in the same molecule), it should still give a correct result. E.g. if you process $\ce{H2N-CH2-CH2-NO2}$, you should get oxidation state -III for the amino nitrogen, and +III for the nitro nitrogen.

Interesting question about the different charged status; I'll let you figure it out for yourself - the method it the same, except that of course you will have covalents bonds in different places.


If ionic compounds are strongly bonded, why are acids with a more polar central atom the stronger types of acids? If the central atom is more polar, it pulls electron density away from the H+, making it more detachable. But why, then, are ionic compounds (which are even MORE polar than acidic covalent compounds) held tightly together?

Ionic compounds make a network of electrostatic attraction. Molecules have covalent intramolecular bonds which are very strong, but each molecular unit is weakly attracted to one another (relatively) due to lack of ionic attraction.  So, essentially, in the solid form, each $\ce{NaCl}$ has more attraction to another $\ce{NaCl}$ than does each molecule to another in a molecular system. That does not, however, translate to each $\ce{Na+}$ ion of an $\ce{NaCl}$ unit being more tightly bound to a $\ce{Cl-}$ ion than two atoms of a covalent bond are to each other.

The idea that ionic bonds are "stronger" than covalent bonds is confusing to early chemistry student (and maybe more so to chemists, but it's tough to convey concisely what is actually occuring) but, for some reason, seems to be nearly ubiquitous in general chemistry courses. It is much easier for compounds like water to stabilize $\ce{Na+}$ and $\ce{Cl-}$ than to dissociate something like paraffin wax into ions in order to solvate the molecules as ions.

So I guess what I'm trying to say is: the ions of ionic compounds aren't usually held more tightly together than two atoms of a covalent bond.

As for the idea that it is a higher electronegativity difference in $\ce{H-A}$ bonds that allow for higher acidity, that can be disproven by looking at $\ce{HF}$ vs $\ce{HI}$. $\ce{HF}$ has a higher electronegativity difference but is a weaker acid. This is because $\ce{I-}$ is larger, so it can spread the negative charge out, essentially stabilizing the negative charge on its own. $\ce{HF}$ produces $\ce{F-}$, and because fluorine is more electronegative than iodine, that may be expected to be a stronger acid than $\ce{HI}$. However, because $\ce{F-}$ is much more of a point-charge rather than a diffuse charge like $\ce{I-}$, it has to be stabilized by solvation effects to a larger extent.

So what causes an acid to be more acidic is the thermodynamic stability of the resultant conjugate base. Because the charge in $\ce{I-}$ is more stabilized by the ability to spread out its own charge, it more readily donates $\ce{H+}$ than does $\ce{HF}$.


metal oxide analysis I am looking in to analytical techniques for a future project. I may have a mixture of metals, i.e. an alloy, which I am thinking XRF spectroscopy would be the best quantitative technique. I could even purchase a handheld XRF analyser.

However, I believe that the XRF technology does not detect light elements such as oxygen. Therefore, if I have a mixture of metal oxides in a sample; what is the best quantitative analytical technique for me to identify each oxide?

I have had a look at XPS but I am confused about how I can work out which oxide I have in my sample. For example, if my sample contains Lead oxide and Copper oxide. How do I know whether it is Cu(II) oxide or Cu(I) oxide?

If the oxides are separate grains or compounds (a mixture, not a solid solution), you can find this with SEM microprobe (a geology technique).  For tricky new mixtures, may be better than powder x-ray diffraction (find small amounts of new compound, than resolve stoichiometry and do XRD on phase pure samples.  It will not give you the oxidation level though.

You could just do visual examination.  $\ce{PbO}$ is yellow and $\ce{PbO2}$ is black-brown.  $\ce{CuO}$ is black and $\ce{Cu2O}$ is red.  You can use a metallography microscope (high powered optical) to look at individual grains.  [May also be able to see the crystal habit.]

In general, people working with oxides know the expected oxidation state of the materials they have processed given the conditions.  E.g. cerium will always be $+4$, not $+2$ under atmospheric ceramic processing conditions.  Not always possible, but often the case.

If you could narrow down a little more what you are trying to do, might be easier to answer.  What substances, processing conditions and what info do you need on the end result (and why, really)?

ICP or AA are techniques helpful for trace metal analysis but you won't get oxidation state.

sorry, I can't give a more precise answer, but really your question is very broad.

If you can't discuss your project details on the net, consider to talk to the director of a research university chemistry department shared analysis lab.  I really don't think there is a single answer to your question and it will depend more on what you are looking at and how precise an answer you need, which technique is best.


Why is the effect of beta-branching neglected in this comparison of SN2 reaction rates? Which compound would give a faster SN2 reaction?

  2-bromobutane
  1-bromo-2-methylpropane

The answer is given (2).

While I agree that the primary position is preferred as compared to the secondary position (for an SN2 attack), but won't beta branching hinder the nucleophile's attack, hence, reducing the reactivity for SN2?

Yes, the beta branching will hinder the nucleophile a bit, but the substrate is still a primary halide with 2 hydrogens on the carbon that is attacked. The steric effect of having a methyl group and a hydrogen on that carbon vs two hydrogens is very much larger.


Is there a simple and inexpensive way to generate nitrogen gas? It's fairly easy to generate a significant yield of hydrogen gas simply by reacting zinc metal with $\ce{HCl}$ or aluminum with $\ce{NaOH}$ solution. But is there a simple and inexpensive chemical reaction that generates a significant yield of pure nitrogen gas?

Per Wikipedia, we have this laboratory method:

$$\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$

Also per Wikipedia, this method generates small quantities of impurities of $\ce{NO}$ and $\ce{HNO3}$, which can be eliminated by passing the gas through potassium dichromate $\ce{(K2Cr2O7)}$ dissolved in sulfuric acid $\ce{(H2SO4)}$

Both ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the $\ce{Cr2O7^{-2}}$ to $\ce{Cr2O3}$). But otherwise, this looks like the best approach.


What's the difference between a nucleophile and a base? Obviously it depends on the context whether you would call a particular species a nucleophile or a base but are the two terms largely synonymous or is there a difference?

The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior?

  Nucleophilicity is a kinetic phenomenon.

Nucleophilicity is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent.

For example, a standard reaction might look like:

$$\ce{CH3I ->[Nu-][H2O] CH3Nu}$$

The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$).

  Basicity is a thermodynamic phenomenon.

Basicity is based on the position of equilibrium:

$$\ce{B + HSol <=> BH+ + Sol-}$$


Will CO2 bubbles ever fully dissolve in water? Background:
Many aquarists keeping freshwater planted tanks add $\ce{CO2}$ from a pressurized canister to their aquaria to promote plant growth. Injection is at a slow rate to get dissolved $\ce{CO2}$ in the neighborhood of 30 ppm. Methods vary, but a ceramic discs to release a very fine mist of bubbles are popular, as are $\ce{CO2}$ reactors which trap the bubbles in a bell or canister and stir them around with a water pump.
In any case, within the hobby there is an odd fixation on getting the gas bubbles to dissolve completely, I suppose on the reasoning that a bubble that pops at the surface is $\ce{CO2}$ that escaped to air rather than being dissolved in the water.
I'm skeptical of this logic -- I'd think while the bubble is dissolving, water vapor and whatever other dissolved gasses will be diffusing into the bubble. At some point the bubble becomes just humid air in equilibrium with the surrounding water, and no amount of stirring will get more $\ce{CO2}$ out of it. This is just my guess, based on casual observations where the bubbles will diminish to less than 10% of their initial volume within 30 seconds, and thereafter they don't seem to get any smaller.
Questions:
Is that correct, and does it mean the aquarist's quest for a "100% efficient" $\ce{CO2}$ bubbler that never lets bubbles pop at the surface is doomed to eventual failure?
And if so, what are the basic principles that describe this equilibrium and how can we figure how much $\ce{CO2}$ can be extracted from a bubble, and how big the bubble will be after it's become just humid air?
If not, why does it seem the bubbles will rapidly dissolve in the water at first, but never fully dissolve even though the solution is far from saturated?

To improve your CO2 intake and not lower dissolved oxygen levels, an apparent solution is to make the water in your fish tank more alkaline and also consider increasing the surface turbulence.
The logic is that there is an equilibrium based dissociation of CO2 in water creating Carbonic acid. However, in the presence of OH- ions, the formation of an insoluble carbonate (like CaCO3) will continue to drive the reaction to the right.
On the effect of turbulence from an air pump agitation of the water or perhaps a waterfall creating mist, will capture the carbon dioxide.
So, as solutions consider adding alkaline stones. There is also a supporting article: Limestone Rocks in Aquariums and Water pH, to quote:

Have you set up an aquarium and then noticed that the pH of the water continues to rise? Could rocks or gravel that were added as decor in the aquarium affect the water pH? Yes. If your rocks are actually limestone, they are the cause for the pH elevation in your aquarium water. Limestone is calcareous (contains calcium) and is known for its ability to both harden the water and increase the pH. In fact, if someone asks how to raise water pH, one method we suggest is to place crushed limestone, coral, oyster shell or any highly calcareous material in the filter.

On the waterfall idea a reference, to quote:

Speaking of pond health and waterfalls, they're a great way to aerate your pond, but you might not want to overestimate how much oxygen is getting into your pond from the waterfall. Specifically, the larger your pond is, the higher the waterfall would need to be to effectively aerate it alone.

Hope this helps.


Can you lower the alcohol concentration of a drink using vegetable oils? The log POW of ethanol is −0.18 according to Wikipedia, i.e. in a two-phase system ethanol would preferrably accumulate in the aqueous phase, but not by much.

Suppose I wanted to lower the alcohol concentration in a drink, how well would adding some vegetable oil, mixing and removing the top layer with a straw work? What other parts of the drink are likely to accumulate in the oil? Which (unwanted) side effects can I expect?

Edit: Definition of Drinks

I'm interested in the effects on different types of drinks/alcoholic beverages from low-alcohol content beers to wines, whiskeys, vodkas and Stroh.

PS: While this may sound like a question for the cooking subsite, I am only interested in the chemistry.

Liquid-liquid extraction is a widely used method, but it's normally used the other way round: You add e.g. ether to a solution of your valueable substance X in water, separate the phases, and evaporate the ether to get pure X. Add some more ether to the aqueous phase, etc, two-three times.

Now you can be quite sure that you have recovered all of your substance X, but also your aqueous phase will be saturated with ether. No problem if you'd discard it anyway, not so good for drinking. :-/

It's also bad if X, like ethanol, is well soluble in both phases. For hard liquor, the whole mixture is likely totally miscible, and for beer, you'll need a lot of oil to get any significant effect. And then what do you do with the ethanolic oil afterwards?


Protein Data Bank: Asymmetric Anit/Biological Assembly For a schoolproject I have to interpret the 3D structure of an imatinib-BCR ABLC kinase complex. When I visited the PDB website I downloaded the PDB file corresponding to this complex. Once I opened it on my 3D modelling program it struck me that the protein complex was portrayed twice, looking almost the same but not completely. So after that I stumbled upon something called the Asymmetric Unit/Biological Essembly on the PDB site. I read the description  down this post here, but didn't fully understand it. Could anybody explain me what this is and how I'm able to import the right complex in my program?

Thanks in advance.

http://pdb101.rcsb.org/learn/guide-to-understanding-pdb-data/biological-assemblies

Most of the structures in the Protein Databank are determined with X-ray crystallography. Like inorganic compounds and small organic molecules, under the right conditions proteins can form crystals - ordered, repeating patterns of protein molecules, filling a 3D space. That is, you can take a basic building block and rotate and translate this building block to completely fill the three-dimensional space. (2D example).

So what's being modeled in an X-ray crystallography experiment is that basic building block, the one which undergoes symmetry operations (rotation and translation) in order to fill out 3D space. This basic building block is called the asymmetric unit, because it is without the crystallographic symmetries which result in filling space.

But even in the asymmetric unit there can exist noncrystallographic symmetry. These are symmetries (rotations and translations) which relate portions of the asymmetric unit, but aren't lined up quite right to fill space. Because they don't contribute to the overall symmetry of the crystal, normally they're not exactly symmetrical. Usually there are slight differences, based on how the noncrystallographic symmetry fits into the crystallographic one.

For a 2D example, think of a triangle which is tiled on a square grid. The square grid is the crystallographic symmetry. The triangle has noncrystallographic symmetry, because the three-fold rotational axis of the triangle does not contribute to tiling the plane. This means the corners of the triangle are not exactly equivalent, as each corner is in a different position with respect to the neighbors in the next square.

So when an X-ray cryptographer goes to solve the structure of a protein crystal, the primary unit of interest is the asymmetric unit - even if there's noncrystallographic symmetry in the asymmetric unit. So that's what the PDB records as its primary entry - the contents of the asymmetric unit, including the slight differences between any repeated units.

In your case (1iep), that's what happened. Because of the specific way the protein crystallized, there just happens to be two copies of the protein in the asymmetric unit, almost - but not quite - identical to each other. This representation of the asymmetric unit - including both proteins - is what you've downloaded.

Because this might not be what you're interested in, the PDB also provides one or more "Biological Assembly" files. These are processed files which attempt to model what the experimentalists believe* is the actual biologically relevant unit. In your case, the relevant biological unit is a monomer, so they provide two biological assembly files - one for each (distinct) monomer in the asymmetric unit. For multimeric proteins, it sometimes happens that one of the symmetry axes for the multimer lines up with the crystallographic axis. In this case, you'd only get part of the protein in the asymmetric unit download, but the biological assembly file would contain a reconstructed representation of the multimer.

If you want further, extensive detail, I would second the recommendation made here to look at the textbook Crystallography Made Crystal Clear, by Gale Rhodes.

(*The biological unit provided by the PDB is a guess. Sometimes you'll see proteins which should be multimeric represented as monomers, or monomeric proteins represented as multimers. It depends in part on how accurately the depositors annotated what the relevant biological unit was.)


Stability of carbocations We know that if a structure has all equivalent resonance structures, then it will be highly stabilised. So then why benzylic carbocation is more stable than other cyclic allylic carbocations, though allylic carbocation has all equivalent resonance structures?

More the number of resonance structures (alias mesomers) more is the stability of the carbocation so benzylic carbocation with 4 mesomers is more stable than 2 structures of the allylic carbocation.


What is the meaning of the prefix "per-"? The prefix "per-" appears in the names of many compounds, such as in perchlorate, persulfate, peroxide. What does it mean?

Per- can mean a number of different things, depending on the context.
It can mean that there is a peroxi bond(oxygen-oxygen bond) in the system, persulfate is a good example of this.

More generally it tends to mean that the compound is the most completely altered product of a certain(possibly hypothetical) reaction. For example perchlorates and permanganates are the most completely oxygenated compounds of chlorine and manganese, respectively. This also describes another common use of per-, perhalogenated compounds. In that context, it usually means that all hydrogens present in the parent compound, that were bonded to a carbon, have been replaced with a halogen, for example perfluoroethylene (C2F4, also known as tetrafluoroethylene, the monomer of Teflon(PTFE))


Mechanism of a gold catalyzed addition to triple bond I needed to draw the reaction mechanism for this reaction:

So, after consulting some literature I drew this mechanism:

But, I am not sure about the last step. Does the $\ce{H+}$ just substitute the $\ce{Au+}$ ion? If so, is my last step correct?
Or, something else happens in the last step?

In their 2005 The Journal of the American Chemical Society publication, Genêt, et al. have reported a gold-promoted cyclization reaction of bis-homopropagylic diols (see Graphical Abstract below), which afforded bicyclic ketals at room temperature in very good to excellent yields (ref. 1). This cycloisomerization reaction was catalyzed by either Au(I) or Au(II) chloride in methanol:

Abstract: A highly efficient gold-catalyzed cycloisomerization reaction of bis-homopropargylic diols is described. The cyclizations are conducted in the presence of either $\ce{Au^I}$ or $\ce{Au^{II}}$ catalysts in $\ce{MeOH}$ at room temperature in a very short time. The reaction conditions are compatible with functional groups, such as n-butyl, phenyl, allyl, benzyl, and alcohol groups, leading to original strained dioxabicyclo[2.2.1], -[2.2.2], or -[3.2.1] ketals in good to excellent yields.

There was a sequence of compounds has been used as starting material, which consist of different R- groups, one of them been Ph- and $n = n' = 1$ as in this example. A plausible mechanism for the cyclization is given in the paper, which is shown in  following scheme:

In there, 1a and 2a represent the appropriate structures with $n = n' = 1$ and R- = cinnamyl- group. The authors have given the following explanation for the mechanism:

The reaction may be initiated by the formation of the π-alkynyl complex A through the complexation of the unsaturated triple bond to the $\ce{Au}$ catalyst. The coordination of the triple bond, therefore, enhances the electrophilicity of the alkyne as usually advocated for metal-catalyzed reactions of alkynes (ref. 2). The addition of one alcohol may be favored by an intramolecular complexation of the $\ce{OH}$ group to the gold catalyst, as already envisaged by Teles and co-workers (ref. 3) for the addition of methanol to alkynes. The enol vinylgold intermediate B may then be protonolyzed leading to an enol ether, which then undergoes another intramolecular addition of the remaining alcohol leading to the cyclic ketal 2a. This latter step may also be catalyzed by gold, considering the proposed mechanism for the $\ce{Au(II)}$-catalyzed cyclization of α-hydroxyallenes (ref. 4).

In reference 2, it is stated that that the activation of the carbon−carbon triple bond toward nucleophilic attack can be performed by metals with high Lewis acidity (alkyne activation), as illustrated in Equation 1 of the following scheme. It is also easy to realize that an external nucleophile could make an anti-addition as a result and thus, the product should be obtained with trans-stereochemistry (Equation 1). Yet, as envisioned by Teles and co-workers (ref. 3), syn-addition is also possible by an external nucleophile, through complexation with metal, first by alkyne and then by the external nucleophile (see Last Scheme below). However, in many other cases (specifically where a nucleophile would be available intramolecularly as the case in here) the picture is quite different (Equation 2). Here, metal can activate the nucleophile first (nucleophile activation) and the reactions may proceed as a result of the insertion of the alkyne moiety into M−H or M−Nu bonds of “H−M−Nu” complexes, typically with syn-addition as illustrated in Equation 2 of the scheme (This is the case illustrated in A and B).

Last Scheme:

References:

S. Antoniotti, E. Genin, V. Michelet, J.-P. Genêt, “Highly Efficient Access to Strained Bicyclic Ketals via Gold-Catalyzed Cycloisomerization of Bis-homopropargylic Diols,” J. Am. Chem. Soc. 2005, 127(28), 9976–9977 (DOI: 10.1021/ja0530671).
F. Alonso, I. P. Beletskaya, M. Yus, “ransition-Metal-Catalyzed Addition of Heteroatom−Hydrogen Bonds to Alkynes,” Chem. Rev. 2004, 104(6), 3079–3160 (DOI: 10.1021/cr0201068).
J. H. Teles, S. Brode, M. Chabanas, “Cationic Gold(I) Complexes: Highly Efficient Catalysts for the Addition of Alcohols to Alkynes,” Angew. Chem. Int. Ed. 1998, 37(10), 1515–1418 (https://doi.org/10.1002/(SICI)1521-3773(19980605)37:10<1415::AID-ANIE1415>3.0.CO;2-N).
A. Hoffmann-Röder, N. Krause, “Gold(III) Chloride Catalyzed Cyclization of α-Hydroxyallenes to 2,5-Dihydrofurans,” Org. Lett. 2001, 3(16), 2537–2538 (DOI: 10.1021/ol016205+).


Concentration uncertainty due to volume measurement with cylinder (All volumes were measured using a measuring cylinder which has a $\pm\pu{0.5 cm3}$ uncertainty)

You have a $\pu{50 cm3}$ of $\pu{0.5 M}$ solution to begin with
You then make a  $\pu{0.4 M}$  solution by mixing $\text{40 cm}^3$ of the above solution with $\pu{10 cm3}$ water.

If I'm not mistaken, the uncertainty for the volume should be $\pm\pu{1.0 cm3}$ because the second solution is the result of two measurements (and also uncertainties).
My question is: what is the uncertainty for the concentration?

1. Let's assume that you're talking about random error. If you specify the error in the volume with $\pm$, this is probably the case---systematic error in this case would have a single sign; for example, if the scale marks on the side of cylinder were shifted by $\rm 0.5\ cm^3$ down from what their true values should have been, you'd have a systematic error of $-0.5\ \rm cm^3$ (the negative sign tells you that your readings are $0.5\ \rm cm^3$ lower than they should have been).

You can say that the error in a sum will be less than or equal to the sum of errors. It's better to say that the uncertainty in a sum is the square root of the sum of squared uncertainties:

$$\delta(V_1+V_2) = \sqrt{\delta(V_1)^2 + \delta(V_2)^2} = \sqrt{\rm(0.5\ cm^3)^2 + (0.5\ cm^3)^2} = 0.70_7\ \rm cm^3$$

For a product or quotient, you can say that the relative error will be less than or equal to the sum of the relative errors in the measurements. It's better to say that the relative error in a product or quotient is the square root of the sum of squared relative errors:

$$\begin{array}{rcl}
\delta(C_{dilute}) &=& \delta\left(\frac{C_{stock}V_1}{V_1+V_2}\right)\\
\frac{\delta(C_{dilute})}{C_{dilute}} &=& \sqrt{\left(\frac{\delta(C_{stock})}{C_{stock}}\right)^2 + \left(\frac{\delta(V_{1})}{V_{1}}\right)^2 + \left(\frac{\delta(V_{1}+V_2)}{V_{1}+V_2}\right)^2
}
\end{array}$$

...can you take it from there?

2. Let's assume you have systematic error, that is, you actually meant something like "The cylinder always reads $\rm 0.5\ cm^3$ too high". You will need a definite sign on a systematic error, not a $\pm$.

There are several ways to proceed here. Let's take the easiest route, which is to estimate the systematic error in the concentration as the difference between the concentration and its perturbed value (the value with all of the volumes having the systematic error added in). If your systematic error in volume was $\delta V = +0.5\ \rm cm^3$, you can estimate the systematic error in the concentration as

$$\begin{array}{rcl}
 \delta{C_{dilute}} &\approx& C_{dilute}^{perturbed} - C_{dilute} \\
&\approx& \frac{C_{stock}(V_1+\delta(V)}{(V_1+\delta(V))(V_2+\delta(V))} - C_{dilute} \end{array}$$

The sign on this error will be important; it tells you whether the systematic error in volume makes the concentration higher or lower than its "true" value.

Note: The above calculation is just an estimate of the error. You can do a better job using derivatives here. I'll show you that if you want to see it.


How to alkalize moist fabric Here's some practical application of chemistry for ya.

In short:
I have a moist, hard-to-remove carpet in a car that's hard to get dry.
I treated it with an anti fungal agent containing 1/5 hydrogen peroxide supposed to kill the mold. Measured it - it's very acidic - a condition mold will like in the near future.

How do I alkalize the carpet without creating a mess? I'd like to do this w easily obtainable everyday substances.

(Explaining "mess": E.g. pouring calcium hydroxide on it would probably be a hard-to-remove mess?)

The long story for context:
I had a leak between the interior of my car and the engine compartment let rainwater moisten the carpet.
The leak was fixed, but in the past few months had really the wrong weather for getting it dry, silica absorber packs didn't help.
So, it started to mold.

To stop it, I bought an anti-mold agent fluid. I tested its pH value and it's off the scale acidic for my litmus paper.

Mold likes acidity, hence, as soon as the mold killing by oxidation wears off, the mold will like it even more, which is why I think I need to alkalize the carpet.

If you really want to neutralize the acid of your mould removal product and don’t want to leave solid deposits, you can try a dilute aqueous solution of ammonia.

You can find typical solutions containing for example about 25 % ammonia (sometimes labelled “ammonium hydroxide”) in hardware stores. Such solutions are corrosive; they can cause severe skin burns and eye damage. You should dilute the solution to well below 5 %.

Ammonia is volatile; you will notice its characteristic pungent smell and you should thoroughly air the treated area. For the same reason, excess ammonia will evaporate and not leave solid deposits. However, it can form salts with the acid of your mould removal product and with carbon dioxide from the air. Such salts are water-soluble and, if necessary, they could be removed easily.

Ammonia is not corrosive to the metal parts of your car (ammonia is actually added to the feedwater of industrial boilers in order to keep the pH high and to reduce the risk of corrosion), except for open electrical contacts made of copper. However, you should test the ammonia solution on an unobtrusive spot to see whether it affects the colour of your carpet.


Can nitrogen dioxide react with plastics of any kind? I have searched a lot on the internet but couldn't find any helpful thing. My question is that can nitrogen dioxide react with plastics of any kind?

Nitrogen dioxide is a free radical but its energy is rather low so it will be unlikely to be able to abstract a hydrogen from most polymers. If we consider polypropylene as the "standard plastic" then I would not expect nitrogen dioxide to be able to corrode it by abstracting hydrogens from the plastic.

But if we choose something else such as an epoxy as the standard plastic then nitrogen dioxide when combined with water could start to chemically alter the plastic. It can react with amines to form things such as N-nitroso compounds.

When nitrogen dioxide dissolves in water it forms nitric and nitrous acids. It can be regarded as a mixed acid anhydride for these two nitrogen oxy acids. The nitric acid could also start to degrade and alter plastics such as epoxy plastics by nitration of electron rich rings.

The acid formed when nitrogen dioxide dissolves in water could degrade plastic by increasing the rate at which esters such as the phthalate diesters react with water. It is important to keep in mind that many plastics are a combination of a polymer (or several polymers) and some additives such as plasticisers. If you degrade the plasticisers some plastics will stop being "plastic" and become brittle.


Can the nitrogen of an amide displace a primary chloride in a SN2 reaction in the presence of a strong base? I was looking at the following reaction and am confused.

It appears that $\ce{N}$ is attacking the $\ce{C-Cl}$ and $\ce{Cl}$ is leaving in a $\mathrm{S_N2}$ reaction. I know that $\ce{Cl}$ is a good leaving group, but $\ce{N}$ isn't a particularly soft nucleophile so I wouldn't expect it to be great at attacking a saturated carbon.

Additionaly there is $\ce{MeO-}$ in solution, even with intramolecular cyclisation being quick wouldn't there be a realitive amount of attack by $\ce{MeO-}$.

So am I missing something? Is this not as simple an $\mathrm{S_N2}$ as I have made out?
Why is sodium methoxide ($\ce{NaOMe}$) a good choice of base?

Well it is difficult to compare an intra-molecular reaction with an extra-molecular one because it really depends on the conditions. However, unless you have a very high concentration of one reactant, the intramolecular way will always be favored.

In your question, you consider that "N" is not a very good "attacking group". Well this is wrong. "N" is always happy to attack any nucleophilic group, even if it becomes "N+". However, a "CON-" is much more nucleophilic than just "CONH".

Sodium methoxide here just serves at deprotonating CONH to activate it, but MeONa (sodium methoxide) is not nucleophilic enough to interfere.


How can I increase the viscosity of dilute hydrochloric acid I'm cleaning the mortar off a new clinker brick wall. Walls actually. I hate this job. The builder told me to wash down the wall with weak HCl, leave it for a minute or two, and then spray it off with a pressure hose (eg Kärcher or Gerni). There are lots of problems with this. The main one is that the HCl gets painted on with a coarse bristled brush and splashes everywhere you don't want it to be. The protective clothing is a huge impediment to being able to do the actual job. The safety eye protection fogs up and you can't see how well you've cleaned. Once the bricks are wet, the mortar residue becomes transparent, so it looks clean, only to dry out and turn white again. Then there's the problem with blasting weak acid all over the place. I really hate this job.

So I was thinking that in order get some control over the application of the acid, it would be great if it was the consistency of paint. Then I could apply to only those areas that it needs to be. Question is, how do I make my dilute HCl the consistency of paint? Please feel free to correct me if viscosity is not the right word.

If the solution is not too concentrated and you dont need it to be stable for a prolonged period, many viscosity modifiers can be used. I would start trying something simple, like CMC based wallpaper paste.


Why do the atoms in molecular models have the colors they have? Molecular models are a great tool for visualizing a molecule in three-dimensional space. These models even appear on the web (practically every chemical's Wikipedia page), and there is a very specific color coding for atoms. After some searching I found that it was called the CPK coloring system. It seems rather intuitive for some elements, like for $\text{carbon}$, $\color{\green}{\text{fluorine}}$, and $\color{\yellow}{\text{sulfur}}$, which use the colors for the elemental forms of these atoms (diamonds aside).

On what basis, though, did Robert Corey and Linus Pauling choose the colors for $\color{\red}{\text{oxygen}}$, $\color{\turquoise}{\text{nitrogen}}$, and hydrogen (I'm not going to do white on white). Were these arbitrary choices that had to be made for these colorless elements, or is there some reasoning behind them?

There's an interesting article here - Illustrating Atoms and Molecules - that discusses atom colors:

  In 1865, the chemist August Hoffman gave a Friday Evening Discourse at London’s Royal Institution on the “Combining Power of Atoms.” In order to demonstrate chemical bonding of atoms, he drilled holes in croquet balls and connected them with metal pipes (Figure 2). His choices were limited by the available colors of croquet balls and he relied on many of the same color conventions that had been adopted centuries earlier

These colors are listed as :

  Carbon is colored black because it’s the color of charcoal.
  Oxygen is red because it’s necessary for combustion.
  Nitrogen is blue because it’s the most abundant element in the
  Earth’s atmosphere and the sky appears blue.
  Hydrogen is white because it forms a colorless gas.
  Chlorine is green because it forms a greenish gas.
  Sulfur is yellow because that’s its color in mineral form.
  Phosphorus is orange because it glows orange in a flame.
  Iron is reddish brown because it rusts.

Which is interesting, but I've no idea where this list comes from...


Finding the sensitivity of zero-order ultra sensitivity and its parameter dependence I am provided a one-dimensional reaction system showing zero-order ultra sensitivity.
$$\ce{X_1 ->[f_1] X_2}, \quad \quad \quad \ce{X_2 ->[f_2] X_1}$$
The reaction rate function is the reduced enzyme reaction according to Michaelis-Menten's law.
$$f_1(x) = E\left(\frac{x_1}{K_1+x_1}\right)$$
$$f_2(x) = \left(\frac{x_2}{K_2+x_2}\right)$$
$$K_1=K_2=K$$
$$x_1(t) + x_2(t) = x_0$$
where $x_0$ is a constant. Let the steady state of $x_2$ be denoted by $x_2^{\text{st}}$. I need to find the value of $E$ that satisfies
$$x_2^{\text{st}}(E) = \frac{x_0}{2}$$
and the logarithmic sensitivity. I tried to look into some published papers/textbooks but could not find similar problems. Can anyone give some suggestions on directions to look into?

I found the answer.
First, we need to find balanced equation for reaction. Then express the concentration of $xˆst$ in the steady state through E.
Then we can arrive to this kind of solution for the value of E.
$\displaystyle{E\frac{x_0 - \frac{x_o}{2}}{K + x_0 - \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} = 0.$
$\displaystyle{E\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}}= 0. $
E = 1


Is the XRF spectroscopy equally sensitive for every element? I am curious whether or not every element can be equally well detected by using X-Ray-fluorescence-spectroscopy. Might there be any contitions or circumstances that make it hard/impossible to detect a certain element in a probe?

In a very very broad reasoning, XRF is much more sensitive to the "mid-range" elements.

Elements below Na are generally not detectable.

The x-ray tubes must have a beryllium window to seal the tubs which absorbs the low energy x-rays.
The lower the atomic number the less the fluorescent yield.
Low energy x-rays don't penetrate from the sample efficiently.
Low energy x-rays tend to be absorbed in the detector windows.

For the very heavy elements the detectability is poor too. Say anything about lanthanum.

The K x-rays are so energetic that a high voltage on the x-ray tube is needed to excite them.
The lanthanides and actinides tend to behave similarly chemically so geological samples tend to contain a mixture which creates peak overlaps.
Higher energy detection of x-rays is poor. EDS because the x-rays penetrate the detector. WDS because the available crystals don't diffract high energies well.
The L x-rays are not quite as useful because the greater number of x-rays leads to more peak overlap.


Rules to identify the most stable resonance structures While studying resonance from this Chemistry LibreTexts article, I found the rules to follow in order to decide which structure is the most stable. Among them:

The structure with the least number of formal charges is more stable
The structure with the least separation of formal charges is more stable.

Firstly I'm not sure how to interpret the "number of formal charges" in rule 1: does it refer to the number of atoms with non-zero formal charge or to the magnitude of each formal charge? It confuses me a bit also because the total number of formal charge (the sum of them) is clearly constant (and equal to the total charge or molecule).

Secondly I don't understand which of these two rules (1 or 2) is more important and can dominate the other one.

For example two possible structures of $\ce{CNS^{-}}$ are
$$\ce{C=N=S}\qquad\ce{C#N-S}$$

The formal charge for the first one are (in order) -2, 1, 0 while for the second one are -1, 1, -1.

The first is preferable according to rule 1 (interpreting it as the number of atoms with non-zero formal charge) while the second is preferable according to rule 2 since the separation of charge is not as strong as in the first structure (I hope I'm understanding correctly rule 2).

Am I reading incorrectly the two rules or is there one that is actually "stronger" than the other?

You will encounter situations such as these when the two rules seem to contradict each other, or they confuse you.
However, when this happens, you must use another criteria to determine the best resonance structure. Specifically, you must take electronegativity into account.

What this means is that the most electronegative element should have the lowest formal charge, and the most electropositive element should have the highest formal charge.
This is because the electronegativity "rules" explained above are what are likely to happen (that is how thiocyanate is likely to appear). This makes sense because the higher EN and element has, the more likely it is to "hog" the electrons and have a lower formal charge.
When you use these rules to evaluate the resonance structures of the $\ce{CNS^{-}}$ ion, you can figure out why the first resonance structure is preferable.


What are aryl chlorides currently used for? In a recent paper by Gustafson et al. (DOI: 10.1021/acs.orglett.6b02650) the authors open with the argument that

  aryl chlorides are both versatile synthetic handles and common functionalities in drug discovery.

However, they fail to give examples of current use in industry and pharmaceutical research.

I know of only one reaction that aryl chlorides are used for, and that would be the Ullmann biaryl (ether) coupling. Are there any other ones for which aryl chlorides are a good starting material?
And then, about the second part of the sentence: Why are aryl chlorides a common functionality in drug discovery? Is it simply that compounds are chlorinated and then compared to their unchlorinated analogues?

All in all, please provide examples for the quoted statement above.

Synthetic handles: for organometallic couplings (Sonogashira, Suzuki...), there is a selectivity ($\ce{aryl-I} > \ce{aryl-OTf} >\ce{aryl-Br} >\ce{aryl-Cl}$) that can be crucial in the total synthesis. You can make the other position ($\ce{aryl-I}$ or whatever) react first, then use modified conditions suitable for aryl chloride couplings (plenty of examples on this website). Or make an halogen exchange before coupling your chloro- position.
Common functionalities: halogen bonds can change binding affinities. Even if they are weaker than hydrogen bonds, they are specific (see C. Bissantz et al., J. Med. Chem., 2010, 53 (14), 5061–5084.). They can make interactions with electrophiles, nucleophiles, or with themselves (ex: $\ce{C-X...O}$ $\ce{sp^2}$ or $\ce{X...X}$). One example of molecule designed with chloroaryl moieties is
represented here (see Furet et al., Bioorg. Med. Chem. Lett.
2012, 22 (10), 3498–3502, which shows the docking model). This is an inhibitor which blocks the binding pocket of the
regulator protein MDM2. The 6-chloroindolyl moiety enables to fill a
subpocket (TRP 23), while the p-chlorophenyl fills another one (Leu 26). The chlorine-chlorine interactions between the 6-chloroindolyl moiety and the triptophan residue in the protein was a key point in the design of this inhibitor, strongly enhancing the binding affinities.