text:
Rearrangements; Carbocations; E1 Mechanism
If the leaving of the leaving group creates a secondary carbocation, and a tertiary carbocation has the possibility of being formed through a 1,2 methyl shift:
1) Will the methyl shift? I'm guessing the answer is an emphatic yes. Tertiary carbocations are more stable than secondary carbocations.
2) Will there still be some secondary carbocation left over? I.e. will there be both major and minor products? I'm guessing yes.
I ask because I was watching a video on E1 elimination and the narrator ignored the possibility of a shift, and I was wondering if this was because he forgot or because it was irrelevant.

1) Will the methyl shift? I'm guessing the answer is an emphatic yes.
  Tertiary carbocations are more stable than secondary carbocations.

Yes, if that is the only way to generate a tertiary carbocation and the C-Methyl bond can more or less line up with the carbocation p orbital, then the methyl would likely shift.  If there are other alkyl groups attached to the carbon initially bearing the methyl group, they could shift too.

2) Will there still be some secondary carbocation left over? I.e. will
  there be both major and minor products? I'm guessing yes

An equilibrium exists between the various carbocations.  At 300 K, an energy difference of 1.4 kcal/mol results in a power of ten difference in the equilibrium (e.g. from 50:50 to 90:10).  So it really depends on the energy difference between the various carbocations involved, as well as the rate they are captured or proceed to the next step.
Finally, if there is a hydrogen on the carbon adjacent to the carbocation center, then elimination to create a double bond becomes a competing pathway,


text:
Does ice melt faster when submerged in water?
Suppose you have a container with a drain at the bottom full of frozen H2O which you leave a room temperature to melt. Will the ice melt faster if you drain the water as it melts?
I would expect that the water, being warmer than the ice, will speed the melting of the ice. However, I would also expect that draining the water would remove allow more air to enter the system and heat the ice faster. Which explanation is correct?
Related question: if the container is a cooler to fill with warm beverages, will the beverages cool faster in a wet or dry ice-filled cooler?

Assuming the water and air are the same temperature (and that the air is mostly stationary) then the ice will melt quicker in the water. This is due to water having a higher thermal conductivity and heat capacity. In fact, this is easy to test: place a cocktail stick through an ice cube and lay it on top of a glass filled right to the top with (room temperature) water. The submerged half will melt quicker than that on top.
Regarding the second question: A warm object will cool quicker in icy water than in just ice for the same reason above but also because of increased surface area contact. However, icy water will be warmer than just ice. So the item will potentially be able to achieve a lower temperature in just ice (slow and steady wins the race).

I think the main property at play is the thermal conductivity, rather than heat capacity. The faster the medium can conduct heat to the ice, the faster it will melt.

Assume that the water is colder than the air because it is recently-melted ice and no room-temperature water has been added to the ice.

@buckminst You're right, both are relevant - I've corrected the answer.

Actually, the water is the same temperature as the ice, at the melting point. Any heat transferred into the water gets transferred to the ice for it to melt. So any ice in the container will melt where the heat is applied. Your example of an ice cube on a cocktail stick is not correct since the water in the glass is not at the freezing/melting point. So the ice melts fast to cool the water down to the freezing/melting point.

The ice in the drum will melt faster if the water is not draining out.
This is because any air space between the drum and the ice slug will be a good insulator.
As mentioned in a comment all the water will remain at 0 °C until all the ice is melted.  Heat from the outside will cause convection currents in the water and transfer heat faster to the slug and melt more.  Keeping the water in will allow for a maximum of drum surface in direct contact with the internal cold water and maximise the external heat transfer.
There is an added bonus if the drum is full, the ice slug will soon float free and rise up to touch the lid and gain further external surface area to assist with melting at the lower bulk ice temperature witch will have a slightly better thermal gradient and so be faster.
If the air is cold but the floor is hot then heat is mostly transferred through the bottom then it will be faster to drain the water.
Bonus answer:
In a cooler you want the maximum of cold thermal mass but the minimum of heat transfer.  As the sides are an insulator the thermal mass is an advantage.  If the sides were not an insulator like a tub or a drum then draining the water away from the warm tub walls is advisable but will only help while the ice has not melted to conform to the tub shape.  The problem is that after the ice has melted you no longer have the small benefit of the cold water (that you drained).  This means that you need to insulate to slow the heat flow and gain the benefits of the ice and the water.  Using a foam yoga mat inside or outside the tub will improve things a lot in conjunction with some wood under it.
Keep in mind that melting ice is MUCH better at keeping drinks at 0 °C than water at 0 °C.  Also ice water slurry is ideal for fast cooling of a warm beverage thrown into a tub or cooler as the water absorbs the heat faster due to better surface contact.

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text:
How to detect if a beverage has Sucralose or aspartame?
There is a drink that claims to be sweetened with stevia, but I think I taste Sucralose in there even though it is not listed on the ingredients.
Is there any easy way I could test for the presence of Sucralose to confirm my hunch?
How about Aspartame?
By easy, I mean not needing access to something like a gas chromatograph.

If it's something with gas (e.g. Coke), try inserting the Mentos to it, if it does (containing Aspartame), if it'll blow up:)

I'd very much like to find a "field test" for aspartame.  I have an allergy of sorts to aspartame and I've observed that I often develop symptoms after consuming a "fountain" drink of a supposedly sugar-sweetened beverage.  I suspect that through carelessness or perhaps intent the pipes are often crossed in beverage dispensers, resulting in a sugar/aspartame mix which is difficult to detect by taste.

I have heard that you can separate diet coke from regular coke by putting cans of each in a bucket of water. The regular coke sinks while the diet coke floats. From this it might be interesting to measure the specific gravity of the drink you are suspicious of. Also my wife brought home Capri Sun claiming to have a reduced amount of sugar perhaps 8 grams vs 16 grams in a 6 oz serving. I made two 6 oz glasses of water put 8 grams of sugar in one and 16 grams in the other. Guess what the Capri Sun tasted exactly like the water with 16 oz of sugar. Also there is a sweetener called Neotame. It is

@AllanRydberg This won't detect which artificial sweetener is used. The density is a function of the sugar content and all artificial drinks with artificial sweeteners contain a lot less of the sweetening ingredient (they are all hundreds of times sweeter than sugar). So they will all pass the test vs sugar-sweetened drinks.

@kenorb This is hardly a reliable test. The mentos effect is worse in aspartame drinks because aspartame is a surfactant. But so are some alternatives. And even sugary coke will show some effect.

If your drink contains a sweetener other than the one it claims to have, then someone has broken the law. Labelling regulations are very strict.

Here's the structure of sucralose

and here is the structure of aspartame

They are very different one from the other and should be easily separable.  You can do a "poor man's" separation using paper chromatography.  Aldrich sells both of these materials, but I'm sure you could find them much cheaper at a local Health or Health Food store.  Get the samples and some filter paper.  You'll need some solvents, a tank (preferably glass) with a lid, some filter paper coated with a fluorescent material (as an easier, but more expensive alternative, you could always buy some TLC plates from a chemical supply house) and a black light.  Put one of the solvents in the tank so that it is 1/8 - 1/4" deep.  Put the top on the tank and let it equilibrate. Make dilute solutions of your known samples using the same solvent that's in the tank.  Using the tip of a needle or a toothpick place a small spot of each of the dissolved materials on the filter paper, put the spots about 1" apart and high enough above the bottom edge of the filter paper so that they will be above the solvent level when you place the filter paper in the tank.  Give the spots a moment to dry and then put the spotted paper in the tank.  Watch as the solvent level climbs up the filter paper (or tlc plate) and remove the filter paper when the solvent is about 3/4 of the way up the paper or plate.  Let the paper or plate dry and then examine it with the black light.  The paper should fluoresce under black light except where the two spots have migrated up to.  The spots cover the fluorescent material and will appear as dark spots under the black light.  If you got a good separation between the two spots stop here and repeat the experiment putting a spot of your diluted drink on a new piece of filter paper along with spots of the two known materials.  Rerun it in the tank and see if there are any spots in your drink that match either of the spots from the two known compounds.  If there is no match, then sucralose or aspartame are not in your drink.  If there is a match, then whichever spots matched might be in your drink - more advance experiments will be needed to further analyze the situation.  If you didn't get a good separation between the spots for the two known compounds to begin with, start over and repeat with different solvents until you do get a separation.

This seems like a good idea; TLC didn't even occur to me. That said, I would imagine the plates would actually be pretty difficult to interpret, especially with the multitude of ingredients that might be present, and if the concentrations of aspartame and sucralose are as small as I would expect them to be.

@Greg E You're probably right.  I once had to use a 3 foot strip of paper to separate some compounds I was interested in.  Wasn't it Archimedes who said something like - Give me a big enough piece of paper and I can separate anything in the world?

I shudder to imagine the size of the developing jar.

Would sucralose even show up under UV in TLC? Otherwise good suggestion.

@Jan  Sucralose won't emit under a typical UV lamp.  That's why I said above to use a plate or paper coated with a fluorescent material.  The sucralose spot will simply cover the fluorescent plate material and appear as a black spot,

I’m assuming you’re thinking of the typical $\lambda = 254~\mathrm{nm}$ lamps. Under those, I could typically only see my products (some of which included silyl-protected and otherwise modified monosaccharides) if they had something in the molecule that absorbs at that wavelength. My sugars would simply be transparent, i.e. let enough light through that the background fluorescence seemed unaffected; they had to be stained to show up (e.g. permanganate).

I think it would be not an easy task. Aspartam does not contain any special groups, just a carboxyl, amide, phenyl... Nothing that you would not find in the beverage ingredients. In case of sucralose maybe you can search for some tests for chloroalkanes.
"the reaction between silver nitrate and certain organic halides (those compounds containing chlorine, bromine, or iodine) results in the formation of a silver halide precipitate as a positive test for organic halides." (http://www.britannica.com/EBchecked/topic/108529/chemical-analysis/80788/Classical-qualitative-analysis)
But in your case I would expect the product to contain free chloride ions and therefore giving false positive result.

I agree. This would be very difficult (if not impossible) without analytical equipment (HPLC, [Py]GC-MS, etc.). One problem is that most artificial sweeteners (sucralose and aspartame included) are two-to-three orders of magnitude sweeter than sucrose, so they're present in low concentrations even when added deliberately. Furthermore, there are probably numerous other compounds present in the mixture (flavoring agents, preservatives, pH buffers, etc.).

A trick my mother uses is to weigh it, but this only works in some very select cases and tends not to be conclusive in all but the most extreme cases.
Because some modern soft drinks are up to 1g sugar per 10mL volume, they tend to be a fair bit denser than water. Since aspartame is much, much sweeter than sugar, you usually only need a tiny bit (according to Wikipedia, only 180mg per 12 fl oz) and it doesn't increase the density as much.
Of course, this only has a hope of working if you're comparing very sweet beverages (e.g. Mountain Dew and its 0.13g/mL of sugar), and isn't terribly conclusive on iced teas or lightly sweetened beverages. Still, it's an interesting technique that doesn't rely on trying to chemically separate the two substances.
Note: If you actually do try this, make sure that both beverages are completely flat (decarbonated) and at the same temperature.

The artificial sweetener aspartame contains 2 amino acids, aspartic acid and phenylalanine. aspartame is hydrolyzed by heating it up in hydrochloric acid and the hydrolyzed product is then analyzed  using paper chromatography and the standards of these specific amino acids to demonstrate  their identity. A UV lamp or nynhydrin spray to locate.
first hydrolyze the aspartame by heating it in dilute hydrochloric acid until turns black... then mix it with charcoal and filter mixture in order to make a colorless solution. next ethanol:water:ammonia chromatography for 1 our under glass not touching the walls. Finally hang in a fume cupboard the paper sprayed with the locating agent nynhydrin and baked in an oven at 110 celcius for 5-10 minutes. finally the spots can be seen if the locating agent was used, if not they can be seen using a uv light.

I’ld be pretty sure that whatever the drink is there will be other amino acids in there. Unfortunately, that renders this solution impractical.

Stevia is an extract from a plant and contains the sweeteners stevioside and rebaudioside. Both of these compounds are very sweet (about 300 times and 400 times sweeter than sucrose), but stevioside has a bitter after taste. It is very likely that what you think is sucralose is rabaudioside, since it leaves a very sweet after taste.

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Calculating Ka values from std. state thermodynamic data
My professor showed us a sample calculation of the Ka value for a strong acid. He used standard state thermodynamic data.
I have an issue with this:
1) Wouldn't this give us an understated Ka value?
2) Standard state conditions start with 1 molar solutions of solute. So in writing the reaction equation of an acid with water and by using that equation to find the delta G std. of reaction and then using the relationship between delta G and K to find K, wouldn't we be running into a sort of "common ion" effect issue? There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions. Finding the K of an acid dissolving in water that has a pH of 0.0 doesn't give us a Ka value (or does it?)
EDIT to address Martin:
This is what he did:
$\ce{HNO_3 + H2O ->H_3O^+ +NO_3^-}$
Find ${\Delta G^o_r}$ for above reaction.
Use ${\Delta G^o_r=-RTlnK}$ to solve for K.
My question: Is this K that you find a $\ce{K_a}$ value?

It's worth remarking that the common ion effect is a consequence of the constancy of the equilibrium constant.  If you add nitric acid to an already acidic solution, the result is higher in H+ and lower in NO3- than if you added the acid to water, but the product of these two concentrations divided by the concentration of undissociated nitric acid will be the same in the two cases.

Ohh I understand now! Thanks; that made a LOT of sense.

The definition of $K_a$ for $\ce{HNO3}$ is $$\frac{[\ce{H3O+}][\ce{NO3-}]}{[\ce{HNO3}]}$$ (concentrations in mol/L), which is precisely the $K$ that you get from the $\Delta G_r^0 = -RT\ln K$ formula, assuming you use the right free energies (which is not a trivial matter since they must take into account the interaction of each species with surrounding water molecules, an effect known as solvation).
You may be wondering why the concentration of water does not appear in the equation for $K_a$ (or more generally the equation for $K$ in a reaction in aqueous solution).  Because that concentration remains essentially unchanged throughout, it is kept out of the defining expression for $K$, for simplicity. From the point of view of $G^0$ this is correct since the standard state to which the $G^0$ of water refers is pure liquid water, while the standard state for the other three species is 1M concentration.

There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions.

You're correct that for specifying the $G^0$ of $\ce{H3O+}$ the standard state is 1M. But the standard state of $\ce{H3O+}$ is not where the reaction takes place. The same $K$ will be found whether you add the $\ce{HNO3}$ to pure water or to a solution of $\ce{HCl}$ of pH zero -- the final concentrations of $\ce{H+}$, $\ce{NO3-}$ and $\ce{HNO3}$ will be different in the two cases, but $K$ will be the same.

You can find a good discussion of the "standard state" for ionic species in solution here:
http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set10/lecture10.doc
I'll quote a couple of paragraphs.  You can see from this discussion that the calculation of these standard free energies is based exactly on the equilibrium constants, so the formula your professor used works almost by definition!

As always, we can talk about the Gibbs free energy, enthalpy and entropy of such reactions. However, if we try to break these reactions down into the Gibbs energy, enthalpy and entropy of formation of each species in the reaction, we end up defining things like the Gibbs energy of a sodium cation in solution. Yet there is no way to generate just a sodium cation in solution (that would violate charge balance). Thus, we can only determine the Gibbs energy of formation of pairs of ions. This does not cause any real physical chemistry problems, but it causes a book keeping problem: how do we write down the Gibbs energy of formation of a Sodium cation in a book so that we can use it in any reaction where sodium cations are formed?
The answer is that we come up with a sort of standard ion, $\ce{H+}$ (really something more like $\ce{ H3O+}$) and we define the free energy of formation of all other ions relative to this one. For example, I can measure the Gibbs energy change when I dissolve HCl in water to form $\ce{H+}$ and $\ce{Cl-}$.
  $$\ce{HCl}\text{(g)}\to\ce{H+}\text{(aq)}+\ce{Cl-}\text{(aq)}
 \quad\Delta G^0{}_1
$$
  I can now just define the Gibbs energy of formation of $\ce{H+}$ as zero and then determine the relative Gibbs energy of formation of $\ce{Cl-}$.
  $$\begin{align}&
\Delta G^0{}_1=\Delta G^0_{\mathrm{F}}(\ce{H+})+\Delta G^0_{\mathrm{F}}(\ce{Cl-})-\Delta G^0_{\mathrm{F}}(\ce{HCl})
\\&
\Delta G^0_{\mathrm{F}}(\ce{H+})=0\quad\text{(definition)}\\&
\Delta G^0_{\mathrm{F}}(\ce{Cl-})= \Delta G^0{}_1+\Delta G^0_{\mathrm{F}}(\ce{HCl})
\end{align}
$$

Standard state for solutions means having 1 molar solutions. Or in terms of activities, that means activity = 1.

Quite right; I had misunderstood your comment.  I've edited the answer to clarify what I meant.

They were standard free energies.

From the national bureau of standards.

So does this have any bearing on the found K value?

@Dissenter - I guess what I meant was whether he explained how these constants are derived.  I've added text to my answer to explain that.

@Dissenter - Yes it does have a bearing on the value of K.  It means that the definition of K (which is the same definition as K_a) is consistent with the definition of the standard free energies.

@Dissenter -- I think it's up to you now to explain why you think there is a problem.

My issue is that this delta g and K are for the dissociation of nitric acid in a solution already at pH of 0.0. That's not how Ka is defined.

@Dissenter -- I think I got what's worrying you! The same $K$ will be found whether you add nitric acid to pure water or to a highly acidic solution (or a basic one).  That's the "magic" of equilibrium constants -- they don't change much for a wide variety of concentration values.  (They change a bit -- then you have to use activities instead.)

I am puzzled a little bit. The equilibrium constant is defined via $$\Delta G^\circ = -RT\ln{}K,$$ which resolves to $$K=\frac{a(\ce{H3+O})a(\ce{NO3-})}{a(\ce{HNO3})a(\ce{H2O})}.$$ Now if we assume reasonably diluted systems we can write concentrations $$K=\frac{[\ce{H3+O}][\ce{NO3-}]}{[\ce{HNO3}][\ce{H2O}]} = \frac{K_a}{[\ce{H2O}]}.$$ Assuming that the concentration of water does not change, because it is diluted enough, wouldn't that mean that the $K_a$ is off by a factor of $(55.5~\mathrm{mol/L})^{-1}$?

Hi @Martin - I addressed this briefly in the answer. Solvent concentration is not part of the equilibrium constant, by convention. The standard values of $G^0$ for aqueous species are calculated using the same convention, so everything balances out.

Can you point me to the source of this convention, please. It is not given in the goldbook page for equilibrium constant.

Sure it is. It says there $K^0=e^{-\Delta_tG^0/RT}$. So the two are
consistent by definition. The rest follows from the observations in my
answer: the standard state for water is pure water, for the other
species it is 1M conc; for ions, there is a further
normalization relative to $\ce{H+}$. That is, tabulated
values of $G^0$ assume the fiction that the species
of interest is $\ce{H+}$ not $\ce{H3O+}$, and the reaction might
just as well be $\ce{HNO3 \leftrightharpoons H+ +NO3-}$. See problem 10
at http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php
for an explicit example.

Interestingly (and coherently), I see that the Golden Book respects the same fiction in
its definition of the acidity constant: "The equilibrium constant for
splitting off the $n$th proton", rather than, for example, "for
transfering the $n$th proton to the solvent".  See http://goldbook.iupac.org/A00080.html


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Can very dilute acids exposed to air become more concentrated as water evaporates?
If a very minute quantity spilled acid (e.g. $\ce{H2SO4}$, though the question would apply to other acids as well) is cleaned up with water (e.g. yielding a $0.0001~\mathrm{M}$ solution), and some of the water (dilute acid solution) is left behind, will the resulting solution become more concentrated as the water evaporates?  Would there be any danger that a drop of an initially-harmless concentration might become a visually-unnoticeable quantity of acid that was strong enough to pose a contact hazard?  Would wiping an area down with a moderately-dilute solution of a weak base such as sodium bicarbonate be an appropriate safety precaution, or would it be unnecessary if any acid droplets that might exist on a surface have been thoroughly diluted?

Sulfuric acid (and other acids) can be concentrated through evaporation of water.  It would be a good idea to wash the surface down with dilute, basic solution.  That said, if I'm reading your post correctly, your droplets are already 0.1M.  That is still a strong acid and care should be taken in its clean up.  I was once working with a 0.01M solution of sulfuric acid and when cleaning the flask a few drops of the dilute solution splattered onto my foot, I didn't even notice it at the time.  An hour or so later my foot started to burn.  When I looked, my sock was gone in the affected area.  If you're dealing with anything more than drop or two use a spill kit and let someone know what you are doing.

Yeah, I probably meant something more like 0.001M or maybe 0.00001M [it's been ages since I've taken chemistry].  Basically, the question is not so much about a spilled quantity which is large enough to be visible, but rather the possibility that wiping down a surface where acid was handled, and were it would be possible that a drop might have fallen without being noticed, might end up leaving behind a highly-concentrated residue.

Yes, if anything is left behind it will re-concentrate.  A mild base wash would make good sense.

That's what I figured, but I would have expected that if re-concentration was a hazard the instructions that came with the battery and acid for my sump pump would have suggest that as a reason for why use of baking soda was a good idea.

"If you aren't in a hurry, you can concentrate sulfuric acid by allowing the water to evaporate naturally. This takes several days."  http://chemistry.about.com/od/makechemicalsyourself/a/How-To-Make-Sulfuric-Acid-At-Home.htm

Is there any minimum quantity or concentration below which one wouldn't have to worry?  If a drop were placed in a bathtub full of distilled water and left alone long enough for the water to evaporate, could that drop still pose a hazard?

Tap water, depending where it comes from, can have a pH between about 6.5-8.5 due to dissolved minerals and gases.  I suspect the long-term risk is low given all the stuff in the air for it to react with as well.

Is there any general principle for what quantities and concentrations are worth worrying about, if one wants to err on the side of caution but not paranoia?

See  https://www.phys.unsw.edu.au/safety/documents/SOP_acid_spill.pdf  as far as I'm aware they're all handled the same unless it's a large spill (~1 liter).  If any of this has been helpful, please mark the answer as accepted, thanks.

I've marked the answer as accepted.  Do you have any thoughts about whether things like whether the possibility of re-concentration is sufficient that lead acid battery instructions should mention it?

Instructions can only go so far.  Rather than mentioning re-concentration, I'd focus the instructions on the seriousness of all acid spills and the need for complete cleanup and proper disposal of waste.

Applying the stated level of protocol to a single drop that falls from the hose while moving from one opening to the next would seem excessive.  I would expect that wiping down a single-drop spill with a disposable rag that's damp with a solution of NaHCO3 would suffice, but the baking soda should not be considered optional.  Would that seem an appropriate level of care?  Suggesting a protocol that might seem excessive for single-drop quantities leaves unclear what level of care is required for such quantities.

I agree, even for a few drops.  The only additional thing I would do if the drop was on an easily accessible, flat surface would be to first put a little bicarbonate powder on the drop(s), let it sit for a minute or two and then wipe it up with a damp (bicarbonate solution), disposable rag.


text:
Are non-ionic compounds generally as dehydrating as ionic compounds in drinking water?
In order to turn sea water into drinkable water, virtually all of the salt must be removed.  By definition, salts are ionic compounds.  I am assuming that there are other non-ionic compounds devolved in the sea water, must these compounds be removes as well before water becomes safe to drink?  In general, do non-ionic compounds present the same dehydration dangers as salts when consumed in drinking water?  Why or why not?
For this question, lets not consider other toxic effects besides dehydration.

Whether the solute is ionic or not actually has little to do with the danger of drinking sea water. The amount of solute is the problem. If the oceans were made of syrup (a non-ionic sucrose solution in water) with the same molar concentration as salt in seawater, then I think they would be just as deadly, by about the same mechanism as seawater; solutions more concentrated than blood pull water out of cells by osmosis, killing them from dehydration.

It's not as simple as that.  If the ocean were made of syrup (0.6M glucose instead of 0.6M ionic salts) it would be food.  You could live on it, together with some protein and other nutrients. Consider also that many creatures live in the sea, and they don't lose all water.

Actually, the salt in seawater doesn't kill you directly. It dehydrates you and you die of thirst. If you mixed 1 part of seawater with 2 parts of fresh water, it would not dehydrate you, but you would need to eliminate the excess salt so you would need to drink a lot of this salty water.
There are many living things in seawater that can kill you. Red Tide (excuse me, harmful algal bloom) is an algae common to warm water that produces a toxin. Very little of this toxin is needed to kill a person.

Yes, non-ionic but solvable compounds, if present in high enough concentration in water, can dehydrate human. For example, sugars, if present in high concentrations, dehydrate living organisms, and this is the reason why honey can be stored for years without significant loss in quality. Sugar sirup is a well known preservative, traditionally used in slavic countries to preserve fruits and berries.
However, seawater contains very little amount of non-ionic compounds, and most of them belongs to living things of various nature and, to some extent, their wastes. Moreover, traditions schemes of desalination filters off this impurities as well. This is actually a problem, as perfectly clean water is not healthy to consume, so generally it is preferable to add some carefully balanced amount of salts to such water.

If all of the ionic substances are removed, would salt water still contain a toxic level of dissolved substances?

@Hoytman To my knowledge, no.

@Hoytman To be precise, some volumes of seawater may contains toxic compounds on their own, but these are special cases. For example, deep waters of Black Sea contains (mostly not dissociated) hydrogen sulfide.


text:
Difference between torsion, out of plane, coplanar and perpendicular bends?
In Infrared spectroscopy, stretches are easily understandable. But, how do I visualize (or conceptually understand) the difference between dihedral (torsions), out of plane bends, coplanar bends, and perpendicular bends? Can you give examples?

A molecule with N atoms has 3N degrees of freedom
3 translational modes
if it is linear, it will have 2 rotational modes; if non-linear 3
This leaves 3N-6 (3N-5 if linear) vibrational modes.  Vibrational modes are further subdivided into stretching, bending or torsional motions.

Stretching motions can be symmetric or asymmetric
Bending motions are often referred to as scissoring, twisting, wagging, and rocking
Torsional motions refer to rotations about bonds

This diagram illustrates these various stretching and bending vibrations

Depending on the geometric relationship between the transition moment for a specific vibrational mode and the symmetry axis for the mode, vibrational modes may be further classified as parallel or perpendicular.
Here is a link to a 31 page article (slide show) that has a lot of background and pictures on this subject. Section 4.11 (page 19) might be particularly interesting to you.
One final example that you might be interested in, formaldehyde has 4 atoms and is non-linear so it will have 6 vibrational modes.  These are all pictured on page 2 in this link.

The link to the 31-page article is no longer useful (redirects to MacKenzie's academic home page).

@dat Thanks for pointing that out.  I was able to find a new link to the full 31 page article using the Wayback Machine.


text:
Reaction mechanism for reduction of nitrobenzene to aniline in catalytic and acidic media
Can anyone explain the reaction mechanism of the two reactions given below?
They are the conversion of nitrobenzene to aniline in catalytic ($\ce{H2/Pd}$ in ethanol) and acidic ($\ce{Sn/HCl}$ or $\ce{Fe/HCl}$) mediums respectively.
$$\ce{Ar-NO2 ->[\ce{H2/Pd}][\ce{EtOH}] Ar-NH2}$$

$$\ce{Ar-NO2 ->[\ce{Sn + HCl} \text{ or } \ce{Fe + HCl}] Ar-NH2}$$

What are your own thoughts on the matter? Do you have done any research or have any clue what might happen?

The first reaction is called a catalytic reduction.  Here is a picture of how it works with an olefin like ethylene.

You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form.  The same things happen with nitrobenzene.  As the following diagram shows, the reduction passes through the nitroso and hydroxylamine intermediates.

The mechanism for the Sn\HCl reduction involves reductive electron transfer of the nitro group while the tin is being oxidized. This diagram shows the electron transfers involved

and here is the mechanistic scheme.

How did you come to know the mechanism...?

@user70856 The catalytic reduction I knew from some time ago.  For the Sn\HCl reduction mechanism I looked on the internet.

What would be the product if we use Sn/NaOH ?

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


text:
E2 Elimination; Anti-periplanar beta hydrogen
I'm confused by this picture I found on masterorganicchemistry.com—a very popular organic chemistry website:

I agree that the product is a product.
However, can't there be another product? Specifically, can't the methyl group by the bromine have an anti and beta hydrogen relative to the bromine?
So shouldn't there be two products?
Specifically, this one?

For some odd reason, introduction of $sp^2$ center into six- and five- rings is strongly unfavored. I never cared much to find the exact reason, but guess that single $sp^2$ center induces more strain than two of them. ]

Would that have to do with angle strain? I mean in any case the author of the graphic says that a double bond in a 6-membered ring is a "major" product; I don't see how what I drew can't also be a "major" or at least "minor" product; my product only introduces one sp2 carbon into a ring.

A reasonable chemist would be on the lookout for two products, especially given that E2 elimination from 1-bromo-1-methylcyclohexane produces some methylenecyclohexane

In the transition state for the E2 elimination the pi bond is already partially developed.  This means that the relative thermodynamic stabilities of the products will play a role in determining the product distribution.  Since more highly substituted olefins are thermodynamically more stable than less substituted olefins, one would predict that the product with the double bond internal to the ring would predominate over the analogue with the external double bond.

Reaction temperature, base size, steric constraints around the various hydrogens that might be involved and the numbers of the various hydrogens all factor into the final product ratio, but these effects are usually secondary unless you change circumstances so much that the reaction mechanism changes and shifts away from E2.

So the picture I posted is incorrect?

Your picture is correct, I would have expected some of that product based on the methylenecyclohexane example I provided in my post, albeit I would have expected it to be the minor product.  Despite our predictions, according to the link you provided, the isomer with the exo-methylene group apparently doesn't form.  That's why we make predictions and then run experiments.

Oops I meant the first one - the one from the site. I think we can both agree with the site that the non-exocyclic double bond containing molecule is probably not the "only product!" At least from our predictions. Perhaps the author of the site actually ran the experiment and found that it was the "only product!"

Yes, that's what I was referring to - shouldn't have called it "your" picture. I assume it is correct - that is what the link clearly says.  That's why I said, given my example I was surprised by their result, but I assume it is correct.

I think on a test I'd probably still put down the other, less substituted alkene. In any case I doubt the author actually ran the experiment because that's not noted anywhere on the page.

So would I.  I'd add that I expected it to be the minor isomer.

Why is the more stable geometry favored (anti) while the more unstable would react faster (syn geometry-higher energy) ?

If the elimination occurs via an ${\mathrm{E_2}}$ mechanism, the other endocyclic double bond (i.e., the one between the two methyl bearing carbons) cannot be formed, as the $\ce{C_2}$ $\beta$-hydrogen will not be anti-periplanar to the $\ce{Br-}$ leaving group. That said, the other possible elimination product you've identified, with the exocyclic double bond, is entirely possible. However, this would be the Hofmann product, as it contains the less-substituted double bond. The major product of an elimination reaction is typically the Zaitsev product, which contains the most-substituted double bond. The Zaitsev product is the thermodynamic one (as the more-substituted alkene is typically more stable), while the Hofmann product is the kinetic one (as it results from a base abstracting a more sterically accessible or otherwise more acidic hydrogen). The Hofmann product can be obtained if a sterically bulky base is used, as this raises the energy of the transition state due to steric hindrance between the approaching base and the substrate molecule, and puts the reaction under kinetic control (assuming reaction temperature is controlled such that the $E_a$ for the thermodynamic pathway is prohibitively high). $\ce{NaOEt}$ is not bulky, so we expect the Zaitsev product (as given in the reaction scheme you cite).
If an $\mathrm{E_1}$ mechanism occurs, a tertiary carbocation intermediate is generated, and the double-bond is likely to form between the two methyl bearing carbons, as this would give the maximally stable tetrasubstituted alkene, and the requirement for anti-periplanarity is lifted. That said, $\mathrm{E_1}$ reactions are not favorable given a high concentration of strong base, so I would expect $\ce{E_2}$ to predominate, assuming at least a stoichiometric amount of $\ce{NaOEt}$ is present.
(Note that, with cyclohexanes, the issue is slightly more complex, as the anti-periplanarity requirement of the $\mathrm{E_2}$ mechanism requires the leaving group to be axial, which in turn necessitates a ring flip that has an associated energy barrier. Hence, in theory, the reaction with a cyclic molecule may be more likely to be kinetically controlled under a wider range of conditions.)

The ring flip is very important, very much agreed! Pre equilibria can mess up a lot of reaction mechanisms...

These E2 reactions are under kinetic control (since it isn't plausible that H-Br adds back onto the alkene, so the reaction is irreversible), it's just that the TS resembles the product (as @Ron says above) according to the Hammond postulate. This means the thermodynamic (more substituted alkene) product is favoured even though the reaction is kinetically controlled.

Yes that should be correct. Your product is less favoured but still likely. In fact I guess, ( I am not near a database to confirm that but you could try it) in reality you get the E1 product too. So a "nice" mixture of at least 3 products

What's the 3rd product?

Oh nvm you mean the SN1 product right?

@Dissenter The third product via E1 forms the double bond between C1 and C2 as greg also pointed this out.


text:
SN1 reaction of ethanol at tertiary alkyl halide
I know this undergoes an $\ce{S_{N}1}$ reaction:

Ethanol is the solvent. Weak nucleophile. Neutral nucleophile. This points toward $\ce{S_{N}1/E1}$ (competing pathways).
The leaving group is on a tertiary carbon. This precludes backside attack, $\ce{S_{N}2}$.
The professor labeled this reaction as $\ce{S_{N}1}$.

My question: where's the substitution part of this reaction? I don't see any substitution. Is there supposed to be no reaction here? From what he wrote it looks like a bromine left as bromide ion and a hydrogen appeared out of nowhere.
My expected transition state is a tertiary carbocation, which will not rearrange, and then the nucleophilic oxygen part of the ethanol molecule attacks the carbocation. Bromide ion deprotonates the oxonium ion, and I get an ether.
Who's right?

You're description is right.

Sheesh, there was something so obviously wrong with this I was confused. Thanks for clarifying.

Yep, it's hard enough to learn this stuff without the teacher creating obstacles.  You did well figuring it out!

It's the reason I'm here on this site!

Just to leave this question not unanswered. As ron already pointed out, you reasoning is correct. Based on the information given, the ethanol may perform a nucleophilic attack at the carbocation. To a very minor extent you might find the elimination product, but I guess you will have to cook it quite well. The elimination pathway is more likely to occur, when there is a stronger base than ethanol.

Just to get your terminology right. The carbocation is a reactive intermediate, not a transition state. The $\ce{S_{N}1/E1}$ reaction have always (at least) two transition states. The first one corresponds to the carbon leaving group bond break and the second one corresponds to the nucleophile carbon bond formation, proton abstraction, respectively.

The product drawn is a reduction product. You can use $\ce{LiAlH4}$ or homolytic cleavage of the $\ce{C-Br}$ bond (think radicals) to prepare it (but good luck isolating this, isopentane's b.p. is 28 $^\circ$C ). But not $\ce{EtOH}$ hehe.
To the point, I think a more appropriate example would be to use something like adamantyl bromide instead, as in this case you can not have competing elimination:


text:
Finding the value of stoichiometric coefficient x
$\ce{2A_{(s)} + xB_{(g)} <=> 3C_{(g)}}$
$\mathrm{K_p}=19.3, \mathrm{K_c}=0.45, \mathrm T=523\pu K$
I work $\Delta n$ to be $1$, So would $x$ be $1$ or $2$? I'm not sure if the coefficient is part of $\Delta n$.

The relationship between $K_p$ and $K_c$ is
$$K_p=K_c(RT)^{\Delta n}$$
where $\Delta n$ is the change in the number of moles of gas molecules over the course of the reaction. The only way you could know this change is from the stoichiometric coefficients. The $\Delta n$ term derives from the laws of mass action in which all partial pressures and/or concentrations are raised to the stoichiometric coefficients.
Consider:
$$\ce{2NO(g) + O2(g) <=> 2NO2(g)}$$
$$K_c = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$
$$K_p = \dfrac{P^2_{\ce{NO2}}}{P^2_{\ce{NO}}P_{\ce{O2}}}$$
$$P_x = \dfrac{n_x}{V_x}RT=[\ce{X}]RT$$
$$K_p = \dfrac{([\ce{NO2}]RT)^2}{([\ce{NO}]RT)^2([\ce{O2}]RT)}$$
$$K_p = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}\dfrac{(RT)^2}{(RT)^3}=K_c(RT)^{-1}$$
$$\Delta n = -1 =2-(2+1)$$

Thank you for pointing out the error it seems to make a lot more sense.

Assuming you calculated $\Delta n$ correctly, the value of $x$ would  be 2, because $\Delta n= 3 - x$.  You only count the coefficients of gases in determining $\Delta n$.

Okay, thank you, that was the main thing i was confused about, though I did make an error in calculating the delta n Value.


text:
-ic -ous nomenclature
I recently came across a practice problem in my textbook asking me to name a few compounds using -ic and -ous endings. The exact wording is:
Write the name of each of the following ionic substances, using -ous and -ic endings to indicate the charge of the cation.
The first one was $\ce{CoCl2}$. However, cobalt has more than just two oxidization states (3, 2, 0, and -1).
So if it is one of the states in the middle, how do I decide whether to use -ous or -ic

Cobalt may possibly have all those other oxidation states (and others too) for this nomenclature system, we only care about the two most common oxidation states in ionic compounds: $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$. Cobalt(II) compounds would thus be named  cobaltous and cobalt(III) compounds would be cobaltic. Which is yours?

Chloride is -1 and it's Cl2 so Cobaltous. Thanks!

-1 as in the charge, not your answer. That was really helpful! +1

Is there a reference you would recommend for finding the most common oxidization states? I can't find one in my book and online.

$\ce{Co^3+}$ oxidizes $\ce{Cl-}$.  I know this from personal experience.

Depending on its ligands Co3+ might oxidize Cl- is closer to the truth


text:
Steric Hindrance and Brønsted Basicity
Does steric hindrance affect Brønsted basicity?
I understand that basicity is a thermodynamic quality. Not a kinetic factor. So it doesn't matter if it takes a million years for the proton to reach the site of basicity; time is nothing in thermodynamics. We only care about the initial and final state. So this suggests to me that steric hindrance does NOT affect Brønsted basicity
In addition, the hydrogen proton is quite small. So I wouldn't expect it to be particularly sterically hindered by anything (unless we're in solution and the proton is heavily solvated). Still, once the proton reaches the site of basicity, it's reached it, and if I understand thermo correctly, thermo doesn't deal with anything in the middle of the energy graph. Only the points on the two ends.

Also it would be nice if one could comment on the difficulty of this problem in general. Just curious. Personally I think it's a fascinating problem. (On the other hand I'm not the one actually doing the problem).

You've got it right.

That's very interesting because my professor asked us to consider steric effects on basicity. Picture of problem posted. Something's fishy here.

Waiting for the intellectual smackdown.

@SilvioLevy I disagree (a little). Steric effects change the geometry of the molecule, and therefore also the bonding of each atom to another. This will also affect the lone pairs, that bond to the proton. Based on steric reasons that bond can be weaker or stronger than ordinarily expected. Since the bond dissociation energy is also a thermodynamic property, it should not be neglected. Consider dimethyl and diethyl amine, the latter is slightly (tiny!!) more basic mostly because of steric considerations.

Why is the latter more basic, specifically? From what I understand alkyl groups are electron releasing (slightly).

So this suggests to me that steric hindrance does NOT affect Bronsted basicity
|||
Steric hindrance for hydrogen - maybe, but steric hindrance may prevent stabilization of ion, or contrary, stabilize it a bit or destabilize initial state.

Hint 2: consider conjugation.

So the only way sterics would affect basicity is through solvation of the conjugate base?

Anything that stabilize one of the species in an acid-base equilibrium influences the free energy difference between the reactants and products, which alters the equilibrium constant:
$$\ce{HA <=> H+ + A-}$$
$$\Delta_r G^\circ = -RT\ln K_{eq}$$
You forget the solvent. Acidity and basicity depend on the interaction between the acid/base and the solvent. While steric hindrance will decrease the rate solvent access, it more importantly also decreases solvation of the acidic or basic site. More steric hindrance prevents additional solvent molecules from reaching the site to stabilize an ionic conjugate acid or base.
Consider the difference in $\mathrm{p}K_a$ between methanol, ethanol, 1-butanol, and tert-butanol. These are equilibrium constants, so they most certainly derived from thermodynamic phenomena!
Methanol $\ce{CH3OH}$ has a $\mathrm{p}K_a = 15.5$ (in water)
Ethanol $\ce{CH3CH2OH}$ has $\mathrm{p}K_a = 15.9$ (in water)
1-Butanol $\ce{CH3CH2CH2CH2OH}$ has $\mathrm{p}K_a = 16.1$ (in water)
t-Butanol $\ce{(CH3)3COH}$ has $\mathrm{p}K_a = 17$ (in water) (I've also seen 18 as its $\mathrm{p}K_a$)
The difference between methanol, ethanol, and 1-butanol (all primary alcohols) is small, while t-butanol has a larger pKa. If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups.
However in methanol, ethanol, and 1-butanol, more water molecules can solvate the alkoxide anion conjugate base, stabilizing that conjugate base more than the conjugate base of tert-butanol, which due to sterics cannot have as much solvation of the anion.
Now, gas phase acidities, in which there is no solvent, should have no solvent and solvation effects, and thus no (or at least reduced) steric effects.

And although my answer focuses on acidity, acidity and basicity follow the same rules.

Damn, that is a good angle of looking at this problem. You know, I doubt my professor thought of this himself. I'll have to ask him.

"If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups." Similar doesn't mean the same.  Consider that $pK_a$ of something in $\ce{H2O}$ is different from $pK_a$ of the same thing in $\ce{D2O}$.  Hard to argue sterics there. (link to an example, variation up to 0.6 pH units but dependent on the exact solute: http://pubs.acs.org/doi/pdf/10.1021/j100631a015)

I'm guessing the extra proton in D has to do with the differing acidities?

Correct; the simple fact that the mass changes causes  electron densities to change slightly in the various molecules. (There is an $m$ in the kinetic energy term of the Hamiltonian in the Schrödinger equation.)  I don't know much about this effect and would not be able to predict its magnitude a priori, but it is real and there are people who devote their research careers to it.  The isotope effect can give important hints in mechanistic studies, for example.

Makes sense. I was thinking the same thing - the effect of added positive charge on electron density and implications for bonds, etc. In any case, do you think the question is flawed in its asking for "steric" effects?

I think words like "steric effects" are not precise enough to warrant too much ink spilled in arguments.  That said, @Ben Norris has a point that there could be an influence of sterics on $pK_a$ due to solvation.  I don't agree that his examples prove that effect, because with 3 attached methyl groups, each slightly electron-donating, the tert-butoxide ion is more destabilized than the 1-butoxide ion anyway.  If there were an example running the other way, the case would be altered -- for instance if $\ce{CF3OH}$ dissociated less than $\ce{CH3OH}$, the sterics argument would be persuasive.

So the question is vague? Vague to the point of possibly being a valid question?

Sterics can affect basicity not only through solvation but also through inhibition of resonance.

Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoin acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.

In the case of meta-xylene with an $\ce{-NH2}$ group between its methyl groups,

resonance is inhibited because the methyl groups prevent the amine group from adopting a perfectly planar conformation. This inhibition of resonance decreases the electron-withdrawing capabilities of the aromatic ring. As a result, electron density is concentrated on the amine group, and meta-xylene with such a configuration is unusually basic.
Further reading:
http://books.google.com/books?id=my3fqYtt0hwC&pg=SA1-PA16&lpg=SA1-PA16&dq=steric+inhibition+of+resonance+benzene&source=bl&ots=-ZFR_PcpMD&sig=hoE2NQxBP-KjpEgDQIlvRIT2gwQ&hl=en&sa=X&ei=mDnoU7K7DovMsQSsnoHQBA&ved=0CGwQ6AEwCA#v=onepage&q=steric%20inhibition%20of%20resonance%20benzene&f=false


text:
How does propylene glycol/glycerin form formaldehyde/acetaldehyde when vaporized?
So apparently, e-cigarettes produce formaldehyde/acetaldehyde because they vaporize propylene glycol/glycerin. What is the chemical reaction through which propylene glycol/glycerin forms formaldehyde/acetaldehyde? Why does propylene glycol form more carbonylated products than glycerol?

I think the reaction is essentially an incomplete combustion reaction.
In a vaporizer, the idea is to get the propylene glycol, glycerine, and and "flavor" molecules into the vapor phase (or into aerosol droplets) without chemically degrading them.
However, the way this is accomplished is with very high-temperature resistor coils - the idea is that if you have a small amount of heat in a confined space, and the liquid is distributed throughout an absorbent (like cotton, for example), then the heat transfer will happen very quickly and a large amount of liquid will be vaporized.
The problem is that when you have a very high temperature heat source, and there is lots of oxygen in comparison to the liquid, there is a good chance that at least some of the liquid will reach the combustion temperature instead of just vaporizing.
What happens is that the hydrocarbons in the propylene glycol and glycerin molecules are partially oxidized - oxygen reacts with them to "steal" electrons. If the combustion was complete, you would wind up with just carbon dioxide and water. When combustion is incomplete, you can get any number of compounds, two of which happen to be formaldehyde and acetaldehyde. Carbon monoxide is another one I would be concerned about.
To see how this works, take a look at proplyene glycol (all pictures are from wikipedia):

and glycerine:

Notice how they both have three carbon atoms in the "backbone" and either one or none $\ce{O-H}$ groups (called hydroxyls or alcohols) attached to each one.
Oxygen can react with any one of those carbons and will try to form $\ce{CO2}$ and water, the most thermodynamically stable products that are possible:

If there is plenty of oxygen, it will. However, if there is not enough, you will get some intermediate species - formaldehyde or acetaldehyde, for example.
Let's look at formaldehyde:

In this case, oxygen has reacted with the the other two carbons on the backbone, leaving one with a hydroxyl. The hydrogen on the oxygen goes somewhere (either water or one of the other pieces of the original molecule) and the leftover electrons end up in the oxygen-carbon bond, forming a double bond. This molecule is more thermodynamically stable than proplyene glycol or glycerol, but less stable than carbon dioxide.
Now let's look at acetaldehyde:

Notice the similarity - it's almost the same, there is just a $\ce{CH3}$ (methyl) group in place of one of the hydrogens. You could get this by reacting the first carbon on propylene glycol with oxygen, and then removing the hydrogen from the hydroxyl on the second carbon. This is again more stable than propylene glycol, but less stable than carbon dioxide - it is also an intermediate.

Why does propylene glycol form more carbonylated products than glycerol?

This is a guess - but I think it likely has to do with the reaction kinetics. The reason I think this is true is that all of the intermediate products are less thermodynamically stable than carbon dioxide and water. Therefore, if a propylene glycol (PG) gives more of a given intermediate product than glycerine does, that product is "easier" to make for PG. This would happen if the activation energy for forming it is lower, or if the activation energy to go from the intermediate to carbon dioxide and water is higher compared to the intermediates formed by glycerine.
Note: Many people in other answers are pointing out that there are other mechanisms that could lead to these products, and that combustion is not likely at these temperatures. I think that given the mechanics of heat transfer we can't just rule it out, but I don't have any evidence for that. My answer is meant to show you how the reactant molecules contain the same structures as the products - I wouldn't put much faith in my mechanism (or anybody else's) without some real evidence.

What is the chemical reaction through which propylene glycol/glycerin forms formaldehyde/acetaldehyde?

Propylene glycol is NOT the same as glycerin. Propylene glycol does not decompose on heating, it simply boils. In theory, under certain conditions (pressure/encapsulation/catalysts) it could dehydrate to either acetone or Propionaldehyde, probably the former.

As for glycerine (= glycerol = VG = animal glycerine), that decomposes below its boiling point by a simple dehydration mechanism, to acrolein and water.
$$\ce{C3H8O3$\xrightarrow{\Delta}$C3H4O +2H2O}$$
Acrolein could further degrade by a retro aldol condensation to acetaldehyde and formaldehyde. This could be catalysed by acids/bases (both of which are present in eliquids) or even by hot metal catalysts (such as the heating coil):
$$\ce{C3H4O\xrightarrow{H^+}H3CCHO +HCHO}$$
or
$$\ce{C3H4O\xrightarrow{hot~metal}H3CCHO +HCHO}$$
These are well known reactions to any organic chemist, and are far more likely to happen than any vague/spurious 'oxidation' reactions as suggested by a few.

Why does propylene glycol form more carbonylated products than glycerol?
Because it has a dual mode of dehydration.
Really, this is a 'Have you stopped beating your wife?' question: PG and glycerol can form multiple carbonyl compounds, depending on what further side reactions happen:

Formaldehyde is stable, other than the possibility of oxidation to formic acid or carbon dioxide.
Acetalydehyde can dimerize to crotonaldehyde.
Acrolein can both dimerize, polymerise and undergo addition reactions, in all cases forming a variety of carbonyl compounds

One of the most bizarre things about some studies (including those of Dr Fasolinos), is that while formaldehyde and acetaldehyde are observed, there is insignificant acrolein. And he makes little comment about the latter!
Don't you wonder where it's all going?
I'd suggest that the bulk of the acrolein is decomposing to those two other carbonyl compounds.
Most aldehydes undergo oxidation over time, but this is not combustion taking place.


text:
Can one reduce solubility of NaCl by adding CO2 under pressure? Will the salt fall out of solution?
One can salt CO2 out of solution by adding salt to carbonated water.  Can one do the reverse?  If one adds more and more CO2 to the water under pressure, can you drive out the salt?

NaCl is vastly more soluble than carbon dioxide, so this method won't work.  You can get some NaCl to precipitate (assuming you have a concentrated solution) by bubbling HCl in, thanks to the common-ion effect, so you could for example get pure NaCl from rock salt this way.  But somehow I don't think that's the situation you have in mind :-)

The OP mentioned pressure, so technically speaking you can put as much CO2 into the solution as much you want. I think his question is primarily about if this equilibrium can be pushed into the reverse direction with enough pressure. The answer is yes, and you are completely right that it is not much practical.

Good point, but "as much as you want" is not correct.  BTW, pressure alone will drive out the salt, since at 10,000 atm or so water will freeze at room temperature (http://www1.lsbu.ac.uk/water/phase.html)  I'm not sure how much CO2 you can dissolve before that; best I could find is that at 100 MPa (1000 atm) the mole fraction of CO2 in H2O is about 0.03, and the curve is quite flat already; see http://pubs.acs.org/doi/pdf/10.1021/ie980768g
-- Anyway, you have a good point and I'll modify my answer after waiting to see if there are more comments.

Very good points! Thank you for cleaning it up.


text:
How do you calculate enthalpy for a reaction at non-standard conditions?
I understand how to calculate the change in enthalpy and entropy for a reaction in standard conditions, but is there a way to calculate these values at non-standard conditions?
Also, is there any way that a reaction with positive change in gibbs free energy will occur?

Do you know how to calculate the change in enthalpy and entropy for a substance as you change the conditions (T, P, ...) in a thermodynamically reversible fashion? Now take the reactants at "standard" conditions, transform them to reactant(s) at the non-standard conditions, react them, and take the product(s) back to the standard condition.  Compare with the reaction occurring at standard conditions. Figure out the enthalpy and entropy differences between reacting at 'standard' and 'non-standard' conditions.
As for your second question, ponder why some reactions do not go to completion.

1) Yes.
The change of entalpy of one mole with cnange of temperature is known as $C_p$, a heat capacity at constant pressure. It is, unfortunately, a function of temperature, so one has to perform integration of
$\Delta H= \int\limits^{T_{final}}_{T_{standard}}{C_p(T)}dT$
AFAIK, $C_p$ is usually approximated as cubic polynom $C_p=aT^3+bT^2 +cT +d $, where $a,b,c,d$ are found in reference literature. However, the situation is further complicated by phase change. In case one calculates enthalpy of water at $120^oC$, he will have to perform integration both for liquid and gas phases in their respective regions of stability and then add heat of evaporation of water.
The change of entropy is calculated in similiar manner, using the fact that
$\Delta S= \int \frac {dQ} T$
during heating to different temperature
$\Delta S= \int\limits^{T_{final}}_{T_{standard}}\frac{C_p(T)}TdT $
After obtaining entalpies and entropies of reagents and products at different temperature, one can obtain $\Delta G$ of reaction at same temperature.
As you can see it requires a lot of math, so such excercises are almost universally hated.
2)Technically speaking, most reactions move to point of equilibrium, where concentraions of reagents and products are guareded by constant of equilibrium, like
$K=\frac{[A]^a[B]^b...[E]^e}{[W]^w[X]^x...[Z]^z}$
for reaction
$\ce{w W + x X + ... z Z = a A + b B + ... + e E}$
so, if one start from products, i.e. when
$[W]=[X]=...=[Z]=0$
a sligh amount of reagents will appear as equilibrium is achieved.


text:
Nomenclature of organic molecules with more than one functional group
I need some help naming compounds with more than one functional group based on their priority order. Here are some examples:

A 4-membered carbon chain with an aldehyde at one end and a ketone right next to the aldehyde
A 4-membered carbon chain with an aldehyde at one end, and the ketone and aldehyde having 1 carbon in between.
A 5-membered carbon chain with a $\ce{C=C}$ double bond at carbon number 2  and a $\ce{C#C}$ triple bond at $\ce{C}_4$ (maybe it is the reverse)

I think that for the one with a double bond and a triple bond, the triple bond gets higher priority and so it would be pent-4-en-2-yne but I am not sure about it.

ketone has lower priority than aldehyde. Okay that makes sense since OH has higher priority than C-C and that is basically priority-wise what ketones and aldeydes are. But why does the double bond have priority over the triple bond? Surely there are some where the triple bond is lower than the double bond, in fact I know so. Is it just that E comes before Y and that in the actual numbering you number it from the carbon closest to the triple bond?

There are some priority rules about naming compounds with more than 1 functional group.

Aldehyde has higher priority than ketone (I don't know why but I think it is because of aldehyde always take place at the end of carbon chain)
Alkene has higher priority than alkyne, because -ene suffix comes before -yne

So the answer for your question is:

2-oxobutanal
3-oxobutanal
pent-1-en-4-yne

how did you get that pent-1-en-4-yne from "A 5 carbon molecule with an C-C double bond at the 2 carbon and a C-C triple bond at C4". I mean the pent and 4-yne parts make sense. What doesn't make sense is the 1-en part. and don't you number closer to where either the triple bond or double bond is and if it is a tie, then choose the double bond? I mean that would explain why sometimes the alkyne has a lower number than the alkene, even though in the naming the alkene always comes first.

@caters sorry for late reply. This is because alkene starts with "e" and alkyne starts with "y", so alkene has the higher priority.

but I still don't see how this part "a C-C double bond at the 2 carbon" lead you to 1-en.


text:
Is this disaccharide sucrose?
I want to know if the structure I drew is sucrose or an enantiomer of it.
The One I Made:

The one on internet:

Those are the same molecules - the fructose part has just been rotated about the glycosidic bond (the oxygen between the two).
Enantiomers are molecules that have the same formula, but are non-superimposable - there have to be differences in chirality. If you can superimpose the structures by rotating about a bond or by rotating the molecule, it is not an enantiomer.

Aha! you mean rotating the fructose ring on a horizontal axis by $180^o$, I didn't notice that...

I agree that the drawn structure is sucrose


text:
Why do all medicines require storage temperatures below 25 °C?
Almost all medicines I buy have storage requirements saying $T\le25^\circ C$.
But do they really spoil at higher temperatures? What if they are stored at e.g. $30^\circ C$, which is common at summertime?

As an informal rule of thumb, reaction rates double for every 10 degree C change in temperature. So a medicine at 30C should degrade twice as fast as at 20C. If I had more data to back this up I'd make it an answer, but I don't. Mostly, I think the warning is put there out of an abundance of caution and not wanting to get sued.

People do not realize that the quality requirements for a medicine is rather high. We are talking about a complex mixture of more or less reactive organic compounds that are potentially sensitive to air, moister and light. Any reaction of the original ingredient is potentially biological active and harmful, and even simple things like recrystallization or coagulation can destroy the effect of the original mixture.
The question "will they really spoil" makes no sense in the given context: if 99.999% of the cases not spoil, but there is a handful cases when things goes bad then the public goes banana. It took only one (purposefully false) paper to convince tens of thousands of people to go anti-vaccine, and decades till it dies out.
If it happens with a medicine, it potentially kills the product, so easier to err (by huge margin) on the safe side.


text:
Why is up spin in the px orbital and down spin in the py orbital considered an excited state?
Why is this form of p orbital occupation an excited state? I understand that there should be one up spin and one down spin in one orbital box by Pauli's exclusion rule. But in the p orbitals, it doesn't matter, whether we put up spins first or down spins first according to Hund's rule. I thought unless up and down spin are in same box, they don't affect each other. But as what I thought is wrong, I would like to know the exact reason for this. Why is it considered to be an excited state?

Hund's first rule states that the lowest energy atomic state is the one that maximizes the sum of the S values for all of the electrons in the open subshell. So, a state with unaligned spins in the p orbitals will be higher in energy than the one with aligned spins and thus it will not be the ground but an excited state.

see http://chemistry.stackexchange.com/questions/15215/how-can-two-electrons-lie-together-in-an-orbital/15223#15223

Hund's rule of maximum multiplicity states that for a given electron configuration, the term with the lowest energy has a maximum multiplicity $2S+1$. $S$ is the total spin angular momentum of all electrons in the partially-filled subshell.
For the application of the rule, we consider the 2 electrons in the $p$ subshell and add up their spin quantum numbers to obtain $S$ and the multiplicity. For the ground state configuration, with both electrons "spin-up", we get
$$S=+\frac{1}{2}+(+\frac{1}{2})=1$$
$$2S+1=3$$
For your proposed configuration with one $p$ electron "spin-up" and the other one "spin down":
$$S=+\frac{1}{2}+(-\frac{1}{2})=0$$
$$2S+1=1$$
Since the multiplicity value is smaller, this configuration is not the one which is lowest in energy, and therefore cannot be regarded as the ground state.

This is a weird effect resulted from Pauli principle.
Naive oversimplificated version: since no two electrons with same spin may occupy the same place, if two electrons have opposite spin, a part of their time they spend very close to each other, and this results in increased electrostatic repulsion term in energy of the state, while in case two electrons have the same spin, they are always away of each other, so the global energy is lower.
For this effect to manifest, a a significant overlapping of orbitals occupied by electrons in consideration must occur. This usually is observed on atomic or molecular level, and sometimes on larger scale


text:
Carbon Oxides and stability
If $\ce{C2O4}$ is very unstable no matter the configuration(that is the structure based on the bonds) and especially unstable in the form that has an oxyanion and a carbocation than how come other carbon oxides are much more stable despite the charges like $\ce{CO}$ for instance.
There are 2 possible structures with octets for this. They are
1) $\ce{C=O}$ with the carbon having 1 lone pair and the oxygen having 2 lone pairs. No charges here.
2) $\ce{C≡O}$ with $\ce{C}$ having 1 lone pair and $\ce{O}$ having 1 lone pair making the oxygen positive and the carbon negative.
This one is an important but very unstable structure.
Why is it so unstable? It is because while carbon isn't so determining as to what charge it prefers, Oxygen is and it really hates being positive because it wants to be either neutral or negative since it is more electronegative than everything else except fluorine. This is why $\ce{H3O+}$ isn't really $\ce{H3O+}$ but rather an ion in a complex of ions is because of how oxygen hates being positive.
So why are other carbon oxides more stable than $\ce{C2O4}$ when they too have very unstable structures like $\ce{CO2}$ having as a resonance structure a single bond to 1 oxygen making it negative, a triple bond to the other making it positive, and the carbon staying neutral? And if every carbon oxide is either unstable no matter the structure or has unstable resonance structures than why do carbon oxides with just carbon and oxygen exist?
And would adding hydrogens to a carbon oxide make the carbon oxide even more stable? By this I mean that with the fact that carbon oxides tend to be unstable and hydrocarbons tend to be stable would a carbon oxide with hydrogens added to it be less stable than a hydrocarbon but more stable than a carbon oxide?
Is this how come glucose is so stable and 6 $\ce{CO}$s bonded together would be extremely unstable just like $\ce{C2O4}$ is?
Sorry if it is too many questions but it is all about 1 thing, Carbon Oxides.

On the bonding in $\ce{CO2}$ is partly an answer by me here: http://chemistry.stackexchange.com/q/10976/4945

thats just about the standard 2 C=O bonds. CO2 also has an unstable resonance structure like this: O-C-=O(-= means triple bond). 1 oxygen being positive and the other being negative with the carbon staying neutral is unstable. It is sort of like how in CO the triple bond resonance structure is unstable because of how the carbon is negative and the oxygen is positive.

First thing you need to understand is, that the real bonding situation lies between all possible resonance structures. A resonance structure is only a concept, so that the human brain can begin to understand the bonding in such molecules. These structures do in no way represent any physically observable quantity. The bonding especially in these very small molecules is very complicated and complex and hard to comprehend. That's why we need simpler models to explain them. Btw, do you mean $\ce{^{\ominus}O-C#O^{\oplus}}$?

yes that is what I mean for the unstable resonance structure of CO2.

There are plenty of other carbon oxides - for example mellitic anhydride and graphite oxide have been known since the 1800s.

Remember,positive charge on oxygen is formal, not actual.... in carbon monoxides.

Why is $\ce{C2O4}$ unstable?
Have you tried drawing it?
I get four different kinds of structures for it:

$\ce{(O=)O^{.}C-CO^{.}(=O)}$
$\ce{(O^{.})2C-C(O^{.})2}$
$\ce{O#C-O-CO^{.}(=O)}$
$\ce{O#C-O-O-C#O}$

They are unstable for the following reasons:

Biradical
Tetraradical (This won't really exist anywhere near anything else.)
Radical
Peroxo bond, very easy to break homolytically.

All of these compounds can easily decompose to give (the thermodynamical Mariana trench) carbon dioxide:
$$\ce{C2O4 -> 2CO2}$$
That is why it is so unstable.
Reduce 'em!

And would adding hydrogens to a carbon oxide make the carbon oxide even more stable?

Yes. If you start reducing the compounds above, you will end up with ketones and water (because geminal hydroxy groups tend to give water and ketones), or even alcohols (if you continue reducing).


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Hybridization in PH3
Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?
Arguments against hybridization:

$\ce{PH3}$ is less basic than $\ce{NH3}$.
This jibes with the supposition that $\ce{PH3}$ keeps its lone pair in what is essentially an unhybridized $\ce{s}$-orbital.
$\ce{s}$-orbitals are symmetrical and therefore do not concentrate electron density anywhere.
This lack of concentrated electron density as opposed to $\ce{NH3}$ which is $\ce{sp^3}$ hybridized makes $\ce{PH3}$ a poorer base.

However, my professor writes in his book that it can be argued that $\ce{PH3}$ exhibits $\ce{sp^3}$ hybridization.
Another professor told me that the wave function "blows up" when putting an electron pair in an s-orbital. What does that mean?

Short answer: Not really. I am sure there are many comments about this molecule on the site. There are various reasons why there is only little to none hybridisation.

I'm quite sure that I read in "Organic Chemistry" by Clayden et al. that $\ce{PH3}$ is unhybridized.

See also here: http://chemistry.stackexchange.com/q/14087/4945 I might be able to boil it down for you later, but it is written there between the lines.

Pre: My views may be clashing with some people. I express them as far as I know.

To start with, hybridization is a hypothetical concept$^{*1}$ just to explain all facts and make our work easier, otherwise we would have to take into account every single interaction of electron-proton and also other phenomenons; many theories developed which explained all the major/important facts at a macro-level. And what I have been saying to take into account was all micro-level analysis. Attributing hybridization is purely personal choice because neither does it account all facts (for some molecules) nor it can be dropped altogether (for its $\cdots$), if then we would look for another theory$^{*2}$ to explain observed phenomenon. Parallels similarity between this and the situation of Ideal-Gas equation, in general no perfect theory has been developed, but the most easy, practical and useful are the hybridization$^{*3}$ and van-der-Waals equation.
As we move downwards in the periodic table the hybridization concept fails at many places due to overpowering of other factors. I wonder you didn't mentioned the bond angle$^{*4}$ ($93.5^\circ$) for a $\ce{sp^3}$ hybridized (ideally $109.5^\circ$).
In general atom/molecule doesn't seek to follow a theory, it does what it is best comfortable with, it is us, who mould the theory to the observations.
As a general rule, for sake of convenience, assume all molecules show hybridisation and the tendency to hybridise decreases down the group.
In conclusion, I would say it depends on the need of the situation and such dubious statements such as "Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?" should be replaced with "Does hybridization explain all (actually not all) properties of $\ce{PH3}$"
Also there are weights in $\ce{sp^\alpha}$ where $\alpha$ ranges from 1 to 3(even upto 4$^{*5}$ and maybe beyond) and is in between these for $\ce{PH3}$ and some may continue to treat it has a hybridised molecule with suitable weights.

$^{*1}$(as far as I know)
$^{*2}$(probably the MOT)
$^{*3}$(many may argue for other theories which is genuine, but in context this follows)
$^{*4}$(One may argue as we move down the group electronegativity decreases and atomic size increases. In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)
$^{*5}$[$\ce{O}$ in $\ce{H2O}$ which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals]

How would you rationalize the 93.5 degree bond angle?

@Dissenter my point $*4$ might answer

@Dissenter - Hybridization is a fine model for the smaller atoms, but breaks down in the third period. See my answer to this question: http://chemistry.stackexchange.com/questions/33818/bond-angles-for-the-hydrides/33824#33824

I think in NH3 N-H bonds are very close to each other so if lone pair does not participate in the hybridisation then repulsion will take place so it participate in the hybridisation to decrease the repulsion but not completely because hybridisation increases the energy of the s orbital because P orbitals give it direction and it is also far than s orbital so bond angle is not 109.5 it is approx 107 but in the case of PH3 the P-H bonds are far from each other and H is also a small atom and its electronegativity is also near the P atom so hybridisation almost does not take place and bond angle is near 90 degree.but in PCl3  Cl is a big atom and its electronegativity is also very high than P because of this it decreases the size of P and hybridisation take place.

The gist of this is correct, it is just badly phrased and not well presented.


text:
Calculation of mass from mmol/L
Sorry I understand this looks like a "homework question", (I'm actually a 46yr old dentist looking to see what amount of calcium phosphate should be effective in a proprietary toothpaste!). However, if the actual question below can not be worked out can you let me know what the process of the calculation would be to determine mass of a salt to add to gain a mmol/L concentration?
Original Question:
I wonder if you can help on this? I want to calculate how much calcium phosphate to add to 1g of toothpaste to have between 18mmol/L to 1.5mmol/L of calcium and phosphate on the tooth surface? Also if you are able to work this out can you let me know what assumptions you have made in this calculation?

If you need a concentration of Calcium phosphate on the surface of a tooth, we need to know how thick this layer is and what the surface area of teeth are. What is the volume of 1g of toothpaste? Should we assume that 1g of toothpaste will cover the entire surface of the teeth?

There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)
The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from
http://en.wikipedia.org/wiki/Monocalcium_phosphate although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.
To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known volume of paste.  Instead you have a known weight, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.
Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg.
So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .
I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!


text:
Is the activity coefficient always a constant?
The activity of a species $i$ is defined as $a_i = e^\frac{\mu_i-\mu_i^\ominus}{RT}$, with $\mu_i$ the chemical potential, and $\mu_i^\ominus$ the chemical potential under standard conditions.
Very often, for a species in a non-ideal solution, the activity is written $\gamma_i x_i$, where $x_i$ is the mole fraction and $\gamma$ is a constant known as the activity coefficient for that species. This implies that $\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i$.
What I don't have a good idea of is whether the $\gamma_i x_i$ formula is exact (under some assumptions) or an approximation. Are there situations in which the activity depends on the mole fraction in some non-linear way, or upon the concentrations of other species in addition to species $i$? Is the formula $a_i = \gamma_i x_i$ an exact one under some circumstances, or are there always (in principle) nonlinear correction terms that would make it more accurate?

In my copy of Atkins Physical Chemistry, 6th edition (it's old!) there is a chart that shows a calculation of the activity and the activity coefficient for an example system (chloroform and acetone). It shows the results using a calculation based on Henry's law (approximating ideal-dilution) and again with one based on Raoult's law (approximating a pure substance).
In both cases, the activity coefficient is not constant over the range of concentrations.
For some evidence from a less-reputable but more easily accessible source, check out the wikipedia article on activity coefficients:

$\gamma_B$ is the activity coefficient, which may itself depend on $x_B$

($x_B$ is the mole fraction of the solute).
Now this means that the activity coefficient is not constant for a given substance in a given mixture at all concentrations, but that isn't the same as saying that the definition of activities is not exact - it is exact, because it is defined that way.
In other words,
$$
\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i
$$
is an exact relationship, even though $\gamma_i$ is not constant.

Ok, that sounds convincing. I'll ask the following as a separate question if you don't know the answer, but is it possible to have a situation where $\gamma_i$ depends not only on $x_i$ but also on $x_j$, the concentration of a different substance?

I believe that is possible, but I haven't found a reference for it yet. I think I remember looking at that case in a stat mech class several years ago.

Actually - this has to be true, at least in a binary mixture. If $x_j$ changes, $x_i$ changes as well. The only question is whether the change is large enough to matter. For a multi-component mixture, I believe it is also true, simply because the non-ideality is a result of intermolecular interactions, and all molecules in a multi-component solution "see" each other.

I meant in a non-binary mixture of course. It'd be nice to have a specific example, but this is more than enough help for now. Thank you!


text:
Ethyl alcohols use on polycarbonate and acrylics
I have an interesting question about the use of ethyl alcohol on polycarbonate windshields.  I have had friends asking me about the use of Pledge Multi-Surface cleaner on their windshields. Some insist it will degrade, yellow, or even crack and soften their windshields. Others have been using it for years with no apparent effect.
Pledge does have a small amount of ethyl alcohol, but does it have enough to degrade and ruin motorcycle windshields or does ethanol do any damage anyway?
Pledges' MSDS states ethyl alcohol weight at 0.01-1.00  (http://www.scjohnson.ca/pdfViewer.aspx?pkMSDSId=367&language=en).
So what are the effects, if any, on using this cleaner on polycarbonate windshields?
Thanks

I'm not an expert in polymer chemistry by any means, but if you do a search on chemical compatibility of solvents with polycarbonate, most sources indicate no effect or only minor cosmetic effects from (near pure) ethanol on polycarbonate. Keep in mind that, in windshields, the outer layers are glass, so there shouldn't be direct contact with the polymer interlayer if the windshield is structurally intact.

And if you're ever concerned about a solvent's affect on a material, you can try taking a small amount of solvent and applying it to a hidden corner of the object. Not sure what would be best place on a windshield, maybe a lower edge near the dashboard.

Not able to comment....
I think the main problem is propyl alcohol (not present in this cleaner). Propyl alcohol (propanol) is present in a lot of cleaning solutions and known to damage polycarbonate and acrylics.
Check this chemical resistance sheet:
Sheet
30% solution containing ethanol (ethyl alcohol) lightly attacks the plastic. However, a 10% (propanol) propyl alcohol already does the same damage. That is a factor 3!

the link is down?

authors: علاء الاسدي علاء


text:
Why is the deprotonated form of dopamine more reactive than the protonated/neutral form of dopamine?
See this (from the Wikipedia article on dopamine):

Dopamine, like most amines, is an organic base. At neutral or acidic
  pH levels it is generally protonated. The protonated form is highly
  water-soluble and relatively stable, though it is capable of oxidizing
  if exposed to oxygen or other oxidants. At basic pH levels, dopamine
  becomes deprotonated. In this free base form it is less soluble and
  also highly reactive and easily oxidized

I'm very curious about this since I'm curious about dopamine auto-oxidation, as it is often a cause of neurodegeneration observed in those who take amphetamines (and seems more likely in the cytosol, than the vesicles it would otherwise be in). Though I'm also curious how much auto-oxidation would happen for extracellular dopamine as well.

Just to give some concrete numbers (which appear to be lacking in the Wikipedia article), one can calculate the concentrations of the various species. I crunched the numbers in Mathematica. Assuming aqueous solution, physiologic $pH \approx 7.35$, and using $pK_a$ values from this article, the mole fraction of the fully protonated form is roughly $0.98$.

Thanks for the calculation! I'm curious: Even if the deprotonated form was a very tiny minority of all the forms of dopamine (which would be < .02 because most of the non-protonated form would be the neutral form of dopamine), would that tiny minority be enough to explain the auto-oxidative damage of dopamine if it were reactive enough?

I can't answer that with any certainty, sorry. I just felt it might be useful to actually quantify some of the relevant parameters; the chemistry section of the Wikipedia article is sparse in that respect.

@IK: You are partly correct. Even a 0.02 fraction of active compound could induce activity. Imagine that the active 0.02 would be continuously removed from the equilibrium and new active "0.02" would take its place until all the inactive form is consumed. Obviously the active concentration would also be 500 times weaker. But then the oxidation should not be directly connected to the $-NH_2$ group (see my post below). "Autooxidation" is a set of interconnected reactions with multiple paths and products and $-NH_2$ could play some role downstream. Only way to answer is experiment (or literature)

In general the free amine will react as a nucleophile i.e. by "donating" an available lone pair of electrons. In its protonated form, this lone pair is "blocked" and thus it has to lose a "blocking" proton to react (I use the words in "quotes" more freely - although I'm a chemist and not biochemist but hopefully it will help).
However re oxidation of dopamine, the reactive site is not the nitrogen (at least not the primary one - but will come back later) but the catechol part. Because of the electron donating effect of the 2 neighboring OH's, the aromatic ring is electron rich and more prone to oxidation to give o-quinone products. This oxidation reaction depends to an extend to the pH but mainly to the presence of (obviously enough) reactive oxygen species (ROS) and also enzymes (oxidases) or even Fe ions that can catalyse this oxidation.
Coming back to the nitrogen, after oxidation occurs with creation of reactive sites on the quinone part, it can act now as a nucleophile and attack, creating indolines that will further oxidize to the more stable but toxic indole products.


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Why no thermal expansion for camping tents?
I know that solid materials expand when heated, which is called thermal expansion.
But what is happening to tent square or tent canvas of (plastic?) camping tents? I have noticed, that during the day, when it’s hot, there is a lot of tension in the skin of a tent. But when it gets colder (when raining or during the night), the tension is less and the fabric gets sloppy.
What is the reason?

I failed to find proper term, so I'll write what I recall from my brief course of polymer science. Polymers, including polymer fibers, consist of long entangled molecules . When under strain, the molecules straightens and the material elongates in direction of stress. When heated, this order is destroyed and molecules once again entangle, shortening the fibers.

some more info on the subject http://en.wikipedia.org/wiki/Rubber_elasticity

If the tent poles are made of metal, perhaps they are expanding during the day when it is hot and contracting at night when it is cold. This could increase the tension of the canvas during a hot day and make it loose and sloppy on a cooler night.

theo, consider writing your comment into an answer, I think that explanation is perfectly reasonable given the information. Polymer deformation seems less likely to contribute to this observation significantly.

@theo: To me the effect of elongating tent poles can not explain this, because this happens with every kind of tent poles (aluminium, carbon fiber, …). And if you’ve ever watched a tent, you will notice, that the whole tent canvas is under tension. Also the parts, that have no tent poles inside.

Check the ropes on your tent. Twisted natural fibers tend to shorten considerably when they take up water (swelling of the fibers), and get longer when they dry in the sun. I've seen many tent collapse or even get ripped apart in beginning rain. Of course canvas tents are made of twisted fibers, and also shrink.

The surface of a tent consists mostly of water-resistant polymers such as latex or plastic. These polymers are disordered molecules such as spaghetti, so they have unlike crystals a high entropy.
Gough (1805) and Joule (1859) studied the thermal behavior of rubber. They made the following observations, which are known as Gough-Joule effects:
• a taut with constant weight rubber shrinks when it is heated
• rubber emits heat when it is stretched
The cause of the rubber elasticity is fundamentally different than the elasticity of crystalline materials. In contrast to a crystal, the atoms are not regularly arranged in a rubber, but they form long chains (polymers) which are interconnected and form a network. Clearly, one can think of it as a plate of cooked spaghetti. This difference in microstructure are reflected in the mechanical properties of rubber. When the rubber network is loose, very many different configurations (configurations) of the chain between two connection points are possible - more disorder (larger entropy S).

Does your answer explain, why this happens again and again, every day when the tent is heated it has a high tension, but in the evening, when it gets colder, it has no tension at all. Wikipedia says: The Gough–Joule effect is the tendency of elastomers to contract when heated if they are under tension. Elastomers that are not under tension do not see this effect. In the evening it seems to not be under tension, but when it heats up the tent canvas seems to contract. Maybe this effect is not the reason for this to happen.

When heated up the material shrinks - when cooled the length of the material increases.

When heated up the material shrinks - when cooled the length of the material increases.
The tension is given from 2 effects:
a) from inside the structure of such a polymer, as the result of the inner energy, given by enthalpy and entropy.
b) from outside by the tension due to the weight of the material of the tent, the gravity and from the metal construction that builds the tent.

That Gough in 1805 did his experiments with unvulcanized rubber?

The air inside the tent heats up and expands, the tent, while not completely airtight still acts like a balloon expanding with the heated air inside it and as the air cools down it shrinks, deflating the tent/balloon.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. || I don't have much insight to the problem, but your answer seems misleadingly illogical to me.

Welcome to chemistry.SE , Tents always let some air interchange with outside, if not, people inside would get out of oxygen, so it can not work as a balloon if it takes a reasonable time to heat the air.


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Cows milk protein in lactose
Does anyone know if the casein/whey cows milk protein can be found within processed lactose?
Some background: our 4-month-old baby was given teething granules. She was then presented with a rash, screaming, etc. When we checked the teething granules (called teetha) we found they contain lactose.
However, from reading it appears that lactose intolerance is extremely rare in the UK and cows milk allergy is much more common.

Yes, process lactose could still contain protein from the milk. It's equally likely that the casein could contain lactose. Whey is much more likely to contain lactose since the whey and lactose are in the water when the cheese is made.
It would be extremely rare for a baby to be lactose intolerant since lactose is the major sugar they consume. As the child ages, the ability to digest lactose decreases. I think I heard that 5% of Caucasians adults can still digest lactose. Only a few people develop lactose intolerance.