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Observing crystal of water
By just freezing water and using an optical microscope, can I observe water crystals?

Or do I have to use a special light to see?

No i dont think so. Read here for more informations about snowflakes formation : https://en.wikipedia.org/wiki/Ice#Snowflakes

Maybe if you use polarisation of light you may have the possibility. This is well known and use for crystal or emulsions so why not.

No special light is required, although you will want to keep the light source distant to avoid melting your ice. You will also want to experiment with light angles to get the best resolution of the different angled facets of the ice crystal.
Without the right equipment and experience it will be difficult to get an image quite as spectacular as the crystal you've shown. The keys to getting the best image are avoiding heat, from a lamp or the surrounding air, which causes sintering of the finest faceted crystals.
A good way to view ice under a microscope is to place crushed dry ice under a dark covering (depending on your equipment, getting something that fits and still focuses may be your greatest challenge). Whether you use glass or plastic or whatever else isn't very important. The dry ice is cold enough that ice will nucleate on the surface of your choice and grow from there. Be prepared to watch the whole show, from freezing to melting!

You can certainly "just freeze" some water and look at it - either through a microscope or directly. This type of frozen water will consist of one or, much more likely, many crystalline regions, but will have the shape of its container; what you have pictured is a single crystal of ice. In order to grow a single crystal, special methods must be used.
This link:http://www.its.caltech.edu/~atomic/snowcrystals/project/project.htm contains directions for making ice crystals at home but note the author incorrectly claims dry ice (solid CO2) is "completely safe" but it is both capable of burning skin upon direct contact (especially if your skin is moist) and enough of it can kill if it sublimates (vaporizes) in a (very) small enclosed space. You must keep the concentration of $\ce{CO2}$ gas below the toxic limit, which is generally not a problem in an average size room. Use it only with gloves and adequate ventilation. To microscopically observe an ice crystal, you need a platform (or slide) to rest it on which has a temperature below freezing (otherwise it will melt). You also need a light source which is cool enough to avoid heating and melting the crystal. This is fairly easy outdoors when temperatures are below the middle 20s (°F) (-5 to -10°C). You should allow your microscope to slowly cool to ambient temperature or water may condense on the lenses and/or mirrors. Overnight in your garage, say, would be good (if your garage is unheated).
A snowflake, like the one you have pictured, is very difficult to make. They are formed in the atmosphere, and generally, you have no means of replicating that with common household items/equipment. There are a number of known chemicals which can be used to nucleate ice crystals, as well as some proteins and bacteria. The easiest way to create ice crystals is to cool a glass microscope slide (possibly with very small amounts of the nucleation materials on it) to well below 0°C and then expose it to humid air at about 2 - 5°C (35°-40°F). The water vapor will condense onto the slide. Don't allow the slide to warm up, however.

A wall of text like this is very hard to read.

Great link and experimental details. Maybe you should format it, as previous comment suggested, to better readability.


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Are reduction potential and stability related?
Consider the half reactions:
\begin{align}
\ce{Na^+ +  e- &->  Na} &\quad E^\circ &= -\pu{2.71 V} \tag{1}\\
\ce{Cu^2+ + 2 e- &->  Cu} &\quad E^\circ &= \pu{+0.34 V} \tag{2}
\end{align}
Can I say that $\ce{Na+}$ is more stable than $\ce{Na}$ since $\ce{Na+}$ has negative potential? If so, can I also say that $\ce{Cu^2+}$ is less stable than $\ce{Cu}$ so $\ce{Cu^2+}$ has a positive potential?

Remember that reduction potentials do not have any meaning by themselves. They are measured against a hydrogen cell to which arbitrarily we assign the value $E^\circ=0$. You can compare between different potentials. For example, in your case, you could say that that $\ce{Cu^2+}$ has a greater tendency to be reduced than $\ce{Na^+}$  because its reduction potential is more positive. However, I do not think the potential itself tells you anything with regard to the stability (and, in any case, what do you mean by that) of the species involved in a particular half reaction.

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What is the difference between a 500W high pressure mercury lamp and a 400W high pressure mercury lamp
I know that the Wattage of a UV lamp is only one factor in the UV output of a light, and therefore cannot be used as a measure of how powerful a light is. Also it doesnt have any effect on the produced wavelength. Is it like this that a lamp with higher wattage can cause higher temperature on substrates that are being irradiated ?

My guess is that a higher wattage would create a wave with a higher amplitude

Only that one is brighter than the other but probably with a bigger focussed spot size which may necessarily not give a larger number of photons/area. (A low pressure Hg lamp is far less intense and just at one wavelength)

Generally speaking, the wattage rating on devices simply tells you how much power it will draw. Whether or not it will put that power to good use is dependent on the design. I would agree with Bob and say that you'll get the same wavelength from both lamps but the 500 W lamp is likely to give you higher amplitude light.

I don't like the wording "higher amplitude light." A 500 watt light and a 400 watt light would probably have very similar wavelength distributions. At every wavelength of the spectral output the 500 watt light would have 5/4 the intensity (photons per second) of the 400 watt light.

The difference between the two is $\pu{100W}$ …

I would say all other things being equal, i. e. Spectral distribution,  quantum efficiency,  that more wattage = more photons per second. And yes, more heat.

Exactly. The difference is: 100W.

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HDPE bottle for gasoline?
What are the risks of storing E10 gasoline in a gallon HDPE container for a few months, vs storing in a proper plastic fuel can? Assuming the HDPE gallon is thick (used for Hydrochloric acid before, and was properly rinsed and dried). I'm interested in knowing what are the technical differences that make the legal plastic can so much different that a regular container, since they are both made of the same material with similar thickness.

I wouldn't - mixing acids and organics is bad. And a real gas container is cheap and legal.

Never store liquids in a container that could lead anyone to mistake them for something else. Not only food containers, Imagine a small fire and the only helpful thing someone see is a bottle of dilute HCl. You don't want that to be gasoline.

Thanks for the comment. Let's assume the bottle is repainted in red with big letters 'GASOLINE' all over it. I'm really interested in understanding the technical difference between the two containers, from a chemical and physical perspective.

HDPE fuel containers are surface treated to prevent fuel penetrating the plastic, other hdpe containers do not have this treatment

I wouldn't recommend it because the water content in the gasoline can react unfavorably with residual acid and/or the plastic is not meant to be in contact with a chemical like gasoline. You should only transport and store flammable liquids in specific containers meant for that purpose.

I agree that this is a really bad idea for some of the reasons you bring up. I'm unclear on one thing though. When you say "the water content in the gasoline can react unfavorably with residual acid", I'm not sure what kind of reaction you are talking about.

I was mainly getting at the fact that mixing concentrated acid and water generates excess heat and that could make the container unsafe to handle; I am making a conservative assumption that 20 Baume is concentrated because I have never worked with this convention for acids.

Good point.  And for that matter water, acidic or not, is a bad idea in gas.

I agree overall, use a proper gas can. // If the HCl container has been rinsed a couple of times and dried there is very very little water left in/on the container. There will be more water in the gas itself. There certainly won't be enough acid left to cause the added gasoline to get hot.

Thanks for the comments. Indeed there would be very little - if any - HCl left in the bottle to react with water/gasoline/alcohol. So no significant increase in temperature of HCl reacting with water, thus no explosion. So from this perspective the two containers are equivalent. Any other potential risk (from a chemical/physical perspective)?

The main concern I would have is if the particular HDPE plastic formulation for HCl containers wasn't compatible for solvent storage (gasoline in this context can be seen as a solvent). There could be a greater degree of crosslinking, fewer plasticizers, etc. for a regulation fuel can.

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(At what temperature) can salt evaporate?
Related to my previous question, can sodium chloride evaporate, and if so, at what temperature? Since salt is a solid, and apparently all substances can at various temperatures exist as liquids and vapors as well as solids, salt should also have a melting and a vaporization points. At what temperatures are these points if they occur, and if they don't occur, why? (This is assuming the pressure is 1000 mb)

Remember in asking "what temperature?" you have to assume a fixed pressure. For example, if you put the salt in a vacuum, it would evaporate at a lower temperature than an 1.0 atm.

1000 mb = 1 bar, or one atm. In astronomy (my proffession) we use bars,  not atm's.

Do you mean evaporate or vaporize (boil)? Those are different things.

@jpmc26 I mean evaporate, as seawater does.

Note 1 atm = 1.01325 bar; they're not the same

All substances may have a melting and boiling point. I say "may" because in some instances there can be decomposition, for example, before reaching either temperature.
Sodium chloride has a melting and boiling point. Check Wikipedia:
https://en.wikipedia.org/wiki/Sodium_chloride
The sidebar lists the melting point at 801 °C (1,474 °F; 1,074 K) and the boiling point at 1,413 °C (2,575 °F; 1,686 K). The values are relatively high being consistent with the ionic nature of NaCl.

Some substances do not have a stable liquid at 1 atmosphere. They sublimate rather than melt.

True. In fact, dry ice is a good example. At 1 atm the solid goes directly into the gas phase.

At any temperature there is a certain fraction of a liquid to evaporate, because the temperature in the liquid is not homogeneous across its molecules. You know that from a puddle drying out, even the temperatures are below the boiling point of water. This fraction gives rise to a gas pressure above the liquid called the vapor pressure Check Wiki link.
As also stated in the article, such a pressure exists even above a solid, though much smaller. So in that sense, a partial evaporation of salt takes place at any temperature.
A much higher evaporation rate at room temperature can be seen in sprays of saline solution, for example at the coast, or in so called graduation towers.

What is the vapor pressure of sodium chloride at room temperature? Does it correspond to a significant number of particles per liter of gas?

@TannerSwett See answer by Bernard J. Sandman.

The vapor pressure of sodium vapor is approximately 1.08 E-27 Pascals or 1.08 E-32 Bar at 25 C or 298.15 K, slightly above room temperature.  Vapor pressure is so low that the saturated vapor density is 2.55 E-35 grams/cu. cm. with 99.99 % ClNa and 0.01% Cl2Na2 dimer fraction in the vapor.  There are only 2.63 E-10 molecules of ClNa Vapor with 3 E-14 moles of Cl2Na2 dimer molecules per liter of gas at this temperature.  One molecule of sodium chloride will occupy 3.80 billion liters at this temperature, which is 3,800,000 cubic meters.  Therefore, the pressure is so low at 25 C that it will take many times the current age of the universe, 13.8 billion years, just to sublimate one atomic layer of sodium chloride.
Calculations.  1.08 E-32 Bar/83.14462618/298.15K*58.449 molecular weight vapor = 2.55E-35 grams/cm^3.  Used NIST 2018 Constant for R molar gas constant 8.314462618 J/(mol-K)*10 = 83.14462618 to get R into correct units when using bar and g/cm^3.
2.55E-35 grams/cm^3 * 1000 cm^3/liter*6.02214076E23 (2018 NIST Avogadro Constant)/58.4426 (molecular weight of salt) = 2.63 E-10 molecules per liter.  This is assuming ClNa monomer vapor at room temperature. There is only about 1 part per 10,000 dimers Cl2Na2 at this temperature and can be neglected.  However at temperatures above 500 K, the saturated sodium chloride vapor starts to dimerize and polymerize.  In addition to monomers, Sodium Chloride however has a significant fraction of dimers and even some higher polymers such as Cl4Na4 in the vapor at its boiling point of approximately 1727 K (1454 C).

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Where does the water in an egg go on boiling?
Our textbook says that boiling an egg causes coagulation and denaturation of the proteins present in it. Where does the water in it disappear?

Do you mean where does the water inside the egg go? If so, most of it is incorporated in the boiled egg.

If you are wondering why the egg doesn't blow up when the water expands upon heating to it's boiling point, that's a good question.  That's exactly what would happen if it weren't for a fair-sized air pocket within the egg that compresses during boiling.

Then again, the egg does blow up. Try to heat it in a microwave oven and you'll see.

even more the egg may crack if over boiled or over heated

@ToddMinehardt, I want to know what chemically occurs. I know that it gets incorporated in thee boiled egg.

See my comment here.

The way chemically stays water. The coagulation/denaturation simply leads to less dissolved material and therefore a less fluid egg.


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Why does a gas' ability to absorb infrared radiation affect its potency as a greenhouse gas?
My book says that the ability of a gas to absorb infrared radiation affects its potency as a greenhouse gas, but I don't understand why. So what if a gas is able to absorb a lot of IR? How does that affect its potential as a greenhouse gas?

Here is a good definition of IR radiation, which is basically heat. Understanding the definition of IR is the first step in really understanding the answer to your question.

What is occurring in the greenhouse effect? If a gas does not absorb IR does that matter?

There is only one difference between a greenhouse gas and a non-greenhouse gas: a greenhouse gas has a greater ability to absorb radiation in the infrared (IR) region of the electromagnetic spectrum than does a non-greenhouse gas.
Once a greenhouse gas absorbs IR radiation, that energy must be converted to another type of energy such that the the energy is conserved. The only thing that can happen at IR energy levels is that the absorbed IR energy gets transformed into thermal energy, resulting in an increase in heat in the atmosphere.
From there it follows that the more strongly a particular gas absorbs IR radiation, the more heat is retained in the atmosphere, which might have otherwise just radiated off into space as IR radiation. This means a gas absorbing strongly in the IR is a more "potent" greenhouse gas than one that only absorbs weakly or not at all in the IR.

I'd add that, even if a gas doesn't absorb the IR and warm up but just reradiates it, it will still warm the earth. This is because the radiation absorbed from the earth's thermal emissions will be re-radiated in random directions so some will head back towards the earth, warming the earth rather than being lost to space.


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Why does the distance between atoms in a hydrogen bond impact the strength of the bond?
Wouldn't the distance be irrelevant as after they are attracted they will just close the gap? E.g. once 2 magnets are close enough to be attracted, isn't the distance irrelevant to the strength of the bond between the magnets as they will just become flush against each other, making the distance between any 2 magnets that are attracted to each other equal.
Or is it perhaps talking about the other groups stopping the hydrogen and other atom becoming close?

Context:
Of the following, the factor that has the least effect on the strenght of a hydrogen bond in a molecule is the

type of atom to which the hydrogen atom is covalently bonded.
distance that the hydrogen atom is from the atom on the other part of the molecule
number of other hydrogen atoms atttracted to the atom on the other part of the molecule
type of atom in the other part of the molecule to which the hydrogen atom is attracted

(answer is 3, hence 2 must have some effect).

To extend you magnet analogy: Try holding two magnets a few centimeters apart, where you can just start to feel the magnetic attraction. Then try holding them 1 millimeter apart.  Is the attraction stronger when the magnets are separated by a shorter distance at any given time?

You didn't account for the fact that the protons and electrons in different atoms repel each other. Your analogy is way over simplified.

The problem, as others have suggested, is that atoms aren't magnets that are free to snap to each other. Other forces and other interactions can and will hold them apart, and atoms also repel each other if they get too close (or else life would be in huge trouble because everything would collapse into a singularity).

The main attractive force between the nuclei and the electrons is electromagnetic, positive and negative attracting. When two atoms are "pushed" together each one shares its own electrons with the other one's nuclei, and they bond. So why wouldn't they nuclei just get as close to each other as possible, so that they each would feel as many electrons as possible? At the same time, the electrons are repelling each other, and the nuclei repel (same charges repel). The bond length is a function of this "equilibrium" that is reached between the attraction of the nuclei to the electrons, and the repelling of the nuclei  and the repelling of the electrons. Therefore, if they begin to "close the gap" as you say, there will be more repelling than attracting, and they will move farther away from each other, and return to the lowest energy distance, the original bond length.


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What are the ideal pair of (everyday) elements to pair in a thermocouple
I want to create a thermocouple (to carry out experiments on thermoelectricity).
I want to use the pair of everyday* elements (e.g $\ce{CuO}$ and $\ce{Al}$ wires) to create the thermocouple. The temperature range of the heat source will be between 300° C - 500° C
My (little remaining) knowledge of physical chemistry informs me that the periodic table would provide useful clues as to which elements to use (that's why I posted this question here, rather than in physics or engineering) - but unfortunately, it's not clear to me how I may use the periodic table in this case, to select the two elements that meet my twin requirements of:

Maximum voltage and current generated (via the Seebeck Effect) within the operating temperature indicated above
Both elements are "everyday" elements

Note: By everyday elements, I mean elements that I can buy off the shelf (preferably, in wire or sheet form), and are relatively inexpensive.

Have you checked a thermoelectric series? See if you can find common materials in it.

My book has a few metals/alloys listed in a Seebeck Thermoelectric Series:

Antimony, Nichrome, Iron, Zinc, Copper, Gold, Silver, Lead, Aluminum, Mercury, Platinum-Rhodium, Platinum, Nickel, Constantan, Bismuth

There's also the Seebeck formula for emf of a thermocouple:
$$E_{\text{AB}} = a_{\text{AB}}\theta + \frac{1}{2}b_{\text{AB}}{\theta}^2$$
Here, $a_{\text{AB}}$ and $b_{\text{AB}}$ are constants that depend on the pair of metals $\text{A}$ and $\text{B}$ chosen. $\theta$ is the temperature difference between the junctions of the thermocouple.
There's another defined quantity called thermoelectric Power, which is defined as $\frac{dE_{\text{AB}}}{d\theta}$.
The graph of $E_{\text{AB}}$ with respect to $\theta$ would be a parabola. If $a_{\text{AB}}$ and $b_{\text{AB}}$ are of opposite signs, then the parabola crosses the $x$-axis twice, and thus, the thermo-emf becomes zero for two values of $\theta$ and maximum for one value of $\theta$.
From the above thermoelectric series, a iron-nickel might seem like a good option to consider building a thermocouple with, considering they're quite far away in the series and are easily accessible. For this couple, we get the values of thermoelectric constants as $a_{\ce{Fe-Ni}}=\pu{2.5 \mu V/°C}$ and $b_{\ce{Fe-Ni}}≈0$.Thus, you get:
$$E_{\ce{Fe-Ni}}=2.5\theta$$
If you're using your temperature range of 300°C-500°C for the hot junctions and room temperature (25°C) for the cold junction, you would get your thermo-emf in the range $\pu{0.687-1.1875mV}$.

If you don't want to reinvent the wheel, use iron / constantan. Good for those temperatures and voltage/temperature tables are readily available.

Completely agree. I primarily used Fe - Cu/Ni for casual stuff ($\pu{−180 - +800 ^\circ C}$), and Pt/Rh - Pt for higher temperatures (up to $\pu{1550 ^\circ C}$). These two were enough practically for all applications.


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High Degrees of Symmetry in Molecules
I recently encountered a question on a test:

Substances whose molecules have a high degree of symmetry will have:

A. a low specific heat
    B. a high melting point
    C. a high heat of fusion
    D. a low heat of vaporization
    E. strong Van der Waals forces

Looking through a few compounds, I found that $\ce{CO2}$ is symmetric (it's shape is a straight line), $\ce{CH4}$ is symmetric (tetrahedrons seem symmetric).
Through these examples, I was able to eliminate A, B, and C, and have a suspicious feeling about E, so I guessed D and was right.
I was wondering what exactly "high degree of symmetry" meant. Through my math courses, I knew about rotational symmetry (where if you turn any 2-D shape in a plane by n degrees, it will look identical to the shape before the rotation).
Is a high degree of symmetry when the lowest possible value of n for a compound is over some constant, lets say 90 degrees, or is it something else?

Answer B also can be defended durene has one more plane of symmetry in comparison to its isomers and much higher melting point.

Symmetry in 3D is much more complicated than in 2D. So "high degree of symmetry" is an informal expression with hardly any rigorous meaning. If the question asked about central symmetry in particular, then B would be the right answer indeed.

Although the term 'high degree of symmetry' is somewhat subjective the only molecules with dipoles are those that belong to the point groups $C_1$ (no symmetry elements other than rotation by 360 degrees), the group $C_s$ which has only one mirror plane and the groups $C_n$ ($n \ge 2$ ) which have rotation axes of $2\pi/n$ degrees and $C_{nv}$ which in addition have a (vertical) mirror plane along a rotation axis. An example of a $C_{2v}$ molecule is $\ce{H2O}$, and  $\ce{NH3}$ is $C_{3v}$. All these point groups do perhaps have 'lower' symmetry than say an octahedral group and so the question suggests that molecules with low symmetry have a dipole moment and the rest don't. (You can see 3D models of many molecules of all common point  groups at molecule-viewer.com and add/remove their symmetry elements)
As @Dan Burden suggest the question is therefore most probably to do with the presence or absence of a dipole moment. One might suppose that a dipole will cause the molecule to have a larger van der waals interaction energy due to dipole-dipole interaction, than a similar molecule without such a dipole and so have higher boiling and melting points. Isomers for example, could be used to test this.
Three such could be 1,2-, 1,3- and 1,4-dichlorobenzene; I chose at random. The 1,2- and 1,3- have a dipole ($C_{2v}$ point group) but the 1,4- does not ($D_{2h}$ point group). The data (from Wikipedia pages) are below
boiling point (centigrade)  $1,2 = 180.5, ~~~~ 1,3 = 172,~~~~ 1,4 = 174 $
 so not much difference
melting point(centigrade) $1,2 = -17, ~~~~1,3= -22, ~~~~ 1,4 = 53$
not what one would expect if dipole interaction is important, quite the opposite in fact.
vapour pressure $1,2= 1.0, ~~~~ 1,4 = 1.3$ mmHg so again, not much difference here
To me the question does not have clear answers and it seems possible to choose molecules to support whatever the 'official' answers are or to show the opposite.

I think that the question meant to test your knowledge about the polarity of simple molecules.
The fact that symmetric molecules contain identical bond polarities between the central atom and surrounding atoms, the net dipole moment usually tends to be zero, creating a nonpolar molecule.
In other words, I think symmetry in this case refers to the fact that all of the surrounding atoms are identical.


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What happens to a greenhouse gas after it absorbs infrared radiation?
We say that a greenhouse gas increases the temperature of the surface of the earth but how does it do that ? Only by re emitting the absorbed IR towards the earth ?

Actually a greenhouse gas absorbs light in the infrared wavelengths of light (longwave in the picture). One of two things will happen after the gas absorbs the IR radiation: 1) the molecules to move around and vibrate faster, which is really the definition of temperature, so the air gets warmer, and 2) the gas molecules can re-emit the IR radiation. In the second case, the radiation is emitted isotropically, so that some of it just goes out into space as it would have had the gas not been there in the first place, but some of it will be directed back towards the surface of the earth, warming the surface by absorption of the IR (heat).

Source: https://commons.wikimedia.org/wiki/File:Atmospheric_radiation.png
Note that most of the heat being absorbed by greenhouse gases is coming from the ground after the ground absorbs sunlight and warms up, not from the sun shining through the air and being absorbed by the greenhouse gases directly.

The 'greenhouse' gases such as $\ce{CO2, H2O, CH4}$ absorb infrared radiation and become vibrationally and rotationally excited. When excited, their temperature is effectively greater than that of their surroundings.
Two things can now occur; the vibrationally excited molecules may radiate, some of this energy will be lost to space, and some absorbed by the Earth and can therefore be re-radiated. Secondly, a collision with other air molecules will compete with radiating and thus transfer some, or all, of the extra vibrational/rotational energy to the collision partner until, after several collisions, all gases are at the new equilibrium temperature. This increase in temperature can only occur because the majority gases in the atmosphere, nitrogen and oxygen, cannot radiate infrared radiation as they have no electric dipole. Thus, the energy is trapped as vibration and rotational energy in the ground state of these molecules and therefore after collisions with other molecules transfers this energy and so heats up the atmosphere.

The last 2 sentences are confusing and inaccurate. The heat is contained in the increased KE of the various molecules. The various states are populated dependent on the temperature for O2 and N2. Look up the Raman spectra of these gases.

No, You are wrong in this assertion about the accuracy of these statements. Raman is a scattering process, it does not occur spontaneously but only via a photon already there and it is not efficient anyway. A molecule with a dipole can radiate and absorb (ir) radiation to come into equilibrium with its surroundings. As there is no dipole on O2/N2 they cannot do this and so get hotter and spread their energy to other molecules via collisions. Hence a hotter atmosphere.

I have to agree that at least the final sentence isn’t quite right. The temperature is a function of the translational KE only, not the rotation and vibration energy levels, though equipartition means that all three increase equally as the system equilibrates (neglecting vib modes that are inaccessible)

Raman spectroscopy is a means of examining the rotational and vibrational levels occupied by O2 and N2. The increase in temperature comes from collisional deactivation of excited CO2 molecules by other molecules increasing the temperature. If appreciable energy is stored in higher excited states of N2 and O2 then Raman analysis should reveal it.  I am not aware of a study showing transfer of energy from an excited CO2 molecule to excite an O2 or N2 molecule[interesting possibility].  The comment specified ground state. Energy in excited states does not increase T until they are deactivated.

A few points in reply to comments. I was not talking about excited states just vibrational/rotational levels in the ground state. Raman has both stokes and anti stokes transitions so effect will be small on population overall, and as mentioned Raman probability is small. Collisions can transfer energy all over the place, i.e. not restricted as are dipole transitions. There is a vibrational and rotational temperature ( as described by Boltzmann distribution) in a molecule before thermalisation that can be different to the average; this is why energy gets spread. I have clarified last sentence.

Comments are not for extended discussion; this conversation has been moved to chat.

In addition to the point made by airhuff, which is that greenhouse gases essentially prevent the earth from emitting infrared radiation into space, this slight effect shifts the equilibria of greenhouse gases which actually amplifies the effect.
That is to say, when people speak of $\ce{CO2}$ being a big problem, this mostly  because of an indirect effect. $\ce{CO2}$ slightly warms the earth which increases the amount of water vapor present in the atmosphere (in an average sort of way because this fluctuates quite a bit). This is a problem because water is a much more potent greenhouse gas than $\ce{CO2}$ and it absorbs at different wavelengths in the IR than $\ce{CO2}$ does. This matters because the amount of IR which $\ce{CO2}$ absorbs is fairly close to saturation. That is, doubling the amount of $\ce{CO2}$ does not mean doubling the amount of IR radiation absorbed and redirected back towards earth. The relationship is much closer to logarithmic than linear. But, this is only true for the wavelengths at which $\ce{CO2}$ absorbs, so introducing more water vapor allows a wavelength of IR to be absorbed which may not be saturated.
This is only one mechanism by which greenhouse gases warm the earth. The atmosphere is an extremely complicated system. None of this mentions how well certain gases mix in the atmosphere and how long they stay there and anything about methane for that matter. All of these things have been written about in scholarly articles and books, so those are gonna be the places you find the nitty-gritty details and data.
Also a note on what I mean when I say carbon dioxide is roughly saturated in terms of absorbance. It is true that carbon dioxide quite effectively absorbs IR light, but it can never truly be saturated. This is because the relationship is logarithmic, so it always increases, but just more slowly as you add more $\ce{CO2}$. Second, it has been said that convection can play an important role here because one shouldn't really think of bulk concentration of carbon dioxide but really the concentration in different layers of the atmosphere. This complication actually allows $\ce{CO2}$ to absorb more radiation than you might otherwise expect.

The high level effect of green gases is, they cause Earth as a planet, at the same surface temperature, emits to space less IR energy (at affected wavelengths) than without them.
With addition of GH gas, this IR is emitted to space more by high altitude cold gas and less by low altitude warm surface. That leads to disbalance of net energy until Earth surface temperature in mean levels increases.
As surface has to emit at IR windows from surface the extra energy, than is not emitted by GH gases. ( my meteorology background speaking)

Good overview!   The details sometimes just get in the way.


text:
dU/dV in van der Waals equation of state derivation
Previous, I proved that
$$\frac{T_2}{T_1}^{C_V/(C_p-C_V)}=\frac{V_1}{V_2}$$
I've been asked to derive a similar expression for a gas that obeys the equation of state
$$p=\frac{RT}{(\overline{V}-b)}$$
I started with the first law and made the assumption that
$$\mathrm dU=\frac{\partial{U}}{\partial{V}} + \frac{\partial{U}}{\partial{T}}  + \frac{\partial{U}}{\partial{n}}$$
I know that for an ideal gas, we can say that $\frac{\partial{U}}{\partial{V}}$ goes to zero... but would this also be true for my given equation of state?
Making that assumption, I ended up with something like
$$\frac{T_2}{T_1}^{\overline{C_V(t)}}=\frac{\overline{V_1}-b}{\overline{V_2}-b}$$
Any suggestions would be immensely appreciated.

First, you need to qualify your derivatives better. if you say $\partial{U}/{V}$, what are you keeping constant? It is important to know because that defines the result of your derivative.
Second, du/dv)T goes to zero only for an ideal gas. This term is a measure of the intermolecular forces. For other equations of states the result will be different

For an arbitrary gas, $$\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial P}{\partial T}\right)_V-P$$

That makes sense. However I seem to run into a bit of an ugly expression..

@PedroO'Verde , that makes sense.. however my expression seems a pit ugly.. so let's say I say that $ \left(\frac{\partial{T}}{\partial\bar{V}}\right) = \pi_T $, I end up having to integrate the expression $$ C_\bar{V}= \frac{-RT}{(\bar{V}-b)} - \pi_T $$

if you calculate the above derivative it should give you 0. Remember that V is constant so the term $\frac{R}{(V-b)}$ is a constant.

The "above derivative" means the one provided by @Chester Miller
By the way, note that this is NOT a Van der Waals gas because it lacks the $\frac{a}{V^2}$ term.

Firstly, assuming that you want the results for a reversible and adiabatic process where composition is constant throughout the process. (Evident from your expression for the ideal case)

$U$ is a state function depending on $V$ and $T$. Thus, we can write,
$$\mathrm dU=\left(\frac{\partial{U}}{\partial{V}}\right)_T\mathrm dV + \left(\frac{\partial{U}}{\partial{T}}\right)_V\mathrm dT \tag{1}$$
For ideal gases the value of $\left(\frac{\partial{U}}{\partial{V}}\right)_T$ is always zero and thus we can ignore it. But for non-ideal gases, it may or maynot be zero and thus we have to find an expression for it using the equation of state, $P=\frac{RT}{(V-b)}$ in this case.
Now, we know that for a reversible and adiabatic process where composition is constant throughout the process,
$$
\begin{align}
\mathrm dU &= T\mathrm dS - P\mathrm dV\\
\left(\frac{\partial{U}}{\partial{V}}\right)_T &= T\left(\frac{\partial{S}}{\partial{V}}\right)_T - P
\end{align}$$
Now, by Maxwell's relations we know that $\left(\frac{\partial{S}}{\partial{V}}\right)_T=\left(\frac{\partial{P}}{\partial{T}}\right)_V$. Thus,
$$\left(\frac{\partial{U}}{\partial{V}}\right)_T = T\left(\frac{\partial{P}}{\partial{T}}\right)_V - P \tag{2}$$
Now, it is easy to find $\left(\frac{\partial{P}}{\partial{T}}\right)_V$ with the help of equation of state.
$$\left(\frac{\partial{P}}{\partial{T}}\right)_V = \frac{R}{(V-b)}$$
Substituting this back in $(2)$,
$$
\begin{align}
\left(\frac{\partial{U}}{\partial{V}}\right)_T &= \frac{TR}{(V-b)}-\frac{TR}{(V-b)}\\
\left(\frac{\partial{U}}{\partial{V}}\right)_T &=0
\end{align}
$$
Surprisingly this results in ideal gas like behaviour. If we think about it in retrospect, it was expected. Because the $\left(\frac{\partial{U}}{\partial{V}}\right)_T$ term is related to the internal interactions of the gas molecules, which we still haven't considered. Notice, how there is no Van Der Walls' constant $a$ in the equation of state.
So, the answer to your question will be same as that for ideal gases, i.e,
$$\frac{T_2}{T_1}^{C_V/(C_p-C_V)}=\frac{V_1}{V_2}$$


text:
Will ammonia react with gasoline mixed with ethanol?
Recently, a few South Koreans invented a car that runs on an 70% ammonia 30% gasoline fuel mix.

Trevor Brown, The AmVeh – an ammonia fueled car from South Korea, June 20, 2013.

In America, we already have instituted policies that require refueling stations to mix ethanol produced from corn stills with gasoline to extend our fuel supply.
If you added ammonia to an ethanol/gasoline fuel mix, would it cause a reaction?  If not, would it still burn and be able to be used as fuel?

It wouldn't react, and I guess it could be used, but don't see much sense in adding ammonia.

Also, the US rule is that in the winter gasoline must contain 2.7 wt/wt% oxygenated fuels as ethanol, where several different oxygenates are allowed.  Technically, a fuel pump that says something like "Contains 15% ethanol" may actually have little to no ethanol and use ethers like ETBE instead (MTBE has mostly been banned, but used to compose 95% of California's oxygenates, which were frequently reported as ethanol).

Would be concerned at the increased risk of generating nitrogen oxides which are already a problem in atmospheric pollution in cities

The biggest problem with using ammonia is going to be the need to store it differently. It is, after all, a gas under normal conditions and standard tanks won't be suitable for containing it. Also, the engine needs to be altered and many components in the fuel system will need to be different to avoid corrosion/degradation. So just adding ammonia to a conventional car's fule sounds like a really bad idea.

A fuel that is mostly ammonia is an altogether different fuel formulation that would not reap the benefits of the oxygenated fuels program in the US for gasoline. In other words, there would be no reason to mix in ethanol or any of the other approved oxygenated species used in gasoline (the specification requires 1.5% - 3.5% oxygen by weight during the winter (state dependent) and is reported as ethanol, regardless of the oxygenated species actually used).
According to this US Environmental Protection Agency document:

Oxygenates are fuel additives that contain oxygen, usually in the form of alcohol or ether. Oxygenates can enhance fuel combustion and thereby reduce exhaust emissions. Some oxygenates also boost gasoline octane. The Clean Air Act requires use of oxygenated gasoline in areas where winter time carbon monoxide levels exceed federal air quality standards. Without oxygenated gasoline, carbon monoxide emissions from gasoline-fueled vehicles tend to increase in cold weather. Winter oxygenated gasoline programs are implemented by the states.

The octane levels of ammonia fuels are already very high at around 120. There would seem to be little benefit in the form of carbon monoxide emission reductions by adding oxygenated hydrocarbons to a fuel that is 70% ammonia.
So, the bottom line is that there is no reason to expect ethanol and ammonia to be in the same fuel formulation. Furthermore, if there were ethanol in the ammonia based fuel, there is no reason to expect the two to react in any significant way.

I'm glad to know people are still reading my old article about the South Korean AmVeh car!
No, ammonia wouldn't react with gasoline or ethanol, as far as I'm aware. Yes, it would burn as a fuel. Ammonia is hard to combust, so to reap the benefits of its high 120 octane rating, you'd need to make certain engine adjustments. Also, you'd need to make sure no copper/zinc/brass alloys were in use anywhere the ammonia is - it'll corrode those very quickly and that would not be good.
Using ammonia as a drop-in fuel is a totally different idea to using ethanol as a drop-in fuel. Different motivations, different results. The idea with ammonia is that, if/when produced from renewable feedstock, it is a carbon-free, energy-dense liquid fuel. If you don't think carbon is a problem, you won't think ammonia fuel is a good idea.
For actual engine performance using a gasoline-ethanol-ammonia blend, see this presentation: https://nh3fuel.files.wordpress.com/2014/10/nh3fa-2014-shehan-haputhanthri_1.pdf. See pages 13 and on: dynamometer tests demonstrating horsepower/RPM for various % ethanol blends; note the ethanol is used here as an emulsifier as well as a fuel, to stabilize the gasoline-ammonia blend. The same team of researchers at Texas Tech have published other work on the subject but I don't have my hands on their most recent publications, although I know they've continued to develop the work.
Regarding NOx emissions, ammonia is the reagent used in most SCR clean-up systems, so it's an onboard solution to its own problem, just takes a bit of engineering design to address.
For those interested, the actual application of ammonia as a fuel is more likely to begin in stationary generation (and certainly not in the US). Here's an article I wrote earlier this week regarding two utility-scale demonstrations of low(er)-carbon power generation in Japan: both employ ammonia in a dual-fuel combustion process, one in a coal plant the other in a natural gas plant: http://www.ammoniaenergy.org/industrial-demonstrations-of-ammonia-fuel-in-japan/


text:
Hydrogen Bonding in Water Phase Changes
How is the disparity between the heat of fusion and the heat of vaporization for $\ce{H2O}$ best explained?
A) It takes more hydrogen bonds for water to fuse than it does to vaporize.
B) Water molecules are moving farther apart during fusion than during vaporization.
C) Water molecules are moving closer together during fusion and farther apart during vaporization.
D) Vaporization occurs at a higher kinetic energy than fusion.
E) More hydrogen bonds are broken during vaporization.
At first, I tried this problem via Process of Elimination. I looked at (B) and immediately knew that that fact was false, because water molecules move closer together during fusion, and farther during vaporization. Next, I saw (C), and though it was true, it had no involvement with the problem. (How far the water molecules move does not have anything with the disparity between heat of fusion and heat of vaporization). Then, I saw (D), and though it was also true (vaporization occurs at a higher temperature than fusion, and kinetic energy is directly proportional to temperature), I felt that it did not relate to the problem. So I have ruled out 3 of the 5 answers by now. It was the options (A) and (E) that stumped me. I did not know the relationship between the temperature and Hydrogen bonds in water, and conducted some research on the internet.
I found a useful link http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html , and saw that answer choice (A) is in fact true, if you look at the graph towards the end of the page. It has more hydrogen bonds in the solid state than in the vapor/gas state. So, I thought the answer was (A). However, I looked in the answer key provided with this question, and the answer was (E): more hydrogen bonds are broken during vaporization. I don't really see how this is, and would like assistance and answers why. Also, I would greatly appreciate an explanation for why the number of bonds increases when temperature is low, and decreases when temperature is higher. Even an explanation on how hydrogen bond numbers (approximately) change as phase changes would be great.

This might help:
Why Everybody is Confused About Hydrogen Bonding in Water?
https://www.youtube.com/watch?v=RfNuWJDJvRw

The molar heat of fusion for water is 6.02 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Fusion.html), while the molar heat of vaporization for water is 40.7 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Vaporization.html). That is equal to 6 times difference, even a little more. I suggest this energy is used to break existing hydrogen bonds and barely for anything else. The energy of hydrogen bonds is "6-30 kJ/mol" according to this source: http://book.bionumbers.org/what-is-the-energy-of-a-hydrogen-bond/.
Let's consider for simplicity, that breaking 1 mole of hydrogen bonds will require 6.02kJ of energy.
So, when water melts, there is maximum one mole of hydrogen bonds is broken per one mole of water. That is totally different for vaporization process: during vaporization of 1 mole of water it looks like 6 moles of hydrogen bonds are broken.
This invariably leads one to answer E:
More hydrogen bonds are broken during vaporization.


text:
analytical chemistry standard deviation
Analysis of a bottle of $100$mg of vitamin C, yields an average vitamin content of $99.8\pm0.3$ mg. Assuming Gaussian statistics, which of the following is true and the answer is $68\%$ of tablets contains between $99.5$ and $100.1$ mg of vitamin C how is that?

You should look into what is meant by Gaussian statistics or normal distribution. In the case of a normally distributed random variable, approximately 68% of the set should be within one standard deviation of the mean. Particularly look at the section entitled "Standard deviation and coverage".


text:
How does alcohol keep smell and taste flavors in wine?
Nowadays alcohol free wine is also sometimes available. Now is seems that this wine has less taste and flavors in it. The cause for this is that alcohol seems to the capability to keep these flavors longer inside the wine. But how can alcohol do this?

First of all, ethanol has its own smell (which we perceive as taste), it also can stimulate pain sensors in the mouth. Therefore alcohol free wine will taste different.
The process of separating alcohol from wine is quite invasive: the water/alcohol is separated from flavour and colour compounds by reverse osmosis, then the water is distilled to remove alcohol, the water is added back, to the concentrated wine flavour. This process is surprisingly good at not removing anything but the alcohol, but it does change the flavour. Other small light volatile compounds are lost by this process. Tannins are also lost in this process, resulting in reds that lack body
Next the presence of alcohol helps some aroma compounds to evaporate. This means that the aroma of the wine changes. The scent chemicals can't get into the air without alcohol to carry them.
Finally, of course, the best wine is sold with alcohol. When you drink non-alcoholic wine, you are usually drinking a lower quality source wine.
source: http://winefolly.com/tutorial/non-alcoholic-wine/

I don't think it is necessarily the ethanol (alcohol) that keeps the flavor in wine. It is more about the wine-making process. Oenology, is the study of winemaking in case you're curious.
During wine fermentations, yeast is often added to a fruit-must, such as that from grapes. Sugars like fructose and sucrose are often fermented to give a sweet flavor. And during fermentations, other biproducts like acetate, acetaldehyde, or glycerol are also produced to give wines a distinct flavor.
In an alcohol-free wine, I would imagine that there is no fermentation, and therefore no "wine-chemicals" are produced. Or, if the ethanol is removed, other sensory chemicals are probably altered or removed in the process.

Alcohol free wine is made by removing the alcohol from wine after a fermentation.


text:
NMR spectroscopy
i have a question regarding this nmr spectrum, after looking at HMQC & HMBC & COSY & 13C & 1H Spectrums I'm 100% sure of everything except of the hydrogens at position number 3 ( i marked it in th image ) it should be a triplet but it sounds as a doublet and i don't know why ?
thanks in advance

H3 is never going to be a triplet. If just looking at 3J, then it will be coupled to H4 and H8, which would make it a doublet of doublets. This substituted pyrrolizine is a very sterically constricted fused ring; the possibility that you do not observe one of these 3J couplings is not unreasonable. In fact, if your assignments are correct, then you see the same pattern with your H6 protons, with the 2J (geminal) coupling apparent, but only one of the H6 protons showing coupling through to H7.
Remember that in restricted ring systems, it is often possible to see 4J and 5J as well.
However, if you look closely, you will see that the H6 signals are wanting to split into smaller doublets. I'm sure you'd see it if you reprocessed this data using a suitable window function; from the data you've got it looks like it was run on an Avance 500, so using Topspin you could set gb 0.3 and lb -0.3 and reprocess using gfp, or just set lb -.3 and process using efp. Alternatively, as this is a MNova printout, you can use interactive processing to see (using similar parameters)


text:
Ka relating to Dissociation of Weak acid and Water
I am studying $K_a$ currently and there is a passage in the book that doesn't appear to make sense.
It states that from the equation
$$\ce{HA <=> H+ + A-}$$
that $$K_a = \frac{[H^+][A^-]}{[HA]}$$
The following passage says When dealing with weak acids, you can assume that dissociation of the acid is much greater than the dissociation of water.
It continues to say that you can assume all the H+ ions come from the acid.
hence $$K_a = \frac{[H^+]^2}{[HA]}$$
What I don't understand is the dissociation of water.
The book gives the equation of dissociation of water as the following -
$$\ce{H_2O(l) + H_2O(l) <=> H3O+ + OH-}$$
but where does the H+ come from from this dissociation - it's made a Hydroxonium ion?
Is it because the $H_3O^+$ can act as the conjugate acid and release the proton?
Please help!

Yes, water autodissociates according to the equation $\ce{2H2O <=> H3O+ + OH-}$.
But I should point out that the equilibrium constant is written as though the concentration of water itself is a constant, so the reactions becomes:
$\ce{H2O <=> H+ + OH-}$.
and the equilibrium is expressed as:
$K_\rm{w} = 1\times10^{-14} = \ce{[H+][OH-]}$
The pH is $\text{-log}\ce{([H+])}$ thus the pH of pure water is 7.
This also has to do with significant figures. Few pKa's are known to better than two significant figures. So:

For weak acid with pKa < 5 you can ignore water contribution.
For acid with pKa > 9 you can assume all the  $\ce{H+}$ is from water and that all of the acid is essentially present as the HA form.
For 9 > pKa > 5 you have to consider both ionization of the acid and water.

I think MaxW's answer covers the gist of the question from the calculation side, but its also worth noting something else about your question. $\ce{H^+}$ and $\ce{H3O^+}$ are often used interchangeably for the acidic proton, but $\ce{H3O^+}$ is more accurate; free protons are unlikely to exist in solution. $\ce{H3O^+}$ is one example of how the proton could be passed into the solution and so is used to represent the general case of the proton being solvated by a water cluster. For more information on these water clusters, go the solvation section of https://en.wikipedia.org/wiki/Hydronium.


text:
Can plastic be used as fuel for vehicles?
I don't have much of a background in chemistry, but I know that plastics are largely made of carbon, as is oil. Is there any way to "covert", i.e. a plastic water bottle, into a substance that can be used as fuel for ? Thanks.

Even lots of ways...

Is it often done? I feel like this would really help the problem with what to do with all the garbage that pollutes the earth. Although, emissions aren't particularly great either...

There are lots of recycling methods.

For this you can google plastic pyrolysis and see

There are many ways to recycle different types of plastic, and conversion to liquid fuels suitable for internal combustion engines is currently done on an industrial scale. The key to converting the carbon contained in long-chain polymers that make up the plastics is called pyrolysis, which is essentially a way to thermally decompose these products into shorter chain hydrocarbons suitable for use as liquid (or even gaseous or solid) fuel.  According to this Wikipedia article:

Anhydrous pyrolysis can also be used to produce liquid fuel similar to diesel from plastic waste, with a higher cetane value and lower sulfur content than traditional diesel.

An example of a current commercial waste-plastic-to-fuel operation is the Canadian corportion Plastic2Oil, Inc., who claims to "convert waste plastic to ultra clean oil" in their 250,000 gallon fuel production and blending facility.
The following is an excerpt from the description of the patent-pending process used by Plastic2Oil:

The processor accepts unwashed, unsorted waste plastics, composites and commingled materials. Although many sources of feedstock are available, we are focusing initially on post-commercial and industrial sources, since these are readily available in large supply, and present a cost-effective solution for companies. We currently have several companies providing plastic feedstock.
The hopper is loaded with up to 4,000 lbs. of plastic feedstock per machine per hour using a forklift. The plastic is loaded into the pre-melt reactor by a continuous conveyor between the hopper and the reactor. The plastic is then heated using the off-gases produced by the process. After the plastic has been liquified in the pre-melt reactor it passes through a solids-liquids separator before going into the main reactor.
In the main reactor, the liquified plastic hydrocarbons are cracked into various shorter hydrocarbon chains and exit in a gaseous state. Plastic2Oil's proprietary catalyst and unique process engineering enables us to capture an average of 86% of the hydrocarbon content of plastic. The Petcoke residue produced at this stage (approx. 2-4%) remains in the processor chamber and is automatically removed while the processor is in operation.
From the main reactor, the gases that are fuel oil and diesel are condensed and separated, then proceed into temporary fuel tanks. All of the gaseous "light fractions" or off-gases (approx. 10-12% of process output), such as methane, ethane, butane and propane, exit the system and are used to fuel the furnaces.

Note: I have no affiliation whatsoever with Plastic2Oil or their industry. I chose their company as an example largely at random, or more accurately, due to their very usable and easily found website.


text:
GOC allylic,vinylic, benzylic positions, carbocation stability
Carbo-cations may be stabilized by:
(a) π-bonds only at allylic position

(b) π-bonds only at vinylic position

(c) π-bonds at allylic and benzylic position also
(d) -I effect

While the answer is obviously not (d), I am really confused about what 'allylic', 'vinylic' and 'benzylic' positions actually mean. I have heard about allylic/vinylic/benzylic carbons, but positions..?
Please, help.
Thank-you.

The question has been taken from ADVANCED PROBLEMS IN ORGANIC CHEMISTRY by HIMANSHU PANDEY

allylic position explanation
benzylic position explanation
vinylic position explanation

For carbocation stability, think about which case has opportunity for electron resonance, i.e., which case(s) allows the positive charge to be delocalized to adjacent pi bonds. This resource might help.

Allylic position is the same as allylic carbon....similarly for vinylic and benzylic...so the stability of carbocations goes as
benzylic > allylic > vinylic
vinyl carbocations moreover are highly unstable...whereas the other two are rather stable ones...and the answer is (C)


text:
Decreasing order of basic strength
I need to find the decreasing basicity of the following compounds(I, II, III, IV, respectively):

I know that to compare the basic strengths, we need to find the stability of their conjugate acids.
Following are the conjugate acids, I drew (I don't know if they're correct):

According to me, due to resonance of the benzyl groups or lack of resonance, the order of basic strength should be:
$$\mathrm{III > II > I > IV}$$
But the answer given is:
$$\mathrm{IV > I > II > III}$$
Where am I going wrong?

Resonance doesn't occur here because N can't expand its octet. Now I am not sure thereon, but I think that due to a lack of resonance the conjugate acids are roughly the same stability so we look at the stabilities of the original bases. Now here the lone pair is in resonance in three structures and you can see the order of stability. Thus the most stable base will have poor basic strength and you will get the answer. However, I am not sure of this. Maybe someone can give a definitive explanation.

@Sawarnik , yes. That could be a method. But, why can't N expand its octet here?

It has no d orbitals, so with only the s and p orbitals it can accomodate a max. of 8 electrons. If you drew resonance structures, it will have 5 bonds aka 10 electrons which is not possible.

To be more precise: nitrogen does have virtual d orbitals, like sulphur, phosphorus or whatever. None of them, however, expand their octet because the d orbitals are energetically not accessible.

Basicity is how well an atom can donate electrons. More is the electron density more is the basicity of the compound.
In Aniline the lone pair of electrons present on the nitrogen are involved in resonance, so the electron density on N is less than expected.
In compound II (Diphenylamine), since two benzene ring is present so the electron density will be less as now lone pair are involved in resonance of both the ring.
Thus electron density on Nitrogen is less than as compare to Nitrogen atom of Aniline
Similarly, in III compound (Triphenylamine), since three benzene ring is present the lone pair are more localized, and, thus electron density will be much less (least) as compared to the other compounds
In 4th structure, there is no resonance so, lone pair are present with nitrogen itself, and the electron density will be high, also due to inductive effect of the alkyl groups the electron density is more, and thus it can easily donate the electrons as compared to all other compounds. So the basicity is the highest for compound IV.
Therefore, the order of basicity is IV>I>II>III.

Exactly what i thought .. i have seen a few questions which were solved by the same logic (where N was acting as a base). Why we don't care about the conjugate acid is something i don't fully understand though.

You are thinking in advance, like conditions after the H+ is accepted, but in order to accept H+ atom, first there should be some electron density present around the donor atom. If there is no (less) electron density then H+ will not get attached, and this is where you go wrong.
I suggest you to solve the basicity problems of this type by considering the electron density, and problems based on acidity by conjugate acid-base method.

As we know if conjugate acid is more stable then their parent compound is least stable.so in 4th option their is no any resonance so it is least stable therefore it's parent compound is more stable.so correct option is iv>I>ii>iii

The aryl group is electron withdrawing in nature, so as the no.of of aryl groups increase, the acidity increase while $\ce{CH3}$ being electron donating in nature leads to +I effect and thus to basic strength..
That's why (IV) will come first then (I) then (II) and then after (III).
So the order as per decreasing basic strength is (IV)> (I)>(II)> (III)

i guess i could help out on this one..as i have completed the book(for JEE)
The reason for the abnormal order is that the presence of more than one phenyl groups dosent actually increase the stability of the conj acid...rather it just increases the steric crowding around 'N'....The reason why more phenyl groups dont add to stability is that the phenyl grups due to repulsions between themseles keep rotating about the axis and are not always in the same plane for conjugation...thus do not really help in stabilising the N+ ion.....and i guess youd know that cyclohexyl amine is more basic than anilene...so for that simple reason.....4>1>2>3
P.S:The answer was provided to me by a teacher at FIITJEE...so you can be sure its authentic

POINT 1. If you think resonance will play a role try drawing resonance structure. Got stuck hmmm, because nitrogen can not form 5 bonds or can not expand its octet. Hence, resonance will not play any role.
POINT 2. Now, coming to conjugate method, add $\ce{H+}$ to the structures and then try to compare stability by considering inductive effect ( phenyl group shows -I effect ). You can not use resonance effect for comparing because of reason already discussed in point 1, that is, -I effect will be maximum in the case of triphenyl then di phenyl then phenyl and so on so the correct order will be 4>1>2>3.
POINT3. Now consider steric factors. The greather the steric hindrance, the less will be the basicity because $\ce{H+}$ will face more difficulty in coming close. So again, according to steric factors, the order will be 4>1>2>3.

I will clear your doubt.
In the first, $\ce{NH3+}$ is in conjugation with the benzene and also it is a $+M$ group which increases the negative charge and decreases the $K_\mathrm a$ value and in turn increases the $\mathrm pK_\mathrm a$ value, meaning it increases the pH of the solution (this is the same for the 1st 2nd and the 3rd).  The fourth one doesn't have conjugation with the ring so the fourth has the highest acidity.
Now you have to give the order according to the steric hindrance
so the correct answer would be:
$4>1>2>3$


text:
Rate of reaction and dependence on concentrations
Consider the decomposition of ozone as follows

(fast)
$$\ce{O3  <=> O2 + O}$$

(slow)
$$\ce{O + O3  -> O2 }$$

The question comes here if the concentration of $\ce{O2}$ is increased , then comment on the rate of the reaction.

increases
decreases
constant
cannot be predicted

My try: I have a confusion that the rate law is not given so we cannot tell about the actual dependence of the rate on the concentration of $\ce{O2}$, I know that the rate law will depend upon the slow step as it is the rate determining step. Answer given is 2) decreases , can we apply Le Chatelier's principle here? Increase in $\ce{O2}$ , backward reaction so less rate?

For future reference, please don't use math markup $...$ to get italics.  Use Markdown emphasis *...* instead.

@zwol I will take care of that in future questions. Thanks.

If we carefully see them ozone formation is a complex ( not elementary) reaction.
Yes the slowest step would be rate determining .
so for slowest step ;
Rate =$   k [O] [O _3] $
where $k$ would be rate constant.
but  final reaction is :
$$\ce { 2 O_3 -> 3O_2} $$
and so ,
equillibrium constant K(for 1st reaction)
$K= \frac{[O][O_2]}{[O_3]}$
thus substituting $[O] $from above equation to the rate equation .
We get
Rate =$\frac {  k × K [O _3]^2 }{[O_2]}$
And hence if $[O_2]$ is increased it causes rate to decrease.

Here is an intuitive way to see that the rate of decomposition should decrease.
The second reaction only happens when an $\ce{O}$ radical collides with a ozone molecule. If you increase the concentration of $\ce{O2}$, the proportion of $\ce{O3}$ must go down (because the proportions of all species must add up to 1) and therefore the frequency of $\ce{O + O3}$ collisions must go down as well.

Intuitively, yes, increasing the concentration of $\ce{O2}$ will drive the (quasi-)equilibrium of the fast reaction $\ce{O3 <=> O2 + O}$ towards the left, decreasing the concentration of the intermediate species $\ce{O}$, and thus slowing down the rate-determining slow reaction $\ce{O + O3 -> 2 O2}$.
A minor complication here is that the rate of the slow reaction is also proportional to the concentration of $\ce{O3}$, which driving the fast reaction towards the left side will increase.  Indeed, in principle, the rate of the slow reaction would be maximized when the fast quasi-equilibrium concentration of $\ce{O}$ was equal to that of $\ce{O3}$, with any deviation from this in either direction slowing down the reaction.
The missing piece of the puzzle is that the combined reaction is called "the decomposition of ozone", not "the dimerization of monatomic oxygen".  It is thus reasonable to assume that $\ce{O3}$ is at least initially present in considerable excess of $\ce{O}$.  Indeed, given the extreme reactivity of monatomic oxygen, it's never likely to be present at more than trace quantities under typical conditions.  Thus, converting some $\ce{O}$ back into $\ce{O3}$ by reacting it with $\ce{O2}$ will reduce the (already very low) $\ce{O}$ concentration by a much larger factor than it will increase the (comparatively high) $\ce{O3}$ concentration, and will thus slow down the rate-determining step of the reaction.
In any case, even if we did somehow (e.g. by running the reaction in near vacuum) contrive a situation where the quasi-equilibrium of the fast reaction was shifted so far to the right that the concentration of $\ce{O}$ became comparable to that of $\ce{O3}$, then we would also need to consider the additional reaction $\ce{2 O -> O2}$ whose rate depends (quadratically!) only on the $\ce{O}$ concentration. I would expect that a more thorough analysis (which would, of course, have to take into account the relative rate constants of the different competing reactions) should turn up no situation where adding $\ce{O2}$ would actually speed up the overall decomposition of $\ce{O3}$.  Of course, I could be wrong; these things are not always intuitive.


text:
Steroisomerism in cyclic compounds
What about stereochemistry of 1,2,3,4-tetrahdroxy cyclobutane
How will we decide the stereocentre it's eating me can anybody help.
Answer is 4 stereoisomers

Related: IUPAC name for 1,2,3-trichlorocyclopropane? and 1,2,3-Trimethylcyclopropane stereoisomers

In the RULES FOR THE NOMENCLATURE OF ORGANIC CHEMISTRY
SECTION E: STEREOCHEMISTRY, published in Pure Appl. Chem., 1976, 45, 11-30, you find:

E—2.3.3. When one substituent and one hydrogen atom
  are attached at each of more than two positions of a
  monocycle, the steric relations of the substituents are
  expressed by adding r (for reference substituent), followed
  by a hyphen, before the locant of the lowestnumbered
  of these substituents and c or t (as appropriate),
  followed by a hyphen, before the locants of the
  other substituents to express their relation to the reference
  substituent.

This means: pick one centre as a reference (r-1) and express the relative stereochemistry of the other centres, that is the relative orientation of the $\ce{OH}$ groups with respect to the reference, using c (if the substituent is on the same face of the ring) or t (if the substituent is on the opposite face of the ring).

From what i see...at a basic point of view...there is no carbon with 4 different functional groups...hence i guess its optically inactive...as in each carbon is bonded to 2 identical carbons which are further bonded to the same carbon ...so the functional group dosent matter...Hopefully thats it

But it has 4 stereoisomers

To solve this first we number the carbons as

Now when we consider cis form of 1 and 2 carbon there will be two stereoisomers cis and trans for 3 and 4 carbon
Now when we consider trans form of 1 and 2 carbon there will be two stereoisomers cis and trans for 3 and 4 carbon
Hence total 4 stereoisomers


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Do dendrites effect lithium-ion polymer batteries?
Do dendrites effect lithium-ion polymer batteries?
I recently came across this video which shows dendrites at a microscopic level durring the charge and discharge cycles of a standard Li-Ion battery.
I have come to understand that the primary difference between Li-Ion and Li-Poly batteries is the electrolyte used between electrodes - Li-Ion using a liquid electrode, and Li-Poly using some type of polymer chain.

Do dendrites form across the polymer the same as the do through a liquid electrolyte?

According to this Berkley Labs publication, "Roots of the Lithium Battery Problem: Berkeley Lab Researchers Find Dendrites Start Below the Surface":

Using X-ray microtomography at Berkeley Lab’s Advanced Light Source (ALS), a team led by Nitash Balsara, a faculty scientist with Berkeley Lab’s Materials Sciences Division, observed the seeds of dendrites forming in lithium anodes and growing out into a polymer electrolyte during cycling. It was not until the advanced stages of development that the bulk of dendrite material was in the electrolyte. Balsara and his colleagues suspect that non-conductive contaminants in the lithium anode trigger dendrite nucleation.
  “Contrary to conventional wisdom, it seems that preventing dendrite formation in polymer electrolytes depends on inhibiting the formation of subsurface dendritic structures in the lithium electrode,” Balsara says.

So, it would seem that dendrite formation through polymer electrodes is indeed a problem, though it may require a "non-conductive contaminant" to initiate nucleation if these authors are correct.

What is the definition of dendrite?

@Mockingbird , in this context it refers to a kind of crystalline growth. There is a Wikipedia article here that addresses the kind of dendrites discussed above.


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What is wrong with the rection
I had a reaction in which I have to find the product .
I figured out something .

But my sir told me that this is a wrong product . I could understand why , as the reagent used act as decarboxylation .

There's no stabilized anion. Decarboxylation seems unlikely.

Decarboxylation proceeds through free radical mechanism...and we see scope for rearrangement of radical...to obtain more substituted product(thermodynamic product....as we are supplying heat)and hence we can obtain the product
2-Methylbutan-2-ene...Hopefully thats your product

Do you have a reference that the reaction proceeds through free radical mechanism?

I don't see why this reaction would proceed through a free radical mechanism. The reaction could just decarboxylate and pick up a proton on the far left carbon through resonance.

sir,im still a student, and have learnt that thermal decarboxylation proceeds through a radical mech....in case you think otherwise...please do enlighten me too..regards@SendersReagent

moreover...whatever charged intermediate it forms...it can rearrange to give a thermodynamic product


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Can an improvised torch be made to last for a prolonged period of time?
In the movie First Blood, there is a scene where Rambo is trapped in a cave after escaping a bazooka rocket that caused the mine he was hiding in to collapse. After realizing the entrance has been blocked, he decides to head down into the cave, and using the fire he built, a branch and some cloth he has, he prepares a torch (I can't remember now if he dips the cloth in some flammable liquid or not). Then he wanders through the cave for some time until he finds his way out, and the torch is holding up rather nicely... (he is seen feeding it with more strips of cloth from time to time)
My question is: is it realistic? Can a torch made from a branch and a burning cloth (whether or not dipped in gasoline or something else) be used for a prolonged period of time while remaining useful (i.e. not being burned-down, too hot to handle, etc.)?

The key is whether or not he added some kind of fuel to the cloth. Without that, a flammable cloth would burn itself out very quickly. If he had kerosene or some other fuel to soak the cloth, then the cloth would act somewhat like a wick in an oil lamp (except that it would not be "wicking" fuel up from any fuel source) in that mostly the fuel would burn rather than the cloth.
If he had a liquid fuel source, even a non-flammable cloth like his clothing would have been made of would have sufficed. In fact, he probably wouldn't have had to keep adding cloth in that case, but he would have had to add more fuel periodically.
So, the bottom line answer is: Without a liquid fuel source to soak the cloth, no chance. With a liquid fuel source to soak the cloth, no problem.


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Why the product is wrong
In the followig reaction .
I predict a product .
That is
but that is wrong I could not understand why .

The most acidic proton is attached to the carbon bearing the nitro group so treatment with $\ce{OH-}$ will remove that and then eliminate $\ce{F-}$ giving the nitro-alkene.

Elimination of a poor leaving group such as F proceeds through E1CB mechanism.
Now here, the base $\ce{OH-}$ would abstract the proton from the site where the negative charge would be best stabilised (i.e. most acidic proton) which happens to be from the carbon bearing the nitro group, as $\ce{NO2}$ group is a strong electron withdrawing group, hence after abstraction of proton by the base, the negative charge attacks the nearest electron deficient carbon (i.e. the one bonded to $\ce{F}$) as $\ce{F}$ is also a good electron withdrawing group (Note: $\ce{F}$ pulls electrons through -I effect only, whereas $\ce{NO2}$ can do the same through -R effect when the negative charge is on the carbon bonded to it). Thus the negative charge expells the $\ce{F}$ atom as $\ce{F-}$ ion, which results in formation of a double bond between the two carbons.
Giving you the final product as $\ce{CH3-CH=CH2-NO2}$


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Possible ammonia anomaly?
In class we have learned that an acid + a base/alkali --> salt + water.
For example:

hydrochloric acid + sodium hydroxide --> sodium chloride + water

However, when I try the same thing with ammonia, I get something like the following:

hydrochloric acid + ammonia--> ammonium chloride

Since there is no oxygen present on the left side, I cannot possibly put water as an extra reagent on the right. However, doesn't this contradict the fact that this type of reaction always forms extra
water? (I do realize there will always be water present, as either the acid and/or the base are in aqueous solution, but I'm talking about the formation of new water) Is this an anomaly of ammonia, or is there something fundamentally wrong about my equations?

Is this an anomaly of ammonia, or is there something fundamentally wrong about my equations?

Fundamentally wrong would be a bit harsh ;)
But notice that the reaction
$$\ce{HCl + NaOH -> NaCl + H2O}$$
is just a special case of a general reaction between an acid and a base. It is special in the sense that the base on the left side of the equation is a hydroxide.
The more general equation for this kind of reaction is
$$\ce{HA + B -> A- + HB+}$$
With other words, the Brønsted-Lowry acid donates a proton to the base. As a result, the acid $\ce{HA}$ is converted to its corresponding base $\ce{A-}$ while the base $\ce{B}$ is converted to its corresponding acid $\ce{HB+}$.
In your first equation, $\ce{OH-}$ is the base, while $\ce{H2O}$ is its corresponding acid.


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How to calculate the remaining mass in the mother solution after multiple extractions?
During an extraction, a solute ($\ce{S}$) will partition itself between the two solvent phases with a partition coefficient ($K$) defined as:
$$\ce{S(solvent 1) <=> S(solvent 2)}$$
$$K = \frac{[\ce{S}]_2}{[\ce{S}]_1}$$
Solute $\ce{S}$ is found to have a partition coefficient of 15 between solvent 1 and solvent 2; i.e., with 15 times as much in solvent 2. Suppose $\pu{150 mL}$ of a $\pu{0.02 M}$ solution of $\ce{S}$ in solvent 1 is extracted with \ce{\pu{50 mL}} of solvent 2. What fraction of $\ce{S}$ remains in solvent 1? If three $\pu{50 mL}$ extractions are performed, what fraction of $\ce{S}$ remains in solvent 1?

In old lecture notes of mine I found this formula
$ g_i = g\left(\frac{V_1}{V_1 + V_2K_2}\right)^n$
where $g_i$ is the remnant mass of a product initially dissolved in volume $V_1$, to be $n$-times extracted by volume $V_2$ with a given Nernst dissipation coefficient $K = \frac{a_2}{a_1}$.  Here, $g$ refers to the mass of the product initially dissolved in $V_1$.
Applied to your case, assuming a molar mass of 180 g/mol, this would read like 0.54 g initially dissolved in you 150 mL solution 1 (3 mmol).  After a single extraction with 50 mL of 2, only 90 mg were remaining (0.5 mmol) in $V_1$, after the third time 2.5 mg (about 14 umol).
I speculate a textbook like P. W. Atkins' Physical Chemistry may contain an example to confirm such a calculation.


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Multidimensional potential energy scans - picking the ideal dimensions to scan
I am relatively new to electronic structure calculations, and I am trying to figure out the most appropriate way of scanning a potential energy surface (PES) to locate an appropriate transition state.
The proposed reactants and products can be found at the bottom of this post (only the relevant fragments are shown).
What is the most appropriate PES scan to perform? It seems logical to scan through possible values of the bond length between atoms 9 (blue) and 6 (red). But should I actually be doing a two-dimensional scan, the other dimension being the bond length between atoms 9 (blue) and 4 (orange)? Or perhaps the bond angle between atoms 9-6-5 is better? I read that doing a two-dimensional scan over two parameters that are related to the same structural parameter may not be ideal.
Any suggestions would be greatly appreciated!
Reactant:

Product: Note that N2 is also the product but is infinitely separated.

Why not search for the TS first and then decide what the best coordinates are (most likely the bonds broken/build up)?

Hi @DSVA. Yes, that is what I ended up doing. It seems that the 1-D scan of the bond length between atoms 6 and 9 produced a maximum that was close to the transition state, which I then found and confirmed via an IRC calculation. I suppose it was best to search first then decide, as you suggested. I just wasn't sure if there was a general rule of thumb or flow of logic that I was missing.


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Deflection of Alpha Particles in Rutherford's model of atom
Rutherford stated that deflection of alpha particles is due to repulsive positive charge of nucleus. Why can't alpha particles be deflected by attraction of so many electrons revolving around nucleus?

http://chemistry.stackexchange.com/questions/10390/why-werent-the-electrons-attracted-to-doubly-charged-helium-particles-in-the-ru http://chemistry.stackexchange.com/questions/64309/rutherfords-alpha-ray-scattering-experiment

also http://chemistry.stackexchange.com/questions/51736/why-did-the-alpha-particles-in-rutherfords-experiment-not-collide-with-the-elec

See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles.
Figure 1 shows a sketch of the apparatus.

Figure 1: depiction of the apparatus (courtesy of Wikipedia).
Now follows a simplified rationalisation.
Assume you have an atom-thick sheet of some material.
By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $n$ are there, and knowing the sheet's surface area $A$ allows you to calculate the atomic surface density $\sigma = n/A$.
Now if the atoms have radius $\rho$, an area $A_\rho = n \pi \rho^2$ of the sheet should be covered by atoms.
This means a ratio
$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$
is actually covered by atoms and $f = 1 - r$ is the ratio of "free space".
Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through.
Imagine $M$ particles were thrown and $m$ of those passed through.
If $M$ is big enough we can set $f_R = m/M$ as Rutherford's "free space" estimate.
Going backwards we may set the coverage ratio
$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$
were $\rho_R$ above is the measured radius of the collision cross section of an atom.
The experiment has shown that $r_R \ll r$ and thus $\rho_R \ll \rho$, which implies that most of the atomic mass is buried inside the atom.
In fact (Wikipedia),

The diameter of the nucleus is in the range of $1.75$ fm ($1.75 \times 10^{-15}$ m) for hydrogen (the diameter of a single proton) to about $15$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $23,000$ (uranium) to about $145,000$ (hydrogen).

In Rutherford's own words (p. 68),

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
--- Ernest Rutherford

Thus, nuclei are simply very small, but very dense, as stated by Enigma and Ilmari Karonen.

Don't you mean "simply very small" (but dense)? A big low-density nucleus would match the Thomson plum pudding model, which is precisely what Rutherford's experiment disproved.

You are right! I'll change the wording, thanks!

The simple reason for this being the ratio of masses of alpha particles and electrons.....Both experience equal electrostatic forces which leaves no option for the electron but to get displaced from its current position ( it gains potential energy to do so) whereas in the case of alpha-nucleus interaction,the size of a nucleus is comparable (in the case of the gold foil he used...way more than that of an alpha particle) to that of the projected alpha particle hence causes a measureable deflection on the alpha particle....Hopefully you got your answer

For comparison, if a 747 runs into a person, which one is deflected more? I actually ran this analysis, and the ratio of rest masses is comparable.

I was also wondering upon the fact that it could be the electrons which caused the sway, but you see if a certain force were acting on two particles, one lighter than the other, the lighter one will be the one feeling the "force". For more clarity refer to Newtons second law in acceleration format. So the only left out thing was a nucleus, which is way more massive than the alpha particle, again apply our first logic here(stated above) and you see, this time its the He(2+) that has to do the displacement.
Hope it resolves your issue.

Well gentlemen we can't deny that fact that there might be some forces of attraction between the alpha particles as well as the electrons,because the alpha particles passed through the atom by small distance margin from nucleus and stroked the zinc sulphide coating but it wasn't clearly stated by Ernest Rutherford that whether the deflection was also caused due to alpha particle attraction and election attraction. Also, attraction (interaction)is between comparable things like a heavy nucleus and heavy alpha particle but it's silly to think about alpha particle and an electron , the alpha particle is approximately 4000 times heavier than electron and it is travelling with a high speed (energy) so we can say that the interaction time might be very much less in order to show its effect and we can underestimate it . though most of the physics practicals are based on assumptions and we have to underestimate those assumptions that's way because the reason for performing the experiments was to propose a model of an atom (accidentally) , to find where is the mass of the atom is concentrated. that's why we got the  so we don't have anything to do with the alpha particle and electron attraction.


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Claisen ester condensation
Why is it a requirement that the ester should have 2 alpha hydrogens in this reaction although we actually need just one for the reaction to occur?

Great question!
The Claisen reaction happens under equilibrium conditions, and the addition of the ester enolate is highly reversible. Your mechanism omits the very important end of this reaction, which is that the β-keto ester product is highly acidic and easily deprotonated to yield the corresponding enolate.
The enolate formation drives the reaction forward according to Le Châtelier's principle. Very importantly, your question about the α-hydrogen. The deprotonation of the β-keto ester is not possible without two hydrogens on the original ester that will serve as the nucleophile.

This photo could use cropping... Also it's better to post drawings of reactions made with software.

I don't have a good reaction editor anymore. I looked at Marvin sketch but their free license doesn't seem to apply to me. Suggestions?

Chemsketch, ISIS Draw...

@Zhe I took the liberty to redraw the reaction scheme with chemfig; in case you find it unacceptable, feel free to revert my edit. If you are interested, the $\mathrm{\LaTeX}$ code can be found at Pastebin.

@andselisk You'll get no objections from me. Thanks!

Two hydrogen  are use , because if only one H is used then again the product form having acidic H which is abstracted by ethoxide ion then this ethoxide ion convert in Ethanol and final product is obtained.


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How does O-benzoyl-N-methylhydroxylamine hydrochloride oxidise this ketone?

How does the mechanism of this reaction with BzONHMe work? Can part of it act as a base and abstract the alpha proton? Would be great if someone could help me!

This can't act as a base. It is a hydrochloride salt.

The first step is the formation of the Me,O-Bz imine of the ketone. This tautomerises to the enamine which then does a pericyclic rearrangement with the carbonyl of the benzoyl group to install O at the alpha position. Finally  the imine hydrolyses to reveal the ketone. See Aldrich Chemfiles (2007) 7.1, page 7.

So the HCl doesn't really do anything? Can't it protonate the ketone and then the alpha-proton gets abstracted by the amine resulting in a positive charge on the amine, letting the neighbouring oxygen get taken by the double bond between carbonyl carbon and alpha-carbon?

Remember that the substrate you are reacting with has a basic centre which is more basic than the benzoylhydroxylamine

Thank you very much! I figured it out now, your mechanism makes a lot of sense!

@ClaraMarie If an answer has helped you, it's a good idea to upvote and/or accept it (green tick next to voting buttons). Please read What should I do when someone answers my question? for further details.


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How do thermodynamics and kinetics control this reaction?
The answer is A, but how is that? I know B is an intermediate, so that after short time it will be bigger than C, but what does "thermodynamic control" and "kinetic control" mean?

Initially no B and C exist. At short times as $k_1 > k_2$ production of B is favoured over C. (The initial amounts of B and C are so small that back reactions can be ignored wrt. forwards ones.) At long times  equilibrium constants matter and as the equilibrium constant $K_2$ is greatest this is favoured to produce C over B.

A kinetic product arises when the reaction is not supplied any additional conditions..therby proceeds through a stable intermediate.
A thermodynamically controlled reaction is one where you supply heat or any such necessary condition to get a specific product ( mostly the more substituted one...ie saydseff products)...thus goes for a stable product even if a not so stable intermediate forms in the mechanism.
I guess thats about it
P.S:Saydseff for organic reactions..idk what you call it in inorganic chemistry


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Tosyl Chloride? why not anything else
Why do we encounter only Tosyl groups and not anything at the end of the benzene group (ie not to replace the SO2Cl part..rather the Me part at the other end)
I mean something like a CF3 or NO2 would help make it a better leaving group right?

The NO2 version is sometimes used, it's called nosyl chloride iirc.

I would guess that the 4-CF3-benzene sulfonyl chloride is significantly more expensive

Part of the reason why p-tosyl chloride is seen so often is not only the ease of its preparation, yet that is a side product in the preparation of o-tosyl chloride towards saccharin.  Ease in storage and addition / use are beneficial.  While a p-nosyl chloride eventually may yield a better nucleofuge than the p-tosyl chloride, the very same nitrogroup may ease the hydrolysis of the chloride as well.


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Why does bleach feel slippery?
My high-school chemistry teacher taught us the mantra "bases are slippery-soapy-slimy!"
This question and this answer in Quora say this is due to saponification - conversion of lipids from the top layer dead skin cells into a soap-like substance.
During some household cleaning I ended up with a small amount of a fairly strong solution of bleach in water between my thumb and forefinger and noticed it felt quite slippery, and it took several seconds of rubbing under water for it to stop.
In this case I would guess that saponification is not the explanation for the very slippery feeling. Is there another explanation?

Bases also hydrolize proteins, in oder words, they disolve slowly your skin therefore they feel slipery.

@Greg that sounds more like what I remember also - the same reason a caustic drain cleaner will "dissolve" or break down hair in a clogged drain.

note: Info from the CDC as well as from phys.org, from Oregon State University and from Lenntech all have a lot to say on the chemistry of bleach as a disinfectant.

I've just asked the related question Is it known for sure that bases feel slippery because of the production of soap/surfactant?.

Maybe it needs to be clarified that the salt of a strong base and a weak acid can conduct saponification. Therefore the fact that bleach reacts with fatty acids creating soap, does not necessarily mean that bleach should be all just base (nor that something else other than saponification should be happening).

Household bleach is mainly sodium hypochlorite ($\ce{NaClO}$) dissolved in water (~<5%). One reason it works as a disinfectant is that it reacts with fatty acids of living organisms' membrane and turns them into soap.
$$\ce{NaClO (bleach) + R-COOH (fatty acid) → HClO + R-COONa (soap)}$$
(There are other mechanisms by which hypochlorite is known to perform disinfection, though the focus of this answer is to address how the slippery feeling comes about)

Why is soap slippery

The non-polar side of the soap molecule is less interactive with solid surfaces than polar substances such as water. Therefore soapy water flows with less friction on solid surfaces compared to water and is more slippery.

I remember being taught in high school about protein hydrolyzation (as mentioned here) by strong basic solutions as well. Could interaction of bleach ingredients (hypochlorite or an added base) be doing this as well?

Yes, it can destroy proteins as well. One way is sodium hypochlorite reacting with amine groups on protein, making amine chlorides and base.

I've done some searching based on answers here, and just added some helpful links to discussions of the chemistry of bleach as a disinfectant in a comment below the question.

There are two highly voted answers that are affirmative to saponification, but yours supports the idea that hypochlorite can lead to saponification without needing to invoke an additional base ingredient.

@Kinformationist op asks as to why it is slippery , I,e why are soaps slippery but you have not mentioned about kindly look into it

@Chemist good point. The assumption here is that we know soap is slippery because the non-polar side of the soap chain is less interacting with solid surfaces than polar substances such as water. Therefore soapy water flows with less friction compared to water and is more slippery.

Gee, thank you for continuing to refine your answer, this is real dedication!

Actually it is saponification.  Bleach has alkali added to it, to stabilize it against decomposion to chlorine gas.
To wash your hands after contact with bleach was a wise move.

I didn't know that. It must be quite basic to feel so slippery. Fingers where not designed to to measure pH, I should look for some litmus paper to check my bleach some day.

@uhoh Cabage will do it, too, and is easier to find (even a chemists household...)  http://www.coolscience.org/CoolScience/Teachers/Activities/CabbageJuice.htm

@Buttonwood Home titration - excellent! Another day of productivity lost forever in the name of science :) I will be extremely careful and use only tiny amounts of bleach, ammonia, or any other strong household substance, and work in a well ventilated area, as titration involving bleach mixtures could release chlorine gas.

Info from the CDC as well as from phys.org, from Oregon State University and from Lenntech all have a lot to say on the chemistry of bleach as a disinfectant. The last two say that NaOH${}^-$(aq) occurs naturally in hypochlorite solutions, rather than as an additive.

@uhoh It would appear to depend on the specific formula you bought as to whether NaOH is added or not. The SDS for "Clorox Regular Bleach(1)" doesn't include any mention of sodium hydroxide, but the SDS their so-called "High Effiency Regular Bleach" and the SDS for "Splash-Less Regular Bleach" both include it.

@uhoh Naturally you want to use bleach only in a well ventillated area with proper protection for yourself.  As an example, using it as dilute warm aqueous solution to clean the shingle-covered walls of your housing (maybe typical not only in New Jersey ...), you hold your brush with a hand covered by a rubber / nitrile glove to protect your skin and you wear googles to protect your eyes from .  And secondary, you won't mix two cleaners together, as some are able to "crack" the hypochlorite present in bleach, and eventually liberate toxic chlorine gas.

We should settle the "Bleach has alkali added to it..." issue. @SendersReagent 's comment suggests all bleach solutions may not necessarily have an added base, and two sources linked in my comment suggest that NaOH is naturally present in aqueous hypochlorite solutions. As a side note, I've just asked this related question.

@Oscar But if the slippery comes from the production of soap, what about baking soda (which also feel slippery)? If it was producing soap too, it would be used to disinfect (it would break down the membrane of bacteria), which doesn't. I don't understand.

If acetic acid has the strength of a typical fatty acid (should be comparable since we gave carboxyl groups), you still get a reaction.  Vinegar + baking soda + teenager working on a science project --> volcanic lava all over the house.


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Can I combine two galvanic cells as one and produce infinite current
I thought if I combined 2 galvanic cells so that the middle one serves as the anode for one cell while being a cathode for the other due to difference in standard reduction potential and the middle one get reduced from one cell and get oxidized at the other creating endless current.
So is my "idea" right?

Yeah..your idea is technically right as long as you have the resources to keep up both the anode and cathode of the two cells which are not at the common side....ie if ure using a daniel cell and a Cu/Cu+||Ag+/Ag cell...as long as you suppliment Ag+ and Zn to the cell...youll have the cell up and running....
P.S:I wouldnt regard it as practical though