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Friedel-Crafts synthesis - How is HCl produced?
In the mechanism for the acetylation of bromobenzene, how is $\ce{HCl}$ gas produced?
The reactants are acetic anhydride, bromobenzene, $\ce{AlCl3}$ and dichloromethane.
Initially, 4-bromoacetophenone is made, accompanied with (as far as I know) acetic acid and the regeneration of $\ce{AlCl3}$.
How then is $\ce{HCl}$ produced?
To remove the $\ce{AlCl3}$ at the end, 3 $\ce{H2O}$ are added and $\ce{HCl}$ gas is produced then.
I'm trying to think of ways that $\ce{HCl}$ gas is produced prior to that, however. Could a chlorine of the aluminate ion deprotonate the arenium ion?
Could dichloromethane (solvent) donate a $\ce{Cl-}$ to $\ce{AlCl3}$?

I suspect your suggestion is correct. $\ce{[AlCl3(OOCCH3)]-}$ loses a chlorine instead of the acetate when deprotonating arenium ion to form $\ce{HCl}$ as a byproduct

You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where $\ce{HCl}$ comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by $\ce{AlCl3}$, which generates in situ the "acylium" ion ($\ce{H3C-C#O+}$) - the actual acylating agent. The second step is the electrophilic aromatic substitution, in which the electrophile is preferentially "attacked" by the aromatic ring at the ortho- or para-position (ortho- and para-orienting effect of halides). The driving force for the reaction is the last step, which leads to restore the aromaticity and it is assisted by $\ce{(AcO)(Al)Cl2}$ with liberation of $\ce{HCl}$.
Here I have drawn the mechanism:

I have a question..Are you sure $AlCl_3$ doesn't attack on the middle oxygen?


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How to partially hydrolyze triglycerides?
How to partially hydrolyze triglycerides for food application?
I want to make a mixture of mono- and diglycerides starting from triglycerides.

Over the last decades, partial hydrolysis of triglycerides has been examined in various contexts.
Note that this isn't my field of work and that there might hundreds of better and/or more instructive recipes, but a cursory search yielded the following freely available references:
THE DIGESTION IN VITRO OF TRIGLYCERIDES BY PANCREATIC LIPASE,
F. H. Mattson, and L. W. Beck,
J. Biol. Chem., 1955, 214, 115-125 (PDF)
A stereospecific analysis of triglycerides,
H. Brockerhoff, Journal of Lipid Research, 1965, 6, 10-15
(PDF).
The products from lipase-catalysed hydrolysis of bovine milkfat kill
Helicobacter pylori in vitro,
Cynthia Q. Sun, Charmian J. O’Connor, Alastair K. H. MacGibbon, and Anthony M. Roberton, FEMS Immunology & Medical Microbiology, 2007, 49, 235-242
(DOI)
It seems that the digestion of triglycerides by lipases in buffered solutions is a viable approach.

Thanks Klaus for the answer. But wouldn't glycerolysis (glycerol + triglceride in a 2:1 ratio and heat) be a cheaper alternative than using Lipase?

@Denatured You're probably right. Another non-enzymatic approach would be the reaction of fatty acid methyl esters with gylcerol (under alkaline conditions).

A big problem in partial esterification as well as hydrolysis is miscibility. Glycerol does not mix with fats and diesters. If water is present, this becomes even more complicated. The reactions are then two-phase and this can lead to get total hydrolysis along with unreacted fat. Hydrolysis of fats is done with superheated steam in practice, maybe this reaction can be used in an appropriate reactor. (I guess tube reactor and only small conversion) The process must be rather cheap, there is a lot of this partially hydrolysed stuff in fat-reduced food.

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Removing trace amounts of water from agricultural grade anhydrous ammonia
Agricultural grade anhydrous ammonia (liquefied gas) contains 2000 ppm $\ce{H2O}$. What would be the process to remove this water to obtain refrigeration grade (<100 ppm $\ce{H2O}$) anhydrous ammonia?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour]. Is it possible you are talking about ammonia chloride, as 100% pure ammonia would be a gas at room temperature.

Hi Martin:  No, I am talking about drying 20 ton truckloads of anhydrous ammonia. This would make agricultural grade anhydrous ammonia suitable for use in industrial refrigeration systems.  Thanks.

We haul it as a liquefied gas (under pressure).

Ah I see now, thanks for clearing this up. I added this information to your question just to make sure it does not get lost.

Calcium oxide can get water down to 4ppm in anhydrous ammonia according to http://www.google.com/patents/US4075306.

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Why are sp hybridized carbocations high energy?

You don’t see carbocations on double bonds very much, and here's a good reason: compared to sp3, there is more s character in the orbitals, so the empty orbital is held more closely to the nucleus. This is destabilizing if we’re talking about positive charge.

(source)
This doesn't really make sense. Positive charge arises from the lack of electrons. How does holding an empty orbital (regardless of whether that makes any physical sense) closer affect your stability if the empty orbital is ... empty?
So, what is the exact reason sp2 (aren't most carbocations sp2 though?) and sp carbocations are generally less stable than their sp3 counterparts?
Also, how can we have an sp3 carbocation? Huh?!

The only reason I can think of is that it's energetically expensive to make an sp carbocation ... pulling electrons which are held more tightly to the nucleus away is probably harder than pulling away electrons which are not as close to the nucleus.
I can't think of any reason why the end products would be any less/more stable ... if anything an sp carbocation should be more stable than an sp2 carbocation in that in the sp carbocation the electrons are held closer to the nucleus and therefore better stabilizing the nucleus.

Why are sp hybridized carbocations high energy?

Here are links to 2 earlier answers where this question has been addressed for vinylic and aryl carbocations.
link 1
link 2
When vinylic carbocations are generated, the empty orbital (the orbital holding no electrons) is intitially $\ce{sp^2}$ hybridized.  This carbonium ion orbital being $\ce{sp^2}$ hybridized is using some s character, the remaining orbitals around the carbonium ion carbon will therefore have less s character available for use in their bonds.  The more s character in a bond, the lower the energy of the electrons in that bond because s orbitals are lower in energy than p orbitals.
If the vinylic carbocation rehybridizes such that the empty orbital becomes a p orbital, then the remaining orbitals that contain electrons will become lower in energy as they rehybridize and incorporate the s character that was initially used in the empty $\ce{sp^2}$ carbocation orbital.  Once rehybridized the carbocations overall energy has been lowered and the carbocation is more stable.
The high energy of the initially generated $\ce{sp^2}$ hybridized vinylic carbocation is why it is so difficult to generate vinylic carbocations.
Note that some vinylic carbocations cannot rehybridize to an $\ce{sp^2}$ carbocation carbon with a p orbital due to geometrical constraints, and they are even harder to generate.  Examples would be the phenyl carbocation and vinylic carbocations in small rings, like the cyclohexenyl carbocation (see figure below).

Also, how can we have an sp3 carbocation?

They exist, but again, realize that since the empty orbital is not p hybridized they will be higher in energy and harder to generate then carbocations in which the empty orbital is p hybridized.  Examples of carbocations in which the empty orbital is roughly $\ce{sp^3}$ hybdridized would include bridgehead carbocations.

The bridgehead in the bicyclo[2.2.1]heptan-1-ylium cation is only about 11 degrees out of plane. The empty orbital will have negligible s character.

@Martin  Really, only 11° out of plane?  Why then does the corresponding tosylate solvolyze 10^17 times more slowly than bicycle[3.3.3]undecyl-1-tosylate?  In any case, just make the ring system smaller to illustrate my point - right?

I was also very surprised. Since you raised the point I was curious how much it would be like sp3, I expected an out of plane between 30-40 degrees. I honestly did not think about this bridgehead carbons. Decreasing the ring size should work illustrating the problem, like bicyclo[1.1.1]pentane, tetrahedrane or cubane cations. If I have the time I run some calculations and add them to my answer.

I checked the out of plane angle (opa) for methane as an ideal sp3 hybridised molecule and it is only 19.5 deg. So 11.7 deg for bicyclo[2.2.1]... is only little smaller and absolutely reasonable. However, NBO analysis predicts only 1% s character in the empty orbital. In bicyclo[1.1.1]... the opa is 25.5 deg, but NBO says only 7% s character in the empty orbital. In cubane the opa is 27.9 deg, but again NBO only wants 1% s character. Tetrahedrane is impossible to converge and fragments immediately. All this leaves me more confused than before.

I guess it is safe to assume, that in any case, the s orbital will be used to the maximum possible extend, even if that means significantly distorting the geometry, or non-optimal overlap. But we are here on the verge of guessing anyway, since it is not really possible to observe empty orbitals. And then again, there is hyperconjugation and long range effects that play a much bigger role.

Let's just tackle the obvious part first: There is no sp3 carbocation. One reason for this is the same, as for why higher substituted carbocations are more stable than lower substituted. Electrons are more stable in orbitals with high s-character.
In $\ce{\overset{\oplus}{C}(CH3)3}$ the cationic carbon is sp2 hybridised while the empty orbital is 100% p character. All s character is used in occupied orbitals.
In $\ce{H2C=\overset{\oplus}{C}-CH3}$, one would assume from the common depiction, that the central cationic carbon is also (roughly) sp2 hybridised. My first draft of answering did also assume this. A quick DF-BP86/def2-SVP calculation revealed something different. The $\angle(\ce{CCC})$ bond angle comes out to be almost exactly $180^\circ$. The $\pi$ bond is obviously only of p character. The underlying $\sigma$ bonds can now contain the whole s character, making it sp hybridised, and as a result the p orbital of the cationic carbon atom is empty. There will be interactions with some carbon hydrogen bonds. (Hyperconjugation, see below.) This effect is most likely much less than in the prior case, causing it to be less stable. In addition to this, the molecule is much more constrained in its geometry. This linear arrangement is probably also only true for this very simple arrangement. Dependent on the moieties attached to it, a bent structure might be more stable and a different bonding angle will be observed. When that happens, there will be some s character in the empty orbital.
In the benzyl cation, due to symmetry restrictions, $D_\mathrm{2h}$, the empty orbital must be almost exactly sp2. Hence the underlying $\sigma$ bonds will roughly take away 67% s character. The empty orbital will therefore have a very high s character, about 33%, as a result. There is almost no bond widening possible to decrease this and successively increase s character in the $\sigma$ bonds. So the s character found in empty orbitals effectively destabilised the remaining occupied orbitals. (One could also think about it as some kind of ring strain.)
In $\ce{H3C-C#\overset{\oplus}{C}}$, symmetry is even higher, $C_\mathrm{\infty{}v}$ for the triple bond. The underlying $\sigma$ bond will have slightly more s character than in an ideal sp orbital. The empty orbital will, however, be binding a lot of s character. This is effectively destabilising the occupied orbitals again. Since the empty orbital contains the most s character in the orbital, it will also have the most destabilised $\sigma$ system of the series.
In summary, the more s character is in unoccupied orbitals, the less s character can be in occupied orbitals. This is destabilising these orbitals, causing the carbocation also to comparably less stable.
The quote expresses this feature in a quite confusing way.

Note
There are of course other reasons for higher substituted carbocations to be more stable than lower. One of the main effects is hyperconjugation of the $\sigma(\ce{C-H})$ bond with the empty p orbital.

so we can apply Bent's rule even to empty orbitals? @Martin

@Dissenter I believe that Bent's rule serves a different purpose. There might be similarities, but as it is only an observed concept one should not stretch the boundaries of it too far.


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Which solvent is good for using SU-8 particle in Optical tweezers?
I'm studying Optical tweezers. I've used Polystyrene for samples. Nowadays, I've got SU-8 pillar. I wanna use that for sample, but its density is heavier than water. I can't use that. So I wanna know some liquid for replacing water.

I've done some experiments recently with SU-8 and polistirene and figured it out, that SU-8 solvent melts polistirene. Just wanna say, that this won't work out well.
But to use something instead of water: why not using alcohol (ETOH) or isopropyl. I use the last as much as possible since it's not leaving any stains on stuff I do.

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P-phenylenediamine Production Process
How p-phenylenediamine (PPD) is produced? PPD is used as raw material for Kevlar (Para amid Fiber).

React benzene with chlorine in the presence of $\ce{FeCl3}$ to yield chlorobenzene.

According to Ullmann's Encyclopedia of Industrial Chemistry:

Most monochlorobenzene is now produced from benzene and chlorine in continuously operated plants. Depending on the ratio of benzene to chlorine chosen, one can achieve either a low rate of benzene conversion and little dichlorobenzene
formation, or almost complete conversion of the benzene with a higher degree of dichlorobenzene formation. [...] The composition of a chlorination mixture containing the highest possible proportion of monochlorobenzene has been given as
4–5% unreacted benzene, 73% monochlorobenzene, and 22–23% dichlorobenzenes.

Nitrate chlorobenzene to obtain a mixture of 2-nitrochlorobenzene and 4-nitrochlorobenzene. Separate the isomers.

In Ullmann's Encyclopedia of Industrial Chemistry, we find:

Nitration of chlorobenzene with mixed acid (30/56/14) typically gives an isomer
mix in 98 % yield consisting of 34–36% 2-chloronitrobenzene, 63–65 % 4-chloronitrobenzene, and only ca. 1% 3-chloronitrobenzene.

"Mixed acid" refers to a mixture of nitric acid, sulfuric acid and water.

React 4-nitrochlorobenzene with ammonia (Nucleophilic aromatic substitution) to obtain 4-nitroaniline.
According to Ullmann's Encyclopedia of Industrial Chemistry, the reaction conditions as as follows:

4-Nitroaniline is produced in an agitated titanium reactor in 99.3 % yield by reacting 4-chloro nitrobenzene with a tenfold excess of aqueous ammonia at 175 °C and 4.2 MPa (42 bar) for 10 h.

Ullmann's Encyclopedia of Industrial Chemistry cites a patent by MONSANTO (DE1768518) as the relevant reference.

Reduce the nitro group with iron.

Is this the actual industrial practice?

@Georg Yes, I have updated my answer and added some information from the Ullmann.

Mr. Klaus Warzecha, (i).Your answer describes synthesis of para phenylenediamine(PPD) by Reduction of Para Nitroaniline(PNA) with iron.
(ii). Thus obtained PPD contains lots of iron impuries with other isomers, and therefore cannot be used as raw material(Monomer)for production of Para amide fiber(Kevlar).
(iii). For this purpose catalytic Hydrogenation of PNA is carried out to produce PPD of desired quality(Grade).
This is my simple understanding. I am open for edition/addition, please come back.

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Why do certain groups withdraw/ donate electron to double bond or conjugated system in resonance effect?
Why for the Resonance effect certain group of substituents/functional group donate electron from the double bond or to a conjugated system while some withdraw electron from the double bond or to a conjugated system? What causes the groups/substituent to withdraw or donate electron causing +R and -R effect?

Generally, it is observed that if there are lone pair(s) present on the atom of function group directly attached to to the conjugated system, then $+R$ effect is observed.
If the same atom has double bonds or dative bonds with another atom, then $-R$ effect is observed.
Example: –$\ce{OH}$ shows +R due to 2 lone pairs on Oxygen.
— $\ce{NO2}$ shows -R due to one double bonds between Nitrogen and Oxygen, and one dative bond from Nitrogen to another Oxygen.

Withdrawing an electron from a conjugated system is usually favoured when the withdrawing group has a higher electronegativity than the group from which the electron is withdrawn This is also termed as -M effect. For example : double bond O attached to carbon withdraws electrons

resonates to

But an atom or group containing lone pairs may also donate it's lone pair to form a coordinate (dative) bond and here elctronegativity does not matter (if the atom can accomodate another bond (like double bond O does not form a dative bond with carbon) (to help conjugation) This is also termed as +M effect. For example : OH attached to carbon donates it's lone pair to form a coordinate bond

resonates to

some groups show both +M,-M effect (which have both a double bond (to show -M effect) and a lone pair (to show +M effect)), like NH=NH, phenyl, et cetera.

This answer unfortunately suffers from the same issues as the others you recently posted. The concept is called resonance, which has nothing to do with resonating.


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the particle in 1-D box momentum and total energy(P.E + K.E (x))
Employing the particle in 1-D box can momentum and total energy(P.E + K.E (x)) can be measured simultaneously?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

@MARamezani The particle in the box is a classical model in quantum chemistry and found in standard textbooks on physical chemistry by Moore & Hummel or Wedler or .... From the point of view of a slightly more experienced chemist, the matter itself is definitely on-topic. Whether this could/should be closed due to the lack of personal effort is another question.

My bad then @KlausWarzecha. I just told myself that "momentum", "potential energy", "kinetic energy" are more physics, without further digging into the problem.

As this is obviously a homework problem, I'll just give you a hint:
Observables can be measured simultaneously with arbitrarily small uncertainties if their operators commute. Compute the commutator of the momentum and particle-in-a-box Hamiltonian operators; if the commutator is zero, the answer to your question is "yes".


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Writing atoms of functional group backwards

Can the $\ce{HO}$ bond be written as $\ce{OH}$ and still be correct?

They mean the same thing. If the OH group on the left were written as OH, it would look like there was a carbon-hydrogen bond. By inverting the order, it makes it clear that the oxygen is bonded to the carbon and the hydrogen is bonded to the oxygen. $\ce{NH2}$ might also be written as $\ce{H2N}$ in a structure drawing if a bond was coming from the right, but the functional group is the same in both cases—the group is connected to the rest of the molecule by the nitrogen atom.


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In ionic solids such as NaCl, do the cations fill interstital holes in the cation's cubic close pack structure?
Are binary ionic solids like $\ce{NaCl}$ comprised of a cubic close packed structure of the anions with the cations filling interstital holes or do the cations and the anions contribute to the cubic close packed structure equally? I am really struggling with solid state chemistry - any help would be gratefully received. Also any suggestions for good learning resources on this topic would be great. Thanks.

This might be helpful. When we talk about close packed structures, we're really only describing the positions of atoms relative to one another, not their bonding and there are often multiple equivalent ways to think about them, though which ions are involved do of course influence which crystal structures are stable.

The relative radii of the ions, and the stoichiometric ratio of the ions needs to be considered.
For NaCl, the radius of Na+ is 0.695 that of Cl-.
Na+ ions occupy octahedral holes in the cubic close packed lattice of Cl-.
If the radii are more equal (ratio greater than 0.732) then each type of ion will form a simple cubic lattice, with each ion of one lattice filing a cubic hole of the other lattice. For example, CsCl.
If the radii are very different (ratio less than 0.414) the smaller ions will fill tetrahedral holes in the cubic close packed lattice of the larger ions. For example ZnS.
See http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf for more information.


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The relationship between energy and air pressure?
When boiling water at room temperature, what is the relationship energy has with the air pressure that is making the water boil cold instead of hot?

Boiling occurs when the vapor pressure of a liquid reaches ambient pressure.  In the example you asked about, water can be made to boil by removing the air in the chamber in which the vessel holding the water is kept.  When the air pressure gets low enough, there is not enough pressure to keep bubbles of water created by the water's vapor pressure from expanding.  We call this boiling.  At atmospheric pressure the water still exerts vapor pressure but the air keeps the bubbles from expanding.  The only way for water to reach the vapor phase here is for it to evaporate from the surface so we see no bubbles.

Think of it as the air pressure holding in water, preventing the water in the container from reaching a lower energy state. Without pressure, the fastest-moving water molecules fly off, leaving behind slower (i.e. colder) water.
As water boils, disregarding superheating, http://en.wikipedia.org/wiki/Superheating, the liquid stays at the boiling point and energy from the heat source goes into breaking the bonds (e.g. hydrogen bonds) holding the liquid together. If pressure is lowered, the heat in the water can vaporize more of the liquid, removing the fastest molecules, thereby cooling the remaining liquid to the new boiling point.
Removing air pressure is not adding energy to the system, though work is done against atmospheric pressure. If you have a piston between two cylinders, as it moves away from one side, pressure drops, but work must be done to compress the fluid on the other side.


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Equilibrium Concentrations of Products/Reactants
Does increasing the concentration of reactants mean the equilibrium concentration of the products will be greater (than the concentration of the reactants?)

See this post.

Adding a reactant increases the concentration of products at equilibrium, but it doesn't necessarily mean that the product concentrations will be larger than the reactant concentrations at equilibrium.
Suppose you have a reaction $\ce{A <=> B}$. Then
$$K = \frac{[\ce{B}]}{[\ce{A}]}$$
where $[\ce{A}]$ and $[\ce{B}]$ are the concentrations at equilibrium and $K$ is a constant. The ratio $[\ce{B}]/[\ce{A}]$ is constant at equilibrium, so if you add some $\ce{A}$ to the system, some $\ce{A}$ must react and become $\ce{B}$ to maintain the ratio. The equilibrium concentration of the products will increase.
But that increase in product concentration won't necessarily mean that the equilibrium product concentration becomes higher than that for the reactants. If $K$ is bigger than one, the equilibrium concentration of the products will be greater than that for the reactants; if it's less than one, the equilibrium concentration of the reactants is bigger.


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How does hyperconjugation lead to the directing properties of alkyl group?
I have explained how hyperconjugation leads to the directing properties of alkyl groups below.
Carbon joining substituent now need only single bond. So if we remove one bond between $\ce{H-C}$ along with electron from hydrogen we have a free $\ce{H+}$ ion. The bond which was removed formed a bond with carbon joining the substituent(second picture in second row). From the those two electron (on the left of the second picture in second row) we see one come from electron joining substituent(from hydrogen) and other from the carbon atom at ortho position.
How did one bond between Hydrogen and Carbon move to $\ce{C-C}$ along with one electron from hydrogen?

Now we see the electron from hydrogen jumping from ortho to para position to methyl substituent. How does this jump actually happen?
Is it that the electron( from hydrogen of $\ce{CH3}$) on Carbon at ortho jump(Tell me in detail how this jump occur) to meta position due to formation of double bond between the carbon at ortho and meta position. Due to the double bond, electron( from hydrogen of $\ce{CH3}$) on meta came to para position and so on.
How does the double bond between the carbon at ortho and meta position send electron to para position?
The explanation given above is what I think. Is it right?
In my book the following was written:

Hyperconjugation involves delocalization of $\sigma$-electron through the overlap of $\pi$ orbital of double bond with $\sigma$ orbital of the adjacent single bond.

Please explain the above extract from book using the above example of hyperconjugation of toluene.

Why methyl group is 2,4-directing? This question describes in detail why the methyl group is o,p directing.
However, I think the major point of your misunderstanding is in the interpretation of resonance structures. Resonance structures are not stable structures in their own right.
A set of resonance structures can be used to represent bonding that cannot be represented using classic Lewis structures. The structures do not exist on their own but the true structure can be viewed as an 'average' of all the contributing resonance structures.
The electrons do not jump between different atoms but are instead delocalized over all of the atoms with a slightly higher chance of being found at the ortho and para positions in the case of toluene and a slightly lower chance of being found at on the methyl carbon. This gives rise to the increased electron density of the ortho and para positions and therefore the increased negative charge and similarly the decreased electron density on the methyl carbon and the positive charge.

The electrons don't "jump" from one position to another:

when the $\ce{H+}$ leaves the methyl group, a full p-orbital is formed on the carbon. this p orbital is on the same plane of the pi bond of the aromatic ring.
when the p-orbital is on the same plane, the electrons can delocalize (move) along the new formed conjugated pi molecular orbital.
so, you get 8 electrons (2 from the methyl group + 6 from the aromatic ring) on 7 carbons all on the same oriented p-orbital (or just $\pi$ bond). that means you have a total of -1 charge over this bond, so where it will be more localized? (or on which carbons there will be a negative charge?)
the picture you posted are resonance structures (they describe the delocalized electrons). There are 3 resonance structures: 1. minus on the carbon of the methyl group 2. minus on orto carbon 3. minus on para carbon
The electron "moves" from each structure by moving the double bond.
the "minus on the carbon of the methyl group" will contribute to the overall structure the least, because its a primary carbon-anion.

@user13186 When drawing resonance structures, you can only move electrons, you cannot move nuclei.  When you write, "when the H+ leaves the methyl group, a full p-orbital is formed on the carbon", this would change the H-C-H angle from around 109 to around 120 degrees - you have moved nucelei which is not allowed.

@ron the contributing structures for the resonance exist only after the H+ leaves. thus, the carbon (now carbo-anion) in the methyl group is already in SP2 hybridization meaning it has a near 120 degree angle. this form is stable in this hybridization because it has more stable resonance structures where the methyl group is bonded by double bond with the 6-ring.

@user13186  That's not my understanding.  The H+ does nor "leave".  It stays in its exact position, only the electrons in the C-H bond move and spend more of their time around the ortho and para positions in the aromatic ring.  The methyl carbon remains sp3 hybridized.  The "double bond" drawn between the aromatic carbon and the methyl carbon involves p-sp3 pi-like overlap.


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Standard curve of absorption and concentration
I just wanted some quick help with a lab I'm doing, based on absorption spectroscopy.  I'm given an aqueous stock solution of Yellow Dye #5 with a concentration of 54.5 micro moles. I'm asked to create a standard curve.  It gives me the diluted volume as 50 mL, and has a data table with the volume of stock solutions and absorbance values for each.  It asks me to create a standard curve (absorbance versus concentration) in excel for the Yellow Dye #5.  First off, would I just use the constant concentration of 54.5 micro moles for this standard curve graph and the absorbance values in the data table?  I'm not familiar with what a standard curve is, or if the volume of stock solution affects the concentration value in the graph. If anyone replies I'll probably comment on the right answer with a couple more, thanks for the help!  It means a lot!

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Supposed that Yellow Dye #5 refers to the FD&C numbering, this is a food colorant more known as tartrazine outside the U.S.
A standard curve correlates absorbance at a particular wavelength with the concentration of the dye.
Supposed that no distortion (e.g. due to aggregation effects) is present, the relation is given by the Lambert-Beer law: $$E_\lambda = \epsilon_\lambda \cdot c \cdot d$$
where $E_\lambda$ is the absorbance, $\epsilon_\lambda$ is the molar absorption coefficient (specific for a particular dye at a particular wavelength $\lambda$ in a certain solvent), $c$ is the concentration of the dye and $d$ is the path length of the cuvette. Please pay attention to the units in your calculations!
In order to create a standard curve,

Record the uv spectrum of the stock solution. If you have two-channel spectrometer, put a second cuvette with the neat solvent in that channel.
Get a couple of volumetric flasks and dilute the stock solution.
Record the uv spectra for each of these solutions.
Plot absorbance (probably at around 420 nm) vs concentration.

Perfect thank you!  And just to verify, the concentration of the tartrazine given is a single value, meaning when I plot the standard curve of absorbance vs concentration I will have a straight, flat line since concentration doesn't change?

@soccer66 When you dilute by taking 1 mL of the stock solution and fill it up to 10 mL, 25 mL or 100 mL in volumetric flasks, the concentration does change! That's the idea behind the dilution ;) For each of these solutions, you'll measure a different absorbance. Plotting these data vs concentration will give a straight line, but not a flat one.

Oh!  That makes sense!  Should I just use the idea of initial concentration times initial volume equals final concentration times final volume to calculate the concentration, where the final volume is the amount of dye in the dilute mixture?  And thanks again for giving me all of this extra help!

Think practical ;) If the assignment does not require a particular concentration range, measure the stock solution first and decide how low you can go with the concentration before the uv spectra start to look crappy. For the lower concentration range, remember that diluted solutions can be diluted again ;) Don't use all of the stock to prepare samples of very low concentrations, you cannot "undilute" them ;)

And, as a rule of thumb, try to get your absorbance to be between 0.1 and 1.


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Actinides and Lanthanides group
I would like to ask which group do actinide and lanthanide belong?
I had a question in my assingment that which group has most number of elements?
if they belong to 3rd group then the 3rd group should be the answer.
Am I wrong or Right? Please Help Me.

They technically do not belong to any column, or they belong to their own 2-element columns.

The periodic table actually looks like this:

...you can see that the first 14 elements in either series don't belong to any group from the main body of the table.
You could say that the lanthanides and actinides form fourteen new 2-element groups, with Lu and Lr belonging to group 3. But IUPAC does not number these groups:


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Chirality on Propranolol
I'm a bit confused when asked to specify the chiral carbons on this sketch of Propranolol as I'm not sure how the $\ce{H3C}$ and $\ce{CH3}$ play in if they are reversed in the name like that...Does that even matter?
I'm pretty sure the only chiral carbon is the one in the middle of those and the $\ce{NH}$. I've heard that you need 4 different groups/atoms surrounding the carbon for it to be chiral but how much truth is there in that, as I've found several with only 3?

Note that the nitrogen is not chiral due to amine inversion.

As stated in user208322's answer, $\ce{CH3}$ and $\ce{H3C}$ are exactly the same, they're only written differently for ease of reading. As per your question about how much truth there is to having 4 groups, it's 100% truth. You cannot have a chiral carbon that doesn't have 4 groups attached to it (i.e. must be sp3 hybridized). If you think you've found a chiral carbon that only has 3 groups attached to it, you're probably forgetting the unwritten hydrogen.
In this example specifically, there is only one chiral carbon, the one attached to the secondary alcohol. Although it may look like the carbon is attached to the same group twice, you need to look further than just the first atom. Yes it is attached to two carbons, but one of those carbons is attached to a nitrogen and the other is attached to an oxygen, so the groups are different. Two groups are  only the same if you can follow it down every single atom and they match exactly.
Edit: In regards to your question about following the chain all the way down, realistically you only need to follow it until you see that it differs from the other chains. So for chains 1 and 3 once you get to the nitrogen and oxygen atoms you see that they're different and there's no point continuing. And the unwritten hydrogen has to do with the fact that organic chemists use shorthand in their molecular structures. Basically, carbons and hydrogens are extremely common in organic molecules so they just don't write them. All the points are carbons unless otherwise labelled, and they must have 4 bonds, so where you see that the carbon is only bonded to 3 things, it is also bonded to a hydrogen as well. See this wikipedia article for more info (specifically, this section).

When you say "follow the carbon chain all the way" do  you mean follow it until you see another atom other than carbon? Cause if I follow the carbon chain all the way from your specified atom I see 3 chains:
1 goes through: carbon, NH, carbon, CH3 or H3C.
2 goes through OH
3 goes through carbon, Oxygen, carbon, carbon, carbon carbon etc.
Also there is a 4th there (has to be)? An unwritten hydrogen? I can't seem to visualize that.

I.e. I thought that they had to be attached group like directly attached to the carbon. Not 4-5 chains away.

@BjarniJóhannsson I've added more info, but think of it this way: that whole $\ce{CH2-NH-CH-(CH3)2}$ chain is one big group, and the chain with the oxygen and benzene rings is another group. When you look at it that way, they're clearly different, and should be treated as such.

@BjarniJóhannsson, chirality is a gnarly beast at first. I strongly recommend you do lots of examples and read up on it to get comfortable with it. It'll come up a lot if you decide to continue with chemistry later on.

@wes3449  You wrote, "You cannot have a chiral carbon that doesn't have 4 groups attached to it (i.e. must be sp3 hybridized)."  That is not true, consider 1,3-dichloroallene.  It is a chiral molecule, none of the carbons are sp3 hybridized and none of the carbons have 4 different substituents attached.  Conversely, having 4 different groups attached to a carbon does not guarantee chirality, consider the example of certain meso compounds.

@ron It is true. You cannot have an sp2 chiral carbon. 1,3-dochoroallene contains no chiral carbons, though it is a chiral molecule. In my mind that is outside the scope of this question and would only serve to confuse the OP.

$\ce{CH3}$ and $\ce{H3C}$ mean the same thing — methyl — But are written that way to show the C is attached directly. It is obvious, but people prefer writing it that way to show attachment.
And I can see only one chiral atom. These are are such that none of the groups directly attached are same.
Example: $\ce{CH3-CH(OH)-C2H5}$  (the $\ce C_2$ is the chiral carbon).

I don't understand your example. What do you mean attacked? Also am I right in my answer?

Your answer is correct.

@Bjarni Jóhannsson No, your not - chiral is atom connected to OH group. Attach means connect.

I'm confused here, the Carbon connected to the OH group has 2 of the same group connected along with the OH group so how can it be chiral?  Could you draw which one you mean to be correct if mine isn't?

@BjarniJóhannsson, I've updated my answer to explain this.

Bjarni you're talking about your question's molecule? If yes, then just write the groups attached to the carbon, which are :1.OH. 2. CH2-O-.... 3.H 4. CH2NH.....

You must compare the whole group not just the atoms attached directly. Even though, CH2 is same for both but both have NH2 and O different. Therefore , both are different group


text:
Is Pauling electronegativity well-defined?
Pauling's definition of electronegativity defines the difference of electronegativity between two elements.
Suppose three elements A, B, and C. Then we can calculate $ϰ_a - ϰ_b$, $ϰ_b - ϰ_c$, and $ϰ_c - ϰ_a$ independently. Is the sum of the three values always zero? Why?

can you explain the symbols? What is "$ϰ$" in $ϰ_a - ϰ_b$?

If you have a single, unique value of $\chi_i$ for each element $i$ then yes, the sum of the three values is always zero.
If you are assigning electronegativities that differ with oxidation state or chemical environment, the answer can be no (for example, if element B has a different electronegativity in an AB bond than it has in a BC bond).
The Pauling electronegativity difference is
$${\chi}_{a}-{\chi}_{b}=(1\ eV)^{-\frac{1}{2}}\,\sqrt{{E}_{d}\left( a,b\right) -\frac{{E}_{d}\left( a,a\right) +{E}_{d}\left( b,b\right) }{2}}$$
where $E_d(i,j)$ is the dissociation energy for a single bond between element $i$ and element $j$.
The definition assumes that you can consistently and accurately say that the square of the electronegativity difference between A and B is equal to the difference between the dissociation energy for AB and the average of the dissociation energies for AA and BB. If this isn't consistently true for each combination of three elements A, B, and C, your sum will be nonzero. Pauling fit his electronegativity values to the data to average out these inconsistencies.


text:
Spin spin coupling in a proton NMR of an ester?
I am learning about proton NMR and spin-spin coupling, and am confused about whether splitting occurs over an ester bond. Specifically, in the case of ethyl methanoate, HCOOCH2CH3, if I were to number the proton environments going from left to right as 1,2 and 3, would 1 and 2 cause splitting of each other?
If they would not cause splitting, then I would expect a singlet, a quartet and a triplet respectively. If they do cause splitting, then would you get a triplet, a quartet of doublets, and a triplet for 1,2 and 3 respectively? I know that with -OH and -NH the protons do not couple (although I am not sure why), so I was wondering whether the same principle may perhaps apply here...

In the most common case you can only observe 3J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds.
In this case the spins are too far away from each other to cause a visible splitting, you are correct that it would result in a singlet a quartet and a triplet.
In the case of OH and NH protons, chemical exchange can play a role. This is a bit more complicated, but the simple version is that protons that can exchange quickly with the solvent are often not observable in NMR (or just as a very broad signal) and due to that also don't couple. Though in the most common cases the OH proton is usually too far away from the other protons in the molecule to show a coupling anyway.

Thank you for your reply! I just looked up 3 J coupling as I was unfamiliar with this, and from what I understand in most cases, although there will be exceptions, you will only get coupling if the protons are on adjacent carbon atoms. Ketones are another homologous series I often deal with in class. Does this mean that there will also be no coupling of the protons on the carbons on either side of the carbon with the =O ?

@21joanna12 Just count the number of bonds between the two protons, if it is larger than 3 you won't see a coupling in most cases.


text:
What is the maximum extent of a hydride shift to form the most stable carbocation?

What is the maximum extent of a hydride shift to form the most stable carbocation?

If we consider the reaction between 2-methylpropan-1-ol and $\ce{HBr}$, then of course, the carbocation formed after a hydride shift will be more stable (and lead to the major product). 3-methylbutan-1-ol will also undergo hydride shift to some extent to form the more stable carbocation, but is there a limit to it?
Like can the number of carbons be so much that there is almost no chance to make the product with hydride shift possible? Like consider the scenario below:

What can be the maximum $n$ possible here to make the product with hydride shift possible?

Advances in Physical Organic Chemistry, volume 19, discusses the relative transition state barriers for 1,2-, 1,3-,1-4-, and 1,5-hydride shifts.
The specific cations considered are 2,3 dimethyl-2-butyl, 2,4 dimethyl-2-pentyl, 2,5 dimethyl-2-hexyl and 2,5 dimethyl-2-heptyl.
Interestingly, they report that whereas the barrier for 1,4 shift in 2,5 dimethyl-2-hexyl is 12 kcal/mole, this decreases dramatically to less than 6 kcal/mole for 1,5 shift in 2,6 dimethyl-2-heptyl
The reference draws the transition states as rings, with the H being transferred as a bridging member of the ring, a five membered ring for the 1,4 shift and a six member ring for the 1,5 shift.
The reference also makes a comparison to intermolecular hydride shift between isobutane and t-butyl cation, which has a barrier of 3.6 kcal/mole.
While the reference does not discuss the possibility of longer distance than 1,5 shifts, it seems that because the ends of a long ion can meet each other, long distance (in terms of number of bonds) shifts are possible because through space the distance is short.

There could be multiple 1,2 shifts as well, as long as it is downhill or close to thermoneutral, right?

Yes, like this http://butane.chem.uiuc.edu/jsmoore/chem232/notes_current/Carbocation_RARs/NOTES-12R-An_Elementary_Step_for_Carbocation_Rearrangements.pdf

I don't know whether I should accept this answer. I think it is a partial answer, but I also think that it is the best answer one could come up with. What should I do?

And by the way, the link you have just given says that the shifted hydride doesn't directly travel to the carbon, it goes step by step. So the main reason that there must be a limit is that if the carbon chain is too long or the barrier energy is very less. But can it be too drastic that the probability of its formation is nearly zero?

Yes, the link in the comment says it "usually" doesn't occur in the way the Advances in Physical Organic Chemistry review article is describing.  Advances in Physical Organic Chemistry draws the transition state as the ends of the molecule approaching each other to form a ring. Somebody could definitely write a better answer, going through the literature trying to determine to what extent the one transition state mechanism occurs vs. the step by step mechanism.  I personally think as distance increases the one-step mechanism would be increasingly favored.

authors: นายกิตติพงษ์ ติยารัชกุล


text:
Periodic table for printing
I there a Good Version of a printable Periodic table Which is minimalistic and gives necessary information only such as -

At. Number
Element Name
At. weight
Shell configuration
color coding of various groups or Series

Whatever version you wind up finding, I would strongly suggest that you keep good track of the image file you wind up printing.  I found a really nice periodic table that I like to give to my students that includes necessary information for my courses as well as a small chart of common polyatomic ions and fundamental constants.  I've been copying the thing for years now because I can't find the original file anywhere, and the print quality is starting to suffer.

The IUPAC periodic table (2016-pdf) is good because it quantifies the accuracy of the natural abundance atomic weights, but it doesn't give electron configuration.
The NIST periodic table has electron configuration, and probably is closest to what you want.

The NIST table is also good from a copyright point of view, because you can reproduce it without worrying about copyright infrigment, as US government works are generally not subject to copyright protection.

These are both great tables; I've given my students the NIST one for a few years now. Maybe you can suggest a table that's good to include on genchem tests (American group numbers + IUPAC numbers, element name, element symbol, atomic weight, atomic number, and nothing else). I use the California Dept. of Education table (http://www.sciencegeek.net/tables/CA_CST.pdf), but it's copyrighted and it's a bit out of date.

My original answser had the old version of the Virginia periodic table http://www.doe.virginia.gov/testing/test_administration/ancilliary_materials/science/2003/2003_sol_periodic_table.pdf, because it has electron configuration (but I deleted becasue it's out of date and has wrong/less-likely configuration for Lawrencium). The newer Virginia table is very similar to California http://www.doe.virginia.gov/testing/test_administration/ancilliary_materials/science/2010/2010_sol_periodic_table.pdf (no electron configurations).

Any post-1977 work is copyrighted in USA, even if it doesn't say so, except federal works.

The tables at Ptable.com might well fill the bill. Here is an example:

FWIW, this is the table I use as the background for my lab computer.

Ptable.com was already suggested, but this PDF fulfills all your requirements, and further, it has features that make it ideal for printing such as:

Customizable information using PDF layers (supported in Acrobat)
Low ink usage, outlines only
Category information retained even if printed in black and white

http://www.ptable.com/print/


text:
Why does the opening of an epoxide occur via an Sn2 like mechanism when using a methanoate ion as a nucleophile and methanol as a solvent?
Below is a picture of the mechanism:

I understand the mechanism but what I don't understand is why it's $\mathrm{S_N2}$. The solvent used is polar and protic, there is a tertiary carbon that would be suitable for $\mathrm{S_N1}$ as well. For what reason does this reaction occur exclusively via $\mathrm{S_N2}$?

unfavourable leaving group

Its exactly the same with either mechanism

unfavorable for Sn1

Quotation from Clayden:

Without protonation, the epoxide oxygen is a poor leaving group, and leaves only if ‘pushed’
  by a strong nucleophile: the reaction becomes pure SN2. Steric hindrance becomes the controlling
  factor and methoxide attacks only the primary end of the epoxide.

The nucleophile isn't strong in this solvent though is it?

I'm afraid you're strongly overestimating effects of solvation.

Does this mean only in the presence of a strong acid (protonation of oxygen), in SN2, the nucleophile will attack the more substituted side? Otherwise (neutral/basic conditions), it will attack the less substituted side as normal SN2 does? [Due to the hydroxyl group is a better leaving group (more stable than its conjuagte base)]?

@234ff Things are more complicated when the epoxide gets protonated. The mechanism is somewhere between SN1 and SN2. Often the SN1 nature prevails and the more susbtituted end gets attacked, but there are numerous examples where the attack occurs on the less substituted. This depends on the nature of the nucleophile as well as sterics of the epoxide.


text:
How do I know which atoms react during hydrolysis?
In the hydrolysis of propanamide ($\ce{CH3CH2CONH2}$) how do I know which molecules react with the water?
$$
\ce{CH3CH2CONH2 + H2O -> ?}
$$
I don't know how to split the molecule. In the videos I've watched (example) it seems as they just assume you know this.

I think you mean which atoms react with the water, as clearly the molecule that reacts is propanamide (it's the only non-water molecule in your question).
The product of the reaction is propanoic acid and ammonia.
$$\ce{CH3CH2CONH2 + H2O -> CH3CH2COOH + NH3}$$
The carbonyl carbon in amides (and esters and acyl halides) has a partial positive charge because of the electron withdrawing effect of the carbonyl oxygen.  (Oxygen is very electronegative.)  This makes the carbon atom is activated toward nucleophilic attack.
This Wikipedia page has a good explanation and different mechanisms for acid vs. base conditions.


text:
Why 6.023x10^23 is not written as 6023x10^20?
We know that 1 mole is equal to $6.023\times10^{23}$ atoms. But why it is not written as $6023\times10^{20}$ ?

How about we write it $ 602300000000000000000000 \times 10^0 $ instead?

For the same reason for which people write $2.015 \times 10^3$ instead of 2015. :|

@MichaelKjörling how did you count the zeros in what you've written? :D

Dissenter's answer is not quite right, since, strictly speaking, as defined there, the scientific notation will work for positive numbers only. Obviously, it can be easily fixed to work for negative as well, but the case of 0 will continue to be problematic. Just try to follow this rule to understand why.
All in all, it is the matter of definitions, but usually, scientific notation is defined as just a way of writing numbers in the form $a \cdot 10^b$ without any restrictions on $a$ which can be any real number. So, both $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are considered to be numbers written in scientific notation.
However, in the normalized scientific notation, indeed,

the exponent $b$ is chosen so that the absolute value of $a$ remains at
  least one but less than ten ($1 ≤ |a| < 10$),

and thus, only $6.023 \cdot 10^{23}$ is written in this notation. Normalized scientific notation  is just one special case of scientific notation. And yeah, you could not write down zero in this notation, which is kind of awkward.
Another special case of scientific notation is the so-called engineering notation in which the exponent $b$ is restricted to multiples of 3. Neither $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are written in engineering notation, while $602.3 \cdot 10^{21}$ is. Although engineering notation is a special case of scientific one as defined above, is rarely called scientific notation to avoid an ambiguity.

Looks like Dissenter has updated the answer to account for the comment in your first paragraph, making that obsolete. You may want to revise.

You could write it as either. But the former is scientific notation while the latter is not considered scientific notation.

In scientific notation, you move the decimal place until you have a
  number [whose absolute value is] between 1 and 10. Then you add a power of ten that tells how
  many places you moved the decimal.

http://www.factmonster.com/ipka/A0876783.html

To add to all the other answers, in case you're wondering, the reason we like to use the one standard for scientific notation is because it makes comparing sizes very easy. A lot of times in making approximations you only care about the order of magnitude (this is especially true when looking at things like equilibrium constants). Having this standard essentially allows you to compare multiple values/quantities very quickly by disregarding the coefficients and just having a quick glance at their orders of magnitude. If we all used different standards, we'd have to take into account how many digits are before the decimal place.

There are a couple of advantages to using normalized scientific notation (a number less than ten and greater than or equal to one times a power of ten):

Counting significant figures is easier.  All of the digits written in the number before the power of ten are significant when you're using normalized scientific notation.
Writing it other ways isn't incorrect, but it's not as convenient. For example, if you write $1.000\times 10^3$ it's absolutely clear that you've got 4 significant figures, but if you write $1000\times 10^0$ it's not so clear; you can only say "I have at least one significant digit".
Comparing orders of magnitude is easier. As other answers here note, you don't have to move the decimal point around to compare the powers of ten for a pair of numbers if you're using normalized scientific notation.

If you let the magnitude of the coefficient / significand vary outside the [1,10) interval, then you have to count digits and add that to the exponent to see the magnitude.
With normalized scientific notation, you can see the magnitude at a glance.  This is WHY normalized notation is standard practice.


text:
Does hyperconjugation and induction always have the same effect?
My professor does not believe in the hyperconjugative effect. Instead, we only consider inductive effects.
So do both effects have the same general effect? I know for methyl groups both have the same effect; both hyperconjugation and induction can be invoked to explain the electron donating tendencies of the methyl group.
Are there any cases in which the hyperconjugative effect and the inductive effect have opposite effects?

The inductive effect occurs through overlapping sigma bonds, whereas the hyperconjugative effect occurs through overlap of a sigma bond with a p orbital.

There are many different types of hyperconjugation and hyperconjugation has been used to explain many different chemical phenomenon.  A nice overview can be found here.

Are there any cases in which the hyperconjugative effect and the
  inductive effect have opposite effects?

Since inductive effects are transmitted through the sigma system, they effect should be independent of rotation around the bond(s) involved.  On the other hand, the pi type of overlap used in hyperconjugation is very dependent upon rotation about the bond(s) involved.  In the diagram above, rotation about the $\ce{C-C}$ bond by 90° shouldn't affect any inductive effect of the methyl group upon the carbocation center, but such rotation would completely destroy the overlap required for hyperconjugation to occur.
As you might expect, hydrogen/deuterium isotope effects can be used to assess hyperconjugation $\ce{C-H}$
[Note: isotope effects are a subject unto themselves; a nice overview can be found here.  See p. 12 for their application to hyperconjugation.]
Here is a "thought experiment" where inductive effects and hyperconjugation should have different effects (I don't know if this or a similar experiment has been done).  Take the following two carbocations.

One would expect them to both have similar inductive effects upon the carbocation center.  The tris-ethyl carbocation, free to rotate about its various $\ce{C-C}$ bonds could also exist in conformations where the $\ce{C-H}$ bonds alpha to the carbocation center are correctly positioned for hyperconjugation to occur - it should show a secondary $\ce{K_{H}/K_{D} >1}$ since hydrogen is more electron donating than deuterium (see here for the reasoning behind this statement).  However, the $\ce{C-H}$ bonds alpha to the carbocation center in the rigid bicycloctyl carbocation cannot adopt a conformation that would allow effective hyperconjugative overlap of these $\ce{C-H}$ bonds with the cationic center - it should show a secondary $\ce{K_{H}/K_{D} =1}$.  Different secondary isotope effects would be expected in the solvolysis of these two systems.
Another case where inductive effects and hyperconjugation do not provide equivalent explanations relates to rotational barriers.  The origin of the rotational barrier in ethane has been a subject of investigation for a long time.  Inductive effects cannot be used to explain the barrier because, as discussed above, inductive effects are rotationally invariant.  An argument using hyperconjugation to explain the rotational barrier has been advanced and is the subject of much discussion (see the "Rotational barrier of ethane" section).  Whether the hyperconjugation argument is correct remains to be proven, but it is another case where hyperconjugation can be used to explain observations that the inductive effect cannot.

Nice answer! Unfortunately, the first two links to Princeton University's website, in your answer aren't working. Also, these pages were not archived in the Wayback Machine (1, 2). So, I guess you might want to replace them with an alternate link. Thanks!

The effect is similar to the positive mesomeric, it appears by substituting the hydrogen in the alkyl group of multiple bond. This effect is directed toward multiple bond:

Hyperconjugation reminds positive mesomeric effect, because it gives sigma-electrons in the conjugated pi-system:

Inductive effect and hyperconjugation usually occur simultaneously. Inductive effect will undoubtedly outweigh, when the molecule is easily polarized. In addition, the spatial effect may mask the hyperconjugation effect.
In benzene homologues hyperconjugation enhances the inductive effect.

I'm afraid Dissenter knows well how hyperconiugation works and asks about unusual cases.

What however is the spatial effect?

@Dissenter spacing effect


text:
Why isn't hybridization done in America?
Warning: This is a question from my teacher.
In America hybridization isn't done, while in Europe it is done around the 10th grade. Is there any reason or advantage for doing(not doing) it as part of the school program?

you mean orbital hybridization like sp, sp2, sp3? Yes, this is taught in America.

Hybridization is taught in American high schools and colleges. I learned hybridization first in 10th grade. I think the main issue is that the quality of education varies greatly across America. Good schools will almost definitely teach hybridization in HS. After all, it's tested on the AP Chemistry Exam (a HS examination students can elect to take).

@MikhailTal You can't generalize the entire USA high school system.  Each state can be different.  Here is the Arkansas chemistry curriculum: http://www.arkansased.org/public/userfiles/Learning_Services/Curriculum%20and%20Instruction/Frameworks/Science/chemistry_9_12_2005_060508.pdf Hybridization is a required part of the curriculum.

AP student here, yes hybridization is taught.

In Russia hybridization of carbon learns at the beginning of the 10th grade, before the main organic chemistry. Hybridization in the atoms of metals (complexes) are studied in grade 11. These two topics are difficult for students to understand. But they can be on the exams.
Advantages:

It is easier to understand the spatial structure
Explanation of coordination numbers in complexes
For an explanation of the facts, when the atom forms a large number of bonds than the number of unpaired electrons in the ground state

Disadvantages:

Difficult to understand
There are many different theories
Rarely used in the study follow-up program

I don't see how this is at all relevant to the question, it really doesn't answer it...

OPs asking about the US and Europe. If you have an idea about one country, wouldn't it be better if it was a comment? (I may be overreacting!)

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post.

@MARamezani what's wrong with this answer? It's just fine. Russia is in Europe.

Indeed, but Russia isn't the whole Europe, is it? (The problem is with the question itself, honestly)

@Michael, I'd just like to point out that the initial question (when the comments were made) had much less information. The problem with it was that all it said was that they teach it in russia, which was not at all what the question was asking. Now that he has added advantages and disadvantages it is more suitable.


text:
What are solvolytic displacement reactions?
High school student here, I recently purchased a book titled: "Solvolytic Displacement Reactions" by Andrew Streitwieser published in the early 1960's. I found it at a book store and it looks like one of those incredibly valuable (in terms of knowledge) books. I tried reading the first few pages (have a basic background in chemistry) but I do not see "the big picture". What are solvolytic displacement reactions, how are they used and what are they used for?

A book could be written on this subject!
A solvolysis reaction is an $\ce{S_{N}1}$ reaction where the solvent acts as the nucleophile and replaces the leaving group.  The following reaction is a solvolysis reaction run in ethanol.

I suspect there are some industrial uses of solvolysis reactions and they have been used in labs and by drug companies as a step in a synthetic sequence to prepare some complex molecules.
But I suspect the biggest contribution from the study of solvolysis reactions is the impact they have had on our understanding of carbocation chemistry.  From our study of solvolysis reactions we have learned how and why some carbocations rearrange before being captured by solvent.

While studying such rearrangements the solvolysis of the norbornyl system was examined.

It turns out that

norbornyl compound 1 solvolyzes several hundred times faster than compound 3 - why?
if compound 1 is prepared optically active, the products are racemic - why?
If compound 1 is radiolabeled, the label is scrambled in the products - why

It took decades to answer these and other questions.
These investigations led to a whole new field of carbocation chemistry: the classical - non-classical carbocation debate (see the norbornyl link above for a discussion of this topic).  These investigations added a whole new section to physical organic chemistry textbooks.  What was learned from these solvolysis studies completely changed our thinking about carbocation chemistry.


text:
Adding lemon to lentils changes color from black to pink, why?
I noticed that when I cook black lentils and add lemon juice to the water that's left, the water turns from black to pink!
What can be the reason for that?

In the black lentils used in Indian cooking (Vigna mungo), the natural water-soluble dye that causes the color change has been identified as quercitin-3-O-pentoside:

which belongs to a class of compounds called flavonols. When you pop hydrogen ions on and off some of the -OH groups, you can get a pure benzenoid form (colorless), a quinonoid form (pink), and an extensive quinonoid form (yellow). The color changes
are pH dependent, and you get pale pink at pH < 7, yellow at pH > 7, and
colorless at pH 7. When you add your lemon juice to the lentil water, you're getting the quinonoid form.
Other species of black lentils probably do something similar.
The pH-dependent color change occurs because when you have $-\mathrm{OH}$ groups on the rings instead of $-\mathrm{O}^-$ groups, the way electrons are confined changes; that changes the spacing between the molecule's energy levels, and changes the colors of light the molecule can absorb. (See my article here for more about how this works).
Many plants and household materials contain molecules that can change color or otherwise indicate changes in pH (See my article here for a list of some of them). Almost any flower, fruit, or plant part that is red, blue, or purple contains a class of chemical compounds called anthocyanins that change color with pH. The color of a flower or fruit depends on which anthocyanins are present, the pH of the pigment-bearing tissues, and the presence of other pigments, like yellow flavones.

Nice article about household indicators.  Should update the ExLax part since phenolphthalein was banned from over-the-counter laxatives in 1999.  http://www.fda.gov/ohrms/dockets/98fr/012999d.txt

Thanks, DavePhD---I didn't know that. I'll fix it; that site is long overdue for an update and I'm going to launch a complete makeover this summer. That particular article was written in 1998.

Black lentils contains a natural pH indicator. These dyes change their colour depending on how acidic (or alkaline) the solution is. The dyes are water-soluble, consequently you can observe the colour change in the remaining water too.
The natural dyes, which often belong to the class of anthocyanines, are found in other food too, such as in red cabbage.


text:
Non-Newtonian substance - Tahini
Raw Tahini seems to me like a non-Newtonian substance. When I mix it fast, it gets tougher. When I mix it slowly, it gets softer.
Is there any truth to that?

Yes, tahini (and cornstarch-water mixtures) are dilatant, non-Newtonian fluids, with viscosity increasing with shear strain. See http://en.wikipedia.org/wiki/Dilatant for definition and https://www.youtube.com/watch?v=D-wxnID2q4A, https://www.youtube.com/watch?v=BleCJJAKkgw for demo.
Tahini is much better in hummus than in a swimming pool, though. ;-)


text:
Liquid to solid when exposed to air
Is there a liquid mixture that will become a solid when exposed to air/oxygen. Preferably a spongy material that is difficult to tear.

Urethane foam from isocyanates, commonly sold at hardware stores for insulation.
Click links for general information and for its chemistry.


text:
How are atoms manipulated?
I was wondering how scientists are able to handle atoms. They are very small, but surely people are able to interact with them somehow? The Large Hadron Collider is one example. Also, they try and make new elements by mixing atoms together. How are they able to handle something so small?

In general, one usually doesn't work directly with single atoms, though there are ways to do that as well, usually involving scanning probe microscopy (basically poking at atoms with very sharp probes to either manipulate them or just image them).
Atoms become much easier to work with when they are are ionized and can be manipulated by electric or magnetic fields (atoms have neutral charge so the Lorentz force can't be used to control them). Even then, most techniques deal with small groups of ions rather than individual ones. Ions are relatively easy to manipulate and we can easily trap them, accelerate them into beams, and focus them. The LHC is, of course, a very impressive (and expensive) example of this. The way it works is that packets of ions are accelerated by electric and magnetic fields to form two ion beams, which are then focused into a very small area and made to intersect, allowing the ions to collide and all the cool particle physics stuff to happen.
Manipulation of ions is also used in countless other scientific applications: semiconductor manufacturing, surface analysis, medical analysis, radioisotope dating, to name a few.


text:
Does NaOH react with K to form NaOK or KOH?
I realized that the $\ce{-OH}$ group when it reacts with some metal like $\ce{Na}$ or $\ce{K}$ will replace the $\ce{H}$ element to form $\ce{-OK}$ or $\ce{-ONa}$:
\begin{align}
    \ce{H2O + Na &-> NaOH + H2}\\
    \ce{CH3OH + Na &-> CH3ONa + H2}\\
    \ce{H2O + K &-> KOH + H2}\\
\end{align}
So I am just wondering is that always true? Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$? Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?

The question: Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?
This is a redox reaction regardless of what species $\ce{K}$ would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).
Answer to situation (a) is potassium would essentially react with water (other ions in the solution are $\ce{Na+}$ and $\ce{OH-}$) based on standard potentials of following possible half-reactions.
$$
\begin{align}
\ce{K+ + e- &<=> K} &E^\circ &=\pu{-2.931V}&\\
\ce{Na+ + e- &<=> Na} &E^\circ &= \pu{-2.71V}&\\
\ce{2H2O + 2e- &<=> H2 + 2OH-} &E^\circ &=\pu{-0.8277V}&\\
\end{align}$$
Therefore, the only ionic species in the solution would be $(\ce{Na+ + K+ + OH-})$ in $\ce{H2O}$.
Answer to situation (b) is a little complicated than this. Nowadays, in electrochemical research, use of molten $\ce{NaOH}$ is the choice of the electrolyte. The choice of molten $\ce{NaOH}$ over molten $\ce{Na2CO3}$ is based on a number of advantages of the molten hydroxide electrolyte over others. The first advantage is its higher electrical (ionic) conductivity, which is important in the question at hand. The ionic conductivity of molten $\ce{NaOH}$ exceeds that of molten sodium carbonate by a factor of 1.5 [1]. Thus, it is safe to assume that molten $\ce{NaOH}$ functions as $\ce{Na+}$ and $\ce{OH-}$ ions in molten state. Additionally, $\ce{Ni}$ has already been shown to be a chemically stable material in a molten $\ce{NaOH}$ environment, and hence, nickel containers are used to produce anhydrous $\ce{NaOH}$. Concentrating an aqueous $\ce{NaOH}$ solution in $\ce{Ni}$ containers by heating up to $\pu{500 ^\circ C}$ will produce the desired $\ce{NaOH}$ melt.
According to [1], molecular oxygen is extremely oxidizing in fused hydroxides and reacts with hydroxide ions to produce superoxide and peroxide ions. The concentration of these ions is determined by the following equilibria (Note also that increasing the water vapor pressure would reduce the peroxide and superoxide ion content by favoring the equilibria to LHS, but it's not our concern here):
$$
\begin{align}
\ce{O2 + 4OH- &<=> 2O_2^{2-} &+& 2H2O }\\
\ce{3O2 + 4OH- &<=> 4O_2^{-} &+& 2H2O }\\
\ce{2OH- &<=> 4O^{2-} &+& H2O}
\end{align}$$
For example, see following Abstract:

A thermodynamical discussion of the $\ce{Pt/O2}$ electrode in molten $\ce{NaOH}$ shows that the redox potential depends on $p_\mathrm{O_2}$, $p_\mathrm{H_2O}$, $a_\mathrm{O_2^{2-}}$ and $a_\mathrm{O^{2-}}$. Potential measurements have shown that the reaction $\ce{O2 + 2e− = O2^{2−}}$ is potential-controlling. From the electrode materials investigated—$\ce{Pt}$, $\ce{Au}$, $\ce{Ag}$, $\ce{Ni}$ and $\ce{Fe}$—the last one only exhibits a positive potential shift by comparison to the other metals; this shift has been interpreted in terms of an $\ce{O2^{2-}}$ impoverishment following corrosion.
The peroxide which is required for the corrosion reactions can be formed only by chemical reaction of the melt with $\ce{O2}$ , not, however, by anodic oxidation, because peroxide oxidation essentially precedes the oxidation of hydroxyl ions yielding oxygen and water vapour. With anodic and cathodic polarization, peroxide is reacted with limiting current flow, the anodic reaction being less inhibited. Water vapour is cathodically reduced to $\ce{H2}$ and $\ce{OH-}$ —with limiting current flow—before $\ce{NaOH}$ is reduced. $\ce{Pt}$  corrosion is anodically enhanced by water vapour; in this context, current fluctuations are observed. In the pure $\ce{NaOH}$  melt the cathidic reduction may be assumed to involve formation of $\ce{H2}$ and $\ce{Na2O}$ ; not, however, $\ce{Na}$  deposition. [2]

Thus, in a molten hydroxide, one can expect ionic species such as metal cation ($\ce{Na+}$ in this case), $\ce{OH-}$, $\ce{O^{2-}}$, $\ce{O_2^{2-}}$, and $\ce{O_2^{-}}$ anions, and molecular $\ce{O2}$ present in this so-called basic melt in equilibria. Thus, when $\ce{K}$  reacts with any of these (most probably with $\ce{H2O}$ and $\ce{OH-}$ based on their concentrations at equilibrium) to oxidize to $\ce{K+}$, it will stay as mixture of $\ce{KOH}$, $\ce{K2O}$, $\ce{K2O2}$, and $\ce{KO2}$.
Example for ionic nature in molten salt solution molten $\ce{NaCl}$:

References:

Direct electrochemical power generation from carbon in fuel cells with molten sodium hydroxide: Chemical Engineering Communications, 2005, 192 (12), 1655-1670. DOI: 10.1080/009864490896241.
Elektrochemische messungen in NaOH-schmelzen: Electrochimica Acta, 1968, 13 (3), 625-643. DOI: 10.1016/0013-4686(68)87031-8.

The last question (Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?) is answered in Wikipedia: yes. Similarly, $\ce{KOH + molten K}$ gives $\ce{K2O}$ (Wikipedia again).
The first question (Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?) suggests two products, but the real question is whether any reaction will occur. If a reaction occurs, either hydrogen will be released or sodium metal will be released.
If the reaction proceeds by hydrogen evolution, the product will be $\ce{NaOK}$ or what is essentially a mixture of $\ce{Na2O + K2O}$.
If sodium metal is released, it will be in contact with some of the original $\ce{NaOH}$, which has been demonstrated to react to form $\ce{Na2O}$. If the reaction proceeds in this manner, at the halfway point it will be: $\ce{2 K + 2 NaOH -> K + KOH + NaOH + Na}$. Then the $\ce{Na}$ can react with $\ce{NaOH}$ to form $\ce{Na2O}$ and the $\ce{K}$ can react with the $\ce{KOH}$ to form $\ce{K2O}$, giving a mixture of $\ce{Na2O + K2O}$, or what could be represented as $\ce{NaOK}$. (Interestingly, at the halfway point, you have $\ce{NaK}$ alloy, which is extremely reactive!)
The essential reaction is elimination of hydrogen, which is not energetically favored, but which drives the reaction by eliminating the reverse reaction ($\ce{H2 + Na2O -> 2 NaOH}$ or $\ce{H2 + K2O -> 2 KOH}$). From data in the CRC Handbook, for the reaction: $\ce{2 NaOH + 2 K -> Na2O + K2O +H2}$, the heats are $2 \times (-102) + 0 < -99 -86 + 0$, so entropy of mixing $\ce{Na2O}$ or $\ce{K2O}$ must be factored in, or just consider that removing hydrogen, however slow, will drive the reaction to the mixture of oxides.
The comment about using molten potassium to react with $\ce{KOH}$ (and presumably liquid sodium to react with $\ce{NaOH}$) suggests that simply pushing a solid hydroxide against a solid metal might not give a rapid reaction. So, a little heat might be needed, but the reaction will almost certainly give the mixed oxide.

Hey, thanks for the answer! So, you are suggesting that both pathways - removal of sodium and removal of hydrogen - will give $\ce{NaOK}$ - or as you say - the mixture $\ce{Na2O}$ and $\ce{K2O}$?

Could you please elaborate that the final "mixture" state, homogeneous or heterogeneous?

Mush. That's probably what it would turn out to be. NaOH melts at 318 C, KOH at 360 C, Na2O at 1132 C and K2O at 740 C. After you heat the mass to get it started, and get some more heat from the reaction and get mixed melting point depression, and then higher melting products - mush. Maybe you could stir to get a molten mass that solidifies into a uniform solid but it might be a mix of crystals, like granite. I have no data to lean on.

Two good references to address this question are:
DETERMINATION OF THE SOLUBILITY OF OXYGEN BEARING
IMPURITIES IN SODIUM, POTASSIUM AND THEIR ALLOYS J. Phys. Chem., 1959, 63(1), pp 68–70;
and
US Patent 1872611 Process of making potassium metal or sodium-potassium metal alloy
According to the journal article:
At $\ce{400 ^\circ C}$, under vacuum distillation conditions:

$\ce{NaOH + Na -> Na2O + NaH}$
$\ce{ 2NaH -> 2Na + H2}$

It is also stated:

The equilibrium oxide in a sodium-potassium ($\ce{NaK}$)-oxygen system is sodium monoxide.

According to the patent:
Sodium metal can be reacted with $\ce{KOH}$ to yield a $\ce{NaK}$ alloy plus a mixture of $\ce{NaOH}$ and $\ce{KOK}$. This is presented as a cost-effective way to make the more desirable potassium metal from the less desirable sodium metal.

That replacement takes place when oxygen in the hydroxyl group does not have any charge, but in a covalent compound.
Take ethanol for instance. The hydroxyl hydrogen will be replaced by $\ce{Na}$ or $\ce{K}$, if you add corresponding metals on pure ethanol. Same through for water, at first, oxygen has two covalent bonds with two hydrogens. Alkali metal comes in, oxygen gets one electron from its outer shell and displaces the hydrogen homolytically. Then there will be a negatively charged hydroxyl group left and plus charged alkali metal, they will have an ionic association in solution. Say it $\ce{M-OH}$, this will be $\ce{M+ + OH-}$ in solution, and adding another alkali metal will not have any effect on them, $\ce{OH-}$ is quite stable in this condition, likewise $\ce{M+}$.
This is inherently a redox reaction, the word "replacement" instead is confusing. In short, I expect sodium displacement by potassium, not another oxygen reduction. Potassium ionic radii ($\pu{280 pm}$) and sodium ionic radii ($\pu{227 pm}$) seem quite different. It makes unlikely to see its solid as $\ce{NaKO}$ or $\ce{NaOK}$, and there would be no such product in liquid reaction if lattice energy does not stabilize possible end product.

Thanks for the answer! Though, your middle paragraph seems to be answering the question that what happens when an alkali metal is added to water, not when potassium metal is added to sodium hydroxide. Or did I misinterpret?

You understood right, but I wanted to begin from that to explain why what was asked is not likely. I don't understand why I got -2 for this answer. In the case of question, when potassium liquid is added to sodium hydroxide pure liquid, sodium will be displaced. There is no NaKO compound to the best of my knowledge.

Having both sodium hydroxide and potassium in aqueous solution would result in the following reaction:
$$
\begin{align}
\ce{NaOH + H2O &-> Na+ + OH- + H2O}\\
\ce{K + H2O + OH- &-> K+ + 2OH- + H2}
\end{align}
$$
Essentially, the potassium is going to react with the water long before anything else. The only way I can think of to get your reaction to go forward would be to have the molten forms of $\ce{NaOH}$ and $\ce{K}$, which would react like this:
$$\ce{NaOH + K -> KOH + Na}$$

Are you sure that hydrogen in molten $\ce{NaOH}$ is "dead" and won't get engaged with $\ce K$ to yield $\ce{H2}$? That's what I'd expect to happen.

In aqueous solution, I'd expect the overall reaction of potassium and water to be: $$\ce{2K + 2H2O  -> 2K+ + 2OH- + H2}$$

$\ce{NaOH}$ reacts with $\ce{K}$ to form $\ce{KOH}$
there is no compound which is called as NaOK
Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?
Yes of course

$\ce{NaOH + Na  -> Na2O +  ½ H2}$

$\ce{Na2O}$ is Sodium oxide, which is like anhydrous form of $\ce{NaOH}$. When you add water to Sodium oxide it will give you back $\ce{NaOH}$.
Note - Answer edited completely

Wikipedia article about potassium oxide says otherwise

Answer edited and improved

Are you sure? There is no compound like NaOK?

KOH because K is highly reactive in the metal reactive series and metals don't react with metals as both are elctropostive


text:
Where to find pre-optimized files for Quantum Chemistry Softwares?
Many times Geometry optimization gets very time consuming ( even with access to super-computers ). I have seen some pre-optimized libraries for biomolecules like proteins that you can easily download machine optimized files from.
I use Avogadro,  GAMESS and McMolPlt for my quantum calculations.
I am thinking such libraries should exist for some complex chemical structures ( usable on softwares mentioned above ) but couldn't find any.
At this point I am specifically interested in " Hydroxyapatite Crystals ".
Does anyone know such resources ?

Short answer: we're building one for molecules
In your case, you don't actually want molecules. You're mentioning crystals. In that case, I suggest the Crystallography Open Database or if you have access, try the commercial Cambridge Crystal Structure Database. The COD is planning a theoretical crystal database, but I don't think it's online yet. These are experimental x-ray structures, but should be good starting structures for solid-state simulations.
For molecules, my group is building one. It's the Pitt Quantum Repository There are a few related efforts that I'm aware of.
With the PQR, you can either search through the website (even on mobile). The database is keyed by InChI keys, so if you have an existing structure you can generate the InChIKey and then get the structure, e.g.:

D-(-)-Morphine -> .mol2 file at http://pqr.pitt.edu/api/mol2/BQJCRHHNABKAKU-KBQPJGBKSA-N

The long answer is that some commercial software vendors consider such libraries as violations of the license. No joke, I've planned this for years to augment Avogadro, but the company behind a leading program told me that they would void my license to distribute output files in this fashion.
With the rise of community and open source quantum chemistry software, this is moot. Use programs that let you distribute output files. What remained was:

Gathering the 3D coordinates (i.e., a low-energy conformer) of X thousand/million molecules
Obtaining the computer time to optimize all of them
Devoting a server and disk space to serving them
Having a great website to help view, browse, and search them

Thank you !  We have done some simulations in Colorado and I will be more than happy if I can contribute to that repository project.

Incidentally, we only have ~64k molecules right now at the PM7 semiempirical level and growing, so PQR may not be right for every purpose.

@GeoffHutchison, could you clarify, please. Did the "company behind a leading program" told you they would void your license if you will distribute output files created by their program? If so, why not choose a competing product with a much more liberal license?

@Wildcat Of course the company indicated was implying that their output files cannot be distributed. The PQR does not use this product and my group no longer uses this product.

@GeoffHutchison, you never know what to expect in terms of licensing from that "company behind a leading program". :( For me even the fact that they prohibit the distribution of the result obtained with their program is quite surprising. So, you pay for the program, but you don't even own the results, since you can't do whatever you want with them...

@GeoffHutchison, anyway, I wish your group good luck with the PQR project.

@Wildcat It's further ironic, since the founder of the "unmentionable program" actually wanted to have something exactly like PQR. We'll be adding more model chemistries for people to compare and contrast later this year.

What is your guys' favorite probability distribution?  Lorentzian?  Binomial?  Boltzmann?  It's got to be one of those, right?  No?  Which one did I leave out?

@GeoffHutchison Probably you should add info on how to enable WebGL in order to access PQR, otherwise many people are not going to see structures due to the default behavior of Chrome (WebGL is disabled). And thank you very much for your hard work!

You could try figshare.com for comp chem files and http://www.ccdc.cam.ac.uk/pages/Home.aspx for crystal structures.


text:
Best ways to start a thermite reaction?
I'm trying to do a thermite reaction with rust and aluminum and am trying to figure out the optimal way to get the reaction started. My understanding is that I need a temperature of around $\pu{3200 ^\circ F}$ to start it. I am currently aware of two methods to achieve that:

Potassium permanganate and glycerin: But apparently this requires an ambient temperature of $\gt \pu{70 ^\circ F}$, and it's winter where I live, $\pu{40 ^\circ F}$ at most some days.
Magnesium strip: But this requires a blowtorch (trying to avoid buying a blowtorch) and I've read is unreliable for starting this reaction.

Does anyone have any alternative suggestions? Has anyone done this before? All suggestions are welcome.
I am aware of the need to do it far from anything flammable/destructible, and am aware it can damage your eyes, and am aware to keep it away from water and keep my distance. Other tips are welcome.

I think technically, ignition mixture is used to initiate the reaction which consists of mixture barium peroxide and aluminium powder. But never tried it personally.

The method I've used is a simple propane torch, which you can get at Home Depot for $14 and magnesium powder. The mixture of powdered aluminium, iron oxide, and magnesium is thoroughly mixed and then lit using the propane torch. It lights all the time with the magnesium embers burning brightly and then the main reaction takes hold. Getting powdered magnesium is cheap as well. Using this method, the reaction takes all the time and I haven't had a single problem with it the numerous times I've done it for an audience.

"Sparklers" – literally the same pyrotechnic sticks handed to children for celebrations – are what railroad welders use to ignite iron thermite.  These are essentially flash powders mixed with a binder to slow their burn rate.
One pyrotechnician created electric thermite igniters by binding a 50/50 mix of CuO and 325-mesh graphite with nitrocellulose lacquer, coating the exposed tips of an unbridged wire pair, and firing it with 12V.  I have tested those and found them to work reliably as well.
Another alternative is a small charge of copper thermite (9:1 CuO:Al, using not finer than 325-mesh Al), which can be ignited using fuse or electric match.

By blowtorch anyway. It is cheap and extremely useful for similar experiments.
Personally, I used $\ce{Al}+\ce{S}$ mix as a starter, that was initiated by blowtorch (note, this mix produces aluminium sulphide that stinks with hydrogen sulfide.) - it is easy, cheap and reliable starter.
Also, a magnesium ribbon can be ignited by common lighter - I did it myself many times by keeping a piece of ribbon just above visible part of the flame for about 15-30 seconds.

But will the magnesium strip light the thermite? I have heard it is unreliable.

@Aerovistae I successfylly initiated thermite using magnesium ribbon many times. True, it sometimes fails, but mostly because the ribbon falls from the hold. So, yeah, using initiator is better if you need completely reliable option. The initiating mixture composition may be different depending on available chemicals. Also, please note: the easy of ignition is heavily dependent on the particle size in thermite. I recommend the smallest flakes you can find. Also, the rust your are using must be dried, so it contained zero water.

My son did thermite reactions for a 3rd. grade science project.  We were able to easily initiate the reactions by using commercially-available thermite "fuses" which appear to be regular fuse that is about 2 inches long which then has a small, shiny square of metal about 1/2" x 1/2" and fuse product at the bottom.  We just inserted the fuse, shiny metal square into the thermite, lit the fuse, and retreated to a safe distance.  We had no problems with failed ignitions, and we got a reliable burn time.
We found sites that sold thermite kits usually had these fuses, as well as small, pre-fused thermite "pots" that could be used alone to create a couple of ounces of iron, or set on top of larger piles of thermite to initiate full-scale reactions.  Survival prepper sites frequently sell these self-contained thermite fuses and pots as survival accessories to go along with thermite mix kits.

You might want to wrap the magnesium ribbon around a sparkler. The latter can be ignited with an ordinary lighter.

Have you done it this way before? Can you elaborate on how you wrapped it? That doesn't quite make sense to me described so simply.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is the pure tungsten (W) melting point higher than that of its oxides (either WO3 or WO2)?
Why is the pure tungsten ($\ce{W}$) melting point way higher than that of its oxides $\ce{WO3}$ or $\ce{WO2}$? And why does pure $\ce{Al}$ react differently? I.e. the oxides have a higher melting point than pure $\ce{Al}$.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

I think along with these, $\ce{W87O13}$ exists as well.

The phase diagrams available at the ASM Alloy Phase Diagram Database (if your institution has access) show varying numbers of 'stable' phases.  Most are likely various orderings on the oxygen sublattice (my guess). I can find no recent (last 20 years) work on the thermodynamics of the oxides.

There is no reason to think that the oxide shold have a higher melting point. The bonding mechanism in the metal is totally different, you cannot reasonably compare their physical properties.

As far as I know,
$\ce{WO2}$ and $\ce{WO3}$ have a covalent characteristic, and thus lower melting point than the metal.
while, $\ce{Al2O3}$ have more ionic characteristic, and thus higher melting point than the metal.
$\ce{W}$ is a d-block metal from the $\ce{3^{rd}}$ row which make it possible to have covalent materials rather than ionic materials.
this article might prove this point
http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk

Well, except that it doesn't explain why the titanium oxides, cobalt oxides, and chrome oxides, amongst others, have higher melting temperatures than their metals.  Bottom line is: thermodynamics of solid phases may not be easy to predict based on simple heuristics.


text:
Is it possible to extract magnesium from epsom salt?
Is it possible (at home) to extract magnesium from $\ce{MgSO4·7H2O}$, also known as Epsom salt (or, alternatively, from $\ce{MgSO4}$)?

Do you want to extract magnesium ions, or are you actually looking for a way to get magnesium metal?

Not really.  IIRC they use the chloride for that.

you should melt magnesium acetate in a vacuum, then its only 130 degrees or so to do it.

Supposed you are trying to obtain magnesium metal from Epsom salt, $\ce{MgSO4.7 H2O}$:
Is is worth the effort?
Magnesium metal in the form of chips (particle size 4-30 mesh) in decent purity (99.98%) is available from suppliers such as Sigma-Aldrich for about 150 Euro per kilogram.
Can I use electrolysis to generate $\ce{Mg}$ metal?
Yes, but you wouldn't want to do that at home. An older industrial process uses electrolysis of molten magnesium chloride, $\ce{MgCl2}$ (mp. is around 700 °C). While $\ce{Mg}$ metal is formed at the cathode, chlorine gas is released at the graphite anode. Did I mention that this isn't meant to be done at home?
Can I convert $\ce{MgSO4}$ to something more usable to make $\ce{Mg}$ metal?
Yes, in principle. A group of Chinese researchers examined the conversion of  $\ce{MgSO4}$ to  magnesium oxide, $\ce{MgO}$ by heating in the presence of natural gas at temperatures around 500 °C and outlined their results in
Geochemical Journal, 2011, 45, 97-108.
The initial pressure by natural gas at room temperature was 3 MPa, but upon heating to the final temperature, the pressure in the reactor was as high as 12-20 MPa! This can't be done at home and it shouldn't be.
What if I would have $\ce{MgO}$? What could I do with it?
Magnesium oxide, $\ce{MgO}$, is indeed the most important starting material in the production of magnesium metal in the Pidgeon process. Here, magnesium oxide is heated in the presence of silicon in an evacuated reactor. Silicon is oxidized to silicon dioxide, magnesium oxide is reduced to magnesium metal. The latter boils off (bp. around 1100 °C) and is condensed. This isn't kitchen chemistry either.
In summary, you're better off buying some magnesium metal from a supplier of your choice!

On the sulfate decomposition, don't forget the toxic sulfur-bearing fumes.  Also the flammability of natural gas, which is needed to reduce the sulfur compounds (otherwise the acid-base interaction between $\ce{MgO}$ and $\ce{SO3}$ is so strong that you'd need a much higher temperature to decompose the sulfate).

You need molten salts of some kind or a mercury cathode to get metal.
Source: http://www.scienceforums.net/topic/36363-extracting-magnesium/

Yes, it is possible. If you electrolyze MgSO4(aq), one of the products will be magnesium hydroxide. If you you add Hydrochloric acid to it you get magnesium chloride 2HCl(aq)+Mg(OH)₂(s)⟶MgCl₂(aq)+2H₂O(l), electrolyze that and you get magnesium and chlorine gas, the last step must be done outdoors
Here's a useful link: https://www.youtube.com/watch?v=6BThiJpbBJQ


text:
Limiting Reagent Stoichiometry
What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?
I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.
My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {->} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$
But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid?
Please help, thanks :)

You're right, it is soluble, given enough water. However, potassium nitrate is WAYYY more soluble, so it will dissociate first. If there's not enough water for both of them to dissociate, the barium sulfide will precipitate out first.

Oh, okay. So in general, when doing these types of questions, what guidelines should I abide by to figure out if a compound in the product is aqueous or not (other than solubility rules)?

Did the question specify which product was the precipitate?

@wes3449 it didn't, unfortunately. the question was what i typed verbatim

See my answer for a moderately detailed explanation.

So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, precipitates cannot be aqueous, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.
In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -> post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google "solubility constants" and "common ion effect".

How do you know which compound is "more soluble" than another?

Use a solubility chart such as this one

Generally speaking, if the problem doesn't give concentration or enough information to find concentration, you can't know whether all or only a portion of the product will dissolve.  You need to know the solubility product constant (Ksp), the final volume, and the amount of each ion to calculate the amount that will dissolve.
However, this particular question has a complicating issue because:
$\ce{2BaS2 + 2H2O -> Ba(OH)2 + Ba(SH)2}$


text:
How to remove epoxy on clothes?
How to remove epoxy on clothes in the perspective of chemistry?
More precisely in my case, how to remove the hardener of epoxy, which consists of two components: hardener and resin.
The clothes is made in nylon, please take the least damage to the cloth into high priorities of consideration.
Does alcohol help?
Thank you very much for your reply.

What hardener - amine, anhydride/amine?

I have checked. It is unfortunately that it is not listed in the container. But, it is a hardener of a steel epoxy if this information helps.

Lestoil, PineSol or other organic-solvent cleaner should get out the unset hardener. Once set, epoxy is hard to remove; you might try methylene chloride-based paint removers [toxic!]. Some hardeners are organic peroxides, which may bleach spots on the clothing. –

Thank you very much for your response. I will try that ASAP and update the result.

@Superuser Did it help?