Expanded structure formula and Lewis dot structure, whats the difference? I am teaching my self the basics of chemistry and I have done lewis dot structure before. Right now I am going into carbon chemistry which often use structure formulas to represent compounds. There is this formula called the expanded structure, and from what I can see, the expanded formula is just the same as the Lewis except we use the lines to represent bonds etc?

So my main question is whats the difference between them and do we have to assign the formal charges in the expanded structure formula? Also, do we show the dots (valence electrons) in the expanded formula? Thank you.

Yes. Extended molecular structures replace dots with line emphasising electron pairs that are shared between two atoms. In other words each line does represent a bond. Most often the EMS omit electrons that are not directly involved in bonds. As such the depiction of electrons between C and O in, e.g., a carbonyl group is simply sketched as C=O. However, whenever the presence of lone pairs is relevant to the discussion, those are explicitly shown.

In case in which the structure does not correspond to a single Lewis structure, the most important one is depicted and the situation is considered to be known to the reader. An archetype example is benzene. It can be sketched as having three double bonds but their delocalization is assumed. Those six delocalized electrons can be depicted as an inner ring, an expedient which is more evocative but must be abandoned when fine details are to be discussed (mechanisms of reaction or mesomeric effects). Moreover a circle in aromatic rings doesn't always represent six electrons.

Finally you can find depiction of delocalized electrons as variously dotted lines enveloping the group of atoms over which they do delocalize, for instance a kind of "cloud" around the two oxigens of a carboxylate anion. Their meaning is usually clear.

To the second question: on the light of the above yes, formal charge are to be assigned. They are usually important while discussing the relative stability of limiting forms which enter the description of the actual hybrid of resonance.

https://en.m.wikipedia.org/wiki/Lewis_structure. Particularly the subsections Resonance and Usage and limitations.

Edit: For sake of clarity I must add that unpaired electrons are always depicted. Indeed they are always of chemical relevance!


Why is the reaction between potassium permanganate and hydrogen peroxide spontaneous? When hydrogen peroxide is mixed with potassium permanganate, oxygen gas and water vapour are formed, according to the reaction (source):

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O + 3O2 + 2OH-}$$

This reaction is spontaneous, and exothermic. It is an example of a redox reaction, with the following half reactions occurring (data from Vanýsek):

$$
\begin{align}
\ce{MnO4- + 2 H2O + 3 e- &-> MnO2 + 4 OH-} &\quad E^\circ_\mathrm{red} &= 0.595~\mathrm{V} \\
\ce{H2O2 &-> O2 + 2 H+ + 2 e-} &\quad E^\circ_\mathrm{ox} &= -0.695~\mathrm{V}
\end{align}
$$

$E^\circ_\mathrm{cell}$ is equal to the sum of the oxidation potential and the reduction potential of the two half reactions; in this case, it would be $-0.1~\mathrm{V}$. A redox reaction is spontaneous if $E^\circ_\mathrm{cell}$ is positive — how can it be, then, that hydrogen peroxide spontaneously reacts with permanganate ions?

Using thermodynamical data (from NIST), I have calculated that the $\Delta G^\circ_\mathrm{m}$ of the reaction is $-463.576~\mathrm{kJ}$. The reaction should indeed be spontaneous. How can it be, then, that the results of the thermodynamical approach and the electrochemical one differ drastically?

For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$.

The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$:

$$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$

Whereas the given reaction of permanganate and the corresponding potential apply to $\mathrm{pH=14}$ or $\left[\ce{OH-}\right]=1$:

$$\ce{MnO4- + 2 H2O + 3e- <=> MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$

The corresponding reaction and potential for $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$ are:

$$\ce{MnO4- + 4 H+ + 3e- <=> MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$


How to perform a total synthesis on paper ? (Think before to make it in lab) I studied organics chemistry during five years. All those years I just made reactions to made them, the only moments where I had to perform a synthesis on paper was during exams or exercises but all steps were given. So, even if, it was very interesting, there were no really fun to apply all this knowledge to think about the synthesis of a molecule. Obviouly with all these lessons I'm not that bad and I'm able to do a little bit of retrosynthesis.

However I know that total synthesis of natural origin molecules is perform a lot for some decades, we can find a lot of examples here, and sometimes it looks pretty weird to think that human people were able to think on how to make a synthesis in more than twenty steps.

For example, the weirdest synthesis I found, is the synthsis of Azadirachtin "only" perform in $79$ steps. That is abosolutely insane.

My question:

  I'm wondering, for every molecules in general, how do chemist do to
  think about the retrosynthesis of such big molecules which contain in
  general a lot of stereocenters which then add more difficulties?
  I mean without to use a computer, is there some rules, such like theorems in mathematics, to think in the best possible way? Because with a bit of luck for $79$ steps at $90$% yield it makes a final yield of $0.024$%...

Retrosynthesis, the way to determine which cuts to make and which steps to use in which order (at least in the general picture), is an art. It is not a science or a craft, it is an art. You learn it by doing it and there is only so much you can get taught by others.

Recently, a multi-group seminar I attended started a ‘molecule of the month’ challenge. We were given a molecule and divided into groups. Each group then had about six weeks to come up with a total synthesis plan, find literature precedence or good arguments for all steps and then present it. Our basic strategy was to draw substructures into SciFinder and rely heavily on its reaction search functionality. With more experience and having read more synthetic journal articles, we probably would have known what we would want to go for.

Finally, I feel like adding that sometimes it is not the molecule whose total synthesis you are publishing that piques your interest. There are also those that primarily do method development and then try to find out ‘which molecule could be a good target for the new and super-cool method I just established?’ Thereafter, the synthesis is modelled to fit the desired novel key step. This paragraph is just to show an alternative approach to total synthesis.


Alcohol (ethanol) distillation: mystery congener I regularly home-distil small amounts of potable alcohol (ethanol), mainly from fermented sugar(s) solutions (the 'wash').
My still is a box standard, $4.5\mathrm{L}$ electrically heated (safety!), air-cooled pot still, which produces exactly $1$ single stage distillation ('one Theoretical Plate') Starting from a wash of about $17\text{ %}$ ABV, I can distil that to (initial ABV) about $65\text{ %}$ ABV.
About $1.0\mathrm{L}$ is distilled off from $1$ gallon that way and the distillate then diluted with water to about $37\text{ %}$ ABV.
Distillers (home, pro, moonshiners) use a system to get the most and the best out of their distilled washes and it goes as follows:

The "Heads": the first part ($5 - 10\text{ %}$) of the distillate is often discarded because it contains non-ethanol low boiling congeners like methanol, ethanal, acetone and low boiling esters (it's a commonly held but erroneous belief that methanol and ethanol can be separated by means of a single distillation - see the $\text{MeOH/EtOH}$ VLE). These congeners give the distillate a 'synthetic' and undesired taste profile.
The "Heart": this is the good tasting part of the distillate, that is kept as an alcoholic beverage.
The 'Tails' (or 'faints'): the last part of the distillate contains very little alcohol, as well as higher boiling compounds like higher alcohols and $\text{C8 - C12}$-based esters. The high boiling congeners are also referred to as "fusel oils".

Apologies for the lengthy preamble.

Due to the pandemic and restrictions on the hospitality sector I obtained several gallon of out-of-date $4.5\text{ %}$ ABV apple cider of a very well known brand, from a friendly publican.
I first consumed a couple of liters as such and it tasted fine and caused no adverse effects. The product was indistinguishable from 'in date' virgin cider. But my purpose was always to distill the ethanol of out it.
So, using my standard procedure I distilled several gallon of the out of date cider, obtaining about $2\mathrm{L}$ of distillate at about $35\text{ %}$ ABV. Heads were discarded and the distillate was run over activated carbon (a practice highly recommended by home distillers and moonshiners, even though I'm not sure it serves a real purpose)
And here's the rub: the 'heart' smelled and tasted awful. The taste is hard to describe but keywords would be 'spicy', 'hot', 'sharp', 'solvent-like'.
I then took the step of redistilling the $2\mathrm{L}$ of product, discarding the first $300\mathrm{ml}$ and keeping a 'heart' of about $1.1\mathrm{L}$. Alas it changed very little, if anything: the redistilled alcohol was still basically undrinkable!
The contrast between the redistilled alcohol and cheap commercial gin is quite stark.
I measured the paper $\text{pH}$ of both and they were both essentially neutral.
That redistilling didn't remove the offending compound(s?) suggests that  its BP is close to that of ethanol or that it forms an azeotrope with it.
The smell and taste of the mystery congener(s?) doesn't really comply with what is usually suspected to be in the 'heads':

$\text{MeOH}$ smells and tastes very much like alcohol (and in any case should only be present in very small amounts)
acetone and ethanal probably are too low boiling to remain partly in the 'heart'. Ethanal has a pleasant odour anyway.
all the low boiling esters have pleasant, fruity flavours.

My questions are:

What could be the problem congener{s)?
How to further investigate to find out what might be the problem congener{s)? I don't have access to modern analytical techniques like IR, GC, GC-MS etc.

I do have access to $\text{TLC}$ and a $1\mathrm{L}$ lab still with an $800\mathrm{mm}$ Vigreux column.

People turn apple cider all the way to apple cider vinegar, so during the process, there will be a time when the amount and ratio of ethanol and acetic acid are optimally balanced for making ethyl acetate - but it doesn't seem to happen. However, given enough time (and oxygen and bacteria), maybe different things happen. Interesting that the expired cider didn't have the nasty taste, but the concentrate did. I wonder if you actually made the ethyl acetate by your heating and distillation procedure.

Ethyl acetate presence gives nail polish remover/ solvent flavor/aroma. This is not related to volatile acidity. It's known to be produced by Pichia (and probably other microbes) in the presence of oxygen.
Source: https://www.homebrewtalk.com/threads/need-a-little-help-with-cider-flaws.666728/

The CRC Handbook shows a ternary azeotrope of water (9.0%), ethanol (8.4%) and ethyl acetate (82.6%) that boils at 70.23 °C vs the water-ethanol azeotrope that boils at 78.17 °C (96% ethanol). The boiling points are sufficiently close that a one-plate column might not do a significant separation, but the heads still should taste worse.
Suggestion: if the apple cider is acidic from a little acetic acid, neutralize to pH 7 or 8 before distillation ($\ce{NaHCO3}$). This might reduce the possibility of ethyl acetate formation. Another experiment would be to take fresh apple cider and see if you can get good tasting alcohol from it.
TLC of volatile solvents is difficult, but you might be able to detect different spreading patterns for the heads, the nasty heart, and some other heart (which tastes OK).


RO (alkoxide) as a leaving group Why is RO (alkoxide) a better leaving group than OH, despite RO being more unstable due to the electron donating effect of the alkyl group on RO?

I read that the suitability of leaving groups is dependent also on their nucleophilicity, as good nucleophiles will tend to re-attack the molecule it left from. To what extent is this true?

The alkoxide and the hydroxides aren't good leaving groups. Consider an alcohol, the $\ce{OH}$ group never leaves on its own. Oxygen donates a lone pair to the hydrogen of a hydronium ion (considering it to be in an aqueous solution). The water molecule now attached is a good leaving group (oxygen has a positive charge).

Nucleophilicity does not determine the suitability of a leaving group. A group is said to be a good leaving group when it can leave as a relatively stable ion. For example: iodine is a very good leaving group for an $\mathrm{S_N2}$ reaction.


How can I turn LPG from liquid to gas? I am working on a personal project that involves a balloon filled with liquefied petroleum gas (LPG) at a pressure of $\pu{4 - 5 bar}$, and I need some kind of a device/filter that would convert it to gas (instead of liquid) at a low pressure. I used an industrial vaporizer, but it gave me something stronger than I need, and "not clean."

Basically, what i need is to build a very big lighter,
like this:

As you can see, just gas, and wick that turns to a "clean" fire, without all the side effects that the vaporizer creates.

Now, as I am missing a lot of knowledge here, I would like to know if someone could guide me through how I should do this, where I can find the things I will need, what the names and definitions for the stuff needed, etc.

EDIT:

When I used the phrase "home gas," which is common in my state, I meant liquefied petroleum gas (LPG).

Here is the vaporizer I use now (it belongs to someone who knows what he is doing, I do not plan on killing myself :) )

So, I will try to re-describe my problem:

I want to get a "clean, slow flame" such as a lighter has. What I am getting, regardless of the vaporizer pressure, is an unclean flame, as though some gas is not really burning and just comes out as well, or as if something in the process makes the flame burn dirtier.

You don't really need to do anything to get LPG vapour. It is a gas a room temperature and pressure so as long as you just want the clean gas not mixed with air (which would be dangerous as the mixture will be flammable and potentially explosive), all you need to do is allow the output of the canister to vent into a balloon.

LPG is mostly propane and butane which liquefy under mild pressure. Butane will liquefy at about 220 kPa at normal room temperature (this is a little over twice atmospheric pressure). This is what happens in a typical cigarette lighter like the one illustrated. There is no "wick" involved just a valve that allows the gas to vent to the atmosphere and a flint to create a spark and ignite it (which you can avoid doing by careful handling of the lighter activating just the valve and not the flint). Large cylinders are similar and it might be possible to vent directly into a balloon just by carefully controlling the exit valve on the cylinder.

It is possible that standard supplies of LPG for household use also contain stinky additives to make leaks easy to detect, but otherwise the gas should be similar to that found in a cigarette lighter.


When it comes to bond energy or molecular binding energy, do people mean enthalpy? When calculating bond energy or molecular binding energy, which value should I use, enthalpy or total energy?

For example, A+B=C

Should it be C's enthalpy of formation minus (A's+B's)enthalpies of formation,
or C's total energy minus (A's+B's) total energies?

When in doubt, consult IUPAC Gold Book for definitions (italics below is mine).

bond-dissociation energy, $D$

The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> CH3^{.} + H^{.}}$, symbolized as  $D(\ce{CH3−H})$.

bond energy (mean bond energy)

The average value of the gas-phase bond dissociation energies (usually at a temperature of  298 K) for all bonds of the same type within the same chemical species. The mean bond energy for methane, for example, is one-fourth the enthalpy of reaction for: $$ \ce{CH4_{(g)} -> C_{(g)} + 4H_{(g)}} \, .$$ Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

So, yes, conceptually you should use enthalpy. Note, however, that the enthalpy of a molecular specie in gas phase at, say 298K, is calculated as follows
$$H_{298} = E_{\mathrm{e}} + E_\mathrm{ZPE} + H_{\mathrm{trans}} + H_{\mathrm{rot}} + H_{\mathrm{vib}} + RT \, ,$$
where

$E_{\mathrm{e}}$ is the electronic energy;
$E_\mathrm{ZPE}$ is the zero point energy;
$H_{\mathrm{trans}}$, $H_{\mathrm{rot}}$ and $H_{\mathrm{vib}}$ are the temperature-dependent contributions to enthalpy from translation, rotational and vibrational motion, respectively.

And the problem (as Martin mentioned in comments) is that basically only the first term out of this five can be calculated accurately enough when the ultimate goal is to calculate energies (bond energies in our case) with the so-called "chemical accuracy" which is about $\pm 1$ kcal/mol.
For relatively large molecules harmonic frequencies and consequently $E_\mathrm{ZPE}$, as well as a rigid-rotor harmonic-oscillator approximation used to calculate thermal corrections, become inaccurate, and thus, the change of electronic energy should always be used as a reference.

$%edit$


What's the Gibbs free energy of a reaction at zero temperature and pressure? I heard that Gibbs free energy changes with temperature and pressure as

$$ \Delta G_1 = \Delta G_0 + R T \ln Q $$

But this makes no sense when pressure is zero or temperature is zero.

$G^\circ = U(0K) = E^{ele}+ZPE$, i.e. the sum of the electronic energy (electrons and nuclei interacting) and the zero point vibrational energy.

You can't use $K=e^{-\Delta G^\circ/RT}$ at 0K because the derivation assumes that a certain amount of thermal energy is available. But you don't need the equation to figure out the relative amount of two molecules. At equilibrium at 0K all molecules are in the lowest possible state.


Notation of hydrogen $$\ce{e- + H2O -> [H$\cdot$] + HO-}$$

I was a little confused on the notation of the hydrogen in brackets ([H]) in the product side of the reaction. Why is there a dot next to it and what does it mean? Is it saying Hydrogen gas?

Also, what will be the reduction potential of this reaction?

The notation $\ce{H^.}$ refers to the hydrogen radical (a hydrogen atom). The dot represents the unpaired valence electron.

An equivalent notation for the reaction would be

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

However, adding the dot emphasizes the fate of the electron on the reactant side.


What kind of bonds (σ or π) are formed by px, py and pz orbitals? Our chemistry teacher said that two  $\mathrm{p}_z$ orbital only form sigma bond whereas  two $\mathrm{p}_x$ and two $\mathrm{p}_y$ orbitals form only pi bond.

He also stated that only $\mathrm{p}_z$ orbital forms a sigma bond with the $\mathrm{s}$  orbital.

But in our book it is written that $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals can form both sigma and pi bonds.

Please give me the correct explanation.

It's a question of

(1) labeling conventions (what you call x, y and z in the coordinate system defining the position of two bonded atoms)

and

(2) the orientation of the two overlapping orbitals from the bonding atoms

If an overlap maximum occurs on the internuclear axis between the atoms, you call the resulting bonding orbital a "sigma" orbital. If maximum overlap occurs on an axis parallel to the internuclear axis, you call this type of bonding "pi". Either s or p orbitals can form sigma bonding orbitals, but s orbitals generally form sigma orbitals, whereas of the 3 conventional p orbitals, only one has lobes oriented along the internuclear axis (this orbital is usually called $p_z$ because its principal symmetry axis coincides with the internuclear axis, which is also labeled the z axis) and forms sigma bonding orbitals. The remaining two (x and y) form pi orbitals. This is illustrated schematically in the following figure:

(note the AOs don't overlap significantly, focus is on showing their relative orientation and labels)


Analysis of mixture by precipitation and complexometric titrations I've performed an experiment about chloride mixtures. First using $\ce{AgNO3}$ as titrant, I've titrated the mixture to determine the amount of total chloride. Then by direct titration with $\ce{EDTA}$, I wanted to determine the amount of $\ce{Mg}$ ions.

I'm confused about the amount of total chloride ions as there are two sources of chloride ions.

I'd really appreciate if someone can help me here.

Thank you so much.

Let $x$ be the amount of $\ce{MgCl2}$ in the original mixture, and $y$ be the amount of $\ce{NaCl}$.

Both of these salts dissociate completely in aqueous solution to give their constituent ions:

$$\begin{align}
\ce{MgCl2 (s) -> Mg^2+ (aq) + 2 Cl- (aq)} \\
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
\end{align}$$

Therefore, the total number of moles of chloride ions in the mixture is:

$$\begin{align}
\eta_{\ce{Cl-},\text{total}} &= \eta_{\ce{Cl-}\text{(from }\ce{MgCl2})} + \eta_{\ce{Cl-}\text{(from }\ce{NaCl})} \\
&= 2\eta_{\ce{MgCl2}} + \eta_{\ce{NaCl}} \\
&= 2x + y
\end{align}$$

$\ce{Ag+}$ ions react with $\ce{Cl-}$ ions in a $1:1$ stoichiometric ratio:

$$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$

Therefore, we have $\eta_{\ce{Ag+}} = \eta_{\ce{Cl-},\text{total}} = 2x + y$.

$\ce{EDTA}$ only complexes magnesium, and again in a $1:1$ stoichiometric ratio:

$$\ce{Mg^2+ (aq) + edta^4- (aq) -> [Mg(edta)]^2- (aq)}$$

Therefore, $\eta_{\ce{EDTA}} = \eta_{\ce{Mg^2+}} = \eta_{\ce{MgCl2}} = x$.

From your titration, you would have determined both values $\eta_{\ce{Ag+}}$ and $\eta_{\ce{EDTA}}$. Therefore, you have a system of two simultaneous equations in two unknowns, which is extremely simple to solve:

$$\begin{align}
\eta_{\ce{Ag+}} &= 2x + y \\
\eta_{\ce{EDTA}} &= x
\end{align}$$


Will the following complex show optical isomerism So there is this question which asks whether the following complex

  $\ce{[Co(NH3)4Br2]+}$

will show optical  isomerism or not.

According to my knowledge it will show optical isomerism in cis form only and not in trans form so overall the compound shouldn’t be optically active.

This what I hav been taught. And moreover shouldn’t it be optically active in every scenario to be called so?

But the ans says that it shows optical isomerism.
I don’t understand why.. is my understanding wrong?

As already commented by Ivan Neretin, the cis form also does not show optical isomerism. Elaborating a bit further on the comments:

Complexes show optical isomerism when they are asymmetric (also known as chiral). They should not have any elements of symmetry, such as planes/axes/centres of symmetry. The trans form has a clearly visible plane of symmetry. It's a bit trickier to find the plane for the cis form. Here's a diagram showing the planes:

(source: Chemistry libretexts)


How to prove that chloroform is more acidic than fluoroform? How to prove that chloroform is more acidic than fluoroform?

Using the concept that the inductive effect is dominant over the resonance (mesomeric effect) in the case of halide substituents: $\require{mediawiki-texvc}$

Since both have single hydrogen, let's compare stability of carbanions $\ce{CF3-}$ and $\ce{CCl3-}$.
Since fluorine shows more negative inductive effect than chlorine, $\ce{CF3-}$ should have been more stable than $\ce{CCl3-}$, so some greater cause must be acting here.
In $\ce{CCl3-}$ carbon has non bonded electrons in 2p orbital and chlorine has an empty 3d orbital, but since the energy difference is very high, $\ce{p\pi-d\pi}$ backbonding is barely possible. Worst case scenario, even if it's possible, then giving preference to backbonding will violate our initial rule (that inductive > mesomeric).
Bond lengths, $\ce{C-Cl}=1.75\AA, \ce{C-F}=1.35\AA$, but since chlorine is bigger than fluorine, they both will have approximately same electronic $\ce{p-\pi}$ repulsion (of lone pair of halogen and at anionic centre).

Which of my reasoning are wrong? How do I prove the fact that chloroform is more acidic than fluoroform?

Data: $\mathrm pK_\mathrm a (\ce{CHBr3})=13.7$ and $\mathrm pK_\mathrm a (\ce{CHCl3})=15.5$ and $\mathrm pK_\mathrm a (\ce{CHF3})=28$ (source - page 2)

My Views (I'm not entirely sure about these reasons)

(1) $\ce{CF3-}$ is highly unstable due to huge repulsions between the negative charge on carbon and the lone pair of fluorine.

(2)While in $\ce{CCl3-}$ chlorine has an empty d orbital thus the negative will be more dispersed into the this empty orbital  hence the repulsions will be quite less as compared to $\ce{CF3-}$.


Calculating the Radius of a Molecule Assuming the molecule is a sphere, would the radius of amylopectin be $9.17*10^{-22}~cm^3$? I arrived at this answer by deriving the molar volume by dividing the molar mass ($828.72~g/mol$) of amylopectin by its density ($1.5~g/cm^3$). Then I divided the quotient by Avogadro's constant to arrive at my aforementioned answer. However, I don't know if this is correct so please correct me if I am not.

When you divide the mass of the solid by its density then you get the volume of the object which means not only the volume of the molecules also the volume of voids. Now the volume you get is = volume of the particles + volume of voids. So the radius you get is = radius of void + radius of molecule.  So in order to solve the problem you must which kind of packing is this. And then you should know it's packing efficiency. So now the volume you got should be multiplied by the packing efficiency to get the actual volume of molecules. If you say why do they have voids. Then I'll  say you can't fill a particular shaped object with small sphere's so packing efficiency can't be 100%. So all you need to know is the kind of lattice formed by the molecules of amelopectin. And now what you got in your question is the volume of single particle + the volume of void. Well if you wanted to get the radius from it then it would be error as I said before. But in that case also you had to put $V=\frac{4}{3}\pi(r^3)$= the volume of the single molecule. And in this way you get the radius as 6.02*10^-8cm. And then get the radius. But actually taking the molecules as cube gives you a better answer though this would not give the proper answer if amilopectin molecules aren't cubic. But certainly you can fill a shape with cubes more efficiently (without having much voids) so in that case $a^3=volume you mentioned which is 9.17*10^-22 cubic cm. That gives a as 9.7*10^-8cm as the side length.


Writing the overall cell reaction for the Calomel electrode I an studying about electrochemical cells, and came across the Calomel electrode. It turns out that the half cell reactions and the overall reaction are as follows.
Anode half-cell: $\text{Hg}_{2}\text{Cl}_{2}\rightarrow \text{Hg}_{2}^{2+}+2\text{Cl}^{-}$
Cathode half-cell: $\text{Hg}_{2}^{2+}+2e^{-}\rightarrow 2\text{Hg}$
Overall cell reaction: $\text{Hg}_{2}\text{Cl}_{2}+2e^{-}\rightarrow 2\text{Hg}+2\text{Cl}^{-}$
I am not sure why while writing the overall reaction, the electron was not gotten rid of. Besides why is there no net exchange of an electron in the anode half-cell reaction. Clarification would be appreciated. Thanks.

The calomel electrode is not a cell. It can be part of a cell. It is a half-cell. A cell is made of two electrodes, one calomel electrode and another one. The reaction occurring in the calomel electrode is the reaction you present as "overall cell reaction". What you described as "anode half-cell" has nothing to do with electrochemistry. It is the equation describing how $\ce{Hg_2Cl_2}$ gets dissolved in water.


In Energy-Dispersive X-ray Spectroscopy, do higher atomic elements ever generate just K lines without L or M lines? The heavier elements only show L or M lines. Is that because the critical ionisation energy is too high to get K lines in higher atomic numbers? Can you ever get just the K lines, or just the M lines or L lines?

With a tuned X-ray source, e.g. from a synchrotron, it should, at least in theory, be possible to excite any specific line, or cause absorption at a specific wavelength, as in X-ray absorption spectroscopy.


18 Electron Rule For Determining the Stability of Transition Metal Complexes I chanced upon this 'rule' while I was messing around on Wikipedia, and since we just finished studying about transition metal complexes at school, it naturally drew my interest.

Though I did read the article on it on Wikipedia, I don't feel I've really grasped the concept well enough (in short … I'm kinda lost).

So I hope it isn't too much to ask but, is there someone here who's willing to spend some explaining the 18 electron rule, lucidly enough for a 17 year old to understand? Or if that isn't possible, could someone recommend a site that explains the rule decently enough? Also could you touch upon some exceptions to the rule (at least for period 4 transition metals)?

A brief description of the 18 electron rule

A valence shell of a transition metal contains the following: 1 $s$ orbital, 3 $p$ orbitals and 5 $d$ orbitals; 9 orbitals that can collectively accommodate 18 electrons (as either bonding or nonbonding electron pairs). This means that, the combination of these nine atomic orbitals with ligand orbitals creates nine molecular orbitals that are either metal-ligand bonding or non-bonding,  and when metal complex has 18 valence electrons, it is said to have achieved the same electron configuration as the noble gas in the period.

In some respect, it is similar to the octet rule for main group elements, something you might be more familiar with, and thus it may be useful to bear that in mind. So in a sense, there's not much more to it than "electron bookkeeping"
As already mentioned in the comments, 18 electron rule is more useful in the context of organometallics.
Two methods are commonly employed for electron counting:

Neutral atom method: Metal is taken as in zero oxidation state for counting purpose

Oxidation state method: We first arrive at the oxidation state of the metal by considering the number of anionic ligands present and overall charge of the complex

I think this website does a good job of explaining this: http://www.ilpi.com/organomet/electroncount.html (plus, they have some practice exercises towards the end)
Let's just focus on Neutral Atom Method (quote from the link above)

The major premise of this method is that we remove all of the ligands from the metal, but rather than take them to a closed shell state, we do whatever is necessary to make them neutral. Let's consider ammonia once again. When we remove it from the metal, it is a neutral molecule with one lone pair of electrons. Therefore, as with the ionic model, ammonia is a neutral two electron donor.
But we diverge from the ionic model when we consider a ligand such as methyl. When we remove it from the metal and make the methyl fragment neutral, we have a neutral methyl radical. Both the metal and the methyl radical must donate one electron each to form our metal-ligand bond. Therefore, the methyl group is a one electron donor, not a two electron donor as it is under the ionic formalism. Where did the other electron "go"? It remains on the metal and is counted there. In the covalent method, metals retain their full complement of d electrons because we never change the oxidation state from zero; i.e. Fe will always count for 8 electrons regardless of the oxidation state and Ti will always count for four.

Ligand Electron Contribution (for neutral atom method)
a. Neutral Terminal (eg. $\ce{CO}, \ce{PR_3}, \ce{NR_3}$) :  2 electrons
b. Anionic Terminal (eg. $\ce{X^-}, \ce{R_2P^-}, \ce{Ro^-}$) :  1 electron
c. Hapto Ligands (eg. $ \eta^2-\ce{C_2R_4},  \eta^1-\text{allyl}$): Same as hapticity
d. Bridging neutral (eg. $\mu_2-\ce{CO}$) : 2 electrons
e. Bridging anionic (eg. $\mu_2-\ce{CH_3}$) ( no lone pairs): 1 electron
f.Bridging anionic (eg. $\mu_2-\ce{Cl}, \mu_2-\ce{OR}$) (with 1 lone pair)): 3 electrons
(with 1 lone pair)
or, $\mu_2-\ce{Cl}$(with 2 lone pairs): 5 electrons
g. Bridging alkyne 4 electrons
h. NO linear 3 electrons
i. NO bent ( lone pair on nitrogen): 1 electron
j. Carbene (M=C): 2 electron
k.Carbyne (M≡C): 3 electron
Determining # Metal-Metal bonds
Step 1: Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence electrons of the metal plus the number of electrons from each ligand and the charge)-- I'll call this T (T for total, I'm making this up )
Subtract this number from $n × 18$ where $n$ is the number of metals in the complex, i.e $(n × 18) – T$ -- call this R (R for result, nothing fancy)
(a) R divided by 2 gives the total number of M–M bonds in the complex.
(b) T divided by n gives the number of electrons per metal.
If the number of electrons is 18, it indicates that there is no M–M bond; if it is 17 electrons, it indicates that there is 1 M–M bond; if it is 16
electrons, it indicates that there are 2 M–M bonds and so on.
At this point, let's apply this method to a few examples
a)  Tungsten Hexacarbonyl (picture)
Let's use the neutral atom method, W has 6 electrons, the carbonyls donate
12 electrons and we get a total of 18. Of course there  can be no metal metal bonds here.
(b) Tetracobalt dodecacarbonyl (picture), here let's figure out the no. of metal-metal bonds. T is 16, R is 12, Total # M-M bonds is 6, # electrons per metal is 15, so 3 M-M bonds.
A few examples where the "18 electron Rule" works
I. Octahedral Complexes with strong $\pi$ - acceptor ligands
eg.  $\ce{[Cr(CO)_6]}$

Here, $t_{2g}$ is  strongly bonding and is filled and,  $e_g$ is strongly antibonding, and empty. Complexes of this kind  tend to obey the 18-electron rule irrespective of their coordination number. Exceptions exist for $d^8$, $d^{10}$ systems (see below)
II. Tetrahedral Complexes
e.g. $\ce{[Ni(PPh_3)_4]}$ ($\ce{Ni^0}$, $d^{10}$ 18-electron complex

Tetrahedral complexes cannot exceed 18 electrons because there are no low
lying MOs that can be filled to obtain tetrahedral complexes with more than 18 electrons. In addition, a transition metal complex with the maximum of 10 d-electrons, will receive 8 electrons from the ligands and end up with a total of 18 electrons.
Violations of 18 electron rule
I. Bulky ligands : (eg. $\ce{Ti(\text{neopentyl})_4} $ has 8 electrons) Bulky ligands prevent a full complement of ligands to assemble around the metal to satisfy the 18 electron rule.
Additionally, for early transition metals, (e.g in $d^0$ systems),  it is often not possible to fit the number of ligands necessary to reach 18 electrons around the metal. (eg. tungsten hexamethyl, see below)
II. Square Planar $d^8$ complexes (16 electrons) and Linear $d^{10}$ complexes (14 electrons)
For square planar complexes, $d^8$ metals with 4 ligands, gives 16-electron complexes. This is commonly seen with metals and ligands high in the spectrochemical series
For instance, $\ce{Rh^+}, \ce{Ir^+}, \ce{Pd^2+}, \ce{Pt^2+}$) are square planar. Similarly,  $\ce{Ni^2+}$ can be square planar, with strong $\pi$-acceptor ligands.
Similarly, $d^{10}$ metals with 2 ligands, give 14-electron complexes. Commonly seen for for $\ce{Ag^+}, \ce{Au^+}, \ce{Hg^{2+}}$
III. Octahedral Complexes which disobey the 18 electron rule, but still have fewer than 18 electrons (12 to 18)
This is seen with second and third row transition metal complexes, high in the spectrochemical series of metal ions with $\sigma$-donor or $\pi$-donor ligands (low to medium in the spectrochemical series).
$t_{2g}$  is non-bonding or weakly anti-bonding (because the ligands are either $\sigma$-donor or $\pi$-donor), and $t_{2g}$ usually contains 0 to 6 electrons. On the other hand, $e_g$ are strongly antibonding, and thus are empty.

IV. Octahedral Complexes which exceed 18 electrons (12 to 22)
This is observed in first row transition metal complexes that are low in the spectrochemical series of metal ions, with $\sigma$-donor or $\pi$-donor ligands
Here, the $t_{2g}$ is non-bonding or weakly anti-bonding, but the $e_g$ are only weakly antibonding, and thus can contain electrons. Thus, 18 electrons maybe exceeded.
References:
The following weblinks proved useful to me while I was writing this post, (especially handy for things like MO diagrams)
http://www.chem.tamu.edu/rgroup/marcetta/chem462/lectures/Lecture%203%20%20excerpts%20from%20Coord.%20Chem.%20lecture.pdf
http://classes.uleth.ca/201103/chem4000b/18%20electron%20rule%20overheads.pdf
http://web.iitd.ac.in/~sdeep/Elias_Inorg_lec_5.pdf
http://www.ilpi.com/organomet/electroncount.html
http://www.yorku.ca/stynes/Tolman.pdf
and obviously, the wikipedia page is a helpful guide https://en.wikipedia.org/wiki/18-electron_rule


Dehydration of carboxylic acids using P2O5 Why isn’t concentrated $\ce{H2SO4}$ used for the dehydration of carboxylic acids to anhydrides? What is special about $\ce{P2O5}$?

$\ce{P2O5}$ is a solid. Hence, it is easy to separate from reaction products. Moreover, $\ce{P2O5}$ is a stronger dehydrating agent than $\ce{H2SO4}$


How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion? How do I know when I should add an electron to a single electron? Do I just always do this? I've came across this in the textbook in the form of the amide ion $\ce{NH2}$, among others. It was shown to have 8 electrons, despite nitrogen only having 5 valence electrons, and each hydrogen only having 1. The last came from some surrounding atom to complete the outer shell and thus become more stable. So, do I always do that when there is a single electron not part of a pair?

I think I know where your confusion comes from. Firstly, I would want to note an important point:

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Chemwiki.UCDavis

This is the characteristic that makes them different from ionic bonds. The surrounding atom will not become an ion in its full sense.
I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.
$\ce{H2O}$:
We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:

Source 1
Follow these instructions for "normal" molecules (the ones that abide octet in a perfect sense):

Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: $$6+(2*1)=8$$
I prefer to call this $\text{n}_v$. V represents valence shell.
Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get:
$$8+2*2=12$$
I prefer to call this $\text{n}_f$. F represents "final". You can name it whatever you want.
If you want to know how many electrons are bonding and are shared between atoms, subtract $\text{n}_v$ from $\text{n}_f$:
$$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$
For an ease in calculation, I call this $\text{n}_b$. As you might have just figured out,
If you want to know how many es are nonbonding and "belong" to a single atom, subtract $\text{n}_b$ from $\text{n}_v$ to get:
$$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$
Call this whatever you want. I rather calling it $\text{n}_{nb}$.

With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:

Source 2
$\ce{NH4+}$
In ions, we have to take the charge of the ion into an account. Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from $\text{n}_v$.
So let's get on with this thing's lewis structure:

$\text{n}_v = 5+(4*1) -1=8$ (Notice (-1) that was because of a 1+ ion charge.)
$\text{n}_f = 8 + (4*2)=16$
$\text{n}_b = 16-8=8$
$\text{n}_{nb} = 8-8=0$

There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as dative covalent bonding. We get this structure in accordance to the calculations and thoughts:

Source 4
Finally, $\ce{NH2-}$
Now this case is easily solvable through the explanations for ammonium.

$\text{n}_v = 5 + 2 +1=8$ The +1 is because of the charge of the ion.
$\text{n}_f = 8 +2*2=12$
$\text{n}_b = 12-8=4$
$\text{n}_{nb} = 8-4=4$

Put some thinking, and you should get this (ignore the other ones if you like):

Source 5


Are these chemical reactions with amino acids possible, and if so, what are they called? In high school, I am taught that the synthesis of two amino acids of the form $\ce{H2NR1COOH}$ and $\ce{H2NR2COOH}$ is a condensation reaction that produces a polypeptide $$\ce{H2NR1COOH}+\ce{H2NR2COOH}\to\ce{H2NR1CONHR2COOH}+\ce{H2O}$$ where $\ce{R1}$ and $\ce{R2}$ are arbitrary components within each acid. Then I started experimenting with different positions that yield isomers of the polypeptide.
Examples

Water is formed though the pairing of the two $-\ce{OH}$ groups $$\ce{H2NR1COOH}+\ce{HOOCR2NH2}\to\ce{H2NR1COOCOR2NH2}+\ce{H2O}$$ which forms something that resembles an ester?

Water is formed through the pairing of the $-\ce{OH}$ and $\ce{H2N}-$ groups but the oxygen atom is taken instead from $\ce{C=O}$ $$\ce{H2NR1COOH}+\ce{H2NR2COOH}\to\ce{H2NR1C(OH)NR2COOH}+\ce{H2O}$$ where essentially the hydrogen atom from $-\ce{CONH}$ has shifted one place.

Are these reactions directly possible? If so, are there special names for
a) the reaction and
b) the type of product formed?

I am at an elementary level in chemistry so more detailed explanations would be appreciated.

Reaction 1

The product is called carboxylic acid anhydride. If you make sure the amino groups don't react (protecting and deprotecting) and get rid of water to prevent back-reactions, this reaction is possible.

Reaction 2

The product could be called a keto-enol tautomer (here, the enol form). The keto form is much more stable, so this product - even if made as an intermediate - would isomerize to the standard peptide bond.


Rationalizing enamine stabilities The solutions manual said that enamine B with its roughly sp2 methyl group can force the nitrogen out of the plane and this messes around with the conjugation.

How about the fact there is a significant degree of resonance donation by the nitrogen, which places a significant partial negative charge on carbon. In enamine A, this carbon is secondary. In enamine B, this carbon is tertiary. Extra inductive donation from the methyl group destabilizes the product.

Conjugation effects are not very important in this case.
The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.

In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.

But there are exceptions, for example:

In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.


Wave function of a perturbed system In perturbation theory, wave function of a perturbed system can be expressed in a power series of some $\lambda$ as $\psi^{(0)} + \lambda\psi^{(1)}+\lambda^2\psi^{(2)}......$.

Then we express Hamiltonian operator as $H^{(0)} + \lambda H^{(1)}$.

Introduction of $\lambda$ in $\psi$ can be understood but why we introduce $\lambda$ in Hamiltonian operator and what is the significance of this $\lambda$ factor.

The method is only outlined below, the full details should be in your textbook or lecture notes.

Your $H^{(1)}$ is just the (small) perturbing potential energy added to the original hamiltonian so you could write $H^{(0)}+\lambda V$.  ($\lambda$ is dimensionless and varies from 0, no perturbation, to 1, full perturbation ).

The Schroedinger equation is now $(H^{(0)}+\lambda V)\varphi_n=E_n\varphi_n$  where $\varphi_n,\; E_n$ are the new wavefunctions and energy. These are expanded as a Taylor series in $\lambda$, e.g.  $\displaystyle \varphi_n \approx \varphi_n^0 + \lambda \frac{d\varphi_n}{d\lambda}+\frac{\lambda^2}{2!} \frac{d\varphi_n^2}{d\lambda^2}+ ..\equiv \varphi_n^{(0)}+\lambda\varphi^{(1)}_n +\lambda^2\varphi^{(2)}_n+..$ and similarly for $E_n$.

Next this result and that for  $E_n$ are substituted back into the Schroedinger eqn and powers of $\lambda$ factored out. These remaining equations are solved but none now contain $\lambda$; for example the first order term is $\lambda\left((V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}\right)$ where in this new notation  $\varphi^{(0)}$ is the original unperturbed wavefunction and similarly for the energy.

The equation to solve is now $(V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}=0$.

After (formally) integrating the equation some fiddling around you find that $\varphi^{(1)}$ is needed. The important step now is to expand this wavefunction in terms of the original basis set of unperturbed wavefunctions $\phi$. This can be done because these form a complete orthonormal set, thus $\varphi^{(1)}= \sum_k a_k\phi_k$ where $a_k$ are the expansion coefficients and these have to be found.  Again after some more fiddling with equations and using the orthonormality of the $\phi_n$ the perturbation energy $E^{(1)}_n = \int \varphi^{(0)*}_nV\varphi^{(0)}_nd\tau = \int\phi^{0*}V\phi^0 d\tau$.

Notice how only the unperturbed wavefunctions contribute to the energy.  The new wavefunction is $$ \varphi^{(1)}_n=\sum_{n\ne k} \frac{E_n^{(1)}}{E_n^{(0)}-E_k^{(0)}}\varphi^{(0)}$$

which is a linear combination of the unperturbed wavefunctions with the $a$ coefficients as ratio of energies.


Pyro- Prefix in Chemistry I was wondering is someone could please help answer this question for me. I was recently reading the ingredients on my toothpaste container, and one of the compounds in it was 'Sodium Pyrophosphate'; what does the 'Pyro-' prefix mean? I see it elsewhere but I've never come across such a nomenclature such as this. What does it mean in terms of Chemistry?

It means that there are two phosphates connected together under loss of water, i.e.:

If you have any questions about the definitions of chemical nomenclature, have a look at the IUPAC gold book. See here. It basically says, that a pyro compound is one that is formed by heating another compound, usually under the loss of simple molecules like $\mathrm{H_2O}$ or $\mathrm{CO_2}$.


Comparing heat of hydrogenation per mole of butenes and 1,3-butadiene The question asks us to select the compound with the smallest heat of hydrogenation per mole out of the following:

a) 1-butene

b) trans-2-butene

c) cis-2-butene

d) 1,3-butadiene

Since all four compounds form the same product (butane), a comparison without actual data should be possible. 1,3-butadiene is conjugated so it is the most stable compound and therefore its heat of hydrogenation should be the smallest.

But it requires 2 moles of hydrogen for complete conversion to butane. And the answer is actually trans-2-butene.

So would we consider each heat of hydrogenation "per mole of hydrogen" or "per mole of the given compound"? The question isn't clear about this and also, if we do consider it as "per mole of hydrogen" then is it correct to simply half the value heat of hydrogenation for 2 moles?

a comparison without actual data should be possible. 1,3-butadiene is
  conjugated so it is the most stable compound and therefore its heat of
  hydrogenation should be the smallest

No, 1,3-butadiene is not the most stable of these compounds, either for the whole molecule or on a per double bond basis (see below).  The actual numbers (see table below) are too close, we need to look at some data in order to answer the question correctly.

  The question asks us to select the compound with the smallest heat of
  hydrogenation per mole out of the following:

  So would we consider each heat of hydrogenation "per mole of hydrogen"
  or "per mole of the given compound"?

Here is some data that will help answer the question.

(image source)

On a "per mole of given compound", trans-2-butene has the lowest heat of hydrogenation, 27.6 kcal/mole.

On a "per mole of hydrogen", 1,3-butadiene has a heat of hydrogenation between 28.2 (56.5/2) and 28.6 (57.1/2) kcal/mole, so trans-2-butene still has the lowest heat of hydrogenation at 27.6 kcal/mole.

So in this particular list of compounds, it doesn't matter whether the question is asking per mole of hydrogen or per mole of given compound, the answer is the same, trans-2-butene.

At first I thought "trick question", but then later thought, not really.  The questioner did his homework and wanted the student to work out the answer both ways and see that since the you get the same answer whether you answer on a per mole compound or per mole hydrogen, you don't need to know whether the question is being asked on a per mole compound or per mole hydrogen basis.


Do Biuret test always means proteins presence in a substance? I am a student and I got this answer I cant really say for sure if I understood correctly :

"A substance reacts positively to Biuret's test. ¿ Can you say with 100% accuracy that such substance is a protein? Justify"

Is this correct to say that the substance is a protein?
Is there a chance that the test reacts but the substance is NOT a protein?

Additionally, I am being asked why we are using whole milk instead of skim milk in this test.

Biuret test react with a peptide bond, so nylon can react with.( https://en.wikipedia.org/wiki/Biuret_test, https://twitter.com/jane_essex/status/921359485939134465)


Separating gold and silver from gold foil using hydrochloric acid and hydrogen peroxide I used  $\ce{HCl}$ and $\ce{H2O2}$ to recover gold foils by dissolving all other metals. My solution I assume contains a large amount of silver. How do I drop the silver out of the  $\ce{HCl}$  and peroxide solution.)

Hot oxidizing acids, such as nitric acid od sulfuric acid, are the more canonical methods to separate gold from less rare metals.

    My solution I assume contains a large amount of silver.

If so, shouldn't you have noticed a precipitate of silver chloride ($\ce{AgCl}$) under these conditions?


How to calculate number of each type of atom in a residue in PyMOL? I have a 3D model of PDB 4UDC (dexamethasone bound to human glucocorticoid receptor), and I'd like to calculate how many atoms of each element are in the dexamethasone molecule (which has the residue name DEX) in this model, just to make sure it is correct (as I have added hydrogen atoms with h_add and if I missed a double bond it will be off). So how do I determine how many atoms of each element are present in this residue?

The sort of output I'd be looking for, from whatever answer you provide, would be something like:

C22H29FO5

if I got the bond order and everything else right. Just so you know, as I would dislike it if you thought I meant if we were to count all the atoms up altogether (which would be $22+29+1+5=57$).

Not super elegant, but this should work:

select resn dex and symbol C

Pymol will respond,

Selector: selection "sele" defined with 22 atoms.

Repeat for H, F, O.


How should I modify my molecular dynamics simulation so that known chemical reactions can be modelled? I have written some molecular dynamics code that I want to use to model known chemical reactions.

By known I mean that it has been observed in a lab that $\ce{A +B->C}$.

I am not interested in finding out if it is feasible that two things react (i.e. is there a favorable energy configuration?), rather I am interested in finding out what the concentration over time looks like given a volume, concentration, temperature and a pair of chemicals that I know will react.

At the moment I just check if they are very close to each other. This produces reasonable looking concentration curves.

Is this is a correct way of doing things? I am looking for publications so it has to be defend-able, not just OK.

I am also interested in modelling large bio-molecules in a coarse grained simulation.

Are there any particular potential functions that I should use for large biomolecules, or will LJ do?

Since you've written your own code, you will probably be able to judge this best, but I'll give you some pointers as to where I'd start.

As far as I've gathered, you simulate a NVT ensemble of a pair of chemicals.

First of all you have to decide whether you want explicit representation of the solvent or not. Hint: The latter is computationally far more expensive and I doubt that you have implemented all the optimizations that you'd find in larger MD software.

This is because this choice can drastically influence the diffusion properties of your chemicals. In a coarse-grained force field they might still be able to diffuse around pretty much unhindered, while when using explicit atomistic representation of the solvent, the properties may change to a larger degree.

The way I'd do it would be to check whether the two areas of the molecules that react with each other are close to one another, meaning that some distance criteria $(d<d_\text{rxn.})$ has to apply. When this criteria is met, you can make the reaction happen by creating new bonds and changing the atom types that are bonded on-the-fly. It wouldn't be so tricky to implement this part, I guess. Caveat: This will only work if you have a reaction whose mechanism is close to the proposed scheme. You might introduce nontrivial errors otherwise.

Then you can model the "real" concentration curves of all the substances, and simultaneously change your system to reflect the change in number of molecules. I wouldn't be surprised that you'd find that the pressure would slowly decrease as the reaction takes place (since you keep the volume constant).


What are the metamers of penta-1,2-diene? First of all, are there any metamers for dienes? If there are any, how can I find them?

Yes. Consider metamer to be a different word for isomer, as noted by Lighthart:

  The term metamer is equivalent to isomer...As an aside, I have never heard the term metamer before, I expect it is archaic.

And, according to Dictionary.com, consider a diene to be:

  a hydrocarbon that contains two carbon-to-carbon double bonds in its molecules.

There are too many diene isomers to be listed here ($\ce{C5H8}$ has many atoms), but you can begin by changing the numbers in the name to generate different Lewis structures.

Penta-1,2-diene:

Penta-1,4-diene:

This is just one structure. Try cis vs. trans, position isomers, etc. You may think of many more.


What does "electrochemically active species" mean in the expression for the mass-transfer limited current? The expression for the mass-transfer limited current $i$ is given by
$i = nFAmC$.

where $C~[\mathrm{mol\, cm^{-3}}$] represents the concentration of the electrochemically active species that gets converted at the electrode; if the entire analyte gets converted, then $C$ is equal to the initial bulk concentration of the analyte.

I am confused as to what electrochemically active species mean? Is that just saying the ions in solution? Or is it the oxidized ion and the reduced ion?

It is the species that can get oxidized or reduced in the system.

It is specifically written out in order to make sure that it doesn't get confused with supporting electrolyte or any other component in the system.

So, for a reduction reaction, it is the concentration of the oxidized species.


What's the difference between Carvacrol and Thymol? What is the difference when it comes to separating Carvacrol and Thymol on HPPLC with reverse phase - C18 column?

Carvacrol elutes first, is that because it is ortho?

On a reverse phase C18 column  (which itself is hydrophobic, it is an octadecyl hydrocarbon chain bound to the support material) the more hydrophobic the molecule, the more strongly it will bind to the stationary phase [like dissolves (or in this case binds to) like], and the longer it will take to elute with an organic solvent. Below are the structures of carvacrol and thymol.  The more exposed the hydroxyl group, the less hydrophobic the molecule.  The hydroxyl group in carvacrol (ortho to a methyl group) is more exposed than the hydroxyl group in thymol (ortho to an isopropyl group) because of the relative steric sizes of a methyl group vs. an isopropyl group. Therefor thymol is more hydrophobic than carvacrol.  Consequently we would expect thymol to bind to the column more strongly than carvacrol, and therfor we would expect the carvacrol to elute first which matches your observation.


Does the Dieckmann Reaction proceed via the conjugated enolate of the product? In the following mechanism of the Dieckmann Condensation, the product has been formed in the fourth step, but still the reaction proceeds and $ \ce{RO^{-}}$deprotonates the product. Then $\ce{H^{+}}$ rejoins to form the same product. Won't this result in extra expenditure of energy? Also, if the reaction proceeds through this mechanism, shouldn't there be an infinite loop?

Can anyone explain why is there an extra step?

The first four steps (i.e. the main reaction) are conducted under basic conditions due to the presence of alkoxide. When that reaction is complete, the result is the deprotonated ketoester--the formation of that stable species is what drives the reaction to completion. Note that the first 3 steps are shown as equilibria, whereas the fourth step is essentially irreversible.

Later, during workup (shown above as the fifth step of the process), acid is added to neutralize that anion and form the final product.


What is the difference between the structure of nitrito-o and nitrito-n? I am preparing for chemistry olympiad and I am stuck nitrito-o and nitrito-n. Can anyone explain difference between the structure of nitrito-o and nitrito-n or point a resource which is suitable for a beginner?

The last letter of nitrite-o and nitrite-n depicts the atom which is attached to the other atom. This phenomenon is called linkage isomerism and is an important concept in both organic and inorganic chemistry.

inorganic chemistry

This complex exhibits linkage isomerism which results in two possible ways of attachment of ligand to the central atom. First one, the $\ce{NO2}$ ligates with the central atom through the nitrogen atom and thus is called pentaammine nitrito-N-cobalt (III) chloride and the second one,  $\ce{NO2}$ ligates with the central atom through the oxygen atom and thus is called pentaammine nitrito-O-cobalt (III) chloride. In this case, $\ce{NO2}$ is called ambident ligant and is represented as nitrito-κ-N and nitrito-κ-O, replacing the old system of trivial names such as nitro and nitroso.

organic chemistry

Alkyl nitrite forms two isomers:- $\ce{RNO2}$ (called alkyl nitrite-N) and $\ce{R-O-N=O}$ (called alkyl nitrite-O). [For preparation of these two isomers, see this answer].


How to convert benzene to N,N-dimethylbenzamide? This is my solution to convert benzene to N,N-dimethylbenzamide. If you think is the right way, even just saying the equations are correct would help me! Thanks in advance.

The scheme you have written will satisfy your teacher in my opinion but it can be shortened. The suggestion of @Eli Jones of the V-H formylation to give benzaldehyde is a good one. However, it can be done in 2 steps using reasonably well-known chemistry:

Step 1 - Brominate with $\ce{FeBr3/Br2}$ to give bromobenzene

Step 2 - Either lithiate by Li-Halogen exchange using $\ce{n-BuLi}$, or form the Grignard (directly with Mg or by Knochel exchange with $\ce{i-PrMgBr}$) then react the resulting benzene organometallic with dimethylcarbamoyl chloride ($\ce{(Me)2NCOCl}$) to give the benzamide.

There are a couple of ways of doing this in one step from benzene by the use of biscarbamoyl diselenides/Lewis acids$\ce{^{[1]}}$ and by $\ce{(Me)3SiOTf}$ activated carbamoyl chloride$\ce{^{[2]}}$.

References

Biscarbamoyl diselenides as new carbamoylating reagents. Lewis acid promoted carbamoylation of aromatic compounds by Shin-Ichi Fujiwara, Akiya Ogawa, Nobuaki Kambe, Ilhyong Ryu, Noboru Sonoda, Volume 29, Issue 47, 1988, Pages 6121-6124, DOI: https://doi.org/10.1016/S0040-4039(00)82282-0
Jin-Wei Yuan, Qian Chen, Chuang Li, Jun-Liang Zhu, Liang-Ru Yang, Shou-Ren Zhang, Pu Mao, Yong-Mei Xiao and Ling-Bo Qu, Silver-catalyzed direct C–H oxidative carbamoylation of quinolines with oxamic acids, Organic & Biomolecular Chemistry, 10.1039/D0OB00358A, (2020).


Reduction of N,N-dimethylbenzamide with LiAlH4 Consider the reaction $$\ce{C6H5CON(CH3)2->[\ce{LiAlH4}]$\:\:$?}$$

What will the product be?

According to me the product should be $\ce{C6H5CH2OH}$ but I am not sure. Also, I don't have any reasoning for why this must be the product.

Here's the mechanism. The key is the second step, it's the lone pair of nitrogen that attacks the carbon, and not the lone pair of oxygen.

The reasoning is that the lone pair of nitrogen is more reactive because of the two electron-donating methyl groups attached to the nitrogen atom. The electron density is higher on the nitrogen atom than the oxygen atom.

The second reason is that the oxygen anion as a leaving group is much more stable than the nitrogen anion R2N- you would have gotten if the reaction led to your product.


What is the composition of the oil-like fluid that seeps out of dry-cells over time? An experience a lot of us might've had; a dry cell (not completely depleted) when left in an electrical device for long periods of time, has this viscous fluid coming out that usually corrodes the terminals of the battery.

I almost always find this fluid, present as a sorta thin film over the terminals and the casing of the battery.

I'm told that the Ammonium Chloride in the cell, being slightly acidic reacts with the zinc casing of the cell, which is why dry cells don't last long (My teacher mentioned this during our Electrochem. class, he didn't mention anything about that fluid I was talking about though). Could this fluid possibly the result of that reaction? What are its chemical constituents?

There are various kinds of dry cells.I will give some examples as to what is that oil-like fluid:

In zinc-carbon cell:

  Zinc carbon cells have a short shelf life as the zinc is attacked by
  ammonium chloride. The zinc container becomes thinner as the cell is
  used, because zinc metal is oxidized to zinc ions. When the zinc case
  thins enough, zinc chloride begins to leak out of the battery. The old
  dry cell is not leak proof and becomes very sticky as the paste leaks
  through the holes in the zinc case. The zinc casing in the dry cell
  gets thinner even when the cell is not being used, because the
  ammonium chloride inside the battery reacts with the zinc. An
  "inside-out" form with a carbon cup and zinc vanes on the interior,
  while more leak resistant, has not been made since the 1960s.

In alkaline cells:

  Technically, that white fluffy corrosion that develops at the ends of
  the battery (most often the negative end) is called potassium
  carbonate.

  All batteries will slowly gradually self-discharge over time. This
  will occur whether they are setting on the shelf (a much slower
  process) or installed in a device (which often occurs much quicker) –
  and dead batteries will eventually leak.

  Very high temperatures can also cause batteries to rupture and leak
  (e.g. in hot car during the summer).

  The “alkaline” of the battery is potassium hydroxide (it’s the alkali
  equivalent of acid’s hydrochloric acid), and this will leak out,
  forming a white “fluff” of potassium carbonate, typically on the
  negative end of the battery cell – because apparently the positive end
  is vented better.

So the reason for the cell-corrosion varies with the type of the cell.

Read the Wikipedia page to know about the composition of electrolytes of different dry cells.https://en.wikipedia.org/wiki/Dry_cell


How does the cyclisation of the 3,6-dimethylocta-2,6-dien-4-yne occur? The given molecule is treated with $\ce{H+}$ and an acid, that is all. I know that the alkene part reacts with the $\ce{H+}$ first, but have no idea on how to proceed further. I don't see how a cyclisation takes place as the final answer given is 3,4,5,6-tetramethyl-3-cyclohexen-1-one. Any help on how this happens?

First add water to the triple bond a ketone would be formed as the molecule is symmetrical hydroxy group can be attached yo any of the two carbons after that it would undergo tautomerism to form a ketone. Now the second step cyclisation occurs as follows:


Benzylation and acetylation of amines and alcohols Question

Attempt
Since the nitrogen site is more nucleophilic it should react first. But still Oxygen can react and hence I believe A should be right. But the book says C. Where did I go wrong? Please help.

Phenol is a pretty good leaving group so phenyl acetate is an acetylating agent and fairly labile in aqueous base. The reaction conditions are not clear enough IMO to say whether you would get a or c, but with a basic work-up the OAc can hydrolyse back to phenol. If the amount of Ac2O is limited then any PhOAc formed will acetylate the aniline giving back PhOH.


Does CH3COOH + H2O (vinegar) lose its antibacterial and antiviral properties when exposed to air? I have been reading about using CH3COOH + H2O (vinegar) as a mild antibacterial and antiviral agent.  For example: https://pubmed.ncbi.nlm.nih.gov/15698693/
Note, before anyone gets confused (or concerned), this has nothing to do with the SARS-CoV2 virus which causes COVID-19.  I strongly doubt vinegar is effective at killing a coronavirus, and everything I have read confirms that doubt (the pH is too high, for starters).
Now, back to the topic at hand.  Does CH3COOH + H2O lose its antibacterial and antiviral properties when exposed to room temperature air for a long duration (over 8 hours, but less than 500 hours)?
I've been trying to find scientific articles regarding this, and so far, have not found anything helpful.
Ideally, I would love to see one or more graphs that plot effectiveness along one axis and time duration of air exposure along the other.  Of course, tables with data points would be equally welcome.
Note that I'm not looking for anything exact, as there are too many variables to be precise without getting into minute details.  I'm looking for a general idea of the changes (if any) over time as CH3COOH + H2O is exposed to room temperature air.

Acetic acid boils at higher temperature than water, at 118 °C. Acetic acid does not react with the air.
It is more likely that acetic acid antimicrobial properties will improve as more water evaporates, because it evaporates faster.


How are alkanes, alkenes, and alkynes homologous series since they have no functional group attached? I understand that a homologous series is a group of compounds with the same functional group attached to it.

I also know of an alternate definition of it as being a series of compounds with the same general formula.

But if the first definition is to be considered then how can alkanes, alkenes or alkynes be considered homologous series since there is no functional group attached to any of these, as opposed to, e.g., the homologous series of alcohols?

For the linear alkanes, the internal methylene groups $\left(-\ce{CH2}-\right)$ are considered to be the functional group defining the homologous series:

$$
\ce{H3C-\underline{(\ce{CH2})_n}-CH3}~,~~n\geq 0 \tag{n-alkanes}
$$

Alternatively, this series is defined by the absence of any particular functional group other than methyl and methylene.

Ethane, $\ce{C2H6}$ is the member of the series for $n=0$. Technically, one could exclude methane, $\ce{CH4}$, from the series, since it doesn't have two capping methyl groups. However, to my knowledge it's usually considered part of the series as a special case.

For the 1-alkenes and 1-alkynes, the double and triple bonded carbons are the functional groups defining the series:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH2}}} ~,~~n\geq 0 \tag{1-alkenes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C#CH}}} ~,~~n\geq 0 \tag{1-alkynes}
$$

Parallel to the alkanes, the shortest "official" members of the two series are technically propene $\left(\ce{H3C-CH=CH2}\right)$ and propyne $\left(\ce{H3C-C#CH}\right)$, respectively. As with methane above, though, ethylene/ethene $\left(\ce{H2C=CH2}\right)$ and acetylene/ethyne $\left(\ce{HC#CH}\right)$ are typically grandfathered in.

The case of non-terminal linear alkenes/alkynes is less straightforward. Depending on one's philosophy, they can either be lumped in with the terminal alkene/alkyne series, or they can be considered as distinct homologous series:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH-CH3}}}~,~~n\geq 0 \tag{2-alkenes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C#C-CH2-CH3}}}~,~~n\geq 1 \tag{3-alkynes}
$$

$$
etc.
$$

Note that $n\geq 1$ is strictly required for the 3-alkynes, since $n=0$ would give 2-pentyne $\left(\ce{H3C-C#C-CH2CH3}\right)$, a 2-alkyne. This requirement is general: e.g., for the x-alkenes, $n \geq x-2$ must hold for $x\geq 3$.

Branched alkanes/alkenes/alkynes complicate things even further. There are at least two common alkane series that I can think of off the top of my head:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH(CH3)2}}} ~,~~n\geq 0 \tag{iso-alkanes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C(CH3)3}}} ~,~~n\geq 0 \tag{neo-alkanes}
$$

You can mix and match branching, double bonds and triple bonds in countless ways to make new homologous series. I would personally argue you could define a homologous series to contain multiple variable-length segments:

$$
\ce{\underline{\ce{(H3C)2C=C}}-(CH2)_n-\underline{\ce{C#C}}-(CH2)_m-\underline{\ce{CH(CH3)}}-(CH2)_r-\underline{\ce{CH=CH2}}} \tag{?}
$$

Of course, the practical utility of any correspondence of properties among the various species in such a convoluted homologous series may be ... limited.


ATP break down and energy release? I have a very basic question. Energy is absorbed to break bonds. Bond-breaking is an endothermic process. Energy is released when new bonds form. Bond-making is an exothermic process.

But ATP breaks down to ADP and releases energy right? Is there anything wrong in my understanding?

A bond is formed between the oxygen of water and the phosphorus of the gamma-phosphate.  Here is a good link.

Bonds are both broken and made in chemical reactions but many biology teachers and textbooks state that "Breaking ATP bonds releases energy."  In reactions bonds are broken and made.  If the strength of the bonds formed exceeds the strength of the bonds broken the reaction is exothermic. Under physiological conditions (with respect to pH and magnesium concentration), the reaction is exergonic as well (i.e. negative Gibbs free energy of reaction).


Non-uniform sampling (NUS) in 2D-NMR The NMR technicians where I work have recently started replacing the standard 2D NMR experiments (COSY, HSQC) with NUS variants (25% sampling). For the uninitiated, the rationale behind the technique is summarised by Claridge (who does a better job of explaining than I would):

  The classical sampling of 2D (and more generally multi-dimensional) NMR experiments requires the uniform sampling of data in the indirect dimension(s) that allows for the processing of the data by the discrete Fourier transform. This means a sequential, stepwise increment of the t1 period of a 2D data set is made to the limit t1 which dictates the resolution in this dimension. The number of such t1 increments employed ultimately defines the total duration of the experiment. The method of non-uniform sampling (NUS) seeks to reduce the number of data points collected in the indirect dimension(s) and so reduces the total experiment time.

  Claridge, T.D.W. High-Resolution NMR Techniques
  in Organic Chemistry, 2016

My experience of these experiments has been overwhelmingly positive so far: shorter experiment times and greater resolution, even when acquiring spectra of small quantities (0.1 - 0.5 mg) of large molecules (ca. 1000 Da).

The challenge is knowing what to do when the spectrum doesn't come out looking quite right. With 'traditional' experiments, running more scans often fixes a multitude of sins (up to a point), but with random point sampling I can't see this holding up so well.

The processing of data is also causing some confusion - TopSpin and MNova give (often very) different results, despite both having NUS processing capability. This processing issue leads me to suspect that many of the issues I'm having with spectra from NUS pulse sequences are actually due to the processing (which isn't terribly well covered in the TopSpin manual).

My question, therefore, is what considerations should be taken into account with NUS spectra: are there instances where its better to take the hit and run a long HSQC? What do I need to be careful of when processing data to ensure spectra come out looking okay?

I think a really full answer will require delving into the theory of NUS, which (just like the Fourier transform) goes way beyond NMR as a field: the idea of undersampling a signal and then reconstructing it is of great interest in a variety of engineering areas. I can't give that level of detail, so I don't pretend that this is really an answer, but maybe having some of this written down will help.
I'm really not an expert in this, but I can backup Mad Scientist's comment that for 2D data, TopSpin's reconstruction mode (the MDD_mod processing parameter) generally should be set to cs and not mdd. I suspect this may require a license from Bruker, so it might be best to check with the relevant NMR manager. Mnova, on the other hand, uses a different reconstruction algorithm entirely, so it's not overly surprising that there might be some differences in performance.
The amount of undersampling you want to do will also affect the quality of your results. I think a good rule of thumb is to make sure the ratio of points measured in $t_1$ to number of peaks in each column of the spectrum is somewhere around 2 to 5. The reasoning is simple: if you need to detect $k$ peaks in the frequency domain, then you need to measure some multiple of $k$ signals in the time domain. In other words, the more data you want to pick up, the more signals you need to measure. For a HSQC where each column has very few frequencies, you can get away with very sparse sampling, and 25% is likely to be fine (as long as your TD1 is not too small). For other spectra like COSY, TOCSY, HMBC where there is less empty space in the spectrum, this may not work quite as well.
As far as I'm aware, increasing the number of scans should have a similar effect for both uniform and non-uniform sampling. The main changes are the increase in S/N, as well as possibly better artefact suppression achieved via phase cycling. However, with modern gradient-selected experiments (especially the standard experiments like HSQC, COSY, etc.) you should not really need extremely long phase cycles to achieve good artefact suppression. In my experience the difference in spectral quality between NS=2 and greater is already pretty minimal. Bumping up the number of scans is mainly useful for increasing S/N. [Obviously, if the reason why the NUS reconstruction is failing is because of too poor S/N, then increasing the number of scans will help.]
Lastly, this is a trivial point, but I'd also suggest double-checking the other processing parameters (window functions, zero-filling etc.) after the reconstruction is done. It's quite easy to miss something out (for example, changing QSINE to SINE) and inadvertently get ugly peak shapes. I learnt (and am still learning) it the hard way.
Having said all this, what should you do if you find your NUS data is not good enough? I don't think there are really many answers to this, although someone more knowledgeable will be able to make better suggestions. One is to play around with the reconstruction algorithm, but unless you are very well-versed with the theory, this is essentially limited to trial-and-error with a dropdown box. The other is to just reacquire the spectrum with less undersampling, especially if you suspect you may have pushed it a bit too much.


How does the activity of Sn chemical potential vary as a function of composition at 900 C? This is referring to the Cu-Sn phase diagram in which composition varies along a horizontal line at a certain temperature. I understand that as you move along that line different intermetallic or liquid phases form, but from a activity point of view does the activity go up because of different crystallographic arrangements that are present at different compositions?

Just above the line, in the two-phase field where $\epsilon$ is in equilibrium with the liquid, the chemical potentials of both Cu and Sn are constant (very likely not equal to each other mind you).  This is a consequence of the tie-line on the Gibbs free energy diagram - check out a book on binary alloy thermodynamics such as Porter and Easterling, Phase Transformations in Metals and Alloys.

At the peritectic point itself, the tie line connects all three phases, $\epsilon$, $\eta$ and the liquid, so the chemical potentials for each element are equal. Dropping the temperature just a smidge leads to two separate two-phase regions, one of $\epsilon$ with $\eta$, and one with $\eta$ and liquid.  The chemical potential of Sn (or Cu) is different in each of those two-phase regions.

Again, Porter and Easterling would be a good reference.


What indicators can be used when substituting iodine with other halogens in the Winkler method? I'm a chemistry student who is doing an investigation with respect to the Winkler method. The Winkler method is one which is used to find the oxygen content in water. Primarily, the halide which is used is iodine. However, I want to make use of other halogens such as fluorine, chlorine and bromine, together with their respective halides. However, I am stuck on how I will actually do the titration, since the indicator used for the case of iodine is a starch indicator. I don't exactly know which indicator I should use for the other halides.

My question is this: What indicators should I use in the titrations involving the other halides?

It will likely not work with other halides. As far as I understood the winkle method, the oxygen is first fixated with $\ce{Mn^2+}$ in basic conditions:
$$\begin{align}
\ce{Mn^2+ (aq) + 2 OH- (aq)&-> Mn(OH)2 (aq)}\tag{1}\label{alkalinemanganese}\\
\ce{2Mn(OH)2 (aq) + O2 (aq) &-> 2 MnO(OH)2 (s) v}\tag{2}\label{bindingoxygen}\\
\end{align}$$

The sample is acidified and iodide is added. This is oxidised to $\ce{I2}$, which forms a complex with remaining iodide.
$$\begin{align}
\ce{MnO(OH)2 (s) + H+ (aq) &-> Mn(OH)3 (aq)}\tag{3}\label{acidicmanganese}\\
\ce{2 Mn(OH)3 (aq) + 2 I- (aq) + 6 H+ (aq)
&-> 2 Mn^2+ (aq) + I2 (aq) + 6 H2O}\tag4\label{iodide-ox}\\
\ce{I2 (aq) + I- (aq) &<=> I3- (aq)}\tag5\label{triiodide}\\
\end{align}$$

The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions.
$$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$

The reason why this works is because you have the equilibrium $\eqref{triiodide}$ which produces the  triiodide ion, which is relatively stable in aqueous solution. Iod by itself does not dissolve in water (much). This complex again forms a very colourful complex with starch, the switch can easily be detected by the human eye.

Polyatomic ions of other halogens are not unheard of (ref. Trichloride ion (c.f. triiodide ion) and therein), but they are a lot less stable. I am not aware that they form a complex with starch so they won't be that easily detected.

Another aspect, I don't want to go much deeper here, is that the redox reactions necessary $(\ref{iodide-ox},\ref{triiodide-reduction})$ for this procedure to work might not work at all with other halides.

The original method was proposed in the 19th century and is still applied today. It is a quite accurate and robust method. There is not really a reason to tamper with it.


Ignoring initial H+ Concentration For Addition of Strong Base to a Buffer Solution When adding a strong base to a buffer solution, with a generic weak acid HA and generic weak conjugate base A-, is the initial amount of H+ ignored because it is so small? Specifically, at least in my text book, we only consider what happens to HA and A- when a strong base, say NaOH, is added.

Suppose you have a weak acid and its corresponding base in solution. Suppose the solution has the pH = 5.

HA + H2O = A- + H3O+

Now we add a strong base to the solution. What will happen?

The pH will increase, suppose it will increase to pH = 6.

The weak acid-base system will immediately respond to the pH change of the solution and a new equlibrium will be established defined by the new pH and the equilibrium constant of the weak acid-base system. The equilibrium will be forced to the right, i.e. more A- will be fomed at pH = 6 compared to what we had at pH = 5.

Ok, what has happened to the hydrogen ions present at pH = 5 or those that were formed up to

pH = 6?

They have reacted with the strong base forming water.

Do we think the concentration of the hydroxide ions from the strong base has decreased at pH = 6?

No.

Why?

At pH = 6 the hydrogen ion concentration is lower than at pH = 5. This is equal to say that the hydroxide ion concentration is higher at pH = 6 than at pH = 5. The difference in the amount of hydroxide ions present at pH 6  and 5 will be equal to the amount of hydroxide ions consumed in the transfer from pH 5 to 6. What you put in, you get back in this case. The net effect is that the hydroxide ion concentration of the strong base has not changed at the equilibrium of the process.

I suggest you also look at my answer for another question "Buffer and strong acid".


Change in H+ ion concentration of water on boiling I know that on boiling water, pH scale decreases but water remains neutral. At 100°C the pH=6.14 is neutral.  And at this temperature also, the H+ ion concentration and OH- ion concentration are equal.
My question is what happens to the  H+ ion concentration on boiling  relative to the concentration at 25°C? Does it increase or decrease or remain same?

It is an interesting question, however there is a misconception. You are giving too much attention to the word neutral. The word neutral is just a qualitative version. Numbers and equations are more accurate way to discuss science.

You have to see how water's $\pu{pH}$ at 100 °C is calculated. How do you get those numbers? Lets us do it algebraically, so that the word neutral does not matter.

Autoionization of water at temperature T:$\ce{ H2O <=> H+  + OH-}$, if the conc. of $\ce{H+}$ is x, the quantity of $\ce{OH-}$ is also x, if you have accepted the ionic product of water is

$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = x^2=
51.3\times{10^{-14}}$

Solve for $x = [\ce{H+}] => -log [\ce{H+}] = 6.14$

So the numbers prove that $\pu{pH}$ has changed. Water has become slightly acidic. If you associate the word neutral = 7.0 at 25 °C. However, if you use the word neutral with respect to water temperature, then $\pu{pH}$ of 6.14 is neutral at 100 oC. Ultimately, the numbers Kw and pH matter, nobody will quibble about the word neutral in serious research.

Water is the most amazing solvent in Nature. People are still exploring its properties and it is yet to be fully understood. For example, I just heard in a conference that someone was trying to understand what happens on the surface of micron sized water droplets. "Pure" water showed very odd chemical effects which have never been shown before.

Read this well written Wikipedia article: https://en.wikipedia.org/wiki/PH#cite_note-2


Where does the 9th electron go in a N=O bond? In the first resonating structure you can see 5 unpaired electrons and 4 shared electrons on nitrogen, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving?
In the second resonating structure you can see that oxygen has 5 unpaired electrons on nitrogen and 4 shared electrons, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving?
Can anyone explain completely about the above questions? If this question is already been asked please comment me the link of that question. Thank you in advance.
Please explain with boxes if you can.

Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a different bonding picture of $\ce{NO}$ to answer your question about which orbital the unpaired electron is in.

  The Lewis Structure of $\ce{NO}$

The mistake you made in drawing those Lewis Structures is that you used all of the electrons in the nitrogen and oxygen atoms and not just the valence or outer-shell electrons.

A Nitrogen atom has 7 total electrons in the following configuration:
$$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$

Of these seven electrons, only the $5$ in the second shell (energy level) are available for bonding. These are the valence electrons. The $1s$ electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.

Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: $$2s^2 2p_x^2 2p_y^1 2p_z^1$$

The bonding in $\ce{NO}$ thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.

$$\begin{aligned}
&.. \ \ \  .. &&\  &&&\ ..\ \ \ ..\ \\
\ce{.} &\ce{N=O:} &&\ce{<->} &&&\ce{:N=O.}
\end{aligned}$$

  Resonance

The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures inter-converting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.

  Where is the unpaired electron?

To answer this question, we need molecular orbital theory. A molecular orbital diagram describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: $\sigma$ or $\pi$, bonding or anti-bonding. Since this answer is running long already, I won't go into how to construct such a diagram.

Here is the MO diagram for nitric oxide:

The unpaired electron (highlighted in blue)is in a $\pi$ anti-bonding orbital that is formed from the destructive overlap of out of phase $p$ orbitals, one from nitrogen and one from oxygen.


Chirality on Propranolol I'm a bit confused when asked to specify the chiral carbons on this sketch of Propranolol as I'm not sure how the $\ce{H3C}$ and $\ce{CH3}$ play in if they are reversed in the name like that...Does that even matter?

I'm pretty sure the only chiral carbon is the one in the middle of those and the $\ce{NH}$. I've heard that you need 4 different groups/atoms surrounding the carbon for it to be chiral but how much truth is there in that, as I've found several with only 3?

As stated in user208322's answer, $\ce{CH3}$ and $\ce{H3C}$ are exactly the same, they're only written differently for ease of reading. As per your question about how much truth there is to having 4 groups, it's 100% truth. You cannot have a chiral carbon that doesn't have 4 groups attached to it (i.e. must be sp3 hybridized). If you think you've found a chiral carbon that only has 3 groups attached to it, you're probably forgetting the unwritten hydrogen.

In this example specifically, there is only one chiral carbon, the one attached to the secondary alcohol. Although it may look like the carbon is attached to the same group twice, you need to look further than just the first atom. Yes it is attached to two carbons, but one of those carbons is attached to a nitrogen and the other is attached to an oxygen, so the groups are different. Two groups are  only the same if you can follow it down every single atom and they match exactly.

Edit: In regards to your question about following the chain all the way down, realistically you only need to follow it until you see that it differs from the other chains. So for chains 1 and 3 once you get to the nitrogen and oxygen atoms you see that they're different and there's no point continuing. And the unwritten hydrogen has to do with the fact that organic chemists use shorthand in their molecular structures. Basically, carbons and hydrogens are extremely common in organic molecules so they just don't write them. All the points are carbons unless otherwise labelled, and they must have 4 bonds, so where you see that the carbon is only bonded to 3 things, it is also bonded to a hydrogen as well. See this wikipedia article for more info (specifically, this section).


Solubility of salt in heavy water vs. "regular" water I would like to know if salt can be dissolved in heavy water ($\ce{D2O}$) as it would be in normal water ($\ce{H2O}$).

If so, what will the maximum salinity be in $\ce{D2O}$? I know the maximum in $\ce{H2O}$ is about 26% of salt by mass.

Well, consider a salt as $\ce{AB_s}$ in water you have :

$$\ce{AB}_s\rightleftharpoons \ce{A_{aq}^{q+} + B_{aq}^{q-}}$$

And $\Delta H_{s}=-\ce{U +\Delta H_{aq}}$ where $\ce{U}$ is the lattice energy.

When the salt needs to be solvate, $\ce{O-H}$-bonds of water molecules around this salt need to be rearranged. More, the "$\ce{DO}$-bonds" in $\ce{D_2O}$ are stronger than "$\ce{HO}$-bonds".$^1$ Then it will be harder to rearranged hence it will cost more energy to solvate the salt which will be then less soluble in heavy water than in water.

I found an article in the JACS and as you can see on the image below the maximum solubility of $\ce{KCl}$ in heavy water is a little less than in water (you can make a linear extrapolation) the difference is not quite importante but it looks it exists.

Look at the second reference $^2$.

$^1$ : https://arxiv.org/ftp/arxiv/papers/0706/0706.1355.pdf, p-5

   Nuclear quantum effects are particularly important in the different
  properties of light (H2O) and heavy water (D2O) where the more restricted atomic vibrations in D2O
  reduce the negative effect of its van der Waals repulsive core so increasing its overall hydrogen bond
  energy.

$^2$ : For the image, http://pubs.acs.org/doi/abs/10.1021/je60070a011


Orthogonal Wavefunctions My current understanding of orthogonal wavefunctions is: two wavefunctions that  are perpendicular to each other and must satisfy the following equation: $$\int\psi_1\, \psi_2\, \mathrm{d}\tau =0$$

From this, it implies that orthogonality is a relationship between 2 wavefunctions and a single wavefunction itself can not be labelled as 'orthogonal'. They must be orthogonal with respect to some other wavefunction. However I have seen some textbooks refer to single wavefunctions as being orthogonal.

Am I missing some crucial thing in my definition of orthogonal wavefunctions? Could anyone provide me with a more complete definition of orthogonal wavefunctions from a chemistry point of view?

The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.

First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if
$$
  \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, ,
$$
where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1

So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$
$$
  \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, .
$$
In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.