Markovnikov's addition to halogenated alkene? The justification or rather intuition behind Markovnikov's rule is the stability of intermediate carbocation. The more substituted the carbocation the more stable the product. But based on this ideas how can we satisfactorily explain the following reaction?
$$\ce{CH2=CHCl + HCl -> CH3-CHCl2}$$

Markovnikov's rule is usually stated in terms of subsitution on the alkene, for example:

  In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms).

This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{HCl}$ end up on the carbon atom that already has two hydrogen atoms. The chlorine atom ends up on the other carbon atom:

$$\ce{CH2=CHCl +} {\bf \ce{H-Cl}} \ce{-> {\bf H}-CH2-CHCl-{\bf Cl} }$$

A better version of the rule has to do with the stability of the carbocation intermediate (and can be generalized to electrophiles other than $\ce{H+}$.

  An alkene reacts with electrophiles in such a way as the electrophile attaches to the carbon atom which will result in the more stable carbocation.

The order of stability for simple carbocations is $3^\circ >2^\circ >1^\circ >\text{methyl}$, however there can always be more complex carbocations (de)stabilized by heteroatoms and/or resonance. The original wording of Markovnikov's rule may fail in these more complex cases.

For example, the addition of $\ce{HX}$ to $\ce{CH3O-CH=C(CH3)2}$. In this case the major product (before it decomposes into the aldehyde) is $\ce{CH3O-CHCl-CH(CH3)2}$. Two possible carbocations form, and the one where the proton attaches at the more substituted position is the one that is more stable.

First carbocation, $2^\circ$, but stabilized by resonance (more stable):

$$\ce{CH3-O-CH+ -CH(CH3)2 <-> CH3-O+ =CH-CH(CH3)2}$$

Second carbocation, $3^\circ$, but not stabilized by resonance (less stable):

$$\ce{CH3-O-CH2-C+(CH3)2}$$

The first carbocation goes on to form the product. Can you extend this to what happens to your molecule?


Does corrosion follow the law of conservation of mass? For e.g .Due to rusting , iron becomes weak .So it should become weak only if the amount of iron is reduced. Even if we say the reduced iron is now the rust ,the amount of rust practically is very less comparing to iron.

The answer to the question in the title is yes.

Iron oxide flakes away from the iron so just looking at the iron would make you think that the mass has disappeared; it's actually on the surface below, unless someone has moved it.


What is the pKa of methylamine I completed a practice (not for marks) assignment for an organic chemistry course, and one of the questions asked for the pKa of methylamine.

I looked on PubChem, and it gave a result of $10.6$. Marking myself, I found that the answer key states the $\mathrm{p}K_\mathrm{a}$ of methylamine is $33$! Especially on a log scale, that is not a trivial difference.

I talked to my professor about it; he was confident the true $\mathrm{p}K_\mathrm{a}$ is in the low thirties and stated that I need to be skeptical of material online. His point in part was that there was no way methylamine was more acidic than water.

I am not personally convinced on either number yet.

What is the $\mathrm{p}K_\mathrm{a}$ of methylamine, and what evidence supports that number?

Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.

$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$

The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.

Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.


How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2 The relative leaving group ability of iodide is better than that of bromide but steric crowding is more important as the neucleophile will attack the back lobe of carbon atom.
So how should I order the steric hinderance produced by t-butyl and neopentyl groups?

The order will be $\text{neopentyl bromide} > \text{t-butyl bromide}$. This is just an addition to @user377386's answer. From Solomons and Fryhle's Organic Chemistry, the relative rates of reaction (SN2) of different kinds of halide are (for the same halogen):
$$ \begin{array}{c|c}
 \textbf{Type of halide} & \textbf{Relative Rate} \\ \hline
 \text{Primary Halide} & 1 \\
 \text{Secondary Halide} & 0.03 \\
 \text{Neopentyl Halide} & 0.00001 \\
 \text{Tertiary Halide} & \approx 0 \\
\end{array} $$

You can easily see that neopentyl halides and tertiary halides are almost unreactive towards SN2 reaction, so it makes little sense to compare them. Neither of them is going to react by SN2 ever, they will instead react through SN1 mechanism. The slightly increased leaving group ability of iodine does not matter much. If you insist on comparing though, neopentyl halide will be better at SN2.


Constant concentrations in solution I'm a high school student considering the concentration of solutions, and whether the concentration is constant at the bottom or towards the surface. Even then, how would I know that the solution is consistent and not more concentrated in places compared to others? Is there any visual way or method for identifying non-consistent concentrations in solution i.e. Copper (II) sulphate, Iron (II) sulphate

A solution by definition is a homogeneous mixture, so its "concentration" is same throughout the bulk. Let us have a look as to how one would prepare a solution of a given concentration, say in 1 L. We would weigh out a known weight of a solute and dissolve it  ~ 500 mL of solvent, with mechanical stirring or shaking. We would add more solvent until the total volume is 1 L. We would stir it sufficiently, again, until we believe that the stirring has made the solution uniform.

a) If it a colorless solid, like NaCl, there is no visual way to tell if the mixing is complete and the solution is uniform. You have to stir it very well.

b) A colored solid, like copper sulfate or iron sulfate is more or less visible if it is not homogenous. Different colors in the bulk.

If you are still not (mentally) satisfied, you can take portions of the solution from surface, middle and bottom. Measure their refractive index and they should be identical. If they are not not, the mixture needs more stirring.

Edits:
Some people may wish to differentiate microscopic vs. macroscopic homogeneity regarding the definition of solutions. IUPAC does not offer such a distinction. For an analytical chemist or any researcher interested in measuring or making known concentrations of solutions- a solution has to have identical concentration throughout the bulk. Ultimately, it will boil down to semantics or the problem of "how many angels can land on the head of the pin" - a debate which has no end! en.wikipedia.org/wiki/…


Why does a substance with an endothermic heat of solution dissolve? How does a substance with an endothermic heat of solution dissolve?

Dissolution happens in three steps.

Solute-solute attractions must be broken (consumes energy, endotherm), solvent-solvent attractions must be broken (also endotherm), and finally solute-solvent attractions form. This results in a lower energy state and is exotherm.

Dissolution will be endotherm if it takes more energy to break the mentioned attractions, than is released in the last step. You are asking then, why is this reaction spontaneous? In other words, why does it occur? To find out we need chemical thermodynamics. $$\Delta G=\Delta H - T\Delta S$$

$\Delta G$ = Change in Gibb's free energy for a reaction. $\Delta G<0$ = reaction is spontaneous. Otherwise it is NOT spontaneous (ie. will not occur on its own).

$\Delta H$ = Change in enthalpy. $\Delta H <0$ = reaction is exotherm. If larger than 0, the reaction is endotherm. Already now we can see that an endotherm reaction will make $\Delta G$ for the reaction more positive, and thus the reaction is less likely to be spontaneous.

$T$ = temperature in K.

$\Delta S$ = Change in entropy. $\Delta S > 0$ = reaction leads to a state of higher entropy. Entropy describes disorder in a system. Dissolution of a solute in a solvent will always lead to a state of higher disorder, since we go from having all the solute concentrated in a cluster (more ordered) to being spread evenly throughout the solution (more disordered). We can see from the equation that if the reaction leads to higher entropy, then higher temperatures increase likeliness of $\Delta G < 0$, and the reaction will be spontaneous. On the other hand, if we're going to a more ordered state, lower temperatures increase likelihood for a spontaneous reaction.

So how do we apply this to dissolution?
You know that the dissolution reaction is endotherm. That means that $\Delta H > 0$ for this particular reaction. Yet we know that the reaction is spontaneous, since you report that it does indeed occur! This means $\Delta G < 0$. The only way to make that happen is for $\Delta S$ to be positive. So the reaction must go to a more disordered state. Since it's a dissolution, we know that it does. The solute goes from being concentrated in a cluster (ordered) to being spread throughout the solvent (disordered). Pay attention to where temperature is in the equation. It is quite easy to see, that for a dissolution reaction where  $\Delta S > 0$, you can increase the chance of the reaction being spontaneous by increasing temperature!

Bottom line
Your dissolution reaction is endotherm, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless! If it wasn't, heating the system could make it so.


Are there some similarities between J-coupling and j,j-coupling? I'm a bit confused having heard two different terms in different lecture parts:

J-coupling

There is the J-coupling (= indirect dipole dipole coupling = scalar coupling) which arises from a hyperfine interaction between the nuclei and the interaction. The J-coupling is a coupling between the two nuclear spins of two nuclei (e.g. a C-H bond). Because of the Pauli principle the two binding electrons are anti-parallel located so the nuclear spins $I_1$ and $I_2$ also have an anti-parallel nuclear spin.

j,j-coupling

There is as well the jj-coupling (= spin orbit coupling) which describes the coupling of the different $j_i$. This coupling scheme is valid for heavier atoms. (For leight atoms the ls-coupling is valid). In atoms with bigger nuclear charges, spin-orbit interactions are frequently as large as or larger than spin-spin interactions or orbit-orbit interactions.
$\mathbf J = \sum_i \mathbf j_i = \sum_i (\boldsymbol{\ell}_i + \mathbf{s}_i).$ Wikipedia

I don't see any commonalities. Why is the term "J-coupling" similar to the term "j,j-coupling"?

I don't believe there are any commonalities, and the terms have been derived independently.

The term J coupling in NMR was first proposed in 1951 by Erwin Hahn in his paper  describing the spin echo experiment, in which two frequencies are observed; one at the chemical shift (which they called delta, as was convention, even back then) as expected, and a second frequency which they labelled as frequency J. There is no explanation for why the choice of J is given. Gutowsky and McCall  who observed the same phenomenon independently did not give it a label, and the term J splitting stuck. J coupling describes the relationship between nuclear spin, the intrinsic angular momentum of atomic nuclei.

The term jj coupling relates orbital angular momentum of electrons. I'm not going to pretend to know anything about the origins of the term, nor anything about its application in physics and astronomy. However, the term J is sometimes used to represent total angular momentum, which is made up of individual components ji. It is the interaction between these individual j terms that give rise to jj coupling.

Other than being types of interactions between entities of angular momentum, the two couplings are quite unrelated, as are the terms used to describe them.


Reason for the formation of azeotropes Why is it that some liquid mixtures (that exhibit positive/negative deviation from Raoult's law) form azeotropic mixtures at certain compositions? What is the physical reason behind this; are there any extra forces holding the two components together? If so, please elaborate on this.

Before getting to the azeotropes, it's worth reminding ourselves of the relevant assumption that we make about ideal mixtures: that the intermolecular interactions have constant energy regardless of the mixture.

Given the important role that intermolecular interactions have in determining the boiling point of a pure liquid, it's surprising that so many mixtures behave approximately ideally, in line with this assumption. There are two main reasons for this.

First, liquids are only miscible if they have somewhat similar types of intermolecular interactions with similar magnitudes of energy. For example, water and hexane have very different intermolecular interactions, so surrounding water with hexane or hexane with water is quite unfavorable, and the two do not mix. Thus, any actual mixture has interactions between the two types of molecules that are not hugely different from the interactions within the pure liquids.

Second, substances that are liquid are room temperature generally do not have especially strong intermolecular forces, so the small differences in the intermolecular interactions of two miscible liquids is usually dwarfed by the impact of the difference in molecular weight that is a major determinant of boiling point. Water is quite unusual in having such a low molecular weight and still being a liquid at room temperature, a symptom of the exceptionally strong role of intermolecular interactions in determining its boiling point.

With that in mind, we can talk about azeotropes. To get a minimum boiling azeotrope, the deviation from Raoult's Law requires that the vapor pressure of the lower boiling component be reduced from what is predicted for ideal behavior and that the vapor pressure of the higher boiling component be increased. In order for that to happen, we need two liquids with substantially different intermolecular interactions whose mixing changes those interactions in opposite directions.

In the example of the water:ethanol azeotrope, the water is able to increase the amount of hydrogen bonding between ethanol molecules by increasing the density of hydrogen bond donors and acceptors per unit volume. At a low water concentration, the water also does not interfere greatly with the hydrophobic interactions between the ethyl ends of the molecules. As a result, the average intermolecular interaction for the ethanol molecules is stronger than in pure ethanol, and we have the necessary decrease in the vapor pressure of ethanol. For the water, when it is at low concentration in mixture, each water molecule is surrounded by ethanol molecules, so it has fewer hydrogen bonding interactions than in pure water. As a result, its average intermolecular interactions are weaker than in pure water, and we have the necessary increase in vapor pressure. Many low molecular weight alcohols form azeotropes with water for the same reasons.

When the water concentration gets below what is required for the azeotrope, enough of the alcohol molecules interact only with other alcohol molecules that the deviation from ideality decreases.

For the maximum-boiling azeotropes, the requirement is that the vapor pressure of both components is decreased. This is quite common in acid:water mixtures where the degree of ionization of the acid in the mixture is higher than when it is in pure form, which means that there are more ionic interactions, which are quite strong. If the water concentration gets below what is required for the azeotrope, the amount of ionization decreases and the deviation from ideality decreases with it. When the water concentration is higher than in the azeotrope, there may be a great deal of ionization of the acid, but there is enough water that is not interacting with the acid (and conjugate base) molecules that the deviation from ideality is less.


Selective alkylation of carboxylic acid versus phenol Starting from the ketone shown, I plan to use a haloform reaction to transform the acetyl group into a carboxylic acid.

Following that, there are two alkylations to be carried out: one esterification on the carboxylic acid (using propargyl bromide) and one etherification on the alcohol (using 6-bromohexan-1-ol). In what order should these be performed?

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that.

In 2, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the Rubber Book), you know that

the carboxylic acid is more acidic than the phenol
the difference is acidity is quite large

Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as N,N-dimethylformamide (DMF) should be sufficient to deprotonate both centres.

Back to your question on the order of events:

I would add 1 equivalent $\ce{K2CO3}$ of to a solution of 2, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.

Monitoring the reaction by TLC will show whether the esterification is quantitative.

If this is the case, I would dare to run a one-pot reaction, add another equivalent of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.

It might be a good idea to try this first with very small amounts, just enough to check via TLC.


Can I remove flesh from seeds with hydrochloric acid? I have some crenellated seeds with which I make jewellery and I need to remove the remaining fruit flesh from them.  I have tried lots of things including ammonia, vinegar, bleach and a ultrasonic jewellery cleaning machine.  the only thing that works is a wire brush which takes forever and results in grazed fingers.  I wondered if some solution of the hydrochloric acid I use in the pool would work.  If so, what concentration should I use and how long should the seeds be immersed in it?  If not, what do you think would work?

It depends highly on the seed and the fruit flesh, but using hydrochloric acid to clean off the "jelly" from freshly collected tomato seeds is a recognized laboratory protocol. That protocol uses a 1:1 mix of concentrated HCl and water for 20 min., followed by a 20 min wash in 10% w/v trisodium phosphate to neutralize the acid.

Depending on what sort of flesh is covering the seeds, it may take different amounts of time and different strengths of acid to accomplish the job. You will likely have to experiment a bit. Start "gentle" (as in, less concentrated acid) and work your way up. - You probably don't want conditions that results in complete flesh removal, but rather ones where gentle agitation (e.g with an old toothbrush) after neutralization will dislodge the flesh. - Viability of the seeds will also vary depending on the exact protocol and how tough the seeds are, though it doesn't sound like you need the seeds to be able to germinate afterwards.

As always, when working with concentrated acids and bases, use of proper safety equipment is highly recommended - even for acids and bases they sell in the hardware store.

For the chemical perspective, concentrated acid solutions catalyze the hydrolysis of peptide, ester and glycosidic bonds of the polymers that make up the flesh of the fruit surrounding the seeds. With the polymers degraded, they more easily dissolve in solution. (This is also the reason that our stomachs have hydrochloric acid in them.) The cellulose that makes up the seed coat is recalcitrant to degradation, and so remains untouched to the "mild" acidic conditions.


Ion chromatography of anions Using an eluent of 1.00 mM NaHCO3 and 8.00 mM Na2CO3 in Milli-Q water, chromatographs were taken of the following anions: fluoride, chloride, nitrate, sulfate And phosphate. Why were the  retention times of sulfate and phosphate longer than those of chloride, nitrate, and fluoride? The retention time of phosphate is not longer than the retention time of sulfate, why? What data do I use to find pH

In ion chromatography there are multiple retention mechanisms, the ion-exchange is the primary one. However, let us keep it simple.

You should be able to explain why monovalents retain less than divalents? Use Coulomb's law! The force of attraction between the ion and the ion-exchange on stationary phase is proportional to the product of their charges and inversely proportional to the hydrated radius. Think about it why fluoride comes out before chloride, almost universally on all IC columns.

Another complication, this explanation breaks down with large anions like iodide, thiocyanate and perchlorate. They retain extremely long on most IC columns. They are called polarisable ions. These ions are sticking because of some other reasons.

Coming to divalents and trivalents: The retention of divalents and trivalents are very complicated. It depends on the chemistry of the stationary phase and its cross-linking and the very nature of the charged site on the ion-exchanger. However you can do a ball-park at an elementary level. Hard to predict in general.

Is phosphate ion always 3(-) or it can have 2(-) or 1(-) charge? Is the ionization pH dependent? The pKa (third proton) is >12. Is the mobile phase pH above 12?


What factors affect the chemical stability of a compound? My question is essentially about the factors that make certain atoms in certain compounds leave the compound they are currently in in order to bond with another atom/compound.

For example, I've been taught that if you react a halo-alkane with a strong base, like KOH, a substitution occurs in which OH and the halogen change places. Why does this happen (if it does)?

Would an electronegative atom rather want to bond with another extremely electronegative atom, or one that has extremely little electronegativity? Basically, would fluorine want another fluorine, or rather something like caesium?

$$\ce{Cs + F \to CsF}$$
$$\ce{F + F \to F2}$$

My thought is that fluorine 'loves' electrons so much, and there is nothing else that loves electrons more than this guy, so this has to be the most stable compound ever that is not dissociable by any other compound/atom.

Any change which is spontaneous, takes place to lower the usable energy present in that system (This principle is so important, that even universe obeys it), which depends upon the conditions in which change in taking place. To demonstrate this, lets take an example of a stone at height h, one can do a lot of work with that configuration, it can do work by hitting someone on their head. But in contrast, when it is at ground, it does not do anything much on its own, the reason being it has used up some of its energy stored in the form of gravitational potential energy. Now of course, if conditions change, the stone can again do some work, in fact it is always loosing some of its usable energy energy, in the form of increasing entropy since it is breaking down by the natural forces of wind.
Coming back to your question, why does substitution takes place in case of halo alkane with hydroxide anion. Now, we can see that haloalkane has transformed into an alcohol. If one calculates the usable energy (Gibbs free energy, this encompasses several other factors  which include enthalpy, entropy and thermal energy) stored in haloalkane and corresponding alcohol, it will be clear that usable energy has decreased going from haloalkane to alcohol. To understand more about that, you can look up gibbs energy which quantifies that usable form of energy. Will fluorine bond with fluorine or Cesium- The answer, which I think is obvious to you, is surely cesium. why? Looking at their atoms, imagine cesium has no or very little pull on its valence electrons, while fluorine nucleus has so much pull, that electrostatic force permeates farther than atom. Hence, when two fluorine atoms come close to each other, there is some attraction between both atoms due to their individual pulls, but because both pulls are of same strength, it results in a covalent bond, but in case of cesium, the electrostatic force of fluorine nuclei pulls the valence electrons of cesium taking them away from the cesium nucleus, and thus fluorine's thirst for electrons is satisfied (the rock is at surface now!), while cesium is totally cool about it, since it had very little attraction for those electrons. After this process fluorine has negative and cesium has a positive charge, which attracts both of them and also it decreases the energy stored in fluorine-cesium systemIs CsF so stable that it cannot be dissociated into its components? No its not, put in water and water will dissociate the compound easily and energy is supplied by hydration and increasing entropy.


Why are amidines more basic than amines? Amidines can be thought of as carboxylic acid derivatives with the carbonyl oxygen replaced by a nitrogen and hydrogen atom; amidines exhibit conjugation.

Amidines are more basic (less acidic) than amines and amides by a factor of around ${10^{3}}$ and ${10^{12}}$, respectively (from pKas).

I would naturally think that conjugation would lower the energy level of MO containing the lone pair; thus, the lone pair would be more bound to the nucleus and less available to deprotonate.

This is an argument similar to explaining why pyridine is less basic than piperidine; $\ce{N}$ is $sp^{3}$ hybridized in piperidine but $sp^{2}$ hybridized in pyridine. Since $sp^{2}$ orbitals have less p character than $sp^{3}$ orbitals (2/3 vs 3/4), the lone pair on $\ce{N}$ in pyridine is at a lower energy level than the lone pair on $\ce{N}$ in piperidien, and pyridine is less basic than pyridine.

Although I realize that the positive charge on the amidine is stabilized by conjugation (as explored by a previous post comparing the basicity of amidines and amides), wouldn't my reasoning with MO theory outlined above suggest a lowering of the basicity of amidines?

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

Note that when you are comparing amidines with amines, you are comparing 2 nitrogen atoms with 1 unless you are comparing it with diamines.  Amidines which have 2 ionizable nitrogen atoms are stabilized  by 2 resonant structures, therefore making them more likely to accept protons (that is more basic).  Similarly, guanidine (not to be confused with guanine which is a nucleic base) which has 3 ionizable nitrogen atoms are stabilized by 3 resonant structures thereby making it even more basic.
For additional reading, check out the basicity of the following drugs.
Betrixaban (which has 1 amidine group)
Pentamidine (which has 2 amidine groups)
Guanethidine (which has 1 guanidine group)
Metformin (a biguanide, that is, having 2 guanidine groups fused together)
Altretamine (which has 3 guanidine groups fused together into a cyclic structure)
Famotidine (which has 1 amidine group and 1 guanidine group).


Why are halogens ortho para directing even though deactivating? Halogens bonded to benzene ring has three lone pairs. These three electron pairs can cause resonance in benzene ring. But, halogens are also highly electronegative and thus they have strong -I effect. So, they are deactivating groups. But, why are they ortho para directing?

Let's take as an example a typical halogen, chlorine. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through the inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution

Image courtesy: NCERT Chemistry Class 11th

Through resonance, a halogen tends to stabilise the carbocation, and the effect is more pronounced at ortho- and para- positions. However, the inductive effect is stronger than resonance here, and causes net electron withdrawal and thus net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack.

Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.


Why is the bond angle H-P-H smaller than H-N-H? $\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.

But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?

Starting point: 2s orbitals are lower in energy than 2p orbitals.

The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.

In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.

So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.

There are two choices for the central atom:

remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result,

substituents will be arranged closer together and destabilizing steric interactions will be increased
due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or

hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result,

steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:

Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!

Back to our problem:

Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.

In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.


antisymmetric wavefunction Why can't we choose any other antisymmetric function instead of a Slater determinant for a multi-electron system? Why do we choose our wavefunction for a multi-electron atom as a product of single-electron wavefunctions?

1. A pragmatic view
As the comments suggested you can use different functions than slater determinants.
But this would be more or less as bad as an idea
as describing a movement in a circle without using polar coordinates.
If you include the constraints of your system implicitly in the coordinate system or basis of your description, nearly everything gets better.
More elegant equations, faster convergence... .
Determinants are a fast and elegant way to include constraints like the Pauli Principle in your description.
2. A bit more mathematical view
Let's go back to the definition of a Determinant in general.
From Wikipedia:

[...] Another way to define the determinant is expressed in terms of the columns of the matrix.  If we write an $n \times n$ matrix $A$ in terms of its column vectors
$$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$$
where the $a_j$ are vectors of size $n$, then the determinant of $A$
is defined so that
$$
\begin{align}
  \det \begin{bmatrix} a_1 & \cdots & b a_j + c v & \cdots & a_n \end{bmatrix}
          &= b\det(A) + c \det \begin{bmatrix} a_1 & \cdots & v & \cdots & a_n \end{bmatrix} \\
  \det \begin{bmatrix} a_1 & \cdots & a_j & a_{j+1} & \cdots & a_n \end{bmatrix}
          &= -\det \begin{bmatrix} a_1 & \cdots & a_{j+1} & a_j & \cdots & a_n \end{bmatrix} \\
  \det(I) &= 1
\end{align}
$$
[...]

The last constraint comes from the normalization of the atomic orbitals and the second constraint is just the Pauli-Principle.
To the best of my knowledge you don't have to force the linearity in the first constraint, but if you do, you end up with a determinant.


How can we say that one mole of a substance contains avogadro number of particles? I just want to know, what was the procedure through which the value of one mole of a substance was considered as a an aggregate of $6.023\cdot10^{23}$ particles.

Secondly, how do we calculate the volume occupied by one mole of gas?

There's a nice Scientific American article that provides an interesting historical perspective on Avogadro's Number.  They go on to point out that one of the early accurate determinations was based on the first accurate measurement of the charge on an electron.  The charge on a mole of electrons (1 Faraday) had been known for some time.  Dividing the latter by the former yielded Avogadro's Number.

As to the volume of 1 mole of gas at STP, imagine the following experiment:

1) weigh out 1 mole of oxygen gas (32 gm)

2) fill a balloon with this weight of gas

3) attach a pressure gauge to the balloon and expand or contract the balloon's volume until the gauge reads 1 atmosphere

4) Immerse the balloon in a water tank and measure the volume of water displaced - 22.4 liters


Mixing Supplements - Will the Distributions Remain? I am combining various vitamins and minerals into a food supplement for my dog and I currently measure each powder individually for each batch of food each week.  The supplements are in powder form but have varying grain sizes.  Examples include copper, zinc, dicalcium phosphate, kelp powder, table salt, Vitamin E, Vitamin B-12.

If I were to make a giant batch of the supplements above using the proper proportions can I safely measure out of that large batch the appropriate volume each week or will the various consistencies / densities of the powders unevenly distribute making the weekly scoops proportionally unbalanced?

Is there some piece of equipment that could be used to mix the powders guaranteeing even distribution for batch usage?

Thank you in advance for any advice.

It's hard to maintain homogeneity for the mix of powders. The bigger the batch, the less homogeneous it's going to be. Ideally you want to disperse solid ingredients; there are several techniques to do that. Two of the most popular ones are:

dissolving all the components in a non-reactive solvent and they dry out or precipitate the powder;
using a ball mill.

With supplements, however, it's somewhat trickier, as they may react with each other over time, with the solvent, or be contaminated in grinder. Also they are often granulated for a good reason: this reduces the contact with atmosphere and prolongs shelf life.

If you decide to avoid the dispersion part, then due to Brazil nut effect the largest powder particles will remain on top of the batch, the smallest – on the bottom. This is especially noticeable when the batch has been shaken, e.g. after being transported in a car trunk, or just being kicked a few times lying on the kitchen's floor.

All in all, I wouldn't recommend doing this. The probability of suffering from an imbalanced diet is too high. There are some multivitamin and mineral complexes designed for humans, maybe look for something similar for your pet.


Which attack occurs first: nucleophilic acyl substitution or nucleophilic substitution? If there is a compound with a site for nucleophilic acyl substitution (i.e. an ester group or the like) as well as a nucleophilic substitution (i.e. a bromine group or the like), will a nucleophile preferentially attack one of them? How do i decide which one a nucleophile would prefer attacking?

The answer is - it depends. On the conditions, on the solvent, on the nucleophile being used, on which ester is being used. You may get selectivity, you may get attack at both sites which generates a mixture.

In the real world of chemistry you often don't know for certain until you run the experiment unless someone has published an exact procedure for the transormation which you follow.


3-step process to make propyl ethanoate from propene I was asked to make propyl ethanoate from propene in three steps. I have access to the reactions below.

Nucleophilic substitution (Both SN1 and SN2)
Electrophilic addition
Hydrogenation
Dehydration
Hydration
Condensation
Free radical substitution
Oxidation of alcohols

I'm assuming that you can add whatever reactants you want, since propyl ethanoate has five carbons and you can't just chop a carbon off of a propene with the above reactions. I know that the final step has to be condensation, since that's how you make esters. So essentially, I need a two-step process to make a primary alcohol that becomes the propyl in propyl ethanoate.

If you use hydrogenation on propene first, you get propane. Once you have that, it takes two steps to make a primary alcohol: one in free radical substitution (to get 1-bromopropane) and one in nucleophilic substitution (to get propan-1-ol).

If you use hydration first, you get a secondary alcohol in one step. However, you can't make propyl ethanoate with propan-2-ol.

If you use electrophilic addition first, you get a secondary haloalkane (2-bromopropane). That can also only get you a secondary alcohol.

Does anyone know how to do this? I'm convinced that it's impossible to do.

My proposal is hydroboration followed by Fischer esterification. It's 2 or 3 steps depending on how many you count hydroboration to be.

Alternatively if you are intent on going through a haloalkane pathway, you could try anti-Markovnikov addition of $\ce{HBr}$ by reacting it in the presence of $\ce{H2O2}$ to prepare 1-bromopropane. Then you could do nucleophilic substitution of $\ce{OH-}$ followed by Fischer esterification. This method would have a lower yield though due to the competing E2 reaction.


What is the physical basis for Hund's first rule? According to Hund's first rule, a set of degenerate orbitals are singly occupied first, before the second slot in any of the orbitals are populated. This is quite intuitive because electron-electron repulsions would make an atom more unstable if the electrons start filling two at a time in a single orbital.

However, the rule also states that in the ground state, electrons that singly fill the orbitals will adopt the same spin. Why is it necessary for the electrons to have the same spin?

Disclaimer

My following answer is the "traditional" explanation of Hund's first rule, which is based on a smaller value of $V_\mathrm{ee}$ (electron-electron repulsions) in the triplet state arising from Fermi holes. According to Levine's Quantum Chemistry 7th ed.:

  This traditional explanation turns out to be wrong in most cases. It is true that the probability that the two electrons are very close together is smaller for the helium $^3S$ $1s2s$ term than for the $^1S$ $1s2s$ term. However, calculations with accurate wave functions show that the probability that the two electrons are very far apart is also less for the $^3S$ term. The net result is that the average distance between the two electrons is slightly less for the $^3S$ term than for the $^1S$ term, and the interelectronic repulsion is slightly greater for the $^3S$ term.

  The calculations show that the $^3S$ term lies below the $^1S$ term because of a substantially greater electron nucleus-attraction in the $^3S$ term as compared with the $^1S$ term. Similar results are found for terms of the atoms beryllium and carbon.

Wildcat has an answer which briefly mentions this. Aside from that Levine gives several references:

Katriel, J.; Pauncz, R. Theoretical Interpretation of Hund's Rule. Adv. Quantum Chem. 1977, 10, 143–185. DOI: 10.1016/S0065-3276(08)60580-8.
Shim, I.; Dahl, J. P. Theor. Chim. Acta 1978, 48 (2), 165. The DOI doesn't resolve properly, but here's the link to the publisher's website.
Boyd, R. J. A quantum mechanical explanation for Hund's multiplicity rule. Nature 1984, 310, 480–481. DOI: 10.1038/310480a0.
Oyamada, T.; Hongo, K.; Kawazoe, Y.; Yasuhara, H. Unified interpretation of Hund's first and second rules for 2p and 3p atoms. J. Chem. Phys. 2010, 133, 164113. DOI: 10.1063/1.3488099.

With that said, here is the old-fashioned argument for why parallel spins are favoured over paired spins (all else being equal).

Introduction

The origin of Hund's first rule lies in "exchange energy", which is a way of saying that electrons with like spin repel each other less than electrons with unlike spin.

To see why, let's consider an excited state of the helium atom: $\mathrm{1s^1 2s^1}$. (The ground state configuration $\mathrm{1s^2}$ is of no use to us because those electrons have to be paired with unlike spin.) In the excited state, the electrons can either have parallel spin (the triplet case) or paired spin (the singlet case).

The total electronic wavefunction comprises a spatial part, which describes which orbitals the electrons are in (i.e. it specifies $n,l,m_l$), and a spin part, which describes the, well, spin ($m_s$). According to the Pauli exclusion principle, the total wavefunction must be antisymmetric with respect to interchange of the electron labels: $\Psi(1,2) = -\Psi(2,1)$.

Spin wavefunctions

The spin wavefunctions are simultaneous eigenfunctions of the $\hat{S}^2$ and $\hat{S}_z$ operators. The total spin quantum number, $S$, can be found from the eigenvalue of the operator $\hat{S}^2$:

$$\hat{S}^2\psi = \hbar^2 S(S+1)\psi$$

For more detailed discussion, refer to any QM text on the coupling of angular momenta.

There are three possible triplet spin wavefunctions, which are degenerate.

$$\psi_{\mathrm{spin,triplet}}(1,2) = \begin{cases}
\alpha(1)\alpha(2) \\
\beta(1)\beta(2) \\
\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \alpha(2)\beta(1)]
\end{cases}$$

It turns out that $\hat{S}^2\psi_\mathrm{triplet} = 2\hbar^2\psi_{\mathrm{triplet}}$, which means that these have the quantum number $S = 1$. This corresponds to a multiplicity of $M = 2S + 1 = 3$, hence "triplet". These wavefunctions represent the case where the electrons have parallel spin. (Loosely speaking, each electron has a spin of $1/2$, and if these are aligned parallel the total spin is $1/2 + 1/2 = 1$.)

The singlet state is

$$\psi_{\mathrm{spin,singlet}}(1,2) = \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)]$$

and as you might guess, $\hat{S}^2\psi_\mathrm{singlet} = 0$, meaning that $S = 0$ and $M = 2S+1 = 1$. This represents the case where the electrons are paired - one can think of it as the two $1/2$ electrons spins “cancelling each other out”.

The important thing to notice is that the triplet spin wavefunctions are symmetric with respect to permutation of the labels, whereas the singlet spin wavefunction is antisymmetric. For example:

$$\begin{align}
\psi_\mathrm{singlet}(2,1) &= \frac{1}{\sqrt{2}}[\alpha(2)\beta(1) - \alpha(1)\beta(2)] \\
&= -\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)] \\
&= -\psi_\mathrm{singlet}(1,2)
\end{align}$$

Spatial wavefunctions

The only way of constructing these are the linear combinations below.

$$\begin{align}
\psi_\mathrm{space,symm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) + \mathrm{1s}(2)\mathrm{2s}(1)] \\
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)]
\end{align}$$

Remember that the atomic orbitals above are functions of the coordinates of the electron: $\mathrm{1s}(1) = \mathrm{1s}(\vec{r}_1) = \mathrm{1s}(r_1,\theta_1, \phi_1)$. Now, the antisymmetric spatial wavefunction has an interesting property. Let's explore what happens when the electrons get very close together, i.e. $\vec{r}_1 = \vec{r}_2 = \vec{r}$.

$$\begin{align}
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r}) - \mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r})] \\
&= 0
\end{align}$$

If the wavefunction is 0 when the two electrons are on top of each other, that means that there is no probability of this happening. The average distance between the two electrons $\left\langle\left|\vec{r}_2 - \vec{r}_1\right|\right\rangle$ will therefore be larger, and on average the electrons will repel each other less, since they are on average further away from each other. This is known as a Fermi hole.

In fact, one can also show that the symmetric wavefunction has a larger probability amplitude when $\vec{r}_1 = \vec{r}_2$. This is known as a Fermi heap.

The total wavefunction

As I mentioned earlier, the Pauli exclusion principle dictates that the total wavefunction be antisymmetric. This means that the triplet spin wavefunctions, which are symmetric, must be paired with the antisymmetric spatial wavefunction. Likewise, the singlet spin wavefunction, which is antisymmetric, must be paired with the symmetric spatial wavefunction.

$$\begin{array}{c|c|c}
\text{Spin} & \text{Spatial} & \text{Total}\\
\hline
\text{Triplet (S)} & \text{AS} & \text{S } \times \text{ AS} = \text{AS} \\
\text{Singlet (AS)} & \text{S} & \text{AS } \times \text{ S} = \text{AS}
\end{array}$$

So, there is a decrease in electron-electron repulsion when the electrons are parallel - because this means they have a triplet spin wavefunction, and hence an antisymmetric spatial wavefunction, which has a Fermi hole.

Exchange integrals

There is another, perhaps more mathematical, way of looking at it, and that is to calculate the expectation value of the electronic energy for both wavefunctions. For the helium atom, under the assumption of an infinitely heavy nucleus (essentially an atomic version of the Born-Oppenheimer approximation), the total Hamiltonian (in atomic units) is

$$\hat{H} = -\frac{1}{2}(\nabla_1^2+\nabla_2^2) - \frac{2}{r_1} - \frac{2}{r_2} + \frac{1}{r_{12}}$$

where $r_i$ is the distance of electron $i$ from the nucleus, and $r_{12}$ is the distance between the two electrons. The expectation value of the energy is simply $\left\langle\Psi\middle|\hat{H}\middle|\Psi\right\rangle$ (assuming our wavefunction is already normalised). It turns out that the only important term in our present discussion is the electron-electron repulsion term $1/r_{12}$.

Let's look at the triplet spin wavefunction (of course paired with the antisymmetric spatial wavefunction) first. The operator doesn't depend on spin degrees of freedom, so we can separate those, and the inner product of the spin wavefunction with itself is just unity.

$$\begin{align}
\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle
&= \left\langle\psi_\mathrm{spin,triplet}\middle|\psi_\mathrm{spin,triplet}\right\rangle\left\langle\psi_\mathrm{space,antisymm}\middle|\frac{1}{r_{12}}\middle|\psi_\mathrm{space,antisymm}\right\rangle \\
&= \frac{1}{2} \left\langle \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1) \middle| \frac{1}{r_{12}} \middle| \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)  \right\rangle \\
&=
\frac{1}{2} \Biggr\{ \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle  \\
&\quad\quad\quad - \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle \\
&\quad\quad\quad - \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle \\
&\quad\quad\quad + \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle\Biggr\}
\end{align}$$

Ugly, but that's what it is.

We can simplify it however by noting that the first and the fourth brackets are the same (since the variables in the integral are just dummy variables, interchanging the electron labels does not affect the value of the integral). This integral is traditionally called a Coulomb integral, since it simply reflects the average Coulombic repulsion between two independent (uncorrelated) electrons, one in the 1s orbital and the second in the 2s orbital. It is given the symbol $J_\mathrm{1s2s}$ (the subscripts indicating the orbitals that the electrons are in). Because this is an expectation value for a potential energy arising from repulsion, it is necessarily a positive quantity.

The second and third terms are actually also the same, again because the variables are just dummy variables and you can interchange the labels. This integral, however, does not have a physical interpretation; it is a purely quantum mechanical effect. It is called the exchange integral, and is given the symbol $K_\mathrm{1s2s}$. It's a bit harder to show, but the exchange integral is also a positive quantity (there is a question on it here). However, its magnitude is not as large as $J_\mathrm{1s2s}$.

Putting it all together (leaving out subscripts for clarity) we have

$$\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle = \frac{1}{2}(J-K-K+J) = J-K$$

and if you were to perform the same calculation on the singlet spin wavefunction (not very difficult, just replace the minus signs with a plus!) you would get

$$\left\langle\Psi_\mathrm{singlet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{singlet}\right\rangle = \frac{1}{2}(J+K+K+J) = J+K$$

Since $J > K > 0$, the expectation value of the energy for the singlet state is higher.

But I have more than two electrons. What now?

Well, you just have to do a ton more ugly maths and perhaps a few hundred more integrals. But the idea is the same. For every pair of electrons with parallel spins, there is a corresponding exchange integral that reduces the expectation value of the energy.

Therefore, in a $\mathrm{3d^5}$ manganese atom (we ignore the closed shells) there are

$$\frac{5!}{2!(5-2)!} = 10$$

different pairs of d-d electrons with parallel spins. Therefore if you evaluate the expectation value of the energy, there will be a $-10K_\mathrm{3d3d}$ term, which is stabilising. This is precisely the exchange energy effect that Geoff describes in his answer.


Electronic configuration of excited states of iron The electron configuration of the ground state of $\ce{Fe}$ is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$. What are the configurations of its first and third excited states? The first excited states seems to be $\mathrm{[Ar]4s^1 3d^7}$, but isn't the angular momentum not conserved because $l=0$ for $\mathrm{s}$ and $l=2$ for $\mathrm{d}$?

The electronic configurations of the excited states of neutral iron are listed on this NIST Database page:

3d64s2 — a 5D — ground state
3d7(4F)4s1 — a 5F — 6928.268 cm–1
3d7(4F)4s1 — a 3F — 11976.238 cm–1
3d7(4P)4s1 — a 5P — 17550.180 cm–1
3d64s2 — a 3P — 18378.185 cm–1
3d6(5D)4s14p1(3P) — z 7D0 — 19350.892 cm–1

The reference give for this data is: G. Nave, S. Johansson, R. C. M. Learner, A. P. Thorne, and J. W. Brault, Astrophys. J. Suppl. Ser. 94, 221 (1994).

So, while in general predicting electronic configurations for excited states is a tricky business, you are right: the first excited electronic configuration is to “promote” one 4s electron into the 3d orbitals. And the second excited state configuration (regardless of the different spectroscopic terms and sublevels) is to promote the 4s electron into 4p.


How do I decide whether ring will expand in this reaction or not? Attempt:

After forming the carbocation adjacent to the cyclopropyl ring I had two options,

$\ce{Cl-}$ then and there without any changes
Expand ring, form a new carbocation and add $\ce{Cl-}$ there.

I chose the first option because of exceptional stability of cyclopropyl methyl carbocation. However, answer given is option C which follows mechanism  (2).

Can someone explain how to judge and decide how to proceed after the carbocation formation?

Explanation

In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane

In case 2. You can see it is 3° carbocation (or tertiary carbocation)
with angle strain of ~10° on cyclopropane

It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product that is C.

Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.

The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.

Edit2: See image below for reference

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel

In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.

To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.

Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.


How are the LCAO ansatz and Slater determinant ansatz for the wavefunction connected? Say we have a system of 2 electrons, an ansatz for the wavefunction is then a Slater determinant consisting of two orbitals.
$$
\left|\Psi_{Slater}\right>= \frac{1}{\sqrt{2!}}\left(\left|\phi_1\phi_2\right>-\left|\phi_2\phi_1\right>\right)
$$

How is this related to the basis expansion ansatz:
$$
\left|\Psi\right>= \sum_i c_i \left|\phi_i\right>
$$

The basis expansion formula would give me a function of 1 coordinate x_1 if i projected it on the position space, while the slater determinant ansatz would give me a function of two coordinates x_1, x_2. x_i stands for a set of x,y,z
coordinates. Spin is neglected in this example.
I understand that for the basis set expansion in an orthonormal basis, one can obtain the coefficients by the inner product
$$
\left<\phi_i|\Psi \right>=c_i
$$

but how can i use this on a slater determinant ? I only understand how this works if everything relies on one coordinate x_i, then i can calculate this:
$$
\left<\phi_i|\Psi \right>=\int\phi_i(x)\Psi(x)dx=c_i
$$

But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?
$$
\left<\phi_i|\Psi_{Slater} \right>=\int\phi_i(x_1)\Psi_{Slater}(x_1,x_2)dx= c_i(x_2)?
$$
This expression makes no sense, since the remaining object would still be a function of $$x_2$$

To my knowledge only basis functions of one coordinate are used in calculations, like 3D Gaussians. Assuming that i have a wavefunction given in slater determinant form, how would i find the coefficients of the basis functions ?

I am afraid i am mixing up things that don't belong together, clarifications would be welcome.

I found the question somewhat confusing. I think it is because mixing up some concepts.

  I understand that for the basis set expansion in an orthonormal basis,
  one can obtain the coefficients by the inner product  $$
 \left<\phi_i|\Psi \right>=c_i $$

Please, think about what you mean by the coefficients and where the functions live, before continue reading.

As you thought this expression as an inner product (which is right), both functions must live in the same space so they have the same number of coordinates. What you are doing with that inner product is projecting $\Psi$ over the $\phi_i$. If $\{\phi_i\}$ is an orthogonal set, then you can make the expansion you wrote. But, if the are not, then such expansion is meaning less. You still can make expansions that recover $\Psi$  (in fact, an infinite number of them). So, I get confused about which coefficients are you referring to. As for example, $\phi_1$, appears in both: many Slater's det. and in many terms in one (or more) Slater's det., to ask  which is its coefficient in the Slater det. makes not sense, except if you want to expand the Slater det. into a LCAO which is not possible (but you still can project it).

See below.

  But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?

First, note that even for 1-electron problems the problem is not mono-dimensional. So, the case of Slater's determinants is exactly the same. You only need to integrate over every coordinate.

  To my knowledge only basis functions of one coordinate are used in
  calculations, like 3D Gaussians.

3D-Gaussians have three coordinates. It is obvious using Cartesian coordinates. Under spherical coordinates they only depends on $r$, but they still are functions of $\theta$ and $\phi$. Although you can take only one coordinate ($r$) for its evaluation, it is not true for computing an arbitrary inner product.

  Assuming that i have a wavefunction given in slater determinant form,
  how would i find the coefficients of the basis functions ?

At first, take in mind that those coefficients are for expanding the Slater det. into the subspace spanned by your $\{\phi_i\}$. That is, it they are not an complete set, you only will get an approximation. You just have to perform those integrations. In your example, the $c_i$ will not be a function of $x_2$, because you have to integrate over every coordinate.

I am not sure if I understand correctly your doubts. Please, do not hesitate in ask for clarification. Maybe this answer could be useful.


What roles do neutrons play in an atom? An atom typically consists of electrons, protons, and neutrons. Electrons are negatively charged and participate in bonding to stabilize the atom.

Conversely, protons are positively charged and balance the charge of the atom. In addition, their positive charge attracts the negatively charged electrons.

What role do neutrons play in an atom?

Neutrons bind with protons and one another in the nucleus through the strong force, effectively moderating the repulsive forces between the protons and stabilizing the nucleus.$^{[1]}$

$\ce{^2He}$ (2 protons, 0 neutrons) is extremely unstable, though according to theoretical calculations would be much more stable if the strong force were 2% stronger. Its instability is due to spin–spin interactions in the strong force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the $\ce{^2He}$ nucleus a negative binding energy. $\ce{^3He}$ (2 protons, 1 neutron), on the other hand, is stable, and is also the only stable isotope other than $\ce{^1H}$ with more protons than neutrons.$^{[2]}$

$^{[1]}$ Wikipedia, Neutron, Beta Decay and the Stability of the Nucleus
$^{[2]}$ Wikipedia, Isotopes of Helium


What is the proper Lewis structure for HCOOH? I tried to draw the Lewis structure of $\ce{HCOOH}$ but I didn't get the correct representation.

As you can see in the picture below, at no. 1, I put $\ce{C}$ as the central atom because it has less electronegativity than $\ce{O}$ ($\ce{H}$ cannot be a central atom). I drew the structure, but it is wrong because it violates the octet rule.

Then I came up with another structure as shown at no. 2,
and it seemed correct to me, but when I searched on Google for the correct structure, I found out that I am wrong.

Why is the structure that I drew in no. 2  wrong, even though the formal charges are zero?

How do I figure out the correct order of atoms in such molecules like the one above. What I mean is how do I know that the $\ce{C}$ will have $\ce{H}$ on the left and one $\ce{O}$ up and $\ce{O, H}$ on the  right side?

With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$.

It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have  involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$.

Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming four bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.

Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example.

The above, in addition to keeping in mind that a carboxylic acid is a Bronsted/Lowry acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure.


What exactly happens while washing clothes I am reading about soaps , micelles and all that. I understand how soap behaves in water contaminated by oil. Soap helps in emulsifying the oil into water.

Let us suppose now we have dirt(i.e oil, grease etc) not alone in water but present on a some cloth. Why can't we put the cloth just in water to get it clean ? What precisely is the need for agitation in a washing machine or by any other method ?

I searched and found that agitation produces foam but according to this post foam does not alter the washing process anyhow.

When you wash a piece of cloth, even without soap, it's better to agitate. The vibration, friction, etc. will help speed up the process of mixing the dirt/grease with water.

Even the very definition of "agitation" uses the example of washing. From Vocabulary.com

  Agitation is the act of stirring things up, like the agitation of a washing machine that moves the water, detergent, and clothes around and gets the dirt out

And as you rightly pointed out from another question, "the froth has little or no effect on the detergent action".


Schrodinger's Equation and Wave Function So I understand that there exists the shrodinger's equation, which on solving,gives the wave function of an electron. The wave function as I understand, gives all possible information about an electron(any extra information on wavefunction will be appreciated).
Could someone actually solve a shrodinger's equation for an electron, and vaguely describe the process involved? Also after the wavefunction is obtained, could you explain to me what exactly does the wavefunction depict, and actually graph it?
I am having a lot of trouble grasping the concept. Any help will be greatly appreciated

I suspect the problem is not mathematical. It is in the meaning of the wave function. Well. I will now show you how I explain it qualitatively in my high school classes. I state that the electron is like a violin string, but a string with three dimensions that is vibrating in the fourth dimension. As I know this last words are not understandable, I start with one dimension and pass slowly to two and three dimensions.
Let's start from a horizontal string whose end is fixed on a wall, and the other end is in your hand. If you put some energy to the string, it will vibrate in a stationary vibration. The string looks like half a sine curve, with the maximum in the middle of the curve. If you want to describe the curve with words and without drawings, you can say that the curve is like a half-sinusoid whose amplitude has the same sign all over its length, part time positive, part time negative.
Now if you put more energy in the string, it will vibrate in another mode, with a node in the middle. When one half of the string is vibrating upwards, the other half is vibrating downwards. With still more energy you can have it vibrate with two nodes. To describe the shape of the string as before without drawings, you can say that the string is a sinusoid with one or several nodes. So that you need one number $n$, the number of nodes, to describe its movement : $n$ is the first quantum number. $n = 1$ for the first vibration, because the only point not moving is the end of the string in the wall. $n = 2$ for the case with one node in the middle of the string.
Now let's consider the string in two dimensions. This is not a string any more, but a two-dimensional plate like a drum. If you hit it softly in the middle, it will vibrate with all its points moving simultaneously upwards then downwards. The only points of no vibrations are the periphery of the drum. In two dimensions, points of no vibration are replaced by lines of no vibration. Here there is only one such line : the number of nodal lines is $n = 1$.
If you put more energy by hitting the center of the plate, you may produce a new vibration mode if the center of the plate is moving upwards (or downwards) when the outer part is moving in the opposite direction. In this mode no vibration occurs on the small circle between these. This circle is a nodal line. There are two concentric circles on this mode. So  $n = 2$. We may describe its amplitude by drawing curves of equal amplitude on the plate, like level curves on a map.
But it is going to be more difficult, because there is another type of vibration on the plate. If you hit the plate not in the center, but elsewhere, you may produce a plate vibration where one half of the plate moves upwards and one half downwards. Here the nodal line is a diameter. And there are two nodal lines : the diameter and the periphery. Here $n = 2$. As a consequence, two vibration modes exist with the same quantum number. We must introduce a second quantum number, which is called $l$, and which is the number of straight nodal lines. So this type of vibration is given by $n = 2$ and $l = 1$. Hopefully you understand that other vibration modes do exist with the drum, with arbitrary values of $n$ and $l$ < $n$
Let's summarize. An object with one dimension vibrates in another dimension, and one quantum number is enough to describe its vibration. An object with two dimensions is vibrating in the third dimension, and two quantum numbers are required to describe its vibration. The number of dimension is equal to the number of quantum numbers.
Now let's go to the electron, which has $3$ dimensions, and which may be considered as if vibrating in a fourth dimension, called $\Psi$. As this dimension cannot be represented in our world, all we can do is to give different numerical values to three quantum numbers for all vibrating modes, and represent surfaces which correspond to equal numerical values of the local "vibration" of the electron. These surfaces of equal value of $\Psi$ are spheres for $l = 0$, the famous dumb-bells with two lobes for $l = 1$, because of $1$ nodal plane. Here is how it goes.
Here is how I explain the wave function to my high school students. They usually have some difficulty understanding. But all have told me that this explanation had helped them sudying quantum mechanics later on in college.


What is the name for a reaction with a solid product rising from its container? The reaction between sugar and sulfuric acid is well known to produce a solid column of rising carbon which may leave its container.  A very similar rising column may also be produced from high temperature pyrolysis of certain organic substrates.

Is there a name exclusive to the type of reaction that produces a rising column?

In pyrotechnic chemistry, they might call it a pharoh's snake (or serpent) reaction. To a degree, (or rather 330 °C) pyrolysis of the sugar is occurring (starting at 170 °C).

In cooking, they would call it caramelization, although the term is also applicable in general chemistry. Less specifically, you can call it a "browning reaction"... note- caramelization (non-enzymatic) should not be confused with a Maillard reaction (enzymatic). More specifically, you could call it dehydration or decomposition of sugar with sulfuric acid. Somewhere in-between you might find the term carbonization.

You may also find the terms carbon soufflé and carbon foam interesting.


Oxidation number for nitrogen in ammonium sulfate I'm trying to find the oxidation number of $ \ce N$ in $ \ce{(NH4)2SO4}$.

The answer is supposed to be $-3$ but I can't figure out the logic behind this.

Ammonium sulfate is an ionic compound, composed of two ammonium ions ($\ce{NH_{4}^{+}}$) and one sulfate ion ($\ce{SO_{4}^{2-}}$). As such, you can neglect the sulfate entirely from your consideration of the oxidation state of the nitrogen, as well as treating each ammonium ion as a separate entity. Given the total charge on the ammonium cation, it should be easy to see why the nitrogen has the specified oxidation state.


How can I get a hand sanitizer solution to form a gel? I am trying to formulate a gel hand sanitizer for a thesis project. The sanitizer that I have right now is watery and will not gel.  My teacher told me to change our formulation because it is not good. Our ingredients are ethyl alcohol, aloe vera gel, and lavender scent. What else could we add to this mixture to really "gel" the hand sanitizer?

According to the wikipedia article on hand sanitizers, polyacrylic acid and polyethelyne glycol are commonly used thickeners.

You could try any hydrophilic polymer if those aren't available. Cornstarch might even work if you dissolved it in the alcohol before adding the other ingredients.


Rate of hydrolysis of ester in presence of an acidic medium Does the rate of hydrolysis of an ester depend on the concentration of both the ester and the acid, or only the ester?

I was told by my professor that hydrolysis of ester follows a pseudo-first order reaction: as the amount of water present for the reaction is very large, the change in concentration of acid/base will result in very negligible changes in concentration of $\ce{H+}$/$\ce{OH-}$ ion. So, the rate is only proportional to the concentration of ester taken.

Can someone explain if the above is in presence of basic medium or acidic medium?

We need to consider 4 quantities for the equilibrium reaction:

$$\ce{R1COOR2 + H2O <=> R1COOH + R2OH}$$

The equilibrium constant
The reaction quotient
The forward reaction rate
The reverse reaction rate

The forward reaction above is of the pseudo first order, as the professor says.

But its rate constant depends on acidic  catalytic factors and the net reaction rate is affected by the reverse reaction rate

The concentration of products- the organic acid and the alcohol - affects the net reaction rate by increased rate of the reverse reaction
and by shifting of the reaction quotient.

The content of a mineral acid  catalyzes the reaction, what means speeding up both reactions, therefore keeping the same equilibrium constant.(required as consequence of the energy conservation law).

The reaction is still the pseudo first order, but the rate constant depends on concentration of the catalysing acid.

The alternative approach is considering the mineral acid as self-regenerating reagent. But it must be then evaluated as reaction chain.

The content of a mineral base decreases reverse reaction rate and shifts the reaction quotient by reacting with the organic acid. It also directly reacts with the ester.

$$\ce{R1COOR2 + OH- -> R1COO- + R2OH}$$

This reaction is of the 2nd order.


What is meant by 'electrons of like/unlike rotation'? Explaining the reasons behind electron repulsion,
Radiochemistry asserts that,

  Interelectronic repulsion in related not just to electron pairing, but also to [the] angular momentum of the electrons. [---]

  e.g., in $\ce{Pr^2+ (4f^3)->Pr^3+ (4f^2)}$ ionisation removes repulsion between [electrons] of like rotation [---]

  $\ce{Pm^2+ (4f^4)->Pm^3+ (4f^3)}$ ionisation removes repulsion between [electrons] of unlike rotation [---] [emphasis added]

What exactly is meant by 'electrons of like/unlike rotation'?

You may glance at these links, but only if you want to.

Why don't we see these lanthanide species?
Howcome orbitals become 'core-like' when electrons are removed?
What's up with this quarter / three-quarter rule?

Original topic: Predominance of III oxidation state for lanthanides

It simply refers to electrons in orbitals with the same sign of $m_l$.

The quick and easy way to determine the ground-state term symbol of a lanthanide is to fill the 4f orbitals from most positive $m_l$ to most negative. This is in order to get the largest possible value of $M_L$:

$$M_L = \sum m_l$$

This is not valid by quantum mechanical arguments due to the indistinguishability of the electrons. However, it always leads to the correct ground-state term symbol because of Hund's second rule, which essentially states that the term with the largest value of $L$ is the most stable (after the first rule has been taken into account). Since the value of the projection quantum number $M_L$ ranges from $-L, -L+1, \ldots, +L$, a ${}^{2S+1}L$ term necessarily contains a term with $M_L = +L$. Therefore, by maximising the value of $M_L$, you will automatically generate the term with the largest possible $L$.

For example, consider the $\mathrm{f^3}$ ion $\ce{Nd^3+}$. You would want to "fill the orbitals" in this way:

where you have first and foremost put the electrons all with the same spin in order to satisfy Hund's first rule, and then filled them in decreasing order of $m_l$ in order to satisfy Hund's second rule.

The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.

Therefore, removal of an electron with "like rotation" (as you would if you were to ionise $\ce{Nd^3+}$ to $\ce{Nd^4+}$) is marginally more difficult than expected. Do not imagine it as a huge effect. It is very small and only leads to minor deviations from the ionisation energy trends (quarter-shell and three-quarter-shell effects).


Why is diazobenzene named so and not phenyl diazobenzene? I have searched for this answer for a long while but haven't got it. Why is diazobenzene named so and not phenyl diazobenzene or something? The group with two bonded nitrogens is the diazo group. Why does the term diazobenzene refer to a molecule with two benzene rings?

Diazo group is $\ce{=N2}$.

Example: diazomethane $\ce{CH2=N2}$
(detailed structure: $\ce{H2C=N+=N- <-> H2C^{-}-N+\bond{#}N}$)

$\ce{-N=N-}$ is azo group.

Example: azomethane $\ce{CH3-N=N-CH3}$
However, preferred names are based on diazene $\ce{HN=NH}$, so better name for azomethane is dimethyldiazene.

So, note that “diazobenzene” does not exist. You obviously mean “azobenzene”, $\ce{C6H5-N=N-C6H5}$, whose preferred name is diphenyldiazene.


At what temperature will proteins and fats boil in a vacuum? A question was asked on another stackexchange site:

  This is not nice perspective, but eventually it will happen. An astronaut falls out of spaceship because of damage caused by collision with other object, or because of suit decontamination. The fluids from the body would evaporate, and if any bacteria would survive, than only as spores. Does it mean the perfect mummification of the body? Or there will be some decay, caused by enzymes from damaged cells, for example?

Given that the water in the body will outgass taking with it much of the rest of the body, are there any cited sources that specify the pressure at which proteins, fats and bones will break down and become vapour?

You're talking about the human body being exposed to outer space and asking about what would happen to it's internal structure. Bones, fats and proteins are within the enclosed system that is the human body, they are not subjected to the low pressures of space. The answer to the pressure at which proteins, fats and bones will break down and become vapour is not related to the answer to what would happen to them in the human body if the body is exposed to the vacuum of space. Nevertheless, I'll try to provide both answers.

The answer to the second question (short-term) can be found on NASA's website at "ask an astrophysicist". It's actually quite surprising for someone used to sci-fi movies:

  You do not explode. Your blood does not boil. You do not freeze. You do not instantly lose consciousness.

The question you linked talks about "decay", or the effects of long time exposure. That's hard to predict, but at some point the body would be attracted by a planet and collide and you wouldn't have anything to boil anymore. Now to the chemistry part of it:

There isn't much (if any) data on the boiling point or vapor pressure of the compounds you're asking - it's hard to reproduce and it's more a curiosity than anything else. So let's talk about what would happen if you have fat, bones and proteins directly exposed to outer space pressures and temperatures, which seems to be your point anyway. Temperatures in space go from -270°C (in outer space, unexposed to sunlight) to +120°C (near the earth, exposed to sunlight). Since you're asking about boiling point let's talk about what will happen at the highest temperature.

Let's start with the easier part - the bones. It is safe to assume that the inorganic components will be tougher than the organic ones. The inorganic part of bones is mostly a specific type of apatite, which is a rock. It can be found in moon rocks. It will not boil. So at least the inorganic part of the bone would remain somewhat intact.

Fat is made of triglycerides. It'll probably break down to glycerol and the correspondent acid. Then it would decompose and boil.

Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance.


Does potassium aluminum sulfate have any effect on sexual performance/sperm counts or anything relating to the functionality of the sexual organs? I recently read about potassium aluminum sulfate that it has astringent property and has the ability to constrict body tissues and restrict blood flow. Since, for penis, the erection happens with the flow of the blood being directed to penis' blood vessels, does this effect of potassium aluminum sulfate cause an erection problem or any other effect on male/female sex organ?

Oral ingestion of simple aluminum compounds is not absorbed.  Topical genital application is definitely not recommended.  There is no obvious connection between aluminum and tumescence, one way or another

Soluble aluminum is toxic, likely cumulatively neurotoxic.  It is not absorbed from the GI tract unless complexed.  Aluminum hydroxide antacids as foaming tablets( citric acid plus sodium bicarbonate) gave aluminum citrate that is bioavailable.  Bad idea.  Pickles are made with alum to keep them ever so crunchy. The soluble aluminum is claimed not to be absorbed. Antiperspirant contains aluminum chlorhydroxide or glycinate to stopper sweat glands.  Topical application is not absorbed.  Exposing mucus membranes to soluble aluminum is definitely not recommended.


Mechanism for condensation reactions of amino acids It is well known that peptides are synthesized by a so called "condensation" reaction between an amine and a carboxylic acid group to form the final amide moiety (because it releases water). In its general form, it resembles the picture below.

However, what is the actual chemical mechanism behind this reaction? My thoughts are that the amine can deprotonate the carboxylic acid and form a carboxylate - which is a very bad electrophile - so how would it form the amide from nucleophilic attack? I am aware that in the body, it is possible that this reaction is mediated by biochemical catalysts but I am still not sure of the overall mechanism. If this reaction does indeed involve biochemical catalysts, is it possible to complete the synthesis without them?

Here is a proposed mechanism in a 2005 paper. One thing to point out is that this is a proposed mechanism but it is all I could find:

link to pdf: https://www.pnas.org/content/pnas/102/35/12395.full.pdf
Typically, these reactions involve the nitrogen attacking the partially positive carbonyl carbon as a form of nucleophilic substitution. However, OH is a bad leaving group so often, reagents such as DCC can be used to make the OH into a good leaving group before reacting. This is why it is easier to make nylon using a diacid chloride instead of a diacid.


Positive vs. negative heat of solution confusion Given that I have absolutely zero formal (or informal) education in chemistry (sorry about that), I'm struggling to get my head around what appears to be repeated confusions of the term "heat of solution" with regards positive vs. negative. As a specific example, the Wikipedia entry on Erythritol states:

  Erythritol has a strong cooling effect (endothermic, or positive heat of solution) when it dissolves in water

However, many other sources, including a number I found via Google Books suggest that Erythritol has a negative heat of solution. For example, this table extract from Food Science, an Ecological Approach:

  Erythritol: Noncaloric (0.2 kcal/g), high digestive tolerance, zero glycemic index, does not promote tooth decay, good heat and acid stability, high negative heat of solution.

This is frustrating as I'm trying to find something that has the opposite heat of solution to counteract erythritol's cooling effect, but since many sources seem to contradict each other, I'm struggling to find what I'm looking for.

Which is correct? Or am I using ambiguous terminology?

The heat of solution (aka enthalpy of the solution and $\Delta\mathrm{H}_{\text{sol}}$) is negative if heat is released. The reason is that energy is leaving the system. The reason you got two different answers is that the sources you referenced are describing two different things. The $\Delta\mathrm{H}_{\text{sol}}$ is positive (endothermic) when dissolved in water, but Erythritol is commonly combined with Inulin which produces an exothermic reaction (cancelling out the cooling effect of the Erythritol).


Realistic chemical spill accident in high school chemistry class In a story I'm writing, a teenager spills chemicals on herself in a high school chemistry class, causing her to be put under the safety shower. The spill destroys her clothes and causes minor burns. I would like this to be realistic and since I know nothing about chemistry, I have two questions:

Is there a chemical used in high school chemistry that when spilled could cause this situation to occur?
If so, do you know of an experiment that would use that chemical?

Collecting together the responses from the comments and Ben Norris's answer and adding one of my own, individual solutions in water of the following chemicals are realistic options for a liquid spill in a high school chemistry lab, leading to destroyed clothes and time under a safety shower:

Potassium permanganate $(\ce{KMnO4})$
Sulfuric acid $(\ce{H2SO4})$
Hydrochloric acid $(\ce{HCl})$
Nitric acid $(\ce{HNO3})$
Sodium hydroxide $(\ce{NaOH})$

As for an experiment leading to a spill, one possible example would be in the course of filling a buret in preparation for a titration. Burets are usually quite tall and narrow (click image for larger version):

      (Marketing image from Microtech Scientific)

A slip while filling one could easily result in spilling a few hundred milliliters or more of liquid on oneself.


Isopropanol in contact lens solution plastic bottle I have a rather large plastic bottle of isopropanol ($99.7\%$), and I wanted to store some of it in a smaller, more handy container. I have two empty plastic bottles from contact lens solution that I wanted to use, but I'm not sure if it is safe for storing them in there for longer periods of time.

One of them has a 01 symbol inside a triangle, from which I concluded that this means PET and it also has a number 3 inside a circle (I'm not sure what that means).

The other one is HDPE (2 in a triangle) and has a number 5 elsewhere. Are both of those safe for long term storage of isopropanol?

I do intend to mark them very clearly.

Based on your description, it looks like you've determined both plastics correctly. In general, they both would work fine for storing isopropanol; however, according to the Thermo Fisher's Labware Chemical Resistance Table, HDPE is preferred:

$\ce{i-PrOH}$ – PET

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: N Immediate damage may occur. Not recommended for continuous use.

$\ce{i-PrOH}$ – HDPE

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: E Little or no damage after 30 days of constant exposure.


How to raise pH without generating CO2 What buffering chemical/supplement will reduce lactic acid for runners but not generate $\ce{CO2}$?  Sodium bicarbonate does so but causes gastric upset.

I don't think this question has an answer.  First, muscle soreness is not caused by lactic acid.   Second, distance runners run aerobically, meaning very little lactic acid would be generated by running in the first place.  (Sprinters are the exception here.)  Third, even if lactic acid buildup was a problem (which it isn't), both the cellular cytoplasm and human bloodstream are very well-buffered.  The cytoplasm is primarily buffered by potassium glutamate and the circulation by bicarbonate (created from respiration).  Production of "lactic acid" would really be production of lactate salts.


Isothermal paradox(kind of) I have recently read that for an isothermal expansion ∆u =0 which implies that work done =heat released as temperature  remains constant ....but also I read that q=m×c×∆T which in this case should be zero as ∆T=0,,,then this implies work done = 0 ..but this is not true

Where am I committing mistake

In freshman physics, we were taught that $q=mC\Delta T$, but in all the cases we considered, no work was being done.  When we got to thermodynamics, we learned that q represents energy in transit, and is a function of process path.  We also learned that, when work is being done, q is not equal to $mC\Delta T$.  Instead, we learned about the internal energy of a system U, which is expressed as $$\Delta U=q-w$$  When no work is being done, $$\Delta U=q=mC\Delta T$$All this misunderstanding is resolved and made consistent in thermodynamics when we change the definition of heat capacity slightly, defining it strictly in terms of U, rather than in terms of q:$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$Then, whether or not work is being done, we get the correct answer for the heat transferred and the change in internal energy.  The advantage of this definition is that C is properly recognized as a physical property of the material being processed rather than a function of the process path.


Why does the dissolution of calcium hydroxide have a negative entropy? We had an experiment on the dissolution of calcium hydroxide and we obtained the experimental entropy value $\pu{-203 J mol^-1 K^-1}$ with a percent error of $26.7\%.$ Doesn't dissolving a solid in a solvent increase entropy?

When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation.  The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed.
Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction.  In some cases, complex ions may be more structured than the crystals of undissolved ions.       In such a case, the entropy change of solution will be negative.


How do I calculate the equilibrium constant of a reaction from two related reactions' constants? At a certain temperature the following reactions have the constants shown:

  $$\begin{align}\ce{S(s) + O2(g) &<=> SO2(g)} && K'_\mathrm{c} = 4.2\times 10^{52}\\
\ce{2S(s) + 3 O2(g) &<=> 2 SO3(g)} && K''_\mathrm{c} = 9.8\times 10^{128}\end{align}$$

  Calculate the equilibrium constant $K_\mathrm c$ for the following reaction at that temperature:

  $$\ce{2SO2(g) + O2(g) <=> 2 SO3(g)}$$

I already have the answer (which is given in the back of the textbook): $$K_\mathrm{c}=5.6\times10^{23}$$

But whenever I complete the problem (I've done it thrice) I get $2.3\times10^{76}$.

My process for solving this is as follows. I reverse the first equation and double the coefficients. Since I've done this, I use the inverse of $K'_\mathrm{c}$, $(K_\mathrm{c}')^{-1}$. I then multiply both $K$ values and the result is $2.3\times10^{76}$. Where is my mistake?

Equilibrium constants depend upon the choice of reaction equations. I have noted the extended way to calculate $K''_\mathrm{c}$ below:

$$\begin{align}\ce{2S + 3 O2 &<=> 2 SO3}\tag{2}\\[0.4em]
K''_\mathrm{c} &= \frac{[\ce{SO3}]^2}{[\ce{O2}]^3}\end{align}$$

(Sulfur does not take part in the equilibrium constant because its activity, as a solid, is $a(\ce{S}) = 1$.)

Having this, we can approach the question. First, we need to add particles in an equivalent way (i.e. the same to reactant and product side) to get from equation $(3)$ to something that can be expressed as a linear combination of $(1)$ and $(2)$ ($a(1) + b(2)$):

\begin{align}\ce{2SO2 + O2 &<=> 2SO3}\tag{3}\\[0.4em]
\ce{2SO2 + 3 O2 &<=> 2 SO3 + 2 O2}\tag{$3'$}\\[0.4em]
\ce{2 SO2 + 3 O2 + 2 S &<=> 2 SO3 + 2 O2 + 2 S}\tag{$3''$}\end{align}

   $\displaystyle (3'') = -2 \times (1) + 1 \times (2)$

With that in hand, we can determine what the equilibrium constant of $(3'')$ would look like and how that can be traced back to the other equilibrium constants:

   $$\displaystyle K_\mathrm{c} = \frac{[\ce{SO3}]^2[\ce{O2}]^2}{[\ce{SO2}]^2[\ce{O2}]^3}\tag{4}$$

 $$\displaystyle K_\mathrm{c} = \frac{[\ce{O2}]^2}{[\ce{SO2}]^2} \times \frac{[\ce{SO3}]^2}{[\ce{O2}]^3} = (K'_\mathrm{c})^{-2} \times K''_\mathrm{c} \tag{5}$$

And thus, your final solution is:

   $$(4.2\times10^{52})^{-2} \times (9.8\times10^{128}) = 5.6\times10^{23}\tag{6}$$


Synthesize the trisubstituted benzene I am trying to synthesize 1-(2-amino-4-chlorophenyl)ethan-1-one from benzene. The answer I came up with is that the benzene must undergo nitration first, then halogenation, then hydration for $\ce{NO2}$ to be $\ce{NH2}$, then acylation. But from the readings I've made acylation cannot happen with $\ce{NH2}$. Is there another way to synthesize this to benzene?

You can synthesize your desired compound from Benzene in the following way.

First, you can synthesize m-dinitrobenzene, and perform selective reduction to have 3-nitroaniline. Now, you can acylate the ring by first protecting the amine group. After acylation you can convert the amide to amine by acidic hydrolysis, diazotize the amine group and perform Sandmeyer's reaction. Then just reduce the remaining nitro group to amine to have your product.

The synthetic scheme is given below,


I think elements can be broken down, but their definition says otherwise, why? Elements are made up of quarks, right?
and physicists have smashed atoms into one another to get quarks, right?

Then if the first two statements are true then elements can be broken down into simpler substances.

If I am right, then can someone give me a better definition?

What definition, where? The short answer is given in the IUPAC Gold Book:

  chemical element

  A species of atoms; all atoms with the same number of protons in the atomic nucleus.
  A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

A longer answer is: Yes, physicists have managed to split atoms further down than to nuclei and electrons. It is also possible to "transmute" an element into another, such as creating Plutonium from Uranium. This involves a lot of machinery and energy, whereas splitting a substance into its constituent elements can be as simple as heating or dissolving in an acid and could be done in an afternoon with a few glass beakers.

Looking at it from a more theoretical perspective, one can (snarkily) define Chemistry as "the Physics concerned with the outer electron shell". Once you work on other areas of the atom (that is, the nucleus) or an electron shell of an atom can no longer be reasonably defined (plasma as in the sun), you tend to leave the area of chemistry (and the applicability of the term "element").


Steps in mechanism of pyruvate dehydrogenase I am having some trouble with certain steps in the mechanism below (from Mike Lu on Proteopedia) detailing the mechanism of action of pyruvate dehydrogenase converting pyruvate to acetyl-CoA. This reaction happens under the physiological conditions present in the human body ($\mathrm{p}H = 7.4, 37^{\circ}C$, 1 atm, various intracellular ionic concentrations: http://www.ld99.com/reference/old/text/2878909-566.html).

In the first step, how does the ylide of TPP form?
Why is the deprotonation of the hydroxyl group to form a ketone in the 6th step favoured, especially given that the $\ce{-OH}$ group normally has a very high $\mathrm{p}K_\mathrm{a}$?

The ylide of thiamine pyrophosphate does not ‘form’; rather it is there. There are two major mesomeric structures for the five-membered ring: One including a double bond from nitrogen to the carbene carbon, the other one with a single bond (lone pair on nitrogen, sextet on the carbene carbon). As with mesomeric forms of benzene, the truth is somewhere between these two extremes.

In the carbene elimination step, you seem to assume just by the existence of the arrows that the deprotonation of the hydroxide must come first. That is not the case. In fact, I would assume the first arrow to be the recreation of the keto unit, followed by the removal of the thiamine. Only then would the proton be lost, since we are now dealing with a protonated ketone which is a very strong acid.