text:
O-carboxylation in the Kolbe–Schmitt reaction
In the Kolbe–Schmitt reaction, we know that the pi electrons of the benzene ring attack the carbon in carbon dioxide, but why doesn't the reaction occur on oxygen instead (as shown in the diagram below)?

Try to think the other way around: If such an attack would occur, what is the fate of the resulting intermediate, a carbonate half ester? With other words: Such a reaction might take place, but it is reversible and does not lead to a product that con be isolated.

The resulting product would be a carbonate hemiester monoanion, specifically monophenyl carbonate.  Similar species have been observed in carbonated alcoholic beverages.  (Instead of phenol, the alcohol involved is ethanol so the carbonate hemiester is monoethyl carbonate.)  There is an equilibrium between bicarbonate and the hemiester:
$$\ce{HCO3- + ROH <=> RCO3- + H2O}$$
According to the paper I linked the equilibrium constant is about 2.0, i.e.
$$K_{eq} = \frac{[\ce{H2O}][\ce{RCO3-}]}{[\ce{HCO3-}][\ce{ROH}]} \approx 2.$$
Under aqueous conditions, the concentration of water is very large and the concentration of bicarbonate is much lower, so usually the equilibrium concentration of the carbonate hemiesters is very low, on the order of a few mM.
In contrast, under the conditions of the Kolbe–Schmitt reaction, the carbon-carbon bond formation is effectively irreversible.  There is no equilibrium, and the desired product can be obtained in high yield.

Carbon in the benzene ring are more prone to give their electrons than oxygen atom to CO2 because oxygen is more electronegative than carbon and thus carbon gives electron to carbon dioxide more easily. Comparing these effects with the help of electronegativity is not bad, because carbon and oxygen belongs to same period.

Sorry, but this is too simplistic. You're not comparing carbon anions with oxygen anions; both the carbon and the oxygen are part of a single pi system which can react at two points (or even three, if you consider ortho/para).

@orthocresol Thanks for suggestion . Can you please elaborate more the problem so that this answer can be improved?

@orthocresol Please elaborate which carbon and oxygen you are talking of : of the benzene or the CO2

First, you should consider whether you are addressing the kinetic product or the thermodynamic product: the latter is already covered by the accepted answer, so there's nothing more for me to elaborate there. And probably the question is more interested in the thermodynamic product anyway, because that is what is observed in practice. And if you are trying to address the kinetic product, I'm afraid it really requires a rather deeper analysis than what you've posted.

@orthocresol Thanks, it looks like I am not much experienced to answer so deeply. I will edit it as soon as I will learn such higher chemistry.

However the reasoning I gave is also valid for reimer tiemann reaction

The phenoxide doesn't attack on oxygen because the oxygen on the carbon dioxide is electron withdrawing leaving an electron deficient region on the carbon atom.

You misunderstood the question.


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Adsorption and distribution chromatography difference?
Well, yes, that is the question. What is the difference between the absorption chromatography and distribution chromatography?

Adsorption Chromatography = Utilizes a mobile liquid or gaseous phase that is adsorbed onto the surface of a stationary solid phase. The equilibriation between the mobile and stationary phase accounts for the separation of different solutes.
Distribution Chromatography (aka Liquid-Liquid extraction) = Due to its liquid stationary phase, the ingredients are only separated by their respective polarity, i.e. according to their distribution coefficient KD. Distribution chromatography is also ideal for the concentration of trace quantities. This is more closely related to Flash, Column, and Gas Chromatography.

Basically, the first is solid-liquid, the second liquid-liquid?
And if it is so, why does RP-chromatography belong to distribution chromatography and not to adsorption chromatography?

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text:
Number of water molecules to contribute one molecule to every milliliter on earth

How many millilitres of water would it take so that, when they are thoroughly dispersed around the world, it would yield an average of one molecule of the original sample of water in each millilitre of water in the world?  For the sake of argument, assume that the H and O atoms do not dissociate, and the water molecules remain intact.

I calculated it once and got something around 40 ml, but I forgot how I made that calculation.

Well, you need to find out how many mL of water are there in the world, and then take that many molecules.

The answer I calculated is very close to the 40 mL calculated by the OP.  Thus I think it is safe to assume that we do know what he asking and that we should reopen this question.

A very quick Googling tells me that the volume of the oceans is 1.3 billion $\mathrm{km}^3 = 1.3\times10^9\;\mathrm{km}^3 = 1.3\times10^{18}\;\mathrm{m}^3 = 1.3\times10^{24}\;\mathrm{cm}^3 = 1.3\times10^{24}\;\mathrm{mL}$.
To get one molecule into each of those mL, we'd need $1.3\times10^{24}$ molecules of water, or $\frac{1.3\times10^{24}}{N_A} = \frac{1.3\times10^{24}}{6.02\times 10^{23}}$ moles of water, about 2.2 mols.  Since water is 18 g/mol, that comes out to 39 grams of water or about 39 mL.

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Hypothetical concerted process feasibility
How feasible is the following hypothetical reaction, where both C and D are awful leaving groups (so the reaction must happen in one step)?

I don't see a way how the orbitals could interact, so I think that such a transformation could only occur intermolecularly. I need a second opinion.

One example of this could be the reaction of fluorine gas with oxygen gas to form FOOF.  C and D are both F and A and B are both O.  This occurs by a free-radical mechanism I believe.  https://en.wikipedia.org/wiki/Dioxygen_difluoride

But it can't be concerted and intramolecular, right?

The process you describe seems very improbable. The reverse reaction will be synchronous syn- addition, which does exist, but I don't think it is true concerted process in a sense of orbital control.
Other limit is the [2+2] cycloaddition, but there is a C-D bond even in reactants.


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Can the properties in a tomato help make hair curl?
Is it possible for the active properties of a tomato to turn your hair from straight to curly? And what are the active ones that do this and why?

Do you mean by eating tomatoes?  Or putting them in your hair?  Or???

What are active properties?

I'm not really sure this is on-topic here. Unless you are interested in specific chemical processes then I will vote to close this as off-topic.

I would have guessed it to be the other way around, that the acidic components found in tomatoes would break the disulfide bonds in hair protein allowing the curls to "relax".

Tomatoes do contain interesting compounds! One of them, responsible for the red colour, is lycopene, a carotenoid. These compounds are known to be radical scavengers and antioxidants. It seems that mixtures of squashed tomatoes and olive oil are used for hair treatment, but that's not my league. However, I find it unlikely to achieve  the effect of a permanent here.

I suggest to leave the question open . Let's be a tad more relaxed with questions concerning [tag:everyday-chemistry] and sink more do-my-homework instead ;)

Nice question. Hair care is a multi-billion dollar industry globally, and one that is very much hinged around chemistry. Can't see how this question is not about chemistry.

From what I know, hair straightening/curling, both require the reduction of disulphide bridges using some form of reducing agent and then allowing them to reform once the hair has straightened/curled so that the fibres hold their shape. Now, I don't know what you are referring to when you say "active" properties of tomatoes, but if the fruit contains compounds which can do what I've described above, then yes it could be possible.


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Photolysis of N-bromo compound in acidic conditions (radical cyclisation)

For the proposed mechanism, I believe that the N-Br bond is firstly broken by the absorbed light hv, then the radical formed on the N will perform an intra-molecular attack on the end of the carbon chain to form a 5 membered ring. Then the sulphuric acid becomes deprotonated by the nitrogen lone pair? Just want to check whether this is correct.

This is the Hofmann-Löffler-Freytag reaction, which was later examined by E. J. Corey (J. Am. Chem. Soc., 1960, 82, 1657-1668,
DOI).
When the reaction is performed in sulfuric acid, it is conceivable that the nitrogen atom of the N-Bromoalkane is protonated. Homolytic cleavage of the $\ce{N-Br}$ thus probably leads to a radical cation.
The latter then abstracts a $\delta$-hydrogen atom, resulting in a radical centre in the side chain. It has been suggested that this centre might recombine with the bromo radical previously released. With other words: $\ce{H}$ and $\ce{Br}$ have changed their position. Deprotonation of the ammonium cation and a subsequent nucleophilic substitution, releasing $\ce{Br-}$ would then result in the formation of the indolizidinium skeleton.

Thank you so much! I do however have a question. How can one determine the decomposition pathways in this reaction?

Interesting and difficult question! If there would be any pi-system at all, I'd suggest time-resolved uv-vis spectroscopy. All the radicals and ions involved here are spectroscopically invisible, just like a DABCO radical cation. Is there any chance to trap an intermediate? Intermolecular vs intramolecular often isn't very promising either and sulfuric acid as a solvent limits the options too. Maybe you can find some more recent mechanistical papers that cite the old Corey paper? Did anybody perform CIDNP or ESR spectroscopy on this reaction? Honestly, I don't know.


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Breaking bonds with energy
I'm learning about breaking molecular bonds currently and I'm wondering what are the methods to actually use the amount of bond energy to break the bond. What I'm saying is, knowing the bond energy, how can you actually use this information to break the bond? Do you heat it up the equivalent amount as the Kj/mol or something similar to that?

I guess you deleted earlier question and reposted it... Never do such thing!

Sorry I really need the question to be answered

It's stupid, simply editing it bumps it to the top and could actually improve it.

I'm new here. I didn't know that

At least now you'll know ;)

Yes, the energy showed that is needed to break bonds is usually in the form of heat. I don't believe there is another common way to do it, but don't quote me on it. I haven't learned about lasers and other such technologies :). Keep in mind that as bonds are broken, they release energy themselves. As bonds are created, they take in energy on their own. Often, you will just need to start a reaction with some outside energy source (A bunsen burner for example) and it will be able to continue on its own because it produces energy from breaking bonds. If you look at the formation of Magnesium Oxide, all you really have to do is light a small end of a piece of Magnesium ribbon, and the whole thing burns because it creates energy as it reacts. I hope that helps!


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How to balance the following equation?
In the following unbalanced chemical equation the oxidation state of nitrogen ($+5$) is both decreased ($+5$ to $+2$) and unchanged ($+5$), so I am unable to balance it .
$$\ce{Cu + HNO3(dil) -> Cu(NO3)2 + 2NO + H2O}$$
My attempt: I tried by making equal decrease in oxidation number and increase in oxidation number but I am  not getting the given answer.

Possible duplicate of Simplest way to balance any chemical equation

There are two half reactions:

elemental copper to cupric ion
nitrate ion to nitric oxide

The copper half reaction should be easy to balance, but the reduction of nitrate to nitric oxide is more complicated. start with what you know:

Nitrate goes to nitric oxide, one to one.
Then balance oxygen with water since it is aqueous, presumably.
Then balance hydrogen with protons since it is acidic from nitric acid.
Finally, balance the half reaction with free electrons, which should be three electrons for a single nitrogen atom in the reactants to be consistent with nitrogen going from +5 to +2.

There are different numbers of of electrons in the copper oxidation and nitrogen reduction half reactions at this point, so you have to multiply each half reaction to get the numbers of electrons in each equal to cancel out. The free protons should be converted into nitric acid by adding the appropriate amount of nitrate ion to each side. your final balanced equation should have a ratio of elemental copper to nitric acid of 3 to 8, respectively.

Consider those nitrogens two different elements for a while.
$$\ce{Cu + \underbrace{HNO3}_{for\;redox} + \underbrace{HNO3}_{not\;for\;redox}->}\dots$$
This you supposedly know how to balance. The "first" nitrogen is an oxidant and should be balanced as such; as for the "second", you need 2 of it per each $\ce{Cu}$.


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Is Acetanilide toxic to fish?
I understand that acetanilide can hydrolyze to aniline and assume that will happen if that gets added to an aquarium, thus causing toxicity?
I have a solution that is $3\%$ hydrogen peroxide and $0.25\%$ acetanilide that I'd like to add to my fish and shrim aquarium to kill some algae.

If there is any miracle mixture of chemicals that would kill algae but not fish, then I'm sure that it is sold in aquarium stores. Rather than experimenting on your fish, I suggest you buy some commercial product if it is available.

When I checked the SDS of acetanilide on Sigma-Aldrich website, I saw that it is harmful to aquatic life.
So, there is no exception whether it is an alga or fish to kill. I suggest you not use acetanilide in the aquarium.


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What is the relation between dielectric constant and hydrating tendency?
What is the relation between dielectric constant and hydrating tendency?
Could someone also explain both terms?

Completely different things...

What Mithoron commented is not actually true. They are related in a a few ways.
First of all, I'll tell you about dielectric constant. The way I learned it, a dielectric constant is the ratio of the capacitance of a capacitor filled with a certain material (Let's say $C$), to the capacitance of the exact same capacitor in a vacuum (Lets call this $C_0$). The dielectric constant is just $C$/$C_0$.
Hydration tendency is a little more complicated, but after researching it, it seems that it is the ability of a substance to make another substance more soluble.
Water has a very high dielectric constant. If you look at water, the O-H bonds are polarized because they have a high electronegativity difference. This means that water has strong interactions with the ions of salts, creating hydration shells around them (This basically means the salt particles won't be attracted to each other as much). Stronger dipoles of water result in a greater dielectric constant, and a higher hydration tendency. I must admit though, this is a strange question. If you don't mind, you could you explain why you need it?

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Comparing Methods of Determination of Dissolved Oxygen Content in Water Sample
I am currently doing an IB Chemistry Individual Investigation on dissolved oxygen in water samples by using Winkler method. However, my teacher suggested that I have to use oxygen probe as a standard and comparing the efficiency of dissolved oxygen content determination between Winkler method and other methods.
My problem is I don't know what method I have to compare with the Winkler method that I had done. Are there any suggestions or improvements on my investigation to let it have more exploration aspect?

Let's abstract your task to determine dissolved oxygen (DO) to something more general:

You have to tie an observable to a concentration.
It would be nice if the method would work over a broad range of concentrations.
The method should be specific, or, alt least, very selective.
If possible, this relation should be linear.
It would be nice if the method could be packed into a sensor that works on its own (permanently or in cycles) and does not require a lot of personal interaction (= autonomous probe)

What properties of molecular oxygen dissolved in water can be measured?

In water, $\ce{O2}$ can be reduced electrochemically: $\ce{O2 + 4e- + 2 H2O -> 4 OH-}$→ Have a look for galvanic and polarographic oxygen sensors
Molecular oxygen is known to quench the luminescence of electronically excited states of dyes. The observed luminescence intensity (and lifetime) is reciprocally proportional to the the concentration of the quencher(Stern-Volmer equation)
→ Have a look for luminescence-based oxygen sensors


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How do I analyse this data? (optimization experiment)
I conducted an experiment to find the optimal concentration for sugar in yeast respiration, and this is the data I found:

What do I do from here? Do I just fit a linear model on the increasing portion, another one on the decreasing portion, and see where they meet and say that's the optimal value? If not, do I try to fit another type of model? How do I justify that model?

From the looks of it, you can try to fit a quadratic $-ax^2 + bx +c$ using least square or some such fit. The vertex of such a quadratic is $\frac{b}{2a}$ which is your optimal value

Short answer: It depends on how accurately need that value and what do you use it for. Someone longer: Fitting this data is almost meaningless. You need at least 1-2 points between 50 and 100 to have at least  a guess what you should fit.

Since you say 'average' concentration of sugar solution and 'average' volume of water displaced, do you have more data? As @Greg said, having more points will make it clearer where the optimal value is.

@Greg I don't have any values for between 50 and 100. And yes, I can see what you mean by almost meaningless. I previously just had a linear line of best fit through the two sections (marked with different colours) and calculated where they met. I'm now considering instead to run a one-way ANOVA and then Tukey, but I'm not sure if that's correct. I'll ask in math stack exchange as well I suppose

Thank you all for answering! I will try those things and check with my teacher

You really do need to give some more details of what you did otherwise its not possible to give a sensible answer. Assuming that your data is correct then a rising and falling profile can be analysed by an equation of the form $a.exp(-b.x)-c.exp(-d.x)+f$ where $a, b, c, d, f$ are fitting parameters. However, you do need more data points if you are going to fit.

Some materials (e.g. ionic salts) in certain solvents (e.g. water) actually do decrease volume of the solution in certain concentrations (obviously at some point it inevitably increases).
So you certainly can end up with non-linear functions of displacement from $\Delta_{conc}$.
That said, a negative quadratic may fit your data well, e.g. $ax^2 + bx + c$, with $a < 0$. Maybe even a damped exponential, e.g. $ae^{-bx}$. You can use Microsoft Excel etc. for a quadratic fit or fancy math skills for other functional forms.
Another Ben describes in more detail here but the image he used is worth repeating. Here the concentration of an added substance actually shrinks/increases volume displacement "oddly".

As you add sugar the displacement will only increase. In other words, if you add 300 grams of sugar to a liter of water, the volume will have increased (or been displaced) by about 150 ml (guesstimated)... the displacement is certainly not 25 millileters!
For what it's worth, you should have two (practically) straight lines. One relatively straight line before the saturation point (at around 200g sugar per liter at room temperature) and another relatively straight line after the saturation point. But believe me, the more sugar you add, the greater the volume will be (until the volumetric cylinder is full).
Displacement in a nutshell:

Isn't the water displacement a separate measure of the yeast respiration and not referring to the displacement of water due to sugar addition i. e. Not related to the density increase?

@Beerhunter At first, I wondered if he added the yeast and allowed the sugar to be fermented before measuring... then I figured that if he added 300g of sugar, it should be converted to around 180ml of alcohol... so however he came up with ~25ml it's simply a mistake. Part 2- Technically, displacement is related to density (see the story behind Eureka). Displacement is the measure of volume-increase when stuff is added to a liquid. So, I'm not sure what he did, but J.Ma should (re)measure of how much the water rose.

It does not say anything about the volume to which the sugar is added to. Your statement in the first paragraph is maybe based on a litre of water. But if I add 30 g sugar to 100 mL water, I could very well have a displacement of 25 mL.

@Martin-マーチン okay, + for thinking outside the beaker... but the chart says 300 grams... so that's gonna add more than 25ml. 50g of sugar displacing 25 ml makes sense... 100g displacing 50ml makes sense... then you add more, and viola it collapsed?

The chart specifies 300 g/L, which is a concentration, it does nowhere specify in which volume the displacement is measured. It seems fairly obvious that the base volume is not one litre, because as you mentioned that would cause a higher value of displacement.

@Martin-マーチン I suspect that, after 100g was added to 1 L of water, another 50 g was added, and the difference between the new and previous measurement was 45 mls... which was recorded and not added to the previously measured displacement (50 mls)... so the displacement for 150g might be 95, not 45; and 200g ~ 137ml, not 42. Also, the chart doesn't say whether or not sneaky ninjas drank some... I think it's distinct possibility.

I feel like you are mocking me now and I can't really appreciate this. I am simply saying that you are basing your first statement on an assumption, you are over interpreting the given data. There's really no need to involve ninjas here.


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What are some Internal Standards used for Ion Chromatography?
What are some internal standards (ISTDs) used for ion chromatography (IC)?
For experienced IC users, what ISTDs have you used for your sample analysis?

This might be related: http://chemistry.stackexchange.com/q/14085/4945

Thank you Martin for your kind advice! I gotta edit the question. :)


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Reaction of enols with electrophiles
In a reaction between an enol and an electrophile which will attack the electrophile, the oxygen of the $\ce{-OH}$ group or the pi electrons of the the double bond?
This is my attempt at a solution: Since, the oxygen atom has two lone pair of electrons it should be more electron rich than the double bond and therefore it should attack the eletrophile, Am i correct?

It depends and number of lone pairs doesn't matter.

How have you got a measurable concentration of enol in the first place? 2) think about stability of the resulting particle and 3) this actually strongly depends on the electrophile in question. Furthermore, often the attack and the whole reaction are reversible and it the product is defined not but preferable attack position but by thermodynamical stability of the alternative products.

The functional group in question is a vinyl alcohol, and it is a mistake to consider these groups as isolated. The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking. There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.
This is almost the same as your other question:
Hydrolysis of 3,4-dihydropyran


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Hydrolysis of 3,4-dihydropyran
In the 3,4-dihydropyran shown below will the oxygen or the pi electrons of the double bond attack the hydrogen cation during hydrolysis and why?
This is my attempt towards a solution: The oxygen lone pair clearly can attack the hydrogen cation better than the pi electrons being a better a nucleophile, but according to my textbook (T.W.Grahams Solomons) the pi electrons of the double bond attacks, but shouldn't the oxygen atom be a better nucleophile than the pi electrons?

The functional group in question is a vinyl ether, and it is a mistake to consider these groups as isolated.  The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking.  There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.

Could you please rephrase your last sentence? I didn't really get it.

If you protonate the oxygen, you are not really using a conjugated system any more, and the energy gain from conjugation is not helpful for driving the reaction


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Software for compound database creation\enumeration
The only program I have found is Plexus Suite from chemaxon. (I asked for trial, but no answer to me yet).
The task is: create variation of compounds, where is scaffold and different substituents:

And I need to create dataset/database of all possible variations.
How to deal with this problem? Or I need to use some scripting to make the database?
UPD1: best option - creation of SMILES (Simplified molecular-input line-entry system), 3d structure is automated and easy step if there is correct SMILES/connectivity of molecule (which is the same, as smiles)
Using XYZ - it means, you should make substituent database in correct position for adding.  I am asking for software with functionality
https://www.youtube.com/watch?v=s4CgtUgGuzk
Vega ZZ has the module for database enumeration, but it is working as Python script.

The Plexus Suite looks absolutely awesome! Is it available under an academic or academic commercial license? If not, it still is the most elegant, but probably expensive solution. In addition, it might take some time and money for a dedicated machine to to set it up. Is it worth the effort for some 48 derivatives? A workflow, consisting of a 3D molecule editor, that is able to expand superatoms (Ph, Ac)  to real atoms, followed by a tool that generates InChI and InChI Keys (to be used as keys in a DB) looks like the next best approach to generate the structures.

@KlausWarzecha Just got the trial - it is really the only one. I can't find anything, similar to this. About how it works better ask support team - website-support@chemaxon.com. I believe, that academic licence will be not very high, otherwise I will create my own script, without any allegiance, but working.

If such an editor isn't available, generating the skeleton with Avogadro, marking the substituted positions by repacing $\ce{H}$ with different halogen atoms and replacing all these manually with the real substituents is the cheapest option sans the labour costs. From the files thus generated, 'openbabel' can generate the InChIs. The rest is setting up the respective tables in a RDBMS of your choice.

I have used MarvinSketch in the past under an academic license at no cost, thanks to ChemAxon, but the Plexus Suite is definitely another league :)

@KlausWarzecha also, this is server/web-based application, so they are hiding the source code of the chemistry magic. :)

@XuMuK I don't think I understand your "update." You're saying that you want software that does all of this with XYZ files, not with SMILES?

@GeoffHutchison I am still looking for more  useful GUI database creation tool, and Vega ZZ has almost the same python script built-in.  (sometimes SMILES have problems with interpretation)

This is typically called library (or scaffold) enumeration. Doing it in SMILES is usually pretty easy by script, but there are a few other options:

KNIME
SMILIB

But it's very easy to write a script like this, e.g. (in python)

r1 = [ "", "C", "N" ]
r2 = [ "H", "C", "c3ccccc3", "C(=O)O" ]
r3 = [ "", "C", "C#N", "O" ]
scaffold = "Xc(cc1c2)ccc1c(Y)cc2-C(C=C1)C=C1Z"
for x in r1:
    for y in r2:
        for z in r3:
            print scaffold.replace('X', x).replace('Y', y).replace('Z', z)

You can then process the resulting SMILES strings with your program of choice (e.g., Open Babel).

the python smiles is much more interesting - will play with substituents tomorrow. Sorry, I don't understand how to load scaffold molecule to the KNIME, haven't worked with it.

@XuMuK - Indeed, it's very easy to do this kind of thing with SMILES.

I have tested there are some problems with correct input for scaffold and substituents, but it is working!!!

Very nice and instructive. Regarding the openbabel smiles->xyz, it tend to produce molecular knots for even small macrocyclic structures.

@ssavec If you have some examples, please e-mail me. We are working substantially on coordinate generation algorithms.

There are no doubt many software tools out there to do this kind of scaffold enumeration, however the problem is surely simple enough to do by scripting - or even by hand!
eg:

Compound 1 = {R1=H, R2=H, R3=H}
Compound 2 = {R1=H, R2=H, R3=Me}
Compound 3 = {R1=H, R2=H, R3=CN}
...
Compound N = {R1=NH2, R2=Ac, R3=OH}

So just the cartesian-product : R1 x R2 x R3 except with unordered output ... not sure what this is called. There is this math stackexchange question about this.
Oh, and this stackoverflow python code to do this.

The xyz not needed, I need 2d structures or SMILES, or 3d with correct connectivity records. Now I am thinking about creation my own script to deal with the problem. "are no doubt many software tools out there to do this kind of scaffold enumeration" - I am not sure, could you give the list?


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Reaction involving markovnikov's rule

Question:

Attempt:
The first reaction is addition of hydrogen halide. The reaction follows markovnikov's rule. The two possible products of the first reaction are:

For each product, the carbocations formed are:

After the formation of carbocation, $\ce{Cl-}$ attacks the carbocation forming a dihalide.
The final product after second reaction depends on the above intermediates.
The second carbocation faces electron withdrawing inductive effect and electron donating resonance effect from chlorine. But for halogens, inductive effect is more than resonance effect. So the carbocation is destabilized. So I chose option (A) as the answer since the first carbocation is formed.
But the answer given is (C) which means the second carbocation is formed.

What is the question?

@Lighthart the answer given is C. But I got A.

Clearly the resonance effects for pi-bond formation with chlorine are stronger than the inductive effects.  Most of the time resonance is a stronger effect that induction, unless fluorine is involved.

@Lighthart but that's not the case in chlorobenzene is it?

Chlorobenzene is an electrophilic aromatic substitution (EAS), which is a different mechanism that the electrophilic addition you have presented here.  That being said, in EAS, chlorine's pi-donation directs ortho and para, so yes, the resonance effects are dominating the inductive effects in this case too.

"But the answer given is (C) which means the second carbocation is formed." Aldehydes are primary...and so is the geminal dichloride from which it was formed. The answer is C, but your reasoning is incorrect. The inductive effect is not stronger than the resonance effect. Consider what kind of director halides are in electrophilic aromatic substitution.

Indeed the second carbocation is formed.  It's favored by the chlorine atom being able to delocalize the charge through resonance, draw resonance hybrids for the second cation and you should see it.  In a sense you might say that the carbon bonded to chlorine is more effectively substituted for stabilizing the cation and so the Markovnikov rule is "bent" accordingly.
The usual version of the Markovnikov rule is for hydrocarbon substrates.


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Testing for ammonia by smelling it?
My chemistry teacher taught us that one of the ways to see if we had produced ammonia gas in a reaction was if it released a pungent smell (Even BBC supported that here! http://www.bbc.co.uk/schools/gcsebitesize/science/triple_edexcel/qualitative_analysis/qualitative_analysis/revision/3/ ). However, as far as I knew, Ammonia was poisonous. Is it okay to breathe in small amounts of Ammonia, or is this a hazardous practice that can potentially harm us?
This question What chemical compounds used in qualitative chemistry can be distinguished by smell? did say that Ammonia could harm you, but it doesn't say how much is needed to cause the damage, which is a big part of my question.
Thanks!

Yes, large amounts of ammonia are poisonous. However ammonia is a common household cleaner. Very small amounts are ok. Your teacher should have taught you the technique to use the smell test too. You don't stick a nostril over the test tube and deeply inhale, but rather fan the top of the tube with your hand towards your nose to just get a whiff.

First, using the smell test for chemistry is dangerous - don't do it!
Second, for this an similar questions, consult a Safety Data Sheet (ex-MSDS) for exposure limits.
The copy that I looked at indicates:
TWA of 25ppm (from the ACGIH) or 50ppm (OSHA) -- these are based on a 8 hour/day, 40 hour/week exposure.
STEL of 35ppm (short term exposure limit). Exceeding the STEL may result in 'upper respiratory tract irritation, eye damage'.
Now, the yield from your lab bench chemistry experiment is unlikely to be very high, and as you may have noticed ammonia is very noticeable (although of the several SDSs that I consulted, none list an odor thershold, surprisingly).
But, the bottom line is, do not inhale your chemistry experiments.

This answer is an overreaction. The smell test is fine as a conformation for certain chemicals. Ammonia is sold as a common household cleaner.  If getting a whiff was deadly there would be dead folks dropping like flies. Obviously the smell test isn't something that you'd use for nerve gas.

That, of course, depends on the scale of your experiment/bath/whatnot. It is quite likely that his particular case is not deadly. But, it is not made clear if this is a small test tube, or a 30 liter bucket - I've seen 'simple' scale ups result in really bad outcomes involving rescue folks in SCBA. Getting in to the habit of sniffing, particularly without checking an SDS, is a really dumb idea. And, yes, people do drop dead from mixing common household chemicals. And, yes, ammonia could certainly trigger an asthma attack. Use your smell for cooking, not chemistry.

@JonCuster: If a reaction emits dangerous sufficiently dangerous gases that breathing apparatus is required, removing the apparatus to sniff the gases would obviously be a bad idea.  If breathing apparatus would not be required, however, I don't think sniffing is any more dangerous than breathing in fumes without trying to sniff them.

And there is a reason we use fume hoods for real work. Look, I would rather tell people to do things the right way rather than telling random folks with unknown facilities and safety support to go around sniffing their chemistry experiments. Much of the time it might work out OK. Some times it won't and people get hurt or killed. In the real world, in real labs, folks get hurt.

I would advise against placing any importance on smell as a test.

Just because you smell ammonia doesn't mean it's necessarily ammonia. If humans had such a great sense of smell, we wouldn't need NMR and all those fancy machines, we would just confirm our products by their smells. Identifying chemicals by smell and taste is something that you would do a century ago and certainly not in (CURRENT YEAR).
As you progress in your study of chemistry, you will make more and more dangerous compounds which obviously preclude using smell as a test. For goodness' sake, don't do $\ce{[Mo(CO)6] + 3py -> [Mo(CO)3(py)3] + 3CO}$ and try to smell the reaction to tell when it's done. (Not that you would be able to tell anyway!)

Essentially, if you're making ammonia by heating some ammonium salt, please make sure that it also turns damp red litmus paper blue. I know that in some syllabuses you will be required to write "a colourless, pungent gas was evolved" whenever you heat some ammonium salt. If you have to, by all means, write it. Just don't place so much weight on it in your mind.
Safety-wise, if it was going to kill you, it wouldn't be done in a high school lab. You are not going to die if you sniff some ammonia in a high school lab, unless you continuously and voluntarily breathe it in. You'd also have to be either pretty stupid or masochistic to do so, since it stinks.
With that said, the smartest thing to do is to avoid smelling or tasting any of your chemistry experiments. Things that can be smelled or tasted without harm are the exceptions, not the norm.

Thank you. We do also use the litmus test for it. I knew it was not going to kill me, but I was mainly wondering if it was a violation of lab safety rules. It looked like a widespread practice, but it seemed kind of illogical to me.

"[I]f it was going to kill you, it wouldn't be done in a high school lab"? Never underestimate the ingenuity of high school students!


text:
Separating iron from blood
How can I separate iron from blood, without doing any damage to the blood/haemoglobin?
I have thought about using a magnet, but apparently iron in blood is not magnetic, and I have also looked at filtration techniques but not sure if that will affect the haemoglobin.
Updated Dec 2017: Let me add some context, I was/am looking for an alternative for chelation therapy such as Desferaloxamine infusions to remove iron in patients with iron overload as a result of repeated transfusions due to anemia

I vaguely recall some work several years ago from several universities using magnetic nanoparticles coated with something that would bind to the haemoglobin. Outside of the body, they had an apparatus that would inject the nanoparticles, and then magnetically separate them out before putting the blood back in the body. Once separated, the nanoparticles would be treated (can't remember how) to release the haemoglobin.

Once the iron is removed, it's no longer Heme b, but protoporphyrin IX.

According to this study http://www.dtic.mil/dtic/tr/fulltext/u2/a266503.pdf ,
Iron can somehow be released from hemoglobin. I didn't fully understand it as I'm just a student myself, but hopefully it can help you out :)

This seems like an impossible task.  The iron in blood is primarily in the hemoglobin itself, isn't it?  So at the very least you are going to have to remove the iron from its heme ligand in hemoglobin, from iron-sulfur clusters in various enzymes in red and white blood cells, and from ferritin proteins, etc.

There was a question from a casual chemistry textbook which is somewhat related. "A legend says that one fine young man gave his beloved an engagement ring made of iron. But that iron was not the trivial one: the young man obtained it out of his own blood. Calculate amount of blood required for a young man to produce $\pu{2.50 g}$ iron. The content of hemoglobin in the blood of an average young men is $\pu{150 g/dm3}$. The molar mass of human hemoglobin is $\pu{68800 g/mol}$. One hemoglobin molecule contains 4 iron atoms." Too bad there is no info how the young man did the tricky part:)

You can't get the iron out without destroying the haemoglobin and the ability of the blood to carry oxygen. So your premise is wrong. But if you are happy to get the iron anyway, then it can be done but will be messy.

@matt_black But what about after the haemoglobin has done it's job.

@naz do you mean after you stop breathing? Or do you mean how can you use out-of-date blood from bloodbanks?

@matt_black When the person is still alive.

@naz You do get that you need the iron in your blood to supply oxygen to your body every second of every day, don't you? So at which point has it "done its job"?

@matt_black Let me add some context, I was/am looking for an alternative for chelation therapy such as Desferaloxamine infusions to remove iron in patients with iron overload as a result of repeated transfusions due to anemia.

@naz Then you should rewrite the question to add that context as it turns what sounds like a naive and simplistic idea into a good and interesting question.

It is not possible to separate iron from blood without damaging the haemoglobin. Iron is a part of haemoglobin, and if iron is removed, the haemoglobin becomes something else. The blood then becomes incapable of transferring oxygen.

But what about after the heamoglobin has done it's job.

The iron is recycled when the haemoglobin is alternately reduced and oxidised.   It is never used up in normal service.  It may be that your friend has too much iron in their blood that is NOT in the haemoglobin or they have too much haemoglobin but the 4 iron atoms in the haemoglobin are there to stay. https://en.wikipedia.org/wiki/Hemoglobin

Using the reactivity series, just displace the iron with a more reactive metal.  The Haber process uses carbon to do this but as the heme molecule is mostly made of carbon with a central iron atom, this wouldn't do the trick.  Add heat, pressure and the more reactive metal and you'd get the iron.
http://www.bbc.co.uk/education/guides/zqjsgk7/revision

What you are proposing won’t work for a number of reasons.


text:
Is beryllium an alkaline earth metal?
For context: I got a question asking, "Which of the following alkaline earth metals do not give flame colour?". I quickly marked $\ce{Be}$ and $\ce{Mg}$ and got negative marks.
The following is a quote from my book (emphasis in the original):

The elements of Group 2 include beryllium, magnesium, calcium, strontium, barium and radium. These elements with the exception of beryllium are commonly known as the alkaline earth metals. These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth's crust.

As you can see, it's stated that beryllium is not an alkaline earth metal!
Is beryllium an alkaline earth metal or not?

Shriver & Atkins 6th ed: "The elements calcium, strontium, barium, and radium are known as the alkaline earth metals, but the term is often applied to the whole of Group 2." (p 336) It actually excludes Mg too. (And I am not saying that it is an authoritative source on chemical nomenclature.) Greenwood & Earnshaw simply refer to the whole of Group 2 as the alkaline earth metals.

Are you preparing for JEE ? And by the way , I feel yours answers is correct since Be and Mg do not give flame test.

It's not. The picture is from NCERT book @AkshayPratapSingh

Self-teaching worker has already answer the question. Regarding the negative marks you got Be is exceptionally small. And hence hold its electrons tight enough not to be excited by oxidizing flame.

Be is not an alkaline earth metal as its oxide or hydroxide is not 'alkaline'.( Its amphoteric like Al, (diagonal relationship))

There is some disagreement in usage among authors, but IUPAC standard nomenclature approves calling beryllium an alkaline earth metal, as explained on page 51 of IUPAC's last Red Book.
In fact, all the elements belonging to group 2, $\ce{Be,Mg,Ca,Sr,Ba,Ra}$, are called alkaline earth metals with IUPAC's approval.
Other common traditional names approved by IUPAC are alkali metals for the elements of group 1 except hydrogen, i.e. $\ce{Li,Na,K,Rb,Cs,Fr}$; halogens for $\ce{F,Cl,Br,I,At}$ in group 17,whose only member excluded from such a designation is $\ce{Uus}$ (now $\ce{Ts}$), and noble gases for all the elements of group 18 except $\ce{Uuo}$ (now $\ce{Og}$), i.e. $\ce{He,Ne,Ar,Kr,Xe,Rn}$.

If we're going to be fussy about alkaline earth metals being "alkaline earth" in the proper sense, then there are only two of them:  calcium and magnesium.  Beryllium oxide fails to be alkaline and the heavier Group 2 oxides (or, the hydroxides formed upon reaction with water) are too readily soluble to really be "earths".
Compared with the other Group 2 metals, calcium and magnesium are so ubiquitous that they're basically (no pun intended) the namesake for the whole group.  Thus as IUPAC asserts the phrase "alkaline earth metals" has spread to all of Group 2.

Alkaline metals are called so because of two reasons:

They are obtained from the earth in the form of ores
Their oxides and hydroxides show basic (alkaline) natures.

Beryllium fulfills the first criterion but not the second one, since its oxides and hydroxides show amphoteric behavior, rather than alkaline.  So, it is not considered an alkaline earth metal in the true sense, even though it does resemble them in various other properties.
So, if asked which group 2 metal is not considered an alkaline earth metal, then your answer should be beryllium. Otherwise, you should probably consider beryllium as an alkaline earth metal.

Praveen, I made some major edits to your answer for readability. If you don't like anything I did, or if I misrepresented what you were trying to say, please feel free to edit it further. Thanks for posting here!


text:
Predicting changes in pH of water when chlorinating using two different methods
Two methods for chlorinating water are using chlorine gas $(\ce{Cl2})$ or calcium hypochlorite $(\ce{Ca(OCl)2}).$ By using chemical equations how can you predict the pH of the water brought about by the chlorinating process?

In aqueous solution, chlorine can undergo disproportionation: one atom in $\ce{Cl2}$ (oxidation state $0$) is reduced to $\ce{Cl-}$ (oxidation state $-1),$ the other is oxidized to a species in which the chlorine atom has the oxidation state $+1.$

$$\ce{Cl2 + 3H2O -> 2H3O+ + Cl- + OCl-}$$

Ask yourself how that influences the pH of the solution.

What about adding calcium hypochlorite to water? If $\ce{CaCl2}$ is the calcium salt of hydrochloric acid, what about $\ce{Ca(OCl)2}?$

 You may have guessed it: $\ce{Ca(OCl)2}$ is the calcium salt of chloric(I) acid.

You may remember that adding the soluble salt $\ce{MA}$ of a strong acid $\ce{HA}$ to water will not influence the pH of the solution.

 If $\ce{HA}$ is a strong acid, it will fully dissociate in water:
 $$\ce{HA + H2O <=>> A- + H3O+}$$
 This means in return that the corresponding base $\ce{A-}$ is very weak:
 $$\ce{A- + H2O <<=> HA +OH-}$$

Now, is $\ce{OCl-}$ a strong or a weak base? In order to decide, you might want to have a look up the $\mathrm{p}K_\mathrm{a}$ of the corresponding acid and keep in mind that for a corresponding acid-base pair in water $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14.$


text:
Why is the melting point of propane lower than that of both methane and ethane?
I've seen their properties in my textbook and multiple online sources, but I can't seem to find a straight answer as to why. Excluding propane, the trend is increased melting point as the molecule gets bigger, but propane is the weird exception here.

Ok, let's first look at the trend...
$$
\begin{array}{c|c|c|c}\hline
\text{Chemical} & n \text{ in }\ce{C_nH_{2n+2}} & \text{m.p. / K} & \text{m.p.}(n) - \text{m.p.}(n-1)\text{ / K} \\
\hline
\text{Methane} & 1 & 90.7 & \\
\text{Ethane} & 2 & 90.4 & -0.3 \\
\text{Propane} & 3 & 85.5 & -4.9 \\
\text{Butane} & 4 & 133\mathrm{-}139 & +50.5 \\
\text{Pentane} & 5 & 144.1 & +8.0 \\
\text{Hexane} & 6 & 177\mathrm{-}179 & +36.9 \\
\text{Heptane} & 7 & 182.2\mathrm{-}183.0 & +4.6 \\
\text{Octane} & 8 & 216.0\mathrm{-}216.6 & +33.7 \\
\text{Nonane} & 9 & 219.0\mathrm{-}220.0 & +3.2 \\
\text{Decane} & 10 & 242.7\mathrm{-}243.9 & +23.8 \\
\hline
\end{array}
$$
Using the mean for the alkanes with melting point ranges given (e.g. 136 for n-butane and 178 for n-hexane), it isn't clear there is any "great" linear trend. In fact it seems that the even number n-alkanes are on one curve and the odd n-alkanes are on another.
There is a "general trend in that with increasing MW the MP increases. However the exceptions would seem to be both methane and ethane, not propane. Also it is obvious that a simple correlation with MW alone is not sufficient to characterize the trends. All in all I think one must be careful not to read too much into such trends without a very careful analysis.

In general even number of $\ce{C}$ atoms in a chain are much more accommodating to a lattice than odd number of carbons. This is why the increase in boiling point from odd $\implies$ even is much more higher than even $\implies$ odd.

The phase transition towards solid state involves crystalization, i.e. organization of the molecules into the long-range repetitive patterns.
The formation of crystal lattice is reasonably simple for balls (methane) and rigid sticks (ethane), but becomes problematic for flexible triangle (propane), which needs to be properly aligned. Propane is the first of alkanes, which have important internal degrees of freedom, which have to adjust during the solidification.


text:
How estimate E0-0 of an organic dye?
How can I estimate the $E_{0-0}$ value of an organic dye? I have obtained the UV-vis absorption and fluorescence spectra of this dye in solution. This article talks about (in Table 2 notes) the intercept between the normalized absorption and emission spectra, but I don't quite understand what they mean.
Does someone here know? Attached is the (non-normalized) spectra.

You're looking for the point where the normalized absorption and emission spectra cross each other.
The UV/Vis absorption spectra is almost always higher in energy (shorter wavelengths) than the emission spectra, because the molecule goes from the electronic and vibrational ground state to the electronic excited state. Usually this also means some vibrational modes are also excited. Since the molecule is typically in either the vibrational ground state or a low-energy vibrational state, the absorption process can go up several vibrational levels (i.e, the energy absorbed is higher than just the basic electronic excitation).
For emission, the molecule often loses some vibrational energy, so the emission comes from the ground vibrational state of that electronic excited state. The molecule ends up in the ground electronic state, but a vibrationally excited state (i.e., it loses a bit less energy than the $E_{0-0}$ energy difference).
So if you normalize the absorption and emission spectra, the crossing point should roughly correspond to the $E_{0-0}$ difference.

When I have a chance, I'll add some graphics to make these points clearer.

Area normalized to 1, or max intensity normalized to 1?

I would normalize the peak intensity.

An alternative and probably more accurate way to estimate the $E_{0{-}0}$ for a molecule in solution is to take half the difference between absorption and fluorescence peaks. To do this first convert these wavelengths to wavenumbers, then add half the difference to the fluorescence peak (or subtract from absorption) and convert back to wavelengths.
(As an experimental observation you should also try, if possible, to get better quality absorption and emission spectra as negative values are meaningless and this suggest that your background subtraction is not quite correct).
The figure below shows the relationship between absorption and fluorescence spectra when the absorption is the mirror image of the fluorescence. (Note that the spectra on the right of the figure are not on the same scale with the energy levels on the left.) A mirror image does not quite occur with the spectra in your question, but the principle is the same. The reason that the fluorescence comes from the lowest vibrational level in the excited state is that the rate of vibrational relaxation (thin wavy line), caused by energy transfer to surroundings (usually solvent), is almost always far faster than emission. If the potential energy surfaces are not displaced horizontally, indicating different bond lengths in ground and excited state, then by the Franck-Condon Principle almost no transitions will be observed.

Taking the crossing point between normalized and reduced absorption and emission spectra will give you the $E_{0-0}$.
Take care that reduced here means that the absorption spectrum $A(E)$ needs to be divided by the photon energy $E$, and the emission spectrum $N(E)$ needs to be divided by $E^3$. By normalizing the reduced spectra to the maximum of the corresponding peak, the two spectra intersect exaclty at $E=E_{0-0}$.
In the case of mirror-image spectra, $E_{0-0}$ is the midpoint between the absorption and emission maxima, separated by 2$λ$.
You can find more details at:
Vandewal et al., How to determine optical gaps and voltage losses in organic photovoltaic materials. Sustainable Energy & Fuels (2018)


text:
No. of photoelectrons vs frequency and intensity
We know that when the intensity of incident light increases, number of photoelectrons ejected increases irrespective of the frequency (assuming it to be greater than threshold frequency. But does it depend on frequency too in some situations like in case of high frequency or something?

I think it might happen in the case of semi-conductors. If the frequency is quite high, it might result in breakage of bonds, which will cause ejection of more electrons than predicted. I am suggesting this because Si-Si bonds are quite weak (coming in at 52 kcal/mol) and light with wavelength in the region 200 Angstrom is enough to break it.

The number of photoelectrons per incident photon doesn't really vary with photon energy (frequency). Particularly a photon won't eject multiple "photoelectrons." By definition a photoelectron is a loosely bound electron (i.e. not in an atomic or molecular orbital) which is ejected in a one-to-one basis by a photon and thus satisfying the equation $$ \rm{KE}_{electron} = \rm{h}\nu - \phi$$ where:

$ \rm{KE}_{electron}$ is the kinetic energy of the electron
$\rm{h}\nu$ is the energy of the incident photon
$\phi$ is the work function for the substrate (how much energy it takes to remove electron from substrate).

In general the substrate would be a metal and the electron would be in the metallic (electrically conducting) layer.

If the electron is bound in an atomic or molecular orbital then the
technique is Electron Spectroscopy for Chemical Analysis (ESCA).
If the photon is too energetic then it penetrates too deeply to eject an photoelectron.

The gist of the above conditions is that photoelectrons are created with relatively low energy photon just above the needed threshold given by the work function, $\phi$.
Another aspect of this is that a photon can only interact with one electron at a time. However if the photon is greater than $2\phi$ in energy it is possible that the electron can receive enough energy to knock yet another electron out of the metal. These two electrons wouldn't be "photoelectrons" since they don't fit the equation. They would be noise in the experiment and the experiment would need to be designed to minimize counting such electrons. So the ejection of multiple electrons by a photon is possible, but not multiple photoelectrons by definition.


text:
pH of rain with and without thunderstorm
I just came to know that pH of rain water will be lesser when it is accompanied by a thunderstorm than when it is not accompanied by thunderstorm.
Why would it be so?
Will the temperature increase due to excessive friction (of air) and lighting and thus more ions of water will dissociate reducing the pH?
Or will it be because of production of acids in air during thunderstorm?
Help.

Where'd you see this? Do you have a reference or two you can share?

@ToddMinehardt It was a question asked in a national level examination. I don't have much of a reference though.

https://books.google.co.in/books?id=bYb7t00SqkEC&pg=SA5-PA83&lpg=SA5-PA83&dq=ph+of+rain+in+a+thunderstorm&source=bl&ots=jHWUfEYbe-&sig=LK5QW1hghOSQdKngcOSzpuyMnJ4&hl=en&sa=X&ved=0ahUKEwiB_9SInfvKAhVRTo4KHWD9DoM4ChDoAQgwMAU#v=onepage&q=ph%20of%20rain%20in%20a%20thunderstorm&f=false
Here is the question

Though I've never heard of that statement before, it's possible that nitrogen oxides formed by electrostatic discharge (not only lightning, but even glow discharge) dissolve in the rainwater, producing nitrous and nitric acids.
There is a 1997 citation describing comparison of pH in rain accompanied (or not) by lightning.
That said, presence of fly-ash from coal burning, sulfur oxides from volcanic activity and carbon dioxide from natural as well as anthropogenic sources probably have greater influence on pH.

When lightning and thundering occurs, it oxidizes into $\ce{SO2}$ and $\ce{NO_x}$ gases from pollution. They mix with rain water and reduces pH of water.
Ref.: L.Bruce Railsback, Lower pH of acid rain associated with lightning: evidence from sampling within 14 showers and storms in the Georgia Piedmont in summer 1996, Science of The Total Environment,
Volume 198, Issue 3, 1997, DOI: 10.1016/S0048-9697(97)05459-4.

Traces of nitric acid are the natural component of the rain, originated by the reaction $$\ce{N2(g) + O2(g) -> 2 NO(g)}.$$
The reaction occurs at electric discharges, naturally like during lightnings and silent discharges, or industrially in electric arc ( in obsolete, energy demanding Birkeland-Eyde process, replaced later by catalytic oxidation of ammonia. )
It goes farther as:
\begin{align}
\ce{2 NO(g) + O2(g) &-> 2 NO2(g)}\\
\ce{4 NO2(g) + O2(g) + 2 H2O(l) &-> 4 HNO3(aq)}\\
\ce{4 NO2(g) + 2 H2O(l) &-> 2 HNO3(aq) + 2 HNO2(aq)}\\
\ce{2 HNO2(aq) &-> NO2(g) + NO(g) +  H2O(l)}
\end{align}
Therefore, in the end, pollution aside, the rain during active thunderstorms is slightly more acidic because of $\ce{HNO3}$, compared to the natural baseline acidity caused by dissolved $\ce{CO2}$.
Unfortunately, the rain $\mathrm{pH}$ depends these days much more on $\ce{SOx}$ and $\ce{NOx}$ pollution than on natural processes, leading to industrially induced very weak solutions of $\ce{H2SO4}$ and $\ce{HNO3}$.

I'm old...........

It can be advantage, you have a lot of classical knowledge. Natural HNO3 based rain acidity is just a little addition to natural CO2 based rain acidity, which both are just little addition to pollution based acidity.


text:
Peroxide in acidic conditions being reacted with iron sulphate mechanism

For the proposed mechanism, I would like to think the $\ce{O-O}$ bond would be cleaved first, thus creating two radicals (an $\ce{OH}$ radical and the radical situated on the other Oxygen), then radical on the oxygen will go into the ring to form a double bond in which the alkene double bond will donate 1 electron towards that bond and 1 electron to the carbon on the end of the opened ring.
My mechanism would be incomplete as I cannot seem to figure out how this reaction occurs.
However, I have another suggestion in which there is an epoxide formation. Can anyone elaborate as to which one is correct or can give me some insight?

Looks like catalytic O-O cleavage, so rather O+ then radical is created.

After the formation of the oxygen radical, it iniciates a C-C bond cleavage, as two possibilities exist, two products are obtained after subsequent addition of the formed C radical to the double bond.

At least for the second part of the product formation, you have pretty much given the answer yourself.
The methylcyclopentane moiety of the products results from two different 5-exo-trig radical ring closures, that is from the intramolecular addition of two different radicals to $\ce{R-\color{\red}{C}H=CH2}$.
How are these two different radicals generated?
Have a look at the starting materials! Isn't the combination of the hydroperoxide $\ce{R-OOH}$ and $\ce{FeSO4}$ very similar to the conditions of a Fenton reaction?
Let's therefore postulate the following: $\ce{R-OOH + Fe^2+ -> R-O* +\quad{}OH-  + Fe^3+}$
The resulting 1-methylcylohexan-1-ol derived radical can form methyl ketones by cleaving two different $\ce{C-C}$ bonds; the resulting carbon-centered radicals then give rise to the different cyclization products.


text:
Will a copper anode and hydride ion cell work?
Can a galvanic cell with the overall reaction
$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$ work?
$\Delta G$ of this reaction is negative (because $E^{\circ}$ of the reaction is $\mathrm{0.34~V}$ as the standard electrode potential of copper will be $\mathrm{-0.34~V}$ as it is acting as an anode). This shows the cell is spontaneous;
However, I have read that $\ce{Cu(s)}$ cannot reduce $\ce{H+}$ so I was wondering if this cell would work?

TL;DR - What you know from the activity series of metals is still true. You made a mistake in your interpretation of standard electrode potential. The rationale follows:

Standard electrode potentials are written in the reduction direction:
$$\ce{M}^{x+} + x\ce{e- -> M(s)}$$
Standard electrode potentials are measured against the standard hydrogen electrode:
$$\ce{2H+ + 2e- -> H2(g)}$$
Thus for the $\ce{Cu(s)|Cu^{2+}}$ electrode, $E^\circ$ is defined for the followed redox reaction (everything at 1 molar, 1 bar, and in standard states):
$$\ce{Cu^2+ + H2 -> Cu(s) + 2H+}\ \ \ E^\circ = +0.34$$
For the oxidation of copper with acid, the electrode potential would be negative in the direction you wrote it, which translates to a positive $\Delta G^\circ$, since:
$$\Delta G^\circ = - nFE^\circ$$

wouldn't ∆G be negative since copper usually gets reduced and its E˚ is 0.34. However, when you use it as a anode, the E˚ will change signs, so E˚ anode would become -0.34. When you go to solve E˚ for the cell, it would be  E˚cathode - E˚anode. The E˚ for cathode is 0 and E anode is -0.34. Thus, E˚ for the overall cell would be +0.34V (o -- -0.34). Am I doing something wrong in this?

I see your confusion. The equation for $E^\circ_{cell}$ is: $$E^\circ_{cell}=E^\circ_{red,cathode}-E^\circ_{red,anode}$$ The negative sign in the equation accounts for the use of $E^\circ$ for both electrodes in reduction direction.

So if the anode where aluminum, would the cell work? Aluminums standard electrode potential is -1.66V

Yes. Aluminum would work.

Just curious... why isn't the negative sign accounted for if you are using, in this case, aluminum as a anode rather than a cathode (which will have negative electrode potential of -1.66V)


text:
Dinitrophenyl (DNP) derivatives of amino acids (specifically epsilon-DNP-lysine)
I have a pretty basic biochemistry question but am having trouble finding the answer to it:
Normally, DNFB (dinitrofluorobenzene, or 1-fluoro-2,4-dinitrobenzene to be exact) reacts with just the amino terminus of amino acids, creating dinitrophenyl (DNP) derivatives.  However, with lysine, DNFB can react in two places (at the amino terminus and to the epsilon amino group).  The DNP derivative of lysine is referred to as "epsilon-DNP-lysine".
Does "epsilon-DNP-lysine" refer to a product where the DNP is bound only to the epsilon amino group (but not to the amino terminus), or does it refer to a product where DNP is bound simultaneously to the epsilon amino group and to the amino terminus?
Also, can the same thing happen with other amino acids that have an amino group in the side chain?  i.e. can there be epsilon-DNP-arginine, for instance, or is this behavior unique to lysine?

Yes! my mistake.

Does "epsilon-DNP-lysine" refer to a product where the DNP is bound only to the epsilon amino group (but not to the amino terminus), or does it refer to a product where DNP is bound simultaneously to the epsilon amino group and to the amino terminus?

Good question.  ε-DNP-lysine usually refers to a lysine in which the dinitrophenyl substituent is bound only to the ε atom, i.e., to the side-chain nitrogen.
The reason is that in lysine-focused derivatization methods, the reaction with DNP can be done prior to protein hydrolysis.  Thus, during DNP derivatization, the vast majority of lysine residues will have no free α-amino terminus because they are part of polypeptide chains.  Only the ε-amino groups are free to react with DNFB.  Derivatization forms DNP-ylated proteins where the peptide bonds are intact.  Later steps in the procedure hydrolyze all the peptide bonds.  During this step, free α-amino groups are created.  But no DNFB is used in these later steps.
I found a paper from 1963 which describes the derivatization procedure in detail.

...Care was taken to
  deposit the sample at the bottom of
  the flask. Ten milliliters of bicarbonate
  solution were added, care being taken
  that no meal adhered to the side of the
  flask. The contents were then thoroughly
  mixed by gentle swirling, and the
  suspension was permitted to stand for 10
  minutes. A solution of 0.3 ml. of
  DNFB in 10 ml. of ethanol was then
  added and the contents of the flask
  were thoroughly mixed by gentle
  swirling. The side of the flask was
  rinsed with 3 ml. of absolute ethanol
  and then the contents were shaken in
  subdued light for two hours on a shaker
  with a wrist-like motion. Previous
  work (10) has established that 2 hours
  reaction time is sufficient for substantially
  complete reaction between DNFB
  arid the free epsilon amino groups of
  lysine. alcohol and most of the
  water were removed by evaporation
  under an air stream, and the residue was extracted with four 50-ml. portions
  of anhydrous, peroxide-free diethyl
  ether. The ether in each case was
  removed from the residue by decantation
  and the residue in the flask was dried at ambient temperature by
  aeration.

The ether washings will remove the vast majority of unreacted DNFB.

... Two hundred milliliters of
  constant-boiling aqueous HCl nere
  added to the flask, and the resulting
  mixture was boiled overnight at the
  reflux temperature, cooled and then filtered through a sintered glass funnel
  directly into a 250-ml. volumetric flask.
  The filtrate, and washings from the
  filtrate, were made up to 250 mi. with
  distilled water.
  An aliquot (2.0 to 4.0 ml.) of the
  hydrolyzate containing from 0.1 to 0.3
  mg. of epsilon-dinitrophenyllysine...

Now for your other question:

Also, can the same thing happen with other amino acids that have an amino group in the side chain? i.e. can there be epsilon-DNP-arginine, for instance, or is this behavior unique to lysine?

DNFB is not reactive with arginine.  The arginine side-chain is best viewed as a guanidine, not as an amino group.  Guanidine is considerably less nucleophilic than lysine.

And likewise for the amides present in glutamine and asparagine, even though they may formally be $\ce{NH2}$-including.


text:
How to determine concentration of salt in a salt bridge
How do you determine as to what concentration of solution you need for a salt bridge in a galvanic cell? Do you even need a specific concentration in the first place?

I'm not sure whether there is a downside to using a saturated salt solution to in the bridge, but the higher the salt concentration, the greater the bridge's conductivity and the greater the usable potential of the cell.

Ok so how concentrated your electrolytes are does not tell you how concentrated the salt bridge should be?


text:
Lead(II) oxide CRT monitor extraction
I have a large CRT monitor that, by the weight of the monitor, I estimate to contain over two pounds of lead.  I want to know if the lead oxide is a mixture with the glass, or if it is chemically bonded to the glass.
I also want to know if there is any way (preferably extraction using heat) I can remove the lead in pure form or in oxide form. I know for sure you'd crush the glass.
I bring up heat extraction because chemicals are expensive, and I've heard of it being done that way on a large scale.
Edit: I'm not concerned about making a profit because I don't really have the setup to do so. I'm just curious about it because I've seen it done on an industrial scale using heat, so could it be replicated on home scale.
In what form would the lead be if you smelted it out?  Does heating lead oxide reduce it to lead again? If so, would the lead melt out and flow to the bottom?

The lead in the face glass of the CRT would be a lead oxide glass. Smelting lead is nasty. There would be some lead oxide given off as dust which would contaminate the surroundings. Unless you already have a lead smelting setup, you could buy lead cheaper than you could recover it.

You really do not want to try to extract the lead out of the glass. It will be a long, dangerous process that at best will give you and yours a big dose of lead fumes. Recycle it.

The lead glass used in CRTs is not a mixture (at least not in any sense where a physical process could readily separate it): it is a glass that contains a high proportion of lead oxide in its structure. It is about as easy to extract from glass a aluminium would be from a high alumina glass (which is not easy at all).

A simple Google search will find you a number of informative articles.
I went and looked at one of the links and it was an abstract which I will quote below in full.  It looks like a possible method though I strongly advise against proceeding without a lot of forethought into pollution control measures.  There will be various lead compounds formed and keeping the dust and products out of the air and environment will not be easy.  As a business case this is going to be irrelevant soon as the number of CRT displays being recycled is decreasing continuously as 2nd generation LCD monitors have already reached end of life, only a small percentage of industrial control equipment still operates on CRT monitors and they will slowly be recycled over time, the big gold rush is over.
Here is a video from SWEEEP KUUSAKOSKI showing their industrial processing system, the method is not described but this started operation in 2013. The same time as the paper below was published.
Another video by Green Lyon describes a patented process to extract the lead from glass using an aqueous phase process at moderate temperatures with non-scary reagents.  Not sure if this has been reproduced elsewhere, this video was also published in 2013, a good year for CRT glass.
Extraction of Metallic Lead from Cathode Ray Tube (CRT) Funnel Glass by Thermal Reduction with Metallic Iron
Xingwen Lu†, Kaimin Shih*†, Chengshuai Liu‡, and Fei Wang†
Environ. Sci. Technol., 2013, 47 (17), pp 9972–9978
DOI: 10.1021/es401674d
Publication Date (Web): August 5, 2013
"A novel and effective process of thermal reduction treatment with the addition of metallic iron (Fe(0)) to recover lead from cathode ray tube (CRT) funnel glass is introduced. The key technological breakthrough of this process is the use of a relatively lower temperature and an inexpensive reducing agent to extract the metallic lead. The influences of temperature, the reducing agent content, and the holding time for lead reduction were examined to determine the optimal extraction efficiency. The lead extraction efficiency first increased and then decreased with increasing temperature. The maximum lead extraction efficiency occurred at 700 °C. The growth of crystalline lead first increased significantly with an increase in the Fe content, reaching maximum growth at an Fe addition of 50 wt %. The most effective treatment time was determined to be 30 min, as the vitrification of lead back to the glass matrix occurred under longer treatment times. The experimentally derived results indicate that a 58 wt % lead extraction can be achieved with the optimized operational parameters (50 wt % Fe addition, heating at 700 °C for 30 min) in a single extraction operation."


text:
Reaction of alkyne with sodium in ethanol
What is the product obtained on the reaction of but-2-yne with $\ce{Na}$ in ethanol?
I know that $\ce{Na}$ in liquid $\ce{NH3}$ converts an alkyne to an alkene by anti addition, and $\ce{Na}$ in ethanol is the principle reagent in Bouveault Blanc Reduction, but what about this reaction.
The answer given in my book is a trans-alkene. Is this a standard reaction? Also does it depend if it is a terminal alkyne or not? (I guess yes, because the terminal alkyne will try to deprotonate)

Do you correctly decipher the question? The reduction by Na in liquid ammonia is followed by addition of ethanol as a source of protons.
If you insist on your reaction process, addition of sodium metal into but-2-yne in ethanol will lead to formation of sodium ethanoate, which can add on the triple bond, leading to trans alkene. For further references see Houben-Weyl, https://books.google.de/books?id=P0uGAwAAQBAJ&lpg=PA1535&ots=ypiCXy5xNX&dq=sodium%20ethanoate%20alkyne&hl=cs&pg=PA1106#v=onepage&q=sodium%20ethanoate%20alkyne&f=false

the reduction of alkynes using Na, proceeds through the anion radical, which can obtain trans conformation for the stability. Now considering the usage of NH3 and ethanol are proton donors. So, there is no effect on trans conformation altering the proton donors.

Hey, welcome to Chem.SE! Can you please cite some references to back your answer up?

That's actually an example of Birch reduction of internal alkynes.
It's given in detail on this website.
Also, you're completely right in your reasoning that a terminal alkyne will not show this reaction.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.

Also, the citation on that article seems dubious. While the mechanism looks plausible, both the papers by Zimmerman cited therein make no mention of alkynes at all!


text:
Are we wrong about Bakelite formation mechanism?
In the reaction of phenol with acetyl chloride the oxygen of the phenol attacks the acetyl chloride. In Bakelite formation we know that pi electrons attack the carbonyl carbon of formaldehyde.My question is shouldn't the oxygen atom of phenol attack the formalaldehyde just as in the reaction of phenol with acetyl chloride?

I'm having a little trouble understanding what you're trying to ask. Could you try separating this into two sentences? I'd do it, but I don't want to misinterpret.

C-acylation : acylating agent (acyl chloride or anhydride) and $\ce{AlCl3}$
O-acylation : acylating agent (acyl chloride or anhydride).This is the one which you have talked about in the question.

Since the formation of bakelite is by Lederer Manasse reaction which proceeds through electrophilic aromatic substitution on phenol, its path is similar to that in C- acylation and not O-acylation.
Hence, the polymerisation of phenol and formaldehyde takes place either from two places:

For para substitution end:

For ortho substitution end:

Please cite the sources for your pictures, and preferably text.

https://it.wikipedia.org/wiki/Reazione_di_Lederer-Manasse I have learnt about reactions that  proceed through EAS (hence the above reactions: Lederer  Manasse and Friedel Craft Acylation) hence the text is my original passed down by my professor.


text:
Solvent for Norland Optical Adhesive
I'd like to know if there is any solvent that dissolves Norland Optical Adhesive 65 (and Rhodamine 6G), but does not dissolve PMMA that is used in e-beam lithography.
I've tried acetone, methanol, chloroform, n-hexane and toluene; none worked.

Can you provide the solvents that do dissolve your adhesive and the PMMA together?

In Acetone,Methanol,Chloroform, and Toluene both are soluble. Whereas in n-Hexane both are insoluble

Is the adhesive some sort of thiocarbamate do you know? Have you tried ester solvents?

I don't know. I didn't try esters because, both PMMA and NOA65 are soluble even in toluene of polarity 2.4. I need something that has polarity greater than 0.2 (as that of cyclohexane) and less than 2.4 (that of toluene)

Are you using Hansen solubility parameters?


text:
How would the relative rate of reduction of carboxylic acid be if the catalyst used is LiAlH3D?
How would the relative rate of reduction of carboxylic acid be if the catalyst used is LiAlH3D , LiAlH2D2 , LiAlHD3 , LiAlD4 just thought about it while revising the reaction mechanism of carboxylic acid reduction .


text:
Nomenclature alkyl halides
I was solving questions on the MSU website on organic nomenclature.
Out of the following questions, I got only the first and last one right.
Where am I going wrong with the others?Hints would suffice.
EDIT:I got the second one correct as well.
(2)----7,7-Dichloro-4-ethyl-3,3,8-trimethyl-4-propyldecane

Indeed, most of the names proposed by you correspond to the preferred IUPAC names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):
(1) 3-bromo-2,5-dimethylhexane
(2) 7,7-dichloro-4-ethyl-3,3,8-trimethyl-4-propyldecane (your edit is correct)
(3) see below
(4) 2-chloropent-1-en-4-yne
(5) 1-bromo-3-ethyl-5-methylbenzene
For compound (3), you have to consider the relevant rules for the numbering of locants for substituent prefixes:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), compound (3) could be named as 5-bromo-1-methylcyclohexa-1,4-diene (which was your proposal) as well as 1-bromo-5-methylcyclohexa-1,4-diene since both names correspond to the locant set ‘1,5’. However, according to Rule (g), this example is named as 1-bromo-5-methylcyclohexa-1,4-diene rather than 5-bromo-1-methylcyclohexa-1,4-diene since bromo is cited first as a prefix in the name.


text:
What would happen if the reaction of reduction of carboxylic acids by LiAlH4 is carried out in alcohol?
Which is the intermediate compound in the reduction of carboxylic acids by $\ce{LiAlH4}$ and what would happen if the reduction of carboxylic acids is carried out in alcohol?
From my knowledge I know that a geminal diol is formed and then elimination takes place to form $\ce{R-CH=O}$. If this is reacted with alcohol instead of $\ce{LiAlH4}$ again what are the expected products?

If you carry it out in alcohol, the LiAlH4 will decompose, perhaps more violently than you might like.

@ orthocresol would you please give some kind of the mechanism for the decomposition of LAH in alcohol bcoz the answer below is not that satisfying

What? There's no mechanism for it. It's just an acid-base reaction.

@mhchem Your comment %edit uncommented the } and broke the construct. You can always edit out salutations, buzzwords, and unnecessary "statements of weakness" instead, see the edit history for what I mean.

@Martin-マーチン Oops, the %edit should have gone behind the }.

Alcohols are not an ideal solvent for LiAlH4, but the main issue is that there is no carboxylic center to where the hydride transfer from LiAlH4 can take place.

So in this case, the side-reaction between alcohol and LiAlH4 can be regarded to be minimal and the major product will be the alcohol produced from the carboxylic acid.
P.S. : A diagrammatic representation of the reactions for LiAlH4

The side reaction between alcohol and LAH is certainly not minimal. It can be violent and cause your reaction mixture to catch fire.


text:
Why does the conductivity of a conductivity probe decease if you shake the solution in which it is kept?
Figure showing $3M$ solution $\ce{NaOH}$ conductivity while shaking (The circled area is the time of shaking and drop)

When I took the reading again, without shaking, I got the following graph:


text:
Preparing a solution of hydroxylamine hydrochloride in HCl, struggling
I recently came across a paper with a methodology description i cant seem to get the grasp of.
I need to make a particular solution for Fe extraction. The protocol's description is from this paper.

Take a 100 μL aliquot of X (a soil sample previously mixed with 0.5M HCl) and mix it with 2 mL of a solution of 0.25M hydroxylamine hydrochloride in 0.25M HCl and incubate at 60 °C.

What I don't understand is how to make the solution when the volume of the two compounds is not specified? Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl? Any help is highly appreciated. Thanks a lot in advance!

Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl?

No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.
I think the purpose of the hydrochloric acid is to ensure that the pH stays acid.  One way you could make this solution is by:

Decide on a volume of solution you want to use, say $X$ mL.
Prepare at least $X$ mL of 0.25M hydrochloric acid by dilution from a more concentrated stock solution.
Weigh out the amount of $\ce{NH2OH.HCl}$ that you need.  It has a formula weight of 69.49 g/mol, so for $X$ mL of a 0.25 M solution you will need:

$$0.25 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{69.49\mathrm{g}}{\mathrm{mol}} \times \frac{X\;\mathrm{mL}}{1} \times \frac{\mathrm{L}}{\mathrm{1000mL}}=\frac{69.49\;X}{4000}\mathrm{g\;of\;\ce{NH2OH.HCl}}$$

Dissolve the weighed out powder with your 0.25M $\ce{HCl}$ solution in a volumetric flask, first fully dissolving the powder and then slowly adding more HCl until the target volume is reached.

The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.


text:
Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured?
Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured?
I understand that temperature can cause either the forward or reverse reaction to be favoured. But then doesn't that mean that the rate of which the reactants or products are being formed differ from each other? So, what does it mean when they define equilibrium as I have said above.
Please keep answers simple and clear.

The rate must be equal at equilibrium, the products are favoured means that the concentration of products is much greater than the concentrations of the reactants at equilibrium

A chemical reaction can be reversible, and a reversible reaction can be at equilibrium, but they are not the same thing.
Equilibrium is the state where the forward and reverse reactions are the same rate.  This is clearly not true at the start of a reaction because there are no products yet.  As the reaction progresses the forward reaction rate decreases and the reverse reaction rate increases until they are equal.  At that point the concentrations of reactants and products no longer change.
The position of the equilibrium can be far to the left, so that hardly any product is formed before it reaches equilibrium.  This would be because the forward reaction is much slower than the reverse reaction.  If the position of the equilibrium is far to the right, almost all of the reactants have been used up because the forward reaction is much faster than the reverse reaction.
If you do something to change the conditions, like increase the temperature or add more reactants, you break the equilibrium and the reactions are no longer at the same rate.  As above the reaction will progress until the rates are the same again.  Depending on the change, this new equilibrium position may be more to the left or let or more to the right.
It is tempting to think that if you increase temperature, the rate of both reactions will increase the same.  This is not the case.  Although both reactions will increase in rate, because of the usual collision theory, but the same temperature change can have a much bigger effect on the rate of one reaction than the other.


text:
A good textbook to balance chemical equations?
Is there any good textbook which after reading it I would be able to balance almost any chemical equation in inorganic chemistry?

here is a good site to search for any reaction and get it balanced.

Possible duplicate of Simplest way to balance any chemical equation

This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically.
Consider the reaction $$\ce{C4H10 + O2-> CO2 + H2O}$$
To start, you would put a variable in front of each molecule here.
That would become:  $$\ce{aC4H10 + bO2 -> cCO2 + dH2O}$$
Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c
Repeating this process for the other elements, I get:
10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side).
After this step, you have your three equations.
4a=c
5a=d (This is simplified from 10a=2d)
2b=2c+d
Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1.
From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2.
Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields:
a=2 b=13 c=8 and d=10
Now put those values exactly where you originally put the letters.
It becomes $$\ce{ (2)C4H10 + (13)O2  ->  (8)CO2 + (10)H2O}$$
So the steps are:
Assign variables to each molecule in the reaction.
Write out those variables as equations corresponding to the amount of certain elements in each molecule.
Make one variable equal 1.
Solve for the rest of the variables.
Multiply by the denominators of any fractions to leave whole numbers.
Those numbers are your answers. You just plug them in :)
If you want any clarification, or another example done, just comment and I'll be happy to.
Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome.


text:
Photosynthesis: Chlorophyll and color.
As I know, chlorophyll is green as it absorbs all wavelengths but green. It absorbs red, yellow, blue, purple, etc. Thus, the plant does not utilize green light.
i] Why is chlorophyll green? Is there 'chlorophyll' of other colors?
ii] Is the green chlorophyll an 'evolutionary mistake'?
iii] If chlorophyll was black, what effects would there be?
iv] What is the effect on non-visible electromagnetic-radiation?
Thanks!

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Really, this is a homework question @bon ? How so?

Any question which is presented in the manner of 'here is a list of questions, give me the answers' is a homework question. You need to show that you have though about the questions beforehand and explain why you have difficulty answering them.

It's rather too broad IMO

i] There is chlorophyll-b (yellow-green) and carotenoids (yellowish).
ii] The green chlorophyll isn't an evolutionary mistake (unless, of course, you take the viewpoint that evolution IS just mistakes). Plant's get plenty of sunlight. The sun is way stronger than indoor lights used to grow plants. They don't really need more energy from the sun. Plants actually have some pigments to dissipate light because they would otherwise absorb too much.
iii] Plants might absorb too much energy. Think wearing all black on a hot summer day, and standing out in the sun ALL day.
iv] Plants don't really use ultraviolet light as far as I know. With the ozone layer blocking most of the ultraviolet rays coming at us, it would be very strange if they did evolve to absorb ultraviolet light. This might also harm the plant. Infrared rays do reach plants, but they don't really do much to help in photosynthesis. Anyways, light has to travel through the cytoplasm of plants which is full of water, and visible light is best at travelling through water.

For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service.

This was a question I actually asked myself while I was studying biology, which is why I don't feel it is a homework question @bon

red algae are tuned for utilization of bluer spectrum, as redder spectrum is completely adsorbed at moderate sea depth. Not sure if  they are tuned for UV specifically, but wouldn't be surprized.

@bon This isn't a homework question and is not even in my syllabus (9th grade)... My bio teacher can't give me an answer... Decided to ask here.

To give you a quick start and hint to your questions,
(i) Chlorophyll is green, because it reflects off the green light. Yes, there are chlorophyll of other colours. If you look at the absorption and action spectra of photosynthesis, you can see that there are chlorophyll a, b, carotenoid that absorbs various wavelengths of light.
See if you can use this to deduce the next 3 questions.


text:
REDOX spontaneity using electrode potentials

For this question, I thought the correct answer is A, because if we look at the electrode potential, that of iron(III) is higher than that of iodine, therefore the equilibrium for $\ce{Fe^3+}$ will favor the RHS, and shifts to the right, and therefore iodine will go in the opposite direction. Therefore A would be correct.
But the correct answer is D. Using the same logic, I can see it is also correct, but why isn't A correct then?
Thank you for any ideas in advance and sorry for the trivial question.

The reason is simply semantics. Iodi(n)e which is the neutral valence state of the element will not donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.   Iodi(d)e which is the anionic form of iodi(n)e has a negative charge and can donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.


text:
IUPAC nomenclature for an interesting looking organic compound
How can I name this organic compound while adhering to the IUPAC standards?

Congratulations!  This structure breaks my favorite structure editor (ChemDoodle) when I attempt to have it name this molecule...

Not many online editors worked... unfortunately. :/

Hah! This is an awesome test case for chemical editors and layout engines. Layout seems particularly tricky for what should be a quite easy job.

Also, it looks like https://en.wikipedia.org/wiki/Hexabenzocoronene minus some of the internal bonds.

Yea. Removing some of the internal bonds and adding two other phenyl groups gives you this compound.

Congratulations, this structure breaks ChemDraw's structure-to-name feature too! I'll try LexiChem when I'm at work tomorrow.

I can't even search chemspider with the skeleton (non-aromatic rings) with similarity search (tanimoto 99%) - I get an error. This is the devil's own structure...

Sorry, but I can't resist : sigma-aldritch site also fails, as does pubchem. only e-molecules seems to allow searches by drawing it! https://orderbb.emolecules.com/search/#?smiles=Cc1ccc(cc1)c1cccc(c1c1ccc(cc1)c1c(cccc1c1ccc(cc1)c1ccccc1)c1ccc(cc1)c1ccccc1)c1ccc(cc1)C&searchtype=si&similimit=0.8&system-type=BB&p=1

The preferred systematic name in phane nomenclature for the outer ring (without the “1,4-phenylene” bridge) is “1,3,5,7(1,2),2,4,6,8(1,4)-octabenzenacyclooctaphane”. By way of comparison, the similar [8]cycloparaphenylene with the systematic name “1,2,3,4,5,6,7,8(1,4)-octabenzenacyclooctaphane” is described in http://dx.doi.org/10.1039/c2sc20719b.

I get undecacyclo[?.?.?.?.?.?.?.?.?.?.?.?]tetrapentaconta-1,3(9),4(10),5(13),6(12),7(45),8(46),11(49),14(52),15(35),16(24),17(36),18(26),19(37),20(28),21(38),22(30),23(39),25(40),27(41),29(42),31(43),32(34),33(44),47,50(54),51(53)-heptacosaene from LexiChem. The "?" (actually BLAH) comes when LexiChem aborts numbering a complicated ring system.

I actually suggest issuing this as a Twitter challenge to ChemDraw, OpenEye, ChemDoodle, ACD, etc.

Even though this compound has no chiral center, the stereochemistry would be very interesting.

Related https://chemistry.stackexchange.com/questions/73103/does-iupac-nomenclature-have-the-ability-to-name-all-organic-compounds

It indeed can be named quite nicely using the recent "phane" nomenclature from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, here is the example*

By abstracting to the bicyclo phane composed of nine benzene nodes ("superatoms") of three kinds, we get

1,5(1,3,2),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane
Note that the "superatom" (more precise "amplificant") locant ordering matters. If the numbers in the first parentheses were sorted, it would mean a different structure, that is little bit hard to draw in 2D:

1,5(1,2,3),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane
Publicly accessible references:

Phane Nomenclature, Part I: Phane Parent Names
Phane Nomenclature, Part II: Modification of the Degree of Hydrogenation and Substitution Derivatives of Phane Parent Hydrides

*) Please note there is a mistake in the example from the book, the pyridine (4) has wrong internal ring numbering - 4 should be replaced with 5. The name is correct, though. (It is mentioned in errata, for page 310).

"[3.3.1] " What does it stand for?

@Mockingbird, "[3.3.1]" describes that the skeleton composed of two cycles (bicyclo) has three branches (that each begin and end in two distinct nodes - it is always the case; here they are X) (therefore three dot-separated numbers) that have 3, 3 (two identical (X)YZY(X) branches) and 1 ((X)Y(X) branch) nodes ("superatoms") respectively, excluding the X end points, total 9 (nonaphane). This type of naming is also used for hydrocarbons, etc. If X, Y and Z were just carbon atom (with hydrogens), it would be bicyclo[3.3.1]nonane.

@mykkhal what if there are substituents? Like Cl or Br joined to one molecule.

@Mockingbird, numbering is done by indicating the number of the "superatom" and the position on it as upper index, so this structure with chlorine on all peripheral positions would be 1⁴,1⁵,1⁶,2²,2³,3³,3⁴,3⁵,3⁶,4²,4³,5⁴,5⁵,5⁶,6²,6³,7³,7⁴,7⁵,7⁶,8²,8³-docosachloro-1,5(1,3,2),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

"Note that the "superatom" (more precise "amplificant") locant ordering matters. If the numbers in the first parentheses were sorted, it would mean a different structure, that is little bit hard to draw in 2D" Does it only matter for the superatoms by which the "[3.3.1]" is specified?

@Mockingbird, I am not sure I understand your recent question, the bicyclo skeleton name completely defines node ("superatom"), but does not define any "amplificant". Sorry that I could be more explanatory, so I attached reference links.

I mean the sequence in the first parantheses of your example mattered only because those two benzene molecules are the start and endpoint of the 2 cycles.

@Mockingbird, this is not the exact reason, but is related. It matters, because of symmetry or asymmetry. It matters, because the branches are different. It can be vaguely said that it matters because the three numbers in the square brackets are different. If the superatom amplificant would be e.g. pyridine, the order would matter also for those in-branch nodes (with only two locants in the parentheses), if they were bonded asymmetrically, which is not possible for benzenediyl, in this sense.

I don't understand the order in the first parenthese is in basis of what?

@Mockingbird, .. the order is based on several rules - lowest set of locants, bute more importantly, it is also based on the order of the number of superatom nodes in the parent skeleton, boned to the amplificant locations. Please read through the refs,  and draw possible structures yourself, it is no room here for the deep explanation. See rule PhI-2.3.2.2

ChemDoodle 8 can name this via the extended von Baeyer system, although the name is quite long. Thanks for putting up this nice example though!

Undecacyclo[18.18.6.22,5.212,15.221,24.231,34.240,43.06,11.016,44.025,30.035,39]tetrapentaconta-1(39),2,4,6,8,10,12,14,16(44),17,19,21,23,25(30),26,28,31,33,35,37,40,42,45,47,49,51,53-heptacosaene

Although impressive, the name ignores the identity of benzenes and treats the compound as poly-alekene.

Interesting, Marvin Sketch seems to agree with ChemDoodle and Chem Office can not generate a name at all.


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What will be the product of this reaction
I suspect the answer is either one of C or D, but how do I know which one?
Since such a stable carbocation is being formed, I can't choose whether the reaction will be an Sn1 or Sn2 (elimination semms unlikely).

But do I need to transform this fischer projection to some other form for me to easily answer this question?

Neighbouring group participation of the OMe leads to a double inversion, i.e. overall retention of configuration

I just looked up NGP on the Internet and learned something new. Thank you :).and btw using NGP as a method, I am getting the correct answer.


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Why in -COOH group the C atom pulls electron cloud from the OH group O where as the double bonded O pulls the electron cloud towards itself?

Should both the oxygen pull the cloud as O is more electronegative than C . I think that as the cloud density decreases near carbon therefore it pulls . A proper explanation please .

It's wrong - mesomeric effect shouldn't be described as inductive.

In the book it is under inductive effect . We are dealing with electronegativity