The effect of pH and solubility on coagulation in water treatment In water treatment, a specific pH range must be met in order for the process of coagulation to occur properly. Sources have stated this is due to the pH affecting the solubility of the coagulant, but why is solubility significant?

To start, I assume you are speaking of metal salt coagulants such as aluminum sulfate (alum) and ferric chloride, as these are the most commonly used. As you have stated, their solubility is affected by pH, the simplest representation of the reaction for alum being:
$$\ce{Al^3+ + 3H2O <=> Al(OH)3,_s + 3H+}$$

Small particles in the water such as clay or bacteria are too small to settle out on their own. The purpose of coagulation is to either cause the smaller particles to aggregate and settle or to trap them during precipitation. Coagulation occurs via three main mechanisms: (1) sweep - where the metal salts precipitate (i.e., solubility) and capture particles/contaminants like a net; (2) adsorption/charge neutralization - where the positive charges on the Al and Fe cations neutralize the negatively charged e.g., clay, bacteria, etc., causing them to aggregate and settle out; (3) adsorption and bridging - where a polymer coagulant can be used to "bridge" together like charged particles, causing them to aggregate and settle out. All are affected by solubility, but sweep is the mechanism that uses precipitation of the metal salts to remove particulates.

The mechanism you will want to choose will depend on the source water and the coagulant. The mechanisms are controlled by pH and solubility. Below are the solubility diagrams for alum and ferric chloride, with solubility represented on the left y-axis, dose on the right y-axis, and pH on the x-axis. As you will see, the shaded areas are the regions for achieving optimal coagulation, depending on the mechanism you desire. For example, if your laboratory tests showed that sweep coagulation was the best for your source water and you chose alum, then you would want to be at a pH between ~7 to 8 and an alum dose of ~15 to 60 mg/L alum (the "optimal sweep" circle). This pH and dose would result in the desired solubility behavior to achieve optimum sweep conditions.

*Figure 9-11 from MWH Water Treatment: Principles and Design, 3rd Ed (Crittenden)


The reaction of Alkyl Halides with sodium nitrite in DMF How does the reaction of Alkyl Halides with sodium nitrite in DMF proceed?

I was aware of the reaction of alkyl halides with aqueous sodium nitrite which results in the formation of an alkyl nitrite but I came across the following reaction and was confused of how that would work?
$$\ce{R-X + NaNO2 ->[DMF] R-NO2}$$
DMF being the solvent here, I'm pretty sure it's safe to assume that it will be following $\mathrm{S_N2}$ path here but since I have never seen this reaction before, I'd appreciate if someone could enlighten me.
Also, looking at the reaction, I thought maybe DMF makes the sodium nitrate covalent somehow? (I know this is baseless but that's all I could think) which looks quite confusing.

I came across a post very similar to this but I think that didn't answer what's being asked. So I was hoping if someone could explain the mechanism of this (including if its $\mathrm{S_N1}$ or $\mathrm{S_N2}$)?

The $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction of alkyl halide with nitrite ion is complicated one since nitrite is an ambident nucleophile. For instance, it was reported that silver nitrite reacts with alkyl halides to give nitrites while sodium nitrite gives more nitroalkane than alkyl nitrite (based on Hard Acid/Hard Base concept; Ref.1):
$$\ce{AgNO2 + R-Br -> R-ONO + AgBr \tag1}$$
$$\ce{NaNO2 + R-Br -> R-NO2 + NaBr \tag2}$$
However, these results were contradicted by work of Kornblum who compared the reactions of iodoalkanes with $\ce{NaNO2}$ and $\ce{AgNO2}$ (Ref.2):
$$\ce{C7H15CH2-I + AgNO2/NaNO3 -> C7H15CH2-NO2 + C7H15CH2-ONO + AgI \tag3}$$
It was shown by the work of Kornblum and coworkers that both primary alkyl bromides and iodides have given excellent yields of pure nitroalkanes (73-83% as major product) when $\ce{AgNO2}$ is used as the nucleophile (Ref.3). In contrast, primary chlorides are completely inert under the conditions $(\pu{0\!-\! 25 ^\circ C}/\ce{Et2O})$:
$$
\begin{array}{l|r|r}
\hline
\text{Heloalkane} & \% \ \ce{R-NO2} & \% \ \ce{R-ONO}  \\
\hline
\text{$n$-Butyl bromide} & 73 & 13 \\
\text{$n$-butyl iodide} & 74 & 12 \\
\text{$n$-Hexyl chloride} & 0 & 0 \\
\text{$n$-Hexyl bromide} & 76 & 10 \\
\text{$n$-Hexyl iodide} & 78 & 13 \\
\text{$n$-Heptyl bromide} & 79 & 11 \\
\text{$n$-Heptyl iodide} & 82 & 10 \\
\text{$n$-Octyl chloride} & 0 & 0 \\
\text{$n$-Octyl bromide} & 80 & 14 \\
\text{$n$-Octyl iodide} & 83 & 11 \\
\hline
\end{array}
$$
Based on the results obtained, the authors had concluded that this is a good method to prepare primary-nitroalkanes (Note that nitroalkanes, $\ce{R-NO2}$, are thermodynamically more stable than alkyl nitrites, $\ce{R-ONO}$; Ref.1). Some of the further work by Kornblum et al. have noted that use of tertiary alkyl halides to prepare corresponding nitro compounds is useless (0-5% yield) while those reactions with secondary alkyl halides have also given only less than 15% yield of expected nitroalkanes (Ref.4). After extensive  work on synthesis of nitroalkanes and alkyl nitrites, declared that:

In the synthesis of saturated primary nitro compounds, silver nitrite gives nitroparaffins in about 80% yield as against about 60% yields obtained with sodium nitrite. While silver nitrite is, therefore, the reagent of choice here, the lower cost and ready availability of sodium nitrite, the not very large disparity in yield, and the shorter reaction time all combine to make sodium nitrite an excellent second choice*" (p.114, Ref.2; and Ref.5).

In their mechanistic studies, Kornblum et al. (Ref.6) have found that yield of $\ce{R-NO2}$ isomer falls progressively when substrate changes from primary to secondary to tertiary halide. In contrast, the yield of $\ce{R-ONO}$ counterpart increases with the same order of primary to secondary to tertiary. Although there is a steric effect in play, the authors conclude that the reaction proceeds via a transition state, which has both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ character in proportions that vary gradually with the structure of the halide. The greater the $\mathrm{S_N1}$ character of the transition state, the greater is the preferences for covalency formation with the atom with higher electronegativity (e.g., $\ce{R-ONO}$). Conversely, the greater the $\mathrm{S_N2}$ contribution to the transition state, the greater the preferences for covalency formation to the atom with lower electronegativity (e.g., $\ce{R-NO2}$).
In this regard, I'd say the reaction, $\ce{R-X + NaNO2 ->[DMF] R-NO2 + NaX}$ would proceed differently under different conditions. Since the solvent is (assuming) anhydrous DMF, the condition is set for $\mathrm{S_N2}$ reactions. Thus, for instance:

If $\ce{X = Cl}$ and $\ce{R-X}$ is a primary alkyl chloride, the reaction would not proceed.
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a primary alkyl halide, the reaction would
proceed rapidly to give nitroalkane, $\ce{R-NO2}$, as the major product but in lower yield $(\approx60\%)$. There should also be some alkyl nitrites, $\ce{R-ONO}$ as well (Ref.3).
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a secondary alkyl halide, the reaction would
proceed to give both nitroalkane, $\ce{R-NO2}$, and alkyl nitrites, $\ce{R-ONO}$, as the product. The yield would be lower $(\approx20\%)$ (Ref.4).

I think that would be enough of discussion since very little information is given in the question. For example, we cannot decide whether reaction goes in ether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ without knowing the nature of $\ce{R}$ group. Knowing the structure of $\ce{R}$ group is extremely important since even some specific primary halides such as neo-pentyl halide do not react under same conditions where other primary halides proceed rapidly.
Finally, it is noteworthy to mention following reaction, which play a significant role in the product ratio of $\ce{R-NO2}$ to $\ce{R-ONO}$:

Reactions of the blue benzhydrylium salts, $\bf{1}$-$\ce{BF4}$, with $\ce{n-Bu4N+NO2-}$ in anhydrous acetonitrile gives colorless adducts, $\bf{1}$-$\ce{NO2}$ (Ref.5):

The monitoring method in Ref.5 does not suggest the product is direct formation of $\bf{1}$-$\ce{NO2}$ or through the kinetic product, $\bf{1}$-$\ce{ONO}$. It was suggest that when use more reactive benzhydrylium salts for the reaction, the monitored reaction process is the formation of $\bf{1}$-$\ce{ONO}$, which successively rearrange to the observed nitro compounds $\bf{1}$-$\ce{NO2}$. To prove this theory, the authors prepared pure $\bf{1}$-$\ce{ONO}$ in different method (Reaction $\bf{A}$), and monitored its rearrangement to thermodynamically stable $\bf{1}$-$\ce{NO2}$ (Reaction $\bf{B}$):

These experiments clearly suggest that preparation of nitroalkane is not simple as the reaction suggests, but rather complicated process!

References:

Ian Fleming, In Frontier Orbitals and Organic Chemical Reactions-Student Edition; John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2009, p. 122–123 (ISBN 978-0-470-74660-8).
Nathan Kornblum, “The Synthesis of Aliphatic and Alicyclic Nitro Compounds,” Organic Reactions 1962, Vol 12, 101 – 156 (DOI: https://doi.org/10.1002/0471264180.or012.03).
Nathan Kornblum, Bernard Taub, and Herbert E. Ungnade, “The Reaction of Silver Nitrite with Primary Alkyl Halides,” J. Am. Chem. Soc. 1954, 76(12), 3209–3211 (DOI: https://doi.org/10.1021/ja01641a029).
Nathan Kornblum, Robert A. Smiley, Herbert E. Ungnade, Alan M. White, Bernard Taub, and Stephen A. Herbert Jr., “The Reaction of Silver Nitrite with Secondary and Tertiary Alkyl Halides,” J. Am. Chem. Soc. 1955, 77(21), 5528–5533 (DOI: https://doi.org/10.1021/ja01626a028).
Alexander A. Tishkov, Uli Schmidhammer, Stefan Roth, Eberhard Riedle, and Herbert Mayr, "Ambident Reactivity of the Nitrite Ion Revisited," Angew. Chem. Int. Ed. 2005, 44(29), 4623 –4626 (DOI: https://doi.org/10.1002/anie.200501274).
Nathan Kornblum, Robert A. Smiley, Robert K. Blackwood, and Don C. Iffland, “The Mechanism of the Reaction of Silver Nitrite with Alkyl Halides. The Contrasting Reactions of Silver and Alkali Metal Salts with Alkyl Halides. The Alkylation of Ambident Anions,” J. Am. Chem. Soc. 1955, 77(23), 6269–6280 (DOI: https://doi.org/10.1021/ja01628a064).


Obtaining ammonium perchlorate via double displacement reaction I was tasked with figuring out how to make some $\ce{NH4ClO4}$(ammonium perchlorate). I found that one way of making it was to react potassium perchlorate and ammonium chloride in a double displacement reaction. according to:

$\ce{KClO4 + NH4Cl -> NH4ClO4 + KCl}$

Then I got to work on finding ways of making the necessary reactants without having to buy them by doing additional reactions with easier to find chemicals. After some work I came up with these two reactions:

$\ce{KCl + NH4NO2 -> NH4Cl + KNO2}$ (to make the ammonium cloride)

and

$\ce{KCl + HClO4 -> KClO4 + HCl}$ (to make the potassium perchlorate)

My question is (the newbie part) how should I now go about these reactions? Should I be mixing the reactants together in an aqueous solution? In water? Should I be mixing them into an ionic solution in an electrolytic cell? Or hell do I just mix them together as a granulated form and apply a little heat? How do I get these reactions to occur?

The Wikipedia article for ammonium perchlorate gives a hint at how this can be done. The hint is that ammonium perchlorate is curiously less soluble than sodium perchlorate, and thus with the right concentrations, you would get ammonium perchlorate to precipitate by mixing solutions of sodium perchlorate and and, say, ammonium chloride.

$$\ce{NH4Cl(aq) + NaClO4(aq) -> NH4ClO4(s) + NaCl(aq)}$$

That being said, the best advice that can be given on how to make ammonium perchlorate is

  Don't. Especially do not use heat!

Ammonium perchlorate is fairly dangerous explosive that can decompose with not much provocation into a whole heck of a lot of gas for just a little bit of solid.

$$\ce{2NH4ClO4(s) -> N2(g) + Cl2(g) + O2(g) + 4H2O(g)}$$

That this decomposition produces toxic chlorine gas should only make you want to prepare ammonium perchlorate less.

Safely producing, storing, and transporting this compound is going to require a lot of specialized equipment and training (which is expensive). The danger here is not of the type that may lead to funny smells and dizziness in poorly ventilated rooms. The hazards associated with the compound involve loss of life or limb and serious structural damage if you look at it funny. Don't. For more information into the sort of destructive power ammonium perchlorate has, consider that this is the rocket fuel that got the space shuttle into low earth orbit. This stuff has the power to level industrial complexes with the forces of an earthquake.


Mechanism for palladium-catalysed spirocyclisation of conjugated diene I'm trying to work out the catalytic cycle for the following process:

I gather that the palladium will be reduced from Pd(II) to Pd(0) in the process and that the 1,4-benzoquinone is a redoxidant. However, I am not sure where the palladium would be initially inserted (possibly one of the double bonds?) and how the chlorine ends up on the final product.

You don't have to wonder about which alkene coordinates to the Pd(II) centre. In this case, they are conjugated and therefore can coordinate together as a diene, which is a 4-electron donor. Many pi complexes are known ranging from alkenes (2e donor), to allyl (3e), to diene (4e), to cyclopentadienyl (5e), ... so this is not something really surprising.

Coordination of the diene to Pd(II) makes it more electrophilic. The pendant hydroxyl group simply does a 5-exo-trig cyclisation to form the C-O bond, generating a Pd(II)-allyl species. In terms of stereochemistry, the OH group has to attack anti to the metal. Deprotonation can happen with acetate, or the hydroquinone dianion that benzoquinone is later reduced to.

This Pd(II) allyl species is an intermediate in the well-known Tsuji–Trost allylation. In this case, the Pd-allyl species is being generated in a different way, but goes on to react in the same way. Chloride being a relatively soft nucleophile attacks the allyl moiety, again anti to the Pd, leading to a Pd(0)-alkene complex. The stereochemistry of the chlorine isn't indicated in your product but this is my guess at what's going on.

Dissociation of the alkene directly gives the desired organic product and reoxidation of Pd(0) to Pd(II) with benzoquinone regenerates the catalyst.

Looking this reaction up on Reaxys, it seems that it is taken from a 1991 article by Bäckvall and Andersson.1 The mechanism posited in the paper is identical to what I have written above, and on top of that my guess at the stereochemistry was correct: the authors indicate that the product obtained was 99% cis (referring to the chloride and ether oxygen in the product).

Interestingly, with $\ce{Li2CO3}$ instead of $\ce{LiCl}$, the product obtained has a trans acetate. Carbonate isn't a great nucleophile so it doesn't act as one, but it can deprotonate acetic acid. The acetate ion thus produced then forms a bond to Pd(II) before it migrates to the allyl system.

Reference

Bäckvall, J. E.; Andersson, P. G. Stereocontrolled oxaspirocyclization of conjugated dienes via palladium catalysis. J. Org. Chem. 1991, 56 (7), 2274–2276. DOI: 10.1021/jo00007a002.


Formulas for non-bonded interaction energies If one were to calculate the non-bonded interaction energy between two atoms, this would equate to the sum of the vdW + electrostatic potential energies:

$$
E_{\text{non-bond}} = E_{\text{vdW}} + E_{\text{electrostatic}}
$$

Could anyone please explain how I could calculate these potentials providing a formula?

Perhaps given an example with zinc ($\ce{Zn^2+}$) and oxygen ($\ce{O}$), where $\sigma$ and $\varepsilon$ values (defined by Charmm27) are given as:

\begin{array}{ccc}
\hline
\text{Species} & \sigma & \varepsilon \\
\hline
\ce{Zn} & 1.942 & 0.25 \\
\ce{O}  & 3.029 & 0.12 \\
\hline
\end{array}

Ultimately I just wish to understand how they are calculated.

Simply put, I am trying to calculate non-bonded interaction energy between a ligand and a protein. However, after trying to understand each of these terms I am a little lost.

You've got the Lennard-Jones potential parameters for zinc and oxygen from a certain version of the CHARMM force field. The commonly-used notation is $\varepsilon$ for potential well depth and $\sigma$ is the distance where the pair potential is zero, and $r_{min}$ is the distance which minimizes the L-J potential energy function. The linked Wikipedia article gives the potential as:

$$E_{\rm L-J}(r) = 4\varepsilon\left[\left({\sigma\over r}\right)^{12} - \left({\sigma\over r}\right)^{6}\right] = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]
$$

At $r = r_{min}$, $E_{\rm L-J} = -\varepsilon$ and we can map $r_{min} = 2^{1/6}\sigma$ to work with either form of the L-J potential function.

Accurate Calculation of Hydration Free Energies using Pair-Specific Lennard-Jones Parameters in the CHARMM Drude Polarizable Force Field. J. Chem. Theory Comput. 6(4) 1181–1198 (2010) describes (among many other things) how the L-J potential term is calculated for 2 interacting species (as in your case). See equations (1), (2), and (3) in that paper, reproduced below (where I have used $r_{min}$ instead of $R_{min}$ for consistency).

In the paper, the Lennard-Jones potential is given as

$$E_{\rm L-J}(r) = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]$$

For a pair of interacting species $i,j$:

$$r_{min} = {r_{min,i} + r_{min,j}\over 2}$$

and

$$\varepsilon = (\varepsilon_{i}\cdot\varepsilon_{j})^{1/2}$$

For your system, we compute $r_{min}$ from the given values of $\sigma$ (using the mean of the two values of $\sigma$, as specified in the manuscript):

$$r_{min} = 2^{1/6}\sigma = 2^{1/6}\cdot {1.942 + 3.029\over 2} = 2.789\;\unicode{xc5}$$

and

$$\varepsilon = (0.25\cdot 0.12)^{1/2} = 0.1732\;\mathrm{kcal\;mol^{-1}}$$

Now, at some distance $r$, you can compute $E_{\rm L-J}(r)$.

In the linked publication, the authors also go through the process of how the L-J parameters are derived/fit, which may be of interest to you.

Finally, see this vintage web page authored by a familiar guy which goes through a procedure for taking parameters for computed hydration free energies of cations and deriving L-J parameters for use in the AMBER force field. The procedure is very similar to that used for CHARMM.


Chemical composition of petrol What is the chemical composition of petrol? What are the molecular available in petrol? Some time which is marked with 95% octane. So what is the remain element occupied 5%?

The composition of petrol, also called 'gasoline' or 'gas' in the USA, despite it being a liquid at room temperature, contains a complex mixture of hydrocarbon compounds selected for use in internal combustion engines.

The term "95% octane" refers to the 'octane rating', 'octane number' or 'anti-knock index' of the fuel, not the amount of 'octane' in the fuel (since 'octane' is a specific organic compound in the group of alkanes).  The octane number / rating is an 'anti-knock index', based on an arbitrary scale indexed to a liquid mixture of iso-octane (2,2,4-trimethylpentane and n-heptane). Whilst different countries have different standards for anti-knock index, in the USA, the American Society for Testing and Materials (ASTM) defines two popular indices called the 'research octane number' (RON) and motor octane number (MON). Regardless if the specific stanard, the general principle is that the higher the octane number, the more the fuel can be 'compressed' before 'auto-igniting'. Auto-ignition is not desirable in petrol engines as it causes 'knocking' of the pistons which greatly shortens their life.

What petrol (gasoline) actually contains is a blend of refinery products, including reformate (high in aromatic hydrocarbons, low in alkenes), olefinic hydrocarbons, alkylate (high in paraffinic hydrocarbons) and isomerate (high in branched isomers of pentane and hexane). Specific blends vary from manufacturer. Manufacturerers even offer a range of blends at the bowser (gas station) including 10% ethanol and 'premium' blends for high performance engines.

Refer to 'Development of a Detailed Gasoline Composition-Based Octane Model' for more information.


Entropy change when indistinguishable particles suddenly become distinguishable We have that

$$
S = \frac{U}{T} + k\ln Q
$$

Using that for indistinguishable particles in the canonical ensemble

$$
\ln Q_{\mathrm{indis}} = N\ln \frac{q_{\mathrm{tr}}e}{N} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$

Similarly for distinguishable particles we have that

$$
\ln Q_{\mathrm{dis}} = N\ln q_{\mathrm{tr}} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$

Then we assume the indistinguishable particles suddenly become distinguishable. By how much does the entropy of the system change?

$$
\Delta S = S_{\mathrm{dis}} - S_{\mathrm{indis}} = k(\ln Q_\mathrm{dis} - \ln Q_\mathrm{indis}) = Nk[\ln(N)-1]
$$

In the special case that the system contains only one particle ($N=1$), the entropy changes by $-k$.

Intuitively, it does not matter whether one particle becomes distinguishable from the rest if there is only one particle; Assigning a label to just one particle does not really modify the information within the system.

So how can this be understood?

Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.

Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation

$$\begin{align}
Q_\mathrm{tr,indis} &= \frac{q_\mathrm{tr}^N}{N!} \\
\ln Q_\mathrm{tr,indis} &= N \ln q_\mathrm{tr} - \ln(N!) \\
&\approx N \ln q_\mathrm{tr} - N \ln N + N \\
&= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right)
\end{align}$$

Plugging $N = 1$ into Stirling's approximation however tells you

$$\ln (1!) \approx 1\ln 1 - 1$$

or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!

Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result

$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$

which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.


Is there a simple field test for heavy water? Thought experiment

Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment?

Something like the way a piece of litmus paper (yes, it is laboratory equipment and easily available to the layman at that!) may be used to test whether a solution is an acid, or alkali.

There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.


How To Scan "improper dihedrals" in Gaussian 09? I am trying to parameterize Amber force field for a novel molecule, but I couldn't obtain improper dihedrals. Because if I scan impropers like dihedrals; later or sooner the molecule is overlapping and Gaussian exits with errors.
Let's say it is the $\ce{BH3}$ molecule. I tried many things for example with/without  symmetry=none keyword, with O 4 2 3 1 S 10 18.000000 or   D 4 2 3 1 S 10 18.0 (out of plane), but I couldn't get the results for a 360° improper scan.

I got this error at step 6:

 Old curvilinear step not converged, using linear step
 Error imposing constraints
 Error termination via Lnk1e in C:\G09W\l103.exe

What should I do?

Here is my gjf file:

%nprocshared=4
# opt=modredundant am1 geom=connectivity

Title Card Required

0 1
 B                  0.00000000    0.00000000    0.00000000
 H                  0.00000000    1.10873800    0.00000000
 H                  0.96019500   -0.55436900    0.00000000
 H                 -0.96019500   -0.55436900    0.00000000

 1 2 1.0 3 1.0 4 1.0
 2
 3
 4

D 4 2 3 1 S 10 18.000000

After I asked this question, I found answer. I am putting the solution here so maybe somebody else use this.
The solution is this: if gaussian exit from the job (for example) at step 6; change starting geometry and take it 6 step back. How? For example in my question every step size is 18 degree and step 6 is 90 degree. Open your *.gjf or *.com file select the same dihedral and decrease angle 90 by degree (don't forget 90 degree is just for this example, you wil change it according to your step and the size of step) *sometimes you may have to fix one group . And then start job again.

you can see my "wrong gjf file" in the question. And here is the my "right gjf file": I only changed the starting dihedral angles.

%nprocshared=4
# opt=modredundant am1 geom=connectivity

Title Card Required

0 1
 B                 -0.54526231    0.17208210    0.44752404
 H                  0.54967689    0.02475247    0.03303383
 H                 -1.22032311    1.04666245    0.03303383
 H                 -0.96514071   -0.55516861   -1.21043678

 1 2 1.0 3 1.0 4 1.0
 2
 3
 4

D 1 2 3 4 S 9 18.000000

NOTE: it seems I defined different dihedrals at first glance.
in question I defined:
D 4 2 3 1 S 10 18.000000 and here I defined:
D 1 2 3 4 S 9 18.000000 but out of the symmetry and all H bonds are equal two dihedrals are the same indeed. Here is the article about this:  https://pdfs.semanticscholar.org/ab92/5f43824e59dd9e3be3474ebbac57a64998c3.pdf (DOI 10.1002/jcc.21425  - "Symmetrization of the AMBER and CHARMM
Force Fields", 2009, EDYTA M. et all)
in this article in figure 5 shows only 3 different improper can exist for  I–J–K–L but in my question I=L=J and thus actually 1234 dihedral and 2314 dihedral are the same and gave the same results for the same technics.


Does clove oil/eugenol interact with nylon type plastics? So I am aware that the eugenol in clove oil readily melts #6 type plastic (polystyrene). Polypropylene has a higher resistance, while HDPE type plastics seem to have the highest resistance.

My main question is whether the clove oil will interact with nylon. Say, the bristles of a toothbrush.

Nylon 612, primarily used for filaments and toothbrushes, is resistant to clove oil at 20 °C [1, p. 2034]:

Nylon 12, used for coatings, is also resistant to clove oil at 20 °C [1, p. 1829]:

The tech data on nylon at newmantools.com also suggests there is no effect of clove oil on nylon in general.

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.


How do EM waves cause molecules to rotate, move, and cause electrons to move up energy levels? Atoms can absorb electromagnetic waves to cause electrons to jump up energy levels, when electrons go down energy levels they release a photon. But how do EM waves cause atoms to rotate, and vibrate?

But how do EM waves cause atoms to rotate, and vibrate?

In no way do EM waves cause atoms to rotate and vibrate. Atoms (more precisely, nuclei) in a molecule are always moving, so EM radiation can only change the way they move.

The thing is that within the Born–Oppenheimer approximation an energy  state of a molecule is the sum of the electronic, vibrational, and rotational components,
$$
  E = E_{\rm{el}} + E_{\rm{vib}} + E_{\rm{rot}} \, ,
$$
and there are accordingly three types of states and energy levels: electronic, vibrational, and rotational. For each and every electronic state there are few vibrational states associated with it and for each and every vibrational state there are few rotational states associated with it.

A transition can happen not only between different electronic states, but also between different vibrational states associated with the same electronic state or between different rotational states associated with the same vibrational state. Different vibrational and rotational states correspond to different ways atoms of a molecule move in space, so vibrational and rotational transitions change the way atoms move.

And, of course, any transition emits or absorbs a photon, the frequency of which is proportional to the difference in energy levels. Vibrational and rotational transitions spectra are studied by rotational–vibrational spectroscopy.


Selectivity in organic reactions I am finding great difficulty in understanding the differences between stereospecific, stereoselective, regiospecific and regioselective nature of reactions. Can you explain it in simple terms without mixing them with each other?

Yours is a common source of confusion. Reactions are can be selective but they are not specific. What's specific is a mechanism for the reaction. If the reaction only proceeds via one mechanism, then it is extremely selective.

Stereospecific. The mechanism provides a specific stereochemical outcome or proceeds from a specific stereochemical configuration. Examples:

a. Pd-catalyzed hydrogenation and hydroboration proceed via syn addition. These mechanisms are stereospecific because they indicate always adding to the same side of an olefin.

b. $\mathrm{E}_{2}$ elimination processes require that the hydrogen and leaving group are anti-periplanar. This mechanism is stereospecific.

c. Acid-catalyzed hydration proceeds via a planar carbocation intermediate where previous stereochemistry might be lost. This mechanism is not stereospecific.
Stereoselective. The reaction provides more of a product with a specific stereochemistry than any of the other possibilities.

a. Allylic alcohol directed epoxidation reactions frequently yield different amounts of the two product epoxides that result from reacting on the different faces of the olefin. This reaction is selective.

b. Asymmetric reactions that rely on pathways through diastereomeric transitions states are stereoselective.

c. Acid-catalyzed hydration could be stereoselective because the molecule can have other stereocenters that provide diastereoselectivity, even though the base mechanism disregards stereochemistry.
Regiospecific. The mechanism indicates a certain connectivity that should be obtained in the product.

a. Exo- versus endo- intramolecular reactions have regiospecific mechanisms.

b. A bit of a stretch, but carbonyl addition happens at the carbon. The mechanism is regiospecific.
Regioselective. The reaction favors one connectivity pattern over the other.

a. Hydrobromination favors the Markovnikov product, while hydrobromination in the presence of peroxides favor the anti-Markovnikov product.

b. In intramolecular ring closing reactions, 5-exo-tet products are preferred over 6-endo-tet products.

c. Carbonyl addition happens at the carbon. The only mechanism in operation is the one mentioned above, so the reactions are regioselective.

Importantly, reactions should not be stereospecific or regiospecific, even if they give only one product. This is a concrete case of a reaction proceeding exclusively (or almost exclusively) via a single mechanism that is stereospecific or regiospecific, respectively. The reaction itself is, respectively, stereoselective or regioselective.


What is the meaning of the word root "aceto"? What is the meaning of the word root "aceto"?

I believed that "aceto" meant a methyl attached to a carbonyl group. For example, as in acetamide. Then, why do we name $\ce{CH3C\bond{3}N}$ as acetonitrile, even though it does not have any carbonyl group?

The root "acet-", according to wiki, is derived from latin acētum, meaning vinegar, i.e. diluted acetic acid. Consequently, direct derivatives of acetic acid, like acetamide and acetonitrile are called so. Also, acetone called so because it can be produced from acetic acid (directly or by dry distillation of calcium acetate)

What isn't obvious is why acetonitrile  is considered a derivative of acetic acid and why nitriles in general are considered derivatives of carboxilic acids. Partially it's tradition and partially it's because compounds with carbon skeleton and oxidation state of carbon atoms unchanged are often rather easy to convert into each other.


Hydride shift and Ring Expansion Below is a problem I got stuck at in a test .

The first compund in this chain was easy to figure out , it was something like this :
Then I had this intermediate compound for B.(I didn't add the bromide anion because this is where my confusion starts)
Now ,here it shouldn't be correct to do a hydride shift outside the ring as it would result in a much less stable carbocation . But if I proceed using that idea and solving further for part D and E , I am getting option (b) as my answer and the correct answer is option (d).
Although if I do a hydride shift there then that opens a way for a ring expansion and hence a correct answer (according to the answer key) .

Ring expansion takes place in the step $\ce{D->E}$.
You were going right till the intermediate. Your intermediate will be attacked by $\ce{Br-}$ anions (as temp. is normal, therefore $\ce{S_N1}$ mechanism). Product will be B.
Then, it will form grignard reagent as C, which will react with aldehyde to produce alcohol D.
After that, it will react with HI to form a carbocation intermediate, which will rearrange to give product E.
The whole mechanism is as follows,


CHARMM parameter files for NBD (nitrobenzoxadiazole) I have been looking all over the internet for forcefield parameters for a fluorescent probe called NBD (nitrobenzoxadiazole). Here is an example of an NBD labeled fatty.

I have so far only found parameters for AMBER or GROMACS. Also when following the articles from these two reviews, there seem to be no CHARMM parameters anywhere.

My questions are:

What CHARMM forcefield parameters would you use for NBD?
How could I transfer the AMBER or GROMACS parameters for use in CHARMM?

Converting between force fields is always an iffy affair, as different force fields were parametrized with different reference sets, and possibly even with different intent in mind.

You have two options here:

You can extract the raw parameters from the particular GROMACS or AMBER force field used here, and try to translate them into the parameter formalism CHARMM uses. Generally speaking, force fields consist of a set of parameters for each atom type, and a set of equations these parameters are used in to calculate forces and geometries. Those equations represent aspects of the atom's or molecule's properties that have an influence on the calculated forces or the molecule geometry, like bonding interactions, non-bonding interactions, angle contraints, bond length constraints, and so on. You need to find the corresponding equations in CHARMM's force field framework, and then translate the parameters from AMBER or GROMACS so that the force field is the same in CHARMM.
Alternatively, you can look at the various CHARMM force fields and try to find one that works with NBD and any other molecules you got in your simulation. The first article you cited details where the various general parameters they used were from and how their simulation in GROMACS was set up. You should be able to define the same simulation parameters in CHARMM and rerun their simulation there. You can then use their published data to benchark your own simulation against theirs.


Sound absorption in carbon dioxide with temperature I am doing speed of sound measurements in $\ce{CO2}$ using ultrasonic pulses at 125 kHz. $\ce{CO2}$ is particularly difficult because it strongly absorbs ultrasound due to its multiple vibrational modes. The problem we have discovered is that absorption increases substantially as temperature rises. Is this expected, and if so is there a simple relationship between temperature and absorption coefficient?

The acoustic wave is accompanied by a temperature oscillation. To a first approximation, the temperature oscillation is adiabatic and reversible in the bulk, but irreversible near solid boundaries. In many cases, boundary losses dominate (for example, in musical wind instruments), but as the frequency increases, and the surface-to-volume ratio decreases, bulk irreversibility may become important. Bulk losses in single-component polyatomic gasses are almost always due to loss of local thermodynamic equilibrium when the temperature oscillation describing internal degrees of freedom lags the temperature oscillation describing translational degrees of freedom.
A full discussion of these effects is given in the book "Ultrasonic Absorption" by A.B. Bhatia. His figure 5.1 shows that the attenuation for $\ce{CO2}$ peaks at about 30 kHz (STP conditions), and his figure 6.3 shows the velocity dispersion that occurs at about the same frequency. The dispersion is a large effect, with the velocity increasing by about 4% as the frequency goes from below to above the attenuation peak. The attenuation and dispersion occur when the acoustic frequency matches the rate at which the vibrational mode temperature relaxes to the translational temperature. The rotational modes relax much faster than the vibrational modes, so they do not play an important role under these conditions.
The relaxation time for the vibrational temperature will generally decrease with increasing temperature, so the attenuation peak will move to higher frequency. I am not sure what temperatures you are exploring, but if your frequency is above the attenuation peak, then you will see increasing attenuation as the temperature increases.


Find the pH of an amphiprotic salt dissolved in water at room temperature Let's say I have a polyprotic acid $\ce{H2A}$ with the following properties:

$$
\begin{align}
\ce{H2A &<=> HA- + H+} & K_1 &= 1\cdot 10^{-1}\\
\ce{HA- &<=> A^2-+ H+} &  K_2 &= 1\cdot 10^{-4}
\end{align}
$$

Now a 0.5 M solution of $\ce{HA-}$ is prepared; what will be the $\mathrm{pH}$ of the solution?

My textbook (and many other sources) says it should be

$$\frac{\mathrm{p}K_1 + \mathrm{p}K_2}{2}$$

which sounds a little counterintuitive to me.

First of all, it implies that the $\mathrm{pH}$ is independent of the concentration of $\ce{HA-}$ present.

Also, if $1/\mathrm{p}K_1 > \mathrm{p}K_2$, then $\ce{HA-}$ should have the tendency to accept a $\ce{H+}$ more than it has of losing one, then why is the $\mathrm{pH}$ exactly partway between $\mathrm{p}K_1$ and $\mathrm{p}K_2$?

$$\begin{align}
\mathrm pH &=\mathrm
pK_\mathrm{a1}+\log \frac {[\ce{HA-}]} {[\ce{H2A}]} \\
\mathrm pH &=\mathrm pK_\mathrm{a2}+\log \frac {[\ce{A^2-}]} {[\ce{HA-}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + pK_\mathrm{a2} +\log \frac {[\ce{A^2-}]} {[\ce{H2A}]}} {2}\\
\end{align}$$

The last equation is the sum of the equation sides, divided by 2.

We can consider for expected acidic $\mathrm pH$, that we can neglect water dissociation and suppose $\ce{H+}$ is produced and consumed by these 2 reactions:
$$\begin{align}
\ce{HA- + H+ &<=> H2A} \\
\ce{HA-  &<=> H+ + A^2-} \\
\end{align}$$

Therefore:

$$\begin{align}
[\ce{H}]&={[\ce{A^2-}]}-{[\ce{H2A}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \frac {[\ce{H2A}] + [\ce{H+}] } {[\ce{H2A}]}} {2}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { [\ce{H+}] } {[\ce{H2A}]}\right)}  {2}\\
K_\mathrm{1a}&=\ce{[H+][HA-] /[H2A]} \\
\ce{[H2A]}&=\ce{[H+][HA-]} /K_\mathrm{1a}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{1a}} {[\ce{HA-}]}\right)}  {2}\\
\end{align}$$

Therefore, if $\mathrm{p}K_\mathrm{a1}$ is high enough, i.e. if $\ce{H2A}$ is rather a weak acid,
the simplified formula $\mathrm{p}H=(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})/2$ is right enough.

If $\mathrm{p}K_\mathrm{a1}=\ce{[HA-]}$, the $\mathrm{p}H$ correction is cca $+0.15 = \log (2)/2$

And then we can use this equation to  iterative corrections, with $\ce{[MHA]}$ being concentration of the used acidic  salt. The fraction is the classical formula to calculate ion fraction.

$$\ce{[HA-]}=\ce{[MHA]} \cdot \small
\frac {K_\mathrm{a1}.\ce{[H+]}}
{[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}} $$

$$\frac {K_\mathrm{a1}}{\ce{[HA-]}}=
\frac {[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
{\ce{[MHA]} \cdot \small
\ce{[H+]}}$$

For particular calculation, the first shoot is taking $\mathrm pH$ from the simplified  formula:

$$\begin{align}
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}=(1+4)/2=2.5 \\
\end{align}$$

I have realized, the direction of iterations must be reversed, as the formulas lead to divergence. ( It is known mathemathical problem to choose the right iteration with the small derivative $\mathrm{d}f(x)/\mathrm{d}x$.

The simplified equation at the bottom of the post expects  not too acidic or alkaline $\mathrm pH$. It supposes the equilibrium reactions affecting
 $[\ce{H}]$ do not significantly affect ratio of concentrations  $$\frac {[\ce{A^2-}]}{[\ce{H2A}]}$$

As the $\mathrm pH$ is near the both $\mathrm pK_\mathrm a$, it is quite a fair assumption for most cases.

$$\begin{align}
\ce{2 HA- &<=> H2A + A^2-} \\
\frac {[\ce{H2A}]} {[\ce{HA-}]} &=\frac {[\ce{A^2-}]} {[\ce{HA-}]}\\
 [\ce{H2A}] &= [\ce{A^2-}] \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}  \\
\end{align}$$


Linking an electrochemical cell to an electrolytic cell Electrolysis occurs when a non-spontaneous redox reaction occurs when electrical energy is put into it. An electrochemical cell functions by using two redox reactions to facilitate movement of electrons. So for an alkaline battery, the zinc anode will oxidize and release electrons, which will flow through the circuit to reach the manganese ($\ce{MnO2}$) cathode, which will be reduced, allowing more electrons to flow through the circuit without buildup of electric charge. I assume here that both reactions are spontaneous, and the electron buildup in the anode prevents further zinc oxidation when the circuit is incomplete.
$$\ce{Zn(s) + 2OH- (aq) \rightarrow ZnO(s) + H2O(l) +2e-} [E^\theta= 1.28V]$$
$$\ce{2MnO2(s) + H2O(l) + 2e- \rightarrow Mn2O3(s) +2OH- (aq)}\ [E^\theta= 0.15V]$$
If the alkaline battery is linked to an electrolytic cell, won't it just be like two electrochemical cells working in conjunction to each other? The zinc anode will be linked to the cathode of the electrolytic cell, and then the ensuing redox reaction will (probably) be spontaneous. So the "electrical energy" put into the electrolytic cell will be instead chemical energy?
Also, seeing that the $\ce{MnO2}$ cathode has a relatively small $E^\theta$ value, won't it be hard to trigger oxidation of the electrolytic anode? When connected to the electrolytic cell, do the cathode and anode of the electrochemical cell now work totally independently? If that is the case, the voltage of the battery should not be considered, but instead the electrode potentials of the anode and cathode, but that is usually not the case. What am I misunderstanding about this?
edit: I'm asking about the mechanism when an electrochemical cell is linked to an electrolytic cell, not about the reactions taking place in the electrochemical cell itself, so the question suggested doesn't quite fit the bill.
edit 2: Added a picture to better show my doubts. If the galvanic cell is linked to an electrolytic one as shown here, the galvanic anode and the electrolytic cathode are linked and seemingly acts as a separate cell. The same thing can be said about the other two electrodes. If this is the case, doesn't the galvanic anode and cathode each act independently of one another, and therefore the electrode potential of the individual half cell must be considered? If they are actually linked, how?

For an electrolytic cell, a galvanic cell is nothing else but a power source.
For a galvanic cell, an electrolytic cell is nothing else but a powered electronic circuit.
The galvanic cell anode/cathode does not form with the electrolytic cell cathode/anode a respective separate cell, as there is no ionic flow between them. Similarly for a simpler case, 2 half cells do not form a cell if there is no ionic connection between them.
The reason for it is simple, based on the electrostatic neutrality requirement and enormous forces between unbalanced charges. Imagine you managed to push the current 1 A for 20 s through the wire, without being balanced by ionic motion. The comparable unbalanced charge causes in atmosphere a lightning several kilometres long.


In a mass spectrum, how can the sum of relative abundances be greater than 100? I apologise if this seems a very simple question, and also if I am overlooking something that is very simple.

For part of my A Level course, we have to be able to interpret mass spectrometer readings for isotopes of elements. Further on in the course, it is extended to molecules. However, quite often it has come up where a question like this arrives:

Given that relative abundance implies that all values are relative to each other, how can all of the relative abundances sum to a number that is greater than 100?

2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11.  However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the normalized ratio of relative abundance of these isotopes.  It is this normalized value that adds up to 100.


How do I do stoichiometry for 2 limiting reactants The Drano reaction produces sodium aluminate along with hydrogen gas. If a typical container of Drano weighs 500 g which is 35% by weight by weight sodium (l) hydroxide and 2.5% by weight aluminum, what volume hydrogen gas would be produced if (excess) water was added to an entire can of crystal Drano at 22 °C and 1.2 atm?

My equation:

$$\ce{4 Al + 4 Na(OH) + 12 H2O -> 4 Na[Al(OH)4] + 6 H2}$$

If my balancing correct, I am trying to get the excess for water. Would I need to do the moles left over? Or would 500 x 62.5% (2.5% and 35% added together) suffice?

If I do need to do moles left over, how do I manage with two limiting reactants?

Your balance equation is correct except for not simplifying it. Simplified equation should be:
 $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$

I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:

The amount of $\ce{NaOH}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}}  = \pu{4.38 mol}$

The amount of $\ce{Al}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}}  = \pu{0.463 mol}$

Since $\ce{NaOH}$ and $\ce{Al}$ react with 1:1 ratio according to the equation, $\ce{Al}$ is clearly the limiting reagent.

Hence, $\pu{0.463 mol}$ of $\ce{Al}$ reacts with $\pu{0.463 mol}$ of $\ce{NaOH}$ if you added enough water to $\pu{500 g}$ of Drano container. The leftover amount of $\ce{NaOH}$ would be: $\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$

Now, how much water you need? You need more than $\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$ of water, which is more than $\pu{25.0 g}$ (to be excess).

Note: Similar calculations would appied to find out amount of $\ce{H2}$ released by the reaction.


Using gas laws to find density An unknown gas at $\mathrm{63.1\ ^\circ C}$ and $\mathrm{1.05\ atm}$ has a molar mass of $\mathrm{16.04\ g/mol}$.
  What is the density of the gas?

I know I need to use the $pV=nRT$ equation but it gives me g/mol and not moles. I'm not really sure what to do here.

$$1.05\ V = n\ 0.08206 \cdot (63.1+273.15)$$

Ok that's as far as I got. I have no idea how would I even go about converting g/mol back to moles so I can solve for V, and after that I would also need to figure out a way to find grams so I can divide g/v to get the density.

Okay, to find the density of the gas, to need to know its mass and its volume. So to do this, lets say we have one mole of this substance. So its mass will be $16.04~\mathrm g$. Now all we need to do is find its volume.

This can be done by rearranging the ideal gas equation: $$V = \frac{nRT}{p}$$ Now all we have to do is plug in the values. However you have be careful that you use the write units. This is a common mistake, especially when using the ideal gas law however it is good to see that you have used the correct units: $$\begin{align}V &= \mathrm{\frac{1~mol\times0.08206~L~ atm~ mol^{-1}~K^{-1}\times336.25~K}{1.05~atm} }\\
&=\mathrm{26.28~L}\end{align}$$

As you can see, all the units cancel nicely so that you get the final answer in litres which is the correct unit for volume.

Now, density can be calculated: $$\rho =\mathrm{\frac{16.04~g}{26.28~L} = 0.61~g~L^{-1}}$$


Are SN2 reactions in equilibrium? I'm learning about SN2-reactions, and my text book gives the following examples of these reactions:

  $$\ce{HO^- +CH_3Br -> CH_3OH + Br^-}\tag{1}$$
  $$\ce{I^- +CH_3Br -> CH_3I + Br^-}\tag{2}$$

Later I learned that $\ce{OH^-}$ is a worse leaving group than $\ce{Br^-}$, and that $\ce{I^-}$ is better than $\ce{Br^-}$. I can understand the reaction (1) is favored, since the leaving group gets worse (and therefore $\ce{CH_3OH}$ is more stable than $\ce{CH_3Br}$), but I cannot understand the reaction (2). Wouldn't the reverse reaction of (2) be favored instead? Or is (2) in fact an equilibrium?

I'm confused because $\ce{Br^-}$ seems like a better nucleophile than $\ce{I^-}$, which would favor the reverse reaction of (2)?

When we talk about Sn1 or Sn2 reactions (and any other reaction to be honest), not only the reactants are crucial to describe what's going on but also the solvent you used. It is perfectly possible that the second reaction will be favored in the direction shown, as long as you use the right solvent for this.

There are two types of solvents one can use in an Sn2 reaction: polar protic (which has a reasonably acidic hydrogen in its structure) and polar aprotic (has a net dipole moment but no acidic hydrogens).

Examples of polar protic solvents: Water, methanol, ethanol, ammonia etc

Examples of polar aprotic solvents: DMSO, acetonitrile, acetone etc

So what does the solvent has to do with the reaction?

Nucleophiles are negatively charged species or dipoles, they will attack a sufficiently positive center (usually a carbon atom connected to an electron withdrawing group)
The solvent is commonly present in significant excess compared to the
reactants.

When you use a polar protic solvent, such as methanol or water, the nucleophile species (ions such as Cl- or molecules such as ethanol) will be surrounded by an overwhelming number of solvent molecules that will interact strongly with them by dipole-dipole interactions or hydrogen bonding. This creates a barrier around the nucleophile, making it considerably more difficult to approach the positive center and consequently slows the reaction.

Smaller nucleophiles (such as F-) will be surrounded by a bigger number of solvent molecules than bigger nucleophiles (such as I-), so this surrounding effect is stronger in smaller species.

If you consider only the nucleophile, Br- is stronger than I- since it has a smaller ionic radius and, therefore, has a more 'concentrated' electron cloud (less polarizable). If no solvent effect was considered, the reaction as shown in your textbook would be incorrect since the equilibrium would be towards the formation of methyl bromide.

HOWEVER, if you use a polar protic solvent, like methanol, the order of nucleophilicity is inverted (because of that surrounding barrier of solvent molecules) and I- becomes a better nucleophile than Br-.

So, to be consistent and correct, your textbook should also provide the solvent in which this reaction is done. If it is a polar protic one then it is correct.

(I took these pictures from master organic chemistry, they have a good text talking about it here: https://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/ )


Does reaction quotient Q take into account pure solids and liquids? I understand that the equilibrium constant does not take into account the concentrations/ pressures of pure solids and liquids. How about the reaction quotient?

In the following reversible reaction, am I right to say that since water is a pure liquid, it is not reflected in the reaction quotient, hence removing water (without changing the temperature) will not shift the equation left/right after applying Le Chatelier's Principle?

$$\ce{H3PO4 (aq) + H2O <=> H3O+ (aq) + H2PO4- (aq)}$$

It isn't that solids don't appear in equilibrium constants and/or reaction quotients. They are simply assumed to have an activity equal to the value of 1! Hence people don't show them, but they should. I think it is important to understand and remember this assumption.

Also, equilibrium constants and reaction quotients are fundamentally a ratio of activities! The use of concentration and/or pressures in equilibrium constants/ reaction quotients is the results of approximations in themselves and this simplicity permeates chemistry and should be stopped or at least respected.

for the reaction you show, the equilibrium constant is

$$K_\mathrm{eq} = \frac{a_\ce{H2PO4-}a_\ce{H3O+}}{a_\ce{H3PO4}a_\ce{H2O}}$$

See the Chemistry LibreTexts for more details about solid and liquid activities and equilibrium constants.


Reaction of Silicon monoxide with alcohols Elements from the carbon group tend to form compounds with two oxygen's. Let's say I want to produce $\ce{SiO_2}$ out of $\ce{SiO}$ by mixing it with some alcohol, for example:

$$\ce{SiO + CH_3-CHOH-CH_3 \longrightarrow SiO_2 + CH_3-CH_2-CH_3}$$

Is this likely to occur? at least before all the alcohol evaporates...

I doubt very much its likely to occur.  Reducing an alcohol to a alkane is not trivial. strong reducing agents such as $\ce{LiAH4}$ cannot reduce alcohols, and to my knowledge $\ce{TiO}$ cannot reduce alcohols which would be an even better canidate that $\ce{SiO}$.


Why does increasing surface area of the electrode decrease resistance? Which increases immediately if the surface area of the silver electrode is increased?

  overall cell voltage
  rate of change of $\ce{[Ag+]}$
  mass of $\ce{Cu}$ electrode
  change in ratio of electrode masses; ∆$(\frac{\text{ mass of Cu}}{\text{ mass of Ag}})$

The correct answer was (2), but I'm not sure why the reaction rate would go up if the surface area of the silver electrode increased. The rate of reaction is proportional to the current flowing through, which by Ohm's Law is $\frac{V}{R}$. This means that the resistance must have decreased if current increases. But resistance of the electrodes is only related to the length and cross sectional area. The problem does not specifically state any change in length or cross sectional area. It only says surface area, so what am I missing here?

Fluid flow is always an excellent analogy for understanding current flow.

Fluid flow rate -> electrical current

Pipe friction, valves and other throttlers -> electrical resistance

Fluid pressure -> voltage.

In this case, the amount of contact between the electrode and the solution is most likely the bottleneck of the circuit so increasing the area of contact should increase the current.  Similar to how cranking the valve on a faucet or outdoor spigot more and more open increases the flow rate of the water coming out.


Do (simple, inorganic) salts ionize completely in water when they dissolve? Specifically: Suppose I add equal parts (i.e. the same number of moles of each) sodium bicarbonate and potassium sulfate into a container of distilled water large enough to dissolve them completely. In a second container of water, I add potassium bicarbonate and sodium sulfate. Are these two containers now physically identical? (That is, is there any experiment that could tell them apart?) My understanding is that each now contains free-floating Na+, K+, S-, and HCO3- ions. And if I evaporate away all the water, what will be left behind in each container?

If I replace the potassium and/or sodium with, say, magnesium or calcium, and the bicarbonate and/or sulfate with chloride or citrate, do the same principles hold? (Assuming the resulting salts are water-soluble - some, like calcium carbonate, won't be.)

(I'm not sure if the list of cations and anions above have much in common chemically. My interest relates to the fact that they can be added to water to simulate the chemical makeup and flavor of natural mineral spring water.)

(PS: What's a more elegant way to say "the same number of moles of each?" If I say to mix things in equal parts, or in some ratio, I would normally specify whether I mean by mass, by volume or... how do I say "by number of moles?")

When a salt, no matter which salt, or any other polar compound is exposed to enough water, then its polarized molecule parts, depending on its polarity, become attracted to the oxygen or hydrogens of the water, which will result, starting with the outermost molecule layer, in the complete dissolution of the compound. Within this process both the positively and the negatively polarized sections will get surrounded by water molecules (= hydration shell) and thus shielded from one another. In this state the original compound(s) have (theoretically) disappeared completely and the two containers will indeed be identical. Possible back reactions will (mainly) result in the most energy efficient compound for the particular experimental conditions and have the same chance in both containers.

When you now evaporate all the water you may get various compounds and mixtures depending on what was dissolved.

PS: I'd say "of equal molarity" or "they are equimolar", but my mother tongue isn't English, so there might be a more elegant way.


What are some good research topics for a high school student? I absolutely love chemistry and I really want to take part in science fairs. I am also taking an AP chemistry class. Cleaner fuel and energy sources and air pollution (especially due to smog and vehicle exhaust) interest me. My initial idea was to make solar panels work along the lines of Chlorophyll, but that is a bit too advanced for my level. Where could I get inspiration for some science fair ideas or chemistry research topics (high school level)?

Green chemistry is a highly advanced topic. It ain't for the weak of heart, so it is probably not for below university level.

I personally favor pyrotechnics as relatively safe iff every bit of safety rules is strictly followed and stunningly spectacular topic that is pretty easy and cheap to do by own hands (most materials are inexpensive as well as most tools). However, it is extremely time-consuming and has questionable legal status in some countries (and for a good reason: safety may be easily compromised and potential for disaster is high).

Another interesting topic is fluorescence. Many compounds shine under UV light, and there are families of such compounds that are not exhaustively studied and yet are relatively easy to make. However, a deep study is required, but it is probably advanced school level.

With access to advanced materials/tools/labs more topics become available, but a full list of such tools is required to recommend anything.


How does electrophoretic separation work in terms of separating soil into minerals? I found this diagram and would like to know how it works with soil.

Google searches yield protein/biology things, anyone know how this method would work with soil?

The search term: "continuous particle electrophoresis" yielded much more results.

electrophoresis - the movement of charged particles in a fluid or gel under the
influence of an electric field.

source: http://www.youtube.com/watch?v=LNUgcq1M4LI

-

source: http://www.nature.com/articles/srep19911/figures/1

Heard some guy invented it while messing with some clay, but couldint find any video examples of that.


What is the name of this organic compound? The organic compound in question is shown above. This is how I tried to name it:

Start from the left hanging  $\ce{H_3C}$.
$\ce{OH}$ gets the number 2 and $\ce{NH_2}$ gets 6. The lowest numbers possible for the functional groups keeping seniority/priority in mind.
There are two ethyl groups at 3 and 5.

Hence, the compound should be called 6-Amino-3,5-diethylhexan-2-ol.

I tried searching the web to confirm this, but nothing useful showed up.

Is this the correct IUPAC name of this organic compound?

The longest chain has 7 carbon atoms. So you would have an ethyl group in carbon 3 and the amino group will be a ramification in carbon 5. I would stand for:

  5-(aminomethyl)-3-ethylheptan-2-ol


Can I pour a solution into another vessel with no air bubbles? I have been trying to create a method to create a homogeneous agar solution.

What are appropriate techniques to ensure homogeneous solidification of an agar-based aqueous solution?

The new solution consists of distilled boiling water, $\pu{20 g/L}$ agar, $\pu{10 g/L}$ $\ce{NaCl}$ and $\pu{1 g/L}$ $\ce{CuSO4}$.
An issue with the agar is that it only dissolves between 90-100C which makes it difficult to dissolve using a hot plate or other methods as it also needs to be homogeneous.

I've made progress speaking to both the microbiology and chemistry departments of my university. The current idea to make the solution homogeneous is to heat it in an autoclave set at 121C. This will also sterilize the solution making the phantom last longer. Unfortunately we have no 3L beakers to do it all in one go. They have plenty of 2L and some 5L beakers, but nothing that is 3L.

A new idea is to heat a round bottomed flask in the autoclave and then transfer the semi-cooled solution to a perspex container that will have a base on it. (flask and container shown in the photo below)

The problem is how do we ensure the solution is still relatively homogeneous and won't be full of air bubbles? Are there any simple ways to solve this?

There are two aspects to consider for the intended "air bubble free" (or at least, "air bubble less") transfer from your container where the liquid is prepared into the container where the liquid is supposed to solidify:

Your first container shown is a round bottom flask.  Initially, an increase in inclination of the flask will increase the flow of liquid out of the flask.  If however the inclination is too high, the air that aims to refill the inner volume of the container, initially occupied by the liquid, will no longer continously enter the flask.  Instead, there will be an alternation of sorting liquid and entering air; which may in turn yield air bubbles in the sorting liquid.  Depending on the viscosity of the liquid, the air bubbles may remain in the liquid for quite some time.  From the perspective to lower the potential air intake by the liquid during the transfer, a wide mouth bottle, or even a beaker were better suited than a round bottom flask.
The distance between the first and the receiving container.  For your purpose, it is not good to pour the liquid directly into the receiver. To lower splash and splatter in the transfer by decantation, an easy aide is using a glass rod, like shown here:

(picture source)

In your instance, it may be more convenient to use a slab of glass / a pane of glass that is just beneath the mouth of the first container, and bridges all the distance to the receiving container; as schematically drawn below:

Here, intentionally the slab of glass is partially immersed in the liquid already present in the receiving container.  If done in a coordinated team of two, one holding and moving gently the slab of glass, the second gently decantating the liquid onto this "chute", you might cast larger volumes into trough-shaped containers / fish tanks, too.  Once solidified, you would cut-out the block / shape you need.  Overall, it reminds me a little to ballistic gelatin...


Would the carbocation intermediate undergo rearrangement? I have my concerns regarding option (a) in this question.

During the dehydration step, wouldn't the resulting carbocation intermediate formed in option (a) rearrange?
That is, the ring should contract from a $7$-carbon to a $6$-carbon one because the resulting carbocation would be more stable. But the resulting product would have to be an addition product since elimination now becomes impossible.

I know that this situation can arise because I have tackled it before and the answer was an addition product in that case, but that reasoning contradicts with the official answer key in this question. (The answers are a,b and d).

The shift I called for was the bond between carbon (endocylic with nitrogen) and the carbocation.

Unfortunately, I cannot back this with references at the moment but in my opinion there should be no carbocation formed in the first place and therefore no tendency to undergo any Wagner-Meerwein rearrangements.

Secondary carbocations, while being less unstable than primary ones, are still rather unstable. There are no stabilisation mechanisms that suggest themselves in the reaction conditions (concentrated sulphuric acid) in which the nitrogen will be protonated and therefore unable to stabilise a carbocation in α position. I would assume that the elimination proceeds much more $\mathrm{E2}$-like: likely with no intermediate carbocation at all but if there is one it should be extremely short lived and the neighbouring proton should dissociate practically instantly.


How many protons are required to satisfy hydroxyamine's (NH₂OH) lust for protons? I ask because I'm trying to write an equilibrium equation for this reaction and I'm wondering whether an extra 2 protons will be required after one is taken to attack the $\ce{OH}$ group and turn it into $\ce{H_2O}$ in order to convert the freed ammonia (with one proton missing) into a conjugate acid.

Well, hydroxylamine is technically amphoteric, meaning it can act as either a Bronsted acid or base. The acidic proton is presumably the one on the oxygen (although the resulting salts are apparently explosively unstable), while the nitrogen is the proton acceptor. I wouldn't expect the oxygen to be basic (which, unless I've misunderstood, seems to be what your question implies), at least not in any measurable way or for any appreciable length of time, given hydroxylamine's instability and tendency to spontaneously decompose in myriad ways. When treated with acid, it forms hydroxylammonium salts, which disproportionate upon heating.

Note: this book was my primary reference; it contains a concise but informative summary of the reactions hydroxylamine undergoes.


Difference in acid strength of oxalic acid and malonic acid Despite very similar skeletal structures, the difference in acid strength between oxalic acid (ethanedioic acid) and malonic acid (propanedioic acid) is quite significant. What is the reason for that?

It seems that the inductive effect plays a role, but what is the explanation for the inductive effect in oxalic acid as there is same group on opposite sides?

How does the inductive effect plays different roles in both acids?

Linear saturated dicarboxylic acids
$n$ = number of $\ce{-CH_2}-$ groups in chain
 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid
Aromatic carboxylic acids
                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ----
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids
$n$ = number of $\ce{-CH=CH}-$ trans groups in chain
n                            pka1      pka2
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87
3

Straight-chain, saturated carboxylic acids
$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).
n                     pKa
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the  carboxylate ion to solvate, which increases its acidity.
In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization ,  the second ionization


Are there any chemicals with this structural skeleton? Is there a chemical that has the bond structure like below?

There are 5 atoms linking together.
Two of them have 4 bonds. (They are different atoms.)
Another two of them have 3 bonds. (They can be either same or different atoms.)
The last atom has only 2 bonds.

Suggested SMILES for unknown $\ce{X}$ elements is [X]2[X]14[X]3[X]1[X]234

Are there such kinds of chemicals?

CCDC doesn't contain any molecules of such geometry. The closest topology observed is among pnictogens such as $\ce{P, As, Sb}$, which form tetrahedron units $[\ce{P4}]$, $[\ce{As4}]$, $[\ce{Sb4}]$ which are readily coordinating with Au, Rh, Ag and Cu by edge-sharing:

The first publication (1) tetraphosphido-ligand presents the following structure of $\ce{Rh}$-complex:

The most relevant structure best fitting the required connectivity matrix would probably be bis(tricyclo[$1.1.0.0^{2,4}$]tetraphosphane)-gold (2):

Thanks to @TAR86 comment mentioning exotic boranes, I also found a research (3) on ab initio calculations for $\ce{B4}$ cores obtained via the following proposed reaction:

They also obtained crystal structure for bis(tetrahydropyran)-(tetra(t-butyl)tetraborane)-lithium where $\ce{B4H}$ core is present:

(1) Lindsell, W. E.; McCullough, K. J.; Welch, A. J. J. Am. Chem. Soc. 1983, 105 (13), 4487–4489. DOI: 10.1021/ja00351a067
(2) Forfar, L. C.; Clark, T. J.; Green, M.; Mansell, S. M.; Russell, C. A.; Sanguramath, R. A.; Slattery, J. M. Chem. Commun. 2012, 48 (14), 1970–1972. DOI: 10.1039/C2CC15291F
(3) Neu, A.; Mennekes, T.; Englert, U.; Paetzold, P.; Hofmann, M.; von Ragué Schleyer, P. Angew. Chem. Int. Ed. Engl. 1997, 36 (19), 2117–2119. DOI: 10.1002/anie.199721171


Can hydride shift and methyl shift happen one after another in a carbocation? Can a hydride shift be followed by a methyl shift for stabilizing a carbocation?
In case both are possible, which will occur first?

Multiple rearrangements can happen if the resulting species is stabilized by it. In general hydride shift is more favourable over methyl shift because of its smaller size.

The example proposed by James was not known to me but its truly an amazing example.

  James:
  It’s possible for multiple hydride/alkyl shifts to occur. One amazing example is in the biosynthesis of lanosterol.

  Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.
  For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.

Source

From Wikipedia:


Can glucose burn in chlorine? Chlorine is an oxidizing agent, so could it replace oxygen? If so, what would be the products, and how much energy would such reactions release? Do any other carbohydrates burn in chlorine? I cannot find any information on this.

Clarification: How does glucose interact with chlorine chemically?

I have not heard of glucose being burned in chlorine, but I do know that glucose reacts with chlorine in aqueous-buffer medium. See reference 1.
The full paper is behind a paywall.

It says that oxidation of glucose by bromine water was studied previously and the reaction of glucose and aqueous chlorine also happens in a similar fashion although the reaction is slow and products yield is less. Nonetheless, major product of the reaction is D-gluconic acid in both acidic and alkaline medium but unbuffered chlorine may further react with D-gluconic acid to various products like 5-keto-D-gluconic acid and D-glucaric acid.

Reference

J. Am. Chem. Soc., 1955, 77 (7), 1875–1880. DOI: 10.1021/ja01612a050.


Finding pH of acid in water How would you find the pH of an acid dissolved in water? Would you need to take the fact that the pH of water is already 7 into account and go on from there?

Say, for example, that you add 0.0500 mL of 0.00100 M HCl into 50 L of pure water, what will be the pH of the new solution?

I made the mistake of just finding the moles of the HCl placed into the pure water, and finding the new concentration by dividing by the litres of water present. This gave me a pH of 9, which does not seem to make any sense as acids are meant to lower pH!

Some help would be wonderful!

Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case.
Simplification for strong acids  $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$
implies $\pu{[\mathrm{acid}] >> 10^{-7} mol/L}$.
Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$,
neutral water contains $\pu{10^{-7} mol/L}$ of $\ce{H+}$.
That means the very most $\ce{H+}$ ion must come from acid dissociation and not from water dissociation.

If OTOH very most $\ce{H+}$ are from water dissociation as in your case, another simplification says $\mathrm{pH}=7$, no matter how much of acid you added.

For general case:
$$c_{\mathrm{acid}}=\ce{[A-]}=\ce{[H+]} - \ce{[OH-]} = \ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}}$$
$${\ce{[H+]}}^2 - \ce{[H+]}\cdot c_{\mathrm{acid}} - K_\mathrm{w}=0$$
$$\ce{[H+]} = \frac{c_\mathrm{acid} + \sqrt{{ c_\mathrm{acid}}^2 + 4\cdot K_\mathrm{w}}}{2}$$
The limit case for $c_\mathrm{acid} << \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = \sqrt{K_\mathrm{w}}$$
$$\mathrm{pH}=\frac 12 \mathrm{p}K_\mathrm{w}$$
The limit case for $c_\mathrm{acid} >> \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = c_\mathrm{acid}$$
$$\mathrm{pH} = -\log{(c_\mathrm{acid})}$$

Simplification for weak acids $$\mathrm{pH}=0.5 \left(\mathrm{p}K_\mathrm{a}-\log {\ce{[acid]}}\right)$$
implies negligible water dissociation and also negligible relative acid dissociation $\ce{[HA]} >> \ce{[A-]}$.
So one can assume:

The acid concentration $\ce{[HA]}$ is not affected by acid dissociation.
$\ce{[A-]} = \ce{[H+]}$


Why are square planar molecules drawn from the side plane as opposed to the top plane? Simply put, why can't you draw square planar molecules top down (so you don't have to use depth bonds)?

You can do that. For example, a Google Image search for "cisplatin structure" suggests that although the hashed + wedged depiction is more common, the four plain bonds depiction is also in regular use.
For more detail on what's "acceptable" or not, refer to: Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970, DOI: 10.1351/pac200678101897. They recommend the hashed + wedged depiction, but do not prohibit four plain bonds (section ST-2.6, pp 1942–3). There is some rationale for this, which may go some way towards explaining why the hashed/wedged depiction is more common: it explicitly differentiates the square planar structure from other tetrahedral centres in organic structures.

The first recommended depiction should be drawn with two solid wedged bonds and two hashed wedged bonds. The two solid wedged bonds should be positioned below the central atom, while the two hashed wedged bonds should be positioned above. [...] It is also acceptable to draw similar structures with plain bonds. There are some cases where this drawing style would even be preferred, particularly when all four bonds are required to be in the plane of the paper by virtue of other aspects of the structural diagram [...] The similar depiction with the bonds themselves positioned along the axes is acceptable but should be avoided as it is extremely commonly used to represent tetrahedral centers, either if the tetrahedral configuration is nonstereogenic (carbon tetrachloride) or in Fischer projections.


Storage life of 3% hydrogen peroxide I am using 3% hydrogen peroxide for a series of operations, and the common practice is to prepare it on-site from 30% hydrogen peroxide and saline. This is time consuming and somewhat more prone to error than preparing a 3% stock. I was told this is not done because 3% hydrogen peroxide has a short shelf-life (~a week).

Is this true?
If so, why?
Would the shelf-life be extended if the solution were prepared with phosphate-buffered saline or just with deionized water (instead of saline)?

It may pretty well be true, and here's why. Not only hydrogen peroxide is highly unstable by itself, but its decay can be catalyzed by a wide variety of compounds. And by wide, I mean really wide; at times it seems to me that those compounds that don't catalyze it are a minority. Now, I don't know what you have in your saline, but chances are that it works on the peroxide, too.

With deionized water, this should not be a problem... or should it? See, that link by @ron says that the 30% thing is stabilized. Guess what? It means that they've put some additive into it, which somewhat slows down the decomposition. When you dilute your peroxide, the stabilizer gets diluted too, and hence can't work as well as before. This is a possible reason why you may get shorter shelf-life even when diluting with pure water.

I'm not sure what is the stabilizer they use. In any case, if you find that out, then add it during the preparation, and don't add anything that can catalyze the decay, your 3% solution should be fine (in terms of shelf-life) - but I'm afraid these additions and subtractions may stand in the way of your primary objectives.


Does hybridization take place in other atoms apart from central atom in a molecule? I have seen my textbooks always give formula and description of hybridization of central atom of a molecule. But I wonder will other atoms present (not Hydrogen) go in hybridization or only the central atom only uses hybrid orbital and other use pure orbital?

Yes, all atoms with p or d orbitals can hybridize in a molecule, not just the central atom.  Take acetone for example, the carbonyl carbon is $\ce{sp^2}$ hybridized and the carbonyl oxygen is also $\ce{sp^2}$ hybridized.  Another example is methyl chloride, the methyl carbon is $\ce{sp^3}$ hybridized and the chlorine is also $\ce{sp^3}$ hybridized.  As a final example consider dimethyl ether, each of the carbon atoms is $\ce{sp^3}$ hybridized and the central oxygen is also $\ce{sp^3}$ hybridized.


How to differentiate diastereotopic protons by NMR in flexible groups? In (1R,2S)-2-(hydroxymethyl)cyclohexan-1-ol shown below, Ha and Hb are diastereotopic (they will have different chemical shifts), but because of fast rotation, NOESY should show that both couple with Hc, right? How to differentiate diastereotopic protons by NMR in flexible groups?

I don’t think you can distinguish which is which in a freely rotatable group like the one you drew. They will be different, both signals will split up as dd (potentially ddd, if coupling to the hydroxy proton is observed). There will be significant roofing. Careful analysis of the spectrum should reveal two different shifts coupling to each other but very close.

Rotation around the $\ce{C-C}$ axis should be rapid on the NMR timescale resulting in the couplings $^3J_{\ce{HaHc}}$ and $^3J_{\ce{HbHc}}$ averaging out to one common value and the signal of $\ce{Hc}$ having a dddt-type pattern (three doublets to the neighbouring ring protons, a triplet to the side-chain protons).

All NMR methods I know to determine relative stereochemistry (e.g. NOESY, Murata, …) rely on rotation being slow on the NMR timescale.

Potentially cooling your sample down to very low temperatures ($-50~\mathrm{^\circ C}$ or less) could inhibit the rotation and allow for intermittent resolution of the protons. However, that should also produce a more distinct chemical shift difference, so it may actually not help.


What does the IUPAC definition of the emf of a galvanic cell mean? IUPAC's Gold Book defines the emf of a galvanic cell in the following way:

The limiting value of E(cell) for zero current flowing through the cell, all local charge transfer and chemical equilibria being established, was formerly called emf (electromotive force). The name electromotive force and the symbol emf are no longer recommended, since a potential difference is not a force.

Check: https://goldbook.iupac.org/E01934.html

I understand that zero current is required to find the emf since passing a current through the cell decreases its voltage. Moreover, we want to find the emf of the cell at some specific composition of the electrolytes, also ensuring a reversible reaction is taking place.

But what does it mean by "all local charge transfer and chemical equilibria being established"? Isn't the emf of the cell ZERO when the cell reaction attains equilibrium? What "equilibrium" does it refer to? And what does "local charge transfer" mean?

We think of electrochemical processes happening immediately in one step, but that's not really true. In a galvanic cell, you have to have the dissolved ions move from the bulk to the surface, adsorb to the surface, and possibly undergo an activation step before they can reduce or oxidize. The last step is for the electron to actually move from the reduced to the oxidized species. This specific step is the "local charge transfer" reaction.

You are right that when the cell reaches equilibrium, the $E_{cell}$ would be zero. However, here we're interested in  the equilibrium of each local process separately: eg. diffusion of ions from the bulk electrolyte to the surface will reach an equilibrium.

When you first submerge the electrode into the electrolyte, all of these steps are able to reach an equilibrium, and things like concentration gradients between the electrolyte bulk and electrolyte surface can effect the potential experienced by the electrode. In fact, some charge transfer even occurs initially, but because the half-cells are not connected, electrostatic forces build up and the concentrations are such that the chemical potential is quickly opposed and a steady-state is achieved.


Can hydrocarbons undergo combustion without oxygen present? Most of us are familiar with the standard reaction of methane gas combustion:

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

In standard combustion, oxygen is generally used as the oxidant. Would it be possible for hydrocarbons to undergo combustion using alternative oxidants? If so, under what conditions could this be reproduced?

For example, could we achieve anoxic combustion by replacing oxygen in the atmosphere with another highly electronegative element such as chlorine or fluorine? Or perhaps compounds in high oxidation states such as permanganate or dichromate under different pressure/temperature conditions such that they were gaseous?

They would certainly combust if meeting a worse adversary than oxygen. $\ce{ClF3}$ would oxidize them, with extreme certainty, rapidity and violence.

Here is a short excerpt from the Wikipedia article of $\ce{ClF3}$ which amused me:

  It is, of course, extremely toxic, but that's the least of the
  problem. It is hypergolic with every known fuel, and so rapidly
  hypergolic that no ignition delay has ever been measured. It is also
  hypergolic with such things as cloth, wood, and test engineers, not to
  mention asbestos, sand, and water — with which it reacts explosively.
  It can be kept in some of the ordinary structural metals — steel,
  copper, aluminum, etc. — because of the formation of a thin film of
  insoluble metal fluoride which protects the bulk of the metal, just as
  the invisible coat of oxide on aluminum keeps it from burning up in
  the atmosphere. If, however, this coat is melted or scrubbed off, and
  has no chance to reform, the operator is confronted with the problem
  of coping with a metal-fluorine fire. For dealing with this situation,
  I have always recommended a good pair of running shoes.


Who came up with C1V1 = C2V2? I was wondering if there is an original scientific publication showing why $C_1V_1 = C_2V_2$ works.
Or perhaps, who described/used it first?

tl;dr It's so simple it was likely never published as a "research result", since it stems quite directly from definitions.

It's a small step from the conservation of matter and the definition of concentration.
By definition: $C = \frac{n}{V}$, thus $n = CV$. If you dilute with pure solvent, you don't change the amount of solute that you have, in other words: $n_1 = n_2$. Using the previous equation, you get $C_1V_1 = C_2V_2$.


Letter codes in molecular term symbols I'm trying to find an explanation for the letter codes (X, A, B, C, etc) when you get a term symbol like this

$$X\,{}^3{\Sigma}^-_g$$

Can anyone point me to some literature that explains these letter codes?

Quoting from Hollas, Modern Spectroscopy, 4th ed. (p 236):

  There is a convention, which is commonly but not always used, for labelling electronic states. The ground state is labelled $X$ and higher states of the same multiplicity are labelled $A, B, C\ldots$ in order of increasing energy. States with multiplicity different from that of the ground state are labelled $a, b, c \ldots$ in order of increasing energy.

As such, for the oxygen molecule, where the lowest three states are ${}^3{\Sigma}_g^-$, ${}^1{\Delta}$ and ${}^1{\Sigma}^+_g$ (in increasing order of energy), the triplet ground state would be labelled $X$ and the next two states $a$ and $b$ respectively.

A word of caution: it seems that this convention is not always obeyed. On the next page, Hollas writes (for $\ce{I2}$, where the ground state is a singlet)

  The transition $B\,{}^3{\Sigma}_{0_u^+} - X\,{}^1{\Sigma}_g^+$ involves the $\Omega = 0$ component of the $B$ state [...] The labelling of the $B\,{}^3{\Sigma}_{0_u^+}$ state follows general usage rather than convention, which would label it $b\,{}^3{\Sigma}_{0_u^+}$.

Wikipedia also adds, although without a citation:

  In polyatomic molecules (but not in diatomics) it is customary to add a tilde (e.g. $\tilde{X}$, $\tilde{a}$) to these empirical labels to prevent possible confusion with symmetry labels based on group representations.


Volume of nitrogen gas produced during taking off The fuel in the buster rockets of the Space Shuttle is constituted of a mixture of ammonium perchlorate, $\ce{NH4ClO4}$, and aluminum powder. One of the reactions taking place during taking off is given by
  $$\ce{6 NH4ClO4(s) + 10 Al(s) -> 5 Al2O3(s) + 3 N2(g) + 6 HCl(g) + 9 H2O(g)}$$
  The indication (s) and (g) signify solid and gaseous phase. Suppose that these rockets are loaded with $\pu{11.75 ton}$ of ammonium perchlorate and $\pu{2.70 ton}$ of aluminum.

  Calculate the volume of nitrogen gas produced during taking off under normal
  conditions ($\pu{0^\circ C}$ and atmospheric pressure, so that $\pu{1 mol}$ corresponds to $\pu{22.4 L}$ of gas).

I got to the point where it's $\pu{10^5 mol}$ for each of the compounds.
How do you get to the answer $\pu{6.72e5 L}$?

Your next step is to identify the limiting reagent.

Observe that six moles of ammonium perchlorate react with ten moles of aluminum.  So by taking ratios you find that:

1)  The $10^5$ moles of ammonium perchlorate you started with react with approximately $1.7×10^5$ moles of aluminum.

2)  The $10^5$ moles of aluminum you started with react with $6×10^4$ moles of ammonium perchlorate.

Now there is a problem with (1) -- you have only $1.0×10^5$ moles of aluminum available so not all of the ammonium perchlorate can react as indicated in (1).  So we now look at (2) -- it indicates that all of the aluminum can react and you still have $4×10^4$ moles of ammonium perchlorate remaining.

Thus we can realize condition (2) rather than (1) because the amount of reaction is limited by the aluminum you have available.  Thus aluminum is the limiting reagent.

Now that you know the limiting reagent you can infer how much product is formed.  With $10^5$ moles of aluminum all being consumed, you should be able to see that $3×10^4$ moles of nitrogen molecules are formed as gas.  The volume under normal conditions should then follow.


What Affects Tm and Tr In Chromatography? In chromatography, a mobile phase with analytes is passed over a stationary phase. A given analyte has different solubility in the two phases and consequently spends different periods of time in them. A particular molecule while in the stationary phase remains immobile, is swept along when present in the mobile phase.
$T_m$, the time spent in the mobile phase, is the same for all analytes. The reason (I'm not so sure) must be since all analytes travel the same distance finally, with the same speed as the mobile phase, so take the same time in so doing. But since some of these spend longer times in the stationary phase than others, the total time to reach the end, that is, $T_m + T_r$ is different ($T_r$ is the time spent in the stationary phase).

But I have a bit of a trouble understanding the rate aspect of the process, and what exactly determines the $T_r$ and $T_m$ - are they the average times spent by an arbitrary molecule in the two phases? If so, are their values dependent on the rate of the process, or the solubilities?

Think of it like a solvent extract. Each solvent extraction is a plate.

So at time 1 I do extraction 1  and move top layer to extraction flask 2.

At time 2 I have two extractions to do. I extract flask 2 and move top layer to flask 3 and then I extract 1 again and move top layer to 2 again.

At time 3 I have three extractions to do. and so one.

Another analogy.

Think of the whole thing as binomial trials. Each molecule has a different binomial probability. So the absolute difference in the peak positions between the molecules gets bigger as the number of trials increases.


Molecular dynamics after modeling a protein loop. Necessary? Firstly, I am postign this question to both chemistry and physics stacks exchange as I am not sure to which it belongs and to use the benefits from both sites.

I have modeled missing loops in a PDB structure using chimera. Should I run MD or energy minimization on the output file?

Thank you in advance.

To expand on Mithoron's comment, it's absolutely necessary to equilibrate (or at least minimize the energy) of your system once you've introduced an appreciable perturbation of any sort (in this case, you've appreciably changed the system itself!). To give you an idea, some lesser changes that would still justify equilibration are: changes in solvation, temperature, pressure, protonation of amino acids, oxidation state of cofactors, etc.

Note: by equilibration, I mean letting your system relax in the presence of surroundings. By minimization, I am referring to minimizing the total energy of the protein (explicitly solvated, unless you're doing some odd gas-phase or implicit solvation) in the absence of temperature and pressure (0 K, no thermostat or barostat).


Are Anomers Epimers? epimers are the class of stereoisomers which differ along configuration at any carbon in a carbohydrate(generally),

And anomers are the kind of stereo  isomers that differ at the carbonylic carbon, which becomes achiral only after hemiacetal (hemiketal) formation

Websites on search give that Epimers are bigger group where Anomers are a part of them, but in certain books on biochemistry like that of Lippincott(d.ferrier) writers define Epimers as change in configuration at any carbon except anomeric carbon..

So are Anomers Epimers?

IUPAC basic terminology of stereochemistry defines anomers and epimers:

Anomers
Diastereomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing
in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.
Epimers
Diastereomers that have opposite configuration at only one of the two or more stereogenic centers present in the respective molecular entities.

So yes, anomers are a subset of epimers.
(Interestingly, on the contrary, ‘anomer’ term is exceptionally used in much broader meaning than ‘epimer’, in the term ‘anomeric effect’, which became generalized to e.g. conformations of molecules completely unrelated to saccharides.)


Redox reactions KMnO4 changing to brown If potassium permanganate is added to a solution and it becomes brown is the substance a reducing agent?

Yes, the substance is a reducing agent which will reduce $\ce{KMnO4}$ into $\ce{MnO2 }$ in neutral or weakly basic medium.


Does sodium form complexes like transition metal ions? I realise that there is a similar question here Difference between sodium ion and a transition metal ion dissolving in water? and it seems to answer my question, however I was reading about how aluminion ions form complexes in water, and aluminium is not a transition metal. So how can aluminium form complexes like transition metal ions, but sodium does not?

Although less common than transition metal complexes, sodium does form complexes with some ligands, particularly oxygen based ligands.

Aqua complexes are formed in aqueous solution, the most common being $\ce{[Na(H2O)6]+}$.

Sodium forms many complexes with crown ethers, cryptands and other related ligands. For example, 15-crown-5:


Overshooting titration effect on K a. Explain how overshooting the titration of the aqueous layer (ie. using too large a value for the volume of thiosulfate to react with the iodine and of triiodide ion) would affect the concentrations of iodide ion, iodine and of triiodide ion, and therefore of the calculated $K$.

My thoughts regarding this question
overshooting the titration will increase the concentrations of the iodide ion, iodine and triiodide ion
- According to me  a change in $K$ comes only when the temperature is changed.

  b. Explain how using too large a value for the minimum volume of thiosulfate to react with the iodine (organic) in the mineral oil layer would affect the concentrations of iodide ion, iodine and triiodide ion, and therefore of the calculated $K$

I am unsure of the answer for this one too.
My reasoning
Using too large a value for the minimum volume of thiosulfate will increase the concentrations of the iodide ion, iodine and triiodide ion and nothing will happen to $K$ again.

Is my reasoning correct?

Your reasoning is correct that $K$ is independent of concentration. But I think you are misreading the question: it wants you to tell how you will miscalculate K with respect to the correct value if you put in the wrong concentrations in the formula, that is, if there was an overshoot in your titration.


Second-step activation energy in SN1 reactions In SN1 reactions, it's known that there are generally two steps, with the first being the R.D.S step involving carbocation formation and the next having a lower activation energy.
Why does the second step require activation energy at all? It's going from being a carbocation to a stable molecule, which will be more stable than the former, no matter how stabilized a cation it is.

The $\mathrm{S_N1}$ reaction has two steps i) the formation of carbocation and 2) the formation of product.

(image taken from libretexts)
In the second step an unstable carbocation reacts with the nucleophile. You are asking why this step has to have an activation energy. You can rationalize that by considering the principle of microscopic reversibility. Think about what would happen if the $\mathrm{S_N1}$ reaction went in reverse—first the nucleophile leaves the product, forming a carbocation, and then the carbocation joins with the leaving group to give the reactant. Obviously when the nucleophile leaves the prdouct, and carbocation forms, there is an activation barrier for that.
So regardless of the themodynamics of the total reaction, whenever you detach a functional group and form a
(carbocation + anion), there has to be an activation barrier i.e. a transition state that is higher in energy than both (carbocation + anion) and the initial state. Which means that if you wish to reattach an anion to the carbocation, there will also be an activation energy.