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Interaction between a pencil and rubber (eraser)
Is there a interaction between a graphite pencil and and a eraser. For example, if I drew a line on a piece of paper and then erased that, what interaction is occurring?
Is it a Van der Waals interaction with intermolecular forces or is something else?

@Mockingbird, agreed. My first thought is that the Van der Waals interactions between the graphite and eraser are just stronger than those between the graphite and the paper, but I hope someone has a more thorough answer than that ;) Do you have any ideas?

I also share the same thought..

It's called friction which is mechanical and not "chemical". Does the rubber erase the graphite if you just put it on your scribbles and don't rub it? I find that most chemists forget about the simple world of mechanics all too fast.  Apply a thin layer of something on a surface and then rub it -> it vanishes. Must be hydrogen bonding, right? Wrong. Call it Pauli repulsion if you want

Tribologists will tell you that friction and wear are complex phenomena with mechanical, physical, and chemical components. The answer is complex.

@JonCuster : If the initial poster and the commenters try to argue with Van-der-Waals interactions, I think saying friction/Pauli repulsion is sufficient. If you look at anything in detail, things get very complicated ...

@AMT , there are several types of non-abrasive erasers that essentially do work by a wiping action or even just "blotting" the graphite from the paper. After reading about this a bit I can't decide if I should try to answer it or flag it as too broad! It's more interesting than appears at first glance.

@AMT, how do you think friction works? Afterall, is it not theoretically impossible for two bodies of matter to come in to direct contact?

@Bob Maybe it depends on your definition of van der Waals forces. I really do not like this term because it isn't well defined. There are different definitions, some so broad that basically everything becomes van der Waals interaction. I think the key thing is that even the mentioned "non-abrasive erasers" seem to work (I'm not an expert in erasers) with at least a little bit of pressure. I think this makes saying Van der Waals interaction misleading. Yes, friction has something to do with dispersion as well, but that is not the full story. I don't see how "friction" isn't the right answer

@airhuff It certainly is. I would like to read that answer even if the initial poster hasn't been seen since. If you have a really good one, please share it. If you want to look at it in great detail, we might get a very interesting picture. You got charges, surface irregularities (do perfectly even surfaces still erase?), adsorption to some degree I guess and a thousand different things. I wonder how much research has been done with erasers though.

I asked for a school project, however since the various different opinions I might have to make an argument report, stating different opinions. I've been looking and researching into this, but it's difficult to find anything relevant. However I thank you all for these arguments!

Please don't, there are no two opinions. Maybe you misunderstood. The debate was about what forces cause friction, not if friction was the thing to look for. This really shouldn't be a chemistry assignment. I would also be surprised if there was a debate about what causes friction, it's just not something your average chemist can answer on a quantum mechanical level.  Also please read this, even though it's one of the worse articles https://en.wikipedia.org/wiki/Friction#Friction_at_the_atomic_level

There was much discussion in the comments to the OP regarding whether erasing a pencil mark is a chemical process or merely a mechanical process, akin to removing an oil spill with high pressure water (my analogy). But I think the answer to "Is it a Van der Waals interaction with intermolecular forces" is clearly yes.
Although the classic "pencil head" eraser generally requires a vigorous rubbing action to rapidly remove a pencil mark, even this can be done in a manner that shows pencil marks that had been bound to the paper via intermolecular forces are transferred to a material having a stronger affinity for the pencil mark than does the paper.
This Wikipedia article discusses several types of erasers that clearly illustrate a chemical process rather than a mechanical one is dominating the removal of the pencil mark. For example, regarding the "poster putty" type of eraser the article states:

"...poster putty works much the same as traditional kneaded erasers,
  but with a greater tack or lifting strength...Repeatedly touching the
  putty to a drawing pulls ever more medium free..."

A "gum" eraser also clearly works by adsorbing graphite from paper. As the gum is dabbed or lightly rubbed on the pencil mark, the intermolecular forces holding the graphite to the paper are weaker than the intermolecular forces between the graphite and the erasure. The following photo from the above Wikipedia article shows that the graphite in the used eraser has accumulated over time. There is no evidence of paper or other material having accumulated along with the material from the pencil mark.

Furthermore, one can imagine using different materials that should make for good erasers if the pencil marks are simply being removed by the mechanical action of an eraser. For example, if that's the case, shouldn't you be able to remove the pencil mark by rubbing it off with a piece of wood, one's finger, another piece of paper, a fingernail, etc.?

You can erase the pencil marks with your finger. I'm not convinced by your arguments. One type of eraser absorbs graphite quite heavily, so all erasers function via "van der waals" interaction? I've mentioned in the comments that this by itself is an extremely vague term. So why do you need to rub your eraser then? The way I understand it, it is 1) rubbing the eraser thus loosening the graphite 2) some kind of absorption process by the material to prevent smearing, is this correct?


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Will Calcium Carbonate react with Sodium Chloride?
After reading through my school chemistry notes and the answer to the question here at Chem S.E of how do we know whether an $\ce{A + B}$ reaction will occur, I find myself stuck on this very basic question.
I understand that $\ce{CaCO3 + HCl}$ will react as it is an acid + base reaction and that the reverse reaction $\ce{Na2CO3 + CaCl2}$ to $\ce{CaCO3}$ and $\ce{HCl}$ is possible.
I don't think I can use bond enthalpy to figure out if the reaction is spontaneous or non-spontaneous, because the reactants and products are ionic rather than molecular compounds.
I believe that I might need to use some sort of enthalpy calculation to figure this reaction out, but I'm not sure which specific one in particular.
On a whim, I believe that this reaction will not occur but I'm still very uncertain.

Hello and welcome to Chemistry.SE. I suggest that you take the short [tour] to further familiarize yourself with how this site works. As this reads like a homework-type of question (though I could be wrong) you should also read through this discussion. To get you started, do you think it would react? Why or why not? Showing some work and effort on your part will motivate others to invest their time and effort in helping you. Good luck!

Well the reverse is true.

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How is seawater alkaline?
If when water and carbon dioxide react they form carbonic acid, carbonate, and bicarbonate, how does seawater still have a pH of around 8? Doesn't a compound need an hydroxide ion to be a base?

The chemistry of the oceans make for a pretty complex situation. But for starters, think about what could be in the ocean and the geology of the materials "containing" the ocean. The adsorption of carbon dioxide of course affects the pH, but it is far from the only process at play here. What kind of minerals could dissolve to give a pH in the 8.3 range?

It's true that I made a questionable assumption, but as I see it, none of the common salts in the sea could've affected the pH, so I would think you would need to dissolve some atmospheric gases in the ocean, or find some organic process to explain the alkalinity.

How about carbonate mineralogy and carbonate based sea shells? You are right that carbon dioxide is an acidic gas, and as the atmospheric concentration increases, more of it dissolves into the oceans, slightly increasing their acidity, but also dissolving more carbonates, acting somewhat as a buffer system.

yeah, I guess that would be the thing. I'd just really like to figure out how it works.

The short answer to the title question is the ubiquitous presence of geologic and biogenic calcium carbonate $(\mathrm{p}K_\mathrm{a} = 9)$.  The oceans as a whole can largely be thought of as residing on beds of calcium carbonate. The following are some excerpts from this Wikipedia page:

Eggshells, snail shells and  most seashells are predominantly calcium carbonate.
Carbonate is found frequently in geologic settings and constitute an enormous carbon reservoir. Calcium carbonate occurs as aragonite, calcite and dolomite. The carbonate minerals form the rock types: limestone, chalk, marble, travertine, tufa, and others.
Calcium carbonate contributors, including plankton (such as coccoliths and planktic foraminifera), coralline algae, sponges, brachiopods, echinoderms, bryozoa and mollusks, are typically found in shallow water environments where sunlight and filterable food are more abundant. Cold-water carbonates do exist at higher latitudes but have a very slow growth rate.

Regarding your question

Doesn't a compound need an hydroxide ion to be a base?

Here you go:
$$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH-(aq)}$$
Although calcium carbonate is only sparingly soluble $(K_\mathrm{sp} = 3.3×10^{−9})$, as stated above the oceans are simply rife with it.

Also, sea water is less than $25^o C$, so the $k_w$ is less in sea water; therefore, the $H^+$ concentration is also lower, leading to a higher pH.

@DrPepper You are not serious, are you?

@andselisk why do you think i'm joking?

@DrPepper First, temperature of the sea water also exceeds 25 °C in tropical areas, which, following your logic, should result in slightly acidic water near equator, which is not true. Second, yes, $K_w$ depends on temperature, but it doesn't depend on concentrations: with more released protons there is going to be an equal amount of hydroxide-ions in the solution; formally, you get a slightly higher pH, but also the same increase in pOH and it doesn't affect alkalinity at all (as the title mentions it).

@DrPepper Third, there is no way to increase the formal pH value by lowering the temperature to even merely reach the pH 8 mentioned in the question. Even at 0 °C pH = pOH = 7.47!

Fellow collegaes,
The Question: how and why is seawater alkaline? Is very important.
Forget what you have ever learned and read " how to understand acid-base" by Peter Stewart.
Trying to break down his message:
1. First Consider pure H2O (without minerals) and it relation with temperature.
PH of pure H2O changes from 7.00 at 25 degrees celcius to 7.47 at 0 degrees celcius. H2O starts dissociating less as temperature goes down. From 25 to 0 degrees both [H+] and [OH] decrease and pH goes up, although the water remains neutral ([H+]=[OH-])

Second: the only independent factors (apart from temperature) capable of changing pH are: 1. Strong ion difference (alkalising) and 2. PCO2 (acidifying)

Consider all cations and anions in seawater.
[Na+]+[Mg]+[Ca2+]+[K+]+[rest cations]+[H+] = [Cl]+[SO4]+[HCO3]+[rest anions]+[OH-]
- All cations are "strong" (meaning fully dissociated)
- Not all anions are "strong" but a minority is "weak" meaning for a small part in equilibrium with [OH-], for instance:
1. HCO3- + H2O <-> H2CO3 + OH- or
2. HPO4- + H2O <-> H2PO4 + OH-.
H2CO3 and H2PO4 are unable to give of H+ due to a positive SID.
C/ the Strong Ion Difference between cations and anions (=SID) is what makes the ocean alkaline.
CO2 acidicifies but is unable to counter the SID
Consider:
1.If H2O evaporates: SID increases and pH goes up If H2O is added due to rains: SID decreases and pH goes down
2. Stewart demonstrated that a positive SID is the most potent buffer system and determines all because electroneutrality determines all. Formation of CaCO3 is only possible due to a positive SID. Any reasoning the other way around (e.q that CaCO3 could buffer the ocean) is wrong.

Strictly speaking, magnesium ion does show some acidity due to hydrolysis and calcium ion, in the presence of bicarbonate, can consume hydroxide ions via $\ce{Ca^{2+} + HCO3- + OH- <=> CaCO3(s) + H2O}$.  But the former is much less than the alkaline bicarbonate hydrolysis and the latter requires the seawater to become alkaline first, so the basic reasoning passes muster.

The ultimate source of alkalinity in the oceans is the basic character of many of the rocks on Earth. As explained here, many rocks contain alkali-metal and alkaline-earth-metal-bearing silicates, in which the silicate ions are basic thus convert carbon dioxide to bicarbonate. In the case of the ocean the latter species is the predominant (weakly) basic ion in ocean waters. Alkaline-earth-metal silicates are especially important in the ocean environment because much of the rock in oceanic crust is basalt, which contains predominantly mafic minerals such as calcium-rich plagioclase feldspar and pyroxenes (which are rich in both calcium and magnesium).
The minerals in basalt are also susceptible to forming basic hydroxides at high temperature even in the absence of carbon dioxide. Serpentization, a high-temperature between magnesium- and iron-rich silicates and water, occurs in hot spots including the geologically active mid-ocean ridges. The overall chemistry of this reaction is complex, but one of the major results is that the magnesium-bearing components in the rock produce basic products such as brucite (magnesium hydroxide). Where calcium is present in the rock, it woukd similarly produce calcium hydroxide; this dissolves and reacts further as a base (Wikipedia gives dissolved calcium ions as the product in such cases).

The chemistry is complex given the organic content of sea water and the influence of sunlight. For example, here is an article detailing the influence of sunlight on oceans containing dissolved organic matter, to quote:

Solar radiation mineralizes dissolved organic matter (DOM) to dissolved inorganic carbon through photochemical reactions (DIC photoproduction) that are influenced by iron (Fe) and pH [...] Fe raised the rate of photobleaching and steepened the spectral slopes of CDOM in low pH but resisted the slope steepening in neutral to alkaline pH.

where apparently acidic pH is much more effect implying an asymmetric effect with the consumption of H+.
Similarly, other reactions involving iron or other transition metals consume H+ also:
2 Mn(II)/Fe(II) + 1/2 O2 +  H+ --> 2 Mn(III)/Fe(III) +  OH-
Bottom line, asymmetric pH reactions can in heterogeneous systems found in large bodies of water, rich in organic matter and subject to irradiation, can result in a pH shift, including into the alkaline range.

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Solubility of NaCl in water
I learned that solubility of substance increases with increase of temperature. But I came to know solubility of NaCl doesn't increase when temperature is increased . So what is the reason for this phenomena?

Also, there are quite a few different substances out there. Some solubilities increase with temperature, some decrease, some do not change at all.

What textbook are you using?

It is true. This sort of fact can be looked up. At 0 C 35.65g NaCl/100 grams H2O, at 100 C 38.99 grams NaCl/100 grams H2O. https://en.wikipedia.org/wiki/Solubility_table#S

@MaxW i didn't understand what you are saying,it will be good if you answer it in some detail.

Cadmium selenate is a salt for which the solubility decreases with increasing temperature. See Wikipedia table noted above.

@Hexacoordinate-C no,it is not duplicate,if I ask question on what is behaviour of 14th group elements with water then still there will be separate question on reaction of carbon with water.

@Fawad you didn't precise you asked about the 14th group element in your question and the related question is not about carbon or you may read again.

@Hexacoordinate-C oops,that was an example

Solubility of solid in a liquid depends on the nature of the solid.
It depends on whether the dissolution process is exothermic or endothermic. Using Le Chatelier's principle, if the temperature is increased for an exothermic process then dissolution decreases and if the temperature is increased for an endothermic process then dissolution increases.
I am going to stick to the case of $NaCl$, if you want to know more about Le Chatelier's principle, click on the link below and look for the 2nd answer https://physics.stackexchange.com/questions/311409/why-do-most-gases-dissolve-easier-at-colder-temperatures/311429#311429

There are some solids for whom change in temperature doesn't affect the degree of dissolution.
There are some substance which on heating, change on a particular temperature from one polymorphic form to another (like $\alpha$ to $\beta$ as in the case of ammonium nitrate) or from one hydrated form to another ($CaCl_2.6H_2 O -> CaCl_2.4H_2 O$) or from hydrated to anhydrous form ($Na_2SO_4.10H_2O -> Na_2SO_4$).
Generally, dissolution process in a solvent is a disintegration process. This process needs energy. In such cases, energy is absorbed. But in some cases, besides the process of breaking or ionization, there is hydrate formation. During hydration heat is evolved. The net result is that heat is either evolved or absorbed.
There are also cases in which heat of separation of ions in just equal to the heat of hydration and there is very little heat effect as in the case of $NaCl$. The heat of solution of $NaCl$ is very small as the heat of ionization is nearly equal to the heat of hydration. Therefore, temperature doesn't affect the dissolution of $NaCl$ as the heat of hydration almost equalizes heat of dissolution.

I take that NaCl dissociates because it breaks into Na+ & Cl ions which go to the OH- & H+ ions of water. Is that right ? Plus how does sugar dissociate in water. I was particularly weak in all this & particularly organic even an year back when we were giving jee

Yes, in case of NaCl, Na attracts the O of H20 and Cl attracts the two H of H20. Sucrose is also a slightly polar molecular. Check this out : http://www.middleschoolchemistry.com/lessonplans/chapter5/lesson4

Generally the water molecules come toward the NaCl crystal and break it apart.

Sucrose is less polar than salt that's why it's tough to prepare a sugar solution. Right ? Also the the solubility has to increase if I hear the solution ?

Yes you are correct. Usually when you heat the solution the solublility increases because the dissolution process in endothermic. According to Le chatelier's principle, if you increase the temperature reaction proceeds towards the endothermic side in order to prevent the temperature change to take place. Fun fact : Although the dissolution  process is endothermic, for NaCl the lattice enthalpy and the hydration enthalpy are almost the same in magnitude, so we don't observe a significant change in the surrounding temperature.

Thanks for clarification. Sorry I don't know anything about hydration and lattice enthalpy. It was only because of physics and maths that I cleared jee. But yes I can see what you say. On the dissolution of NaCl we don't see any change in the temperature of the surroundings. But that will be because it's so big ?? Will it not be ?

One way to understand solubility is to start with the Clausius -Clapyron equation (CC), which despite approximations involved is a good description the vapour pressure over various solids and liquids. For the ideal solution
$$ \ln\left (\frac{p_2}{p_1}\right) = \frac{-\Delta H_{vap}}{R}\left(  \frac{1}{T_2}-\frac{1}{T_1}  \right )$$
where p is the vapour pressure, T the temperature and $\Delta H$ the enthalpy change between states  $1$ and $2$.
The heat of sublimation is $\Delta H_{sub}=\Delta H_{vap}+\Delta H_{fus}$
and calculating the pressure over the pure solid form of the solvent gives
$$ \ln\left (\frac{p_2}{p^*_1}\right) = \frac{-\Delta H_{sub}}{R}\left(  \frac{1}{T_2}-\frac{1}{T_1}  \right )$$
and so subtracting these two equations gives
$$ \ln\left (\frac{p^*_1}{p_1}\right) = \frac{\Delta H_{fus}}{R}\left(  \frac{1}{T_2}-\frac{1}{T_1}  \right )$$
If it is assumed that Raoult’s Law applies then $ p_1=xp^*_1$ where x is the mole fraction of the solute. Substituting into the last equation gives
$$ \ln(x)= -\frac{\Delta H_{fus}}{R}\left(  \frac{1}{T_2}-\frac{1}{T_M}  \right )$$
where now $T_M$ is the melting temperature of the pure solute. As $\Delta H/T_M$ is a constant then the mole fraction and hence solubility in solution is
$$ \ln(x) \propto \frac{-\Delta H_{fus}}{R}\left(  \frac{1}{T_2}  \right )$$
which shows that the mole fraction at temperature T varies as
$$x_T \propto \exp \left (- \frac{\Delta H_{fus}}{RT}\right )$$
and so the solubility rises with increase in temperature. Different species will rise more or less slowly depending on their heat of fusion as $\Delta H_{fus}/R$.
Using data for NaCl shows that the mole fraction hardly varies between $200$ to $400$ K, whereas there is a huge increase for $\ce{NaNO3}$ under the same conditions.


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Conversion of acetaldehyde to butane-1,4-diol
How to convert acetaldehyde (ethanal) to butane-1,4-diol. I just tried doing aldol, but it leaves an alkyl group at the end. Also it cannot be done by oxidizing ethane with basic permanganate.

Hello and welcome to Chemistry.SE! As this appears to be a homework-type of question, I suggest reading through this discussion. The more work and effort you can show, i.e. writing out possible reaction mechanisms, etc., the better. Good luck!

Cross Aldol and add H+

Oxidise Aldehyde to Carboxylic Acid

use any of the reagents

3.Perform Arndt–Eistert reaction

Use LiAlH4 for reduction of acid to alcohol

And finally you get 1,4-dihydroxybutane.
Image courtesy Wkipedia, SigmaAldrich


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Why does boron add to the less substituted carbon in the hydroboration of an alkene?
In the hydroboration of a typical alkene, the boron fragment ends up attached to the less substituted carbon:

What are the electronic or steric factors that lead to this regioselectivity?

For the hydroboration of isobutene

there is a steric aspect:

And though sterics may play some part, it doesn't explain near 100% regioselectivity. More important in determining regiochemistry, I would say, are the electronics of the transition state:

Boron is electrophilic, and as such will draw the electron density of the alkene π-bond towards itself. This C–B bond formation occurs earlier than the C–H bond formation, which in turn imparts a partial positive charge to the other carbon of the double bond in the transition state of the reaction.
This somewhat resembles a carbocation, which you might remember are stabilized by additional substituents via hyperconjugation. Thus, hydroboration results in an anti-Markovnikov product for mostly the same reason that acid catalyzed hydration results in a Markovnikov product — stabilization of positive charge.

How boron will draw the electron density of the alkene π-bond towards itself while carbon is more electronegative than it?

It is a Lewis acid.

Why C-B bond forms faster ?

How can hyperconjugation explain stability of partial charges.? There is no p orbital on carbon to hyperconjugate

Hyperconjugation doesn't need to go into a pure p orbital, it works with any kind of orbital that isn't full, as long as there is sufficient overlap.


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Why does the trimethylammonium cation have a larger -I effect than the ammonium cation?
Why does $\ce{N(CH_3)_3^+}$ have a larger $-I$ effect than $\ce{NH_3^+}$?
Since the methyl group is good at donating electrons it would stabilize the charge on nitrogen atom, decreasing its potential to withdraw electrons hence its $-I$ effect.
I can't figure out why its the other way around. Can someone help me understand this?

Are the groups in a gaseous phase or in a polar solvent..? The choice of phases in case of these groups changes everything.

@BhavyaSharma nothing as such is mentioned in the question. The question just wants me to find the group with highest -I effect

Methyl groups donate inductively only into empty-p-orbital cations (e.g. carbenium ions). Ammonium cations cannot profit from that stabilisation.

@Jan Oh I didn't know that, thank you! So basically inductive effect from methyl or hydrogen groups doesn't operate here. Then what is the factor that influences the stability of these ions?

@Jan That's not and inductive effect though, that's hyperconjugation.

This question was answered a year ago at https://chemistry.stackexchange.com/questions/88185/justifying-the-tremendous-%E2%88%92i-effect-of-trimethylammonium-group-by-molecular-orbi/135045. The ring donates a pi electron to N and one CH3 group hyperconjugates (in reverse). The effect is demonstrated by the strong meta and deactivation effect typical of pi deactivators; NH3+ is only a weak deactivator and therefore not a pi actor.

$\ce{N(CH3)3+}$ has more -I than $\ce{NH3+}$
, this irony happens because if you consider
$\ce{N-CH3}$ bond and $\ce{N-H}$ bond,  which is more polar?
Of course $\ce{N-H}$ bond will be more polar  due to more difference in electronegativity  value,  so that implies electron density will be more on $\ce{N}$ in $\ce{NH3+}$ than $\ce{N}$ on $\ce{N(CH3)3+}$.
And therefore since electron density is less on $\ce{N}$ of $\ce{N(CH3)3+}$ , it will have greater -I effect.

The reason is that the methyl  though gives electron via +I effect in case of H it is easy to cleave the N-H bond for nitrogen and release proton and take the electron pair from hydrogen but with C it being more electronegative than H the same is not possible .

But why is bond cleavage required?

to disperse the +ve charge from more electronegative atom to less electronegative atoms


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How do you calculate the heat of a reaction given a table of heat of formation values?

So I've been trying to answer this question, and as far as I've learned it's $\Delta H^0_{\mathrm f}$(products)-$\Delta H^0_{\mathrm f}$(reactants), multiplying each by number of moles and such.
Trying to use that on this problem I came up with $\ce{0.008 OH- + 0.008 HCO3- -> 0.008 H2O + 0.008 CO3^2-}$ after (possibly incorrectly) noting that $\ce{HCO3}$ was in excess. However, this does not yield anywhere near the correct answer of C. What am I doing incorrectly?

I can obtain the correct answer using your values. Maybe double check for mistakes and typos?
Here are my quick calculations:
The overall reaction is:
$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$
The equation used is:
\begin{equation}
\Delta H_\text{reaction}^\circ = \sum {\Delta H_{f}^\circ(\text{Products})} - \sum {\Delta H_{f}^\circ(\text{Reactants})}
\end{equation}
$\Delta H_\text{products}^\circ= (\pu{0.008 mol}\times\pu{-286 kJ/mol} +\pu{0.008 mol}\times\pu{-677 kJ/mol}) = \pu{-7.704 kJ}$
$\Delta H_\text{reactants}^\circ= (\pu{0.008 mol}\times\pu{-692 kJ/mol} +\pu{0.008 mol}\times\pu{-230 kJ/mol}) = \pu{-7.376 kJ}$
$\Delta H_\text{reaction}^\circ = (-7.074 - (-7.376))~\pu{kJ/mol} = \pu{-0.328 kJ} = \pu{-328 J}$

Thank you! You helped me realize that I'm a buffoon and that the table is in kJ whereas the answers are in J. :P

No problem. Could you mark the question as answered?

Interesting that the significant figures in the book answer seem to be wrong. I'd agree that the answer should be -328 not -330.


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Is there any convenient way to warm water-based products to room temperature at home?
When it's cold outside, some household water-based products (liquid soaps, wet wipes) become cold to the touch as well. Since these products are used directly on the skin, their coldness may cause discomfort to the user (especially babies and the elderly).
I am wondering if there is any convenient way to warm these products at least to room temperature of 23-25°C to reduce the discomfort. I am guessing the main obstacle is water's relatively high specific capacity (at 15°C, it is approximately 4.19 J/g∙°C, according to this table, taken from this blog entry), which may also be influenced by other ingredients of the soap, but I'm not sure in which direction...
As a possible solution, I thought of putting hot water in some vessel and submerging the soap container in it. I haven't tried it yet, but I wonder how hot should the water be and how much water do I need (i.e. how deep should the container be submerged), when taking into account the volume of soap and the plastic container itself.
And there's still the issue of the wipes, which can't be submerged in water...
Repeating the question: Is there any convenient way to warm water-based products to room temperature at home?

Find a room. Put the objects to be warmed in the room. Wait.

Yes there is a convenient way: a microwave oven. Note, though, that it is pretty easy to overheat things to dangerous temperatures.

I think that is one of the reasons why most people, especially infants and the elderly, prefer to have their bathroom inside a building. It has something to do with rain as well, but let's not get carried away. Btw, most babies don't use soap directly, their parents put it in their hands first and warm it that way. Try it, it works. If I understand correctly, you want to heat soap while living outdoors? If you're stuck in a blizzard in the forest, I don't want to have to use cold soap after washing my hands, yes. Maybe carry them close to your body at all times instead of your backpack?

Breathe on it until its warm to the touch.

Thanks, @AMT, for your comment. Actually, I am not talking about the outdoors. Since bathrooms usually have a window facing outside, it might get pretty cold there when it's cold outside (even if the window is closed). This causes the soap to become cold as well...

Ok, there are some countries where heaters are not commonly used, I understand your question. But I don't think this is a chemistry question but an economic one. Is it cheaper to buy a heater and heat the room or to keep a small water bath at 25 °C degrees at all time. Get your local energy costs, find out what kind of equipment you can afford, consider the warm bathroom as a plus and then do the math. I don't think we can help you in any way better than you can yourself. Economists deal with this kind of math all the time, if you need help, consider writing in one of their forums.

Well, bringing the entire room/apartment to 25 celsius will surely take care of things. I was looking for something more... focused... The chemistry part is the thermochemistry angle - I came up with putting the soap in a vessel with hot water prior to taking a bath according to the commonly used zeroth law of thermodynamics. I was wondering if any of our physical chemists friends might have a better suggestion.

Oh btw think about professional massage people who solved this problem already. I think they heat their stuff up. Maybe ask one of them what exactly they are using.

When my kids were babies, you could buy a little heating pad made to go around a box of baby wipes to keep them warm to the touch. With a baby shower for a colleague coming up we went looking for one as a gift - nobody had them anymore. Not sure why - perhaps the fell out of favor, one got the wipes too hot (never microwave!), who knows. But, ours sure worked wonders...

http://engineering.mit.edu/ask/are-there-materials-can-absorb-heat-without-becoming-hot

I'm voting to close this question as off-topic because it would be more appropriately asked on Lifehacks.SE: http://lifehacks.stackexchange.com/


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How does lansoprazole degrade in acidic conditions (mechanism)?

Can anybody help me and explain the mechanism of reaction here? And also I was wondering when disulfide bond is made, can molecule dissociate, return to the previous state and make a disulfide bond when it gets to acidic environment again?

Exactly which part of the mechanism are you finding difficult? Most of it is relatively straightforward chemistry. In the first step you protonate the pyridine-type nitrogen in the benzimidazole ring, then do a nucleophilic attack with the pyridine. The third step is "just" a nucleophilic substitution on sulfur, etc. It's basically the same few reactions that you learn when starting out organic chemistry except that it's a bigger molecule with some heteroatoms instead of carbons.

@orthocresol I think OP might be confused in the first step; in the diagram, protonation of the N is skipped.  That's what makes the aromatic lose resonance states, and that's what makes the C a good electrophile for any configuration that restorores that dual ring resonance.

@orthocresol it retains aromaticity, but there are fewer resonance states if you can't move that double bond around the ring without knocking off that H+

@orthocresol (an aside not regarding OP) can we have resonance states with such different dipoles and asymmetries, or is that in fact a different species?  And in that picture, the C is a bullseye for a nucleophile with or without an added proton.  Together, this has made the first step confusing for me, and I agree that it's important to know which step OP is referring to.

@orthocresol so we can substitute this resonating thing at that C and expect a randomized chirality if we extract it?  The C(-NH=ring)(-NH+-ring)(-NC7H5OF3-CH2-)(-SO) looks like a floppy chiral center if the product of reaction 1 resonates like that, if the (-NH=ring) and (-NH+-ring) are not required to be symmetrical.

@sqykly (1) The product of reaction 1 isn't chiral anyway? I'm sorry but I'm really quite lost as to what you're describing. Need some diagrams or something. (2) If you have further questions please ask a new question. I don't want to continue replying here (comment section of someone else's question is not the place to do so). I've also cleared my comments

@orthocresol Sulfoxides can be chiral — but I don’t think that was what OP was asking ;)

@orthocresol fair enough


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Why is the overall charge of an ionic compound zero?

My textbook simply says:

Since an ionic compound consists of equal number of positive and negative ions, the overall charge of an ionic compound is zero.

But why is the number of positive and negative ions equal?
Can’t an ionic compound can have an unequal number of negative and positive ions? Why or why not?

Sodium needs to lose 1 Electron to attain stable electronic configuration and chlorine needs to gain 1 electron to stable electronic configuration.
In a big picture, the electron was transferred from sodium to chlorine in the same neutral crystal. No electron was supplied to the crystal from outside, it was already in the same system before and after the formation of NaCl.
If the system before formation of NaCl was neutral then it will be neutral even after the formation of NaCl crystal. That may be the reason.

Well, if the no. Of positive and negative ions were unequal, then the ionic species would have a net charge, i.e., there would be charge separation. Now since electrostatic forces (the forces that bind the cations and anions in an ionic compound) are very strong over short distances (as found in a compound), very large amount of work has to be done to separate the charges. Thus, energy required will be very high.
However, normally a system tends to be in the lowest energy state. So, unless you provide huge amounts of heat or some other external work to separate the charges, the ionic compound will have equal number of cations and anions

An ionic compound can have an equal number of cations and anions. Consider, for example, sodium oxide. Two sodium atoms transfer their electrons to one oxygen atom. Therefore, the ratio of cations to anions is 2:1 but still the total positive charge and the total negative charge of the system is equal.

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


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Retrosynthesis of a lactone
We have been asked to do a retrosynthesis of (6S)-6-[(2S)-2-hydroxy-4-phenylbutyl]oxan-2-one J from 3-phenylpropan-1-ol K:

My first instinct would be to do a functional group interconversion of the alcohol to a ketone, to then do a Baeyer-Villiger oxidation.

The key to this synthesis should be — in my opinion — how to generate the two hydroxy functions with anti stereoselectivity starting from at least one achiral reagent. Most of the time you would attempt building up chirality de novo using chiral auxiliaries, catalysts or reagents as these methods are often more cost-efficient than acquiring a specific chiral reactant — the obvious exception being if a compound exists in the natural chiral pool that you can easily use (which is not really the case here). Having identified our most important problem, we can turn to the other ones:

we need to generate a δ-lactone
we need to form a $\ce{C-C}$ bond, preferably in a chiral manner, to a $\ce{C6}$ body (the bit of the final product that is missing in the reactant).

Scheme 1: General retrosynthesis idea.
In my opinion, the δ-lactone is a non-issue. If you have an open-chain ester such as 3, you can probably form the six-membered lactone simply by $\ce{K2CO3}$-mediated transesterification. Practically all literature methods to synthesise esters should also predominantly give the δ-lactone especially since the unfavourable ζ-lactone is the alternative. (In the unlikely case that it is formed, it can easily be transesterificated as noted.) I would put this step last.

Scheme 2: Transesterification step.
We now have 3, an anti-configured 1,3-diol. When I see this, my mind immediately jumps to the Evans-Tishchenko asymmetric reduction. Thus, we can arrive there from either of the two possible β-hydroxyketones 4 or 5. For this to work best, I would want the carboxy group to be masked as a protected alcohol here, cf. 6.

Scheme 3: Evans-Tishchenko step. $\ce{X}$ would be $\ce{H}$ for the newly-formed hydroxy group and $\ce{CH3CH2CO-{}}$ for the inducting one.
Both of these β-hydroxyketones could be generated from aldol reactions; however, only ketone 4 suggests that it can easily be reached from the desired reactant 2. If we wanted to synthesise 5, we would need an extra carbon atom on 2 which we would need to introduce in an additional reaction. However, the route to 4 only requires selective oxidation to the aldehyde — my personal favourite is Dess-Martin–oxidation, but you should probably check all the others that suggest themselves (Swern and relatives, TEMPO, Ley-Griffith, etc.) as one could be superior over the others due to your lab spirits liking it.
Finally, this also tells us that the aldol addition to generate 4 from 7 and 8 should proceed in a stereoselective manner giving the (S)-configured product predominantly. Again, many methods are known and I would suggest a Paterson aldol reaction. Analysis of the relevant literature (unfortunately, I did not find a website, so a citation it is) reveals that $\ce{(-){-}Ipc2BOTf}$ and Hünig base should give the (S) product preferentially.[1]

Scheme 4: Final aldol disconnection.
This leads us to the following forward synthesis:
$$\ce{\mathbf{2} ->[DMP] \mathbf{7} ->[(-){-}Ipc2BOTf][ (iPr)2NEt, \mathbf{8}] \mathbf{4} ->[SmI2, EtCHO] \mathbf{6} ->[1) TBSOTf; 2) PPTS][3) DMP; 4) {Pinnick}] \mathbf{3} (R $=$ H) ->[DCC, DMAP][K2CO3] \mathbf{1}}$$

Reference:
[1]: A. S. Franklin, I. Paterson, Contemp. Org. Synth. 1994, 1, 317–338. DOI: 10.1039/CO9940100317.

The first idea that comes to mind is an aldol, as you pointed out too:

The only thing left to figure out is what aldol conditions to use to achieve enantioselectivity and the enantioselective reduction that is needed afterwards.
The Bayer-Villiger I don't understand, as:

I guess the regioselectivity should fit, with the alcohol-substituted side having the higher migrating group abilities, but I don't see how you would achieve the donating group to be in 3-position in the resulting cyclopentanon.
Hope that helps...

After the aldol you'd still need to reduce the beta-hydroxyketone to a diol, I think? You'd have to make the aldol enantioselective and the reduction diastereoselective (probably not a difficult task, since there are tons of procedures out there).

yeah right ; )
then its probably bad with two enantioselective transformations needed...

Only the aldol is enantioselective, the reduction can be controlled by the 1st stereocentre. http://www.chem.uky.edu/courses/che538/ac/2006.fall/problems/grad.01/s-2006-926315.pdf (section 4.1.2) I agree with your overall idea anyway, aldol seems the most straightforward choice for a 1,3-dioxygenated compound.


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Synthesis of a spirocycle
I am trying to come up with a synthesis of 7-hydroxyspiro[4.5]decan-1-one;

I think I could use these starting reagents below;

Could I potentially generate the enolate of the cyclic ketone starting material and it would attack the double bond of the alcohol? If this is correct, how would the ring close to create the spirocycle? Thanks.

Enolates won't react with the C=C, it probably just deprotonates the alcohol and then sits there.

Mesylate the alcohol, react that with the enolate, epoxidise the double bond then reform the enolate.

A simple disconnection reveals cyclopentanone and 1,5-dibromopentan-2-ol that can be easily prepared.
To carry out this reaction, first, cyclopentanone is made into a silyl enol ether using TMS with a Lewis acid titanium(IV) chloride. The 1,5-dibromopentan-2-ol is then added and subsequently water to hydrolyse the silyl enol ether. Note the alcohol does not participate in the reaction as it is a bad leaving group compared to the bromide ion.

I may have missed something, but a single enolisation should give a single attachment only, i.e. a linear chain branching off of cyclopentanone?

The silyl enol ether is not nucleophilic enough to accomplish this alkylation.

What is the source of your dibromo-alcohol? This is not a readily available starting material.


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If SOCl2 reacts with alcohols via SNi, why doesn't POCl3?
A bit of a follow-up to Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols?.

The reaction with $\ce{SOCl2}$ is also used instead of $\ce{PCl3}$ and $\ce{PCl5}$ when retention of stereochemistry is required. Phosphorus trichloride and phosphorus pentachloride both lead to inversion.

The mechanism proceeds via a $\mathrm{SN_i}$ pathway, or internal nucleophilic substitution. This step has been highlighed in red; an intimate ion pair$^{[\text{see below}]}$ is given in square brackets. Adding a nucleophilic solvent such as pyridine considerably increases inversion with $\ce{SOCl2}$ via an attack at the sulfur atom. Green marks blocking of the intramolecular substitution; blue is for inversion.

As far as I know, $\ce{POCl3}$ has not been seen following an $\mathrm{SN_i}$ mechanism (even without pyridine). Why?

If possible, please provide a qualitative mechanistic reason. Something that is readily explained by waving hands. Regardless of whether such an approach is achievable, quantitative (calculational) answers are also most welcome.

References discussing intimate ion pair formation

F. A. Carey, R. J. Sundberg. $(2007)$. Advanced Organic Chemistry
Part A: Structure and Mechanisms, $4$th edition, pp 269$-$276. ISBN: 0-306-46242-7
W. A. Hughes, E. D. Cowdrey, C. K. Ingold, S. Masterman, A. D. Scott. 'The Mechanism of Elimination Reactions. Part 1. Unimolecular Olefin Formation from Alkyl Halides in Bulphur Dioxide and Formic Acid'. Journal of the Chemical Society, $(1937)$, 1271$-$1277. DOI: 10.1039/JR9370001271
E. S. Lewis, C. E. Boozer. 'The Kinetics and Stereochemistry of the Decomposition of Secondary Alkyl Chlorosulfites'. Journal of the American Chemical Society, $(1952)$, 74, 308$-$311. DOI: 10.1021/ja01122a005
D. J. Cram. 'Studies in Stereochemistry. XVI. Ionic Intermediates in the Decomposition of Certain Alkyl Chlorosulfites'. Journal of the American Chemical Society, $(1953)$, 75, 332$-$338. DOI: 10.1021/ja01098a024
C. C. Lee, A. J. Finlayson. 'Rearrangement In The Reaction Between Thionyl Chloride And $3$-Methyl-$2$-Butanol. Canadian Journal of Chemistry, $(1961)$, 39(1): 260$-$261. DOI: 10.1139/v61-030
C. C. Lee, J. W. Clayton, D. G. Lee, A. J. Finlayson. 'Rearrangement Studies With $\ce{^14C-XIII}$: The Thermal Decomposition Of $1$-$\ce{^14C}$-$2$-Butyl Chlorosulfite'. Tetrahedron, $(1962)$, 18 1395$-$1402. DOI: 10.1021/ja01098a024
H. Patin, G. Mignani, C. Mahe, J-Y. Le Marouille, A. Benoit, D. Grandjean. 'Ferrocenyltrithiocarbonates: I. Direct access from α-ferrocenylcarbinols by a $\mathrm{SN_i}$ mechanism. Absolute x-ray structure determination of (R)-ferrocenylmethylmethane S-methyl-trithiocarbonate'. Journal of Organometallic Chemistry, $(1980)$, 193, 1, 93$-$103. DOI: 10.1016/S0022-328X(00)86079-9
J. L. Kice, G. C. Hanson. 'Mechanisms of SNi reactions. Effect of aralkyl group structure on ion-pair return in the decomposition of aralkyl thiocarbonates'. Journal of the American Chemical Society, $(1973)$, 38 (7), 1410$-$1415. DOI: 10.1021/jo00947a037
M. B. Smith. $(2013)$. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, $7$th edition, pp 311, 486$-$487, 490, 598, 1316. ISBN: 978-0-470-46259-1
 The $\mathrm{SN_i}$ (substitution nucleophilic internal) mechanism: retention of configuration. Powerpoint presentation, Università degli Studi di Napoli Federico II.
James. $\ce{SOCl2}$ and the $\mathrm{SN_i}$ Mechanism. Master Organic Chemistry, Alcohols. webpage

Hopefully, the subscript letter $\mathrm{i}$ in $\mathrm{SN_i}$ is clearly visible, and so is not to be confused with $\mathrm{SN_1}$. (I for one initially thought it was a typo when I learned about it in December.)

I don’t like the mechanisms as drawn due to the $\ce{S=O}$ double bonds which should be $\ce{\overset{+}{S}-\overset{-}{O}}$, but still $+1$ for a good question.

@Jan I'll see what I can do about it after school tomorrow ;)

@Jan I have updated the mechanism to include more delocalisation and true ionic bond character (both with dashed lines). Sadly, this has made the mechanism a bit harder to follow but is still alright. If there's anything amiss, feel free to ping me.

How is the reaction using thionyl chloride providing the product from retention? The chloride is attacking the activated alcohol from the back side, correct? Whether it's intramolecular or intermolecular is irrelevant.

Furthermore, what is the evidence that thionyl chloride delivers the chloride intramolecularly? That is a difficult mechanistic challenge.

@jerepierre The attack actually occurs from the same side as the oxygen. Probably there is also considerable ionic character to the $\ce{C-O}$ bond but I did not add it because $(1)$ could not draw it with both geometry and ionic character in mind, $(2)$ an ionic intermediary would require a solid reference which I was unable to find. So your counterarguments are definitely on point, and as soon as there's a better citable alternative I will edit :)

In order for the chloride to attack from the same side as the oxygen, it has to attack through the oxygen. Retention of configuration typically arises from two sequential inversion processes.

@jerepierre Does it have to attack through oxygen though ;) (made with all-powerful computational software Chemsketch). I don't think two consecutive inversions better explains the experimental evidence at hand than $\mathrm{SN_i}$.

@jerepierre , No there is no $Cl-$ ion in the reactants. The $Cl-$ that was expelled from $R-(OH+)-SOCl2$ will immediately abstract proton from  $R-(OH+)-SOCl$  thus forming HCl. This is the main reason why $SOCl2$ with pyridine will give inversion pdt (since pyrindine being a weak base will abstract the proton instead of $Cl-$) . The $Cl-$ will then attack by $S_N2$

Is there a source that mentions any reaction by $\ce{POCl3}$ without pyridine? I think pyridine is important catalyst in that reaction. Also with pyridine it can never show $S_ni$ because pyridine helps the chloride ion to act as a leaving group with it's protonated form stabilizing it. And also @Jdeep makes the other point of this argument

The step in the red box is wrong. You probably get dissociation to form an ion pair. The pair quickly collapses, leading to chloride attack on the cation before any dissociation or diffusion occurs so that you get the retention of stereochemistry.
For phosphorus based chlorination, this ion-pair mechanism is probably not favorable, though it's not clear to me why this is the way it is.

Actually in the process of editing that step (or considering at least). A ton of references will be added in a second. (There are also places where the ion pair is said to be 'not completely dissociated', and where it is not included.)

Zhe is right. It is a common misconception to write this mechanism as a direct displacement which is actually impossible on stereoelectronic terms.

There's just no way for the intramolecular $\mathrm{S}_{N}2$ reaction as written to take place, from a stereoelectronic perspective. There's probably some give on how tight the ion pair has to be though it has to be tight enough to explain retention. But we're in a gray area where mechanistic probes are probably going to be quite difficult.

@Zhe Yeah, the final sentence hits the nail on the head. ---- If the chlorine in the red box attached to sulfur has significant ionic character, wouldn't that make the "one-step" method (that I currently have) more probable? (as a last resort, I'm a bit too lazy to rewrite the mechanism again at the moment :D)

It's still probably not one-step. The proposed transition state for a front side $\mathrm{S}_{N}2$ reaction seems unlikely. Therefore, I would think that we have (at least) two steps though the barriers would be quite small...

Once the alcohol attacks the SOCl2 the proton on the oxygen is more acidic because S is more electronegative than P.
What is likely happening is that R-OH-SOCl has a lower pKa than 5.2 (the pKa of the conjugate acid of pyridine), giving the chlorine time to do a nucleophilic attack.
Then R-OH-POCl2 should a pKa close enough to 5.2 that the equilibrium results in a low enough concentration of deprotonated R-O-POCl2 that the Cl- cannot find any. Since the elimination reaction is practically irreversible all of the R-OH-SOCl will be consumed well before the Cl- has a chance to find anything to react with
.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.


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Should compounds in a generic reaction be italicised (typography)?
As of recent, I provided an answer to the question Law of mass action & collision theory. Therein I wrote a generic reaction
$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}.\tag{1}$$
It seemed natural to write stoichiometric coefficents in Italic, and use upright Roman for the species themselves.

Italic would be in line with the fact that the coefficients are numbers, and vary with reactions (are different for different transitions).
Upright Roman is how we usually write chemical species, e.g., $\ce{NaCl}$.

Though if I were to overthink, the $\ce{R_$i$}$ would also seem to be variables as such. Since this is a generic reaction, $\ce{R_$i$} \in \{\mathrm{every\ imaginable \ chemical \ species}\}$. Using this logic, I ought to write $(1)$ instead as
$$r_1R_1 + r_2R_2 + \ldots + r_nR_n \ce{<=>} p_1P_1 + p_2P_2 + \ldots + p_mP_m.\tag{1'}$$

Is there an official example or guideline how to handle this situation?

Personally I wholeheartedly support option $(1)$ but is this what is recommended?

What about the subindices? Upright or Italic?

Note that this qualitatively differs from Proper typography of “$\ce{NO_x}$”. There the number of elements actually changes within a chemical species (marks the number of carbons in alkanes for example) whereas here it is more for labelling purposes.

Since Chem.SE is going through $18$th-turn-$19$th century Vienna in terms of typography expertise, this is probably the best time to ask such a question.

I'd agree with "It seemed natural to write stoichiometric coefficents in Italic, and use upright Roman for the species themselves." Not sure if ACS style guide would agree or not.

Going by the basic rule of a variable (that can hold multiple values) being italic, I would go for the $r_nR_n$ style (4 times italics), but then I found the following in the IUPAC Green Book (printed page 53):

$$0 = \sum_j \nu_j \mathrm{B}_j \quad \text{where }\mathrm{B}_j\text{ denotes a species in the reaction ...}$$

Here, the B is upright although it is definitely a placeholder. I don't understand, why.
For the subindices, I think (hope) there is no doubt that the $n$ and $m$ should be italic, because they clearly are variables standing for numbers (either any number as in the sum, or for an unknown maximum number $1, 2, 3, \ldots, n$ as in the generic reaction). Both form are well established in mathematics (and I hope chemists don't deviate).

The IUPAC Gold Book, R05156 uses the same convention.

For the general chemical reaction :
  $$a\,\mathrm{A} + b\,\mathrm{B} \rightarrow p\,\mathrm{P} + q\,\mathrm{Q} + \dots% not using \ce to have proper formating regardless of mhchem version$$

And they do not mean Boron and Phosphorus.
(Spotted via this answer).

Great find! Luckily it is a recommendation I prefer, so that's nice. (regardless of whether it follows the variable/label rule)

Added 2nd source

If it helps your understanding, simply regard those as contractions or abbreviations. For example when we write a simplified formula for acetic acid, we may write $\ce{AcOH}$. We might abbreviate an aryl-moiety simply with $\ce{Ar\bond{-}}$, or any moiety with $\ce{R\bond{-}}$. I think in this particular case it is better to use upright characters to signify that these are chemical species. But this is just me guessing here...


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Why do laws of thermodynamics "outweigh" any other law ever formulated?

"It is the only physical theory of universal content concerning which I am convinced that within the framework of the applicability of its basic concepts, it will never be overthrown." - Albert Einstein
"Nothing in life is certain except death, taxes and the second law of thermodynamics". - Seth Lloyd
“If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation." - Arthur Eddington
"“Every question or effect has the right to exist if it does not contradict the second law of thermodynamics”. - Boris Pavlovich
“A good many times I have been present at gatherings of people who, by the standards of the traditional culture, are thought highly educated and who have with considerable gusto been expressing their incredulity at the illiteracy of scientists. Once or twice I have been provoked and have asked the company how many of them could describe the second law of thermodynamics. The response was cold: it was also negative. Yet I was asking something which is the scientific equivalent of: Have you read a work of Shakespeare's?” - C.P. Snow

I find these particularly interesting but there are many others which mark the supremacy of the laws of thermodynamics over any other law ever formulated by man.
I have just begun to learn the laws of thermodynamics and it is indeed very different and "interesting" but so was Newton's laws of motion, Faraday's laws, Heisenberg's uncertainty principle, Einstein's theory of relativity, then why are laws of thermodynamics considered so immutable and supreme to other laws. What is it that makes all these great people hold them as so impactful and perpetual and makes these laws outweigh others?

The laws of thermodynamics have been resistant to experimental disproof for much longer than all of the others you mention.

The laws never outweigh each other, nor do they have to. Newton's laws of motion, Maxwell's equations of electromagnetism, Einstein's theory of relativity are all equally fundamental, in that they've never been overthrown within their respective frameworks of applicability. Sure, these frameworks have their limits: Newton fails at near-light speeds where Einstein takes over, Maxwell and Einstein fail at quantum limit. So what? Thermodynamics has its limits as well. If you don't believe this, then good luck trying to apply it to a single molecule.
If anything, the laws of thermodynamics are less fundamental, in that they are true only statistically, that is, with less precision than the rest. With relatively little effort, you can measure the mass of a body to six-digit precision. Can you do the same to its temperature?

Thermodynamics isn't only statistical. Various "Maxwell demons" don't break second law.

This doesn't seem to answer the question. The scientists quoted were remarking that they believed the Second law most strongly of all.

@Mithoron Of course they do (that is, would do if they were possible - that's their purpose in the first place). You can prove their non-existence by assuming the said law, but that would be an example of circular logic. The true reason lies within quantum mechanics. You just can't touch a molecule so lightly as not to change its energy, yet still get the information about it. If the molecules weren't quantum objects... Imagine a "gas" of flying cars, or just a busy traffic in a city, for that matter; the Maxwell demons indeed do exist in this setting, and are working quite successfully.

My point is that there are various implementations of original law breaking idea but they work without actually breaking anything.

What do you mean "Thermodynamics has its limits as well"? What are these limits?

Einstein's special theory of relativy is more fundamental than Newton's laws because the latter is only a limiting case of the former (i.e., $v/c \to 0$). Of course, Newton's laws are still useful because in everyday life, things move much slower than light, but the status as "fundamental laws of nature" is really overthrown by the special theory of relativity. Similar things can be said for quantum vs. non-quantum physics.

"So what?Thermodynamics has its limits as well." I am sure the scientists who said the above things also realized what you have said about the thermodynamics not being applicable to single molecule. Then, why they said what they said if the laws of thermodynamics also fail in specific scenarios. Why not just treat them as the same as other laws??

I don't know why they said what they said; I can't model minds so much greater than my own. I just shared my understanding of the subject. Feel free to agree or disagree.

"Can you do the same to its temperature?" In fact it takes quite a bit of effort to construct a system where statistical fluctuation are measurable in the lab (you have to get the system down to nanomoles in order to see occasional temperature fluctuations in the sixth decimal place). I think this was first done in the late 1970s and my thermodynamics professor described it as "a tour de force" with evident respect.

As Boltzmann taught us, (equilibrium) thermodynamics1 is a natural consequence of the statistical behavior arising in any macroscopic system, regardless of the details of the microscopic laws of physics followed by the constituent particles.
The nature of microscopic laws could be, e.g., non-relativistic, relativistic, classical, quantum. For the sake of thermodynamics, it doesn't matter what kind of laws are more fundamentally valid than others. At the macroscopic level, we always have the statistical behavior and hence thermodynamics. The details of the microscopic laws determine what probability distribution to use when doing statistics, but the fact that we have to do statistics is unchanged.
The above having been said, thermodynamics really has a unique status compared to usual microscopic laws of physics. It doesn't mean that thermodynamics in any way outweighs other laws. Thermodynamics is just in a different ballpark, and in that ballpark, there is really nothing else.

1It should be noted that equilibrium thermodynamics is applicable only to systems that can equilibrate, i.e., whose constituent particles can jiggle around in a sufficiently random fashion during the time scale of observation.

This. The connection between entropy and state counting makes the 2nd law intimately related to ... arithmetic. Which is pretty basic. To be sure there are a few other requirements, but they turn out to be very common. (And Einstein even restricted his statement to cases where those other requirements are met.)

The scientists quoted were remarking that they believed the second law most strongly of all. The reason is that the second law corresponds to real world experience. Scientists for instance believe that perpetual motion machines don't exist and accept that as evidence the second law must be true. (If a perpetual motion machine were possible then you could use such a machine  to create "free" energy by hooking it up to an electrical generator.)
The notion is that humans can't see the lines of force in an electric or magnetic field. We can do experiments to detect relativity, but it outside our day to day experiences. So those theories are abstract in nature.

This describes one downstream implication of the second law, but doesn't answer the question of why the second law is given such deference.

@hBy2Py - I thought it would be intuitively obvious to the casual observer that perpetual motion machines cannot exist. If they did, then they would violate the Second Law and thus prove that the second law was invalid.

Abstract from here: "Perpetual motion will not go away, and it continues to fascinate inventors. It simultaneously appeals to the best and worst sides of human nature: the urge to transcend the limits of the possible, against the desire to get something for nothing. Perpetual motion machines are scientifically impossible, but that does not stop a steady stream of applications attempting to patent them."

Unfortunately, unless one has been trained in the laws of thermodynamics, there is no immediate and obvious reason why they should not be possible. Once one has been trained, the conditioning for reflexive rejection of perpetual motion concepts is so strong that it's hard to imagine it not being instinctively impossible.

@hBy2Py - So we're having a violent agreement? The quotes were from scientists who were trained...

Maybe -- I may have misapprehended your point in posting what you did? Regardless, I'm still not quite sure how your answer addresses the question.

@hBy2Py - Scientists are so confident that perpetual motion machines don't exist, that they accept that fact as evidence that the second law is true.


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What is the quantum mechanical explanation of X-ray diffraction?
One topic in crystallography that I've found a surprising dearth of information on is what the fundamental interaction behind the interaction of the X-ray and atom. Pretty much every book just treats them as classical waves instead of explaining the quantum mechanics behind it. By this I don't mean that interference that leads to diffraction, but what actually happens when the X-ray interacts with the electron cloud.
The only description I've found so far is in Glusker, Lewis and Rossi that talks about:

"When X rays hit an atom, the rapidly oscillating electric field of the radiation sets the electrons of the atom into oscillation about their nuclei. This oscillation has the same frequency as that of the incident radiation. The result is that the electron acts as an oscillating dipole that serves as a source of secondary radiation with the frequency of the incident beam."

and goes on to note that this is coherent scattering and there is no wavelength shift.
Which doesn't make a ton of sense, isn't oscillating about the nuclei an old pre-quantum explanation of orbitals? Is this saying that this is similar to fluorescence/phosphorescence where the photo is absorbed, then another emitted? That is what most of the other grad students I've talked to have assumed, but if that was the case I'd expect a Stokes shift, the direction of emitted radiation to be random, and I don't see why you'd get a $180^\circ$ phase shift.
The other explanation I've heard is that you are actually sloshing electrons around within their orbitals, with the same frequency as the X-ray, similar to when you are dealing with optical properties, but I'm not sure what the mechanism for the interaction would be in this case.

I'm voting to close this question as off-topic because I think it's better located at Physics.SE.

@hBy2Py Really? SCXRD is usually considered to be a part of chemistry.

If it were a question on the interpretation of XRD spectra, then I agree it should be sited here (since there's no Materials.SE yet, AFAIK). For the theory of it, though -- I would think you're likely to get better answers at Physics.SE.  But, there may be some physical chemists here that'll be able to help out. I will not be upset at all if the close vote fails!

Don't close, I think we can deal with it.

Comments are not for extended discussion; this conversation has been moved to chat. I have retained the discussion about closing the question. If there's any relevant info, you can still find it there. Even better, edit them into the question.

There is Duane's hypothesis which is quantum mechanical hand waving, but doesn't really get to the nitty gritty.   https://en.wikipedia.org/wiki/Duane's_hypothesis

There is this reference which has 19 articles that cite it...
Quantum Theory of X-Ray Diffraction by a Crystal // Martin Ashkin, Masao Kuriyama // Journal of the Physical Society of Japan, 21, 1549-1558 (1966) 10.1143/JPSJ.21.1549 // http://journals.jps.jp/doi/abs/10.1143/JPSJ.21.1549

@IvanNeretin Indeed, the community seems to have run well with the question. Close vote retracted.

The explanation is not QM, it is QED, quantum electrodynamics. ;)

The description given in your question pretty much explains things. I would put it in slightly different words which may or may not help, which is that the x-ray photon induces a dipole moment in the atom (a field induced dipole) which then radiates and so the x-ray is scattered. This is pretty much what happens in scattering  by visible photons and we must assume that the photon is distant in energy from any transitions. (In Compton scattering part of the x-ray energy is imparted to the electron, so the frequency of the scattered x-ray is changed).
The field induced dipole moment is proportional to the polarisability which for a single bound electron is
$$ \alpha = \frac{e^2}{m}\left ( \frac{1}{\omega_0^2-\omega^2} \right )$$
where $\omega_0$ is the resonance frequency of the bound electron and $\omega$ the x-ray frequency and so $\omega >> \omega_0$ which in the limit gives
$$\alpha \rightarrow -\frac{e^2}{m\omega^2} $$
Thus the calculation reduces to understanding the QM calculation for the polarisability. See for example Atkins & Friedmann 'Molecular Quantum Mechanics' Ch $12$.
(  By considering that the x-ray is a plane wave the intensity of scattered radiation relative to that incident is given by
$$ \frac{I}{I_0}= \frac{e^4}{m^2c^4R^2}P(\theta)$$
where R is the distance from the scattering centre to the detector, m the electron mass and $\theta$ the scattering angle and $P(\theta)=(1+\cos^2(\theta))/2$ is called the polarisation factor.  This formula is sometimes named after Thomson. The $m^{-2}$ shows why significant scattering occurs only from electrons and not from protons. The book by Flygare 'Molecular Structure and Dynamics' section 8.5 gives a detailed description of x-ray scattering in general and details such as calculating atomic scattering factors. )

Thanks, that makes things make perfect sense, except for "which then radiates": I get the dipole polarization, you've explained that well, but I'd like a little more detail on this. Thank you.

All I can add is that the oscillating induced dipole radiates the scattered radiation at the same frequency as the initial photon.

Explaining the diffraction of a photon off of a crystal lattice quantum mechanically is the same as finding what momenta can be transferred to the photon by the crystal lattice. The simplest way to model this, as Ivan noted, is to describe the momentum states of a particle in this kind of potential as these are the momenta which can be transferred to the photon. Each of the barriers in this is interpreted as one of the atoms in the crystal lattice. Generally, a 1D approach is taken, but this is only because it is explanatory, and extension to 3D is not much harder.
In the interpretation I'm going to give, which I'm mostly reiterating from this link, the lattice itself should be finite as we would never really see diffraction off of an infinite lattice. Thus, the spatial wavefunction is simply a sum of periodic functions such as,$$\Psi(x)=\sum_{i=1}^N\cos(2\pi n\cdot\frac{x}{L})$$where $L$ is the distance between two atoms and $n$ is an integer. It's as simple as Fourier transforming into momentum space and finding the resulting probability density. This density is then interpreted as the diffraction pattern. The Fourier transform takes the form of $$\Phi(p)=\frac{1}{\sqrt{2\pi}}\int_{\frac{-L}{2}}^{\frac{L}{2}}\Psi(x)^2e^{ipx}dx$$
This certainly isn't the most detailed description one could give, but it is probably gonna be the theoretical groundwork for any interpretation of x-ray diffraction. It is also quite illuminating. To quote from the link earlier, the interpretation of this diffraction pattern can be stated as,

According to the quantum mechanical interpretation of diffraction, when this one‐dimensional electron
  density distribution interacts with an X‐ray source the individual photons are temporarily localized
  simultaneously at all lattice sites. The uncertainty principle requires that spatial localization leads to a
  delocalization (scattering) of the momentum distribution and the appearance of interference fringes because
  the spatial localization occurs at multiple sites.

I find that to be quite an illuminating interpretation as it is very similar to the classical idea of the photon making the electrons slosh around, but with the added detail that the electrons were already sloshing around some to begin with.
One can also take the approach of treating this as a particle-particle collision, which the link above also works through. This is more detailed, but you arrive at the same answer. The fact that you can treat the system in this way and get to the right answer basically confirms the original interpretation that the amplitudes of the momentum space wavefunction are a good representation of the diffraction pattern because these are the only momenta that could be transferred to the photon.
To explain how going to the 3-D case is not very hard, we would simply change our spatial wavefunction to be a product of periodic functions x, y, and z. Think particle in a 3-D box wavefunctions. The Fourier transform is almost the same as before because the triple integral could be separated into three identical 1-dimensional integrals. Of course, you don't really wanna do this integral for any more than like two atoms. Oddly enough if you used infinite atoms it'd actually be easier because you could use Bloch's theorem.

There is a well-known exercise in the basic courses on quantum mechanics: calculate a wavefunction of a 1D system with a rectangular potential barrier in the middle. The results are then interpreted as scattering of a wave on the said barrier: some part of it gets reflected, and some penetrates (probably tunnels) the barrier and continues on its way.
Well, the scattering of X-rays on an atom is a similar thing, only somewhat more complicated: we have a 3D setting, with spherical waves in addition to plane waves, and with an atom instead of the rectangular barrier. Other than that, the problem is pretty much the same.

I'd think that model is close. But that model would show a Stokes shift which I really don't think exists. So the diffracted x-ray just doesn't transfer momentum to the crystal, the crystal can also transfer momentum to the x-ray. So there will be a slight angle dependency for the exit x-ray and at the primary angle the energy of the x-rays would be slightly broadened.

Also I'd expect that if the incidence x-rays were in a perfect line that the exit-rays would be sort of a 2-D cone. More elliptical in shape than a pure cone if that makes sense.

The exit rays are spherical waves. It is the interference that shapes them into a series of individual spots. The very idea of a cone is simply not there.

@ IvanNeretin - I understand that is the classical interpretation. But if the x-ray is going one way then changes direction there must be a momentum transfer. The difference in energy is so small that it doesn't make a difference practically. But a true quantum mechanical solution would take the momentum into account.

Yeah, there should be some correction to that effect.

I believe OP is asking what kind of interaction can lead to scattering, since x ray is not necessarily resonant with any electron levels.


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Entropy of surroundings and heat
Apparently the entropy change of a system can be found by
$$\mathrm dS=\frac{\mathrm\delta q_\mathrm{rev}}T$$
I am slightly confused about this because I have seen this being used when the temperature of the system and the surroundings are not the same. However for the heat exchange to be reversible between the system and surroundings, need not the temperature of the system and surroundings be the same?
I have a feeling this might have to do with reversibility with respect to the system and the surroundings. I do not know about this and had not heard of it before I read it in a response to this post:

So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir)

Could someone please elaborate on this?

The temperature of the system and surroundings always match at their interface, irrespective of whether the process is reversible.  I would be interested in seeing an example of when that equation is used in a case when the temperature of the system away from the interface is different from the temperature of the surroundings.

Yes, that is a perfectly acceptable formula for entropy change even outside of thermal equilibrium.
And, as you say, it's related to the irreversibility of heat transfer. The point about ideal thermal reservoirs always exchanging heat reversibly is not necessary to understand this, so let's park it outside for a moment.
Let's imagine a body in an environment. We'll call the temperature of the body $T_b$ and the temperature of the environment $T_e$. If they exchange heat, their respective entropies, $S_b$ and $S_e$, change according to the formula you mentioned:
$dS_b=\cfrac{\delta q}{T_b}$
$dS_e=\cfrac{- \delta q}{T_e}$
Note the sign (the heat flowing into the body is leaving the environment, or vice versa).
Also note, and this is critical, that the magnitude of $\delta q$ is equal in both equations (one receives exactly the same amount of heat as the other loses), but, if $T$ is different, the entropy change won't be the same. The change in entropy (positive or negative) will depend on the temperature of the body or the environment.
Let's have a look at how the total entropy of the system ($S=S_b+S_e$) changes.
If the body and the environment are at thermal equilibrium, in other words, their temperature is the same ($T_b=T_e=T$),
$dS=dS_b+dS_e=\cfrac{\delta q}{T}-\cfrac{\delta q}{T}=0$
In other words, transferring heat between bodies at thermal equilibrium does not change the total entropy of the system.
However, if they are at different temperatures,
$dS=\cfrac{\delta q}{T_1}-\cfrac{\delta q}{T_1}=\delta q \left( \cfrac{1}{T_1}-\cfrac{1}{T_2}\right)$
Now, the sign of the entropy change depends on the relative values of $T_1$ and $T_2$:
If $T_1>T_2 \rightarrow 1/T_1 < 1/T_2 \rightarrow 1/T_1 - 1/T_2 < 0 \rightarrow dS < 0$
If $T_1<T_2 \rightarrow 1/T_1 > 1/T_2 \rightarrow 1/T_1 - 1/T_2 > 0 \rightarrow dS > 0$
Note that I chose the signs of $\delta q$ assuming that the flux of heat went from the environment to the body. So, this means that if the body is hotter than the environment ($T_1 > T_2$), this would imply a reduction in the entropy of the whole system ($dS < 0$) - which the second principle explicitly forbids. And, to the contrary, if the body is colder than the environment, that flux will be spontaneous ($dS > 0$).
So, outside of thermal equilibrium, you still can use that expression to calculate entropy changes - as long as the heat flux goes in the right direction, from hot to cold. It will also apply to heat transfers in equilibrium (and therefore reversible) - although in that case you'll be having some other energy transfer process of equivalent magnitude siphoning that energy off - otherwise the heat would increase the temperature of part of the system, breaking thermal equilibrium (that won't happen).

That formula is used when you have the $Q_{reversible}$ for a system. However, most of the time you don't have $Q_{reversible}$ for the system, but always remember that the $Q$ of the system is actually $-Q_{reversible}$ for the surroundings. So therefore you can use that formula to always find the entropy of the surroundings!


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Why are only elemental halogens used in oxidation of thiols?
I know that purpose of molecular halogen like $\ce{Br2, Cl2}$ is to take away extra electrons on sulfur so that sulfur bridge can form. But can you use any other molecule beside halogen to do the same job, like using nucleophile like $\ce{CN-}$, tosylate etc? Why do you always see examples of halogen only?

Your question is based on quite a bunch of false premises. $\ce{O2}$ isn't a halogen. $\ce{CN-}$ and tosylate (and most other nucleophiles, for that matter) hardly can oxidize anything. Thiols can be oxidized by any agent with sufficient redox potential, which is by no means limited to halogens.

Ok. This was a mistake. I clearly know  O2 is not a halogen. Plus, my book does not have any explanation of Br2 functioning as oxidizing agent here. It just includes Br2 as examples with no explanation. Then can u explain why nucleophiles are generally not good at oxidizing anything?

To oxidize something, you have to grab electrons from it. To be good at that, you have to love the electrons, i.e., to be an electrophile, which is an antonym of a nucleophile.

As mentioned above nucleophiles do not oxidise thiols to disulfides, what you need is an oxidising agent. Many reagents have been recorded as taking thiol to disulfide, simply stirring a thiol solution vigourously in air will give you the disulfide in some cases. It is worth noting that disulfides are relatively easily oxidised further to sulfenyl halide(RSHal), thiolsulfinate(RS(O)SR and beyond. The most reliable method of taking thiol to disulfide without over-oxidation in my experience is to reflux in inert solvent (CHCl3, DCM, THF) over activated Manganese Dioxide.


text:
Help with absolute configuration of a substituted cyclohexane?
How is the absolute configuration decided when it is a cyclohexane and several stereocenters? Up to now we have only worked with simple molecules which are only 1. I believe the stereocenters here are the 2nd, 4th and 5th carbon but what are their configurations?
Thanks!

The most helpful way that I was taught to identify the configuration of stereocenters is a right-hand/left-hand rule. First, assign priority numbers to each of the substituents according to atomic mass, or the atomic mass of their first point of differentiation. Next, identify the direction that the substituent with the lowest priority is pointing in (into the plane/paper, out of the plane/paper, or parallel with the plane/paper). Then imagine that your thumb is the lowest priority substituent. If the fingers of your right hand can curl in the direction of the 1-2-3 substituents, then the configuration is R. If the fingers of your left hand curl in the direction of the 1-2-3 substituents, then the configuration is S.
I found this method to be extremely helpful because no matter the orientation of the stereocenter, this method will work because you can adjust your hands to match the orientation of the stereocenter and identify it as R or S.


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Polarity of alcohols and their miscibility in water
I am an AP Chemistry student and there is a small detail concerning polarity of molecules that I am confused about.
My review book (Princeton review) says that ethanol has stronger intermolecular forces than methanol because it has a large molecular mass and is therefore more polarizable and more soluble in water than methanol.
Yet, my chemistry textbook says that both ethanol and methanol are miscible in water, but the more the carbon chain increases, the less miscible it becomes. Further supporting this statement is the fact that methanol is less soluble in hexane, a nonpolar substance, than ethanol. Therefore ethanol is less polar. Plus, heptanol is hardly soluble in water, but according to the Princeton Review's logic, it should be even more soluble in water than methanol because it has a larger molecular mass.
So, which is it? Does the addition of CH3s make a molecule more or less polar, and more or less miscible in water?

If your book says exactly that, then it says nonsense. Sure enough, the addition of CH2's makes a molecule less polar. As for the solubility in water, methanol is miscible - i.e., infinitely well soluble in it. So is ethanol. How would you compare the two infinities?
That being said, the bigger molecule of ethanol is indeed more polarizable (not more polar, mind you!), and has more intermolecular interactions per molecule than methanol, which accounts for the higher boiling point.

Thank you. My main misunderstanding seemed to be that I thought "polarizable" and "polar" were synonymous. Unfortunately my review book is relaying false information as it explicitly states that being more polarizable makes a molecule more polar.

@Ivan Neretin I don't get it. Wikipedia says,"Polarizability is the ability to form instantaneous dipoles." So you mean that the bigger the molecule is the greater the polarizability. But why? is it because thatthe more covalent bonds the molecule has the more dipoles can be formed?

@Mockingbird Forget the bonds. A lone atom would form instantaneous dipoles just as well. So at the end of the day, it is just sheer numbers of atoms of each type that matter.

My review book (Princeton review) says that ethanol has stronger
  intermolecular forces than methanol because it has a large molecular
  mass and is therefore more polarizable

True
Ethanol does on a molecule by molecule basis have stronger intermolecular forces between itself and water than methanol and water.

and more soluble in water than methanol.

False

Yet, my chemistry textbook says that both ethanol and methanol are miscible in water, but the more the carbon chain increases, the less miscible it becomes.

True
Why?
As you increase the number of carbon-chain linkages the polarizibility of the molecules does in fact increase.
However, the thing that makes small chain alcohols soluble in water is not the fact that you can induce a non-polar bond to have a short lived local dipole (van der Waals forces), but because of the permanent dipole of the alcohol group. The hydrogen on a alcohol (or water) is very positive and the oxygen is very negative, so they attract other polar groups around them in a fairly well ordered manner. For short chain alcohols the alcohols at the end of the chain can take the place of the oxygen and hydrogen on other water molecules and with the entropy inherent with mixing can be miscible with water.
For long chain alcohols the non-polar chains can be polarized, but they do not interact with water nearly as strongly as other waters and the entropy inherent with mixing is not enough to offset the energy difference. And so if there are any other waters "nearby" the water molecules will self-segregate, as water is more attracted to other waters than to the non-polar chains that they are inducing into forming dipoles.
As you add more alkane linkages the molecule becomes less polar but more polarizible - but this does not add much (comparatively) to the solubility of a non-polar molecule in a very polar solvent like water compared to highly polar groups.


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Is it a good idea to build a small scale ozone AOP reactor in my garage?
I have some interesting findings from my work (no conflict of interest), and I have been thinking to build an ozone lab in my garage to verify my ideas. What safety issues should I be aware of?
The purpose of the lab is to study some water/wastewater treatment process with ozone injection and advanced oxidation process. No pure oxygen or hydrogen, the power supply will be 110v, and I don't think there will be any explosive chemicals.
In term of safety, I think I need at least a fire extinguisher, basic safety wears, and an air vent to outside. Anything else?

By "air vent", I hope you mean a certified fume hood suitable for your process ;)

Doesn't sound like a good idea to me. Ozone is very reactive chemically. It wouldn't be good for your lungs and it would corrode/degrade almost anything that it comes in contact with in the garage. So you'd need a good vent hood with a vent stack 10 feet above your house.

The UV light is also rather reactive. You'd need to be particularity careful about eye exposure.

It's an empathetic no from me. How would you feel if your neighbour started venting poisonous gas to your backyard? Even though it decomposes to dioxygen, it is still harmful in low concentrations. Gaseous (in addition to liquid) ozone is also explosive on its own, though not sure at which conditions.

Have you considered whether a domestic swimming pool ozone generator would meet your needs? Very common, safe to install and use, and cost effective.

A general safety rule in any chemical experiment is to keep things dilute and small in quantity. As long as you are below safety limits you should be fine. There are some guidelines in Ozone concentration limit from government directed agencies (you could easily google that). Do not do anything in large scales! Your home is not a manufacturing plant and should not be one.


text:
Enthalpy change of chromate-dichromate reaction
I am a uni study doing chemistry. I have taken a little bit in highschool before, but not in a while.
In a (nonassessed) practical, we have to predict how heating a chromate/dichromate solution:
$$\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$$
will affect its equilibrium position. However we have not been given its enthalpy, and I can't find it online anywhere.
I remember from highschool that when you don't know an enthalpy you can estimate it by using heats of formation, heats of combustion, or bond energies. However I have never had to do this with an equilibrium of ions in solution before. Can the same approaches still be taken?
Please provide an explanation of how I can estimate the enthalpy change. Quantitative data would be greatly appreciated.

It's not a forum but Q&A site (Yes, there is some difference) 2) If you had thermodynamic parameters for this you'd be able  to calculate what happens not predict. 3) It's about how temperature in general affects reactions, in this case condensations.

@Mithoron Could you explain 3. in more detail?


text:
A IChO-2014 problem involving reaction mechanism

This is a problem of IChO-2014(&solution).But I am having trouble working out reaction mechanism of the second reaction or even how the product B formed. This reaction is somewhat new to me. And googling didn't help much.
Just need a little hint on this regard.

And Googling didn't seem much help.

Related: http://chemistry.stackexchange.com/questions/69042 and http://chemistry.stackexchange.com/questions/62177

The key also for the second reaction is the use of a very bulky borane. There are two sites for hydroboration, and we are selecting for the one that is less hindered. Using a hindered borane at low temperature means that we can avoid reacting much at the tri-substituted double bond. Too bad it also takes seven days to react at the less substituted double bond.

@Zhe What does acetone do in forming B to C?

As usual, below the arrow, it's solvent.

@zhe why did the alcohol dirctly converted to acid? Is CrO3 too strong oxidezer?

Very conventional oxidation with acidic chromium trioxide. Too strong is relative. Here, the acid was the desired product, so it was strong enough.

A to B is a classic hydroboration of the least hindered double bond followed by oxidation of the trialkyl borane.


text:
Why is there no reaction?
I am working on a practice problem with p-iodotoluene and $\ce{CH3O-}$ in $\ce{CH3OH}$ at $\mathrm{25^{\circ}C}$. Another reaction is run which is identical except there is heat and pressure. Neither of these scenarios produce a reaction according to the answer key. What am I missing?

Would you mind describing what's on your answer key & some necessary details on 'except there is heat and pressure'?

That's all the question and answer key honestly say. I can say CH3)(-) is in CH3OH with p-iodotoluene and is heated and pressurized. Beyond that, I wasn't told anything else unfortunately.

This is not something we can answer in its present state (excluding those who have the ability of mind-reading).

I am confused by what you mean. How could I change the circumstances to to have enough of a question with clarity? It's still an intro course so I am not well-versed.

@Ryan Wisth But what is 'another reaction'?

I already clarified in my comment above what the second circumstance is. By another reaction it's an entirely separate problem.

It does not react from knowing the answer but is assumed to be asking "what will the product be"?

Okay, now I am getting hold of your question. might give you an answer.

(This answer is made on assumptions made on your question; maybe it's wrong.)
p-iodotoluene is nucleophilic. So, in order to happen a reaction with p-iodotoluene, you need an electrophile. But $\ce{CH3-}$ is not an electrophile, rather a nucleophile.
But, you can get $\ce{CH3+}$ from methanol. But, methanol breaks into $\ce{CH3+}$ and $\ce{OH-}$ only in presence of a strong proton donor (i.e $\ce{CH3OH + H+<--> CH3+ + H2O}$) which is not present here, and I think it's very unlikely to have a strong proton donor in this alcoholic solution.
And the reaction
$$\ce{CH3^-X+ +  CH3OH-> CH3+ + CH3- + XOH}$$
seems bogus to me as alcohol won't have that polarity to break a ionic bond.
So nothing seems plausible. Increase of heat and pressure doesn't change the scenario.

He meant CH3O- No idea why you read his typo from comment in other way then repeat of what he already told.

Oops... But when I answered, his question was in bad shape. so, I followed the comments. But it would be same still.