Is there any possibility of tautomerisation to en-ol form? In the picture, you can see the bridge whose end points are in para relation (for making the image clear).

Now, what my friend said is that there is no possibility of tautomerisation as there is no H to the right of =O and to the left of =O, on removing H, the bridge will be loaded(?!?).

But according to me, positive charge on the bridge end-points may be unstable as the bridge is out of the plane of the planar molecule and making it sp2 hybridized isn't a kind thing to do. But on removing H from the bridge end point (for tautomerisation), there will be negative charge on the bridge end point, which is perfectly normal.

So will it tautomerise or not?

If you're asking specifically about the tautomerisation, then I made this comparative computational study using B3LYP theory at the 6-31g* level, and the results are:

So as is evident, the bridge causes the alkene to be dramatically strained. You can compute the Boltzmann ratio and find that the tautomerisation in the first structure is completely negligible at room temperature.

If you're talking about the formation of an enolate ion, the results shouldn't differ much, because the alpha carbon still needs to become sp2 hybridised, in order to delocalise the carbanion into the carbonyl group. And this, as we know, is the reason for the increase in acidity of protons adjacent to carbonyl groups.

All this is of course to the best of my knowledge.


how to measure dissolved oxygen? How do you measure dissolved oxygen reliably in a culture of cells? This page has info on one sensor: http://water.me.vccs.edu/concepts/domeasure.html

how would you know that this sensor is working - meaning that only dissolved oxygen triggers the electrical change in the sensor, and not other atoms? and how would you calibrate the sensor to translate changes in electric signal to the change in number of dissolved oxygen atoms?

also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

To start, the figure and explanation in the link are almost comically misleading. Potassium chloride solution does not "attract" oxygen. The oxygen flowing through the wire is pretty silly and

  Where the probe joins the wire, oxygen mingles with the electricity.
  Oxygen is not very ionized, meaning that it does not have a negative
  charge as electricity does, so the oxygen dilutes the current at the
  electrode beyond the probe.

is just plain ridiculous. This one is better.

The way this type of probe works is by amperometric reduction of $\ce{O2}$, the first example being the Clark electrode. Essentially, an electrolyte is separated from the measurement solution by a porous film of poly(tetrafluoroethylene) (Teflon). These pores are small enough that only small substances like oxygen can diffuse through. On the electrolyte side, there are two electrodes in one of two possible configurations:

In the original Clark-type electrode, the working electrode is platinum, which is chemically-inert but catalyzes the reduction of oxygen. The counter electrode is a AgCl-coated Ag electrode that acts as a reference for the measurement. (Its potential is supposed to remain constant) To reduce the oxygen, an external potential must be applied across the two electrodes (normally using a potentiostat).
The galvanic cell–type electrode has one electrode made of silver (where the oxygen reduction occurs) and one made of an easily oxidized metal like lead. It works the same way as the Clark electrode, but instead of applying an external potential, the current required to reduce the oxygen is supplied by the oxidation of the lead electrode.

Any oxygen that reaches the working electrode is reduced and produces a current that is measured to produce the analytical signal. The amount of current produced depends on how much oxygen diffuses through the membrane, which depends on the concentration of dissolved oxygen in the test solution.

Now, for your questions:

  how would you know that this sensor is working - meaning that only
  dissolved oxygen triggers the electrical change in the sensor, and not
  other atoms?

This one is a bit tricky. You're right that oxygen isn't the only thing that could produce a signal—amperometry isn't a very selective technique on its own. You can control the redox reactions that occur at the working electrode, to a limited degree, by changing the potential you apply, i.e. different substances require different potentials to be reduced and won't be reduced if an insufficient potential is applied. In the case of this specific type of electrode, selectivity comes mainly from the PTFE membrane. Since it limits diffusion to only a few very small molecules, that greatly increases the selectivity of the electrode, though depending on the solution you're trying to measure, there still may be interferences.

  and how would you calibrate the sensor to translate changes in
  electric signal to the change in number of dissolved oxygen atoms?

In principle, you could calculate the number of reduction reactions happening at the surface from Faraday's laws, but to turn this into a measure of the exact concentration in solution requires a lot of thought of how the transport of oxygen to the electrode occurs, from diffusion from the electrolyte to the electrode surface and transport from the test solution into the electrolyte. In practice, the assumption that the measured current is directly proportional to the concentration in solution is made (and is valid under many conditions), and a calibration is made using a standard solution.

  also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

Calibration solutions seem to be commonly made using tables of oxygen saturation at different temperatures. It's also possible to measure dissolved oxygen through other means, such as titration. Solutions can be made up, titrated, then the value used to calibrate an electrode.

Now, for the measurement specifically in cell cultures, a membrane electrode may not be the best bet, at least not without sample pre-treatment. The PTFE membrane is quite susceptible to adsorption of proteins and the like, which will probably affect the electrode's response. It may be necessary to somehow remove these contaminants (without affecting the dissolved oxygen concentration). Other options that might be viable are colourimetric measurements or titration, though there are problems with these, as well.


Is the resonance depiction of benzene typically portrayed incorrectly? I can understand that the carbon atoms in benzene have delocalized p electrons that are shared amongst all 6 of the carbons in a cloud on either side of the ring. I also understand why this could be represented as resonance structures like so:

I can also understand representing it in this form:

What I cannot understand, however, is why only one of the first form appears so often in depictions of organic molecules. For example, this:

Since this is a "resonance structure," is this not incorrect to list only one of the structures?

One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn't sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw both resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.

Starting with a short and blunt answer to the title’s question: Yes. But you likely already knew that, otherwise you wouldn’t have written the rest.

  One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn’t sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw both resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.

Well, there is a lot more time saved that you seem to be imagining according to your quote.

You seem to suggest that there are only two correct resonance structures for benzene, which is incorrect. Even if you only consider the $\pi$ system, there are a lot more possible resonance structures which contribute to the final picture in a non-zero manner.

The ones I displayed in the picture are just a very small subset. In fact, solely for the $\ce{H2}$ molecule, a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures and also claimed to be able to accurately calculate the energy of the molecule by adequately weighting all those structures. (‘Accurately’ meaning getting an energy comparable to one calculated using molecular orbital theory.)

This actually moves us towards another question: What actually is a bond? There is a lot more to a bond than simply drawing a line between two atoms and saying that ‘they are now sharing an electron pair.’ If you calculate simple-looking molecules like formaldehyde using a program like Gaussian, you are able to see a multitude of molecular orbitals — at least twelve. Two are defined as ‘core orbitals’, namely they represent the 1s orbitals of carbon and oxygen, respectively; all the others somehow contribute to bonding or antibonding interactions somewhere across the whole molecule.

So what would an accurate representation be? Drawing the atoms geometrically correctly, a cloud around them and writing a number 12 inside or next to the cloud? That way, it would be made clear that there are twelve electrons delocalised across the whole of the molecule and somehow keeping it together. But what would be gained from this type of more accurate representation? Or rather: What would be lost? You would lose the possibility to easily discuss lone pairs on the oxygen, bond order between carbon and oxygen and a lot more.

Similarly for benzene and all molecules that contain a six-membered ring of carbons with a $\pi$ system with six electrons (a.k.a. the benzene derivates you’re asking about): If you insisted on drawing ‘the two’ resonance structures, you would still be far from the truth. Furthermore, for most molecules that are not benzene but contain the same ring, the bonds are no longer equal and their bond orders are no longer exactly $1.5$ (true for any molecule that has lost the $\mathrm{D}_{6\mathrm{h}}$ symmetry). So drawing two sets of double bonds and claiming the structures to be equal would be outright wrong.

Finally, though, every chemist has learnt to get a more or less thorough understanding on the concept of a bond and on the concept of aromaticity that it can reasonably be assumed that what I presented is accessable somewhere at the back of the head if really required. Thus, every chemist should have a rather good understading that a line on paper is a line, a bond is not just a pair of electrons and a benzene ring can be drawn in multiple ways. When it really matters, we can either expect the chemist to instantly recognise the mattering due to a special case, or we can bring the mattering matter to the chemist’s attention by explicitly drawing it.

So there is a lot of time gained by simplifying drawings of molecules in the way they are and also cases where it matters can easily be illustrated by drawing more elaborate drawings which aren’t skipped in one’s attention because they are rarely encountered. And we also keep the ability to easily recognise and discuss simple concepts.

P.S: It’s not necessary for anyone to interpret these concepts the same way. In fact, it’s a good thing if an artist takes the picture of benzene and interprets it completely differently. If anything, ‘anyone’ should be restricted to the chemical — or maybe the entire scientific — community, but there I can see the interpretation safely, regardless of whether aromaticity is drawn or not.


Is there any danger of forming explosive reaction products during the electrodeposition in Diethylene glycol? I know that perchloric acid could be potentially explosive and some organic solvents could form dangerous peroxides.

I am going to do an electrodeposition in diethyleneglycol solution containing $\ce{NH4Cl}$, $\ce{ZnCl2}$ and $\ce{FeCl2}$. Since in my case there are all elements to form $\ce{HClO4}$ and I am going to heat everything up to 60 degrees I want to be sure that it is safe.

Is there any danger of forming explosive reaction products during the electrodeposition?

Diethylne glycol reacts readily with perchloricacid, however the oxidation and hazard potential of perchloric acid depends on its concentration in side the solvent.

If concentrations are high enough and more than 10 percent perchloric acid formation is expected with 100 percent reaction yield, then it can be explosive.

What actually happens that as soon as diethylene gylcol meets with perchloric acid, diethylene glycol decomposes and some dangerous peroxides formed. These are very reactive to water, so if you can provide very dry and inert atmosphere it should be safe but you should give extra care to make sure that you do not have humidity and water both in your chemicals and in the atmosphere.

If you are going to do the experiment and use milimolar range chemicals then just do it, because you will not be able to form enough perchloricacid to get problems.


Which group is more suitable for metal nanoparticles functionalization? I have synthesized bimetallic nanoparticles (gold core with a silver shell). I have two choices for functionalizing them, through thiol group or amine group.
Which of them is more attracted to this kind of nanoparticles?

Sulfur containing groups, such as thiols, are in general more strongly attached to silver nanoobjects than amine groups. However, the problem with sulfur and silver is that the strength and structure of sulfur-silver bonds depends critically on the local topology of the silver surface, and in-layer lateral interactions between stabiliser molecules can have a strong impact on this, so there isn't a clear pattern of reactivity of sulfur-containing groups and silver nanoobjects. You can easily get irregular, disimilar or patched coverage.

Amine is much more straightforward, links itself more weakly and is routinely used in the synthesis of silver nanoparticles as a stabiliser. However, if you want permanent functionalisation of the nanoparticle (vía covalent bonding), probably amine-silver bonding will be too labile.


Why isn't pressure filtration popular compared to vacuum filtration? I want to build a DIY vacuum filtration apparatus with a bike pump but it's very difficult to find one the could be easily converted into a vacuum pump. This prompted an idea to simply build a reverse version where above atmospheric pressure is forcing the liquid to flow through the filter. I notice pressure filtration is rarely mentioned in write-ups, unlike vacuum filtration. Is there a reason for this lack of popularity or a reason why such a DIY apparatus would likely not work?

Could it be that vacuum is necessary for drying the paste that forms at first because it enhances evaporation?

It's much easier to assemble a system that's safe under external pressure than one that's safe under internal pressure. Convex flasks and tubes resist external pressure better than internal pressure. If something leaks or breaks, you have air rushing into the system, rather than reactants and products blasting out. For that matter, if you've put things together with ground glass joints or rubber stoppers or corks, internal vacuum (external pressure) works to keep them together, while pressurizing the system's interior tends to force them apart.

Also note that a vacuum system will never be under more than one atmosphere of external pressure. With a pressurized system, you need a way to limit the pressure to safe levels, and you need to make sure you know what those levels are.


Why do NMR coupling constants change with temperature? Why do coupling constants change with temperature in NMR?

Let’s use vicinal 3-bond $\ce{H-C-C-H}$ proton coupling constants as an example.  Coupling is a through bond phenomenon and as such the coupling constant will change depending if we have better (the coupling constant will increase) or worse (the coupling constant will decrease) overlap between the two $\ce{C-H}$ bonds involved.  Overlap is maximized when the two, vicinal $\ce{C-H}$ bonds are aligned in an eclipsed conformation ($\ce{H-C-C-H }$ dihedral angle =0°) or anti ($\ce{H-C-C-H }$ dihedral angle =180°)  conformation.  When the dihedral angle is ~90° the overlap of the two vicinal $\ce{C-H}$ bonds is minimized.  This thinking is embodied in the Karplus equation which relates vicinal coupling constants to dihedral angle.  The following graph illustrates this dependence of the vicinal coupling constant on the dihedral angle.

image source

All of this tells us that an experimentally observed vicinal coupling constant will be dependent upon the relative populations of the eclipsed, staggered and anti conformations. Said differently, the vicinal coupling constant will be an average of each of the individual coupling constants for each conformation, weighted by the relative population of that conformation. Since the relative conformer populations change with temperature, the coupling constant will also change with temperature.


How to measure/know the moles of carbon dioxide (as dry ice) when in KOH solution I am doing an experiment where I am dropping dry ice in a $\ce{KOH}$ solution. I want to know how I can figure out how many moles of $\ce{CO2}$ reacted with the $\ce{KOH}$. I am pouring $25 \,\mathrm{g}$ of $\ce{KOH}$ in $100 \,\mathrm{mL}$ of water and I am using $50 \,\mathrm{g}$ of dry ice. The experiment is conducted at room temperature.

I know one way to do this is by isolating the product of $\ce{K2CO3}$ but because $\ce{K2CO3}$ is soluble, it is going to be very hard to isolate this product and weight it and use simple stoichiometry to figure out the moles of $\ce{CO2}$ reacted.

I was thinking maybe the pH meter will help but again, I do not know how to use the pH to calculate how many moles of carbon dioxide reacted.

Observations

During experimentation, I saw the pH decreased when I added the $\ce{CO2}$.Will this fact be useful to calculate moles of $\ce{K2CO3}$ formed? The pH decreased from 13 to 10.

Can someone please help?

In this particular case weighing would work with limited accuracy (I am going to explain the limitations later in the answer).

Since the only compound leaving the reaction is the unreacted $\ce{CO2}$, it is valid to use weighing. Everything else is either sticking around or is already included in the initial weighing. So by weighing everything prior to reaction and after "all unreacted $\ce{CO2}$ is vaporized", you can calculate the amount of "reacted $\ce{CO2}$".

(weight of reacted $\ce{CO2}$) = (Weight of initial $\ce{CO2}$) - (weight of unreacted $\ce{CO2}$)

(Weight of initial $\ce{CO2}$) = (50 g) - (condensed water from atmosphere [estimate])

(Weight of unreacted $\ce{CO2}$) = (deficit in pre vs post reaction weighing)

pre reaction weight: 50 g $\ce{CO2}$ + 25 g $\ce{KOH}$ +100 g water = 175 g

post reaction weight: weight of the whole solution after all unreacted $\ce{CO2}$ is escaped from the solution (make sure the whole solution is at room temperature and make sure there is no bubbling occurring).

I understand there are some practical limitations, mainly :

1- There might be unreacted/frozen $\ce{CO2}$ still hanging around, Try to shake well. Also when weighing post reaction, make sure your solution is slightly warm to assist unreacted $\ce{CO2}$ to evaporate fully from the reaction.

2-Since you have dry ice, some water from atmosphere will condense into your reaction, this will lead to underestimation of the "leaving compound". The way to overcome, is to keep your reaction vessel as small as possible and control its inlet by a cold trap to make sure no water from atmosphere gets in. Also account for amount of water usually present in 50 g dry ice, by examining the water content on a similar sample beforehand.

pH:
Your initial pH is clearly not right. concentration of your $\ce{OH-}$ is way higher than just 0.1 M. Let's assume that it is due to high pH value being out of the calibration range of the pH-meter and let's assume that your pH=10 after addition of $\ce{CO2}$ is correct.

One will be able to calculate the amount of unreacted $\ce{CO2}$ from reaction balance. You are starting with 25 g (0.446 moles) of $\ce{KOH}$ and 50 g (1.14 moles) of $\ce{CO2}$. For every $\ce{CO2}$ reacting, 2 $\ce{KOH}$ are used up. So IF all(or most) of $\ce{KOH}$ is used up, you should end up with 0.446/2~ 0.22 moles of $\ce{K2CO3}$, which means the reacted CO2 is about 0.22 moles(~9.7 g), and unreacted is 0.92 moles(~40.3 g).
Why do I make the assumption that all the $\ce{KOH}$ is reacted, notice your final pH is 10 , meaning that your [OH-] concentration is about 0.0001 M, your starting concentration was 4.4 M.

What stops the reaction from using all $\ce{KOH}$ to be used up and dropping the pH to neutral zone? You are using excess $\ce{CO2}$, so we expect all the $\ce{KOH}$ to be used up. This has to do with part of $\ce{CO2}$ escaping the reaction as gas, also since the temperature is freezing the water (by dry ice) you may not have complete agitation which hinders the reaction procession to some extent.

I still think that you should take the weighing method seriously. Using weighing methods will provide confirmation for your calculations.


Can carbanion rearrange to attain more stability? We all know that carbocations rearrange to attain more stability, but I have never heard anything about carbanion rearrangements for attaining more stability.

Is it possible for a carbanion to rearrange? If not,why?

Sooner or later anionic rearrangement name reactions should turn up in your organic chemistry classes. I will present two in this post.

The first and arguably most well-known are the Wittig rearrangements; these can proceed as [1,2] or as [2,3] rearrangements.[1] The Kürti/Csakó states that:[1]

  The [1,2]-Wittig rearrangement proceeds via a radical-pair dissociation-recombination mechanism, while the [2,3]-Wittig rearrangement is a concerted, thermally allowed sigmatropic process proceeding via an envelop-like transition state in which the substituents are pseudo-equatorial.

This means that strictly speaking only the [2,3]-Wittig rearrangement could be considered an actual carbanion rearrangement as per your question but many chemists may also consider the [1,2]-Wittig rearrangement one since it effectively does generate a rearranged and thereby more stable anion. Both reactions are shown in scheme 1.

Scheme 1: the [1,2] (top) and [2,3] (bottom) Wittig rearrangements.

The second example has recently gained significant applications in synthetic organic chemistry; it is the same general reaction except it employs a silyl ether instead of an alkyl ether. Here, the stability of both sides of the reaction is similar so both the forward and backward direction are known; these are the Brook and retro-Brook rearrangements (see scheme 2).

Scheme 2: The retro-Brook rearrangement (top) and Brook rearrangement (bottom).[2,3]

West et al. who reported the reactions in scheme 2 note the interesting effect that the amount of butyllithium added influences the equilibrium. If substoichiometric or catalytic amounts are added, the normal Brook rearrangement takes place with the silyl ether being the more stable species. Under excess butyllithium conditions, the retro-Brook rearrangement takes place indicating that under those conditions the alcoholate is the more stable species.[3]

I mentioned that arguably the Brook rearrangement is more common in organic synthesis; this is due to Amos B. Smith’s anion relay chemistry or ARC.[4] Smith uses priviledged synthetic intermediates that can give alcoholates upon an initial reaction. The previously introduced silyl group migrates towards the alcoholate, liberating another carbanion which can subsequently participate in another nucleophilic attack. This allows multiple anion-dependent transformations to occur in a single reaction vessel.

References:

[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 490.

[2]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 64.

[3]: R. West, R. Lowe, H. F. Stewart, A. Wright, J. Am. Chem. Soc., 1971, 93, 282–283. DOI: 10.1021/ja00730a065.

[4]: A. B. Smith III, M. Xian, J. Am. Chem. Soc., 2006, 128, 66–67. DOI: 10.1021/ja057059w.


Why is ammonia so much more basic than water? The only reason I can think of is that the lone pairs in the oxygen in water are in lower energy orbital relative to the nitrogen in ammonia due to increased effective nuclear charge in oxygen. However, as far as I am aware, ammonia is SIGNIFICANTLY more basic than water. Surely the orbital energy cannot account for this.

When it comes to comparing ammonia and water, the acidity or basicity can be explained because of the central atoms electronegativity. When comparing atoms of the same periodic row, the more electronegative an atom is, the less it will want to donate an electron pair.

The electronegativity values from the Pauling Scale for $\ce{N}$ and $\ce{O}$ are

$$\ce{N}=3.04$$
$$\ce{O}=3.44$$

So based off of these values we can see that nitrogen is less electronegative, it less strongly it holds onto its electrons, and thus it is a better electron donor. This allows the addition of a $\ce{H}$ to be more stable.

The atomic reasoning behind this is that, as a general periodic trend, as you move across the periodic table, from left to right, the nucleus is increasing, which increases it's charge, producing a stronger attraction to its valence electrons.


Do tetrachloroaluminate salts hydrolyse? Tetrafluoroborate salts, such as lithium tetrafluoroborate, are reasonably stable in aqueous solutions. Does the same hold for sodium tetrachloroaluminate as well or does it end up as $$\ce{NaAlCl4 + 4H2O -> NaAl(OH)4 + 4HCl}?$$

NaAlCl4 will hydrolyze according to the following reaction $$\ce{AlCl4^- + \ 2H2O ->  Al(OH)2^+  +  2 H+ +4 Cl-}$$and if there is enough water it could goo further away like $$\ce{AlCl4^- + \ 3H2O ->  Al(OH)3  +  3 H+ +4 Cl-}$$ But the reaction will never produce $\ce{NaAl(OH)4}$ as you think. The substance $\ce{NaAl(OH)4}$ can only be produced in very basic solution, and here the solution is acidic.


Would any alkene possibly aromatize with proper conditions? I was reading about 2 specific reactions: polymerisation of alkene and aromatization of alkane (of a sufficient size).
I thought why can't, having a condition that favored both these reactions be created, there exist a chain reaction where an alkene starts polymerizing and when the chain becomes long enough, it aromatizes.

It's just a thought that I had but I can't figure out if it IS possible.

Your thought is actually happening in real life. Petrolium industry uses the same concept except for polymerizing ethene (may be that also used to get higher alkanes). Fluid Catalytic Cracking (FCC) is one of the most important conversion processes used in petroleum refineries. FCC (see the schematic diagram below)) is widely used to convert the high-boiling, high-molecular weight hydrocarbon fractions of petroleum crude oils into more valuable gasoline, olefinic gases, and other products such as aromatics (Ref.1):

For example, a conversion reactor is used to carry out several reactions. A few of them are:
$$\ce{n-C26H54 -> C6H12 + C6H14 + C10H20 + C4H10}\tag1$$
$$\ce{C10H20 -> C6H12 + i-C4H8}\tag2$$
$$\ce{C6H12 + 3 i-C4H8 -> 3C4H10 + C6H6}\tag3$$
The equation $(1)$ (cracking reaction of paraffin) shows the formation of cyclohexane from a long-chain paraffin. The equation $(3)$ (hydrogen transfer reaction between cyclohexane and isobutene) shows the conversion of cyclohexane to benzene.
References:

A. Kayode Coker, In Petroleum Refining Design and Applications Handbook, Volume 1; John Wiley & Sons, Inc.: Hoboken, NJ, 2018 (ISBN: 978-1-118-23369-6).


Soaps come in different colors, but why is soap lather always white? Soap comes in different colors, but why is soap lather always white?

Though @DHMO's answer is quite interesting, you shouldn't take it without a pinch of salt. I've to disagree with it in some aspects.

Soap colorants range from various kinds of Mica to dyes found naturally in plants. Now typically when you dye anything the quantity of dye used in comparison to the quantity of the substance to be dyed is very small. Too much dye, and it could possibly stain your hands/clothes, so soap manufacturers use as little dye as possible, but just enough to give the soap a good color.

Now for the rest of this answer, I will assume (reasonably enough) that by 'colored' soap you are referring to soap that goes by colors other than white.

Now when you lather using soap, only a really teeny-tiny bit actually goes into the water. As @DHMO correctly points out, the dye/pigment used to color the soap is greatly diluted. Why don't you carve off a piece of colored soap (having more or less the same size as as an almond) and dissolve it a mug of water by gently stirring it. I emphasize on "gently" so that you don't stir up a lather.

Have a look at the water. You'll see that the soap, apart from making it more cloudy, has not visibly imparted any particular color to the water. This is not unexpected, as I've already mentioned, the quantity of dye used is very small.

Now if you go ahead and agitate that soap solution, it will give rise to white lather, and this shouldn't be surprising anymore.

Now @DHMO's goes on to mention that Total Internal Reflection (TIR) is what imparts the white color to the foam. But this is grossly incorrect. While TIR can result 'white light', it is not the dominant phenomenon acting here (I'm not saying it's completely absent here either).

What is largely giving foam it's white appearance is another phenomenon called Scattering.

Now you might ask, 'Then why isn't soap water white?'. Well, since the foam is made up of lots of teeny tiny bubbles, light passing through it will have to encounter several surfaces, and it's these surfaces that scatter the light in so many directions.

So to say it straight DHMO's answer is incorrect. (No offence DHMO)

Remember I said you can't see any visible coloration in the water because of the dye is present in really small quantities? Well here's a way to validate that claim: Simply combine a teaspoon of red food-color with a bit of hand-wash (now that's really concentrated). Now you can lather the soap and lo behold! You have red foam.

[Credit to @ACuriousMind from Physics.SE, for that bit on why soap water isn't white. Also he confirmed that the white color of foam is due to scattering]

Edit- More kudos to @ACuriousMind for providing this amazing link on scattering.


What makes water a liquid (when it is)? I'm curious about what the minimal conditions for some collection of water molecules to be liquid water are. What is the minimum number? What sort of bonds must hold? What other sorts of things I don't know to ask about need to be true?

In my peeking around a subject I have little familiarity with, I've seen mention of a tetrahedron, suggesting a lower bound of 4 molecules for water to be a liquid. But then I also stumbled across the following (from p. 23, here):

  Whatever the model concerned, the presence of a fifth
  water molecule is necessary to be considered in explaining
  the molecular mechanism of  defect formation and Hbond
  network rearrangement.

So it seems that although only four molecules are in this bond network undergoing constant change, there must actually be a fifth "neighbor" molecule for the bonds to function appropriately. Is this fifth molecule part of the liquid, or merely a helpful neighbor who makes the four molecules' liquification possible?

In short, what makes water a liquid (when it is)?

Let me venture an answer as well as I can.

I would point out that the article you are referencing is speaking of the arrangement of molecules in bulk liquid water when it is talking about this "fifth water" being present. The key is to note that they say a "fifth water molecule in the first coordination shell (23)." That means they are looking at how water molecules are coordinated with each other in liquid water. They are studying a massive (compared to a single water) amount of water, so the question of its liquidity is no question at all.

The "fifth water" thing is curious because it is common belief that waters arrange themselves in a tetrahedral arrangement. This is known to be true for ice, but is in question for liquid water. For instance, this article about the liquid water hydrogen-bonding network indicates that water, as it is heated, begins participating in only two hydrogen bonds rather than four.

Now, to answer the question of what makes liquid water be liquid water (when it is indeed liquid water). I think from the outset, it should be clear that this is a very difficult question to answer and any answer will likely be debated. I say that because you must define how many molecules it takes to make a liquid which is a very difficult thing to do.

I will tell you, however, that I have some useful information from research I am doing on water clusters and eventually liquid water (as a computational chemistry research assistant). The largest cluster we have studied is called DD*(20,1) which means (distorted dodecahedron with a water inside the cluster).

An idea of what $\ce{(H2O)21}$ can look is this: (taken from this very interesting article) water 21 http://pubs.acs.org.ezproxy.spu.edu/appl/literatum/publisher/achs/journals/content/jpcbfk/2006/jpcbfk.2006.110.issue-38/jp056416m/production/images/medium/jp056416mf00001.gif

The DD*(20,1) arrangement is image C.

As you can see from these different isomers, they are very distinct and no doubt each monomer behaves dramatically different from the next monomer. 21 monomers is quite a lot for a cluster, but even this does not mimic the properties of liquid water at all particularly because one would still consider this a gas-phase cluster. The fact that we even have isomers is indicative of not being near the point where something is considered liquid water. There are water clusters whose energy minima have been identified theoretically all the way from n=2 to (I've seen an article that says) n=60. Read more about some of that here.

One point I would like to make is that in structure C, for the research I'm doing, we have identified the vibrational properties of many of the monomers as being significantly different from those of molecules in liquid water.

So, all that being said about water clusters, we are certain that many more than five or 21 waters must be present to truthfully call something a liquid. We should then start looking for a system which is just large enough to accurately predict the properties of liquid water which are found experimentally.

As it happens, it is very difficult to do this on any scale, theoretically, at least, and identifying how many water molecules one is dealing with during an experiment to a quantifiable amount of precision is quite difficult. There are, however, good approximations of liquid water which are generated using Molecular Dynamics simulation software (here's a wikipedia page about MD software). The various models used for approximating the behavior of water do quite well at mimicking water's properties from an MD perspective. You can read about the TIP4P model of water if you want.

From the paper I just linked about TIP4P, they say that they approximate the properties of liquid water using 360 water molecules in their MD simulations, so that can act as a reasonable baseline for how many monomers are necessary before a system behaves like liquid water.

I say baseline for an important reason. The results found in that paper use MD simulations which pull a clever trick. Rather than having 360 waters that essentially float around in a box with nothing at all beyond the edges of the box, the software will actually mirror the behavior of a molecule beyond the walls of the box in which you are simulating. That means if you pick a single water, you would find that exact same water in the same place a full "simulation box" away from where you originally found it. So, if this says 360 waters, the number of waters for which interactions are being calculated is actually much larger than the amount of interactions 360 waters alone would have.

All that leads me to conclude that water behaves like water probably somewhere around 500 monomers. It must be noted, however, that the monomers behaving like liquid water as we know it will be the monomers on the inside of those 500 (but likely more) waters. Once you get towards a boundary, things start getting weird.

Hope that helps.


Identification of a metal from group 2 A chemist adds $\pu{0.115 g}$ of a group 2 element to water and the resulting solution is made up to $\pu{400 cm3}$ by adding distilled water. The equation for the reaction is shown below. $\ce{M}$ represents group 2 element.
$$\ce{M + 2 H2O -> M(OH)2 + H2}$$
The concentration of hydroxide ions $\ce{OH- (aq)}$ in the solution is $\pu{6.56e-3 mol dm-3}$. Determine the molar mass of $\ce{M}$ to one decimal place. Use your answer to identify $\ce{M}$.

By definition molecular weight is mass $m$ divided by amount $n$, e.g. for the unknown metal $\ce{M}$:

$$M(\ce{M}) = \frac{m(\ce{M})}{n(\ce{M})} \tag{1} \label{eq:1}$$

From the chemical reaction of hydroxide formation one can deduce that $\pu{1 mol}$ of $\ce{M}$ results in $\pu{1 mol}$ of $\ce{M(OH)2}$. However, the amount of hydroxide-anions $\ce{OH-}$ in the solution (assuming complete dissociation) is twice as high as the amount of $\ce{M(OH)2}$. Hence

$$n(\ce{M}) = n(\ce{M(OH)2}) = \frac{n(\ce{OH-})}{2} = \frac{C(\ce{OH-}) \cdot V}{2} \tag{2} \label{eq:2}$$

where $C$ – concentration and $V$ – volume.

Plugging $n(\ce{M})$ from \eqref{eq:2} in \eqref{eq:1}, a final expression for the molecular weight can be established. Remember, before plugging in the numbers in make sure the units are unified.

   $$M(\ce{M}) = \frac{2 \cdot m(\ce{M})}{C(\ce{OH-}) \cdot V} = \frac{2 \cdot \pu{0.115 g}}{\pu{6.56e-3 mol dm-3} \cdot \pu{0.4 dm3}} = \pu{87.6 g mol-1} \tag{3} \label{eq:3}$$
 The unknown metal appears to be strontium $\ce{Sr}$.


What prompts sodium to give up an electron? What triggers sodium to convert from elemental Na to Na+? I know that it wants to have a full valence shell and all that, but how does it just eject the electron out?

To asses the possibility of a chemical change, we need to look into the spontaneity as given by the thermodynamic Gibbs' Free energy change at constant T and P. For your particular case, the change occurs when an electron accepting ion is present to form the crystalline compound after the electron exchange, in solution where the hydration releases energy or in certain solutions such as ammonia, where solvation of free electrons also take part.

Case (1). The changes involved are $$\ce{Na_{(s)}->Na_{(g)}^+ +e^{-}}$$, $$\ce{M_{(g)} +e^{-}->M^{-}}$$ $$\ce{Na_{(g)}^+ + M_{(g)}^{-}->NaM_{(s)}}$$

Where $M$ is an electronegative atom.The ionization energy ($\Delta H_{I}$) and the sublimation energy ($\Delta H_S$) required in the first transformation need to be compensated for by the electron gain enthalpy($\Delta H_{eg}$) released in the second transformation and the lattice enthalpy($\Delta H_{F}\text{lattice}$) released in the final transformation. If it is compensated, the change will be spontaneous. Moreover, the enthalpy released (negative) must also compensate for the lost entropy since gaseous $M$ is converted to an ordered crystal. $\Delta G=\Delta H-T\Delta S$ tells you the compensation required for $\Delta G$ to be negative.
($|\Delta H_{eg}+\Delta H_F|-|\Delta H_I+\Delta H_S|>|T\Delta S|$)

Case (2) Elemental sodium in water might lose electrons to the hydrogen evolving hydrogen gas and forming dissolved sodium hydroxide.
$$\ce{Na_{(s)}->Na_{(aq)}^+ + e^{-}}$$
$$\ce{H2O +e^{-}->\frac 12H2 +OH-}$$
here, the relesed hydration energy compensates the required sublimation energy in the first step and the negative electrode potential in the second step. There is net positive entropy change which favours the reaction, making $\Delta G$ negative.  ($|\Delta H_{hyd}(\ce{OH^-,Na^+})|+|T\Delta S|=|\Delta G|$)

Case (3) very similar to case (2) but here the solvation is both of free electrons and that of metal ions by ammonia (liquid). The equations are similar except that now instead of the hydration enthalpy, solvation enthalpy will be released. the entropy change is also positive here, supporting the change.


Why do salts such as NaCl dissolve? If we look at solubility of salts in water, we are told that they disassociate, because the positive $\ce{Na}$ ion is attracted to the partially negative oxygen in water and the negative  $\ce{Cl}$ is attracted to the partially positive hydrogen. But why does this happen? I mean, the couloumb force should be much stronger between the $+1$ and $-1$ charges on the $\ce{Na}$ and $\ce{Cl}$ than the partial charges on the polar water molecule. Why does $\ce{NaCl}$, then, disassociate? It would make much more sense for the two ions to stick together.

Also, when you cut $\ce{NaCl}$, shouldn't it stick to its pieces? Look at salt crystals: they don't have any force of attraction between them. But, if you have positive sodium ions and negative chloride ions, then they should stick together.

As it happens, the enthalpy of solution of $\ce{NaCl}$ in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, i.e., it is an endothermic process. At a constant temperature and pressure, these kinds of thermodynamic processes are dictated by the change in Gibbs free energy, described by the equation

$$
\Delta G = \Delta H - T\Delta S
$$

Where $\Delta G < 0$ is a necessary criterion for a spontaneous process. Given that $\Delta H$ is positive, $\Delta S$ must be positive as well, otherwise the process wouldn't occur spontaneously (which is to say, not without input of work from the surroundings, or coupling to some other strongly favorable reaction, neither of which is the case for the dissolution of salt). In other words, this is a process that is driven by the increase in entropy, which is entirely to be expected when moving from a highly ordered state (i.e., a crystalline solid) to a less ordered liquid solution. Indeed, in ideal solutions, where intermolecular forces of attraction are taken to be equal between all components, enthalpy change is necessarily zero, and entropy is always positive, so that the process of mixing in ideal solutions is always spontaneous. Of course, in real solutions, this is not the case.

Your intuition that the Coulombic forces between ions should be stronger is correct in this particular instance, as indicated by the positive enthalpy (meaning, the breaking of ionic bonds in the crystal lattice, as well as intermolecular bonds between solvent molecules, requires more energy than is released in the forming of ion-dipole bonds). The input of energy required for this process comes in the form of heat, drawn from the solvent. However, there are numerous examples of salts for which the enthalpy of solution is negative under a wide range of conditions.

The statement "[l]ook at salt crystals: they don't have any force of attraction between them" is inaccurate, though. In fact, there is strong ionic bonding. The very fact that the crystals remain solid except at very high temperatures is sufficient evidence of this, and the strength of the bonding in the crystal lattice can be quantified by calculating the lattice energy. It's a mistake to consider any substance in isolation. When you talk about "cutting" $\ce{NaCl}$, what you're talking about is applying physical force to the crystals, exposing them to the complex mixtures of gasses and mositure in the air, and bringing them in contact with, say, the metal in the knife. If left in relative isolation, the salt remains undisturbed, packed in a crystal lattice. A gas, however, which has weak forces of attraction between individual particles, will expand to fill a container without any work being done to it (at least as far as atmospheric pressure allows). A liquid will experience capillary forces, either climbing the walls of a narrow container, resulting in a concavity at its surface, or pooling away from the walls of the container, resulting in a convexity at its surface (which of the two happens depends mainly on the comparative strengths of the forces of attraction between the molecules of the liquid and those of the substance comprising the container).


What is the difference between protein domains, motifs, and supersecondary structures? I was just wondering if there is any difference between supersecondary structure and motifs. As well, is a protein domain an independently folded region of a protein that has a particular function associated with it - that may or may not contain one or multiple motifs.

All supersecondary structures are motifs, but not all motifs are supersecondary structures.

Motifs, in a biological sense, are very similar to fashion motifs. They are patterns that repeat in a lot of different places. Supersecondary structures are motifs that are made of several secondary structures. Some motifs are much smaller (pockets for holding ions are 3-4 amino acids long), or involve no secondary structure interactions (see SLMs).

Regarding the second half of your question, a protein domain (depending on who you ask) is any region or regions of a protein that do something, regardless of how it is folded. Very often, domains can fold correctly on their own. Generally, the required protease/functional assay to determine whether a possible domain is really technically a domain is not usually done. For example, the kinase insert domain found in some tyrosine kinase receptors is considered a domain, despite being a) quite small b) unable to perform its function in isolation (not entirely fair, as its function is to phosphorylate the rest of the receptor).

In sum: protein domains have a couple of competing definitions, so you cannot assume that something referred to as a 'protein domain' is capable of performing its function when separated from the rest of the protein.

Domains can be of nearly any size (due to their functional and not structural classification), but very small and very large ones are rare. They can be made of any number of motifs, or none at all.


UV-Vis Spectra: Multiple peaks for one component I am working on a project in my lab, and we are analyzing capsaicin using HPLC. I am trying to find an appropriate wavelength for the HPLC using a UV-Vis spectrometer. When analyzing the pure capsaicin standard dissolved in methanol, I of course use methanol as a blank. However, I got two peaks, one around 230nm and one at 280nm. Since the system automatically subtracts the methanol, I can't understand why I am getting two peaks.
Anyway, 230 yielded the maximum absorbance, so I used that in my HPLC. I didn't get great results and decided to read some literature on the project. I saw that most other people had a UV-Vis spectrum like mine, but they decided to use 280nm. Why is that?

Different functional groups may show different absorptions and even a single functional group may exhibit two signals in your UV-VIS spectrum, e.g. corresponding to $\pi,\pi^*$ or $n,\pi^*$ transitions.

While HPLC grade methanol is typically much cheaper than acetonitrile, it might not be the best choice here. You might want to have a look at the cutoff wavelengths of different solvents.

I'd definitely give measurements at $\lambda = $ 280 nm a try.


What do the six polymorphs of chocolate look like? Today I learnt that chocolate tempering works by controlling the temperature to ensure the crystals set to one (type V) of the six polymorphs.

What do these polymorphs actually look like? By that I mean: what is the crystal structure? Also, what gives type V its properties?

The actual polymorphic compound is the fat (cocoa butter). They pure phases will all look like white powders and the 'bloom' referred to in your diagram is white cocoa butter extruded by the solid chocolate, and probably the most obvious  visual difference.

The melt-in-your mouth aspect of Form V is going to be because of its appropriate melting point w.r.t your mouth temperature. The decrease in mechanical stability as the melting point decreases is fairly typical of solids (think of plastics softening as you heat them up).

Afraid I don't know why Form V is shinier.

C.f. https://www.sciencedirect.com/science/article/pii/S0969806X04003019


How to determine the efficiency of intermolecular charge transfer? Lets say we have 2 molecules: A and B. Both of these molecules associate with another molecule C. Upon association two complexes form A-C and B-C. We can experimentally see that upon association a new charge transfer (CT) absorption band appears, additionally intensive CT emission appears. Additional evidence for CT can be obtained by DFT calculations (where HOMO is located on A or B and LUMO is located on C). By determining photo luminescence quantum yields (PLQY) it can be seen that A-C has 4 times higher PLQY than B-C.
Are there any ways to determine whether specifically the efficiency of the CT is responsible for this drop of PLQY?

Indeed, if both complex A-C and complex B-C experience CT emission, the efficiency of CT character present in the complexes may be responsible for the difference in PLQY. This however, will be the case if there is no reverse in CT polarity (for example if in both complexes C is the acceptor (and emitter), this may also the reason for porphyrin's comment). Hence if the criteria above are met and You should generally observe the same (or highly similar) emission from both of the complexes. That would also indicate that one species is responsible for the observed emission.

Calculating efficiency of CT. Although I have not seen extensive literature about CT efficiency, there are few papers which try to quantify the CT. The best example I could find is available here: https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.24822. In that paper the authors are trying to quantify CT within a single molecule. Technically the same methodology should also apply for the intermolecular CT in A-C and B-C. Complaring the values obtained should provide some insight in the CT character between both species. Importantly, to do this calculation You need to know exactly how the complex looks. One way to obtain their geometry is to obtain X-Ray single crystal diffraction representations of A-C and B-C in the solid state (this however works only if A-C and B-C do not dissociate upon crystallization).

Godspeed.


how can i clear that secret Veneer Hi
my friend gave me a mysterious 10 dollars and on the back of it there's a secret code hidden but theres something weird on it covering it what i don't know how to erase it. This is the picture i feel like its thicker than an actual dollar this is all i know.
Could anyone help me solve the problem ? How and with what can i erase it?

Use household tools!

Can you bend or fold the note? Is the layer brittle and does break off?
Can you scratch the layer off with a sharp blade?
Place the note between paper (kitchen roll) and use a hot flat iron. Does the layer melt?
Can you wash the layer off using a cotton bud and

water
methylated spirit
nail polish remover


Electrolysis of PbCl2 and use of Nernst Equation The answer is -0.09V.
And the explanation is that the concentration of  $\ce{Pb^2+}$
will decrease.

My question is how will  [$\ce{Pb^2+}$] decrease?
In my opinion, [$\ce{Pb^2+}$] should remain constant. The cell is a voltaic cell meaning that $\ce{Pb}$ should oxidize to  $\ce{Pb^2+}$ to keep the cell running (since  $\ce{\frac{Pb^2+}{Pb}}$ = -0.13V). As a result, the concentration of $\ce{Pb^2+}$ will increase. But since the solution is saturated, no more lead can dissolve keeping the value of [$\ce{Pb^2+}$] constant.
Can somebody please point out the error in my explanation?

I've finally read this question. I don't know what the source is, but I disagree with both the answer and the explanation.

Consider first the reduction half reaction for lead:

$$\ce{Pb^{2+} + 2e- -> Pb}, E^{\varnothing}=-0.13\ \mathrm{V}$$

Now, this is for the reduction at standard conditions, which means that the concentration of lead ion is $1\ \mathrm{M}$. If you decrease the concentration of the reactant ions (and you are, if you consider that the concentration of the dissolved lead chloride is less than 1), then the reaction should be less favorable. This means that the potential should be more negative, not less negative.

This is easy to check based on the form of the Nernst equation as the lead ion concentration enters into the denominator of $Q$. Since $\Delta Q<1$, $\Delta \left(\frac{RT}{nF}\ln Q\right) > 0$.


Why was the Deville process of Alumina production given up in favor of the Bayer process? I guess this is a sort of historical question. The DeVille process seems fairly simple. Does anybody know the history or practical reasons for the switch?

The product of the Bayer process is purer than that of the Deville process.

Trevor I Williams, Aluminium: latecomer to the metal industry, Endeavour, Volume 17, Issue 2, 1993, Pages 89-93


Does the salt bridge in a galvanic cell deplete over time? I searched for the answer for this question on Wikipedia and Stackexchange  a lot but couldn't find same question or a required answer in some other similar question.

I understand that the function of the salt bridge is to complete the circuit and maintain electrical neutrality in both the half cells of a galvanic cell. This is achieved by movement of cations and anion in the salt bridge. the anions move towards anodic half cell and cations move towards cathodic half cell.

My question is-
Are these cations and anions are provided by salt bridge itself and it depletes over time as the electrochemical reaction progresses?

OR

Is salt bridge is just a medium and passing the anions from solution of cathodic cell to anodic cell and passing the cations from solution of anodic cell to cathodic cell?

OR

Both of this happens?

I will be grateful for insight on this topic.

Consider a salt bridge as a part of the cell electrolyte, that is not allowed to mix with the rest of electrolyte.

For a particular ion, there are 2 cases:

The ion is shared with the electrolyte part, from which the ion migrates toward the bridge. Then the ion gets depleted from electrolyte and not from the bridge.
The ion is not shared with the electrolyte part, from which the ion migrates within the bridge toward the other electrolyte part. Then the ion gets depleted from the bridge.

But there is needed to add the bridge maintains ion neutrality, so ions that leave the bridge are replaced at the other bridge end by the same or other ions of the same total charge.

Bridges are not usually used in power cells and cells where significant portion of chemicals gets depleted or react. If they are used, they usually use at least one of ions shared.

Power cells often use a single electrolyte and a diaphragm, soaked with this electrolytes, so ion depletion does not apply.

Bridges are used mostly in cells where accurate potentials matter, to minimize phase potentials, where the current is minimal. Like calomel ( $\ce{Hg2Cl2}$ ) or $\ce{AgCl}$ reference cells, or in laboratory test cells.


What happens when hydrogen bonds stretch? I'm researching about the Mpemba effect and I came across an explanation saying that the effect occurs because at higher temperatures, the hydrogen bonds between the water molecules stretch and store energy and when the water starts to cool, the hydrogen bonds go back to normal and release the stored energy on top of the normal cooling process, that's why hot water freezes faster than cold water.
I'm quite new to hydrogen bonds, so I don't really understand how hydrogen bonds can store energy. I thought bonding was something like a magnet attracting something, the further you get, the weaker the attractive force becomes. Or am I understanding it wrong?
The article that I saw this in is Hot water freezes faster than cold water; physicists solve the Mpemba effect

Yeah this question is actually surprisingly difficult to to answer.  But let me explain it this way.   In thermodynamics you can't have negative temperature.  Thus anytime you are getting heat, you are actually gaining electrons.  When you gain electrons these chemical bonds change and rearrange themselves, which is how they can store energy.
We do not think about this normally because we look at hydrogen and we just see a proton and electron.  But in reality, unless hydrogen is in an extreme environment, it likely has absorbed energy in some manner.
For instance,  if you put a glass of water under concentrated sunlight it will absorb photons, these photons will be absorbed by the moving electrons inside the atom. Which we understand as "heat", moreover If there are more electrons than water molecules than it will start to evaporate because they run out of room.


When did the realistic typesetting of chemical structures start? In many old chemistry books there were limitations on how chemical structures could be represented driven, presumably, by the desire to use available typesetting methods to save costs.

For example, this is a page from a book on industrial chemicals published in the 1930s:

The distorted hexagons are 6-membered aromatic rings but are typeset from vertical and 45° lines with none of the double bonds explicitly shown.

When and how did chemical typesetting become more "realistic" with, for example, proper hexagons for aromatic rings and the correct description of double bonds where appropriate?

this isn't a definite answer but it might show the direction:

I have an old copy of Namensreaktionen in der Organischen Chemie, 2nd ed. by Helmut Krauch and Werner Kunz, published in 1962 at Alfred Hüthig. First edition must have been published in 1960 or 1961.

Most of the arenes in the sketches have the proper shape. Once in a while they seem to be deliberately stretched to align centres that react.

All through the book there's a consistent notation for saturated centres, hydrogen atoms are mostly not omitted.


Is formaldehyde a carbohydrate? Formaldehyde has the formula $\ce{CH2O}$, and the ratio of atoms in a simple carb is $\ce{1C:2H:1O}$. This fits the formula of formaldehyde. When I researched this, I found some sources saying that formaldehyde is the simplest carb, but other sources saying it is glycolaldehyde ($\ce{C2H4O2}$). Is formaldehyde a carbohydrate?

According to IUPAC definition:

  ‘carbohydrate’ includes monosaccharides, oligosaccharides and polysaccharides as well as substances derived from monosaccharides by reduction of the carbonyl group (alditols), by oxidation of one or more terminal groups to carboxylic acids, or by replacement of one or more hydroxy group(s) by a hydrogen atom, an amino group, a thiol group or similar heteroatomic groups. It also includes derivatives of these compounds [...]

But according to Wikibooks

  the term is generally understood in the biochemistry sense, which excludes compounds with only one or two carbons. Natural saccharides are generally built of simple carbohydrates called monosaccharides with general formula $\ce{(CH2O)_n}$ where $n$ is three or more [...]

Because formaldehyde and glycolaldehyde (not a true sugar) has one and two carbon respectively in its structures we can exclude both as a carbohydrate.


What are the possible concentration pathways for a 3-species system? Inspired by this question, which asks if a multi-species system could oscillate around its equilibrium, I thought about the possible concentration pathways the following more narrow specified system could take:

$$ \ce{A ->[\text{$k_1$}] B} $$
$$ \ce{B ->[\text{$k_2$}] C} $$
$$ \ce{C ->[\text{$k_3$}] A} $$

What I was interested in are the ratio of those concentrations $A:B:C$. The initial conditions are $A(0)=1$, $B(0)=0$, $C(0)=0$. (For simplicity, $[\ce{A}] \equiv A$, and analogous for $B$ and $C$)

  What are the possible concentration pathways for different values of $k_1$, $k_2$ and $k_3$?

Of course, we can't display every pathway there is. We have to select a wide variety of rate constants to get a "good map" of the parameter-space. The approach I took was this:

Construct the differential equations

$$\frac{\mathrm dA}{\mathrm dt} = -k_1 A + k_3 C$$
$$\frac{\mathrm dB}{\mathrm dt} = -k_2 B + k_1 A$$
$$\frac{\mathrm dC}{\mathrm dt} = -k_3 C + k_2 B$$

Solve them numerically for every possible combination of 0.1, 1 and 10 for each rate constant and the above stated initial conditions
This leads to 27 pathways, which are plotted in a ternary diagram

The title of the ternary diagram is $k_1:k_2:k_3$, which is 0.1 : 0.1 : 0.1 in the first one.

(click for larger image)


Orientation of halogenation I was going through my chemistry book (RT Morrison and Boyd) and found a statement:

  " the collision frequency of isopropyl and n-propyl w.r.t. Chlorine atoms is the same."

Is it correct because I think the size will matter in this case while both have the same weight?

The statement is:

  This[collision frequency] must be the same for the two reactions, since both involve collisions of the same particles: a propane molecule and a chlorine atom.

It is clear that n-propyl radical and Isopropyl radical have same carbon structure. In the next step the hydrocarbon radical reacts with a chlorine ($\ce{Cl2}$) and since chlorine is symmetrical the collision frequency cannot depend upon it. It should be easy to judge from the following structures, the very minor difference:

Another reason is: Since the difficult step is the abstraction of a hydrogen and the determination of rates should be done while comparing primary, secondary or tertiary radicals, which have different stability.


How to tell which species has the highest ionization energy? I am preparing for my final exam, and I am very confused about ionization energy. An example question would be:

  Between the species $\ce{Ne, Na+, Mg^2+, Ar, K+, $\&$~Ca^2+}$, which one has the highest ionization energy?

I thought that ionization energy increased from left to right in a period and from down to up in a group in the periodic table, so I thought that $\ce{Ne}$ would be the one with the highest ionization energy. But the right answer is $\ce{Mg^2+}$. I assume that this has something to do with the electrons, but I don't know what.

Like most chemical conundrums you need to simplify the problem using some knowledge of chemistry.

First there are two electron configurations. $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$ all have the configuration ${1s^2 2s^22p^6}$. $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$ all have the configuration ${1s^2 2s^22p^6 3s^2 3p^6}$.

For $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$, $\ce{Mg}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Mg^{2+}}$.

For $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$, $\ce{Ca}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Ca^{2+}}$.

Now it is just necessary to compare the ionization of $\ce{Mg^{2+}}$ to $\ce{Ca^{2+}}$. Since the $2p$ orbitals are closer to the nucleus than the $3p$ orbitals, it will be harder to ionize $\ce{Mg^{2+}}$ than $\ce{Ca^{2+}}$.


What are the pros and cons of using Gilman reagent over Grignard reagent to synthesize a hydrocarbon? As far as I know, in order to synthesize a hydrocarbon $\ce{R-R'}$, Grignard reagent or Gilman reagent could be used.

$\ce{RMgX + R'X -> RR' + MgX2}$

$\ce{R2CuLi + R'X -> RR' + LiX + RCu} $

What are the pros and cons of using Gilman reagent over Grignard reagent? Gilman reagent is significantly harder to prepare and involves a reactive organolithium compound. I'm not looking for the pros and cons in terms of cost and yield, but rather specificity:

Are there any $\ce{RR'}$ groups that must be synthesized via Gilman reagent and not Grignard reagent?
Are there any $\ce{RR'}$  groups that must be synthesized via Grignard reagent and not Gilman reagent?

Someone (outside of SE) pointed out to me how pointless it is for chemists to synthesize hydrocarbons using these methods (as they are readily available to purchase). While I agree with him, I'm still interested to find out "theoretically" if Gilman/Grignard reagent is specific to certain R/R' groups.

I am not sure if Gilman reagents are harder to make than Grignard but they are softer nucleophiles and so they are used to perform 1,4-(conjugate) additions on $\ce\alpha,\beta-$unsaturated electrophiles. Grignard reagents will generally go for 1,2-attack instead. See this Wikipedia article for more info:


"Exactly Equal" and "At Least" in electron excitation When we examine IR spectra, we see troughs corresponding to absorption at exactly a specific frequency that corresponds to the energy needed to stretch certain bonds (although translational motions and intermolcular forces can broaden the accepted frequency).

Coordination compounds absorb light at exactly the frequency corresponding to the crystal-field splitting energy.

However, when we talk about energy from photons required to break a  bond, we say "at least" the frequency instead of "exactly" in the examples above. Take $\ce{Cl-Cl}$ for example, we say that the photon must have at least a frequency of 607 THz (or wavelength of no more than 496 nm). There is still a definite energy involved here: 242 kJ/mol to promote an electron from a bonding MO in the molecule to an anti-bonding MO.

As another example, when we talk about ionizing radiation, we say that there is at least a certain amount of energy needed to ionize a compound. For example, my textbook says that a photon needs "at least 1216 kJ/mol" to ionize water. There is still a definite energy level involved: bringing the bonding electron from its negative potential energy MO to 0.

In all of these examples, definite energies were involved. Why is it that sometimes we say that that a photon needs to have exactly the energy needed, and other times the minimum energy needed?

When you ionize or break something, the resulting particles (an electron and an ion, or maybe two atoms, or whatever) would fly away from each other with arbitrary speed. Their kinetic energy would absorb any excess energy of the photon. This is the "at least" case.

On the other hand, when you excite some particle (molecule or whatever) in such a way that it remains in one piece, then you can't send it flying, because the momentum conservation forbids that. This is the "exactly equal" case.

Surely this is a simplification; in fact, a photon may excite a molecule using only a part of its energy, while keeping the rest of it to itself (Raman scattering), but that's another story.


What is the molar mass of a compound with 0.5 atoms in its molecular formula? I'm trying to calculate the theoretical specific capacity of the $\ce{LiFePO4}$ cathode for a fully discharged state, let's say when the cathode has the following notation $\ce{Li_{0.5}FePO4}$. And I'm using the following formula:

$Q=\frac{nF}{3.6M}$ where F is the Faraday constant

The question is slightly related to this thread. The question is, what is the molar mass (M) of $\ce{Li_{0.5}FePO4}$.

Which one is the correct one for finding the specific capacity:

M = 157.751 g/mol, as if Lithium would be taken as a whole?
or
M = 154.285 g/mol, as if only half of Lithium would be considered?

Can anyone help me with this?

You should use 0.5 lithium atoms in the molecular mass. Alternatively you can multiply the formula indices by two to obtain $\ce{LiFe_2(PO_4)_2}$. This formula describes exactly the same compound but it's much more readable, for instance is now clear that the substance contains one $\ce{Li(I)}$, one $\ce{Fe(II)}$, one $\ce{Fe(III)}$ and two phosphate anions per $2n$ electrons transfered.

Here the value of $n$ is multiplied by 2 because you also have your stoichometry multiplied by 2. If you use $\ce{Li_{0.5}Fe(PO_4)}$ to calculate $M$ then the value of $n$ stays unchanged.


Is Acrylamide carcinogenic? Why? Recently concerned with health effects by common chemicals existent in food I've been rather busy reading article after article; and just an interesting one came around:

  Acrylamide (or acrylic amide) is a chemical compound with the chemical formula $\ce{C3H5NO}$. Wikipedia

  Some evidence suggests exposure to large doses can cause damage to the male reproductive glands. Acrylamide is considered a potential occupational carcinogen by U.S. government agencies. Still wikipedia

  A series of case-control studies have investigated the relationship between dietary intake of acrylamide and the risk of developing cancers of the oral cavity, pharynx, esophagus, larynx, large bowel, kidney, breast, and ovary. These studies generally found no excess of tumors associated with acrylamide intake. Cancer.gov

Well, though the references are generally controversial, you can conclude that though nothing is proven by the studies, take care:

FDA: States that acrylamide is a carcinogen of animals, is a human concern, but "should I worry when eating high-temperature fried or roasted food?" "NO."

Food standards agency: Lab proves acrylamide a carcinogen, but real life didn't give up many victims, though this isn't a new issue.

Along with researches done by FAO and WHO, still the carcinogenic concerns aren't very serious.

As a chemist, what reasons would you have provided that acrylamide is carcinogenic or not? Should I really worry about it when I eat french fries?

Acrylamide is a Michael acceptor. All Michael acceptors are potentially carcinogenic because DNA can act as a Michael donor. This Michael reaction can damage the DNA, which can ultimately develop into cancer. Of course how well a certain Michael acceptor reacts with DNA depends on its chemical structure (i.e. the carbon chain and other function groups as well as its conformation), and hence not every Michael acceptor is equally carcinogenic.

If you worry about eating too much French fries I would first be concerned with my caloric  and sodium intake. For that there is solid evidence that it greatly reduces human health if taken in excess.


Dipole-quadrupole polarizability term for induced dipole For describing an induced dipole, I have usually seen the following equation,
$$
P_{i} = \alpha_{ij}E_{j} + \frac{1}{2}\beta_{ijk}E_{j}E_{k}
$$
where $P_{i}$ is the $i^{\text{th}}$ component of the induced dipole moment, $\alpha$ is the (dipole) polarizability and $\beta$ is the first (dipole) hyperpolarizability.

However, I recently read a paper with quadrupole-coupled terms coming in, like the dipole-quadrupole polarizability as shown below:
$$
P_{i} = \alpha_{ij}E_{j} + \frac{1}{3}A_{ijk} \frac{dE_{j}}{dr_{k}}+ \frac{1}{2}\beta_{ijk}E_{j}E_{k}
$$
where in the second term, $A_{ijk}$ is the dipole-quadrupole polarizability term. This equation is introduced for the case of polar molecules in the condensed phase, and in the related paper it is for water.

It appears that the contribution of dipole-quadrupole polarizability is rather small, but maybe larger than the hyperpolarizability.

Could someone explain or refer some literature so as to describe

the importance of the cross-coupled terms like the dipole-quadrupole polarizability, and
if there are some cases where the cross-coupled terms (the second term in the second equation) become negligible?

Reference: Batista, E. R.; Xantheas, S. S.; Jónsson, H. Molecular multipole moments of water molecules in ice Ih. J. Chem. Phys. 1998, 109 (11), 4546-4551. DOI: 10.1063/1.477058.

This question has been around for more than a year without an answer, so probably you have figured it out by now! But anyway, here goes. The standard literature references in this field are Theory of Molecular Fluids vol 1 by CG Gray and KE Gubbins (Clarendon Press 1984) and The Theory of Intermolecular Forces by Anthony Stone (Oxford University Press 2013) and these books explain the various terms in your equation. The dipole-quadrupole polarizability $A_{i,jk}$ may vanish by symmetry (e.g. for centrosymmetric molecules and isolated atoms) but does not vanish for a molecule like water. The corresponding term in your equation is also proportional to the local electric field gradient, so the importance of this term (relative to the other terms) depends on some of the details of the surroundings. In Appendix C of Gray and Gubbins they state

  For the cases we consider in this book the induction $A$ terms are negligible

but clearly there might be some cases in which they should not be neglected.


Entropy as an intrinsic property of matter I would like to ask how we can see entropy as an intrinsic property of molecules, namely as something depending also on their geometry (and not only as gross disorder). Is this what is called the "shape entropy" or there's no relation between the two concepts? Which are the parameters that determine this "geometric" or conformational entropy?

EDIT:
The answer written by Buck Thorn provides an interesting mathematical insight of entropy considering the conformations or "geometries" available to molecules in thermal equilibrium.

Is there also a qualitative analysis of this phenomenon (in terms of the possible conformations playing a role in entropy - the parameters I was referring to in the question)?

You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium.

For molecules in an ideal gas the molar entropy is
$$S=R\ln z +RT\left(\frac{\partial z}{\partial T} \right) _V - R \ln N_A + R$$
Here $z$ is the molecular partition function
$$z=\sum_i e^{\epsilon_i/kT}$$
where the summation extends over all states available to a molecule and $\epsilon _i$ is the energy of state $i$ (states degenerate in energy are counted separately).

If you can identify all the possible conformations ("geometries") available to the molecule and their respective possible energy states, you can use the formula to compute the entropy.

Qualitatively speaking, if you increase the number of conformations you increase the entropy, since the summation in the partition function now extends over more states.

You might invoke Boltzmann's equation
$$p_i=\frac{e^{-\epsilon_i/kT}}{z}$$
to compute the probability of finding a particular molecule in state i.

Reference

McQuarrie and Simon, Statistical Thermodynamics, University Science Books (1999).


Electrolysis of copper without any copper initially in solution Let's say we have two copper electrodes as shown in the image, and there is a solution with copper nitrate. In this case, I know that copper will deposit onto the cathode and copper from the anode will be oxidized and enter solution when a voltage is applied across the two electrodes. However, what if instead of copper nitrate, there was just NaCl (or any other electrolyte that wouldn't cause any problems like precipitation) in solution instead? Would current still be able to flow, and if so, would the electrodes still lose and gain mass as they did before? I am fairly certain that current does flow, but I am not really sure if copper will still deposit onto the cathode and vice versa.

It is a complex problem. Yes, NaCl solution is a conductor, so electricity will pass through the solution. Now in your situation both electrode are made of copper. Since electricity is passing through the solution, electrolysis must occur at both electrodes. You cannot have electrolysis only at one electrode.

Assuming there is sufficient salt in the solution, what will happen at the anode?

Chlorine gas will evolve. I will let you find relevant equations. Chlorine gas attacks copper very quickly and the anode starts to "degrade" quite fast, but will it pass into solution? Nope.

Let us see what is happening at the cathode?

Sodium ions cannot be reduced in water esp. on copper electrodes (on mercury they do), instead water will be reduced. What happens when water is reduced? It turns into hydrogen gas, which escapes, and leaves hydroxide ions behind. This process makes the solution basic.

Now coming back to the anode, since you have hydroxyl ions around, they precipitate copper ions. All you get is yellow orange junk in such a system. Cu(II) is blue so why an orange yellow mess? It is most likely due to Cu(I) compounds. Anode dissolves pretty fast but nothing happens to the cathode.


Separating nickel coin into copper and nickel components I'm trying to create a high school level lab in which we separate the metals in an alloy. The objective is to understand and apply standard reduction potentials, and also to do some extra redox reactions to reclaim solid metals. Plus I think it might be fun!

My plan was to simply submerge nickels in an acidic solution. Theoretically, as long as I don't use nitric acid, the nickel will dissolve and leave the solid copper. I tried it out in the fume hood (1 nickel, 100mL 5M HCl solution), but didn't see the telltale blue/green of nickel ions. I tried sanding the nickel, in case there was an inert oxide layer. For a few minutes, I saw a tiny bit of hydrogen evolution, then the sanded surface turned black.

I read that magnesium is sometimes added to coins for corrosion resistance, which is probably why I'm having problems. Does anyone have suggestions to make this lab possible? Maybe some electrochemistry?

I recommend to dissolve the copper-nickel alloy in nitric acid and determine the dissolved $\ce{Cu(II)}$ by iodometric titration.

Start by weighing the coin, then dissolve it by placing it in nitric acid solution (under a fume hood). Continue to boil the solution for 20 min after complete dissolution of the coin (to remove any traces of remaining $\ce{NO2}$ in soution. The volume of the solution is about $40 \,\mathrm{mL}$.
$$\ce{Cu(s) + 4HNO3(aq)->Cu^{2+}(aq) + 2NO3^- (aq) + 2NO2(g) + 2H2O(l)}$$
$$\ce{Ni(s) + 2H3O^+(aq)->Ni^{2+}(aq) + H2 (g)  + 2H2O(l)}$$
Prepare a fresh solution of sodium thiosulfate of precise concentration $\approx 0.04\,\mathrm{M}$
Add $1.00 \,\mathrm{mL}$ of the diluted copper-containing solution to $15 \,\mathrm{mL}$ of $10\% (W/V)$ potassium iodide solution.
$$\ce{Cu^{2+}(aq) +  5I^- (aq)->CuI(s) + I3^- (aq) }$$
Titrate the yellow-orange suspension with sodium thiosulfate solution until the solution color has become yellow. To detect the end-point add about $1 \,\mathrm{mL}$ of starch solution.
$$\ce{I3^- (aq)  +  2S2O3^{2-} (aq)->S4O6^{2-} (aq) + 3I^- (aq) }$$
Repeat the titration to precisely determine the end-point.


Why are substituted biphenyls optically active? Why are some substituted biphenyl compounds optically active and what kind of substituted biphenyls are optically active . I read this in a book but no satisfactory reasons were given and no examples were given thanks in advance .

In general, compounds without improper axes of rotations are chiral. More concretely (and less correctly), compounds with less symmetry are more likely to be chiral.

Substituted biphenyl compounds generally have little symmetry. Why is this? Steric hindrance causes the two phenyl rings to be tilted relative to each other and serves as a barrier to rotation. This introduces dimensionality to the compound and breaks symmetry in the plane of the ring. If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. A similar situation is observed for spiro compounds.

BINAP is an example:

Remark. Most textbooks mention chiral centers as a sufficient (but not necessary) condition for chirality, and fail to emphasize other possibilities. Substituted biphenyl compounds, for example, may be chiral even without chiral centers.


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

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

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

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


How is using Jones reagent different than Potassium dichromate in Sulfuric acid for the oxidation of alcohols? I was taught at an institution I study in that using $\ce{K2Cr2O7}$ in the presence of $\operatorname{dil}.\ce{H2SO4}$ will oxidise $1^{\circ}$ alcohols to aldehydes, and $2^{\circ}$ alcohols to ketones. This is supported in here too.

At the same time, I came across Jones reagent ($\ce{Na2Cr2O7}$ in the presence of $\operatorname{dil}.\ce{H2SO4}$) on Khan Academy. But this says that it is an extremely strong oxidising agent and can oxidise Aldehydes to Carboxylic Acids too (thus giving acids on the oxidation of $1^{\circ}$ alcohols).

I do not understand how the above two reagents have different oxidising abilities. Is my premise itself wrong or is there any reason for it to have different oxidising abilities?

Sodium or potassium dichromate or chromium trioxide in dilute aq. sulfuric acid all give rise to Jones reagent. This is used, though less than it used to be due to toxicity and cost issues, to oxidize secondary alcohols to ketones and primary alcohols to carboxylic acids via the aldehyde hydrate. In the case where the aldehyde does NOT form a hydrate with water (e.g. benzylic and allylic aldehydes), then that aldehyde will be stable to the conditions and be the major reaction product. Aliphatic aldehydes will generally give carboxylic acids.

Read more here.


Lyophilization Advantage I've been reading about freeze drying and how it increases the shelf life of certain foods and is overall a good preserving technique.

My question is with regard to the actual process. I understand that freeze-drying involves freezing the actual material and then proceeding to dry it by placing it in a vacuum. This way the ice is turned into vapour without going into any liquid phase. My question is why is avoiding the liquid phase necessary and beneficial? When it comes to vaccines, foods, and polymer storage for example, why is it advantageous? I don't intuitively understand the reasoning behind why lyophilization is so favourable.

I don't think the point of freeze drying is to avoid liquid water -- I mean, there's liquid water already in the substance to be dried -- but to avoid heat.  Ordinarily, to dry something out to preserve it, you need to heat it to speed up the evaporation of water, certainly if you want to get rid of the water quickly (which you would want to do in any commercial setting, where time = money).  But heating causes all kinds of changes to foods, and while some foods are fine with it, others  suffer degradations of perceived quality, e.g. undesirable changes in color and texture.   The point of freeze-drying is that you can get rid of the water at a low temperature.


How many peaks would there be in H1 NMR of ethene? Since the bonds can't rotate, would there be 2 peaks? Likewise, how many peaks would there be in trans-but-2-ene and cis-but-2-ene?

You need to consider whether the protons in ethene are chemically equivalent, magnetically equivalent, or both. The lewis structure of ethene:

Because of the symmetry of the planar ethene molecule, all H atoms are chemically and magnetically equivalent. The spin system is therefore A$_4$. The signal is not split because the environment is the same for all H atoms. The signal would therefore be a single peak. Compared to an "average" methyl group, the ethylene protons are less shielded, and would appear at a higher chemical shift than methyl groups.


Are there general rules for Base + Metal Compound reactions like there are for Acids? Are there general rules for Base + Metal Compound reactions like there are for Acids?

For example, acids have some general rules.

**

Acid + Metal => Salt + Hydrogen Gas

Acid + Metal Carbonate => Salt + Water + Carbon Dioxide

Acid + Metal Oxide => Salt + Water

Acid + Metal Hydroxide => Salt + Water

**

Are there equivalent rules for bases?

Base + Metal => ?

On the same level of generalization, the rule is "no reaction", but this rule has quite a few exceptions (probably more than the rules you quoted above).

Water-soluble bases would react with amphoteric metals to produce hydrogen and the corresponding salt:
$$2\ce{Al + 2KOH + 6H2O \to 2KAl(OH)4 + 3H2}$$


Why is the exo product formed in the Diels–Alder reaction between furan and maleic anhydride? For most Diels–Alder reactions, the major product is endo because there are favourable interactions between the newly forming pi bond and the electron withdrawing groups of the dienophile.

Why is the reaction between furan and maleic anhydride an exception? From what I understand, the carbonyl groups of the anhydride are electron withdrawing.

Great question!  It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels–Alder is a reversible reaction. In this case, the exo product is thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$.

So, even though the predicted endo product is the kinetically favored product (and formed first), thermodynamics eventually takes over and the more stable exo adduct is what you wind up with. More details are given in J. Org. Chem. 1978, 43 (3), 518. What a nice example of competing kinetic and thermodynamic pathways.


Dangers of expired bleach? I have some unopened Clorox bleach bought over a year ago (at $8.25~\%\ \ce{NaOCl}$) and read on Wikipedia that bleach naturally decomposes according to

$$\ce{3 NaOCl (aq) -> NaClO3 (aq) + 2 NaCl(aq)}$$

Now, while there potentially might be some further decomposition as

$$\ce{2 NaClO3 (aq) -> 2 NaCl (aq) + 3 O2 (g)}$$

Would I be at risk of filling the air with toxic chlorate intermediates after using this bleach to clean up an ultrasonic ("cool mist") humidifier?

There is a old paper which discuss the mechanism of decomposition of bleach. It says the major product is chlorate($\ce{ClO3-}$)

  The decomposition of sodiun hypochlorite has been re-examined. The
  results show that Foerster and Dolch's mechanism of the decomposition
  to chlorate and chloride is correct; they postulated a slow
  bimolecular reaction to chlorite and chloride, followed by a faster
  reaction of chlorite with more hypochlorite. [...] The rates are such
  that at 40 C. A solution of sodium hypochlorite will contain about lyo
  as much chlorite as hypochlorite.[...]. No signs of specific catalytic
  effects of sodium chloride, hydroxide, or carbonate could be observed,
  and it seems probable that earlier reports of this were due to
  variations in ionic strength. The decomposition to chloride and oxygen
  has been measured and is a unimolecular reaction, which is possibly,
  but not certainly, uncatalyzed.[...]

  $$\ce{2NaOCl -> NaClO2 + NaCl}$$
  $$\ce{NaOCl + NaClO2 -> + NaClO3 + NaCl}$$
  $$\ce{2OCl- -> 2Cl- + O2}$$

But since this paper is in the 50's, most of the information would  be outdated or misleading. So, here is the result.

Wikipedia says:

  It decomposes (autoxidizes) when heated, forming sodium chlorate    and sodium chloride:
     $$\ce{3 NaOCl(aq) → NaClO3(aq) + 2 NaCl(aq)}$$
  Sodium hypochlorite decomposes with increasing temperature and under    the influence of light and such metals as copper, nickel, or
  cobalt:
     $$\ce{2 OCl⁻(aq) → 2 Cl⁻(aq) + O2(g)}$$

There are two pathways of decomposition of bleach. One is forming chlorate and other one is forming chloride. But the 1st reaction is dominant one i.e chlorate is the major product. But the formation of chlorate actually requires heat(to be precise 303-323 K).

The 2nd reaction also requires heat(343 K) and also some catalysts:

  Typical metals for nickel and copper are in the range of 50 ppb or
  less when sub-micron filtration is provided but many times, levels in
  the range of 300-400 ppb for nickel and 50-60 ppm for copper can be in
  the product which will create $\ce{O2}$ and increased decomposition.(source)

So, the conclustion is that the formation of chlorate or chloride do require heat. But there are 2 reaction that indeed happens when bleach is remained untouched for a long time i.e reaction with carbon dioxide and moisture.

  If not properly stored in airtight containers, sodium hypochlorite
  reacts with carbon dioxide to form sodium carbonate:

  $$\ce{2 NaOCl(aq) + CO2(g) → Na2CO3(aq) + Cl2(g)}$$

  $$\ce{NaClO + CO2 + H2O → NaHCO3 + HClO}$$

  Sodium hypochlorite react with carbon dioxide and water (moisture) to
  produce sodium bicarbonate and hypochlorous acid. Sodium hypochlorite -
  saturated solution. The reaction proceeds at room temperature.(chemiday)

Hypochlorous acid is somewhat unstable so it might decompose to give chlorine fumes or further react with hypochlorite to form chlorate.

$$\ce{OCl- + 2HOCl -> ClO3- + 2HCl}$$


How do these solutions "react"? If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right? But since both are soluble no precipitate would form correct? So does that mean that no reaction occurred? Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

You picked a difficult question because this question involves acid-base reactions. It is no longer a simple "double displacement" reaction where the cations and anions "swap partners".

  If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right?

No, not really. You have to realise that the protonation of $\ce{CO3^2-}$ occurs in two stages: firstly, to $\ce{HCO3-}$, and then only to $\ce{H2CO3}$. Also, the $\ce{NH4+}$ ion is acidic, and can react with water to form $\ce{NH3}$.

On top of that, $\ce{H2CO3}$ will spontaneously decompose into $\ce{H2O}$ and $\ce{CO2}$, the latter of which is released as a gas.

Since these species are all soluble in water, and soluble ionic compounds exist as ions in water, a more appropriate way to think about this question is "what species are present", not "what ionic compounds are present". Species is just a general term that includes ions as well as molecules (since, for example, $\ce{CO2}$ and $\ce{NH3}$ are not "ions").

  But since both are soluble no precipitate would form correct?

That is correct.

  So does that mean that no reaction occurred?

No, there is a reaction happening because of the $\ce{CO2}$ gas formed. You will observe a lot of effervescence if you were to mix these two together.

However, if this question did not involve $\ce{CO2}$... Let's say you are reacting $\ce{NaCl}$ with $\ce{KBr}$. Since all the possible products ($\ce{NaBr}$ and $\ce{KCl}$) are soluble, you could say that in this case, no reaction would occur.

  Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

No. You could say it will never be precipitated out. However, $\ce{NO3-}$ can also react in other ways apart from precipitation. For example, it can be used as an oxidising agent. One famous example is the dissolution of a copper coin in concentrated nitric acid.


If I add some KOH, where will the reaction shifts? I just got a final test and one of the question is:

$\ce {Ni^2+}$ + 6$\ce{NH_3}$ $\ce {<=>}$ $\ce{[Ni(NH3)6]^2+}$
If we add some aqueous $\ce{KOH}$ after the reaction reach equilibrium, where will the reaction shifts?

The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:

$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$

The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, source). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the stability constants of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.


Could dimerization account for weak IR signals? How would you interpret the IR spectrum of your product if it's peaks match the literature spectrum, but the signals are weak? Could the argument that the sample was very dilute and lots of water was in it work as a viable explanation?

What about in a scenario where the experimental IR spectrum has all necessary peaks but is missing a key peak in, for example, the 2000-3000 region?

I synthesized acetylglycine and am trying to see why I am missing a peak.
For the purposes of this lab, it's not necessary to have absolute purity of compound but for my own curiousity, i'm trying to figure it out.
I'm trying to consider all the physical reasons for why my IR may not match the literature, while my melting point ranges are sufficiently accurate, meaning my product shouldn't have too many impurities.

I believe a dimerization could have taken place where for example, a carboxylic acid terminal of one molecule hydrogen bonds with the carboxylic acid terminal of another. Carbonyl oxygen to hydroxyl group, hydroxyl to carbonyl oxygen.
This could lead to a suppression of the hydroxyl (-OH) signal.

Intensity is (in general) proportional to sample concentration, so yes, if you are performing IR in solution, then an overly dilute sample would have less intense peaks. But solution IR is somewhat uncommon; I suspect you are either doing IR on your sample in neat form (using an ATR stage) or in a pressed KBr plate? If the latter, you may not have used enough sample.

Regarding your missing peak: sometimes peak intensities can change, depending on your sample prep. For example, if the peak in question corresponds to proton capable of hydrogen bonding (e.g., -COO-H, -NR2-H, -O-H, etc.) then the sharpness and intensity of the peak will depend on the extent of H-bonding your sample, which, in turn, can depend on things like sample concentration, the presence of water, etc. Sometimes these peaks get broadened to the point of being invisible if your sample is not very concentrated.

If your missing peak corresponds to a C-H, C-C, C-O, C-N, or some other non-labile bond, then there really isn't an explanation other than your compound not being what you think it is...


How to deposit a durable layer of sensing molecules in a PVC matrix on the tip of an optical fiber? Here is the situation that I have been tacking for weeks and I have still not understood.

We are trying to understand how to do this:

  This mixture for the preparation of the "LI–PVC" was obtained from a
  batch of 80 mg of poly(vinyl chloride), 160 μl of tributylphosphate
  and 20 mg of Lophine dissolved in 2 ml of freshly distilled
  TetraHydrofuran (THF).

Paper:
http://www.sciencedirect.com/science/article/pii/S014381661400089X
which refers to :http://www.mdpi.com/1424-8220/8/2/847

We have no idea how to do this, since we dont have any step/by/step instruction, reaction time, environment we need to maintain for this.

How do we go about identifying this process and replicating this mixture. This is very critical to us. Any help on this matter would be of immense help, if you have a resource which may have be able to shed some light on this. It would be of help as well.

Many Thanks

We have no idea how to do this, since we dont have any step/by/step instruction, reaction time, environment we need to maintain for this.

Let's redefine the general task:

How to deposit a layer of 2,4,5-triphenylimidazole (lophine) on the tip of an optical fibre?

Requirements:

The layer must be permanent/durable and does not wear or wash off during measurements of the nitrate concentration in drinking water.

The workflow therefore includes:

How to prepare the fibre?
How to deposit one (or several) layers of lophine?
How to cure the layers to make sure that they do not wear/wash off?

As far as the preparation of the fibre is concerned, your primary paper (Optics and Lasers in Engineering, 2014, 60, 38-43) gives some hints on page 39:

  To deposit the sensitive layer on
  the optical fiber, the coating was removed in a 1 cm section, and
  the exposed cladding was sterilized. Isopropyl Alcohol Level was
  used for cleaning the optical fiber in special cleaning cloths and
  immediately the LI–PVC was introduced [...].

Apparently, lophine is deposited on the fibre in a PVC matrix and you have cited the composition of the mixture used.

How is this actually done?

An earlier publication by the same authors gives some directions! It's reference 19 in the paper that you cited!

In Sensors, 2008, 8, 847-859, the describe authors describe the dipping method in chapter 2.2. and provide further references.

In summary, the prepared fibre is

dipped into the solution (lophine, PVC, TBP, THF) and then
cured by placing the fibre in an oven for 15 minutes at 80 °C.

These two steps are repeated as often as necessary to obtain the required total layer thickness.